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An Introduction to Mechanics of

Materials

©

Vijay Gupta

Lovely Professional University, Punjab

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Table of Contents

1 STRUCTURES, LOADS AND STRESSES 4 1.1 Mechanics of material 4 1.2 Deformation and resisting forces 4 1.3 Other loadings, stresses and strains 5 1.4 The concept of stress at a point 7 1.5 Stress on oblique planes 11 1.6 Notation for stress: double-index notation 12 1.7 Equivalence of shear stresses on complementary planes 13 1.8 Stresses in a thin circular pressure vessel 14 1.9 Summary 15

2 DEFORMATIONS, STRAINS AND MATERIAL PROPERTIES 17 2.1 Fundamental strategy of mechanics of deformable mechanics 17 2.2 Statically indeterminate problems 20 2.3 Lateral strain: Poisson ratio 23 2.4 Shear strain 25 2.5 Thermal Strains 27 2.6 Tensile test 28 2.7 Idealized stress-strain curves 30 2.8 Pre-stressing 32 2.9 Strain energy in an axially loaded members 33 2.10 Calculating deflections by energy methods: Castigliano theorem 33 2.11 Strain energy in an elastic body 38 Summary 39

3 TORSION OF CIRCULAR SHAFTS 41 3.1 Introduction 41 3.2 Relating angle of twist to twisting moment 41 3.3 Stresses and strain in a circular shaft 43 3.4 Hollow shaft 46 3.5 Statically indeterminate shafts 47 3.6 Composite shaft 49 3.7 Torsion of thin-walled tubes 50 3.8 Plastic deformation in torsion 51 3.9 Limit Torque 52 3.10 Strain energy in torsion 53

Summary 54

4 FORCES AND MOMENTS IN BEAMS 56 4.1 Introduction 56 4.2 Sign convention 57 4.3 Loads and supports 58 4.4 Determining shear forces and bending moments 59 4.5 General procedure for drawing shear force and bending moment diagrams by method of sections 61 4.6 The area method of drawing the SFDs and BMDs 64 Summary 69

5 STRESSES IN BEAMS 71 5.1 Introduction 71 5.2 Relating curvature of the beam to the bending moment 72 5.3 Composite beams 78 5.4 Stresses in beams carrying shear forces 82 5.5 Relating shear stresses to the shear force in a beam 83 5.6 Shear flow in beams 86 5.7 Shear centre 88 5.8 Plastic deformations in beams 88 5.9 Strain energy in bending 89 Summary 90

APPENDIX B PROPERTIES OF AREAS 93 B.1 First moments of area and centroid 93 B.2 Second moments of area 93 B.3 Parallel axes theorem 94 B.4 Perpendicular axes theorem 96

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1 Structures, loads and stresses

1.1 Mechanics of material

The subject matter of a course on mechanics of materials deals with structures. A

table or a chair is a structure. A building is a structure. A bridge is a structure. A

TV tower is a structure. So is a printed circuit board, the casing of a fax machine,

or the body of a car. Among the many purposes of the various structures, one

common purpose is to resist and/or transmit forces acting on it. By resisting a

force we mean that the structure would not break under the force. The structure

of a building is designed to resist the loads which include the weight of the

people and things occupying it, the forces of wind acting on it in a storm, even

the load imposed by an earthquake, and the self-load of the building itself. The

structure of an aeroplane resists the aerodynamic loads, the weight of its

occupants (including the dynamic loads during acceleration and deceleration),

the load imposed by the thrust produced by the engines, and of course the weight

of the structure itself.

How does a structure resist loads?

Consider a simple case of a cantilever beam loaded as shown in Fig. 1.1. If the

beam is in equilibrium, the net force or moment on the beam or on any part of it

must be zero. Let us consider the part of the beam within the dashed-line box

shown. This is known as a free-body diagram (FBD).

Clearly, this part of the beam is not in equilibrium with just the external force, P.

We need additional external (i.e., external to this part of the beam) forces and/or

moments. The open arrows in Fig. 1.1b show the external force and moment

required to balance the applied load P. We will for the time being refer to these

as the reaction forces and the moments.

Where do these forces and

moment come from?

As we apply the force P to

the beam and if these

reactions do not kick in,

the beam will tend to

shear from the stump built

into the wall at the left-

end. The distortion of the

beam so produced results

in generation of material

forces within the beam

that resist this shearing action. When we consider the part of the beam shown in

the free body of Fig. 1.1b, these material forces appear as external forces (and

moments) on the beam. Of course, there are equal and opposite reaction on the

stump of the beam built in the wall.

We can summarize the above as:

The external forces acting on a structure result in deformation of the

structural members.

The deformation so caused result in resisting forces within the material

of the members.

When we consider the equilibrium of a part of the member, these

internal forces come into play as external forces and balance the applied

forces or moments.

1.2 Deformation and resisting forces

Consider a vertical rod anchored as shown in Fig. 1.2. It is common knowledge

that when you apply a longitudinal force P to this rod, it elongated a definite

amount (depending on its dimensions and its material). Consider a portion of the

rod enclosed by the broken line rectangle. The free body diagram (FBD) of this

part is shown in Fig. 1.1b. Since the rod is in equilibrium after the elongation,

there must be a force that balances the applied force P. Where does that force

come from? Clearly, there are internal forces which are holding this part from

running away from the rest of the member. These internal forces as seen in the

previous section are the consequence of the distortion produced in the bar. Now

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more the force we apply, more is the

elongation, suggesting that the resisting

force that develops in the rod depends

on the elongation. Robert Hooke, a

British scientist is credited to be the first

to explore the relationship between the

resisting force and the elongation. He

found out in 1678 that for a given

material, the resisting force does not

depend on the elongation but on the

relative elongation produced. He

introduced the term strain to denote the

elongation relative to the original length

of the bar. If l denotes the original

length, and δ the elongation, the strain is

defined by

Strain, (1.1)

Strain is dimensionless and has no units.

Hooke also found out that it is not the force, but the intensity of force measured

as the force per unit area that needs to be considered. He called it stress. If A is

the area of cross-section of the bar, the stress is defined as the resisting force P

divided by A.

Stress, (1.2)

Stress has the dimensions of force per unit area (hydraulic pressure, too, has the

same dimensions) and has SI units of Newton per meter squared (M/m2) which is

termed as Pascal and abbreviated as Pa.

He further found that the stress and strain, in a large part, have a simple linear

relation for bars made of the same material:

Stress Strain, or, (1.3)

The constant of proportionately, E, is termed as the elastic modulus, and depends

on the material of the bar. Strain being dimensionless, the dimensions (and

units) of E are the same as those of stress.

Combining Eqs. 1.1 -1.3, we get:

(1.4)

The value of the elastic modulus E for most construction materials is quite high,

denoting that it takes fairly large forces to produce small elongations. Table B.1

in Appendix B gives the values of the elastic modulus for some common

materials. Steel has about the largest value of the elastic modulus of about 200

GPa1. Cast Iron has about half this value. Aluminium is still lower at 70 GPa.

The summary statements of the previous section can now be recast as:

The external forces acting on a structure result in strains in the structural

members.

The strains so produced result in stresses within the material of the

members.

The stresses, for the most part, are proportional to the strains produced.

The constant of proportionality is termed as the modulus of elasticity.

1.3 Other loadings, stresses and strains

We had in the previous section considered one kind of load that tends to elongate

a member, leading to one type of strain (longitudinal) and one type of stress

(tensile). It is possible to load members in various other ways. Compression

load is a familiar example. Compression results when two bodies are pressed

together. Columns that support elevated highways (Fig. 1.3), water tanks or roofs

are all compression members. A compression member shrinks in length because

of the load, resulting in compressive strains and stresses. The footing of a

machine is also under a compressive load. So also is the boom of the crane.

Compression also results in the situation shown in Fig. 1.4. Here two plates are

riveted together. As a force is applied to the plates that tends to pull them apart,

the rivet compresses the plates at the rivet holes as shown. The compressive

stresses so produced in the plates are also termed as the bearing stresses.

1 A GPa is 10

9 Pa, or 10

9 N/m

2

Fig 1.3. The columns supporting a highway deck, the boom of a

crane and a foundation block are all structural members in

compression

Fig. 1.2 A bar loaded

longitudinally

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Fig 1.6 Twisting of a shaft. The shear stresses in the cross-section of the shaft

give rise to a resisting twisting moment.

Another type of load is the shear load. Consider a block of rubber glued to a

table on one side and a board on the other (Fig. 1.5). If we apply a load P as

shown, the block of rubber will undergo distortion and it will tend to slide off the

table. The distortion results in what are termed as shear strains, which, in turn,

result in shear stresses in the material.

Shear stress and shear strain is also produced in the rivet of Fig. 1.4. As the force

is applied the rivet has a tendency to shear at the middle. Shear stresses develop

to counter this action and keep each half the rivet in equilibrium.

Shear strain and shear stresses are also produced when a shaft is twisted as in Fig

1.6. If we consider the shaft as an assembly of thin slices stacked together, the

twisting action of the shaft tends to make the slices slip on one another. Internal

forces develop which resist these motions. These internal resisting forces are the

shear stresses and the resultant of these is a moment, termed as the twisting

moment.

Another type of distortion occurs when a moment is applied to a bar which tends

to bend it. As shown in Fig. 1.7, the upper fibres of the beam tend to shorten

from their original length, introducing compressive strain and resulting in

compressive stresses therein, while the lower fibres elongate introducing tensile

strain and tensile stresses therein.

The magnitude of these compressive and tensile stresses is such that they

integrate out to zero, which they must, since there is no applied force on the bar.

But the resultant moment is non-zero as is required to resist the applied bending

moment and ensuring equilibrium.

Let us summarize what was learnt in this section:

Forces that tend to reduce the size of a structural member produce

compressive strains which, in turn, produce compressive stresses.

Within limits, the magnitude of compressive stresses vary linearly with

the compressive strain, the constant of proportionality being the same as

the modulus of elasticity, E introduced in the last section with tensile

stresses and strains.

Fig. 1.7 Bending of a beam.

Fig 1.5 Shear in a block of rubber

Fig. 1.4 Riveted plates

Bearing stresses in

the plate

Shear stresses in

the rivet

PP

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Forces that tend to distort the shape of a member, as in Fig. 1.5, produce

shear strains which in turn produce shear stresses. The more the shear

strain, the more is the shear stress.

A twisting moment applied to a shaft produces shear strains in the shaft.

These shear strains give rise to shear stresses across the cross-section of

the shaft, which result in a moment which balances the external twisting

moment. The more the external moment, more is the strain, more the

stress, and more the resisting moment.

A moment tending to bend a member as in Fig. 1.7, produces both

compressive and tensile strains and stresses in the beam. These stresses

give rise to a resisting moment which balances the bending moment.

1.4 The concept of stress at a point

In Section 1.2 we were dealing with a very simplistic case of a straight bar of

area A with a longitudinal load P and we defined stress simply as load per unit

area. This assumes a uniform distribution of stresses – a very severe

assumption. In practice, the deformations due to load and, consequently, the

strains and stresses will vary from point to point. It is, therefore, convenient to

define stress as the intensity of force at a point.

Consider a gear which is meshing with another gear which applies a force P on a

tooth (Fig. 1.8). Let us consider the equilibrium of this tooth. Fig 1.8b shows the

part of the tooth as a separate free body. The force that balances the external

force F is the force which was binding this tooth to the rest of the gear. This is

an external force that arises because of the loading of the gear and its consequent

straining. Let us imagine this straining and the internal forces are not distributed

uniformly on this material. Consider an elemental area2 δA on a face of a

structure exposed by making a cut at that location. Let δF represent the force

that the material that has been removed was applying at the elemental area δA.

This force arises from the deformation of the material under whatever external

load is being applied to the structure.

The stress at a point is defined as the intensity of the internal force of

deformation at this point:

Stress vector on this face at this point is t

The stress vector t depends upon the location as well as the direction of the

surface. If we had made the cut (at the same point) with a different inclination

(i.e., with the outward normal in some other direction), the stress vector would

have been different as is shown in Fig. 1.9

We can take the component of this vector along the normal to the surface and

along the surface:

The component of the stress vector normal to the surface is termed as the normal stress and is denoted by the Greek letter ζ. The component of the stress

vector along the surface is termed as the shear stress and is denoted by the Greek

letter η. As noted before, the dimensions of stress are Force/Area, or FL-2

, or ML-

1T

-2, and its SI unit is N/m

2 or Pascal, Pa. A Pascal is a very small unit and it is

common to have stresses in kPa or MPa.

Fig. 1.10a shows a bar loaded uniformly in tension as shown. Because of this

load the bar elongates setting up axial strains and stresses along any cross

section. Fig. 1.10b shows the resulting free body. It stands to reason that if we

take a section of this bar at any level, the resulting distribution of stresses must

be the same. However, if the loading is not uniform but concentrated at a point in

2 Area is a vector quantity, its direction denoted by the direction of the outward

normal to the surface.

(a) A gear (b) A cut-out portion of the tooth

Fig. 1.8 Stress at a point

(a) A solid object (b) Section through X along bb (c) Section through X along cc

Fig. 1.9 Stress depends on the direction of the cut

FFF

FArea δA

δF FFArea δA

δF

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the bar as shown in Fig. 1.11a the story

changes. There is no reason to expect that

the stress distribution at a section is

uniform. In fact, it is not. Moreover, it

changes as we move from section to

section. If we take a section at different

heights along planes b, c, or d, the

resulting free bodies are shown in Fig.

1.11b, c and d. There total of the internal

forces or stresses at each of these sections

must be equal to P to balance the applied

force. But the distribution of these

stresses is quite different at the three

sections. The distribution of stresses becomes more uniform as we move

upwards, away from the point of application of the force P.

In fact, there is a principle known as St. Venant principle, named after a

nineteenth century French theoretician, which lays down this behaviour:

The difference between the stresses caused by statically equivalent load systems

is insignificant at distances greater than the largest dimension of the area over

which the loads are acting.

The loading in Figs. 1.10 and 1.11 are statically equivalent since they result in

the same net force and moments on the structure. The stresses in Fig. 1.10 (with

distributed loading) is uniform everywhere, and therefore, far away from the

point of concentrated loading in Fig. 1.11, we can take the stresses to be

uniformly distributed.

We shall, in this elementary text, will routinely make the assumption of uniform

stresses across sections, unless the context of the problem forbids it.

Example 1.1 Bearing stresses in foundations

Consider a wooden column (Fig. 1.12) resting on a concrete footing. Determine the maximum value of the load P if the maximum permissible stress in concrete is 60 MPa and in wood is 25 MPa.

Solution:

We assume that the stresses at any cross-section are

distributed uniformly. If P is the external load, the

stress in the wooden column is P (N)/ [π× (0.010 m)

2] = 3,183 P (N/m

2 or Pa). Similarly, the stress in

the concrete footing is P (N)/ [(0.030 m) 2

] = 1,111

P (N/m2 or Pa). Equating these stresses to 25 MPa

and 60 MPa, respectively, we find the value of P

from the first as 7.85 kN, and from the second as 54

kN.

Clearly, the maximum permissible load will be the

lesser of the two, i.e., 7.85 N. This suggests that we

can reduce the footing size drastically, if bearing

this load is the only design consideration.

Further we have considered only the compressive

stresses. Thin columns may buckle under

compressive load much earlier than they fail in compression. We will need to

check that too. We shall deal with this in a later chapter.

P P P P

(a) (b) (c) (d)

Fig. 1.11 The stress distribution becomes uniform as

we move away from the point of loading

d

c

b

(a) (b)

Fig. 1.10 The stress distribution

on a uniformly loaded bar

Fig. 1.12

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Example 1.2 A truss

A truss structure is a load bearing structure assembled such that its members carry only axial loads, either tension or compression. It is made up of straight, two-force members3 joined together by frictionless pins. All the loads are applied at only the joints. Further, there are no external moments applied to such structure.

A truss is a gross approximation to an actual structure. There are no practical

frictionless pins, and most members carry some bending loads as well. But the

advantage of considering only two-force members (in simplifying calculations) is

so enormous that we make the truss approximation wherever feasible.

Fig. 1.13 shows a pinned structure consisting of two bars AB and BC from which a weight P hangs as shown. If the diameter of each bar is 6 mm and the permissible stress in the bar is 80 MPa, find the value of the maximum load P.

Solution:

Consider the

FBD of point B

where the load P

is being applied.

Since AB is a

pinned member,

and hence a two-

force member

there can only be

tension along it. Let us call it TBA. Similarly, there will be tension TBC along the

member BC.

The equilibrium of a body requires that the vector sum of forces and moments

should vanish:

∑ ∑ (1.5)

We need to apply only the force equations here. Taking the component-wise

sum of the forces, we get

3 A member on which external forces act only at two distinct points (and there is

no external torque acting on it) is termed as a two-force member. The forces

acting on a two-force member are collinear and opposed.

∑ (a)

∑ (b)

Solving the two simultaneously, we get

TBA = 0.51P, and TBC = 0.78P

The stresses in the bars are obtained by dividing the

above load by the cross-sectional area (assuming

uniform stress distribution), which is π× (0.003 m) 2

= 2.83×10-5

m2. The stress along BC will be the

limiting stress, since it is higher. Thus,

This gives the maximum load P as 2.9 kN

Example 1.3 A truss with roller support

Consider a structure shown in Fig. 1.14. If the members of the structure are mild steel with a radius of 20 mm, find the stresses in members AB, BC and AC.

Solution:

The support at point A is termed as a hinged support. This means that the support

cannot apply any moment at this point to the structure and the structure can

articulate at this point. The only possible external reactions (from the support) at

this point is a force with components both along and normal to the surface.

There is no reaction moment at this point. The support at point B is termed as a

roller support. A roller support, besides permitting rotation of the structure,

permits translation of structure along the supporting surface. This means that not

only is the reaction moment zero, but the force component along the support

surface is also zero. The only reaction possible at such a support is a force

component normal to the surface.

Fig. 1.13

Fig. 1.14

20kN

1 m

1 m

A

C

B

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To find the stresses in the members, we need to determine the forces on them by

doing a statical analysis. The first step in solving this problem consists of

determining the reactions at the supports A and B.

The reactions at support are found by considering the equilibrium of the FBD of

the whole structure as drawn in Fig. 1.15a. Here all the external forces have

been shown: the applied load of 20 kN, the horizontal and vertical reactions RAx

and RAy at the pinned support A, and the horizontal reaction RBx at the roller

support B. The structure should be in equilibrium under the action of these

forces. There are two conditions for equilibrium: vector sum of all the forces as

well as all the moments must be zero.

Writing the sum of horizontal and vertical forces as zero gives us:

∑ (a)

∑ N (b)

Eq. (b) gives RAy = 20 kN. There is only one equation to determine the other two

unknowns. We need one more equation. That we get from the moment balance.

We can take the moments of forces about any convenient point. Here we take it

about the point A (because then the two force components at this point contribute

no moment):

∑ ( ) ( ) ( ) (c)

This gives = 20 kN, and, then Eq. (a) gives 20 kN, that is, 20 kN in

a direction opposite to that shown on the FBD in Fig. 1.15a, or 20 kN towards

left.

Fig. 1.15b shows the forces on the member AC. It is clearly that the net force on

the member is along the length4 of the member, tensile, and equal in magnitude

to √( ) ( ) The member BC is under a compression

of 20 kN.

Further, since at point B

there is no vertical force,

there cannot be any

force5 in member AB.

The stress in the two

members can be found

by dividing the forces

with the respective cross-

sectional areas. Thus,

the stress in member AC

is 28.3 (kN)/(π/4)(15×10-

3 m)

2 = 160.1 MPa, tensile, and that in member BC it is − 20 (kN)/(π/4)(15×10

- 3

m)2 = 113.2 MPa, compressive.

Example 1.4 Key

Consider the transmission of power by belt and pulleys. Fig. 1.16 shows a pulley

being driven clockwise by a belt. If the pulley is turning at 100 RPM and if the

power transmitted is 1 kW, what is the shear stress in the key? The shaft dia. is

20 mm, and the dimensions of the key are 4 mm × 4 mm and a length of 25 mm.

Solution:

A key is a common device used to couple a pulley with a shaft, so that as the

pulley rotates the shaft rotates with it. It is a metal piece inserted so that a part of

it is inside the shaft (in a slot termed as a keyway), and a part is within the pulley.

As the pulley turns, the key moves with it. The pulley applies a force on the key

towards the right as shown in Fig. 1.16b. The key whose lower part is enmeshed

with the slot within the shaft makes applies a force on the shaft to make it turn

with the pulley. As the key pushes the shaft, the shaft, in turn, pushes the key

back towards the left.

4 This should have been obvious without the calculations since members AC and

BC are two-force members. The forces acting on a two-force member are

collinear and opposed. 5 This is quite interesting. If there is no force in member AB, we really do not

need that member. Can the structure survive without the member AB? The

answer is: yes, if we are concerned only with the equilibrium, and no, if we

worry about the stability.

(a) FBD whole structure (b) Force along member AC

Fig. 1.15 (a) (b) (c)

Fig. 1.16 Shear stresses in a key

(a) (b) (c)

Fig. 1.16 Shear stresses in a key

20kN

1 m

RAx

CRBx

RAy

20kN

1 m

RAx

CRBx

RAyA

C

20 kN

20 kN

20kN

20kN

A

C

20 kN

20 kN

20kN

20kN

A

C

20 kN

20 kN

20kN

20kN

11

Now, if we look at Fig. 1.16b, we notice that the upper part of the key is being

pushed to the right while the lower part of the key is being pushed to the left.

We may say that the key will have a tendency to shear in the middle. Fig. 1.16c

shows the FBDs of the lower and the upper parts of the key. The internal shear

stresses acting on each half balance the externally applied forces.

Since the power transmitted is I kW, and the RPM is 100, the torque is 1000 W/

(2π r d/rev×100 rev/min/ 60 s/min) = 95.5 N.m. The force acting on the key (at

a radius of 10 mm) which transmits this torque is obtained as 95.5 Nm/0.01 m =

9,550 N

The total shear force

acting on the key (the

lower or the upper part)

is, thus, 9,550 N. The

area on which it acts is 4

mm × 25 mm and,

therefore, the shear stress

is 9,550 N/(4×10-3

m ×

25×10-3

m) = 95.5 MPa.

As an aside, note that the

applied load is producing

compressive or bearing

stresses as well. The

area on which the

compressive force (9,550

N) acts is 2 mm × 25 mm

and, therefore, the bearing stress is 9,550 N/(2×10-3

m × 25×10-3

m) = 190.1

MPa.

Example 1.5 Riveted lap joint

Consider two steel plates 2 mm thick joined together as shown in Fig. 1.16(a)

and (b) by two rivets of 10 mm diameter. The centres of rivets are 10 mm from

the edge of the plates. Determine the stresses in the rivets if a force of 2 kN is

applied to the plates trying to pull them apart.

Solution:

As was discussed earlier, the pull force tends to shear the rivets as shown in Fig.

1.17c. Since there are two rivets, we can assume at each rivet sustains half the

total force, i.e., 1 kN. This is the shear force in each rivet acting on the cross-

section l re of π× (0.005 m)2 = 7.85×10

-5 m

2. Therefore, the shear stress in

rivet shanks is 1 kN/7.85×10-5

m2 = 12.7 kPa.

Further, the pull force will tend to shear the plates as shown in Fig. 1.16d. This

shearing action acts on a total area of 2×(10 mm × 2 mm) = 4×10-5

m2

for the

force in each rivet, which is 1 kN. Thus, the shear stress in the plates will be 1

kN/4×10-5

m2 = 25 MPa.

Further the pull force will tend to crush the area of the plate as shown in Fig

1.16e, causing compressive or bearing stresses in the area in immediate contact

of the rivet. Surely these will not be distributed uniformly over the area. But if

we make the assumption that they are, the resulting bearing stresses will be 1

kN/(2 mm × 10 mm) = 50 MPa.

1.5 Stress on oblique planes

Consider again a bar of area A loaded axially as shown in Fig. 1.18a. Let us look

at the stresses on an oblique plane b inclined at an angle6 θ. It is clear that if we

draw the FBD of the lower portion of this bar as in (b), the total internal force

acting on this section is P as shown. But what is the stress here?

The area on which this force acts in not A but larger than A, equal to A/cosθ. If

we assume, as before, that the stress is distributed uniformly over the area (which

shall be quite true if the plane is not to close to the point of application of the

load) the stress vector here will be P divided by ( ) , or t = Pcosθ/A. We

can resolve this stress vector in two components (Fig. 1.18c), one normal to the

oblique area, and the other along it. The normal component is the tensile stress:

( ) ( ) (1.6)

This is also termed as the normal stress. The tangential component is a shear

6 The ngle θ is lso the ngle between the norm ls to the new pl ne nd the

original plane.

Fig. 1.17 Lap joint

P(a) (b) (c) (d) (e)

Fig. 1.18 Stresses on an oblique plane

P

Pθ

b

ζη ζ-η

η-η

12

stress:

( ) ( ) . (1.7)

Clearly, the value of the tensile and shear stresses at a location in the bar varies

with the angle of inclination θ of the plane being considered. The maximum

tensile stress occurs when θ is zero, i.e., the cut is perpendicular to the axis of the

bar, and reduces continuously as θ increases. But the behaviour of shear stress is

quite interesting. If we note that cosθ sinθ is (1/2)sin2θ, we see that the shear

stress η first increases as θ increases, attains a maximum value (equal to P/2A,

i.e., half the maximum value of the tensile stress) when θ is π/4, and then

decreases back to zero as θ increases to π/4. Also note that if the sectioning

plane is inclined in the other sense (with negative values of θ), the shear stress

on it has a reversed sense too. This is also obtained from the expression for η

given above since θ now is negative.

Two planes, one at angle θ, and the other at angle - θ are termed as

complementary planes. Shear stress on complementary planes have same

magnitude but opposite sense.

Please note that the equations obtained above are valid only for the special case

of axial loading of a straight uniform bar. If the geometry or loading were

different, these would no longer be valid.

Example 1.6 A Cardboard tube

A cardboard tube of diameter 70 mm is made from a strip of width 60 mm and

thickness 2 mm wound spirally with the edges glued together as shown in Fig.

1.19a. The cylinder is subjected to an axial load of 100 N, determine the shear

stress in the

glued joint.

Solution:

The angle of

the spiral can

be obtained by

imagining that

as the

cardboard strip

is wound up,

the strip should

advance in the

axial direction

a distance

equal to the

width of the strip as we go around the tube through one circumference, 2πR as

shown in Fig. 1.19b. This angle θ is, therefore,

.

The compressive stress in the cardboard tube along the axial direction is 100 N/

[2π(35×10-3

m)2×(2×10

-3 m)] = 227 kPa.

We can now use the Eq. 1.7 to determine the shear stress along the seam, which

is a surface with the normal in the direction as shown in the figure. It can be

seen that the angle θ here is the same as the angle in Eq. 1.7 (the angle between

the normals to the axial plane and the new plane on which stresses are to be

determined). Therefore,

( ) ( ) = 57.7 kPa

This is the shear stress along the glued seam. The seam would rip if the glue

cannot sustain this level of shear stress.

1.6 Notation for stress: double-index notation

As is clear from the

discussion above, the stress

vector at a point depends on

the orientation of the

surface under consideration.

We can, at the same point,

consider many differently

inclined surfaces. The

orientation of a surface is

denoted by the direction of the outward normal to the surface. Therefore, we

must specify the direction of the normal to the surface whenever we talk of stress

at a point.

We shall see later (in Chapter 6) that if we know the stress vectors on three

mutually perpendicular planes we can determine the stress vector on any other

plane. We, therefore, need to specify stresses on three such planes. Figure 1.20

shows stresses on three mutually perpendicular planes, with normals in x, y, and

z directions. We have shown here, the components of stress vectors on each of

these three planes. The tensile (or the compressive) components are represented

by the symbol ζ, and shear stress components by the symbol η. We use here a

double-index notation, with the first index of a stress component denoting the

plane on which it acts, and the second index denoting the coordinate direction of

Fig. 1.20 Specification of stress components

13

a component itself. Thus, ηxy is a component of shear stress acting on a plane

with the normal in the x coordinate direction, the component itself being in the y

coordinate direction. Similarly, ηzy is a shear stress component acting on a plane

with the normal in the z coordinate direction, the component being in the y

coordinate direction, and ζzz is a tensile stress component in the z direction acting

on a plane with normal in the z direction. It should be apparent that there are:

three tensile components, ζxx, ζyy, and ζzz, one each on the three faces

(note that the indices are repeated in each of them),

six shear stress components: two on each face, ηxy and ηxz on the x-face,

ηyx and ηyz on the y-face, and ηzx and ηzy on the z-face. We shall soon see

that three of them are equal to the other three, so that there are only

three independent shear stresses at a point (see section 1.7).

Sign convention: It is convenient to use the following sign convention:

We first define a face to be positive or negative: a face with the outward normal

in the direction of the positive coordinate axis is termed as a positive face; else it

is termed as a negative face. When a positively directed force acts on a positive

face or a negatively directed force acts on a negative face, the stress is assigned a

positive sign. And when a negatively directed force acts on a positive face or a

positively directed force acts on a negative face, the stress component is assigned

a negative sign.

A stress component is considered positive when a positively directed force acts

on a positive face, or a negatively directed force acts on a negative face7.

In other words, if both the sign of the force and the face are the same, positive or

negative, the resulting stress is a positive and if the two signs differ, the stress is

negative. This is summarised in Table 1.1 below:

Table 1.1 Sign convention for stress

Direction of normal Direction of force Sign of stress

In positive coordinate

direction (+)

In positive coordinate

direction(+)

Positive(+)

In negative coordinate

direction (−)

In positive coordinate

direction(+)

Positive(+)

In positive coordinate

direction(+)

In negative coordinate

direction(−)

Neg tive(−)

In negative coordinate

direction(−)

In negative coordinate

direction(−)

Neg tive(−)

7 This sign convention follows from the consideration of action and reaction

having the same sign.

It may be verified that all the stress components shown in Fig. 1.20 are positive.

Example 1.7 Nomenclature and signs of stresses

Fig. 1.21 shows stress

components on some faces

labelled (a) to (d). Name

these planes on which the

stresses have been shown.

Name the stresses and

determine their signs

according to the sign

convention outlined

above.

Solution:

Plane a is an x-plane, b a y-plane, c a z-plane and d is an x-plane (with the normal

in –ve x direction. The nomenclature and signs of the stresses are tabulated

below8:

Stress Index

of

plane

Index for

force

component

Symbol

for stress

Sign

of

plane

Sign of force

component

Sign

of

stress

1 x x ζxx + ve − ve − ve

2 x z ηxz + ve − ve − ve

3 x y ηxy + ve + ve + ve

4 z x ηzx + ve − ve − ve

5 z z ζzz + ve − ve − ve

6 z y ηzy + ve + ve + ve

7 y z ηyz + ve − ve − ve

8 y y ηxz + ve + ve + ve

9 y x ζzz + ve − ve − ve

10 x x ζxx − ve + ve − ve

11 x z ηxz − ve − ve − ve

12 x y ηxy − ve − ve + ve

1.7 Equivalence of shear stresses on complementary planes

8 Note that the signs of stresses on opposite faces a and d are identical, as they

should be. In fact, the sign convention has been designed to ensure this.

x

y

z

Face a

1

3 2

Face b

4

6

5

Face c9

8 7

Fig. 1.21

Face d

10

12

11

x

y

z

x

y

z

Face a

1

3 2

Face b

4

6

5

Face c9

8 7

Fig. 1.21

Face d

10

12

11

14

Consider a small two-dimensional element of dimensions δx and δy and of unit

depth as shown in Fig.1.22. We have

drawn the stress components on the

four faces of the element. We assume

a two-dimensional state of stress, i.e.,

we assume that there is no loading and,

hence, no stress component in the third

direction. We shall now consider the

equilibrium of the element under the

action of the forces due to these

stresses:

Face

(identified by the

direction of

outward normal)

Area

assume unit

depth

Force

x-component x-component

x δy•1 ζxx• δy ηxy• δy

− x δy•1 −ζxx• δy −ηxy• δy

y δx•1 ζyy• δx ηyx• δx

− y δx•1 −ζyy• δx −ηyx• δx

When we consider the force balance, we verify that there is no net force on the

element: the sum of x- and y- forces sum out to zero independently. We next take

the sum of the moments about the centre point of the element. Note that the

tensile forces produce no moment about that point. However, the shear forces

do. The shear forces on opposite faces are equal in magnitude but opposite in

sign and, therefore, constitute two couples, one anticlockwise (positive) and the

other clockwise (negative). The moments of the two couples are found by

multiplying the magnitudes of the forces and the perpendicular distance between

the two lines of action. The moment balance equation gives:

( ) ( ) ,

which on simplification gives = .

This is an important result and establishes that the shear-stress components on

adjacent (orthogonal) faces are equal in magnitude. We can, similarly, show that

= and = . As much was stated in Section 1.6 without proof. We

reiterate what was stated there:

The state of stress at a point can be established with six components of stresses:

three tensile components, ζxx, ζyy, and ζzz, one each on the three

orthogonal faces, and

three shear stress components ηxy, ηyx and ηzx.

