Beams

Mechanics of Solid (NME – 302) Beams

Yatin Kumar Singh

INTRODUCTION:

Apart from axial and torsional forces there are other types of forces to which members maybe subjected. In

many instances in structural and machine design, members must resist forces applied laterally or

transversely to their axes. Such members are called beams. The main members supporting floors of buildings

are beams, just as an axle of a car is a beam.

Many shafts of machinery act simultaneously as torsion members and as beams. With modern materials, the

beam is a dominant member of construction. The determination of the system of internal forces necessary for

equilibrium of any beam segment will be the main objective of this chapter.

For the axially or torsionally loaded members previously considered, only one internal force was required at

an arbitrary section to satisfy the conditions of equilibrium. However, even for a beam with all forces in the

same plane, i.e., a planar beam problem, a system of three internal force components can develop at a section.

These are the axial force, the shear, and the bending moment. Determining these quantities is the focus of this

chapter. The chapter largely deals with single beams. Some discussion of related problems of planar frames

resisting axial forces, shears, and bending moments is also given. Only statically determinate systems will be

fully analyzed for these quantities.

Extensions to members in three-dimensional systems, where there are six possible internal force

components, will be introduced in later chapters as ceded and will rely on the reader's knowledge of statics.

In such problems at a section of a member there can be: an axial force, two shear components, two bending

moment components, and a torque.

Indian Institute of Technology Madras

5.1 Bending of Beams:

Beams:

Bars Subjected to transverse loads; Planar and slender members.

Supports:

Identified by the resistance offered to forces.

(a) Rollers/Links:

Resists forces in a direction along the line of action (Figure 5.1(a)).

(b) Pins:

Resists forces in any direction of the plane (Figure 5.1(b)).

(c) Fixed Support:

Resists forces in any direction (Figure 5.1(c)). Resists moments.

5.1.2 Classification of Beams:

(a) Statically Determinate or Indeterminate

Statically determinate - Equilibrium conditions sufficient to compute reactions.

Statically indeterminate - Deflections (Compatibility conditions) along with equilibrium equations should

be used to find out reactions.

(b) Cross Sectional Shapes - I,T,C or other cross sections

(c) Depending On The Supports Used.

1) Simply supported - pinned at one end and roller at the other (Figure 5.1.2(a))

2) Cantilever - fixed at one end and the other end free (Figure 5.1.2(b)).


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3) Fixed beam - fixed at both ends (Figure 5.1.2(c)).

Where W – loading acting, L – span.

5.1.3 Calculation of Beam Reactions:

When all the forces are applied in a single plane, the three equations of static equilibrium are available for

analysis.

ΣFx = 0, ΣFy = 0 and ΣM3 = 0

Employing these, the reactions at supports could be found out.

5.1.4 Procedures for Computing Forces and Moments:

For a beam with all forces in one plane, three force components are internally developed.

• Axial force

• Shear

• Bending Moment

Procedures are to be established for finding these quantities.

Direct Method/Method of Sections:

This method is illustrated by the following example. Find reactions at supports for the cantilever shown in

figure 5.1.3(a) subjected to uniformly distributed load. Cut the cantilever along section A-A and obtain free

body diagram as given in figure 5.1.3(b).

The segment of the beam shown in figure 5.1.3(b) is in equilibrium under the action of external forces and

internal forces and moments.

ΣFy = 0 gives V = -wx

ΣM = 0 gives

Sign Convention: Shear Force


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Sign Convention: Moments

As Shown in Figure 5.1.5(a), sagging (beam retains water) moment is positive, other wise bending moment is

negative (Figure 5.1.5(b)).

5.1.5 Shear Force and Bending Moment Diagrams (SFD & BMD):

Plot of shear and bending moment values on separate diagrams could be obtained. Magnitude and location of

different quantities can be easily visualized. SFD & BMD are essential for designers to make decisions on the

shape and size of a beam.

5.1.6 Shear force and BM Diagrams / (Alternate approach):

Beams Element: Differential Equations of Equilibrium

Free body diagram of element of length dx is shown Figure 5.17, which is cut from a loaded beam (Figure

5.1.6).

ΣFy = 0 gives

V+wdx – (V + dV) = 0

i.e.

ΣMP = 0 gives

M + dM –Vdx –M – (wdx)(dx/2) = 0

Substituting equations 5.1.2 in 5.1.1

Integrating 5.1.2

Integrating 2

Me - External moment acting.

5.2 Theory of Bending:

5.2.1 General Theory

Plane sections normal to the axis before bending remain plane and normal after bending also, as shown in

Figure 5.2.1. From Figure 5.2.1, ab, cd efs are sections which remain plane and normal. Beam is subjected to

pure bending (no shear). Longitudinal top fibers are in compression and bottom fibres in tension. Layer of

fibres in between which is neither in tension or compression, is called the neutral surface. Neutral axis is

the intersection of such a surface with the right section through the beam.


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Assumptions of the Theory of Bending:

Deflection of the beam axis is small compared to span of the beam. Shear strains, along the plane xy are

negligible. Effect of shear stress in the plane xy(τxy ) on normal stress (σx ) is neglected.

Note: Even through pure bending is assumed, distribution of normal stresses at any given cross section does

not get significantly changed due to non uniform bending.

For pure bending of a beam, beam axis deforms into part of a circle of radius ρ; for an element defined by an

infinitesimal angle d, the fiber length is given by (refer figure 5.2.2)).

ds = R dθ

where

R - Radius of Curvature

k - Axis Curvature

For a fiber located at radius R' = R – y

ds ' = (R − y)dθ

5.2.2 Elastic Flexure Formula:

By Hooke's Law,

σx = Eεx = −Eky

Σ F = 0 gives

−Ek∫ ydA = 0

i.e. neutral axis possess through the centroid of the cross section (Ref Figure 5.2.3.)


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ΣM = 0 gives

Mz = Ek ∫ y2 .dA = Ek Iz

This gives

5.2.3 Beams of Composite Cross Section:

For beams of composite cross section σx = −Eiky for the ith material in the composite.

y = yi - y0

y0 is the location of neutral axis from the bottom of the beam. yi is the location of neutral axis of the ith

material. In the figure yi = yA, from this, we get

Where A the area of cross section of the corresponding material.

Deflection of Beams:

The axis of a beam deflects from its initial position under action of applied forces. Accurate values for these

beam deflections are sought in many practical cases: elements of machines must be sufficiently rigid to

prevent misalignment and to maintain dimensional accuracy under load; in buildings, floor beams cannot

deflect excessively to avoid the undesirable psychological effect of flexible floors on occupants and to

minimize or prevent distress in brittle-finish materials; likewise, information on deformation characteristics

of members is essential in the study of vibrations of machines as well as of stationary and flight structures.

Deflection of Beams (Solution Method by Direct Integration):

From Analytic geometry, Curvature of a line,

where x and y are co-ordinates of a point on the curve.

For small deflections,


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Since, σx = −My / EI

and k = −ε / y, and ε = σx / E

k = M/ EI

where M = My

Hence,

Since, V = dM/dx

Since, w = dv/dx

Boundary Conditions:

Refer figure 5.2.7(a) – (d)

(a) Clamped Support:

y(x1) = 0; y'(x1)=0;

(b) Roller or Pinned Support:

y(x1) = 0; M (x1)=0;

(c) Free end:

M (x1) = 0; V(x1) = 0;

(d) Guided Support:

y'(x1) = 0; V (x1)=0;

Moment - Area Method for finding Beam Deflections:

This method is used generally to obtain displacement and rotation at a single point on a beam. This method

makes use of the Moment - Area theorems given below.