We shall show in Chapter 6 that we can, from these six components on three

orthogonal planes through a point, determine the stress vector on any plane

through that point.

1.8 Stresses in a thin circular pressure vessel

Consider a thin9 cylindrical pressure

vessel of length L, radius R, and wall

thickness t. Fig. 1.23 shows the

cylinder with end plate removed. A

cylindrical polar co-ordinate system

with r, θ, and z coordinates is ideal for

this geometry. Let the internal excess

pressure be p. Under the action of

these forces, stresses will be set up in

the cylindrical vessel. Following the

development of the previous sections,

there will be six stress components

required to describe the state of stress

in the vessel walls: three tensile

components: ζrr, ζθθ, and ζzz, and three shear stress components: ζrθ, ζθz, and ζzr.

Let us first consider the tensile stress component ζθθ. As indicated by the first

subscript, θ, it acts on a plane with normal in the θ direction. This stress is

exposed (or m de ‘extern l’) by taking a section as shown by a diametrical

cutting plane as shown. Figure 1.24a shows the FBD of the lower half of the

cylinder (ends removed). We can determine the stress component ζθθ by

considering the equilibrium of the vertical component of all forces.

The total vertical force due to the tensile stresses is ζθθ times the area on which

these stresses act, which is 2×(L×t). Therefore, this force is 2ζθθLt. This force is

being balanced by the vertical component of the pressure forces acting on the

inside the half-shell. The integration of the vertical component of pressure forces

acting on this half of cylindrical shell is not straightforward. But it can be

determined quite easily by resorting to a frequently used trick. The pressure force

acting on the shell is equal and opposite to the pressure forces acting on the gas at

the shell wall (the principle of action and reaction). Consider the ‘FBD’ of the

‘g s’ cont ined in this h lf shell s shown in Fig. 1.24b. The vertic l net pressure

forces on the curved surface this FBD (which is the same and opposite to the

9 ‘Thin’ refers to the condition th t the w ll thickness t is much less than the

radius R.

Fig. 1.22 A 2-D infinitesimal element

Fig 1.23 A thin cylindrical vessel with

end plate removed

ζxxζxx

ζyy

ζyy

ηyx

ηyx

ηxy

ηxy

•

δy

δx

x

y

ζxxζxx

ζyy

ζyy

ηyx

ηyx

ηxy

ηxy

•

δy

δx

ζxxζxx

ζyy

ζyy

ηyx

ηyx

ηxy

ηxy

•

δy

δx

x

y

L

R

tz

θr

L

R

tz

θr

15

pressure forces of Fig. 1.24a) is balanced by, and hence equal to the pressure

force on the diametrical plane of this FBD. This is p(2RL). Therefore, the

integrated pressure force on the curved surface of the FBD of Fig. 1.24a is 2pRL.

2 ζθθLt = 2pRL, or,

(1.8)

This tensile stress in the circumferential direction is also termed as the hoop

stress.

We next consider the tensile stress ζzz which is the stress on the plane with

normal in the axial

direction. Such a

face is exposed when

we cut the cylinder

with a r-θ plane

parallel to the ends

as shown in Fig. 1.25

which shows the

FBD of one part of

the vessel so

exposed. The

external forces in the

z-direction acting on

this FBD are due to the tensile stress ζzz acting on area which can be

approximated by 2πRt, and the pressure p acting on the end plate area of πR2. The

z-force equilibrium, then, gives: ζzz×2πRt - p× πR2. This gives:

(1.9)

The third tensile stress is ζrr in the radial direction. Note that on the inner

curved surface the pressure p acts radially and,

therefore, is the stress ζrr at that point. But on

the outer curved surface, the pressure is zero, so

the stress ζrr is zero. Therefore, ζrr varies from

0 to p across the thickness of the cylinder. In

any case, since r >> t, the value of ζrr (between

0 and r) is much less than ζzz or ζθθ. Therefore,

in comparison to the other tensile stresses in the

cylindrical shell, it is common to neglect ζrr,

(1.10)

Let us next consider the shear stresses. We first

discuss ηθr. We shall use a very interesting

class of arguments termed as the symmetry arguments. Consider the two halves

of the cylinder as shown in Fig. 1.26. We have shown the shear stress

components ηθr on the upper half. Note that on the left-hand side, the plane has a

normal in the +ve θ direction (being counter-clockwise), and the stress is in the

+ve r direction. Therefore, the stress is ηθr and is +ve. On the right-hand side of

the upper half, both the signs are negative and therefore the stress is +ve again.

The two stresses having opposite sense is correct, since the horizontal forces

acting on this FBD should some out to zero (the sum of the horizontal component

of uniform pressure forces can safely be assumed to be zero).

Now on the lower half, the Newton third law (the

principle of action and reaction) dictates that the

stresses should have a sense opposite to that on the upper half. This is as shown.

But here we run into a problem. The stresses are both outwards on the upper

half, and both inwards on the lower half. This violates symmetry. It is easy to see

that it is not possible to distinguish between the upper and the lower halves of the

cylinder. If somebody came and switched the two halves while a reader was

away, there is no way by which the reader can tell which is which. This

symmetry requires that the state of stress on the two halves must be the same,

either both inwards or both outwards. Thus, the requirements of third law and

that of symmetry are in contradiction, and the both must hold! The resolution

this contradiction is possible only if the stress ηθr vanishes. This is a sufficient

proof. We can construct similar symmetry arguments to show that the other two

shear components ηrz and ηzθ too must vanish. Thus,

ηθr = ηrz = ηzθ = 0 (1.11)

1.9 Summary

The external forces acting on a structure result in strains in the structural

members. The strains so produced result in stresses within the material

of the members. The stresses, for the most part, are proportional to the

strains produced.

The constant of proportionality is termed as the modulus of elasticity

which is a property of the material of which the structural member is

made.

In general, the stresses vary from point to point, but the use of the St.

Venant principle permits us in many simple situations to assume

uniform stress distribution across a section. This is a useful particularly

at sections fairly distant from the points of application of loads.

Stress vector at a point describes the intensity of internal forces that

develop within a material in response to distortions that are produced

under the application of external loads.

(a) (b)

Fig 1.24 FBD of one half of thin cylinder exposing ζθθ

Fig. 1.25 FBD for determining ζzz

pσθθ

σθθ

ppσθθ

σθθ

ζZZ

p

ζZZ

p

16

The stress vector at a point depends on the orientation of the surface on

which the stress acts. Eqs. 1.6 and 1.7 describe the formula for

calculating the tensile and shear stresses on an oblique plane in the case

of longitudinal loading of a bar.

The stress at a point can be described by stating the stress vectors on

three mutually perpendicular planes with a total of three tensile

components and six shear components.

A double index notation is used to designate stress components, the first

index representing the direction of the outward normal to the surface,

and the second index the direction of the component itself.

A stress component is considered positive when a positively directed

force acts on a positive face, or a negatively directed force acts on a

negative face.

Equilibrium of forces require that the shear stress components occur in

pairs, so that the shear stresses on complementary planes are equal:

= , = and = . Thus, there are only three independent

shear-stress components.

We showed by symmetry arguments that the shear stress components in

a thin walled cylindrical pressure vessels all vanish.

17

2 Deformations, strains and material properties

2.1 Fundamental strategy of mechanics of deformable mechanics

All structures resist loads by deforming. A structure deforms as a load is applied

to it. As it deforms, the stresses build up within the structure to resist the applied

load. More the load more is the deformation and more are the stresses. The

deformation increases till the resulting stresses are sufficient to balance the

applied load.

We had considered in Sec. 1.2 the

deformation of a uniform bar under a

longitudinal load. If we take bars of

various cross-sectional areas and of

various lengths and plot the variations

of deformation with the loads, we

obtain plots as shown in Fig. 2.1b.

The variations are largely linear with

slopes that are different for different

bars. We had seen earlier that if we

convert the load to stress (stress, ζ =

load, P/cross-sectional area, A) and

deform tion to str in (str in, ε = elong tion, δ /original length, L), the lines with

different slopes collapse into one line, as long as the material of the various bars

is the s me. This f ct w s st ted s Hooke’s l w (Eq. 1.3), , where the

constant of proportionately E is termed as the elastic modulus, and is the

property of the material of the bar.

The above discussion can be seen in a slightly different light. The deformation

produced in the bar depends upon the load, the geometry of the bar and its

material. So does elongation. But the geometry of the bar is absent from Eq.

1.3. Once you convert the load to stress and deformation to strain, the

geometry of the structure becomes irrelevant.

The above is a very powerful insight and forms the basis of the fundamental

strategy of analysis in mechanics of deformable materials. This strategy can be

stated in the following manner: To determine deformations in a structure under a

given loading, we first convert the loading to stresses using equilibrium

considerations, convert stresses to strains using the material properties, and the

use the geometry of the structure to determine deformations from the strains so

calculated. This is also stated as a formula: macro to micro, conversion at micro

level, and then micro to macro. Loading is macro, stress and strain are micro

level, and deformation is macro level.

We may, at time, need to go in the reverse direction. We may be given the total

deformation from which we need to determine the loading: we calculate strains

from deformation, convert strains to stress using material properties, and the

integrate stress to find the loading. The strategy here too is macro to micro to

micro to macro.

Example 2.1 Tug of war

Consider a tug of war in which 6 young men are pulling on a manila rope (cross-

sectional area: 6 cm2) with forces as shown. Find the net elongation of the rope.

(a) (b) (c)

Fig. 2.1 Elongation of bars with loads

P

δ

ζ =P/A

ε=

δ/L

Area A

l

δ

P P

δ

ζ =P/A

ε=

δ/L

Area A

l

δ

P

Area A

l

δ

P

18

Solution:

Each section of the rope has a different tension, resulting in different stresses,

strains and elongations. Let us take the four steps of our strategy in sequence.

Loading to stresses:

To find the stresses, we first need to determine the tension in each section of the

rope. There are five distinct sections carrying tensions. (The sections at either

end do not have any tension, and, therefore, do not need any consideration.) This

is done by making appropriate FBDs which will make the required tension force

an external force. Fig. 2.3 shows a sequence of FBDs drawn for this purpose.

The first of these has externalised the tension TAB in the section AB of the rope.

By simple balance of forces we obtain TAB as 250 N. Similarly, from the other

FBDs we get TBC = 500 N, TCD = 800 N, TDE = 550 N, and TEF = 300 N.

If we assume a uniform distribution of stresses, which will be quite true away

from the points where loads are applied (St. Venant principle), we can find the

stresses in each section by dividing its tension by the cross-sectional area of the

rope. Thus, the stress in section AB is ζ = 250 N/0.0006 m2 = 416.7 kPa. The

stresses in other sections can be determined in a similar manner and are given in

column 3 of Table 2.1 below.

Stresses to strain:

Stresses nd str ins re rel ted by Hooke’s l w: ζ =Eε, where E is the elastic

modulus of the material. A search of literature reveals significant variations in

the value of E for manila ropes. A value of 100 MPa appears to be a good

approximation10

. We get the strains by dividing the stress values by this value of

E as shown in column 4 of Table 2.1.

Table 2.1 Calculation of elongation of the rope of Example 2.1

Strain to deformation:

Once the strains are calculated, we can determine the elongation in each section

by multiplying the strains with the length of the respective section. This has been

shown in column 6 of the table. The total elongation is 66.74×10- 3

m, or 6.67

cm.

Example 2.2 A hanging cable

Consider a cable of uniform section hanging as shown in Fig. 2.4. A steel cable

hangs from a roof under its own weight. It is important to see that the tension

along the length of the cable is not constant, but varies. This can be seen by

considering two sections, one at level b, and the other at level c. We have, in Fig.

2.4b, drawn the FBD of the cable up to level b. The tension T1 is clearly equal to

weight W1. The weight W2 in the FBD of Fig. 2.4c is no doubt greater than W1,

and therefore, T2 is greater than T1. Thus, the tension, and hence the stress along

10

The value of E for the rope accounts also for its unraveling as it stretches.

(1)

Section

(2)

Tension, N (3)

Stress, kPa

(4)

Strain

(5)

Length, m

(6)

Elongation, m

AB 250 416.7 4.16×10- 3

1.5 6.24×10- 3

BC 500 833.3 8.35×10- 3

2.0 16.70×10- 3

CD 800 1333.3 13.33×10- 3

1.5 20.02×10- 3

DE 550 916.7 9.20×10- 3

1.5 13.78×10- 3

EF 300 500.0 5.00×10- 3

2.0 10.00×10- 3

Fig 2.2 Tug of war

Fig. 2.3 FBDs for determining tensions in the rope

19

the cable increases from bottom to top.

The diameter of the cable is dictated by

the maximum stress in the cable. There is

a considerable wastage of material, since

the material near the bottom is not loaded

to its capacity.

In situations where the material costs are

heavy, it often pays to reduce the cable

diameter as we go down from the top.

Fig. 2.5 shows a cable where it has been

done. Its diameter at x = 0 is Do, and at x

= L is D1. The variation in diameter is

such that the stress at any given location

is constant equal to ζo. Find the

differential equation governing the

variation of diameter with x. Also set up an equation to determine the total

extension of the cable as a

function of the various

parameters involved.

Solution:

What is important to realize in

the problem is the fact that the

tension at any location varies

with x, the distance from the

roof. The weight supported by

the section at x reduces as x

decreases. We first determine

the tension as a function of x.

To determine the tension at x,

we take a slice of thickness dx of the cable at x, and draw its FBD. This FBD has

three forces, the upward tension T at x, the downward tension T + dT at x + dx,

and the weight dW of this slice of the cable. Clearly equilibrium requires that the

algebraic sum of the three be equal to zero. If the variable diameter of the cable

is denoted by D as a function of x, the tension T is easily estimated in terms of

the stress ζo as ( ), the stress being given as constant throughout.

The weight dW is easily estimated as ( ) , where ρ is the density of

the cable material and g the acceleration due to gravity. To determine the tension

T + dT, we have to note that the area at the bottom of the FBD is different from

that at the top. The tension here is thus,

[ . .

/ /

]

.

/ , neglecting terms of second

order in dx.

The force balance will then give, on simplification:

, where the negative sign indicates that diameter D decreases

as x increases. Using the boundary condition that the diameter D = D0 at x = 0,

we can solve this to get11

How would we now calculate the total elongation of the cable?

We first convert the stress at each section to the strain at that section. Since the

stress everywhere is the same, equal to , the strain is also the same everywhere

equal to , where E is the elastic modulus. And, therefore, the total

elongation is easily determined as .

Example 2.3 Deflection in an elementary truss

Consider again the truss with a roller support (Fig. 2.6) discussed previously as

Example 1.3, where the forces and stresses in the various members AC, BC and

AB were determined. It was shown that there is no force in AB, a tension of 28.3

kN in member AC, and a compressive force of 20 kN in the member BC. The

respective stresses were found to be 0, 160.1 MPa, and 113.2 MPa. Determine

the displacement of point C where the load of 20 kN is applied. All members are

made of steel (E = 210 GPa).

Solution:

The equilibrium part has already been solved. So also the stress determination.

We now need to find the displacement of point C to C1.

For this purpose, we first find the elongation (or shortening) of members AC and

BC. The stresses in the two members have already been calculated. The strains

are now calculated as stress/E. Thus,

εAC = 160.1 MPa/ 210 GPa = 0.76×10-3

, and

εBC = − 113.2 / 210 G = − 0.54×10-3

,

11

What about the second boundary condition: D = D1 at x = L? There is no other

constant to determine. Or is there not? The resolution of this question lies in the

fact that the constant stress ζ0 cannot be specified independently of D1 and L.

How would you determine ζ0 in terms of the other parameters?

Fig. 2.5 Cable of variable diameter

(a) (b) (c)

Fig. 2.4 A hanging cable

20

the negative sign

indicating that the

strain is compressive.

We convert the strains

into elongations. The

lengths of the two bars

are 1.41 m and 1 m,

respectively, and the

member AC elongates

by δAC = (0.76×10-

3)×1.41 m = 1.07×10

-3

m (or 1.07 mm), and

the member BC by δBC

= − (0.54×10-3

)×1.0 m

= − 0.54×10-3

m (or

0.54 mm).

To find the new location of point C, we use the following strategy: We change

the lengths to the rod to their new length and make them rotate in arcs about their

pivot points A and B, respectively. The point where the two arcs intersect is the

deflected location of point C.

Fig. 2.6b shows the geometric construction for determining the displaced location

C1 of point C. Bar AC has been shown elongated to AD (with CD being equal to

δAC =1.07 mm. We draw an arc with centre at A and AD as radius. The

displaced point C should be on this arc. Similarly, we displace point C to E by

an amount δBC = − 0.54 mm. We dr w n arc with centre at B and BE as radius.

The displaced point C should be on this arc as well. The point of intersection of

the two arcs then represents the new location C1 of C.

Determining the intersection

point of two arcs is a fairly

lengthy process. Fortunately,

since the changes in lengths are

very small fractions of the

lengths of the rods, and

therefore the angles of swing

are quite small, we can adopt an

approximate procedure wherein

we replace the arcs by the

tangents to the arc. We draw a

perpendicular to AD at D (to

represent the tangent to the arc

through D. Similarly, we draw

a perpendicular to AC at E (to represent the tangent to the arc through E. The

point where the two perpendiculars meet is the location of C1. We express the

new location by specifying the x- and y- displacement of the point C. The x-

displacement is easy to determine. It is nothing but EC, which is δBC = 0.54 mm.

We note that the y-displacement is EC1 = EF + FC1 = CG + FC1. But, CG =

CD/sin 45o = 1.07 mm/ 0.707 = 1.52 mm. we also see that FC1 = FG = 0.54 mm.

Thus, the y-displacement of the point C is 1.25 mm.

The procedure outlined here is quite a standard procedure for determining the

displacement in pinned structures like this truss.

2.2 Statically indeterminate problems

In the examples solved above

we first obtained loads and

stresses in the members

before moving to the next

step of determining strains

and stresses. There are,

however, problems where we

cannot determine the forces

in members without bringing

in the considerations of

deformation.

Consider as an example a

beam supported on two ends

as shown in Fig. 2.8. Since

the be m’s deflection is

excessive, it is propped up in

the middle by a column. Also shown is the FBD of the beam. There are now

three unknown reactions and only two equations (vertical force balance and

moment balance) to determine them. The problem is, therefore, statically

indeterminate.

It is easy to see that the reaction from the central prop depends on the rigidity of

the beam. If the beam is quite rigid, the beam may not even touch the central

prop and there will then be no reaction there. But as the rigidity of the beam

decreases, the load borne by the central prop increases. Therefore, it is necessary

to bring in the considerations of deformation even to solve the problem of

equilibrium.

The general strategy for solving the problems of the mechanics of deformable

bodies consists of three major steps:

Fig. 2.8 A propped beam

Fig. 2.6 Determining displacement in a truss

Fig. 2.7

20kN

1 m

1 m

A

CB

A

B

D

E C

C1

(a) (b)

21

Consideration of static equilibrium and determination of loads,

Consideration of relations between loads and deformations, (first converting

loads to stresses, then transforming stresses to strain using the properties

of the material, and then converting strains to deformations), and

Considerations of the conditions of geometric compatibility.

In statically indeterminate problems we cannot take these steps in a linear

sequence, because there are not enough equations of equilibrium to solve for all

the unknown loads. We have to consider simultaneously all three steps, even if

we were interested in only one, say, in determining the forces in the system, as in

the problem above.

We illustrate this strategy first with a simple example involving springs, and then

with a little more involved problems.

Example 2.3 A spring board

Fig. 2.9a shows a rigid board mounted on two similar springs, each of spring

constant12

k, and length h. A man with weight P walks out on the board till he

reaches a point a distance x from the second spring when the tip of the board

touches the foundation platform as shown in Fig. 2.9b. What is the value of x in

terms of P, L, h and k? Neglect the weight of the platform.

Solution:

It is clear from the

geometry of the

problem that as the

man walks out, the first

spring elongates

applying on the board a

downward force, while

the second spring

compresses applying an

upward force on the

board. Let the first

spring elongate by an

amount δ1 and let its

tension be R1. Let the

second spring shorten

by δ2 and the upward

12

Spring constant k is defined as the force required for producing a unit

deflection in a spring. Thus, if a force F produces an extension (or a

compression) of δ, the spring constant is given by k = P/δ. Its unit is N/m.

force it applies on the board be R2. Fig 2.7c shows the resulting FBD of the

board.

Equilibrium analysis

There are two unknown forces R1 and R2 in this problem. There is also the

unknown distance x that is to be determined. The equations of equilibrium give:

∑ : (a)

∑ ( ) (b)

The moments have been taken about the location of the first support. There is no

other equation of equilibrium. We cannot solve for three unknowns from these

two equations. This is why this problem is statically indeterminate. We

nonetheless move on.

Load deflection equations

Using the spring constant k:

(c)

(d)

This introduces two more equ tions, but two more unknowns, δ1 nd δ2 as well.

Geometric compatibility

From the geometry of the similar triangles in Fig. 2.9b, we can conclude that the

length of the first spring after deformation will be twice that of the second spring

after deformation.

( ) (e)

The problem is solvable now. There are five equations and five unknowns, R1,

R2, δ1, δ2, and x, the distance to be determined.

Writing R1 in term of R2 from Eq. (a), calculating δ1 and δ2 from Eqs. (c) and (d)

in terms of R2 only, and then substituting for δ1 and δ2 in Eq. (e), we can solve

for R2 to obtain

And then from (a):

Using these values of R1 and R2 in Eq. (c) we get the desired result,

Fig. 2.9 A spring board

22

Example 2.4 A bolt and a sleeve

A brass sleeve of length 50 mm (ID: 14

mm, OD: 18 mm) is held between a nut

and the head of a steel bolt of dia 10 mm

(Fig. 2.10). The nut is brought flush with

the sleeve and then tightened one quarter-

turn. If the pitch of the bolt is 0.7 mm,

determine the tension in the bolt.

Solution:

This is a statically indeterminate problem.

As we tighten the nut, a compressive

force builds up in the sleeve which

shortens in length. Simultaneously, a

tension is built up in the bolt which

elongates. Even though we are interested only in calculating a force, we cannot

do so without bringing in the considerations of stress, strain and elongation.

Equilibrium analysis

This is simple. If a compressive force P builds up in the sleeve, the tension in the

bolt is P.

Force deformation analysis

The force deformation analysis is done using the macro-micro-micro-macro

strategy developed in the previous section. To find the contraction in the length

of the sleeve in terms of the compressive force P, we first find the stress ζ = P/A,

the convert it into the strain ε = ζ/E = P/AE. The total contraction then is δ1 = εL

= PL1/A1E1. The elongation of steel bolt is, similarly, δ2 = PL2/A2E2.

Geometric compatibility

The nut moves through a total distance of ¼ of 0.7 mm, or 1.75×10-4

mm. If the

steel bolt did not elongate at all, this would be the magnitude of δ1. But the bolt

elongates through δ2. This means that the contraction of the sleeve would

decrease by an amount equal to δ2: Thus,

δ1 = 1.75×10-4

m − δ2.

Substituting the values of δ1 and δ2 in this condition, we get:

PL1/A1E1 = 1.75×10-4

m – PL2/A2E2

Here, A1 = π[(18×10-4

m)2 − 14×10

-4 m)

2]/4 = 1.00×10

-6 m

2,

A2 = π(10×10-4

m)2/4 = 0.78×10

-6 m

2,

L1 = L2 =5×10-2

m,

E1 = 110 GPa (for brass), and E2 = 200 GPa for steel.

Using these values, we get P = 225 N.

The resulting stresses will be 225 MPa in brass sleeve, and 282 MPa in the steel

bolt. Please note that these stresses are about the maximum that the sleeve and

the bolt can sustain. Any more tightening would result in failure of either or

both.

Example 2.5 A kinky spring

Fig. 2.11 shows an elementary design of a spring which has one spring constant

for the initial deflection (till the load is compressing only the inner aluminium

tube), and then it increases when the

load plate touches the outer brass

cylinder so that the load deflection

curve looks like that shown in the

figure. For the inner tube of 40 mm

dia and 2 mm wall thickness and the

outer tube of 50 mm dia and 2 mm

wall thickness, find the two spring

constants.

Solution:

Till the load is borne only by the

inner aluminium cylinder

For a load P the stress in the cylinder is obtained by dividing the load by its

cross-sectional area, which is πDt = π(40×10- 3

m)(2×10- 3

m) = 2.5×10- 4

m2.

Therefore, the stress is P/2.5×10- 4

Pa (compressive).

We convert this stress into strain by dividing it by the elastic modulus for

aluminium, which is about 70 GPa. Therefore, the strain is (P/2.5×10- 4

Pa)/70×109 Pa = 5.71×10

- 8 P (compressive).

For a length of 300 mm, this strain results in a shortening of the cylinder by ζ×L

= (5.71×10- 8

P)×(300×10- 3

m) = 1.71×10- 8

P m.

The spring constant13

is load per unit deflection, i.e., P/(1.71×10- 8

P) = 58.3

MN/m

13

One can find the spring constant quite directly by noting that the stress is P/A,

strain is stress/E, or P/AE. Deformation, then, is δ = PL/AE. The spring

Fig. 2.11 Kinky spring

Fig. 2.10 A bolt, sleeve and nut

23

The spring constant changes when the compression exceeds 0.1 mm, or when the

load exceeds (58.3×106 N/m)×(0.1×10

- 3 m) = 5.83 kN

After the outer cylinder kicks in

After the outer brass cylinder also starts bearing the load, the nature of the

problem changing drastically. We no longer can use equilibrium alone to

determine the loads borne by the two cylinders independently. It is a statically

indeterminate problem where considerations of equilibrium are interwoven with

the consideration of deflections.

Equilibrium of forces

Let us assume that after the total deflection exceeds 0.1 mm, the load borne by

the inner cylinder is F1 and that by the outer cylinder is F2. Clearly, the

equilibrium requires that:

P = F1 + F2 (a)

There is no other equation of equilibrium that can help us determine the two

forces.

Force deformation considerations

Let us move ahead and determine the stresses in the two cylinders, which are

F1/A1 and F2/A2, respectively, where A1 and A2 are the two areas.

The corresponding strains will be F1/A1E1 and F2/A2E2, both compressive, where

E1 and E2 are the two elastic moduli. The contractions will be:

δ1 = F1L1/A1E1 and δ2 =F2L2/A2E2. (b)

We have taken all steps according to the procedure outlined in Sec. 2.1 and

illustrated in the examples there: load (macro) to stress (micro) to strain (micro)

to deformation (macro).

Geometric compatibility

We know that the contraction of outer cylinder begins only when the inner

cylinder has compressed by a full 0.1 mm. Thus, the geometric compatibility

requires that:

(c)

The values of the various parameters are: L1 = L2 = 0.3 m, A1 = 2.5×10- 4

m2, A2 =

3.1×10- 4

m2, E1 = 70 GPa, and E2 = 110 GPa. We have used a value of E for

constant, therefore, is k = P/δ = AE/L. Plugging in the values of L, A and E we

obtain the same value as above.

brass as 110 GPa, which is about in the middle of the range of values given in

handbooks for brass or bronze as between 100 and 125 GPa.

Putting in the values of the parameters in Eq. (c), we obtain:

1.71×10- 8

F1 = 8.80×10- 9

F2 + or,

F1 = 0.51F2 + (d)

Eqs. (a) and (d) can be solved simultaneously to obtain14

and

. (e)

Finding the spring constant for this part is a little tricky. Note that we should

subtract 0.1 mm from the deflection δ and 5.83 kN from the load P to obtain the

slope of the second part of the load deflection curve of Fig. 2.11. We note that

. The spring constant after the kink is given

by:

The value of δ2 in terms of F2 is given obtained from Eq. (b) as δ2 = 8.80×10-

9F2. Eq. (e) gives F2 in terms of P. Thus, we obtain

( ), or

( )

,

This is more than three times the value when only the aluminium cylinder was

being deformed.

2.3 Lateral strain: Poisson ratio

Consider a bar of length L subjected to an axial load P as shown in Fig. 2.12.

Under the action of this load, the bar elongates by an amount δL so that an axial

strain is set up equal to . From our experience with elastic materials

like rubber bands we expect that as the bar elongates its cross-sectional area

decreases, i.e., transverse strains εyy and εzz are also set up. Simeon Poisson, a

French mathematician in the early 19th

century proposed that the strain in a

14 One can verify this result by using P = 5.83 kN, the value where the

brass cylinder just kick in and finding that and as it

should be.

24

transverse direction is a fixed

(negative) proportion of the strain in

the axial direction. Thus,

(2.1)

where ν is termed as the Poisson ratio

and is a property of the material15

.

Next, consider the case where a

second normal stress ζyy is also

present. Because the stress–strain

relation is linear, the additional

strains produced due to ζyy will

simply be superposed on the strains

due to ζxx. Further, since the properties of the material are independent of the

direction (termed as the property of isotropy), the strains due to ζyy are:

Similarly, the strains due to ζzz are:

.

Since these three strains in the three directions due to three different stresses are

all additive, the total strains in the three directions will be:

( ), (2.2)

( ), and (2.3)

( ). (2.4)

15

It is interesting to explore what change in volume results when a uniaxial load

is applied. Consider a small cuboidal element of dimensions δx, δy, and δz in the

x-, y-, and z-directions. If it is subjected to a load that causes an x-direction strain

equal to εxx, the resulting y- and z-str ins re − νεxx each. The new dimensions of

the elements now are, δx(1 + εxx), δy(1 − νεxx), and δz(1 − νεxx). Therefore, the

deformed volume is δx•δy•δz• (1 + εxx)•(1 − νεxx)•(1 − νεxx) ≈ δx•δy•δz•(1 + εxx −

νεxx − νεxx) = δx•δy•δz•[1 + (1 − 2ν) εxx], neglecting terms of higher order in ε. The

fractional change in volume is, thus, (1 − 2ν) εxx]. If the material is

incompressible, the v lue of ν, the oisson r tion must be ½.

These are the tensile (or compressive, if the algebraic sign is negative) strains on

the material due to tensile stresses. We can show using symmetry arguments that

shear stresses do not cause any tensile strain. Therefore, Eqs. 2.2-2.4 can be used

to convert stresses into strain. These relations can be taken as extensions of

Hooke law for uniaxial stress, and collectively are known as generalized Hooke

law.

Example 2.6 Compression of a block of rubber

A block of rubber 50 mm×50 mm×30 mm is placed in a cavity and compressed

with a force of 1 kN (Fig. 2.13). Determine the decrease in the thickness of the

block.

Solution:

As the rubber block is confined within a

cavity, its transverse dimensions will not

change and therefore there will be no strain

in the y- and z-directions. The (negative)

stress in the x-direction produces a (positive)

strain in the transverse directions. The walls

of the cavity prevent the rubber from

expanding in that direction, and as a

consequence, compressive stresses in y- and

z- directions result. This is, thus, a multi-

axial loading situation and the generalized

Hooke law (Eq. 2.2-2.4) applies. Clearly, ζyy

and ζzz re equ l (= ζ, s y) bec use of

symmetry. (we cannot distinguish between

y- and z- directions. The stress ζxx is equ l to (−1 kN)/ (50 mm× 50 mm) = −400

kPa, the negative sign indicating that it is a compressive stress. The value of E

for rubber is about 500 kPa with the value of Poisson ratio ν of 0.5 (note that this

value indicates that the rubber is essentially incompressible!). The three

equations then give:

-

( ), (a)

( ) (b)

( ) (c)

Eqs. (b) and (c) are identical (as expected, because of symmetry), and from these

we obtain ζ = −400 k . Using this v lue in Eq. ( ), we get

εxx = 0, i.e., no strain at all.

Fig. 2.13 Compressing a rubber

block

Fig. 2.12 Transverse strain under axial

load

L

P

P

x

zy

L

P

P

LL

P

P

x

zy

x

zy

25

What is going on?

This strange result is the consequence of the value of Poisson ratio being taken as

0.5, which is the same thing as assuming the material to be incompressible (see

footnote 5). Clearly, no strain can be produced in an incompressible material

confined in a cavity!

Let us redo this problem with a different material. Let us assume that the value

of E is 500 kPa and that of ν is 0.4, the other conditions remaining the same. The

three equations now are:

-

( ), (d)

( ) (e)

( ) (f)

As before, Eqs. (e) and (f) are identical, and from these we obtain ζ = −266.7

kPa. Using this value in Eq. (d), we get εxx = -0.37, a very large strain. The

thickness of the block will reduce by 37%.

Example 2.7 A confined pressure vessel

A long thin-walled cylindrical tank is snugly fitted between two rigid walls as

shown in Fig. 2.14. It is

then pressurised to an

excess pressure of p. If

radius of the cylinder is R

and its wall thickness t,

determine the compressive

force exerted by the wall on

the cylinder.

Solution:

The length of the

pressurized unrestrained

cylinder increases because

of the combined action of

the axial and the hoop stresses. This results in the cylinder trying to push the

walls outward, which in turn apply a compressive force F on the cylinder. We

can determine this force by noting that the net strain in the cylinder in the z-

direction should be zero.

There are two sources of axial stress ζzz in the cylinder: one is due to the internal

excess pressure ( ⁄ from Eq. 1.9), and the other is the axial force

applied by the walls on the cylinder. The compressive stress due to this force F

is obtained by dividing it by the cross-sectional area of the cylinder which

sustains this load, i.e., 2 . The two stresses are superposed, and the net stress

in the axial direction is given by

(a)

The negative sign is because the force F causes compressive stresses. The hoop

stress is given by Eq. 1.8 as:

(b)

The force F makes no contribution to the hoop stress.

The strain in the axial direction is given by Eq. 2.4 as

( )

Where ν is the Poisson ratio. Using the above values and the fact that ζrr is of

order p and, therefore, negligible, we get

( )

The value of ν is always less than 0.5.

2.4 Shear strain

A material subjected to shear stresses deforms in

shape. Fig. 2.15 shows an element of a material

that is under the action of a shear stress. We have

seen earlier that shear stresses occur in pairs: If on

the x-faces of an element there is are stresses ηxy

giving rise to a counter-clockwise couple as

shown, there must occur on the complementary y-

faces stress components that result in a clockwise

couple. Please note that all the stresses shown here have the same +ve signs16

. If

the direction of one of the stresses was changed, the direction of all the

components will be changed.

Though the element is under equilibrium under the action of these four types of

stresses and the consequent couples, the element deforms in shape. It can be

shown by some very interesting symmetry arguments17

that these stresses cannot

16

In fact, we have deliberately defined our sign convention for stresses to ensure

this very situation. 17

The reader is referred to Crandall, Dahl, Lardner, Mechanics of Solids,

McGraw-Hill, 2nd

SI edition, Section 5.4 for these fascinating arguments.

Fig. 2.14 A cylindrical tank between two walls

Fig. 2.15 Shear strain

26

result in any linear (tensile or compressive) strains: the linear dimensions of the

element will not change. It is only the shape that changes in this case. The

change in shape is manifested in the change in the included angle between two

lines which were parallel to the axes. Thus, angle at A reduces to α. We define

the shear strain in the element at this point to be the change in angle, i.e., (π/2 –

α), for an infinitesimal element in the limit that the dimensions of the element

tend to zero. The symbol used for the shear strain in the x-y plane is γxy, the

Greek letter gamma. Thus,

.

/ (2.5)

This is dimensionless as the linear strains were. The angles are measured in

radians

As for the tensile stress and strain, shear stress too is linearly related to the shear

strain.

, (2.6)

where G is termed as the shear modulus of the material. The shear modulus has

the dimensions of force/area or ML- 1

T- 2

. It has the same unit as E has, namely

Pascal, Pa. Table B.1 in Appendix B gives some representative values of shear

moduli and Poisson ratios of some common materials. Steel has a Poisson ratio

of about 0.27. Aluminium and brass have a little higher values of 0.32 and 0.35,

respectively. Soft rubber has a value of about 0.5, signifying that its volume does

not change appreciably as it stretches.

The shear modulus of most materials is about a third of their elastic modulus18

.

Shear stress in a plane depends on the shear strain in that plane alone, and not on

any other component of strain, shear or tensile. Similarly,

and (2.7)

Eqs. 2.6 and 2.7, together with Eqs. 2.2 – 2.4 form the complete set of strain-

stress relations.

Example 2.8 Calculating strains and stresses

Fig. 2.16 shows a small element of a steel structure after deformation. The

undeformed element was aligned with the axes and its dimensions were 0.2 mm

× 0.2 mm. The coordinates of its vertices A, B and D after deformation are given

as (in mm): A(0,0), B(0.194, 0.013), and D(−0.012, 0.196). C lcul te the v rious

strains and stresses.

18

It will be shown in Chapter 6 that E, G and ν are related: ( )

Solution:

From the given coordinates, the length of the line

AB is √( ) ( ) . The

original length of this line aligned with x-axis was

0.20 mm. This gives a contraction of 0.006 mm,

or a strain εxx of – 0.006 mm/0.2 mm = − 0.03 or −

3%.

Similarly, the length of line AD is

√( ) ( ) . The original

length of this line aligned with y-axis was also 0.20 mm. This gives a contraction

of 0.004 mm, or a strain εyy of – 0.004 mm/0.2 mm = − 0.02 or − 2%.

To calculate the shear strain γxy we determine the angle DAB. That is obtained by

subtracting from π/2 the angle the line AB makes with the x axis

(

) and adding to it the angle the line AD

makes with the y axis (

). Thus,

The shear strain, then, is

This is the shear strain in the material.

We next calculate the shear stress. From Eq. 2.6,

(- ) , or −150 , where

we have used 26 GPa for the value of G for steel.

We next calculate the linear stresses from Eqs. 2.2 and 2.3:

( ):

( ):

Solving these two simultaneously, we get ζxx = −5.57 G , nd ζyy = −2.70 G .

The negative signs indicate that both these stresses are compressive.

Example 2.9 A vibration isolator using rubber in shear

Figure 2.17 shows the schematic of a vibration isolator. The machine whose

vibrations are to be isolated is mounted on the central steel block and applies a

vertical load of 8,000 N as shown. It is suspended from two vertical walls

through two rubber pads of height 12 cm and cross-sectional area of 10 cm ×10

cm. The two rubber pads act as shear springs. Determine the total deflection of

the central post under the load and the effective spring constant of the two pads

taken together.

Fig. 2.16 Calculating strains

27

Solution:

Fig. 2.18 shows a pad in deflected

configuration. It shares the total load of 8,000

N with the other pad, and therefore, carries

4,000 N as a shear load. This load acts on an

area which is 12 cm×10 cm, or, 0.012 m2. The

she r stress is −4,000 N/0.012 m2, or,

. With G for rubber as 1.0 GPa, the

shear train is = −333.3 k /1.0

G = −0.33. The tot l line r deflection,

then, is −0.33×10 cm = −3.3 cm.

Since a load of 8,000 N produces a vertical

deflection of 3.3 cm, the spring constant of

the two pads taken together is 8,000 N/0.033

m = 242.4 kN/m.

2.5 Thermal Strains

As the temperature of a body changes so

does its dimensions. The change of length

with temperature is generally quite linear

with changes in temperature of several

hundred degrees Celsius. If we heat a rod of

length L through a temperature change of ∆T, its length changes as ,

where α is termed as the coefficient of thermal expansion. This can be seen as

producing a strain εT given by

(2.8)

The unit of α is per . Representative values of α for some materials are given

in Table B.1 in Appendix B. It should be noted that temperature changes do not

produce any shear strains.

These thermal strains are superposed on the elastic strains produced by loading of

members. The total strain produced in the presence of elastic strain is obtained

simply by adding the two strains. Thus, using Eqs. 2.2-2.4, and Eq. 2.8, we get:

( ) (2.9)

( ) , and (2.10)

( ) (2.11)

There is no change in shear strain because of thermal expansion, i.e., a shear

strain arises only due to the respective shear stress component.

If the structure in not permitted free expansion due to geometrical constraints,

stresses may be setup in the structure. Such stresses are frequently termed as

thermal stresses.

In examples below we illustrate how the thermal and elastic strains can be

combined.

Example 2.10 Thermal stresses

Fig. 2.19 shows a 20 cm long 3 cm dia steel

rod held between two rigid supports. The

temperature of steel rod is raised by 100 0C.

What is the force in the bar?

Solution:

The increase in temperature will tend to

increase the length of the rod. But the rigid

supports will prevent it from elongating.

This results in an axial force. If F is the compressive axial force acting on the rod

because of the walls, the stress ζxx = F/A = F/7.06×10−4

m2. Since there is

nothing that is applying a force in the other two directions, ζyy and ζzz are both

absent. Also, since there is no net change in the length of the bar, the axial strain

εxx is 0. Using these values in Eq. 2.9, we get:

, or .

Here α for steel is about 11×10- 6/ 0C, E = 210 GPa, and ∆T = 100

0C. The stress is

( - ) ( ) ( ), or 231 MPa,

compressive. By multiplying this stress with the area of the bar (7.06×10−4

m2), we get

the axial force F as 163 kN, a fairly large force.

Example 2.11 Residual stresses in electroplating

The corrosion resistance of steel is

improved by electroplating it with a

thin layer of copper, usually only a

few hundred microns thick. For bright

electroplating the bath temperature

should be around 50 0C. After

electroplating when the plated steel

returns to the room temperature, say

20 0C, the higher coefficient of thermal

expansion of copper compared to that of steel results in copper tending to

Fig. 2.17 Schematic of a

vibration isolator

Fig. 2.19

Fig. 2.20 Copper plating on steel

Fig. 2.18

Wall

8,000 N

Wall

Rubber blocks

Wall

8,000 N

Wall

Rubber blocks

28

contract more than that what the steel does. The bonding between steel and

copper prevents this, and as a result steel does not let the copper layer contract

more than it does itself. Steel layer ends up applying a tension to the copper

layer, and the copper layer pulls the steel inwards, the total tension force in

copper being equal to the compressive force in steel. However, since the steel

layer is an order of magnitude thicker than the copper layer, the stress in steel

will be negligible compared to that in copper.

Thus, there is a residual tensile stress19

in the copper layer. Estimate this residual

tensile stress.

Solution:

There is no stress at 50 0C. When the temperature decreases to 25

0C (∆T = −25

0C) the net strain in the copper layer will be exactly the same as the net strain in

the steel plate. The steel plate will have only the thermal strain, since we can

assume that the thickness of the steel plate is so large (compared to that of copper

film) that the same force in the two layers (equilibrium requirement) will result in

negligible stress in steel. Thus,

.

/

( )

Using the property values: 11×10- 6/ 0C; 16×10- 6/

0C; and

, we get:

( ) ( )

= ( ) (11 – 16) ×10- 6/ 0C×(− 25

0C)

= 15 MPa (tensile), not an insignificant stress

Example 2.12 Shrink fit

A steel shaft (20 mm dia) is to be fitted with a

steel sleeve (20 mm long, OD 26 mm). In

order to tightly fit the sleeve, its internal

diameter is made as 19.9 mm. It is fitted on to

the shaft by heating it till its internal diameter

is larger than 20 mm, slipping on, and then

letting it cool. As the sleeve shrinks, it presses

on to the shaft creating an interfacial pressure

which results in a good bond. Determine (a) the temperature beyond which the

19

This residual tension can be very detrimental when the structure is subjected to

tensile (or bending) load. The copper layer would crack and the corrosion

resistance will be lost.

sleeve should be heated for fitting, and (b) the resulting interfacial pressure on

cooling.

Solution:

It is convenient to work in the cylindrical polar coordinates, r, θ, and z, as shown.

The relevant equations for strains and stresses are obtained by simply substituting

r, θ, and z for x, y, and z in the equations developed so far20

.

To determine the temperature beyond which the sleeve should be heated for

fitting, we use Eq. 2.8 with εT = 0.1 mm/20 mm = 0.005, and αsteel = 11×10- 6/

0C,

to get

( - ) 455 0C

The sleeve has to be heated through a range of more than 455 0C to slip it on the

shaft without forcing it.

Now on cooling, the ID of the sleeve must be equal to the dia of the shaft. Let us

look first at the stress and strain in the shaft. If the interfacial pressure is p, the

radial stresses in the shaft are of order p. There is no other stress in the shaft. On

the other hand, the sleeve will be subjected to a radial stress of the order of p and

a hoop stress (radial stress ζθθ (refer to Section 1.8). The hoop stress is given by

Eq. 1.8 as pR/t. Since R >> t, the hoop stress, and consequently, hoop

strain are much larger than the compressive strain of the shaft. We can, thus,

neglect the compression of the shaft diameter in comparison to the increase in the

ID of the sleeve. Therefore, we can take the final dia of the shaft to be 20 mm

itself, and the final ID of the sleeve too as 20 mm. As the dia of the sleeve

increases from 19.9 mm to 20 mm, its circumference increases from π×19.9 mm

to π×20.0 mm, giving a hoop strain of 0.005. Since the axial stress is absent,

and the radial stress is of order p, much less than , we get 0.005 = /E.

Using the value of E as 210 GPa, and expressing as pR/t, we get p, the

interfacial pressure as 157.5 MPa

2.6 Tensile test

Tensile test is one of the primary tests that are used to determine experimentally

the properties of matter. A standardized specimen (Fig. 2.22) which uses a strip

or a rod of uniform cross-section between two enlarged ends is gripped and a

20

This is possible since none of the equations we have developed so far has

partial derivatives of physical quantities with respect to x, y, and z. Since θ does

not have the dimension of length, we cannot simply replace derivatives with

respect to y by those with respect to θ, and the equations that have partial

derivatives change in a more complex manner.

Fig. 2.21 A shaft with shrink-

fitted sleeve

29

tension force is applied. An extensometer

(Fig. 2.23) is used to measure how the length

between two specified points on the uniform

portion changes as the tension is increased.

The tension is converted into stress ζ by

dividing it with Ao, the cross-sectional area21

,

and the elongation is converted to linear

strain ε by dividing the elongation by Lo, the

gauge length (the undeformed distance

between the two points on the specimen). On plotting the two we get a graph

like that shown in Fig. 2.24 for ductile materials like steel. The initial part of the

curve is a straight line and

represents the linear elastic

range22

of the material. If a

material is loaded to point A

within this range as shown,

and then unloaded, the

material will return to its

unloaded state (strain will

return to zero). We say that

if the material is loaded and

unloaded within the elastic

range, the material exhibits

no permanent set. The slope

of the stress-strain curve in

the linear elastic range is E, the elastic modulus of the material.

The stress level up to which the linear elastic range extends is termed as the

proportional limit.

As the stress is increased, non-linearity comes into play, but the material still

shows elastic behaviour, i.e., on unloading the material exhibits no permanent

set. The stress level up to which the elastic range (linear or non-linear) extends is

21

Note that the stress here is only a nominal stress since we are not accounting

for the changed cross-sectional area as the specimen elongates. This is not

significantly different from the true stress till the strain becomes quite large. In

fact, the decrease in the stress at very large strains as shown in Fig 2.23 arises

only because we are not using the true stresses. The load does not decrease as

strain increases. 22

The linearity and elasticity are two different phenomena. The elasticity refers

to the fact that there is no permanent set, i.e., the material on unloading returns to

the original unloaded state, along the same curve it followed while loading. This

may or may not be linear. In fact, a rubber band is largely non-linear elastic.

termed as the elastic limit. It is not

possible to determine either the

proportional limit or the elastic limit

with any degree of accuracy.

The strains in the elastic region are

very small, quite insensible without

instruments to amplify their

magnitude. That is why extensometers

are used which amplify the strain,

either mechanically or optically.

When the stress in steel is of the order

of 400 MPa, the relative displacement

of two points originally 100 mm apart

would be of order of 0.2 mm. Strains

in most structural materials are of the order of a few tenths of a percent at most.

As the load is increased further the material stops showing the elastic behaviour,

that is to say that if the material is loaded beyond a certain point and then

unloaded, it will not return to its original unloaded state. We say that the

material now has a permanent set. In Fig. 2.25, the material is loaded up to point

Y. When the load is now decreased, the material follows the straight line path

downwards as shown. The straight line has the same slope E as the original

linear elastic line, but now it meets the strain axis at a point left of origin,

signifying that there is now a strain even when the loading and stress have been

reduced to zero. The material is said to have yielded and undergone plastic

deformation.

It is quite difficult to decide

experimentally the point at which the

plastic deformation begins. It is for

this reason that we, quite arbitrarily,

designate the point Y, where the

permanent set (the residual strain when

unloaded after loading to that point) is

0.2%, (or a strain of 0.002) as the yield

point of the material. The

corresponding stress ζY is termed as

the yield strength of the material.

Consider a material that is loaded

beyond the yield point Y to a stress

level Y1 (> Y). If the material is

unloaded now, it will follow a straight line down from Y1 with a slope of E to

point B, and there will be a permanent set as shown in Fig. 2.26. If, after this

Fig. 2.25 Yielding and permanent set

Fig. 2.24 Typical stress-strain curve for

a ductile material

Fig. 2.23 Schematic of an extensometer used in a

tensile test

Fig. 2.22 Tensile test specimens

30

initial yielding and unloading, the

material is loaded yet again, it will

follow the straight line up from

point B (with the slope E). This

material is now not going to yield

again till it reaches the stress level

of Y1, higher than Y. Thus by

loading the material to a level

higher than the yield strength Y,

we have been able to raise the

effective yield strength of the

material (in subsequent loadings)

to Y1. This phenomenon is known

as strain hardening and is

frequently employed to increase the wear resistance on materials. Shot peening

and sand blasting are examples of processes employed to strain harden structures.

Cold rolling of structural steel also produces usable work hardening.

Loading of the bar beyond the yield strength produces large deflections with

small increases in the stress. The maximum stress that the material is able to

attain is termed as the ultimate stress.

When a material is strained beyond the ultimate stress, the stress appears to be

decreasing as strains increases. As noted

earlier this is a false impression, since what we

are plotting here is the nominal stress obtained

by dividing the load by the initial area Ao. The

area of a ductile specimen decreases

significantly when it undergoes plastic

deformation. This is, usually, a local

phenomenon. The area of cross-section of the specimen suddenly starts

decreasing at some point where there are imbedded dislocations in the material.

This is termed as necking (Fig. 2.27). As the necking progresses, the true stress at

the section increases and the specimen fractures.

We summarize here the sequence of major events in the tensile test of a ductile

specimen:

Linear elastic deformation till the proportional limit: .

Non-linear elastic deformation till elastic limit.

Yield point, where the permanent set is 0.2%.

If we unload the material beyond yield point, and then reload it, the

yield strength increases. This is termed as strain hardening.

As the material undergoes plastic deformation, the stress level increase

up to ultimate stress.

Beyond ultimate stress, a neck develops in the specimen; the area

decreases locally while the nominal stress decreases.

Ultimately, the fracture takes place at the neck.

We have discussed so far the stress-strain curve for a ductile material. The

corresponding curve for a brittle material is significantly different from that for a

ductile material. Glass, cast iron, concrete, carbon fibre, and Perspex are some of

the brittle materials. Many material ductile at room temperature may be brittle at

low temperatures.

Most of the differences between the behaviour of ductile and brittle material

arises from the fact that there is little plastic deformation in a brittle material. Fig.

2.28 shows the stress-strain curves for glass and cast iron.

The first thing to

note is that both

these materials

undergo fracture

at very little

strain, much less

than that for a

ductile material.

Glass, for

example, shows

no plastic

deformation at

all.

Another feature

to note is that the

strength to fracture is much larger in compression than in tension. This is a usual

characteristic of brittle materials.

2.7 Idealized stress-strain curves

The stress-strain curves for the various materials do not admit any simple

mathematical description. However, we do need such descriptions for analysis of

structures. It is for this convenience that we model the materials into a few

classes and use simple stress-strain relations to model the actual behaviour.

Most problems of high school mechanics are solved by assuming the structural

members to be rigid. This means that there is zero strain for any level of loading.

Fig. 2.29a depicts the stress-strain curve for a rigid material.

Fig. 2.26 Strain hardening

Fig. 2.27 Necking of ductile

materials

Fig. 2.28 Stress-strain curves for some brittle materials

31

The stress-strain curve for a perfectly elastic material is as shown in Fig. 2.29b, a

sloping straight line whose slope is the value of the elastic modulus E. We have

used this idealization of materials in this book so far. This is a very useful model

and is used extensively in most structural analysis, because most structures are

designed to work within the elastic range.

Another commonly used idealized model is the elastic-perfectly plastic one

shown in Fig. 2.29c. in this, we assume that once the material yields, the stress

remains constants at the yield strength level. We shall have occasion to use this

in a few problems.

Fig. 2.29d shows the

behaviour of a

perfectly plastic

material. In such a

material there is no

strain till the stress

reaches a critical

level, after which

there is no increase

in stress as the

material deforms.

This idealization

may be used when

there are very large

plastic deflections,

so that the elastic-strain region of the stress-strain curve can be neglected.

The strain hardening model is shown in Fig. 2.29e. Note that the elastic-perfectly

plastic graph of Fig. 2.29c cannot model the strain hardening phenomena, since

the slope of the plastic part is zero. Reloading the material after unloading

following some plastic deformation will not increase the yield strength. But the

non-zero slope of the plastic part of the curve of Fig 2.29e ensures that the yield

strength increases on re-loading after the material has been de-stressed after it has

undergone some plastic deformation.

Example 2.13 Fitting an idealized curve

The solid line in Fig. 2.30 shows the stress-strain curve of cold-rolled 1020 mild

steel. Find a simple mathematical relation for load vs. elongation of a 20 cm

long 20 mm × 20 mm cross-section bar.

Solution:

A look at the actual stress-strain curve suggests that though the material exhibits

a bit of strain hardening, adopting an elastic-perfectly plastic idealization will not

be bad at all. We first determine the yield strength of this material by drawing a

line parallel to the elastic portion of the curve from a permanent set value of

0.2% or from strain ε = 0.002. The light broken line so drawn meets the curve at

a stress of about 580 MPa, which may be taken as the value of yield stress ζY for

this material. Therefore, the idealized curve looks like the broken heavy line

shown in the figure. In other words, we may model 1020 CR mild steel as elastic

up to a stress level of 580 MPa with a value of E = 580 MPa/ 0.003 = 193.3 GPa

(0.003 being the strain at the yield point). Thereafter, the steel may be taken to

deform perfectly plastically.

A stress of 580 MPa

corresponds to a load of

580 MPa×(20×10-3

m)2,

or 232 kN. In this

region, the elongation

will be δ = PL/AE =

P×0.2 m/(20×10-3

m)2

×

(193.3 GPa) = 2.59×10-

9P m, where P is the load

in Newton. We draw a

line through the origin

up to the yield point with

P = 232 kN. After this,

the deflection increases

without any increases in

the load. Fig. 2.31 graphs this load vs. elongation behaviour.

Fig. 2.29 Idealized stress strain curves

Fig. 2.31 Load-deflection curve

Fig. 2.30 Stress-strain curve for cold rolled 1020 mild steel

32

2.8 Pre-stressing

Many brittle materials have much higher yield strengths in compression than in

tension. For example, concrete has a tensile strength of only 10-15 % of its

compressive strength. It is

common to use

reinforcement in concrete to

give it increased strength in

tension. This is used quite

effectively in the design of

beams. We had seen in

Section 1.3 that in a beam

subjected to bending under

vertical loads, its upper layers are in compression while the lower layers are in

tension. It is common to increase the tension strength of concrete by reinforcing

it with tendons of high tensile-strength material, such as steel rods (Fig. 2.32).

But many a times this is not sufficient. One of the very interesting ways to

overcome this problem is pre-stressing. If we build up within the concrete some

residual compressive stresses when it is being cast, subsequent tension loading of

the structure will first relieve the compressive stresses before generating tensile

stresses. In this way, the structure will be able to support enhanced tensile loads

before the tensile stresses reach the critical level.

Fig. 2.33 explains how the pre-stressing works. Tendons, most often steel wires,

are put under external tension and the concrete cast around them. After the

concrete is set, the external tension is released. The tendons now try to shrink to

their no-stress lengths, but the concrete around them prevents them from

shrinking to their original length. In the process, there is a residual tension in the

tendons, while there is equivalent compression in the concrete. This residual

compression in concrete helps increase the tensile load capacity of the concrete.

Example 2.14 Pre-stressed concrete

Let us consider a simple pre-stressed concrete structure to serve as an illustration

of the calculations involved. Let a concrete beam of cross-sectional area 5 cm×5

cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20

kN. The external tension in steel released after the concrete is set. What is the

residual compressive stress in the concrete?

Solution:

A tension of 200 kN in the steel bar results in a stress of 20 kN/[¼π(0.010 m)2] =

255 MPa. (This is quite close to the yield strength of commercial structural steel.)

The strain under this stress is ε = 255 MPa/ 210 GPa = 1.21×10- 3

, and the total

elongation of the bar will be (1.21×10- 3

)×2 m = 2.42 mm.

After the tension is released, the steel rod shortens, but it does not reach its

unstrained length, and there is a residual tension in the steel tendon and an equal-

magnitude residual compression

in the concrete. This force

cannot be determined by statics

alone. We have to make

recourse to deformations and

apply a geometric compatibility

condition.

If δs represents the final

elongation of the steel bar (from

its unstrained condition, and δc

represents the contraction of the

concrete, it is clear from the

construction of Fig. 2.34 that δs

+ δc = 2.42 mm. This is the

geometric compatibility

condition.

Let F represent the force, tension in steel and compression in concrete. Then, δs

= FLs/AsEs = F(2 m)/[π/4(0.010 m)2]×210 GPa, and δc = FLc/AcEc = F(2 m)/(0.05

m×0.05 m)×20 GPa.

Thus, from the geometric compatibility condition we get,

( ) ( ) ,

2.32 Reinforcement in a concrete beam

(a) Tendon as cast

(b) The no-stress length of the tendon

(c) The final configuration on release of tension

Fig. 2.34

Fig. 2.33 Pre-stressing

33

Which gives F = 15.03 kN. This is the residual force which produces a

compressive residual stress of 15.03 kN/(0.05 m×0.05 m) = 6.01 MPa. This

leads to considerable strengthening of the structure in tension.

2.9 Strain energy in an axially loaded members

When a force F is slowly applied to a

deformable body (that is supported in such a

way that no rigid-body motion is possible)

work is done as the body deforms. The work

done U in deformation is

∫ (2.12)

This is represented by the area under the F-δ curve of Fig. 2.35. For purely

elastic deformations, the force-deformation curve is linear and therefore the work

done in deformation is given by

. This work is stored within the body

as mechanical energy and is termed as the strain energy. It is a conservative

energy and is released when the load is removed and

the body returns to its original configuration.

Let a number of forces F1, F2, F3, … ct on the body

(or the structure) and the resulting displacements at

their points of application be δ1, δ2, δ3, …. The work

done in deformation is independent of the order in

which the forces are applied. We apply the forces

simultaneously such that the total work of

deformation is given by the sum of the work done by

individual forces:

∑ (2.13)

Let us calculate the strain energy of a bar subjected

to an axial load as shown in Fig. 2.36. The deformation δ (elongation, in this

case) is given by FL/AE, and the strain energy U is then

.

2.10 Calculating deflections by energy methods: Castigliano theorem

Consider an elastic body which is fully supported in the sense that its rigid body

translation and rotation are not permitted. Let it deform under the action of n

forces, F1, F2, F3, …, Fn. Let the total strain energy be U, which is a function of

F1, F2, F3, …, Fn: ( ).

Let us first load the body with all the forces F1,

F2, F3, …, Fn, and, then, increase one of the load,

say Fi, by a small amount, say δFi. The additional

strain work done can be written as:

(a)

Let us next reverse the order of loading. We first

apply the small load δFi on the body before any

other load is applied. After this load is applied,

all other loads F1, F2, F3, …, Fn are applied on the body. Under the combined

action of all these loads, the point of application of load Fi moves through an in-

line distance of δi. The work done by this small force δFi is then23

dU = δFiδi (b)

Since the order of loading (in the linear elastic case) should be immaterial, we

can equate the two values of work given by Eqs. (a) and (b), to get:

(2.14)

This is known as Castigliano theorem24

. It states that if the total strain energy of

a structural system is expressed in terms of the external loads applied to it, the

deflection at any one loading point (in-line with the load) is obtained by taking

the partial derivative of the strain energy with respect to the load at that point.

We illustrate the use of this theorem, first to two simple cases in Examples 2.15

and 2.16, and then apply it to a more involved problem of a truss. Example 2.17

introduces a process that can be generalized and can also be used for solving

statically indeterminate cases as illustrated in Example 2.18.

Example 2.15 An elementary truss

Consider again the truss with a roller support (Fig. 2.38) discussed previously as

Example 2.3. Determine using the energy method the displacement of point C

23

The factor of ½ is missing in this expression because the load δFi remains

constant while the displacement δi takes place. 24

There are some simple extensions to Castigliano theorem. One of the simple

extensions is its application to non-linear systems where the strain energy

∫ is replaced with ∫ , which is termed as the complementary

energy.

Fig. 2.35 The area under the

force-displacement curve is the

work done in deformation

Fig. 2.36 A bar under

an axial load P

Fig. 2.37 An elastic body

34

where the load of 20 kN is applied. All members are made of steel (E = 210

GPa) and have a circular cross-section of 20 mm dia.

Solution:

Since the energy method involves taking the derivative with respect to the force applied at the point where the deflection is to be calculated, we replace the load of 200 kN with a variable load P. Also, since the use of Castigliano theorem gives displacement only in line of the force with respect to which the derivative of the energy is taken, we cannot find the horizontal displacement of point C since there is no horizontal force present there. However, we can overcome this deficiency by introducing a dummy force Q in the horizontal direction at C, and then taking the derivative with respect to Q before equating Q to zero.

A simple equilibrium analysis of the joint at C using the FBD given in Fig. 2.39b,

gives:

from the vertical force

balance, and

from the

horizontal force balance.

These give , and

. The force in the element AB

is zero as before. The total strain energy,

then, is:

( )

( )

The vertical displacement of point C

(in line with force P) is obtained by

taking the partial derivative of U

with respect to P:

( )

Replacing P with 100 kN, Q with 0,

LCA with m, LCB with 1 m, E with

210 GPa, and A with π(0.02 m)2 /4,

we get

= 0.0058 m, nearly the

same as was obtained in Example 2.3.

To determine the horizontal displacement at C, we take the partial derivative of U

with respect to Q:

( )

, which on substitutions of values gives ,

the negative sign indicating that it is in a direction opposed to that shown for Q.

The amount of work involved here with the energy method is far less than that in

Example 2.3, and the results here are more accurate.

Example 2.16 Tug of war revisited

Consider once again the tug of war discussed as Example 2.1. We determine here

the total elongation of the rope using the energy method. The cross-section of the

rope was given as 0.0006 m2, and the value of the elastic modulus for the manila

rope was taken as 100 MPa.

Solution:

To find the total elongation by energy method, we need to anchor the rope at one

end (say, at point A) and apply a force P at the other end (see Fig. 2.41). The

total energy will be determined in terms of P and partially differentiated with

respect to P. To find the total strain energy of the system, we need to determine

the load in each section of the rope which can be obtained by simple equilibrium

Fig. 2. 38 An elementary truss

Fig. 2.41 FBD for application of energy method

(a) Truss (b) FBD of pin C

Fig. 2.39 Introducing forces P and Q

Fig 2.40 Tug of war

20kN

1 m

1 m

A

CB

35

analysis. Once we obtain the forces we can calculate strain energies by

evaluating . Table below organizes these calculations:

Table 2.2 Calculation of elongation of the rope of Example 2.16

The total strain energy is given as ∑ , where summation is taken

over all the segments of the rope. The deflection at the free end is obtained by

taking the partial derivative of the total energy with respect to P.

(∑ )

∑

( )

where the value of ∑ ( )

has been substituted from Table 2.2. Plugging in the

value of A = 0.0006 m2 and that of E as 100 MPa, and the value of P as 300 N,

we get m, or 6.67 cm, the same as was obtained in Example 2.1.

Example 2.17 Unit force method

Consider the truss shown in

Fig. 2.42. The truss is

pinned at point A and is

supported on rollers at

point B. This implies that

while at A, there could be

horizontal as well as

vertical reactions, there is

no horizontal reaction at B.

All joints are assumed as

pin joints, and as has been

explained earlier, all

members are two force

members and hence carry

only axial loads, tensile or compressive. We will illustrate the use of energy

method to determine the vertical displacement at joint F and the horizontal

movement at the roller support B. All members are made of steel and have a

cross-sectional are of 6 cm2.

Solution:

As before, we need to introduce dummy forces P and Q, respectively, at points F

and B in-line with the displacement desired. The method of solution is exactly

the same as in the previous examples, but because of the larger numbers of

members involved we devise a scheme that will permit reduction of labour.

The total strain energy of the truss will be the sum of the strain energies of the

individual members:

∑

The deflection in the direction of force P is obtained by taking the partial

derivative of U with respect to P:

∑

∑

∑

(a)

Attention is drawn to the fact that the here are the forces in the members in

the presence of all loads, including P and Q. The values of may be imagined to

be made up of two parts, one due to the actual loads on the structure, and the

other due to the

application of dummy

loads P and Q. After

we evaluate the sum

specified in Eq. (a)

above, we plug in the

value of P and Q as

zero. This implies

that in the final step,

the value of ’s will

be the original values

of ’s without P and

Q.

The term

, which

is the rate of change of with P can be interpreted as the additional force in the

ith

member due to a unit load at P. This interpretation is valid because of the

linear dependence on P of the additional load in the ith

member.