Moment - Area Theorems:

Refer fig (a) and (b)

(a) (b) (c)


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Referring fig. (c)

This is the first moment area theorem, Where P and Q are any two sections on the beam. i.e. change in angle

measured in radians between any two point P and Q on the elastic curve is equal to the M/EI area bounded

by the ordinates through P and Q.

Referring to Figure, considering an element of the Elastic Curve,

dh = x . dθ

This is the second moment Area theorem.

If is the distance of centroid of the bending moment diagram between P and Q from P, (Refer Figure) then

Here is called the tangent deviation of the point P from a tangent at Q and A is the area of the BM

diagram between P and Q. and are as shown in figure.

Shear Stresses in Beams:

Shear force is related to change in bending moment between adjacent sections.

(a) (b) (c)


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dM/ dx = v

As shown in figure, consider a cut-out section from a beam.

At section 1 {Area PADS}

where y is the distance of any fiber from the neutral axis.

if M2 = M1 + dM and F2 = F1 + dF

then

Force per unit length,

This force per unit length is termed as the shear flow, q

Substituting for dM/ dx = V ; q = VQ/I

Here I is the moment of Inertia of the entire cross-sectional area around the neutral axis.

Shear Stress Formula for Beams:

The shear stress formula is obtained by modifying the shear flow formula.

(Refer figure)

Macaulay's Methods:

If the loading conditions change along the span of beam, there is corresponding change in moment equation.

This requires that a separate moment equation be written between each change of load point and that two

integration be made for each such moment equation. Evaluation of the constants introduced by each

integration can become very involved. Fortunately, these complications can be avoided by writing single

moment equation in such a way that it becomes continuous for entire length of the beam in spite of the

discontinuity of loading.

For example consider the beam shown in fig below:


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Let us write the general moment equation using the definition M = (∑ M)L Which means that we consider the

effects of loads lying on the left of an exploratory section. The moment equations for the portions AB, BC and

CD are written as follows

It may be observed that the equation for MCD will also be valid for both MAB and MBC provided that the terms

(x - 2) and (x - 3)2 are neglected for values of x less than 2 m and 3 m, respectively. In other words, the terms

(x - 2) and (x - 3)2 are nonexistent for values of x for which the terms in parentheses are negative.

As an clear indication of these restrictions, one may use a nomenclature in which the usual form of

parentheses is replaced by pointed brackets, namely, ‹ ›. With this change in nomenclature, we obtain a single

moment equation

Which is valid for the entire beam if we postulate that the terms between the pointed brackets do not exists

for negative values; otherwise the term is to be treated like any ordinary expression.

As an another example, consider the beam as shown in the fig below. Here the distributed load extends only

over the segment BC. We can create continuity, however, by assuming that the distributed load extends

beyond C and adding an equal upward-distributed load to cancel its effect beyond C, as shown in the adjacent

fig below. The general moment equation, written for the last segment DE in the new nomenclature may be

written as:


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It may be noted that in this equation effect of load 600 N won't appear since it is just at the last end of the

beam so if we assume the exploratory just at section at just the point of application of 600 N than x = 0 or else

we will here take the X - section beyond 600 N which is invalid.

Procedure To Solve The Problems:

(i) After writing down the moment equation which is valid for all values of ‘x' i.e. containing pointed brackets,

integrate the moment equation like an ordinary equation.

(ii) While applying the B.C's keep in mind the necessary changes to be made regarding the pointed brackets.

Illustrative Examples: 1. A concentrated load of 300 N is applied to the simply supported beam as shown in Fig. Determine the equations of the elastic curve between each change of load point and the maximum deflection in the beam.

Solution: writing the general moment equation for the last portion BC of the loaded beam,

Integrating twice the above equation to obtain slope and the deflection

To evaluate the two constants of integration. Let us apply the following boundary conditions:

1. At point A where x = 0, the value of deflection y = 0. Substituting these values in Eq. (3) we find C2 = 0.

Keep in mind that < x -2 >3 is to be neglected for negative values.