We, therefore, organize our calculations in two distinct parts: We first calculate

the actual values of ’s, i.e., the v lues of ’s without P and Q. These are used

as values for the first part of the terms in Eq. (a). We, then, calculate the values

of ’s th t would result when unit force is pplied in pl ce of P, without any of

Segment Length, L Force, F F2L ( )

EF 2 m P 2P2 4P

DE 1.5 m P + 250 N 1.5P2 + 750P + 93750 3P + 750

CD 1.5 m P + 500 N 1.5P2 + 1500P + 375000 3P + 1500

BC 2 m P + 200 N 2P2 + 800P + 80000 4P + 800

AB 1.5 m P − 50 N 1.5P2 − 150P + 3750 3P − 150

Fig. 2.42 Example 2.17

Fig. 2.43

36

the other external forces being present. These values of ’s will be used in pl ce

of the partial derivatives

.

This procedure is termed as the unit force method.

Determination of ’s in absence of forces P and Q:

We first determine the reactions R1x, R1y, and R2y by considering the structure as a

whole bin the FBD shown in Fig. 2.44a. The equilibrium equations are:

∑ : ,

∑ : , and

∑ : .

The three equations can be solved simultaneously to give , ,

and .

Once the reactions are known, we can go from pin to pin to calculate the axial

forces in all the members. Consider, for example, the FBD of pin B shown as

Fig. 2.44b. By writing the two force balance equations and noting that

( ) , we get

∑ : ( ) ,

∑ : ( ).

Solving the second of these first, we get , and then .

We can similarly move from pin to pin and calculate all the forces.

We next calculate the value of L/AE for each member of the truss. For example,

for member AF which is 2 m long, the value of L/AE is given as (2 m)/(0.0006

m2)×(210 GPa) = 1.588×10

−8 N

−1. Table 2.3 gives the values of F and L/AE for

all the members of the truss.

Calculation of through

calculation of the Fi’s for a unit

load of P

Refer to Fig. 2.45 which shows

the FBD of the truss for a unit

load at the location of P and no

other load. Clearly, the reactions

at A and B are 0.5 N, each. We

again go from pin to pin, draw

the relevant FBD and calculate

the force in each member.

These have been shown in the 4th

column of Table 2.3.

The last column of Table 2.3 shows the product

for each member of

the truss. The sum of these over all the members gives the deflection of point D,

in-line with the dummy force P. Thus,

Δv,E = 1.176×10−3

m, or 1.18 mm.

This is quite a general procedure for trusses.

Horizontal deflection at the roller support at B:

Fig. 2.44 Some FBD’s for determining the forces

Fig. 2.45 FBD for unit load at F

Table 2.3 Calculation of vertical movement of pin C

Member Force

Fi, kN

L/AE

(×10−8

N−1

)

(×10−5

m)

AD −16.7 1.24 −0.67 13.9

AF + 13.3 1.59 0.89 18.8

BE −25.0 1.24 −0.67 20.8

BF + 20.0 1.59 0.89 28.3

CD −16.7 1.24 −0.67 13.9

CE −16.7 1.24 −0.67 13.9

CF +5.0 1.59 1 8.00

DF 0 1.24 0 0.00

EF −8.3 1.24 0 0.00

Total 117.6

37

We next calculate the

horizontal deflection

at the roller support at

B. For this purpose,

we place a dummy

horizontal load at

point B.

Here again, we use

the same procedure

as above. The

calculation of Fi’s is

does not depend on the dummy loads and we get the same values as in column 2

of Table 2.3. The values of L/AE for the various members are also unchanged.

We, however, have to calculate dFi’s under the unit lo d Q. For this purpose we

use an FBD as shown in Fig. 2.46 and calculate the resulting tensions in each of

the members. Clearly, the vertical reactions R1y and R2y are zero, and the

horizontal reaction R1x is – 1 N. It is easy to see that there will be tension only in

the members AF and BF, both +1 N.

Table 2.4 organizes the calculation of the horizontal movement at the roller

support at B. Here the values in the second and the third columns (Fi, and L/AE)

are the same as in Table 2.3, but column 4 shows the values of the axial forces in

the various members when only the load Q (equal to 1 N) is applied at point B.

As was noted above, only two non-zero entries (for members AF and BF) result.

The last column of Table 2.4 shows the product

for each member of

the truss. The sum of these over all the members gives the horizontal deflection

of point B (in-line with the dummy force Q), which is seen as 52.9×10−5

m,

slightly more than half a millimetre.

Example 2.18 Application of energy method to a statically indeterminate problem

Let us once again consider

the same truss as in

Examples 2.16 and 2.17 and

shown in Fig. 2.42 with one

change: The roller support is

replaced by a pinned

support. This is as shown in

Fig. 2.47. This will

introduce a horizontal

reaction at support point B.

There will now be four

reactions and only three

equilibrium equations to

determine them. The

problem, therefore, is

statically indeterminate. We illustrate below the general strategy to solve such

problems using energy method.

Solution:

The problem is quite

easily solved by making

the point B free to move

horizontally but applying

a horizontal force Q ( in

kN) at this point to

control its movement.

The strategy consists of

calculating the

horizontal movement of point B in terms of the load Q, and then determining the

value of Q for which this movement is zero. It should be clear that this value of

Q for which the movement is zero is the value of the horizontal reaction at point

B.

Fig. 2.46 FBD for a unit horizontal load at B

Table 2.4 Calculation of horizontal movement of the roller support B

Member Force

Fi, kN

L/AE

(×10−8

N−1

)

(×10−4

m)

AD −16.7 1.24 0 0

AF + 13.3 1.59 1 21.1

BE −25.0 1.24 0 0.00

BF + 20.0 1.59 1 31.8

CD −16.7 1.24 0 0

CE −16.7 1.24 0 0

CF +5.0 1.59 0 0

DF 0 1.24 0 0

EF −8.3 1.24 0 0

Total 52.9

Fig. 2.48 Replacing reaction with an undetermined force

Fig. 2.47 A statically indeterminate truss

38

The vertical components of the reactions remain unchanged from the

determination in Example 2.16: and . The forces in

the various members can be calculated by considering the FBDs of the various

pins. This is quite straight forward, and the values so obtained are tabulated as

column 2 in Table 2.5 below. The values in column 3 and 4 of this table will be

the same as in the corresponding column of Table 2.5, and then column 5 is

modified as shown. The horizontal displacement of point B under the action of a

force Q at that point is thus (52.9 +3.18Q) ×10−5

m, with Q in kN.

This displacement is zero when Q = − 52.9/3.18 = − 16.6 kN. Since the ctu l pin

B is restrained from moving, the reaction at point B must be 16.6 kN, inwards.

We have, thus, solved the statically indeterminate problem. Once this is known,

we can calculate loads in all the members of the truss.

2.11 Strain energy in an elastic body

We have so far considered the strain energy of axially loaded structures. Let us

extend this to a more general state of stress and strain. Consider an infinitesimal

cubical element of dimensions δx, δy, and δz along the three coordinate axes. Let

this be acted upon by a force which produces a stress ζxx in the x-direction. The

force acting on these faces is . If the strain in the x-direction is εxx, the

elongation is εxxδx. The strain energy of this infinitesimal element, thus, is

( )( )

( )

, where δV is

the volume of the element. For a finite body

then we can find the strain energy by

integration over the whole volume:

∫

`

(2.15)

It stands to reason that the stress ζxx does not

do any work with the strains εyy and εzz, since

these strains result in displacements which are

perpendicular to this stress. Similar work will

be done by the other stress components.

Let us now look at the work done by shear

stress components. Consider a shear element

as shown in Fig. 2.50 where the lower

surface is anchored and the upper surface has

a shear stress of ηyx. Let the element undergo

a shear strain of γyx as shown. This implies

that the upper surface with a force of ηyxδxδz

acting on it undergoes a displacement of

γyxδy. The strain energy of this infinitesimal

element, thus, is

( )( )

( )

For a finite body then we can find the strain energy by integration over the whole

volume:

∫

` (2.16)

Similarly for the other shear components. Combining the strain energies for

normal and shear strains, we can write the equation for the general state of strain

for a body of volume V as

∫ ( )

` (2.17)

Table 2.5 Calculation of horizontal movement of the roller support B

Member Force

Fi, kN

L/AE

(×10−8

N−1

)

(×10−5

m)

AD −16.7 1.24 0 0

AF + 13.3 + Q 1.59 1 21.1 + 1.59Q

BE −25.0 1.24 0 0.00

BF + 20.0 + Q 1.59 1 31.8 + 1.59Q

CD −16.7 1.24 0 0

CE −16.7 1.24 0 0

CF +5.0 1.59 0 0

DF 0 1.24 0 0

EF −8.3 1.24 0 0

Total 52.9 + 3.18Q

Fig.2.49 An infinitesimal

element under tension

Fig. 2.50 A Shear element

39

Summary

The general strategy for solving the problems of the mechanics of

deformable bodies consists of three major steps:

1. Consideration of static equilibrium and determination of loads in

members of the structures. This equilibrium analysis invariably requires

drawing up strategically chosen free body diagrams (FBDs),

2. consideration of relations between loads and deformations, and

3. consideration of the conditions of geometric compatibility.

The second step above requires the considerations of the materials as well as

the geometry of the structures. This step in itself may be divided into three

distinct phases:

a. We convert loads to stresses in the various members (macro to micro),

b. the stresses are then converted to strains using the material properties

(micro to micro transformation), and finally,

c. with the use of the geometry of the structure, we determine

deformations from the strains so calculated (micro to macro).

The strength of this method lies in the fact that the first and the third steps

above require the consideration of the geometry of the structure (and are

independent of the material), while the second step requires considerations

of only the material of the structure. Thus, the stress-strain relation obtained

from the tensile test with its most simple loading conditions can be used to

predict behaviour of very complicated structure, since the second step above

is independent of loading and geometry and depends entirely on the

material.

We may, at time, need to go in the reverse direction. We may be given the

total deformation from which we need to determine the loading: we

calculate strains from deformation, convert strains to stress using material

properties, and then integrate stress to determine the loading. The strategy

here too is macro to micro to micro to macro.

There are many problems in which we cannot take the steps enumerated

above in a linear sequence, because there are not enough equations of

equilibrium to solve for all the unknown loads. In such cases we have to

consider simultaneously all the three steps, even if we were interested in

only one, say, in determining the forces in the system. Such problems are

known as statically indeterminate problems. Examples 2.3 to 2.5 illustrate

the methodology to be adopted for solving such problems.

As we apply an axial load to a member and it develops a strain εxx in the

axial direction, there is also a transverse strain in the y- and z- directions. The

strain in a transverse direction is a fixed (negative) proportion of the strain in

the axial direction.

where ν is termed as the Poisson ratio and is a property of the material. With

this, the net longitudinal strain in any direction is related to all the three

tensile stress components.

Change in temperature of a structural member introduces thermal strains so

that the net strains of an element subject to a general state of stress and to a

change in temperature is given by:

( ) (2.9)

( ) , and

( )

We define the shear strain γxy in the element at a point as the distortion in the

shape of the element as measured by the change in angle between two

mutually perpendicular lines (parallel to x- and y- directions).

The shear stress is related to shear stress by the relation: . Shear

strain in a plane depends only on the shear stress in that plane and no other

stress component. Heating of a material too does not introduce any shear

strain.

The stress-strain properties of a material are determined by a tensile test. In

Sec. 2.6 we introduced the following concepts:

Proportional limit: The stress up to which the stress is linearly proportional

to strain.

Elastic limit: The stress up to which the material upon unloading returns to

its undeformed length. Both proportional and elastic limits are difficult

to determine.

Permanent set: The residual strain in a material after unloading.

Yield strength: The stress level at which the permanent set is 0.2% (or the

residual strain is 0.002). A material is modelled as elastic up till this

level of loading.

Strain hardening refers to the increase in yield strength of a material due to

its plastic straining. It is exhibited by materials in which the stress-

strain curve slopes upward in some part of the plastic region.

Ultimate strength is the maximum nominal stress in a material before

necking begins.

Necking refers to sudden and localized sharp reduction of cross-sectional

area.

Most of the differences between the behaviour of ductile and brittle material

arises from the fact that there is little plastic deformation in a brittle

40

material. A brittle material undergoes fracture with very little strain,

much less than that for a ductile material. Glass, for example shows no

plastic deformation at all. Another feature to note is that the strength to

fracture is much larger in compression than in tension.

As a structure is loaded the displacements of the points of application of

external loads result in work being done. This work done is stored as the

strain energy of the body given as

∑ . The work done on a simple

bar by an axial load is calculated as

.

Castigliano theorem states that if the total strain energy of a structural

system is expressed in terms of the external loads applied to it, the deflection

at any one loading point (in-line with the load) is obtained by taking the

partial derivative of the strain energy with respect to the load at that point:

.

The procedure consists of applying a load P at the point where the deflection

is to be calculated and in the direction in which the deflection is desired.

This may involve replacing an existing load with P, or inserting a dummy

load. The loads Fi’s in the v rious members re c lcul te in terms of the

unknown load P, and then the total energy U is calculated. After taking the

partial derivative of U with respect to P, the actual value of P is plugged in

(or P is replaced by zero, in case it is a dummy load). This gives the desired

deflection.

In the case of more involved trusses, it was shown that we could simplify the

calculation of the deflection at the location of a load P by rewriting the

partial derivative as:

∑

∑

∑

.

Here we first calculate Fi’s without the lo d P, and then calculate the value

of as the increment in ’s for unit lo d P. This is known as the

unit load method whose calculations can be organized as in Table 2.3 or 2.4

Statically indeterminate problems can also be solved using the energy

method. The strategy consists of relaxing one of the boundary constraints by

letting that boundary point move. A dummy load is introduced at that

location in place of the reaction that would have been there if the constraint

was not relaxed. The deflection of the boundary point is then calculated in

terms of that dummy load. We next obtain the value of the dummy load

required to make the movement of that point as zero. This value is the

desired reaction at that point.

The strain energy per unit volume of the body is given by

( )

The total strain energy can be obtained by integrating it over the entire

volume of the body.

41

3 Torsion of circular shafts

3.1 Introduction

A slender rod which is subjected primarily to twisting or torsional load is termed

as a shaft. One of the more important uses of torsional members is in the

transmission of power or motion through

circular shafts. A shaft is used to transmit

the torque generated by an electric motor

to the load it rotates. The torque produced

by the engine of a car is transmitted to its

wheels by a shaft. When a coil spring

elongates, the wire constituting its coil is

twisted. Torsion bars are also used as

springs in automobile suspension (Fig.

3.1).

In problems involving torsion of shaft we may be interested in determining:

The stresses that result when the shaft carries a specified twisting

moment. We may consequently determine the maximum twisting

moment that the shaft can carry without failure.

The twist in the shaft when carrying a specified twisting moment. In

application to torsional springs, we may like to determine the spring

constant, which is the angular twist per unit twisting moment.

To obtain the relations between twisting moment, angle of twist, and the stresses

and strains we shall follow the strategy that we outlined in the last chapter, but in

a reversed order. The reason for this is that unlike the situations involving axial

loading treated in the last chapter, it is not possible to make the assumption of

uniform stresses in cases of torsional loading. In fact, it will be shown that the

stresses on the cross-section of a shaft vary with the distance from the axis of the

shaft.

For a constant diameter shaft, we shall first assume a total angle of twist θ for an

applied twisting moment T. We then use the geometric considerations to convert

this macro quantity to the strain at various points in the shaft. This is the macro

to micro stage. We next use the material properties to convert strains to stresses.

This is the micro to micro transformation. In the last stage we convert the

stresses into torque, a micro to macro conversion. This completes the solution to

the problem.

3.2 Relating angle of twist to twisting moment

Fig. 3.2 shows a circular shaft25

of radius R and length L

clamped at one end and

subjected to a twisting moment

T. Let the shaft undergo a twist

through an angle θ as shown.

We first determine the strain

components in the shaft. For this

purpose we make some

simplifying assumptions26

:

Circular cross-sections

of the shaft perpendicular to the axis remain plane, i.e., there is no

warping of the shaft,

cross-sections of the shaft do not deform, i.e., there are no strains within

a cross-section of the shaft. Radial lines within the cross-sections remain

straight and radial, and

there is no change in length of the shaft, i.e., there is no axial strain.

25

We assume a circular shaft because many of the assumptions made here are not

valid for non-circular shafts 26

These assumptions can be shown to be true by using some fascinating

symmetry arguments. See Crandall, Dahl and Lardner, An introduction to the

mechanics of solids,2 ed., McGraw-Hill

Fig. 3.1 Torsion bar used as

suspension in automobiles

Fig. 3.2 A circular shaft under torsion

42

The above assumptions imply that there are no linear strains. Therefore, using the

cylindrical polar coordinates as shown,

No distortion of the sections rules out shear strains and γrθ and γzr also vanish.

The only non-zero component of strain is γrθ which we now proceed to

determine.

Calculation of strain from macro distortion

For this purpose we take a

slice of length dz at a distance

z from the clamped end as

shown in Fig. 3.3. The

detailed geometry of

distortion is shown in Fig.

3.4. Let the twist of the face

at z be θ, and that at z + dz be

θ + dθ. To calculate the shear

strain in the shaft, consider a

‘rect ngul r’ element ABCD

on the surface of the cylinder.

After distortion, the shape of

this element changes to

A1B1CD. The angular twist dθ

of this element is the angle that the

movement from A to A1 subtends at the axis

passing through point O on the axis of the

shaft. What, then, is the strain? We see that

which was originally a right angle has

now changed to . Thus, it has reduced

by . Therefore, the strain here is equal

to which is seen as AA1/DA. Thus,

the strain at r = R is

.

It can be seen that the strain γθz is not

constant with r. Consider the deformation at a radius of r. Here, the point E has

shifted to point E1, so that strain at this radius is given by EE1/dz, or,

( )

. (3.1)

Converting strain to stress

Conversion of strain to stress is simple. Since there is no strain other than the

shear strain γθz (and its complementary shear strain γzθ), there is only one

component of strain ηθz (and its complementary

shear stress ηzθ). These stress components on an

element are shown in Fig. 3.5. The shear stress

component can be obtained by multiplying the

strain with the shear modulus G:

( ) ( )

,

(3.2)

where G is the shear modulus of the

material. Please note that neither

the geometry nor the nature of

loading enters this step. The

variation of shear stress with the

radius r is shown in Fig. 3.6.

Converting stress into loading

Consider an elemental area in the

cross-section of the shaft at radius r

and angle θ as shown in Fig. 3.7. The shear stress at this location is ( )

. If the area of the element is dA, the shear force on this element is

. It is easy to see that the force on an element of equal area which is

diametrically opposite to this element is equal in magnitude but has the opposite

direction. This is true of all elements of the cross-

section and, therefore, the net shear force on the

section is zero. But that is not true for the moment

about the origin. Every elemental shear force

contributes a counter-clockwise (hence, positive27

)

moment, so they add up. The sum of moment can be

found by simple integration. The force on the

element shown at a radius of r is stress times its area,

or,

. Therefore, its contribution to the

twisting moment about O is radius times this force,

27

Consistency of results requires that we define our co-ordinate directions and

sign conventions very clearly. It is common to use only right- handed triad for

the direction of axes: Curl the fingers of your right hand from the positive x- to

the positive y-axis. The thumb then points in the positive z-direction. We shall

use only right-handed triads in this book.

Fig. 3.3 The slice whose distortion is studied

Fig. 3.4

Fig. 3.6 The shear stress distribution on

the cross-section of the shaft

Fig.3.7 Shear force on

an elemental area

Fig. 3.5 Shear stresses on

an element

43

or,

. Thus, the total twisting moment T is

∫

∫

The integral ∫

is termed as the second moment of the area about the polar

axis of the section. Many authors term this as the polar moment of area. It is a

geometric property of the area and is denoted by28

Izz. This reduces the

expression for the twisting moment T to

(3.3)

We have completed the solution of the problem of the torsion of a shaft under a

twisting moment. We can recast Eqs. 3.1-3.3 to obtain any quantity of interest.

Replacing dθ/dz from Eq. 3.3 in the expression for the shear stress η,

( )

(3.4)

The total twist θ is obtained by integrating dθ/dz over the entire length of the

shaft:

∫

∫

(3.5)

The product GIzz is termed as the torsional rigidity of a shaft. The more its value,

the less is the twist produced for a given length and twisting moment

3.3 Stresses and strain in a circular shaft

We now evaluate the value of Izz for a circular shaft of radius R. We replace the

elemental area dA of Fig. 3.7 with its value rdθdr, and integrate over θ from 0 to

2π, and over r from 0 to R:

∫

∫ ∫

( ) ∫

, or

for a circular shaft. (3.6)

The torsional rigidity GIzz of a circular shaft varies as the fourth power of its

radius or diameter.

28

Many texts use the symbol J to denote the second moment of area about the

polar (z-) axis. We prefer here the symbol I with two subscripts denoting the

distance from the specified axis.

The maximum shear stress in a shaft occurs where the value of r is maximum,

i.e., at r = R. Using Value of Izz from Eq. 3.6 in Eq. 3.4, we get

(

)

Thus, the maximum shear stress in a shaft for a given twisting moment T

decreases as D3.

We illustrate below the use of these formulae.

Example 3.1 Diesel generating set

The shaft of a 2 MW diesel generating set rotates at 500 RPM. What is the

minimum diameter of the steel shaft connecting the diesel engine to the electric

generator if the shear stress in not to exceed 40 MPa?

Solution:

We first calculate the torque from the power being transmitted: P = T×ω. Here ω

is the angular speed equal to 2π×RPM/60 = 52.36 rad/s. Torque T then is (2×106

W)/(52.36 rad/s) = 3.82×104 N.m. This is the twisting moment in the shaft. The

shear stress varies across the section of the shaft with the maximum stress

occurring at the surface at r = R, where R is the radius of the shaft.

To calculate the stress in the shaft at r = R, we use Eq. 3.4 with the value of Izz

given by Eq. 3.6:

( )

With T =3.82×104 N.m and η(R) = 40 MPa, we get R = [2×3.82×10

4 N.m/(π×40

MPa)]1/3

= 0.085 m, or 8.5 cm. Thus, the diameter of the shaft needs to be at

least 17 cm.

What is the twist per meter run of the pipe?

To obtain the twist angle θ per meter run, use Eq. 3.5 with L = 1 m:

( ) ( )

( ) , ( )

- , or 0.133

o per meter run,

a negligible amount.

Example 3.2 Torsion bar

Fig. 3.8 shows a torsion bar used as suspension in an automobile. The load from

the wheels is applied to a rigid arm as shown. Calculate the spring constant for

the bar.

44

Solution:

The twist in the bar

then is given by Eq.

3.5 as:

. For

an applied force F, the

torque T is F×(0.7 m),

L = 2 m, G for steel is

80 GPa, and Izz =

π(0.04)4/32. Plugging

in the values, we get

for a load F, a twist of 6.96×10−5

F. This twist results in an upward motion of the

load point of 0.7×θ, or 4.87×10−5

F. Thus, the effective spring constant for the

torsion bar spring is k = 20.53 kN/m.

Example 3.3 Power shaft

Fig. 3.9 shows a power shaft of diameter 4 cm transmitting power to two

machines. The shaft is driven by a belt drive that applies a torque of 40 kN.m at

pulley A. The power take-off is again through belts at pulleys B and C. if the

shaft diameter is 8 cm, determine the total twist of the shaft and the maximum

value of shear stress in the shaft.

Solution:

We first draw a diagram depicting the variation of twisting moment along the

axis of the shaft. To determine the twisting moment T at any location in the

shaft, we externalize the moment there by taking a section at that point and draw

an FBD. Fig. 3.10a shows an FBD for determining the value of T for all values

of z between 0 and 1.5 m. The torque acting on pulley A is 40 kN.m, and using

the right-hand thumb29

rule for directions of torques, it is pointed in the positive

z-direction, and hence, positive. We arbitrarily show the twisting moment at z to

be positive T. The moment balance of this FBD shows that T must be −40 kN.m.

This is valid for all vales of z between 0 and 1.5 m.

If, however, the value of z is larger than 1.5 m, the resulting FBD is as shown in

Fig. 3.10b, and the value of T is −20 kN.m. The v ri tions of T with the value of

z have been plotted in Fig. 3.10c. This is known as the Twisting or Torsion

Moment Diagram (TMD)30

The value of shear stress is

given by Eq. 3.4. The maximum

shear stress will occur in the

section AB where the twisting

moment is the maximum. And

it will be at the surface where

the value of r is the maximum

(equal to R):

( )

( )( )

[The diameter of the shaft is 8

cm throughout. The value of Izz

is πD4/32, or π(0.08 m)

4/32 =

4.02×10−6

m4.]

This is a fairly large value, close

to the failure strength of structural steel. We probably need a thicker shaft to

give it a margin of safety.

The twist of the shaft section AB of length L = 1.5 m which carries a uniform T of

−40 kN.m is given by Eq. 3.5 s

29

Right hand thumb rule states that if you curl the fingers of your right hand in

the sense of the torque, the thumb points in the direction of the moment vector. 30

These TMDs are a big help in the analysis of shafts which do not have uniform

twisting moments along them. We can draw the TMDs directly without drawing

FBDs if we follow the following procedure: (1) start from T = 0 at the left most

end; (2) at every location where a concentrated moment To acts, decrease the

value of T by To, if To is positive, or increase by To if it is negative. We will

come across similar procedures for drawing shear force diagrams (SFDs) and

bending moment diagrams (BMDs) in Chapter 4.

Fig. 3.9 Power shaft

Fig. 3.10 Twisting moment diagram

Fig. 3.8 Torsion bar

45

( )( )

( )( )

A reasonable value of G for steel is 80 GPa.

Similarly, the twist of shaft section BC is exactly half of this, since it carries only

h lf the twisting moment (−20 kN.m), i.e., 0.093 r d.

The tot l twist of the sh ft is the lgebr ic sum of the two: −0.279 r d or −16o.

The negative sign indicates that the end at C twists clockwise with respect to the

end A.

Example 3.4 Twist in a tapered shaft

We next consider a tapered shaft in which the polar moment of area, and

therefore, contribution to the twist varies

along the axis of the shaft. Fig. 3.11

shows a 1 m long tapered steel shaft

subjected to a twisting moment of 1

kN.m. If the base diameter is 10 cm and

the tip diameter is 7 cm, determine the

total twist of the shaft.

Solution:

The contribution dθ to the twist

of the shaft by a section of

length dz at z is given by

(T/GIzz)dz. In this problem the

twisting moment T is constant

along the length of the shaft, but

Izz varies with z. It can be

verified that the linear variation

of diameter D is given by

( ) ( ). The value of Izz varies as fourth

power of the diameter, i.e., as

πD4/32. Fig. 3.12 shows the

variations of the twisting

moment, area, the polar moment

of area, and the contribution to

twist along the length of the tapered shaft. The total twist can be obtained by

integrating the contribution of an elemental length. Thus,

∫

∫ ( )

.

We insert here the value of Izz in terms of the diameter D(z), which, in turn, is

expressed in terms of z as above.

∫

( ) , * + -

= 0.024 rad or about 1.4

degrees.

Where and what will be the magnitude of the maximum shear stress in this

tapered shaft? The shear stress is maximum (at a given section) at the outer most

location, i.e., at r = R. Thus,

( )

Since Izz varies as R4, and T is constant along the shaft, the value of maximum η

along the length of the shaft varies like R−3

, i.e., the maximum value of the shear

stress along the length occurs where R has the minimum value. The maximum

shear stress, therefore, occurs at the tip where D is 0.07 m, Izz is π×(0.07 m)4/32 =

2.36×10−6

m4, and T = 1 kN.m.

( ) ( ) ( )

Example 3.5 Cable in a sleeve

Fig. 3.13 shows a 4 mm dia long cable in

a sleeve used to control a setting

remotely. As we turn the knob the

connecting cable twists inside the sleeve.

The sleeve is provided so that the cable

does not flip. The friction between the

sleeve and the cable applies a torque on

the cable which needs to be overcome.

This torque is estimated to be about 0.3

N.m per meter run of the cable. If the

torque needed at the setting point is 0.5

N.m, determine the maximum length of the cable, if the cable material can

withstand a shear stress of 100 MPa. What would be the play? (The play is the

angle through which you need to turn to effect a change in setting when you

reverse the setting.)

Solution:

Fig. 3.14 shows the TMD of a cable of length z. The maximum torque in the

cable is (− 0.5 −0.3z) N.m, and occurs at the knob-end. The maximum shear

Fig. 3.12 Variations of twisting

moment, area, polar moment of

area, and the contribution to twist

along a tapered shaft

Fig. 3.11 Tapered shaft

Fig.3.13 Cable in a sleeve

46

stress would be at the surface of the cable, i.e., at r = 0.002 m. Eq. 3.4 gives

( )

( m) ( )N.m

( m) ( )

Using the condition that

maximum shear stress is

limited to 100 MPa31, we

get the maximum value of z

as [(100×106/79.6×106) –

0.5]/0.3 m, or 2.5 m.

The maximum twisting

moment, then, is (−) 1.25

N.m, and the twist is given

by Eq. 3.5 as

( ) ( )

( ) , ( m) - or 88

o. This is the play in the cable.

Example 3.6 Geared shafts

Fig. 3.15 show two steel shafts geared together. One end of the first shaft is built

in a wall. Find the maximum shear stress in the shafts and the total rotation of

the free end of the second shaft.

Solution:

The first step in

solving this problem

is to determine the

twisting moments

along the two shafts

and to draw the

twisting moment

diagrams. Clearly,

the torque in the

thinner (second) shaft

is 3 kN.m throughout

its length. To determine the torque in the thicker (first) shaft one has to be

careful. The torques in the two shafts are not equal. (Why? Why does a simple

torque balance not give the correct result?32

) The problem can be solved if it is

realized that the contact force F between the two gears must be equal and

opposite. And therefore, the torque in the first shaft is 10/6 of 3 kN.m, or 5

31

We need to use a negative sign with ηmax here. 32

It is because we have not shown the bearings that will be required to hold the

shafts in place. The bearings will apply forces on the shaft which will produce

moments that should also go in the moment balance equation.

kN.m. This torque is negative, i.e.,

clockwise looking from the right end.

Fig. 3.16 shows the twisting moment

diagram for the two shafts.

The maximum torque in a shaft is

given by ( ), where

R is the radius of the shaft. The

maximum value of the shear stress in

the first shaft is ( ) ( ) , ( ) - . This is quite safe. The

maximum value of the shear stress in

the second shaft is ( ) ( ) , ( ) - . Thus, the

maximum (shear) stress in the shafts is 70.8 MPa.

Let us next find the total twist at the free end. The twist in the each of the two

shafts is obtained from the equation .

Twist in the first shaft, ( ) ( ) ( ) , ( ) - , counter-clockwise

looking from right.

Twist in the second shaft (with respect to the geared end

of the shaft,

( ) ( ) ( ) , ( ) - , clockwise looking from right.

To find the total twist of the free end with respect to the

fixed end of the first shaft we cannot just add the two

twists. Consider the motion of the gears as shown in Fig.

3.17. Angle θ1 represents the clockwise movement of the

larger gear (on shaft 1). It is clear that angle θ2 which

represents the counter-clockwise movement of the

smaller gear due to gearing alone is 10/6 of θ1. This is the rotation of the left end

of shaft 2. Since θ1 is 0.0064 rad, angle θ2 is 0.01 rad counter-clockwise. The

twist of shaft 2, which is 0.03 rad counter-clockwise is superimposed on the

motion of the gear, to obtain the total rotation of the free end as 0.01 rad + 0.03

rad = 0.04 rad or 2.3 degree.

3.4 Hollow shaft

We shall see here that hollow shafts provide excellent rigidity for weight of

material. Consider a hollow circular shaft of inner radius Ri and an outer radius

of Ro as shown in Fig. 3.18. The value of Izz can be obtained by replacing dA in

r2dA with rdθdr and integrating over θ from 0 to 2π and over r from Ri to Ro.

Fig. 3.14 TMD of the cable

Fig. 3.15 Geared shafts

Fig. 3.17

Geometry of twist

Fig. 3.16 Twisting moment diagram for

geared shafts

47

∫

∫ ∫

( )

(

) (3.7)

The maximum shear stress in the shaft will be near

the surface, i.e., at r =Ro. Therefore, the maximum

stress in the shaft ( ) ⁄ varies

inversely as Izz. If we define the torque carrying

capacity of the shaft as the maximum torque the

shaft can carry without the stress exceeding a

specified limit, it is clear that this varies directly as

the value of Izz (for a fixed value of Ro). The curve

labelled (a) in Fig. 3.19 which plots the variation of Izz (as the fraction of the Izz of

the solid shaft of radius Ro) also represents the variation of the torque carrying

capacity. The curve (b) plots the variation of the weight (which varies as the area

of cross-section) of

the shaft. The curve

(c) shows the

torque carrying

capacity per unit

weight of the shaft

showing clearly the

cost effectiveness33

of a hollow shaft34

.

3.5 Statically indeterminate shafts

There are many

situations in which

the twisting

moments in shafts

33This, of course, has not taken into account the increased manufacturing

cost of the hollow shaft. 34

This is quite as expected. An element nearer the surface has larger stresses and

contributes more to the torsional moment. Therefore, the material nearer the

surface is used more effectively to bear the torsional load than an equal area near

the axis. Seen in another way, it is an area nearer the outer surface that makes a

larger contribution to Izz, the value of the polar moment of inertia.

cannot be determined by considerations of equilibrium alone. In such shafts the

geometrical constraints on twists need to be invoked to solve the problems.

We follow the same three-step process that was invoked in the last chapter:

Use equilibrium analysis to write equations for the twisting moments in

the various parts of the shaft. There may not be enough equations to

determine the moments explicitly.

Convert twisting moments to twists in each part.

Write geometrical compatibility conditions to complete the analysis.

We illustrate the procedure with a few examples.

Example 3.7 A shaft built in at both ends

Consider a composite steel

shaft ABC built up of two 1 m

long steel sections, AB of 10

cm dia and BC of 15 cm dia

(Fig. 3.20). A torque of 100

kN.m is applied to the collar

in the middle. The two ends

of the shaft are built into

walls so that there is no

rotation of the ends. What is

the twist produced in the

shaft?

Solution:

Equilibrium analysis

Let the reactions at the two

ends be T1 and T2 as shown.

From the condition of the

equilibrium of moments we

get T1 + T2 + 100 kN.m = 0. It

should be clear that either T1

or T2 or both should have a

negative sign. This is the only

equation we get from the

statics for the two unknown

reactions T1 and T2. Clearly

this is a statically

indeterminate problem. We need to consider deformations as well.

Fig. 3.18 Hollow

cylindrical shaft

(a) Variation of the value of Izz of hollow shaft

(b) Variation of the weight of hollow shaft

(c) Variation of the ratio of strength to weight

Fig. 3.19 Comparison of hollow shaft with solid shaft for

various ratios of the internal and external radii

(a) The composite built in shaft

(b) FBD for determining torques in part AB

(c) FBD for determining torques in part BC

(d) Torsion or twisting moment diagram

Fig. 3.20

48

We need to determine the twisting moments in the two halves to calculate the

twists therein. For this purpose, we take a section first in the part AB and draw an

FBD for the left part as shown in Fig. 3.20b. Here T is the twisting moment in

the shaft. The equilibrium of this FBD gives T1 + T = 0, or T = − T1. Fig. 3.20c

shows the FBD for determining the twisting moment in the second part of the

shaft. From this we get T1 + T + 100 kN.m = 0, or T = − T1 – 100 kN.m. Fig.

3.20d shows the complete twisting or torsion moment diagram (TMD) of the

shaft35

. If we have chosen the sign of T1 correctly, then both halves of the shaft

have negative twisting moments36

.

Relating torques and twists

Now that we know the twisting moments, we can calculate the twist using Eq.

3.5: ⁄ .

The values of Izz are:

for shaft AB: πD4/32 = π(0.1 m)

4/32 = 9.81×10

−6 m

4

for shaft BC: πD4/32 = π(0.15 m)

4/32 = 49.7×10

−6 m

4

( )( )

( )( ) , and

( )( )

( )( )

Geometric compatibility

Since both ends of the shaft are

built in, the net twist in the shaft

must be zero: θAB +θBC =0. This

gives:

−1.92×10−6

T1 –

0.38×10−6

T1 – 0.038 = 0, which

gives

T1 = −16.6 kN

With this value of T1, we can

calculate the twist at point B as

35

The reader can verify that we could have drawn this TMD directly without

drawing FBDs if we had followed the procedure outlined in footnote 5. 36

It is clear that the sign of T1 that we have chosen here cannot be correct

because with this we have negative twisting moments in both halves of the shaft.

This, of course, is not possible since if one part twists clockwise, the other must

twist counter-clockwise. But it does not matter for our analysis here.

θAB = −1.92×10−6

T1 =

(−1.92×10−6) × (−16.6 kN) = 0.032

rad

Example 3.8 Meshed gears

Fig. 3.21 shows two shafts, one of dia

6 cm and the other of 4 cm, carrying

identical meshed gears in the middle.

The length of each shaft is 1 m. The

two shafts are disengaged, the top

shaft A rotated by 10o

in the direction

shown, and then the two gears are

meshed again. Shaft A will untwist a

bit twisting shaft B till the

equilibrium is reached. Determine the

locked in torque at equilibrium.

Solution:

Equilibrium requires that the contact

force between the gears be equal and

opposite. Since the two gears are identical, the torques in the two shafts are

equal but opposite. Let this residual torque be To, positive To in (the left part of)

shaft A and negative To in (the left part of) shaft B.

Let us assume that wall at left applies on shaft A a reaction torque T1, and on

shaft B a reaction torque T2. The TMDs of the two shafts can be calculated and

are as shown in Fig. 3.22. These have been drawn under the assumption that T1

and T2 are both positive.

Now let us look at the deformation of

shaft A. It is clear that the twist in the

left half must be the same but opposite

of that in the right half, for the total

twist from end to end to be zero. Since

the two halves have identical

geometries, it stands to reason that the

torque in the two halves must be equal

and opposite. There is, thus, only one

solution: the twisting moment in the

left half is + To/2, and that in the right

h lf is − To/2. Similarly, we can argue

that the twisting moment in the left

half of shaft B be − To/2, and that in the

Fig. 3.21 Two meshed shafts

(a) TMD for top shaft

(b) TMD for bottom shaft

Fig. 3.23 Actual TMDs of the two shafts

(a) TMD for top shaft

(b) TMD for bottom shaft

Fig. 3.22 TMDs of the two shafts under

the assumptions that the reactions on the

left wall are T1 and T2, respectively.

49

right half be + To/2. These are shown in Fig. 3.23.

Geometric compatibility

Refer to Fig. 3.24. Shaft A was rotated by 10o when

the two gears were engaged. After engagement and

letting go, the shaft A unwinds by an angle θ2

which is the same as winding up of shaft B (since

the gear ratio is one). The residual twist in shaft A

is now θ1. It should be clear that the geometric

compatibility condition is:

θ1 + θ2 = 100 = 0.175 rad

We can calculate the twists θ1 and θ2 using Eq. 3.5:

θ = TL/GIzz. Thus,

( ) ( m)

( G ) , ( m) - , and

( ) ( m)

( G ) , ( m) - .

Using these in the geometric compatibility condition, we get

2.46×10−6

To + 12.45×10−6

To = 0.175

This gives To, the locked-in twisting moment as 11.7 kN.m.

3.6 Composite shaft

Let us next consider twisting of a composite shaft that is made up of inner core of

one material clad with an outer sleeve of another material as shown in Fig. 3.25.

To solve this problem, let us look carefully through the derivation of the torsion

formulae in Section 3.2 and figure out the changes we would need to make on

account of two different materials in the cross-section.

We had assumed a twist angle θ and showed

using geometry that there is only one shear

strain component γzθ which is given by Eq. 3.1

as ( )

. This does not change with

the material. Thus, here too, stain γzθ varies

linearly with the distance r from the longitudinal

axis, independent of material.

We had next converted the shear strain to shear

stress by multiplying it with G, the shear

modulus. Here the material dependence arises.

In the inner core up to r = r1, we multiply the

strain with the value of G = G1 for the core material, and in the outer sleeve for r

between r1 and r2, we multiply with the value of G = G2 for the sleeve material,

which is higher. Thus, the graph for shear stress is kinky with two different

slopes, one for the core and the other for the sleeve. This is shown in Fig. 3.26.

The next step in Section 3.2 was considering the shear force on a small element,

calculating its contribution to the twisting moment about the axis, and integrating

over the entire cross-section:

∫

∫

Here, unlike what we did in Section 3.2,

we cannot take G outside the integral

since it is not constant over the section.

But we can do it part-wise: we divide

the area in to two parts, one, with r

between 0 and r1, which covers the core

with modulus G1, and the other with r

between r1 and r2, which covers the

sleeve with modulus G2.

∫

∫

∫

2∫

∫

3

2 ∫

∫

3

or

( ) (3.8)

where I1 is the polar moment of the core area and I2 is the polar moment of the

sleeve area. Compare this with Eq. 3.3. The term GIzz for a uniform shaft is

replaced with (G1I1 + G2I2). Similarly,

( ) (3.9)

The torsional rigidity of the composite shaft is, thus,

( ) ( ).

Here , and (

) .

Fig. 3.24 Geometric

compatibility condition

Fig. 3.26 Shear strain and shear

stress distribution across a

composite shaft

Fig. 3. 25 A composite shaft

50

3.7 Torsion of thin-walled tubes

We have so far found exact solution to the problem of torsion in circular (and

hollow circular) shafts. It is not possible to find such solutions for shafts of

arbitrary cross-sections. However, we can find quite easily the approximate

solution for a thin-walled shaft of arbitrary section. Consider such a shaft as

shown in Fig. 3.27. We use the n, s, z

co-ordinate system as shown, with

the coordinate axis s along the central

line of the tube wall. The dominant

stress component in this case is ηzs

(along with the complementary stress

component ηsz). We shall assume,

without proof, that the other stress

components are either absent or are

negligibly small.

We show in Fig. 3.27b a small

element of length dz of the shaft.

The shear stress ηzs is distributed over

the top and bottom surface of the

element while the complementary

shear stress ηsz is distributed over the surfaces marked 1 and 2 in the figure. We

introduce here the concept of shear flow q defined as the shear force per unit

length of the tubular surface:

∫ (3.10)

In Fig. 3.27b we have shown the net shear forces on the vertical surfaces as shear

flow q times the length dz of the surface. The first thing to notice is that shear

force per unit length on the vertical surface 1 is the same as on the horizontal

surface. The next thing to notice is that the two forces shown are the only

significant vertical forces on the element, and therefore, q1 = q2, that is, the shear

flow is the same at two arbitrary positions, and hence everywhere, along a cross-

section of the tubular shaft37

.

We can now relate the shear flow q to the twisting moment T. Consider an

element of length δs in the cross-section of the shaft as shown in Fig. 3.28a. The

37

This of course implies that if the wall thickness is constant, the shear stress is

constant too. And also that as the wall thickness increases the shear stress

decreases, and vice-versa. The nomenclature shear stress originates perhaps from

the fact that there is a definite analogy between shear flow in the thin wall of a

shaft and the flow of an incompressible fluid in a channel.

shear force on this element is qδs whose moment about an arbitrary point O is

hqδs, which is 2qδA, where δA is the grey area in Fig. 3.28a. The total twisting

moment, then, is simply the shear flow q times twice the area A of the cross-

section of the tube: , from which we can extract the value of the shear

flow38

as:

(3.11)

Example 3.9: A hollow circular shaft

Let us evaluate the stresses in a hollow circular shaft by the approximate method

outlined in Sec.3.7 and compare the results with those obtained by the exact

method. Consider the hollow circular steel shaft of length 1 m shown in Fig.

3.29. It is subjected to a twisting moment T = 100 N.m. Compare the stresses

obtained by the exact and the approximate methods.

Solution:

Eq. 3.4 gives the stresses in a circular

shaft: ( )

, where Izz for a hollow

shaft is given by Eq. 3.7 as

(

).

For the given values,

,( ) ( ) -

.

38

One interesting fact about this derivation is that this relation has been obtained

without any reference to the material properties, and therefore, is applicable both

to elastic and plastic deformations.

(a) (b)

Fig. 3.27 A thin-walled torsion tube

(a) (b)

Fig. 3.28 Relating shear flow to the twisting moment

Fig. 3.29 Hollow circular shaft

51

The shear stress at r = 20 mm, then, is ( ) ( )

13.5 MPa, and at r = 16 mm is ( ) ( )

10.8 MPa. The shear stress varies linearly between these two values across

the thickness of the shaft.

The shear flow q by approximate procedure is given by Eq. 3.11 as , where A

is the area enclosed by the mean line of the section (here a circle of radius 18

mm). Thus, A = π(0.018 m)2 = 1.02×10

−3 m

2, and ( ) (

) = 49 N/m. The average shear stress can be obtained by dividing the

shear flow q by the wall thickness t to get ( ) ( ) The maximum stress is underestimated by a mere 9%.

Example 3.10 A thin rectangular tubular shaft

Consider a tubular shaft of rectangular

section as shown in Fig. 3.30.It is subjected

to a twisting moment of 100 N.m. What are

the stresses in the various sides of the

rectangle?

Solution:

The shear flow q by approximate procedure is given by Eq. 3.11 as , where A

is the area enclosed by the mean line of the section. The value of A for the given

section is (22 cm – 2 cm)×(12 cm – 2.5 cm) or 0.19 m2. The shear flow is

( ) ( 0.19 m2) = 263.2 N/m. The shear flow is constant

throughout the tube. The shear stress in the shorter (thinner) sides is q/t, or

(263.2 N/m)/(0.02 m) = 13.2 kPa, and the shear stress in the longer (thicker) sides

is (263.2 N/m)/(0.025 m) = 10.5 kPa.

3.8 Plastic deformation in torsion

Let us next look at what happens when a shaft is loaded beyond its plastic limit.

This discussion will serve as a simple model of how problems in elastic-plastic

deformations can be handled. Consider a shaft made up of an elastic-plastic

material whose stress-strain behaviour can be modelled as shown in Fig. 3.31.

The material is linearly elastic till the yield shear stress level39

ηY is reached, after

which the material deforms perfectly plastically with the stress remaining

constant at the yield stress value. If we unload such a material after it has been

39

Fig. 3.30 is similar to Fig. 2.29c, except that linear stress and strain have been

replaced with shear stress and strain. It will be shown later in Sec. 6.10 that the

yield shear stress ηY is related quite simply to the yield stress ζY: ηY = ½ ζY.

deformed up to point A in Fig. 3.32, it follows

a straight line AB with slope G, the shear

modulus.

As a shaft is twisted, and the assumptions

about its behaviour that plane sections remain

plane and do not distort hold, the shear strain

will vary linearly with the radius ( ( ) ⁄ ) as shown in Sec. 3.2. The strains

are then converted to stresses using the stress-

strain relation. Fig. 3.32a-d show the shear

stresses for various levels of straining. When the maximum shear strain is less

than γY, the shear stress distribution is linear as shown in Fig. 3.32a. The

maximum shear stress is less than ηY. As the twisting increases, the value of the

maximum strain increases, the stresses increase till the stress and strain reach the

yield values as shown in Fig. 3.32b. The slope of the stress-radius line increases

as shown. Thereafter, the plastic behaviour kicks in. The yield value of shear

stress is obtained at a radius r = ro < R. The shear stress for r between 0 and ro

increases linearly from 0 to ηY, and thereafter remains constant at ηY for r > ro.

This is shown by the stress distribution of Fig. 3.32c.

For Fig. 3.32c, the shear stress distribution is given as40

:

( ) ( )

, for (a)

As the twisting moment increases further, the value of ro, the radius at which the

yield value is reached decreases and more of the shaft is under plastic

40

Since the value of dθ/dz is not known, γzθ is not determined, and, therefore, ro

is undetermined thus far.

Fig. 3.30 A rectangular tubular

shaft

Fig. 3.31 Elastic-plastic behaviour

(a) (b) (c) (d)

Fig. 3.32 Shear stress distribution in an elastic-plastic shaft

52

deformation. Fig. 3.32d shows the condition when the entire shaft has undergone

plastic deformation.

The shear stress distribution is related to twisting moment by using the same

procedure we followed in Sec. 3.2: We take a small area element rdθdr at radius

r and determine the shear force on it as ηrdθdr. The contribution to the twisting

moment of this elemental force is obtained by taking its moment about origin (by

multiplying the force with radius r) and integrating over the entire area:

∫ ∫ ( )

(b)

If we plug in Eq. (b) the value of η as given by Eq. (a), we can obtain the value of

ro. Once ro is obtained, we can determine the value of dθ/dz and complete the

solution of the problem.

Example 3.11 Twist in a shaft with plastic deformation

A shaft of 4 cm dia is subjected to a twisting moment of 2 kN.m. Determine the

twist per meter length of the shaft. The yield shear stress for steel is 125 MPa.

Solution:

Let us first assume that only elastic deformation takes place. Eq. 3.4 gives the

maximum shear stress as ( ) ( ) , ( ) - , well above the yield value. Following the

derivation and notation of the last section, let ro represent the radius at which

plastic deformation is reached41

. Then,

∫ ∫ ( )

∫ ( )

∫ ( ) ∫

0

∫ ∫

1

0

(

)

(

)1 (a)

Plugging in the values of ηY = 125 MPa, T = 2.0 kN.m, and R = 0.02 m, we get ro

= 0.0112 m, or 1.12 cm. The stress at this radius is the yield shear stress 125

MPa. For the value of shear modulus of 80 GPa, this gives a shear strain of 125

MPa/ 80 GPa = 1.56×10−3

at r = 0.0112 m.

41

We cannot find the value of ro by using Eq. 3.2 because it has been derived

under the assumption that the deformation is linear elastic everywhere.

Using these values and the fact that the shear strain is given by ( ) ⁄ for all r, we get ⁄ ( ) 1.56×10

−3/11.2×10

−3 m = 0.14

m−1

. Therefore, the twist per meter ⁄ is 0.14 rad/m.

3.9 Limit Torque

In the preceding section we have seen the difficulties that arise when we

undertake the analysis of both elastic and plastic deformations. The analysis was

possible at all because we were taking a very simple case of pure torsion on a

uniform circular bar.

We can however work with a simpler model

in much of engineering work with

complicated structures. In this model which

is termed as limit analysis, we take a fully-

plastic stress distribution and omit the earlier

elastic-plastic considerations. The load for

which the structure first becomes fully-plastic

is termed as the limit load.

Limit analysis is based on the idea that any

structure made up of materials with fairly

well-defined yield point will not undergo very

large deformations till all of it is in the plastic

region. And, therefore, deformations will be small as long as the load is less than

the limit load for that structure. Limits load, in general, are much easier to

calculate. Let us consider a shaft in the plastic limit when the whole cross-

section is in the plastic state. The stress distribution is then very simple, as

shown in Fig. 3.33. The calculation of the limit torque42

for such a distribution is

quite simple:

∫ ∫ ( )

∫

As long as the applied torque is below this value, some part of the structure will

be in the elastic range and the twist will not be excessive. Limit analysis is quite

extensively used, at least as a first approximation, in the design of structures.

42

The limit load for this case can be obtained quite easily from Eq. (a) in the

solution of Example 3.10. The limit load corresponds to ro vanishing.

Replacing ro by 0, gives the same result as here.

Fig. 3.33 Fully-plastic stress

distribution

53

3.10 Strain energy in torsion

We had introduced in Sec. 2.9 the concept of strain energy. The energy method

based on Castigliano theorem was introduced in Sec. 2.10 where a number of

examples involving axially-loaded members were solved. The method was also

applied to a statically indeterminate problem. We introduced in Sec 2.11 the

calculation of strain energy for an arbitrarily loaded structure. We now extend

the energy methods to cases involving torsion.

Let us consider a shaft of radius R and of length L subjected to a pure torsional

moment T. The strain energy is obtained from Eq. 2.16 using the stress

distribution given by Eq. 3.4.

∫

∫

∫

.

/

∫

∫

∫

∫

(3.12)

where A is the cross-sectional area and the integral of r2 over A has been replaced

by the polar moment of area Izz.

Castigliano theorem can be quite easily extended for the case of a torque load T

to give the twist θ as:

(3.13)

We now illustrate the application of Castigliano theorem.

Example 3.12 Stepped shaft

Consider the stepped steel shaft shown in Fig. 3.34. It consists of three parts:

part A (dia 20 mm, length

1 m), part B (dia 16 mm,

length 1.5 m), and part C

(dia 10 mm, length 2.0

m). It is subjected to three

twisting torques of 300

N.m, 200 N.m and 100

N.m as shown.

Determine the total twist

of the free end of the

shaft.

Solution:

Since Castigliano theorem requires that the total energy be differentiated with

respect to torque applied at the point where the twist angle is to be determined,

we replace the torque of 100 N.m by T. The TMD of the composite shaft, then,

is as shown in Fig. 3.35. The total strain energy is given by:

( )6( ) ( )

( )

( ) ( )

( )

( )

( )

7,

and

( )0( ) ( )

( ) ( )

( )

1

Using 100 N.m as the value of T, we get θ = 3.9 rad, or 223o, a fairly large twist.

Let us, as a check, calculate the maximum shear stress in the shaft. The

maximum shear stress in a circular shaft is given by . This has

been obtained by substituting the value of Izz in terms of R, and determining the

stress η at r = R. This gives the values 47.7 MPa for the part A of the shaft, 46.6

MPa for the part B, and 63.3 MPa for the part C. These appear to be safe even as

the twist appears to be excessive.

Example 3.13 Coil spring

Fig. 3.36a shows a

closely-wound coil spring

consisting of n turns of a

wire of radius of r wound

into a coil of radius R.

Determine the elongation

of the spring under the

action of an axial force P,

and, thereby, obtain the

spring constant.

Solution:

This example

provides a

dramatic

illustration of the

power of energy

method and of

Castigliano

Fig. 3.34 Stepped shaft

Fig. 3.35 TMD of the stepped shaft

(a) (b)

Fig. 3.36 Coiled spring

(a) (b)

Fig. 3.37 Statically indeterminate shaft

54

theorem. Fig. 3.36b shows the FBD of the spring with the wire cut at an arbitrary

location. We can, by equilibrium of moments conclude that the twisting moment

T on the wire is clearly equal to PR, the product of the load P and the radius R of

the coil, independent of the location of the cut. Thus, the twisting moment

everywhere in the wire is PR. Therefore, the strain energy due to the twisting

moment in the wire constituting the spring is obtained by writing the strain

energy dU of an elemental length and integrating it over the entire length of the

wire, i.e., for n coils of radius R. The total energy of the spring is given by

∫( )

( )

∫

( )

( ), since the total length of

the wire is 2πnR.

We can evaluate the shear force acting on the wire as V = P. These shear forces

also contribute to the strain energy, but it can be shown that their contribution is

of order (r/R)2 of the contribution of twisting moments, and hence can be

neglected.

Use of Castigliano theorem then gives the deflection of the spring in the direction

of application of force P as:

( )

( )( )

The spring constant

Example 3.14 Statically-indeterminate torsion problem

Consider a composite shaft consisting of a solid steel shaft and an aluminium

tube as shown in Fig. 3.37. If a torque of 5 kN.m is applied to the rigid disc

attached to the left end, determine the twist of the composite shaft.

Solution:

Let the torque carried by the steel rod and the aluminium tube be T1 and T2,

respectively. Fig. 3.37b shows the FBD of the rigid end disc. From the

equilibrium of this disc:

T1 + T2 = 5 kN.m (a)

Since there is no other equilibrium equation, this problem is statically

indeterminate. All we have besides Eq. a is the geometric compatibility

condition that the twists θ1 and θ2 of the two shafts are equal. Let us determine

θ1 and θ2 by energy method. The total strain energy is given by:

The twists of the two shafts are obtained by partially differentiating U with

respect to T1 and T2, respectively. Equating the two twists, we get

Here, the value of Izz,1, the polar moment of inertia of the steel rod is πD4/32 =

π(0.050 m)4/32 = 0.61×10

−6 m

4, and that of Izz,2, the polar moment of inertia of

the luminium tube is π(Do4 – Di

4) /32 = π(0.080

4 – 0.60

4) m

4/32 = 2.75×10

−6 m

4.

The values of G1 and G2 are about 80 GPa and 26 GPa, respectively. Using the

value of L as 0.5 m, the equivalence of twists gives:

3.90×10−6

T1 = 6.99×10−6

T2, or T1 = 1.79T2 (b)

Solving Eqs. a and b simultaneously, we obtain the values of T1 and T2 as 3.2

kN.m and 1.8 kN.m, respectively. The (equal) twists in the two shafts are then

obtained as 0.0125 rad or 7.17o.

Summary

When a circular shaft is subjected to a pure torsional load, there is no

warping or distortion of cross-sectional planes, and consequently there

is only one component of strain, namely γzθ (and its complementary

strain γθz).

o We first assume a twist angle θ and determine the value of the

strain γzθ as rdθ/dz using purely geometrical considerations.

o We next convert the shear strains to shear stresses using the shear

modulus G: ηzθ = Gγzθ = Grdθ/dz.

o We next take a small element within the cross-section of the shaft,

write an expression for shear force acting on it in terms of the shear

stresses, take its moment about the axis of the shaft and integrate

over the entire cross-sectional area to obtain the twisting moment T

as GIzzdθ/dz. Here Izz is the second moment of area (polar) of cross-

section, defined as integral of r2dA over the cross-section. The

value of Izz for a solid cylinder was found to be πD4/32, and for a

hollow shaft as (

) .

o The twist θ is given by TL/GIzz.

The quantity GIzz is seen as the torsional stiffness which is defined as

the torque required to produce a unit twist per meter length of a shaft.

Since an area nearer the surface has a larger values of stresses that

contributes more to the twisting moment than an equal area nearer the

axis, hollow shafts are much more effective in resisting torsional loads.

There are many situations in which the twisting moments in shafts

cannot be determined by considerations of equilibrium alone. In such

shafts the geometrical constraints on twists need to be invoked to solve

55

the problems. Such problems are termed as statically indeterminate

ones. We follow the same three-step process that was invoked in the last

chapter:

o Use equilibrium analysis to write equations for the twisting

moments in the various parts of the shaft. There may not be enough

equations to determine the moments explicitly.

o Convert twisting moments to twists in each part.

o Write geometrical compatibility conditions to complete the

analysis.

For composite shafts made up of a core of one material overlaid with a

sleeve of another material, the torsional stiffness is given by ( ).

The concept of shear flow q defined as the shear force per unit length of

the wall was introduced for the approximate analysis of thin tubular

shafts. It was shown that , where A is the area enclosed by the

mean line of the wall of the shaft.

Limit load refers to the load under the assumption that the whole of the

structure has yielded and the shear stress everywhere is ηY. If the

applied torque is well below the limit torque so obtained, we can

presume that the twist of the shaft is not excessive.

The strain energy in torsion of a shaft is given by

.

56

4 Forces and moments in beams

4.1 Introduction

We have so far considered two types of structural elements: axially loaded truss

elements that resisted axial forces, tension or compression, and shaft-like

elements that resisted twisting moments. Beams that resist transverse loads by

bending are other common structural elements. We define a beam as a slender

element (i.e., an element whose cross-sectional dimensions are much smaller

than its length dimension) that essentially resists loads acting transverse to its

length. Roof beams that support the roofs of buildings are beams. The horizontal

deck of a road flyover supported on columns is a beam structure. The leaf

springs of an automobile suspension transfer the weight of the car to the axle

(and further to wheels) through beam action. A wing of an airplane is a beam. So

is the cantilever beam shown in Fig. 4.1.

For the beam to be in equilibrium, the net force or moment on the beam or on

ny p rt of it must be zero. Let us ‘cut’ the c ntilever be m shown in Fig. 4.1a

near the wall and draw the free-body diagrams (FBDs) of the two parts as shown

in Figs. 4.1b and c. Clearly, we need a force V and a moment M to balance the

applied load P on the part shown in Fig. 4.1c. These are the forces and moments

of internal reaction and these arise because of the deformations that are caused

by the applied load. There are, of course, equal and opposite reactions on the

stump of the beam built in the wall as shown in Fig. 4.1b.

We can, in general, resolve the internal resistance forces and moments in to their

components. Fig. 4.2 shows that resolution. The following are the nomenclature

and the actions of the various components:

Fx: the axial force that results in elongation (or compression, if

it is negative) of the member.

Fy and Fz: shear forces that result in shearing at the section. The shear

forces are conventionally assigned the symbol V.

Mz: the axial moment that is the torsion moment that causes

twisting of the member. This was the subject matter of the

last chapter.

Mx and My: transverse moments that are termed as the bending moments

and cause the member to bend. Moment Mx results in

bending the beam in the y-z plane while the moment Mz

results in bending the beam in the x-y plane.

We had in Chapter 2 laid out the general strategy for solving the problems of the

mechanics of deformable bodies as consisting of three major steps:

(a) (b) (c)

Fig. 4.1 A cantilever beam

Fig.4.2 Forces and moments in a beam

57

Consideration of static equilibrium and determination of loads in various

members,

Consideration of relations between loads and deformations, (first converting

loads to stresses, then transforming stresses to strain using the properties

of the material, and then converting strains to deformations), and

Considerations of the conditions of geometric compatibility.

In that chapter we had also studied the structures bearing axial loads. In Chapter

3 we considered slender members bearing twisting moments. These were termed

as shafts. We learnt to determine the variations of torsion moments (Mz in the

notation introduced above) as torsion moment diagrams (TMDs), and then to

calculate the resultant shear stresses and twisting angles of the shafts

In view of the importance of the role of beam members in engineering and other

structures, we devote this chapter to beams, and that too, to only the first step of

the 3-step process outlined above. The loads which are of immediate relevance

to a beam are the shear forces Fy and Fz and the bending moments Mx and My

which are caused by the transverse loads as shown in Fig. 4.3. It can be verified

easily that the shear force component Fz

and the bending moment component My

are caused only by the vertical load Pz,

whereas the shear force component Fy and

the bending moment component Mz are

caused only by the transverse load Py..

This leads us to a very simple stratagem:

we calculate the bending moments and

shear forces for the vertical and horizontal

loads separately (assuming that the other

loads are absent), and then patch up the

results. We shall, in the discussion that

follows, consider only the vertical load

system, which can easily be replicated for the horizontal loads.

4.2 Sign convention

It is imperative from the point of view of consistency that we define the sign

convention for shear force and bending moment at a section. Fig. 4.4 shows a

beam which has been sectioned and the two resulting parts separated. The shear

forces and the bending moments at the two sections are equal and opposite, as

shown. We adopt the convention given in Table 4.1to ensure that shear force (or

the bending moment) at either of the two sections have the same sign:

Table 4.1 Sign convention43

Sign of the

outward normal

to the section

Actual direction

of the force or

moment

Assigned sign to the

shear force or bending

moment

Along the + ve

coordinate

direction

Along the +ve

coordinate

direction

+ ve

Along the − ve

coordinate

direction

Along the − ve

coordinate

direction

+ ve

Along the + ve

coordinate

direction

Along the − ve

coordinate

direction

− ve

Along the − ve

coordinate

direction

Along the +ve

coordinate

direction

− ve

It can be verified that the shear forces and bending moments shown in Fig. 4.4

are all positive according to this convention. This convention44

can be

43

It is interesting to note that this sign convention results in a beam bending

concave upwards with a positive bending moment and convex upwards for a

negative bending moment.

Fig. 4.3 Loads on a beam

Fig. 4.4 Positive shear stress and bending moment

58

summarized graphically as Fig. 4.5. Either

of these two can be used as an icon for the

sign convention employed.

4.3 Loads and supports

One common classification of beams is

based on the kind of supports on which a

beam rest. There are three idealized

support systems for beams as shown in Fig. 4.6. Support A is a termed as a

pinned support. Here the beam is attached to the support in such a manner that it

is not restrained from rotating about this point. Consequently, the reaction from

the support is just a force which can be resolved in to two force components RA,x

and RA,y. The support here cannot apply a moment on the beam. The support

shown at point B is termed as a roller support. Here the reaction from the support

can only be a vertical force RB,y. The roller does not restrain the beam from

moving horizontally, and hence there is no horizontal force applied by the

support. As in the case of pinned support, there is no reaction moment at a roller

support as well. Beams with only pinned or roller supports are termed as simply-

supported.

The support shown at C is called a built-in support. Here the beam is built into

the wall so that it is restrained from rotating about point C. The beam will have a

zero slope at such a point. This is a consequence of the restraining moment My

that the support applies to the beam at that point. Considerations of equilibrium

will require that the reaction RC,y be equal to P, and the moment My applied on the

beam by the support will be PL, where L is the length of the beam. A beam can

have a built-in support at either one end or both ends. The beam shown at right

in Fig. 4.6 with only one end built-in (and no support at the other end) is termed

as a cantilever beam.

The lower two diagrams in Fig. 4.6 show the conventional representations of the

three types of idealized supports. Please note that these are idealizations. Most

practical supports are more complicated, but many can be analysed quite

satisfactorily by replacing them with one or the other of these idealizations. The

dashed lines in these represent the (exaggerated) deflected shapes of the beams.

Note that the cantilevered beam has zero slope at the built-in end. Table 4.2

summarizes these idealized supports.

44

Please note that many current books use a sign convention entirely different

from this. It is, therefore, advisable that the sign convention used be noted in

each problem that you solve. The sign convention can be denoted easily by a

graphical symbol similar to the ones shown in Fig. 4.5

We, in a similar fashion, idealize the loads that the beams carry. The left top

diagram in Fig. 4.7 show a simply-supported beam with a concentrated or a point

load, while the on the right shows one with a distributed load. A point load is

specified as a force P with appropriate force units like N.

Table 4.2 Idealized beam supports

Support type Freedom of motion Reactions present

Built-in support

No degree of freedom A moment as well as

horizontal and vertical

reaction forces

Pinned support

Single degree of freedom

- rotation.