2. At the other support where x = 3m, the value of deflection y is also zero.

substituting these values in the deflection Eq. (3), we obtain

Having determined the constants of integration, let us make use of Eqs. (2) and (3) to rewrite the slope and

deflection equations in the conventional form for the two portions.

Continuing the solution, we assume that the maximum deflection will occur in the segment AB. Its location

may be found by differentiating Eq. (5) with respect to x and setting the derivative to be equal to zero, or,


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what amounts to the same thing, setting the slope equation (4) equal to zero and solving for the point of zero

slope. We obtain

50 x2– 133 = 0 or x = 1.63 m (It may be kept in mind that if the solution of the equation does not yield a

value < 2 m then we have to try the other equations which are valid for segment BC). Since this value of x is

valid for segment AB, our assumption that the maximum deflection occurs in this region is correct. Hence, to

determine the maximum deflection, we substitute x = 1.63 m in Eq (5), which yields

The negative value obtained indicates that the deflection y is downward from the x axis. Quite usually only

the magnitude of the deflection, without regard to sign, is desired; this is denoted by δ, the use of y may be

reserved to indicate a directed value of deflection.

if E = 30 Gpa and I = 1.9 x 106 mm4 = 1.9 x 10 -6 m4 , Eq. (8) becomes

Example 2:

It is required to determine the value of EIy at the position midway between the supports and at the

overhanging end for the beam shown in figure below.

Solution: Writing down the moment equation which is valid for the entire span of the beam and applying the

differential equation of the elastic curve, and integrating it twice, we obtain

To determine the value of C2, It may be noted that EIy = 0 at x = 0,which gives C2 = 0.Note that the negative

terms in the pointed brackets are to be ignored. Next, let us use the condition that EIy = 0 at the right support

where x = 6m.This gives

Finally, to obtain the mid-span deflection, let us substitute the value of x = 3m in the deflection equation for

the segment BC obtained by ignoring negative values of the bracketed terms (x – 4)4 and (x – 6)3. We obtain

For the overhanging end where x = 8m, we have

Example 3: A simply supported beam carries the triangularly distributed load as shown in figure. Determine the

deflection equation and the value of the maximum deflection.

Solution: Due to symmetry, the reactions is one half the total load of 1/2w0L, or R1 = R2 = 1/4w0L. Due to the

advantage of symmetry to the deflection curve from A to B is the mirror image of that from C to B. The


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condition of zero deflection at A and of zero slope at B do not require the use of a general moment equation.

Only the moment equation for segment AB is needed, and this may be easily written with the aid of figure (b).

Taking into account the differential equation of the elastic curve for the segment AB and integrating twice,

one can obtain

In order to evaluate the constants of integration, let us apply the B.C's we note that at the support A, y = 0 at x

= 0.Hence from equation (3), we get C2 = 0. Also, because of symmetry, the slope dy/dx = 0 at mid-span

where x = L/2. Substituting these conditions in equation (2) we get

Hence the deflection equation from A to B (and also from C to B because of symmetry) becomes

which reduces to

The maximum deflection at mid-span, where x = L/2 is

Example 4: Couple Acting Consider a simply supported beam which is subjected to a couple M at a distance 'a' from the left end. It is

required to determine using the Macaulay's method.

To deal with couples, only thing to remember is that within the pointed brackets we have to take some

quantity and this should be raised to the power zero i.e. . We have taken the power 0 (zero)


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because ultimately the term should have the moment units.Thus with integration the quantity

becomes either 1 or 2

Or

Therefore, writing the general moment equation we get

Integrating twice

Example 5: A simply supported beam is subjected to U.D.L in combination with couple M. It is required to determine the

deflection.

This problem may be attempted in the some way. The general moment equation may be written as

Thus,

Integrate twice to get the deflection of the loaded beam.

Beams

Engineering

Transcript of Beams