Horizontal and vertical

reaction forces

Roller support

Two degrees of freedom –

rotation and horizontal

movement

Only vertical reaction

force

Fig. 4.5 Graphically representation

of sign convention

Fig. 4.6 Three type of beam supports and their conventional representations

59

A distributed load, on the other hand, is specified by a load density w as force per

unit length with appropriate units like N/m. What is shown on the beam on right

is idealized as a uniformly distributed load, wherein the load density w per unit

length does not change along the length of the beam.

However, in some situations, the load density could vary, as is shown in Fig.

4.8a. Here the load is distributed, but the loading density varies along the length

of the beam. It is modelled as a linearly varying distributed load. If the load

density is w N/m at the end of the beam of length L, the load density at any x

from the left end is given simply by wx/L N/m.

A beam could also be subject to an externally-applied bending moment as shown

in Fig. 4.8b. Here too, the dashed line represents the (exaggerated) deflected

shapes of the beam.

4.4 Determining shear forces and bending moments

Determination of shear forces and bending moments at any point in a beam is

rather a straight forward procedure. The procedure consists of taking a section of

the beam at that point to expose the shear force V and the bending moment M at

that point and drawing a free body diagram (FBD) of one portion of the beam.

We will first solve a couple of simple examples to illustrate the procedure and

then draw up a generalized scheme of doing things in more complicated

situations.

Example 4.1 A cantilever beam with a load at the free end

Consider a cantilever beam as shown in Fig. 4.9. It is loaded at the free end.

Draw the shear

force diagram

(SFD) and the

bending moment

diagram (BMD).

Solution:

We first calculate

the reactions at the

support. A simple

equilibrium analysis

shows that the end

reaction is a force P

upwards and a

clockwise moment

PL as shown. We

next calculate the

shear force V and

the bending moment

M at an arbitrary

location x from the

support. For this

purpose we take a

section at that

location and draw

the FBD of the

(a) (b)

Fig. 4.8

(a) A simply-supported beam with non-uniformly distributed loading

(b) A simply-supported beam with a concentrated applied bending moment

Fig. 4.7 Two idealized loadings of beam

Fig. 4.9 SFD and BMD for a cantilever beam with end load

(a) A cantilever beam with support reactions

(b) FBD of the section beyond x

(c) Shear force diagram (SFD)

(d) Bending moment diagram (BMD)

60

right-hand portion45

of the beam as Fig. 4.9b. The shear force and the bending

moment on this section are the external loads acting on this part of the beam and

will be shown on the FBD. We do not know their directions. In such a case we

show them assuming they are positive. Since the exposed face on which they are

acting has an outward normal in the negative x-direction, the positive shear force

must be downward and the positive bending moment must be clockwise (as

shown).

A simple equilibrium analysis46

of this FBD gives:

∑ : or (a)

∑ : ( ) , or ( ) (b)

Since the location of the section x is arbitrary, and changing the value of x does

not change either the FBD or Eqs. (a) and (b), these equations give the shear

force and the bending moment at all values of x between 0 and L. The resulting

values have been plotted as shear force and bending moment diagrams in Figs.

4.9c and 4.9d, respectively. These show the variations of the respective

quantities with x.

Example 4.2 A simply-supported beam with a concentrated load in the middle

Consider a simply-supported beam with a concentrated load at the mid-point as

shown in Fig. 4.10a. Draw SFD and BMD for this beam.

Solution:

The first step in the solution of this problem is determining the reactions at the

support. Drawing an FBD of the entire beam (or using symmetry) we can easily

find the two vertical reactions to be P/2 each. To determine the shear force (SF)

and the bending moment (BM) at any location x (measured from the left support)

of the beam, we need to externalize the SF and BM by taking a section of the

45

We could have chosen either pat of the beam but have chosen to draw the FBD

for the right-hand portion of the beam because this results in an FBD with fewer

unknown forces simplifying the equations. 46

M,x in Eq. (b) denotes that moments have been taken about a point at x. This is

convenient because the moment contribution of the unknown force V about this

point is zero.

It may further be noted that in the equilibrium equation, the sign to be used with

the bending moment is the standard sign convention for moments: positive for

clockwise and negative for counter-clockwise. Even though the bending moment

shown is positive (negative on a face with negative outward normal), it is taken

as positive in the equilibrium equation.

beam there and drawing the FBD. It should be obvious that there are two kinds

of FBDs depending on whether the value of x is less than L/2 or greater than L/2.

Fig. 4.10b shows the FBD for x < L/2. We shall show the SF and BM as external

loads on this FBD. We do not know their directions. As before, we show them

assuming they are positive. Since the exposed face on which they are acting has

an outward normal in the positive x-direction, the positive shear force must be

upward and the positive bending moment must be counter-clockwise (as shown).

A simple equilibrium analysis of this FBD gives, for x < L/2:

∑ : or (a)

∑ : , or (b)

Thus, for x < L/2, SF has a constant value of – P/2 while the value of the BM

increases linearly from 0 at x = 0 to a value of PL/4 at x = L/2.

Fig. 4.10 SFD and BMD for a simply-supported beam with a concentrated load

in the middle

(a) Loading diagram

(b) FBD when the section is taken in left half of the beam

(c) FBD when the section is taken in right half of the beam

(d) Shear force diagram

(e) Bending moment diagram

61

This formulation is valid only for x < L/2. For x > L/2, the generic FBD changes

and is as shown in Fig. 4.10c. Here the applied load P too figures in the FBD.

An equilibrium analysis of this FBD gives, for x > L/2:

∑ : or (c)

∑ : ( ) , or

( ) (d)

Thus, for x > L/2, SF has a constant value of + P/2 while the value of the BM

decreases linearly from PL/4 at x = L/2 to a value of zero at x = 0. The resulting

SFD and BMD47

are plotted in Figs. 4.10d and 4.10e, respectively.

4.5 General procedure for drawing shear force and bending moment diagrams by method of sections

On the basis of the method adopted in the last two examples, the following

generalized procedure may be formulated:

Draw an idealized loading diagram of the beam.

Determine the reactions at all supports. If the reactions cannot be

determined, the beam is statically indeterminate and further progress

cannot be made without considering the deflections of the beam. These

will be introduced in Chapter 7.

Determine the number of segments with distinct loading pattern to

cover the entire beam. In Example 4.2, we needed two different types

of FBD: one without the concentrated load P, and the other with it. In

practice, this means that we segment the beam such that the end of a

segment is at the location of a discontinuity in loading pattern. These

could be the concentrated loads or moments, or where the type of

distributed loads changes.

For each of the segments identified above, introduce a cutting plane (at a

location x from the left end) and draw an FBD of either part of the beam

(as convenient). In example 4.1 we had drawn the FBD of the right

portion (since it carried fewer loads), while in Example 4.2, we drew the

FBD of the left portions (since these permitted simpler expressions for

force moments).

47

Since the BM is positive, the beam bends concave upwards, as is physically

apparent from the loading

Introduce the unknown shear force V and the bending moment M at the

cutting plane. These should be shown assuming they are positive

according to the sign convention introduced in Table 4.1 above. Thus,

in Example 4.1 where we had drawn the FBD of the right-hand part of

the beam, the cut face has the outward normal in the negative x-direction

and, therefore, the positive SF is downwards and the positive BM is

clockwise. This is exactly opposite of the case of Example 4.2 where the

outward normal is in the positive x-direction and, therefore, the positive

SF is upwards and the positive BM is counter-clockwise.

Fig. 4.11 SFD and BMD for a simply-supported beam loaded

symmetrically with two concentrated loads

(a) Loading diagram

(b) FBD when the section is taken in the left third of the beam

(c) FBD when the section is taken in the middle third of the beam

(d) FBD when the section is taken in the right third of the beam

(e) Shear force diagram

(f) Bending moment diagram

62

Determine the expressions for SF and BM by equilibrium

considerations, equating the sum of vertical forces and the moment

(about a convenient point) to zero.

Plot the resulting expression to obtain the shear-force and bending

moment diagrams (SFD and BMD). These diagrams are conventionally

drawn exactly beneath the loading diagram of the beam as in Figs. 4.9

and 4.10.

We give below a few more examples to illustrate the general procedure.

Example 4.3 A simply-supported beam with two concentrated loads

Consider a simply-supported beam of length L with two loads acting at L/3 and

2L/3 from the end as shown in Fig. 4.11. Draw the SFD and BMD of this beam.

Solution:

As usual, we begin with determining the reactions at the supports. Through a

simple equilibrium analysis (or by using the symmetry of the problem) the

reaction at either support is a force P upwards..

It is easy to see that we need to segment the beam in this case in three different

ways. One kind of FBD results when the section is taken for with no

loads acting on it (as in Fig. 4.11b). Another kind results when the section is

taken for a value of x between L/3 and 2L/3. In such an FBD one external load P

acts at x = L/3 (as in Fig. 4.11b). And the third kind results when the section is

taken for a value of x between 2L/3 and L. In such an FBD two external loads

act, one at x = L/3 and the other at x = 2L/3 (as in Fig. 4.11c).

We next show the unknown shear force V and the bending moment M at the

newly exposed sections assuming they are positive and using the sign convention

of Table 4.1. In each of the three FBDs, the positive SF is upwards and the

positive BM is counter-clockwise, since all outward normals are along the

positive x-axis.

From the equilibrium analysis of the FBD of Fig. 4.11b, we get for x < L/3:

∑ : or (a)

∑ : , or (b)

Similarly, from the equilibrium analysis of the FBD of Fig. 4.11b, we get for L/3

>x > 2L/3:

∑ : or (c)

∑ : ( ) ,

or , a constant (d)

And from the equilibrium analysis of the FBD of Fig. 4.11c, we get for x > 2L/3:

∑ : or (e)

∑ : ( ) ( ) ,

or ( ) (f)

Fig. 4.12 SFD and BMD for a simply-supported beam with

uniformly distributed load

(a) Loading diagram

(b) Typical FBD for a segment of the beam

(c) Distributed load replaced by a statically equivalent load

(d) Shear force diagram

(e) Bending moment diagram

63

The resulting SFD and BMD48

have been plotted as Figs 4.11e and f.

Example 4.4 A simply-supported beam with uniformly distributed loading

Consider a simply-supported beam with a uniformly distributed load of w N/m as

shown in Fig. 4.12. Draw the SFD and BMD for this beam.

48

Here too, the BM is positive and the beam bends concave upwards, as is

physically apparent from the loading.

Solution:

The total load on this beam is wL. The symmetry of the beam suggests49

that the

reaction at each support is wL/2. Since there is no discontinuity in the loading

with no concentrated load or moment, only one kind of FBD as drawn in Fig.

4.12b suffices for this problem. The presence of distributed load poses a problem

in determining the moment contribution of the loading. To this, we replace the

distributed load with its statically equivalent load50

as shown in Fig. 4.12c. It

consists of the total load wx acting at the centroid of the distributed load, which is

at the mid-point of the beam, i.e., at a distance of x/2 from the left end.

From the equilibrium analysis of the FBD of Fig. 4.12c, we get for all values of

x:

∑ :

or (a)

Thus, the shear force varies linearly, with a value of at x = 0, and a value

of at x = L. It is zero at x = L/2. The variations of shear force are

plotted as SFD in Fig. 4.12d.

∑ : ( ) ( )( ) ,

or (b)

This represents a parabola with M = 0 at x = 0 and L, and a maximum value of

at the mid-point x = L/2. These variations of bending moment along the

length of the beam are plotted as BMD in Fig. 4.12e.

Example 4.5 A simply-supported beam loading with a concentrated moment at the middle

Consider a simply-supported beam of length L loaded in the middle with a

concentrated moment M0 as shown in Fig. 4.13a.Draw the shear force and

bending moment diagrams for this beam.

49

This can also be obtained by drawing the FBD of the whole beam and

replacing the distributed load by the statically equivalent load system consisting

of the total load wL acting at the mid-point of the beam. A statically equivalent

load is the alternate load which produces the same net force and moment at any

point. 50

It can easily be verified that the loading shown in Figs. 4.12b and 4.12c are

statically equivalent as defined above.

Fig. 4.13 SFD and BMD for a simply-supported beam with a concentrated

moment

(a) Loading diagram

(b) Typical FBD for a section in first half of the beam

(c) Typical FBD for a section in second half of the beam

(d) Shear force diagram

(e) Bending moment diagram

64

Solution:

We begin with determining the reactions at the supports. Through a simple

equilibrium analysis the reaction at the left support is determined as a force equal

to M0/L, while t the right support it is − M0/L. The negative sign indicated that

the reaction at the right support is directed downwards, as shown in the figure51

.

We need to segment the beam in two different ways: one, when the section is

taken at (without including the concentrated moment M0 in it) acting on

it (as in Fig. 4.13b), and the other when which included the concentrated

moment M0 in the resulting FBD shown as Fig. 4.13c.

We next show the unknown shear force V and the bending moment M at the

newly exposed sections assuming they are positive and using the sign

convention. In either FBD, the positive SF is upwards and the positive BM is

counter-clockwise since the outward normal in either case is along the positive x-

axis.

From the equilibrium analysis of the FBD of Fig. 4.13c, we get for x < L/2:

∑ : or (a)

∑ : ( ) , or ( ) (b)

Thus, the shear force V is constant in this part, while the bending moment M

increases linearly from 0 to at the mid-point of the beam.

Similarly, from the equilibrium analysis of the FBD of Fig. 4.13c, we get for L/2

>x > L:

∑ : or (c)

∑ : ( ) ,

or ( ) (d)

The shear force V is constant in this part too, with the same value as in the first

part. The bending moment M shows an interesting behaviour: it jumps down

from to at x = L/2 nd then incre ses, g in line rly, from −

at the mid-point to 0 at the end-point of the beam52

. The resulting SFD and

BMD are plotted as Figs. 4.13d and e.

51

It has been assumed here, for simplification, that at least one support is not

restraining the beam horizontally, so that there is no horizontal reaction force.

Anyway, the presence of a horizontal force will not affect the SFD and BMD. 52

Attention is drawn to and interesting feature of SFDs and BMDs as is apparent

to the five examples given so far: the line depicting shear force has a

discontinuity wherever there is a concentrated force as a load, and the BMD is

The deflected (but exaggerated) shape of the beam is shown as the broken line in

Fig. 4.13a. As noted earlier, the beam deflects concave upwards where the BM is

positive and convex upwards where it is negative.

4.6 The area method of drawing the SFDs and BMDs

The method of sections outlined in the last section is a simple method of

determining shear forces and bending moments, but it can get quite tedious in all

but very simple cases. We give here an alternate method known as the area

method.

now seen to be discontinuous at the location of the concentrated moment. We

will elaborate on this later.

Fig. 4.14 Elemental FBDs for three different kinds of loads

65

Consider the beam shown in Fig. 4.14, which carries a concentrated load P at

location B and a concentrated moment M0 at location C as shown. A part of the

beam (segment DE) is subject to a distributed load. The density q(x) of this load

measured in N/m may vary along the length.

Let us determine how the SF and BM change with x along the beam. Let us first

take a small element of length dx at a location x within the segment AB where no

load acts on the beam. We draw an FBD of this element of the beam as shown in

Fig. 4.14b. Here we have shown shear force V and bending moment M at location

x, and shear force V + dV and bending moment M + dM at location x + dx. It can

be verified that all directions have been shown such that SF and BM at either

location are positive. The application of equilibrium to this FBD gives:

∑ : , or (a)

∑ : , or (b)

This implies that the SF as well as BM does not change along a beam segments

which does not carry any load.

We next consider an elemental section of the beam which includes point B, the

location of the concentrated load. Fig. 4.14d shows the FBD of this element.

The application of equilibrium to this FBD gives:

∑ : , or (c)

∑ : ( ) , or for

infinitesimal dx, (d)

These indicate that at the location of a concentrated load, the shear force jumps

up by a value equal to the load (acting downwards), but there is no change in the

value of the bending moment. It can be deduced quite easily that had the load P

been acting in the upwards direction, the SF would have jumped down by the

same value.

Fig. 4.14d shows the FBD of an element of the beam which includes point C, the

location of the concentrated moment M0. The application of equilibrium to this

FBD gives:

∑ : , or (e)

∑ : , or for infinitesimal dx,

(f)

Thus, at the location of a concentrated moment, the bending moment jumps down

by a value equal to the applied moment, but there is no change in the value of the

shear force across such an element.

We last consider an elemental section at a point within the segment DE which

carries a distributed load. The FBD of such a section is shown as Fig. 4.14e.

For x within the beam segment DE, the equations of equilibrium53

give:

∑ : or , or

(g)

∑ : ( ) , or

(h)

This suggests that for a distributed load, the rate of change of shear force at a

location is equal to the (downward) load density at that point. And, similarly, the

rate of change of bending moment is equal to the (negative of) shear force at that

point. We can recast the equations (g) and (h) as:

∫

(4.1)

∫

(4.2)

The area method gets its nomenclature from these two equations which say state

that the change in SF across a beam segment is equal to the area under the

loading curve of that segment, and the change in BM across a beam segment is

equal to the area under the shear force curve of that segment54

.

The eight equations (a) – (h) above (along with Eqs. 4.1 and 4.2) provide a

framework for an very convenient method for drawing the FBDs and BMDs of

the beams directly without drawing FBDs of different segments..

Let us first consider Eqs. (a), (c), (e), (g) and 4.1, all pertaining to changes in

shear force V. Eq. (a) states that V does not change across any element which

does not carry any load. Eq. (c) suggests that at the location of a concentrated

load, the shear force jumps up by a value equal to the load (acting downwards).

Eq. (e) states that V does not change across any element carrying a concentrated

53

Two things should be noted here. First, the contribution of the distributed load

to the moment equation has been obtained by replacing the distributed load by its

statically equivalent concentrated load qdx at the midpoint of the element, i.e., at

location dx/2. Second, this contribution is negligible since this moment is of

second infinitesimal order, while the other terms are of first order only. 54

The difference in signs in Eqs. 4.1 and 4.2 is due to the fact that the load

density q has been defined as positive downwards. If we had taken the usual sign

convention that upward forces are positive, we would have got a negative sign

before integral in both the equations.

66

moment, and Eq. (g) says that the rate of change in shear force is equal to the

density of loading.

In fact, Eq. 4.1 summarizes all this information quite succinctly if we assume a

concentrated load as the integral of the load density at that point. The change in

shear force across a segment of the beam is equal to area under the loading

density curve.

From these we can deduce the following procedure to draw the SFD:

1.1. First draw the loading diagram of the beam.

1.2. Calculate the reaction at the supports.

1.3. Start drawing the SFD at a point slightly left of the left support where

the shear force is taken as zero.

1.4. Travel along the beam to the right, modifying the SF using the rules

given below.

1.5. Shear force does not change over the segment of the beam with no

external load.

1.6. The presence of a concentrated moment does not change the value of

shear force.

1.7. A concentrated force causes a jump in SFD at the location of the force:

go up for every load downwards, down for every load upwards.

1.8. The slope of the SFD at any location is equal to the distributed load

density (load per unit length). Positive slopes for loads downward, and

negative slopes for loads acting upwards.

1.9. Change in SF between any two locations is equal to the area under the

distributed load curve. Positive changes for negative areas and negative

changes for positive areas.

1.10. Proceed till you are slightly right of the right support where the SF

should be zero again. This last is a check on calculations.

This is known as the area method of drawing the shear force diagram. It can be

verified quite easily that the SFDs Figs. 4.9c, 4.10d, 4.11e, 4.12d, and 4.13d

follow from the respective loading diagrams using the above rules.

We, in a similar fashion, can deduce the rules for drawing the BMDs from Eqs.

(b), (d), (f), (h) and 4.2, all pertaining to changes in bending moment M. Eq. (b)

states that M does not change across any element which does not carry any load

or moment. Eq. (d) implies that at the location of a concentrated load too, there is

no change in the value of the bending moment. Eq. (f) states that at the location

of a concentrated moment the bending moment jumps down by a value equal to

the (positive) applied moment. Eq. (h) states that the rate of change in bending

moment is equal to the (negative of) shear force at that point.

Eq. 4.2 contains all of the above points. The change in bending moment while

traversing a segment of the beam is equal to the area under the shear force curve

(but with a reversed sign) over that segment.

The following procedure may be adopted for drawing the BMDs:

2.1. First complete the shear force diagram of the beam.

2.2. Start at a point slightly left of the left support where the bending

moment is taken as zero.

2.3. Travel along the beam to the right, modifying the BM using the rules

given below.

2.4. The slope of the BMD at any location is equal to (negative of) the SF

value at that location

2.5. Change in BM between any two locations is equal to the (negative of)

the area under the SF curve.

2.6. Concentrated moments cause a jump in BMD at the location of the

moment.

2.7. Go up for every negative concentrated moment, down for every positive

concentrated moment.

2.8. Proceed till you are slightly right of the right support where the BM

should again be zero again. This last is a check on calculations.

This is known as the area method of drawing the BMDs. It can be verified quite

easily that the BMDs Figs. 4.9d, 4.10e, 4.11f, 4.12e, and 4.13e follow from the

respective shear force diagrams using the above rules.

The following example illustrates the simplicity of the method of area.

Example 4.6 Beam with an overhang

Consider a 4 m long simply-supported beam with an overhang as shown in Fig.

4.15. The overhang portion AB carries a distributed load of density 20 kN.m.

There are two concentrated loads at locations C and E as shown. It also carries a

concentrated counter-clockwise moment of magnitude 20 kN.m at location D.

Draw the SFD and BMD for this beam.

Solution:

We first determine the reactions R1 and R2 at the supports. With reference to the

loading diagram of Fig. 4.15, we can write the equilibrium equations. Equating

the vertical forces to zero, we get:

67

.

/ ( ) ,

or (a)

We next take the moment of all the forces about the left support and equate it to

zero to get55

:

( ) ( ) ( ) ( ) ( ) ( ) ( ) , (b)

From Eq. (b) we get

Using this value in Eq. (a) we get

We are now ready to draw the shear force diagram.

55

Here again we take the moment of the distributed load by replacing it with the

statically equivalent load, which is the total force (20 kN)·(1 m) acting at the

centroid (of this load) which is located at 0.5 m from the left end of the beam.

We start at zero at the left end. We first encounter the uniformly distributed load.

Using rules 1.8 and 1.9 above, the shear force would increase at a constant the

rate of 20 kN/m, and after one meter it would have increased from 0 at A to 20

kN at B, as shown. A concentrated load (acting upwards) occurs at the location

B. Therefore, following rule 1.7, the value of shear force jumps down by 70 kN

(the value of R1) from 20 kN to − 50 kN. Since there is no lo d on segment BC,

the value of SF through this segment remains constant (rule 1.5). At location C,

the she r force jumps up from − 50 kN to − 10 kN (rule 1.7), and thereafter

remains constant at this value across segment CE, since the this segment does not

have any load except a concentrated moment at D, and rule 1.6 says that SF does

not change due to the presence of a concentrated moment. The downward load at

E makes the SF jump up through 40 kN from – 10 kN to + 30 kN (rule 1.7

again). The shear force remains constant at this value through segment EF, and

then jumps down to 0 at F because of the presence of the concentrated load R2 =

30 kN. The final value of 0 completes the check (rule 1.10).

It should be obvious that this method of area is so much simpler than the method

of sections where we would have needed to draw five different kinds of FBDs

and, then, write ten equilibrium conditions for those five FBDs.

After completing the SFD, we can draw the BMD too. We start at zero for a

point slightly left of point A. The SF increases linearly from 0 at A to 20 kN at B.

By rule 2.4, then, the slope of the B D ch nges line rly from zero to − 20

kN·m/m. The linear change in slope implies that the curve is of second order in

x, that is, it is parabolic, as shown. The change in the value of the bending

moment in going from A to B is the area under the SFD in this segment (rule 2.5),

which is ( ) ( ) ( ) . This area is positive and,

therefore, the change in BM is – 10 kN.m, giving a value of BM at point B as –

10 kN.m. The shear force across segment BC is constant at – 50 kN, so the slope

of the BM curve there is constant at + 50 kN.m/m, and the value of the BM

through this section of length 1 m changes (linearly) through 50 kN.m, the area

under the SF curve in this segment (rule 2.5). The value of BM at point C is 40

kN.m.

From point C to D, the SF is – 10 kN, and therefore the BM curve slopes up at

this rate and the value of BM changes from 40 kN.m to 45 kN.m in 0.5 m. The

presence of a positive external concentrated moment of 20 kN.m at point D

makes the BM jump down to 25 kN.m at this point (rule 2.6), after which the

curve continues its upward march at a rate of + 10 kN.m/m to acquire a value of

30 kN.m at point E. The SF in the last segment EF is constant at +30 kN.

Therefore, the BM there decreases linearly from a value of 30 kN.m to 0 in a

length of 1 m. The end value of 0 indicates that the calculations are in order.

Fig. 4.15 Drawing SFD and BMD using area method

68

Example 4.7 A built-in frame

Consider the frame ABCD shown in Fig. 4.16a which is built in at A. Draw the

SFDs and BMDs of the various segments of the frame.

Solution:

To begin the analysis, we need to consider the FBDs of the various sections of

the frame.

Fig. 4.16b shows the upper segment DC of the frame. There will be a force and

moment acting at the point C due to the restraining action of the segment CB.

Using the requirements of equilibrium we evaluate the vertical force as P

upwards and the moment as PL/2 clockwise.

Using the area method outlined above, we can easily obtain the SFD and BMD of

this segment as shown in Fig. 4.16b.

The FBD of the segment CB is shown in Fig. 4.16c. The load at point C of this

segment is equal and opposite to that at point C of the segment DC. Using

equilibrium of this segment we obtain the force at point B as P upwards, and the

moment there to be PL/2 clockwise. The force in this segment is axial and,

therefore, there is no shear force anywhere. The BMD has been obtained using

the rules 2.1, 2.6 and 2.8. The positive sign of the bending moment indicates that

this segment will bend concave upwards (viewing from right), which agrees with

what is expected intuitively.

We then move on to the cantilevered section AB, which carries a force P

downwards and a counter-clockwise moment PL/2 at point B as shown in Fig.

4.16c. We can obtain the reaction force and moment at end A by equilibrium

considerations. Due to the upward force P, the SFD jumps down to –P at the left

end A and remains so up till the other end where it jumps down to 0. The BMD

jumps down to − PL/2 at point A due to the presence of a concentrated moment

PL/2, and thereafter increases at a constant rate of P because of the presence of

the shear force –P throughout this length reaching a value of + PL/2 at the end B.

It jumps back to zero there because of the presence of a counter-clockwise

moment PL/2.

Example 4.8: An optimization problem

Consider a simply-supported beam of length L carrying a load 2P at the middle

as shown in Fig. 4.17a. It also carries a load P at each of the two ends. The two

supports are located a distance a inwards from the ends. Determine the value of

the distance a required to minimize the maximum bending moment in the beam.

Fig. 4.16 Drawing FBDs and BMDs of the various segments of a frame

69

Solution:

From the symmetry of the beam (or from the two equilibrium equations for the

whole beam) we determine the reaction force at either support to be 2P.

We first draw the SFD using the area method. We start the SFD slightly left of

the left end of the beam with a zero value. A downward force P is encountered at

the left end which causes the SFD to jump up to a value of P (Fig. 4.17b). Since

there is no other loading till the left support, the SF remains constant at P till

then. The reaction at the left support is 2P upwards which causes the SFD to

jump down by 2P to a value of –P at the left support. We continue in a similar

manner and complete the SFD as shown.

The BMD is also drawn in the same fashion (Fig. 4.17c). We start at zero. In the

first segment of length a, the SF is constant at a value of +P, and therefore, the

BM in this section decreases at the rate of P per unit length. The change in value

till the left support is the areas under the SFD of this segment, which is Pa.

Thus, the v lue of B t the left support is −Pa. In the second segment (of

length L/2 – a) the v lue of SF is −P and, therefore, the slope of bending moment

curve here is +P. The change in value of BM is P(L/2 − a) from −Pa. Thus, the

value of bending moment at the mid-point is [−Pa + P(L/2 − a)], or +P(L/2 −

2a). We can continue in the same fashion56

and complete the BMD.

The magnitude of the maximum negative bending moment is Pa, and the

magnitude of the maximum positive moment is P(L/2 − 2a). As the value of a

increases, the first value increases while the second decreases. A little reflection

should convince the reader that the minimum of the maximum value will occur

when these two magnitudes are equal:

( ), or , or .

Thus, we minimize the maximum value of the bending moment by locating the

supports at L/6 from either end of the beam.

Summary

Study of the behaviour of beams is very important because the beams are central

to the design of most structures. We, in this chapter have concentrated on

determining the variations in shear forces and bending moments along the length

of the beam.

We first introduced the types of supports and the loads that the beams carry.

Three idealized supports were identified:

The built-in support which allows no degree of freedom to the beam and

admits both components of the reaction force as well as a reaction

moment.

The pin support which allows one degree of freedom, the freedom to

rotate about that point. This support admits both components of the

reaction force but does not admit a reaction moment

A roller support which restraints the beam only from the vertical motion

and, therefore, admits only the vertical reaction force component.

We next introduced the sign convention for the shear force and bending moment.

The sign is based on two parameters: the direction of the outward normal at the

section, and the direction of the force or the moment itself. If both of these are in

the positive coordinate directions, or both in the negative coordinate directions,

the sign of SF or BM is taken as positive. And if one of them is positive and the

other negative, the sign is taken as negative. This is summarized in Table 4.1 and

as icons of Fig. 4.5.

The method of sections for drawing the SFDs and BMDs uses the following

procedure:

56

Or use the symmetry of the beam.

Fig. 4.17

70

Draw an idealized loading diagram of the beam.

Determine the reactions at all supports.

Determine the number of segments with distinct loading patterns to

cover the entire beam.

For each of the segments identified above, introduce a cutting plane (at a

location x from the left end) and draw an FBD of either part of the beam

(as convenient).

Introduce the unknown shear force V and the bending moment M at the

cutting plane.

Determine the expressions for SF and BM by equilibrium

considerations, equating the sum of vertical forces and the moment

(about a convenient point) to zero.

The other method of determining SFDs and BMDs is the method of areas is

based on the two equations:

, and

.

From the integration of these we concluded that the change in SF across a beam

segment is equal to the area under the loading curve of that segment, and the

change in BM across a beam segment is equal to the area under the shear force

curve of that segment.

The following process for drawing SFD was concluded:

First draw the loading diagram of the beam.

Calculate the reaction at the supports.

Start drawing the SFD at a point slightly left of the left support where

the shear force is taken as zero.

Shear force does not change over the segment of the beam with no

external load.

A concentrated force causes a jump in SFD at the location of the force:

go up for every load downwards, down for every load upwards.

The slope of the SFD at any location is equal to the distributed load

density (per unit length). Positive slope for loads downwards and

negative slope for loads acting upwards.

Change in SF between any two locations is equal to the area under the

distributed load curve. Positive changes for negative areas and negative

changes for positive areas.

The following process for drawing BMD was concluded:

First complete the shear force diagram of the beam.

Start at a point slightly left of the left support where the bending

moment is taken as zero.

The slope of the BMD at any location is equal to (negative of) the SF

value at that location

Change in BM between any two locations is equal to the (negative of)

the area under the SF curve.

Concentrated moments cause a jump in BMD at the location of the

moment. Go up for every negative concentrated moment, down for

every positive concentrated moment.

71

5 Stresses in beams

5.1 Introduction

In problems involving bending of beams we may be interested in

determining:

The stresses that result when a beam carries the specified loads.

We may consequently determine the maximum load that a shaft

can carry without failure.

The deflection of a beam when carrying the specified loads.

We have seen in the last chapter that almost all loads in beam result in

variation of the resulting bending

moment and shear stress along its

length. This inevitably introduces

complications in the determination of

the behaviour of the beam. To obtain a

foothold on the problem we first

consider a very simple case of a beam subject to pure bending57

as shown

in Fig. 5.1.

To obtain the relations between the bending moment Mb and the stresses

and strains we shall follow the strategy that we outlined in Chapter 3

while dealing with the case of torsion of a shaft. This was the reverse of

the generalized strategy outlined in Chapter 2. The reason for this again

is the same as in the case of torsion: unlike the situations involving axial

loading treated in Chapter 2, it is not possible to make the assumption of

uniform stresses in the case of bending of beams.

Before going

further, let us

determine the

nature of

deformations in

beams under

pure bending and

the nature of

stresses that

result from such

deformations.

Fig. 5.2 shows a

beam with a grid

drawn on it. This

beam is

subjected to a

positive bending

moment, and

bends as shown.

It is clear from

the enlarged

portion shown in

Fig. 5.2c that the

beam elements

57

The bending moment can be constant along the length of a beam only if the

shear force V is zero everywhere. This is possible only if there is no other load on

the beam except concentrated moments at the two ends.

Fig. 5.1 Beam in pure bending

Fig. 5.2 Bending stresses in beams

(a) Undeformed beam

(b) Beam after bending

(c) Enlarged portion showing that there is contraction

near the top and extension near the bottom

72

near the top surface of the beam are under compression, while those near

the bottom are in elongation. As a consequence, there are compressive

stresses near the top and tensile stresses near the bottom. It should be

clear that since there in no net tension (or compression) in the beam, the

forces due these stresses must total out to zero. The bending moment at

the section is the integrated moment of the elemental forces due to these

stresses.

The strategy for this problem then consists of:

First using symmetry considerations to establish the nature of

deformation in the beam under pure bending.

Using the geometric considerations to determining the tensile

strain across the beam section as a function of the vertical

coordinate. This is the macro to micro conversion discussed in

Section 2.1.

We next use the material properties to convert these strains to

stresses. This is a micro to micro transformation.

In the last stage we convert the stresses into bending moment, a

micro to macro conversion. We use the fact that the total

bending moment is the given Mb, and the total tensile force at the

section is zero. This completes the solution to the problem.

5.2 Relating curvature of the beam to the bending moment

Consider a beam with a cross-

section that is laterally symmetric

(i.e., it has symmetry about a

vertical axis) subjected to a pure

bending moment Mb as shown in

Fig. 5.3. Under the action of this

positive bending moment, the

beam bends and acquires a

curvature. The coordinate axes

used here are shown in the figure.

By using symmetry arguments we

can show that the following

assumptions are valid in this case of pure bending of a beam of uniform

and symmetrical (about a vertical axis) section:

The beam bends in the shape of a circular arc. This follows from

the argument that a uniform bending moment acting along the

length of the beam must result in a uniform curvature.

Plane cross-sections of the beam remain plane after the bending.

Thus, plane sections 1-2 and 3-4 of the undeformed beam of Fig.

5.4 remain plane in the deformed shape as well.

Initially parallel sections (sections 1-2 and 3-4 of the undeformed

beam of Fig. 5.4) must deform so that they have a common point

of intersection as shown. This common point of intersection is

the centre of curvature of the deformed beam.

Calculation of strain from the geometry of deformation

We now consider the geometry of deformation further to obtain an

expression for strain.

It was argued above that the beam elements near the top are under

compression while those near the bottom are under tension. It stands to

logic, then, that there must be a plane in the beam (near the middle) which

is neither under tension nor

compression. Let AB

represent such a plane for

the beam segment 1-2-3-4 of

Fig. 5.4. This plane with its

extension to the rest of the

beam is termed as the

neutral plane of the beam.

It should be understood that

we do not, as yet, know

where this is located, except

for the fact it must be

somewhere between the top

and the bottom surfaces of

the beam.

Let us consider a plane CD a

distance y up from the

neutral plane AB. The

undeformed lengths of

Fig. 5.3 A symmetrical beam

Fig. 5.4 Geometry of deformation of a beam

73

planes (along the x-axis) AB and CD are equal to begin with. Since the

length of the plane AB is unchanged (this being the neutral plane), the

undeformed length of the plane AB or CD is seen as from Fig. 5.4b,

where ρ is the radius of curvature58

of the neutral plane of the beam. The

deformed length of the plane CD is ( ) . Thus, the contraction in

the length of this plane located a distance y up from the neutral plane is

ydθ. The strain at this location is, thus, given by

(5.1)

It can be seen that in this simple case of pure bending (i.e., with no

loading other than the bending moment which is constant along the length

of the beam) there are no shear strain components. However, there will

be longitudinal strains εyy and εzz due to the presence of ζxx through

Poisson ratio.

Converting strain to stress

Conversion of strain to stress is simple. The tensile stress component can

be obtained by multiplying the strain with the elastic modulus E:

( ) ( )

(5.2)

Note that we know neither

the value of y (because the

location of the neutral plane

has not yet been established)

nor the value of ρ, the radius

of curvature59

. But we have

two conditions that we shall

shortly use.

The distribution of the

stresses across the section is

58

The radius of curvature ρ is ∞ for a flat beam. As the beam bends, the value of

ρ decreases. We introduce the term curvature for 1/ρ. Curvature is usually

denoted by κ (read kappa). The value of κ increases as the beam bends. It is 0 for

a flat beam (corresponding to ρ = ∞). 59

But we have two conditions that we have not used so far. The total axial force

on the section is zero, and the total moment acting on the section is Mb.

like that shown in Fig. 5.5. The trace zz of the neutral plane is termed as

the neutral axis. Note here we have shown the neutral axis to be nearer

the bottom plane of the beam. This results in the top plane of the beam to

be farther away from the neutral axis (NA) than the bottom plane is.

Consequently, the maximum compressive stress is larger than the

maximum tensile stress in the beam.

Since there are no shear strain components, there will be no shear

stresses. Further, there are no other longitudinal stresses, ζyy or ζzz, in this

case of pure bending60

.

Converting stress into

loading

Consider an elemental area

dA in the right face of the

beam as shown in Fig. 5.6.

The force acting on this

elemental area due to the

bending stress is ζdA acting

along the normal to the

section as shown. Note that

we have shown this force outwards from the area, assuming ζ to be

tensile (i.e., positive). The algebra below will automatically take care of

the sign if the stress is compressive. The neutral plane of the beam is

shown as grey61

. The z-axis is the neutral axis of the section under

consideration.

The total axial force on the beam can be found by integration as

∫

. Substituting the value of ζ using Eq. 5.2, we get

∫

∫

∫

where y is the vertical distance measured from the neutral axis.

But there is no axial force Fx in this case of pure bending. Therefore,

60

We can now determine the lateral shear strains εyy and εzz as

.

61 Note again that we do not as yet know the location of the neutral plane.

Fig. 5.5 Stress distribution on a section of the

beam

Fig. 5.6 Calculating loading at a section

74

∫

(5.3)

What does this imply? It simply defines the location of the neutral axis

from which the distance y is measured. Thus, the neutral axis of a beam is

located such that the first moment of area about that axis is zero.

Recall that the centroid of an area is defined as the point about which the

first moment of area is zero. Therefore, the neutral axis of a beam passes

through the centroid of the cross-section62

.

We next calculate the moment (about the z-axis) of these elastic forces

and equate it to Mb, the bending moment on the beam:

∫ ∫ (

)

∫

The integral ∫

is recognized as the second moment of area Izz

about the z-axis, which, in this case, is the centroidal axis. This is a

geometric quantity. Once the shape of a cross-section is known, the value

of Izz can be calculated. See Appendix A to learn some important

properties of the second moment of area and the procedure for its

calculation.

We can, therefore, write an expression for 1/ρ, the reciprocal of the radius

of curvature (which is termed as the curvature κ = 1/ρ) of the beam as

(5.4)

The product is termed as the bending rigidity of the beam. The more

the value of the bending rigidity, the less is the curvature κ of the beam

(for a given value of the bending moment), and more is the radius of

curvature ρ.

We can combine Eqs. 5.2 and 5.4 to obtain

(5.5)

These stresses are termed as the bending stresses.

62

We had chosen the location of the neutral plane in Fig. 5.6 closer to the bottom

of the beam anticipating this result.

The bending formulae (Eqs. 5.4 and 5.5) are sometimes summarized63

as:

(5.6)

It may be noted that the equation above have been obtained for the case of

pure bending, i.e., for the case of constant bending moment in the absence

of any other load. The results for the more general case are very difficult

to obtain. But wherever these have been obtained, it is seen that for

slender members, the resulting stress distribution is quite close to those

obtained above. It is for this reason that we use the results obtained here

in the general case as well.

Since the values of and are zero, the lateral strains and

are given by

(5.7)

Thus, the normal strains in the plane of the cross-section are proportional

to the axial strains, but of opposite sign. Further, since εxx varies with y,

the vertical distance measured from the neutral axis, the strains εyy and εzz

change across the cross-section. This leads to the deformation64

of the

section as shown in Fig. 5.7.

Notice from Eq. 5.6 that the curvature of the beam 1/ρ is given by

63

Notice the similarity of this with a similar equation that we can obtain for the

case of torsion of a shaft from Eqs. 3.2 and 3.3:

.

Note that in this equation for shafts, the z-axis is the longitudinal axis, unlike in

the present case where the z-axis is a transverse axis while the longitudinal axis is

represented as the x-axis.

64

Note, in particular, that the neutral axis (which is the trace of the neutral

surface in the section) is now curved. This transverse curvature of the cross-

section of the beam is termed as anticlastic curvature. The neutral plane now has

double curvature, one in the x-y plane, and the other in the y-z plane.

75

The quantity EIzz can be interpreted as the stiffness of the beam. The

more its value, the less is the curvature of the beam, i.e., less is the

bending.

We give below a few examples of calculation of bending stresses.

Example 5.1 Maximum stresses in a cantilever beam

Consider a 2 m long cantilever MS beam of section 10 mm × 20 mm

loaded as shown in Fig. 5.8. Determine the location and magnitude of the

maximum bending

stress.

Solution:

We need to obtain the

bending moment

distribution first. For

this, we follow the

procedure introduced in

the last chapter, and

first calculate the

reactions at support.

From the balance of

vertical forces we can

see straight away that there will be an upward reaction of 200 kN at the

support. From the moment balance we obtain the moment at support to

be 300 N.m, counter-clockwise as shown.

We draw the SFD by the method of areas. We start at 0, jump down 200

N at the left support (load being upwards), continue at ‒200 N till the first

load is encountered. The downwards load makes the SF jump up to ‒100

N at 1 m. The SF jumps up again at the second load.

The BMD is drawn in the same fashion. We start at 0, jump down to ‒300

N.m at the support because of the presence of the positive reaction

moment. Through the first one meter of the beam, the BM line has a

positive slope of 200 N.m/m (equal to the negative of the constant shear

force). The value of BM at 1 m, the midpoint of the beam, is ‒100 N.m.

The change in value of BM between x = 0 m to x = 1 m is equal to the

area under the SFD over this length. From x = 1 m to x = 2 m, the slope

of the B D is +100 N.m/m, the SF being const nt t ‒100 N over this

length.

We, thus, see that the maximum bending moment occurs at the root of the

c ntilever where its v lue is ‒300 N.m. This, then, would be the loc tion

of the maximum stresses. Since the BM is negative, the beam would

bend convex up. The curvature 1/ρ given by Eq. 5.4 will be negative.

The value of Iyy needed in this equation has been obtained in Appendix B

(Eq. B.6) as bh3/12, where b is the width of the section (10 mm in this

case), and h is the height of the section (30 mm). Thus, the value of Iyy is

(10×10‒3

m) × (30×10‒3

m)3/12, which is 2.25×10

‒8 m

4.

The maximum tensile stress occurs at the top of the beam (y = 0.010 m)

at the root (x = 0), and is given by Eq. 5.5 as

( ) ( )

= 200 MPa

The value of the maximum compressive stress will be the same, except

that it will be at the bottom of the root of the beam. The beam is loaded

almost to its ultimate strength.

Note that given a bending moment, the stresses are independent of the

material of the beam. A beam with lesser elastic modulus will bend more

(giving larger values of curvature 1/ρ and strains) but will produce same

stresses.

Fig. 5.7 Deformation of the shape of a rectangular beam

(a) The undeformed rectangular section (b) variation of longitudinal stress

ζxx with distance y from the neutral axis (c) variation of longitudinal strain

εxx across y (d) variation of lateral strains εyy and εzz with y (e) the deformed

shape of the cross-section (deformation exaggerated). Note that the neutral

axis which is the trace of the neutral surface in the section) is curved.

Fig. 5.8 A cantilever beam

76

It is interesting to compare these results with the case when the beam

section has the longer side horizontal. Notice that the length of the

horizontal side now is a factor of 3 larger, the vertical side a factor of 3

smaller, the value on Izz as now two factors in 3 smaller. The maximum

bending stress that varies like y/Izz is 3 times larger, and the radius of

curvature that varies like Izz is two factors in 3 smaller, i.e., the curvature

(1/ρ) is 1/9th of when the beam section is vertical. These results are

summarized in Table 5.1.

It is clear that when more material of the beam is away from the

centroidal axes (on either side of it), more is the value of the second

moment of area, and the stiffer is the beam.

Example 5.2 Bending stresses in an angle-section beam

Consider a simply supported angle-section steel beam loaded as shown in

Fig. 5.9. Its section is the same as was used in Example B.5. Determine

the maximum tensile and compressive stresses in the beam.

Solution:

To draw the SFD and BMD, we calculate the reactions at the supports.

We determine the reaction at the right support by writing the balance of

moments about the left support. This reaction is found to be 50 N. Then,

by vertical force balance, the reaction at left support is found to be 250 N

as shown.

The SFD of the beam can be drawn easily by the method of areas. The

SFD jumps up through 100 N at the left-end because of the presence of a

downward force, then jumps down through 250 N to ‒150 N due to the

concentrated reaction at 1 m, jumps up through 200 N to +50 N because

of the downward load at 2 m length, and then jumps down to zero because

of the 50 N upward reaction at the right support.

The BMD of the beam starts at zero and has a slope of 100 N.m/m

through the first 1 m length. The value at 1 m is equal to the (negative of)

area under the SFD over this length, which is 100 N.m. We complete the

BMD in the same fashion following the rules given in Sec. 4.6.

There are two extrema in the BMD: a negative 100 N.m value at x = 1 m,

and a positive 50 N.m value at x = 1 m.

The centroid of the cross-section of the beam can be determined by taking

the first moment of the area about an axis passing through the bottom of

the section as was done in Example A.5 in Appendix A where the vertical

value of it was found to be 16.1 mm.

The second moment of the area about the centroidal axis was also

determined there. The strategy was to divide the area into two rectangles,

and then for each rectangle first determining the value of I about its own

centroidal axis, and then using the parallel-axes theorem to transfer it to

the neutral axis, i.e., the centroidal axis of the full section. The value was

determined as Izz = 2.99×10‒7

m4.

We can next determine the stresses. At location x = 1 m where the

m ximum neg tive bending moment of ‒100 N.m acts, the beam bends

convex up so that the maximum tensile stress acts at the top plane which

Fig. 5.9 A simply supported angle-section beam

Table 5.1 Comparison of two different orientations of beam sections

Section orientation

Factor

b 10 mm 30 mm 3

h 30 mm 10 mm 1/3

Izz 2.25×10‒8 m4 0.25×10‒8 m4 1/32

ζxx 200 MPa 600 MPa 3

Radius of curvature ρ (at the

point of maximum bending

moment)

15 m 1.67 m 1/32

77

has a value of y = +33.9 mm measured from the neutral axis. Similarly,

the maximum compressive stress act at the bottom plane which has a

value of y = +16.1 mm measured from the neutral axis.

We use Eq. 5.5 to determine the stresses. These are arranged in the Table

5.2 given below.

At location x = 1 m where the bending moment is ‒100 N.m, the

maximum tensile stress (at y =0.0339 m) is given by:

( )( )

The maximum compressive stress occurs at the bottom plane at y =

‒0.0161 m:

( )( )

We can similarly determine the stresses at the location x = 2 m where the

bending moment is +50 N.m. These are shown in the table above.

Example 5.3 Comparison of rectangular, T-section and I-section beams

Consider three cross-

sections of beams shown

in Fig. 5.10. The three

sections have equal

areas and, hence equal

weights per unit length

of the beams. Compare

the maximum tensile

and compressive

stresses in the three

beams for a bending

moment Mb.

Solution:

The calculations have been organized in Table 5.3. We first calculate the

location of the neutral axes by calculating the y-coordinates of the

centroids of the three sections. This is determined by dividing the areas

into the requisite number of rectangles and then using the formula:

∑ ∑ .

Next, we calculate the second moment of inertia. For this, again, we

divide the cross-section in a number of rectangles, calculating the value of

I for each rectangle about its own centroidal axis, and the transferring it to

the neutral axis using the parallel-axes theorem. Thus,

∑.

( )

/ .

Table 5.2 Stresses at the locations of maximum positive and negative bending

moments

x-location 1 m 2 m

Bending moment, Mb ‒100 N.m +50 N.m

Stress distribution

Maximum tensile stress At top plane (y = 0.0339

m)

11.3 MPa

At bottom plane (y =

‒0.0161 m)

2.7 MPa

Maximum compressive

stress

At bottom plane (y =

‒0.0161 m)

‒5.4 MPa

At top plane (y = 0.0339

m)

‒5.6 MPa

(a) (b) (c)

Fig. 5.10 Three beam sections of equal areas

78

The stresses are then evaluated from Eq. 5.5. These stresses vary linearly

from zero at the neutral axis. The stresses are compressive above the

neutral axis and tensile below it. In each case, the maximum compressive

stress occurs in the top plane, while the maximum tensile stress occurs at

the bottom most plane.

It should be noted that in the case of a T-beam, the maximum

compressive stress is much smaller than the maximum tensile stress since

the neutral axis is closer to the compressive side65

.

Another thing to note is the reduction in the maximum stress in the I-

section. This is a result of a larger value of Izz, which in turn results from

more of the section area (compared to the rectangular section) being away

from the central axis. Since an are ’s contribution in Izz is weighted by y2,

the areas far away from the centroidal axis contribute more, making the

beam stiffer.

65

But the total compressive force is exactly equal to the total tensile force on the

section, there being no net axial force. There is more area of the section on the

compressive side than on the tensile side. It is for this reason that the neutral axis

has shifted upwards causing a reduction of tensile stresses.

5.3 Composite beams

Consider the case of pure bending of a composite beam made up by

bonding two materials of different elastic properties, in particular, of

different elastic moduli. Let us consider a section with symmetry about y-

axis as shown in Fig. 5.11. Let this beam be subjected to pure bending

with a positive bending moment Mb. Let us determine the moment

curvature relations for this beam following the procedure used in Sec. 5.2.

Using the symmetry of the situation, we can again argue that plain

sections will remain plain and that the beam will bend in a circular arc.

The geometry of deformation will be the same as depicted in Fig. 5.4, and

we would get an expression for the linear strain similar to Eq. 5.1

(5.8)

where ρ is the curvature of the neutral plane, i.e., a plane that does not

have any strain, and y is the distance measured from the neutral plane.

Note that, as before, for a given curvature of the beam, the strain varies

linearly the distance from the neutral plane and does not depend on the

material of the beam.

However, when we continue to the next step of converting to stress, the

material properties come in and the stress is given by

( ) ( )

(5.9)

where Ei stands for the elastic modulus of the specific material at that

location. Note, in particular, that while the strain at the junction of two

materials is continuous, the stress shows a discontinuity (see Fig. 5.12).

Note that the strain, as well as the stress, is zero at the neutral plane. The

strain increases linearly on either side of the neutral axis, compressive

above it and tensile below it66

. The stress distribution is discontinuous.

The stress jumps at the location where the material changes,. The strain,

being a purely geometrical quantity in a beam, shows no such jump.

Consider two points close together on either side of the interface of the

66

The bending moment, being positive, introduces positive curvature (concave

upwards) resulting in the strains as shown.

Table 5.3 Comparative stresses in beams of equal weight

Section

Location of NA from the

base

30 mm 38 mm 30 mm

Izz 5.40×10‒7 m4 5.21×10‒7 m4 6.77×10‒7 m4

Stress distribution

y for max tensile stress ‒30 mm ‒38 mm ‒30 mm

y for max comp. stress +30 mm +22 mm +30 mm

max tensile stress 5.56×104Mb Pa 7.29×104Mb Pa 4.43×104Mb Pa

Max comp. stress 5.56×104Mb Pa 4.22×104Mb Pa 4.43×104Mb Pa

Fig. 5.11 Beam with composite section

79

two materials. Since the strains are the same, the changed value of E will

result in different values of

stress on the two sides.

Hence the jump67

.

The location of neutral

plane and the value of ρ are

as yet undetermined. To

evaluate these we, as

before, use the

considerations of

equilibrium. The total

force on the section is evaluated as by integration as ∫

.

Substituting the value of ζ using Eq. 5.9, we get

∫

∫

,

where Ei is the elastic modulus of the

material specific to the area element dA

(and y is the distance of the area element

measured from the neutral plane).

This axial force should vanish, and

therefore, the neutral plane should be

located such that ∫

.

To interpret this equation, consider Fig.

5.13. The distance y is measured from the

neutral axis. Let distance y’ be measured from an arbitrary axis from

which the neutral axis is located at a distance of . Then , and

the equation of equilibrium becomes

∫( )

,

67

Two things may be noted in Fig. 5.12. First, near the interface the value of

stress in the upper material is less than that in the lower material. This indicates

that the value of E for the upper material is less than that for the lower material.

Second, note that all the straight lines depicting the stresses in the various

materials originate from zero value at the neutral axis. Why?

or, ∫ ∫

,

or, ∫ ∫

.

We now consider the integral over the whole area as a sum of a number of

integrals, each over an area of uniform material.

Thus, ∫ ∑ ∫

∑ .

Similarly, ∫ ∑ ∫ ∑

, where

represents the location of the centroid of the ith area in coordinate

system. From these we get

∑

∑

(5.10)

This gives the location of the neutral axis within the section68

.

To obtain the moment curvature relation, we write the equilibrium

condition for the moments:

∫ ∫ (

)

∫

The last integral is again interpreted as the sum of integrals over areas of

different (homogeneous) materials to obtain ∑ . Then,

the curvature κ = 1/ρ can be seen as

∑ (5.11)

where Izz,i is the second moment of the area Ai about the neutral axis of

the section.

Once the curvature is known, we can calculate the strain as

68

It may be noted that in Eq. 5.10 the areas of different materials are weighted by

the elastic moduli of the materials. These happens because this relation is

obtained from the balance of force equation where forces are determined from

the stresses which are E times the strain. The strains, by themselves, are

geometric quantities.

A consequence of this is that the neutral plane shifts towards the area with the

larger modulus of elasticity.

Fig. 5.13

(a) (b) (c)

Fig. 5.12 (a) beam section (b) strain distribution

(c) stress distribution

80

∑ (5.12)

and

∑ (5.13)

where E is the elastic modulus of the

relevant material.

Example 5.4 Stresses in a sandwich beam

Consider a beam of sandwich construction

in which two steel plates are separated by

a block of lightweight rigid foam material

with a negligibly small elastic modulus

compared to that of steel (Fig. 5.14). Find

the stress distribution for a bending moment of 100 N.m.

Solution:

It is easy to see from the symmetry of the section that the neutral plane is

situated at the middle. We can also use Eq. 5.10 to obtain the location of

the centroid.

∑

∑

, where A1 refers to the

area of the lower steel plate with E1 = 200 GPa and = 10 mm, A2

refers to the area of foamy material with a negligibly small vale of E, and

A3 to the area of the upper plate with E3 = 200 GPa and .

Using these values, we get

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

0.40 m

The curvature is given by Eq. 5.11

∑

, E2 being negligibly small.

Here refer to the second moment of the three areas about the neutral

axis (passing through the middle of the section as shown above).

moment of the lower area about its own axis + area×

(distance between the neutral axis and the centroid of

area 1)2

= (0.08 m)×(0.02 m)3/12 + (0.08 m)×(0.02 m)×(0.04 m –

0.01 m)2

= 1.49×10‒6

m4.

Similarly, moment of the upper area about its own axis + area×

(distance between neutral axis and the centroid of area 3)2

= (0.08 m)×(0.02 m)3/12 + (0.08 m)×(0.02 m)×(0.04 m –

0.07 m)2

= 1.49×10‒6

m4.

Then,

( ) ( ) ( ) ( )

= 1.68×10‒4

m‒1

The value of the maximum compressive stress is then determined from

Eq. 5.9 with E = 200 GPa, and y = 0.04 m,

( )

( ) ( )

(1.68×10‒4

m‒1

)

= ‒1.34 ,

The minus sign indicating that it is a compressive stress. We can

similarly determine the maximum

tensile stress (at y = ‒ 0.04 m) s

+1.34 MPa.

The rigid foam serves the purpose of

shifting the two steel areas away from

the centroid increasing the

contribution of each area to Izz,

thereby increasing the stiffness of the

beam. If the foam was not separating the two steel plates, the total height

of the section would have been 40 mm and the value of Izz would have

been (0.04 m)×(0.04 m)3/12 = 0.213×10

‒6 m

4, much less than that of the

composite beam with the resulting decrease in beam stiffness, increase in

curvature and increase in the maximum value of stresses. It is always

Fig. 5.14 A sandwich beam

Fig. 5.15 The stress distribution in

sandwich beam

81

beneficial to keep the section areas far

away from the axis. It was for the same

reason that I-beam of Example 5.3 was

better than the rectangular section.

Example 5.5 Reinforced concrete beam

Reinforced concrete construction is a very

ingenious use of two materials with

complementary properties. Common concrete is a brittle material which

has good strength in compression (about

20 MPa). Structural steel on the other hand

has good strength a tension (about 200

MPa). All beams have compression on

one side and tension on the other. This

suggests that we combine the two

materials to produce beams in which steel

is loaded in tension while concrete is

loaded in compression.

Fig. 5.16 shows the section of such a

reinforced concrete beam. What is the maximum bending moment this

beam can transmit?

Solution:

Let us take the value of Ec, the elastic modulus of concrete as 20 GPa,

while that for steel, Es as 200 GPa.

We will use the following strategy to solve this problem:

1. Locate the neutral axis using Eq. 5.10.

2. Determine the stiffness of the beam using Eq. 5.11.

3. Assuming a bending moment Mb, calculate using Eq. 5.13 the

maximum compressive stress in concrete and maximum tensile

stress in steel.

4. Equate these stresses to the specified limiting stresses. The

minimum of the two maximum moments will be the answer to the

problem.

We first determine the location of the neutral axis for the section using

Eq. 5.10. Here, for steel: Es = 200 GPa, m, and As =

5×π×(0.020 m)2/4, and for concrete: Ec = 20 GPa, m, and Ac

= (0.450 m)×(0.200 m). Using these values in Eq. 5.10, we get

∑

∑

m.

Note that as expected, the stiffer material in the lower half of the section

brings the neutral axis down (from 225 mm for the concrete alone).

The stiffness of the composite beam is obtained from ∑ (refer Eq.

5.11). The values of Izz’s re obt ined by first ev lu ting the I’s bout the

centroidal axis of the concerned area and, then, transferring it to the

neutral axis using the parallel-axes theorem.

( )( )

( )( )(

)

m4

( ) 64

5 7

4 ( )

( )

( ) 5

m4

Therefore, the stiffness of the beam is ∑ =

( )( ) ( )( ) =

3.86×107 N.m

2,

And the curvature is

The maximum compressive load will occur in the top plane of the beam

which is (450 ‒ 205) mm w y from the neutr l pl ne. From Eq. 5.13,

( )( )

Pa, and

Fig. 5.16 Reinforced

concrete beam section

Fig. 5.17 Stress distribution

in a reinforced concrete beam

82

( )( )

Pa

For a maximum permissible compressive stress of 20 MPa in concrete,

the maximum permissible moment is given by , or

Mb = 157.6 kN.m. For a maximum tensile stress of 200 MPa in steel, the

maximum permissible moment is given by , or Mb

= 249 kN.m. The minimum of these two, then, is the permissible

maximum moment that this reinforced concrete beam can sustain.

Example 5.6 Force required to bend a strip

Consider a strip of length l, width

b and height h clamped to a block

with a circular arc surface of

radius R as shown in Fig. 5.18.

On application of a force P to the

free end, a length c of the strip

(AB) comes in contact with the

block. Find the force P.

Solution:

We are given that the length AB of

the strip has a constant radius of curvature. This implies that the bending

moment over this length is constant, which, in turn, implies that the shear

stress on this part of the strip vanishes and that there is no other loading

on this part. Let us consider the shape of the SFD of this strip. The SF

over the length AB is constant at zero. There is no load over the portion

BC. This implies that SF over this segment too is constant. This constant

SF will jump up through P at the end C because of the presence of a

concentrated load P (downwards) at that point.. But the shear force must

be zero after end C. This is possible only if the SF is constant at +P over

the segment BC. Since there no SF over AB, and the SF over BC must be

+P, there must be a concentrated reaction at point B where the strip loses

contact with the radius block. The resultant SFD must be as shown in

Fig. 5.19.

The BM at the free end C must be zero. The presence of a negative SF

over the length BC implies that the BM is increasing over this segment,

and therefore, it must be ‒Pd

at point B. The BM over

portion AB has, thus, a

constant value of ‒Pd. This

would require a clamping

moment of +Pd. The resultant

BMD must be as shown in

Fig. 5.19. It can be verified

from the loading diagram that

this indeed is in agreement

with the other loads on the

strip.

Now we are ready to evaluate the load P in terms of the given parameters.

We know the radius of curvature of the strip over the length AB s ‒R.

The negative sign arises from the fact that the beam is bending convex

upwards. Using Eq. 5.4, we get

( ), or

.

5.4 Stresses in beams carrying shear forces

We have so far considered the

case of a beam in pure bending

only. The only stresses that

exist in such a beam are tensile

stresses, termed as the bending

stresses because they arise from

bending of the beam. In a more

general case, there are other

stresses as well, notably shear stresses. To demonstrate the plausibility of

the presence of shear stresses, consider the simply-supported stack of

planks69

shown in Fig. 5.20. The planks will exhibit relative sliding as

shown. If these planks were glued together, they would tend to slide and

the glue will be in shear trying to hold the planks back from sliding. If it

were a solid beam, this would cause shear stresses to be present which

tend to hold the different layers together.

69

This stack of five planks will have five times the moment carrying capacity of

a single plank. But if we had used a beam of the combined thickness, the

moment carrying capacity would have been 25 times! Can you show this?

Fig. 5.20 Relative sliding in a stack of

planks acting as a beam

Fig. 5.18

Fig. 5.19

83

Consider next the simply-

supported beam loaded in

the middle as shown in

Fig. 5.21. The SFD and

BMD of the beam have

also been shown in the

figure. It is clear that the

beam will have shear

forces acting over its

entire length. The

presence of the shear

forces will cause the

bending moment to vary

along the length as

shown. In fact, this follows from the fact that the slope of the bending

moment line is equal to (negative of) the shear force value.

The presence of a positive bending moment will cause the beam to bend

with a positive curvature resulting in compressive stresses in the upper

part and tensile stresses in the lower part of the beam. But since the

moment varies along the length of the beam, the bending stresses (at the

same value of y) also varies with x. Thus, if

we take a slice of the beam (like the shaded

portion in Fig. 5.21), the bending stress

distributions on its two ends are entirely

different resulting in a mismatch of

horizontal forces.

This fact is illustrated by drawing a free

body diagram of this shaded portion as

shown in Fig. 5.22. The (compressive) bending stresses on the right end

vary linearly in the vertical direction. These would be given by equation

,

where y is measured from the neutral axis of the section. There are no

stresses on the left end since the bending moment there is zero. The

bending stresses alone then give rise to a net horizontal force to the left.

This force tends to slide this slice to the left which results in the

development of shear stresses on the bottom

surface of the slice acting towards right as

shown.

This shear is ηyx with a negative sign (using

the sign convention of stresses introduced in

Chapter 2) . We had seen earlier (Sec. 1.7)

that shear stresses occur in complementary

pairs: . Therefore, a stress component acts on the exposed

x-face of the element (Fig. 5.23). The forces due to these shear stresses

acting on this face give the resultant shear force V at the section.

5.5 Relating shear stresses to the shear force in a beam

The discussion above leads us to the strategy that we adopt to relate shear

stresses to the shear force in a beam. We take a slice of the beam of

length dx extending from a distance y (from the neutral axis) to the top

surface of the beam (see Fig. 5.24a). Let the bending moment on the left

face of this element be M, while that on the right face be M + dM (Fig.

5.24c).

The bending stresses due to these bending moments at the two locations, x

and x +dx (and vertical location y’ from the neutral axis) are then given

by:

( )

, and ( )

( )

The net axial

forces on the two

sections are:

∫

, and

∫( )

where the

integration is to

Fig. 5.21

Fig. 5.22 FBD of a slice of

the beam of Fig. 5.21

Fig. 5.23 Complementary

shear stress on the left face

Fig. 5.24 Bending stresses on an elemental slice of a beam

84

be carried out on area A’, which is the area of the elemental section from

y upwards. Note that at x, the positive ζ is directed in the negative x-

direction (the outward normal there being in the negative x-direction) and,

therefore, F,x (the force, by convention, being positive in positive

coordinate direction) is obtained as the negative of the integral of stresses.

However at x + dx, the positive stresses are in the positive coordinate

direction and, hence, F,x+dx is the integral of the stresses on that face.

The presence of a shear force on this element causes the bending

moments on the two sides of the element to be different, resulting in

different values of the axial forces on either side. The net axial force is

given by

.∫

/

where y’ is the distance measured from the neutral axis and, thus, can be

replaced with y. The integral above is the first moment of area A’ about

the neutral axis, where the area A’ is the area of the section above the

location where we need to determine the shear stress (see Fig. 5.23). This

is given the symbol Q.

(5.14)

This difference in axial forces is balanced by the shear stresses acting on

the bottom surface of the element (Fig. 5.24d). If we assume the average

stress on this surface to be ηyx (= ηxy), the net shear force on this face is

ηyx×(b×dx), where b is the width of the section at y. Since ηyx is positive in

the negative x-direction, this force is in the negative direction. Writing the

axial force equilibrium of the FBD of Fig. 5.24d, we get

( ) ,

or ( )

We replace dM by ‒Vdx (realizing from Eq. 4.2 that

) to obtain

( )

Thus, the average shear stress is given by

( ) ( )

(5.15)

Here V is the shear stress, Q is the first moment (about the neutral axis) of

the section area above y, b is the width of the section at location y, and Izz

is the second moment of the whole area. It can

be shown easily that we can write the following

expression for Q in terms of the distance of the

centroid of area A’ from the neutral axis (Fig.

5.25):

( ) ∫

(5.16)

The following examples illustrate the procedure

for calculating the shear stress distribution across beam sections.

Example 5.7 Shear stress distribution across a uniformly loaded cantilever beam

Consider a cantilever beam loaded uniformly with a load intensity of 100

N/m as shown in Fig. 5.26. The beam has a rectangular section of width

10 mm and depth 30 mm. Find the distribution of shear stresses at the root

section.

Solution:

We first determine the reaction at the built in support. By vertical force

balance the vertical reaction is determined to be 200 N upwards. To

determine the reaction moment at root, we replace the distributed load by

its statically-equivalent

concentrated load of 200 N

at the centroid of the

distributed load, which is

at 1 m point. The moment

balance equation then

gives the reaction moment

as 200 N.m counter-

clockwise.

To draw the FBD, we start

at zero. The shear force

jumps down 200 N due to

Fig. 5.25

Fig. 5.26 A uniformly loaded cantilever beam

and the corresponding SFD and BMD

85

the concentrated force at x = 0. Thereafter, the SFD increases at a

constant rate of 100 N/m due to the uniformly distributed load of the same

density, till it reaches the value of 0 at the free end of the beam.

To calculate the shear stress, we do not need to calculate the bending

moments. However, we do so here just to illustrate again the area method

of drawing the BMDs. After starting at 0, the BMD jumps down to 200

N.m because of the positive reaction moment of the same magnitude.

Then, due to a negative shear force which is constantly decreasing in

magnitude, the bending moment

increases with a linearly decreasing

rate, i.e., the BMD has a parabolic

shape as shown in Fig. 5.26. Total

change in BM (as per Eq. 4.2) over

the length of the beam is equal to the

area under the SF curve over the

same length. This is 200 N.m, and

therefore, the BM at the free end

turns out to be 0, as it should be.

The maximum shear force is V =

‒200 N nd occurs t the root of the be m. The she r stress t dist nce

y from the NA is given by Eq. 5.16 as ( )

. Because the section of

the beam is rectangular (b =10 mm, h = 30 mm), its Izz is given by bh3/12

or 2.25×10‒8

m4, and the NA is located at 0.015 m from the top. The

value of b, the width of the section at a distance y from NA is constant at

0.010 m. The value of Q is obtained from Eq. 5.16: ∫

.

Here A’ represents the area above the location y (shaded area in Fig.

5.26). Here ( ) , and therefore,

( )( ) .

/ (

) . The shear stress, then, is

( )

( ) ( )

( ) ( ) (

) .

It is a parabolic distribution with

the maximum at y = 0, i.e., at the

neutral axis, and going to zero at

the top and the bottom surfaces.

Fig. 5.27 shows graphically the shear stress distribution70

across the beam

section.

This shear distribution results in a similar shear strain distribution across

height of the beam. Fig. 5.28 shows an exaggerated view of the

deformation of the beam under the action of shear forces.

Example 5.8 Shear stress distribution on wide-flanged I-beam

Consider a simply-supported overhang beam with uniformly distributed

load as shown in Fig. 5.29. The beam section is a wide-flanged I-

section71

. Obtain the variation of shear stresses in the beam section.

Solution:

As with all statically-determinate beams, we start with the loading

diagram of the

beam, calculate

the reactions at

the support and

draw the SFD of

the beam using

the method of

areas. Because

of a uniformly

distributed load

of 500 N/m, the

SFD has a line

sloping up at the

same rate. But

70 Let us compare the magnitudes of the shear stresses with those of the bending

stresses in this simple case. The maximum bending stresses are of order ( ) , while the maximum shear stresses are of order . Therefore,

. The maximum value of Q in this case is (b.h/2).h/4, and so

. It is reasonable to suppose that M is of order V.L, where L is

the length of the beam, and therefore, . This ratio is small for

slender beams. 71

The horizontal portions away from the central axis are termed as flanges while

the central vertical portion of a beam is termed as web. A wide-flanged I-beam is

also termed as a W-beam

Fig. 5.27 Shear stress distribution on

a rectangular section

Fig. 5.29 A simply-supported overhang beam with uniformly

distributed load

Fig. 5.28 Distortion of a rectangular

beam due to a shear force

86

the presence of the two reactions (750 N each) makes the SF line jump

down by 750 N at the two supports. The resulting SFD is as shown.

The BMD has also been drawn in Fig. 5.29 even though we do not require

it for the purposes of this example.

The maximum positive shear force of 375 N magnitude occurs just before

the two supports, while the maximum negative magnitude occurs just

after the supports. The maximum shear stresses will also occur at these

same locations. We calculate the shear stresses using Eq. 5.15. We

require for this purpose the value of Izz. This is obtained by dividing the

section area into three rectangles A, B, and C, as shown in Fig. 5.30. The

symmetry of the section ensures that the centroid of the section is in the

middle.

We use the parallel axes theorem to calculate

Izz.

, ( ) - ( ) ,

or Izz =2.93×10‒7

m4.

To calculate the value of Q across the section,

first consider the shaded area at a distance y

from the NA as shown in Fig. 5.31a. The value

of Q for this is given by ( ) ( )( ), ( ) - or ( ) valid for y between

0.010 m to 0.020 m. Thus, the value of Q across the flange varies

parabolically from 6×10‒6

m3 at y = 0.01 m to 0 at y = 0.2 m.

For y between 0 and 0.010 m, the shaded area (Fig. 5.31b) is divided in

two parts. The contribution of the top part is clearly the value of Q for y =

0.01 m, and the contribution for the second part is (0.01 m –y)(0.010

m)×[y + (0.01 m –y)/2], or (0.5×10‒6

‒ 0.005y2) m

3. Thus, the value of Q

in the web is ( ) , which is again parabolic, with

a value of 8.5×10-6

m3 at y = 0 and 6×10

‒6 m

3 at y = 0.01 m. Fig. 5.32b

shows the variation of Q.

The shear stress ηxy can now be

determined from Eq. 5.15. At the

location of the maximum shear force

(= 375 N), the maximum shear occurs

at y = 0, and is given by ( ) ( )( ) )( ) .

The value at y =0.01 m is obtained

by using 6×10‒6

m3 for the value of

Q, and is obtained as 0.77 MPa.

In calculating this last value, we have used

0.01 as the value of b, the width of the section.

A value of 0.77 MPa is actually the value of

shear stress at a point slightly below y = 0.01

m where we can take the value of b to be 10

mm. At a point slightly above this point, the

width of the section suddenly increases by a

factor of 4 to 0.040 m and, therefore, there is a

discontinuity in the value of shear stress which

decreases by a factor of 4 to 0.19 MPa. This has been shown in Fig.

5.32c.

Fig. 5.33 shows the distribution of shear stress ηxy across the section. It

may be noted that the web of the section is the primary shear-bearing

element in a wide- flanged beam. The value of the stress ηxy in the flanges

is quite low.

5.6 Shear flow in beams

It was shown above that the shear stress ηxy in the flanges of the wide-flanged I-

beam of the previous

example is quite low. But

it can be shown that the

flanges carry ηxz as well.

To evaluate ηxz, consider

an area at one corner (at a

distance z from the

vertical axis of

symmetry) as shown in

Fig. 5.34. The FBD of

this area is also shown.

Take a small area element

dA in the back face of the section. Let this area be

located at a distance y from the NA. The tensile

stress acting on this area where the bending moment

Fig. 5.30 I-section divided up

in three rectangles

(a) (b)

Fig. 5.31

(a) (b) (c)

Fig. 5.32 (a) The I-section, (b) the

variation of Q, and (c) the

variation of the shear stresses η

across the section.

Fig. 5.33 Shear stress ηxy

distribution in an I-beam

Fig. 5.34 An element and its FBD to determine the

shear stress ηxz

Fig. 5.35 Shear flow in

an I-beam

87

is M c n be t ken to be ‒My/Izz, where Izz is the second moment of the section

re . The tot l force on this element is ‒(My/Izz).dA. The total force on the back

face is then the integral of this over the area of the face. We can , in a similar

fashion, calculate the force acting on the front face of the area. It will be exactly

the same, except that M will be replaced by M + dM, and that this force is

backwards. The net force due to bending stresses on the two faces is the

difference of the two, i.e., the integral over the area of (dM/Izz)ydA. The integral

of ydA is the first moment of area which has been given the symbol Q. Thus, the

net force due to bending stresses is QdM/Izz. The change in moment, as before,

can be replaced by ‒Vdx, where V is the shear force at the section.

This backward force is balanced by the shear force acting on the front

face (of area t.dx) of this element. If ηxz is the average shear stress on this

area

.

From this we obtain

(5.16)

This is same as Eq. 5.15 with section width b replaced with t, the

thickness of the flange.

It is convenient to use the concept of shear flow q first introduced in Sec.

3.7 as ∫

. Thus, both for flange as well as web, we can

write

(5.17)

Fig. 5.35 shows the shear

flow distribution for an I-

beam subject to a negative

shear force V. If the shear

force was positive, all the

arrows would have been

reversed. The shear flow in

the flanges starts at zero at the

tips and is maximum at the junctions with the web72

.

It is pertinent to point out here that the shear flow in bending has been

obtained from purely equilibrium considerations. We have not used any

other tool for these derivations.

Example 5.9 A built-up beam

Consider the built up beam shown in Fig. 5.36. It is held together with 10

mm bolts as shown. If the maximum shear force acting on the beam is 50

kN and the maximum shear stress a bolt can resist is 80 MPa, determine

the maximum spacing s between the bolts.

Solution:

To externalize the shear forces in the bolts, we consider the FBD of an

element of length dx as shown in Fig. 5.37. Because of the presence of

the shear force, the bending moments, and therefore, the bending stresses

on the two opposite faces of this

element are different. The

difference in the axial forces

because of this difference results

in a net force on the element due

to the bending stresses. This

force is resisted by the shear

action Fs,b in the bolts. But there

are two sides to this element.

The unbalanced force is, therefore, resisted by 2Fs,b.

The unbalanced force is determined by Eq. 5.14 as

Each bolt resists this unbalanced load for a length s of the beam, where s

is the distance between the bolts (i.e., the pitch).

.

72

The stress distributions at the junctions are quite complicated. It is for this

reason that the commercially available rolled-sections are provided with

generous fillets to avoid stress concentration.

Fig. 5.36 A built-up beam

Fig. 5.37

88

The value of Izz is evaluated by breaking up the beam section into a

difference of two areas: a rectangle 140 mm wide, 160 mm deep, from

which is subtracted a rectangle 120 mm wide, 140 mm deep. The value

of Izz is

( )( ) ( )( )

The first moment of the area of the face of the FBD about the NA is

obtained by multiplying the area (10 mm × 50 mm) by the distance of its

centroid from the neutr l xis (80 mm ‒ 5 mm). Thus, Q = 3.75×10‒5

m3.

The maximum shear stress in a bolt is prescribed as 80 MPa. Therefore,

maximum value of Fs,b is . ( )

/ ( ) . Using this

value in the equation above along with V = 50 kN, we get

( )( )

This gives a value of s as 0.136 m. Thus, the spacing between the bolts

should be less than 13.6 cm.

5.7 Shear centre

Let us next consider

beams with sections that

have symmetry only

about one axis, like a

channel section shown in

Fig. 5.38. Let the shear

force on the section be V.

We can easily find the

shear flow distribution

along the three legs of the section.

These are shown in Fig. 5.38a. The

shear flow in the flanges increases

from 0 at the tips to the maximum

at the junction with the web.

The maximum value of the shear flow can be determined from Eq. 5.17 as

VQ/Izz. The Q here is the first moment about neutral axis of the area of

the upper flange, which is (bt)h/2. The value of Izz is found by breaking

up the area into three parts, determining the I of each about its own

centroidal axis, and then transferring it to the neutral axis using the

parallel-axes theorem:

, ( ) - ( ) .

The maximum value of q is then Vbht/2Izz. The total horizontal shear

force F1 in the flange is the area under the shear flow triangle, which is

½qmax.b, or F1 = Vhb2t/4Izz, pointing to left. The lower flange has the

same value of shear force, except that it is pointed to the right. The shear

force in the web F2 is equal the sectional shear force V.

The resultant force on the section is now V, but the pair of forces F1

produces a couple. Fig. 5.38(c) shows a statically equivalent force system

in which the force V is displaced to a point O, a distance e to the left, with

, or . The point O is termed as the shear

centre of the section. As is shown in Fig. 5.39 the channel bends without

twisting only when the applied load passes through the shear centre.

5.8 Plastic deformations in beams

Consider a beam in pure bending

which is loaded beyond the onset of

yielding. Let the beam be made of a

perfectly elastic-perfectly plastic

material whose behaviour is as

depicted in Fig. 5.40. The yield stress

of the material is ζY, and the

corresponding strain is εY. As the

bending moment applied to the beam

increases the magnitude of the strain

at any point increases linearly with the bending moment. The strain for

any applied moment varies linearly with y according to Eq. 5.1:

(5.1)

(a) (b) (c)

Fig. 5.38 (a) shear flow distribution; (b) resultant

forces in the three legs of the section; (c) point of

action of the statically-equivalent shear force

(a) (b)

Fig. 5.39 (a) The channel twists while

bending when the load does not pass

through the shear centre b) The

channel bends without twisting when

the load acts through the shear centre.

Fig. 5.40

89

And as long as the strain

is less than the yield

strength value of εY, the

bending stress variation is

given by Eεxx, where E is

the elastic modulus of the

material. Under these

conditions, the stress

distribution is as shown

in Fig. 5.41a. The value of ρ is found by considerations of equilibrium, so

that the stresses are given by Eq. 5.5 as

.

This is valid as long as εmax < εY, and consequently, ζmax < ζY. The Value

of bending moment MY for which the maximum stress reaches the yield

strength value is obtained from Eq. 5.6 as ( )

, where h is the

height of the section.

For a rectangular b×h beam, this gives

(5.18)

For values of bending moment larger than MY, the material starts

deforming plastically at the either end. Though the strain distribution is

still linear, the bending stress in no longer linear with y. The value of the

stress acquires the value ζY at a value of y less than h/2 as shown in Fig.

5.41b and remains constant thereafter. If the radius of curvature is ρ for

this case, then ( ) , and ( ) till its value is ζY,

after which it remains constant at ζY. The value of y at which ζ first

becomes ζY is

. We denote

by yY. Thus, the beam is elastic

within the r nge ‒yY to +yY, and plastic outside this range. The value of ρ

is as yet undetermined. We evaluate the bending moment by integrating

ζdA over the whole area.

∫ ∫

∫ .

/

∫ ( )

or,

[

.

/ ]

[

.

/ ]

(5.19)

The value of yY is 0 when the plastic

region spans the whole depth of the beam.

The beam in this condition is resisting the

maximum possible moment it can resist.

Thus, the maximum bending moment

Mmax that the beam can support is obtained

by setting yY = 0 in Eq.5.19:

(5.20)

The value of ρ is easily determined in

terms of yY by noting that the strain at yY

is ‒ Y, so that

, or

.

Combining this with the fact that the

radius of curvature ρY at the point where the plastic deformation just starts

is given by

, we get

. Using this in Eq. 5.19, we get

[

.

/ ], for ρ < ρY. (5.21)

Fig. 5.41 shows the variation of bending moment with the radius of

curvature of the beam for a rectangular-section beam. The maximum

bending moment approaches a value 1.5 times MY asymptotically. The

ratio of the maximum bending moment to the yield bending moment

changes with the section shape. For circular sections, this ratio is 1.7, and

for thin-walled circular tubes it is about 1.3. For typical commercially

available I-beams, this ratio varies between 1.1 and 1.2.

5.9 Strain energy in bending

The expression for strain energy in a structure subject to arbitrary stresses

is given by Eq. 2.17 as

∫ ( )

(2.17)

where the integration is carried over the entire volume of the beam. For

the case of a beam in pure bending (no shear force and bending moment

constant over the entire length), there is only one stress, ζxx given by Eq.

5.5 as

. The strain energy for such a beam is, therefore

(a) (b) (c)

Fig. 5.41

Fig. 5.42 Variation of bending

moment with curvature in elastic-

plastic deformation of a

rectangular-section beam

90

∫

∫ .

/

.∬

/ ∫ .

/

(5.22)

For a uniform beam subject to a constant bending moment, this equation

reduces to

.

In a more general case when a beam is also subjected to transverse shear,

Eq. 2.17 gives the value of strain energy as:

∫ ( )

, or

∫ .

/

(5.23)

It was shown in Example 5.7 that the ratio of the maximum shear stress to

the maximum tensile stress in the case of a uniformly loaded cantilever

beam was . It can be shown that the same holds true to

almost all beams and loadings. This ratio is small for typical slender

beams. Since the stresses are squared in Eq. 5.23, the contribution of

shear stresses to the strain energy is quite small compared to that of the

bending (tensile) stresses and can, therefore, be safely neglected.

Example 5.10 Strain energy of a simply-supported beam

Consider the simply-supported

beam of length L loaded at the mid-

point as shown in Fig. 5.43. Find

the strain energy of the beam if the

section of the beam is rectangular

with depth h and width b.

Solution:

The reader can verify that the SFD

and the BMD of the beam are as

drawn in the figure. Since we need

to integrate the squares of shear stresses and bending stresses over the

length of the beam, let us express SF and BM as functions of the

longitudinal variable x.

It can be seen that we can write:

( )

for 0 < x < L/2,

for L/2 < x < L (a)

And ( )

for 0 < x < L/2,

(

) for L/2 < x < L

(b)

The contribution of bending stresses to the total energy is given by

∫

, where A is the cross-sectional area of the beam.

Expressing ζxx in terms of M(x), and replacing the value of M(x) using Eq.

b above, we get

∫ ( )

∫

∫

( )

,

And expressing Izz in terms of b and h, we get

It can be verified that this has the dimensions of energy. Please note that

we have not included the energy due to shear which is an order of

magnitude smaller for slender beams.

Summary

A beam is a slender member which carries transverse loads or

moments which tend to bend it.

o A beam resists bending moments by developing axial

stresses. These stresses arise because curving of a beam

necessarily implies contraction of material on one side

and elongation on the other side.

o There exists a plane within the beam which undergoes

neither contraction nor extension. This is termed as the

neutral plane. The trace of the neutral plane within a

section of the beam is termed as the neutral axis.

To obtain a relation between the curvature of the beam, the

bending stresses and the moment causing the bending we adopt

the following strategy:

Fig. 5.43

91

o We first use the geometry of bending in the case of pure

bending to relate strains to the distance from the neutral

plane.

o We convert the strains to stresses using the elastic

modulus.

o We next use the equilibrium conditions. The balance of

axial forces gives us the location of the neutral axis as

passing through the centroid of the section area. The

balance of moments then relates the curvature in the

beam to the bending moment acting on it. Eq. 5.6 gives

the relation between the moment Mb, the radius of

curvature ρ, the strain εxx, and the bending stress ζxx.

The bending stresses vary linearly with distance from the neutral

axis, zero on the axis and increasing as we move away. There is

compression on one side and tension on the other side of the

neutral axis.

We introduce the concept of bending rigidity as the curvature

produced per unit bending moment. The bending rigidity for a

uniform beam in pure bending is seen to be the product of the

elastic modulus E and the second moment Izz of the cross-

sectional area about a transverse neutral (centroidal) axis. The

rigidity of a beam increases as the third power of the depth of a

beam.

It was shown through some solved examples that for the beam

sections of the same areas the beam with more material away

from the centroid is more rigid.

The method for handling of composite beams was developed

following the same process as for simple beams. It was shown

that strains in composite beams are determined by geometric

considerations alone as in the case of simple beams, and are

continuous across different materials. The stresses, however, are

not continuous. It was shown that the location of the neutral axis

is obtained by a relation that weights the areas of different

materials with their elastic moduli (Eq. 5.10). The flexural

relation is the same as that for a simple beam except that the

bending rigidity is now replaced by the sum of the bending

rigidity of the constituent areas. The process was explained using

the example of a steel-reinforced concrete beam which exploits

beautifully the differing capabilities of the two materials.

The presence of shear forces causes the variation of bending

moments along the length of a beam resulting in an unbalance of

axial forces. This gives rise to shear stresses ηzx or ηxz. The shear

stresses are given by Eq. 5.15 as ( ) ( ) ,

where Q is the first moment of the area of the section above the

location y, and b is the width of the section at y. The shear stress

is the maximum at the neutral axis and decreases towards either

end of the section.

The shear stresses in slender beams are an order of magnitude

smaller than the bending stresses.

It is convenient to use the concept of shear flow q as

∫

. Thus, both for flange as well as web, we can write

.

The concept of shear centre was introduced as a point within the

beam section through which the external load must pass for the

beam to bend without twisting. It is the point where the single

shear force which is statically equivalent to the sectional shear

distribution must be applied.

The strain energy in slender beams is dominated by the energy

due to bending stresses and is given by ∫ .

/

.

92

93

Appendix B Properties of Areas

B.1 First moments of area and centroid

Consider an area as shown in Fig. B.1. Two first moments of areas are

defined: one about the x- axis, and the other about the y- axis.

∫

, and ∫

(B.1)

The values of Qx and Qy depend on the location of the origin O.

The centroid of the area is defined as the

location of the origin such that Qx and Qy

vanish. We can use this property to determine

the location of the centroid.

Let the coordinate of the origin in the given

coordinate system be xc and yc. We calculate the

first moment of area about this point. The

distances y and x in Eq. B.1 are the replaced by

(y – yc) and (x – xc), respectively:

∫ ( ) , or

∫ ∫

, or

∫

(B.2)

And similarly, ∫

(B.3)

Example B.1 Centroid of a T-section

Consider the T-section shown in Fig. B.2.

Find the location of the centroid.

Solution:

The x location of the centroid is

immediately obtained by using symmetry.

Thus, xc = 2.5 units. The y-coordinate of

the centroid can be obtained by using Eq.

B.3.

We introduce here a very convenient

method for determining the centroid of areas that are made up of simple

areas. Thus, we divide the T-section in to two rectangular areas, a 1×4

unit area labelled A, and a 4×1area labelled B. We know that centroid of

area A is located at its mid-point, i.e., at y = 4.5 units. The centroid of

area B is located at its mid-point, i.e., at y = 2.0 units. The location of the

centroid of the composite area is given by

(∑ ) ∑ (B.4)

Using this formula, we get

( )

units

The location of the centroid is shown in the figure. The centroid is

located nearer the top since the area is top heavy.

B.2 Second moments of area

A second moment of area is a moment in which two lengths are used. We

define three second moments73

:

73

The nomenclature of the second moments has two indices, each indicating the

axis from which the distance is taken. Thus, Izz uses the distance y from the z

axis twice, Iyy uses the distance z from the y axis twice, and Iyz uses the distance y

from the z axis as well as the distance z from the y axis. This last is also called a

cross moment of area.

The polar moment of area used in Chapter 3 is also a second moment where the

distance of the area is measured from the polar axis, i.e., the axis normal to the

area. It will be labeled Ixx in the convention used here:

Fig. A.1

Fig. A.2 T-section

94

∫

; ∫

, and ∫

(B.5)

The dimensions of these second moments are L4, and its SI units are m

4.

Example B.2 Second moment of a rectangular section

Determine the second moment of area of the rectangular section shown in

Fig. B.3 about its centroidal axis.

Solution:

The centroid of this symmetric section obviously

lies at the middle of the area. We take the

coordinate axes with the centroid as the origin.

Let us take a rectangular elemental area of

dimensions δy and δz. Then, by Eq. B.5:

∫

∫ ∫

Carrying out the indicated integrations, we get:

(B.6)

Similarly, .

We next calculate these moments about the y-axis

that coincides with the bottom plane of the

section.

∫

∫ ∫

four times the value when the x-axis passes through the centroid. Clearly,

the value depends on the location of the x-axis. We shall show below that

that the value with the x-axis passing

through the centroid is the minimum of all

possible values.

Example B.3 Second moment of a hollow square section

Determine the second moment of area of

the hollow square section shown in Fig. B.5

about its centroidal axis.

Solution:

We can, using symmetry, see that the centroid will be at the centre of the

section.

∫

∫

∫

Recognizing that the first integral is the Ixx of the outer square (=

a.a3/12), and the second integral is the Ixx of the inner square (= b.b

3/12),

we get the Ixx of the section as ( ) .

This is a general method of determining the moments of composite

sections.

B.3 Parallel axes theorem

We now introduce a very important property of the second moments of

area which is very handy in calculations of

the moments. Refer to Fig. B.6. Consider

First a coordinate system x-y with the

origin O and x-axis passing through the

centroid. The moment in this coordinate

system is given by

∫

Next consider an arbitrary coordinate

system x’-y’ with the x’-axis located a distance d away from the

centroidal axis. The second moment in this new system is

∫

∫ ( ) ∫ ( )

or, ∫

∫ ∫

Since y is measured from the centroidal axes, ∫

is immediately

seen as Ixx, ∫

is the first moment of area about a centroidal axis and

hence zero, and ∫

is nothing but the total area A. This gives:

(B.7)

Fig. A.3 A

rectangular section

Fig. A.4

Fig. A.6

Fig. A.5 Hollow square section

95

This is known as the parallel axes theorem. The second moment about an

axis a distance d away from the centroid is equal to the moment about the

centroidal axes plus area times the square of d.

One can immediately see from this theorem that the second moment about

the centroidal axis is the minimum. This theorem provides a very

convenient tool to obtain moment of inertia of a composite area as is

illustrated below.

Example B.4 Second moment of a T-section about the centroidal axis

Consider the T-section of Example B.1. Determine its second moment

about the centroidal axis.

Solution:

The centroid of the section was determined in Example B.1 to be located

on the centre line a distance 3.25 units up

from the bottom of the T. The origin of

the axes is now taken at this location as

shown.

We split up the area of the section into two

rectangles A and B as before (Fig. B.7).

We can calculate the second moment of

area of A about its own centroidal axes (at

a distance of 4.5 units from the bottom) using Eq. B.6 as (5×13)/12 =

0.417. We then use the parallel axis theorem (Eq. B.7) to shift the axis to

3.25 units above the bottom (from 4.5 units above the bottom. Thus, the

shift distance d is 4.5 – 3.25 = 1.25, and the value of becomes

0.417 + (Area = 5×1)×(1.25)2 = 8.23 units. Similarly, the second

moment of area B about its own centroid (which is 2 units from the

bottom) is 1×43/12 = 5.33 units. When we transfer it to the centroid of the

whole section which is a distance d (= 3.25-2.0 = 1.25 units) away, it

becomes (Area = 5×1)×(1.25)2 = 13.14 units. Thus, the

net moment is units.

Example B.5 Second moment of an angle section about the centroidal axis

Consider the angle section shown in Fig.

B.8. Determine its second moment about

the centroidal axis.

Solution:

We first determine the location of the

centroid of the section. We work with the

division of the section in two rectangles A

and B as shown. The area of A is 400 mm2

and its centroid is at (5 mm, 30 mm) and

the area of B is 500 mm2 and its centroid is

at (25 mm, 5 mm). The coordinates of the

centroid are:

∑

∑

mm,

and ∑

∑

mm.

The location of the centroid has been marked on the figure.

Here, too, we calculate the second moment using the parallel axis

theorem. We first calculate the moments of the two areas about their own

centroidal axis, and the transfer them to the centroidal axis of the

composite section:

,( ) ( ) -

0

( ) 1

0

( ) 1 mm

4.

Fig. A.7

Fig. A.8 An angle section

96

Thus, the second moment of this angle section about the centroidal axis

has been determined as 2.99×10÷7

m4.

B.4 Perpendicular axes theorem

Consider an area as shown in Fig. B.8. Let us consider a right-handed

triad of coordinate axes with x- and y- axes within the plane of the section

and the z-axis pointing outwards from the paper. Using the definitions of

the polar moment of area ∫

. But . Therefore,

∫

∫

∫

. Thus,

(B.8)

This is the perpendicular axes theorem which states that the polar moment

of inertia of a section is the sum of the second moment of the area about

two perpendicular axes within the plane of the area.

Example B.6 Second moment of a circular section

Determine the second moment of a circular

section.

Solution:

Consider the circular section of Fig. B.9. We

need to determine ∫

. Direct

determination of this integral is not easy.

Therefore, we use the perpendicular axes

theorem (Eq. B.8):

The symmetry of the circular section gives us ensures that the moments

and are equal. Thus, . The polar moment had been

evaluated in Sec. 3.3 as

(Eq. 3.6). Therefore,

Fig. A.8

Fig. A.9 A circular section

Table A.1 Properties of some areas

Geometry Area Vertical

location

of

centroid

Second moment of

area about

centroidal axis

Rectangle

bh h/2 bh3/12

Isosceles triangle

bh/2 h/3 bh3/36

Circle

πD2/4 D/2 πD

4/64

I-beam

[2k1+(1‒2k1)] bh h/2

,(

) (

)-

97

Index area method, 64

area, cross moment of, 93

area, first moment, 84, 93

area, polar moment, 43

area, second moment, 43, 93

axis, neutral, 73, 91

beam, 56

beam supports, idealized, 58

beam, built-up, 87

beam, composite, 78

beam, plastic deformations in,

89

beam, sandwich, 80

beam, shear stresses, 82

beam, simply-supported, 58

beams, shear flow, 87

bearing stress, 5

bending moment, 7, 56

bending rigidity, 74, 91

bending strain, 73

bending stress, 73

bending, pure, 71

bending, strain energy, 90

built-in support, 58

built-up beam, 87

cable in a sleeve, 45

Castigliano theorem, 33, 40

centroid, 74, 93

coefficient of thermal expansion,

27

coil spring, 53

complementary planes, 12

composite beam, 78

composite shaft, 49

concrete, pre-stressed, 32

concrete, reinforced, 81

cross moment of area, 93

deformation, plastic, 29

distributed load, 59

elastic limit, 29

elastic modulus, 5, 17

electroplating, residual stresses

in, 27

energy methods, 33

energy, strain, 33

first moment of area, 84, 93

gauge length, 29

generalized Hooke law, 24

Hooke law, generalized, 24

Hooke, Robert, 5

idealized beam supports, 58

idealized stress-strain curves, 30

key, 10

keyway, 10

limit analysis, 52

limit load, 52

linear elastic range, 29

load, distributed, 59

load, point, 58

loading density, 59

method of sections, 61

moment, bending, 7, 56

moment, twisting, 6

necking, 30

neutral axis, 73, 91

neutral plane, 72, 91

normal stress, 11

parallel axes theorem, 95

Pascal, 7

permanent set, 29

perpendicular axes theorem, 96

pinned support, 58

plane, neutral, 72, 91

planes, complementary, 12

plastic deformation, 29

plastic deformation in torsion,

51

plastic deformations in beams,

89

point load, 58

Poisson ratio, 23

polar moment of area, 43

pre-stressing, 32

proportional limit, 29

radius of curvature, 73

rectangular section, second

moment of, 94

reinforced concrete beam, 81

right-hand thumb rule, 44

rigidity, bending, 74, 91

rigidity, torsional, 43

roller support, 58

sandwich beam, 80

second moment of area, 43, 93

second moment of rectangular

section, 94

shaft, composite, 49

shaft, statically indeterminate,

47, 54

shaft, tapered, 45

shear centre, 88

shear flow, 50

shear flow, beams, 87

shear force, 56

shear modulus, 26

shear strain, 25

shear strain in torsion, 42

shear stress, 6

shear stress in torsion, 42

shear stresses in beams, 82

shrink fit, 28

sign convention for shear force

and bending moment, 57

simply-supported beam, 58

spring, kinky, 22

St. Venant principle, 8

statically equivalent loads, 8

statically indeterminate

problems, 20, 37

statically indeterminate shaft,

47, 54

strain, 5

strain energy, 33

strain energy, bending, 90

strain energy, elastic body, 38

strain energy, torsion, 53

strain hardening, 30

strain, bending, 73

strain, shear, 25

strength, yield, 29

stress, 5

stress at a point, 7

stress on oblique planes, 11

stress, bearing, 5

stress, bending, 73

stress, double-index notation, 12

stress, shear, 6

stress, sign convention, 13

stresses, pressure vessel, 14

support, built-in, 58

support, hinged, 9

support, pinned, 58

support, roller, 9

symmetry arguments, 15

tensile test, 28

thermal expansion, coefficient

of, 27

thermal strains, 27

thermal stresses, 27

torsion, 41

torsion bar, 41

98

torsion moment diagram, 44

torsion, plastic deformation, 51

torsion, strain energy, 53

torsional rigidity, 43

torsional strain, 42

torsional stress, 42

twisting moment, 6

twisting moment diagram, 44

two-force member, 9, 10

ultimate stress, 30

unit force method, 35

vibration isolator, 26

yield point, 29

yield strength, 29