Unit1 From Som by Bansal

CHAPTER

1.1. INTRODUCTION

When an external force acts on a body, the body tends to undergo some deformation.Due to cohesion between the molecules, the body resists deformation. This resistance by whichmaterial of the body opposes the deformation is known as strength of material. Within acertain limit (i.e., in the elastic stage) the resistance offered by the material is proportional tothe deformation brought out on the material by the external force. Also within this limit theresistance is equal to the external force (or applied load). But beyond the elastic stage, theresistance offered by the material is less than the applied load. In such a case, the deformationcontinues, until failure takes place.

Within elastic stage, the resisting force equals applied load. This resisting force perunit area is called stress or intensity of stress.

1.2. STRESS

The force of resistance per unit area, offered by a body against deformation is known asstress. The external force acting on the body is called the load or force. The load is applied onthe body while the stress is induced in the material of the body. A loaded member remains inequilibrium when the resistance offered by the member against the deformation and theapplied load are equal.

pMathematically stress is written as, a = -

A

where a = Stress (also called intensity of stress),

P = External force or load, and

A = Cross-sectional area.

1.2.1

. Units of Stress. The unit of stress depends upon the unit of load (or force) andunit of area. In M.K.S. units, the force is expressed in kgf and area in metre square (i.e., m2).Hence unit of stress becomes as kgf/m2. If area is expressed in centimetre square (i.e., cm2),the stress is expressed as kgf/cm2.

In the S.I. units, the force is expressed in newtons (written as N) and area is expressedas m

2. Hence unit of stress becomes as N/m2. The area is also expressed in millimetre square

then unit of force becomes as N/mm2

1 N/m2 = 1 N/dOO cm)2 = 1 N/104 cm2

= 10-4 N/cm2 or 10ÿ N/mm2

cm2 102 mm2

STRENGTH OF MATERIALS

1 N/mm2 = 106 N/m2.

Also 1 N/m2 = 1 Pascal = 1 Pa.

The large quantities are represented by kilo, mega, giga and terra. They stand for :Kilo = 103 and represented by.k

Mega = 106 and represented by.MGiga = 109 and represented by.G

Terra = 1012 and represented by.T.Thus mega newton means 10f newtons and is represented by MN. The symbol 1 MPa

stands for 1 mega pascal which is equal to 106 pascal (or 106 N/m2).The small quantities are represented by milli, micro, nano and pico. They are equal to

Milli = 10~3 and represented by.mMicro = 10"6 and represented by.nNano = 10_9 and represented by.rj

Pico = 10~,2 and represented by.p.Notes, i. Newton is a force acting on a mass of one kg and produces an acceleration of 1 m/si i.e.,

1 N = 1 (kg) x 1 m/s2.2. The stress in S.I. units is expressed in N/m2 or N/mm2.3. The stress 1 N/mm2 = 10f N/m2 = MN/m2. Thus one N/mm2 is equal to one MN/m2.4. One pascal is written as 1 Pa and is equal to 1 N/m2.

1.3

. STRAIN

When a body is subjected to some external force, there is some change of dimension ofthe body. The ratio of change of dimension of the body to the original dimension is known asstrain. Strain is dimensionless.

Strain may be :1

. Tensile strain, 2. Compressive strain,3

. Volumetric strain, and 4. Shear strain.

If there is some increase in length of a body due to external force, then the ratio ofincrease of length to the original length of the body is known as tensile strain. But if there issome decrease in length of the body, then the ratio of decrease of the length of the body to theoriginal length is known as compressive strain. The ratio of change of volume of the body tothe original volume is known as volumetric strain. The strain produced by shear stress isknown as shear strain.

1.4

. TYPES OF STRESSES

The stress may be normal stress or a shear stress.

Normal stress is the stress which acts in a direction perpendicular to the area. It isrepresented by o (sigma). The normal stress is further divided into tensile stress and compressivestress.

1.4.1

. I'ensilc Stress. The stress induced in a body, when subjected to two equal andopposite pulls as shown in Fig. 1.1 (a) as a result of which there is an increase in length, isknown as tensile stress. The ratio of increase in length to the original length is known as tensilestrain. The tensile stress acts normal to the area and it pulls on the area.

2

SIMPLE STRESSES AND STRAINS

Let P = Pull (or force) acting on the body,

A = Cross-sectional area of the body,L = Original length of the body,

dL = Increase in length due to pull P acting on the body,

o = Stress induced in the body, and

e = Strain (i.e., tensile strain).

Fig. 1.1 (a) shows a bar subjected to a tensile force P at its ends. Consider a section x-x,which divides the bar into two parts. The part left to the section x-x, will be in equilibrium ifP = Resisting force (R). This is shown in Fig. 1.1 (b). Similarly the part right to the sectionx-x, will be in equilibrium if P = Resisting force as shown in Fig. 1.1 (c). This resisting force perunit area is known as stress or intensity of stress.

x

(d)I

Fig. 1.1

Tensile stress = a =Resisting force (R) Tensile load (P)

Cross-sectional area

Por a = -

A

And tensile strain is given by,

Increase in length dL"

~

L

(v P = R)

,..(1.1)

e = ...(1.2)Original length

1.1.2. Compressive Stress. The stress induced in a body, when subjected to two equal

and opposite pushes as shown in Fig. 1.2 (a) as a result of which there is a decrease in lengthof the body, is known as compressive stress. And the ratio of decrease in length to the originallength is known as compressive strain. The compressive stress acts normal to the area and itpushes on the area.

Let an axial push P is acting on a body in cross-sectional area A. Due to external push P,let the original length L of the body decreases by dL.


>< (a)

pÿ DArictino forrA / Q\ÿ_ nesisung Torce jn)

; <»

Resisting force (R) ->

(c)

P

R r"! R

(d)

|_

Fig. 1.2

Then compressive stress is given by,

_

Resisting Force (R) _ Push (P)Area (A) Area (A)

And compressive strain is given by,

P

A

e _ Decrease in length

_

dLOriginal length L

1.4.3. Shear Stress. The stress induced in a body, when subjected to two equal andopposite forces which are acting tangentially across the resisting section as shown in Fig. 1.3as a result of which the body tends to shear off across the section, is known as shear stress.The corresponding strain is known as shear strain. The shear stress is the stress which actstangential to the area. It is represented by x.


Consider a rectangular block of height h, length L and width unity. Let the bottom faceAB of the block be fixed to the surface as shown in Fig. 1.4 (a). Let a force P be appliedtangentially along the top face CD of the block. Such a force acting tangentially along asurface is known as shear force. For the equilibrium of the block, the surface AB will offer atangential reaction P equal and opposite to the applied tangential force P.

Resistance

R-ÿ

77777777777777777777777

A 4- P B

h- L -H(a) (6) (c>

L __ Fis-14Consider a section x-x (parallel to the applied force), which divides the block into two

parts. The upper part will be in equilibrium if P = Resistance (R). This is shown in Fig. 1.4 (b).Similarly the lower part will be in equilibrium if P = Resistance (R) as shown in Fig. 1.4 (c).This resistance is known as shear resistance. And the shear resistance per unit area is knownas shear stress which is represented by t.

Shear resistance RShear stress, x =

Shear area

P

L x 1(v R = P and A = L x 1) ...(1.3)

dl

D.

Note that shear stress is tangential to the area over which it acts.

As the bottom face of the block is fixed, the faceABCD will be distorted to ABC

1Dl through an angle <>

as a result of force P as shown in Fig. 1.4 (d).

And shear strain ((j>) is given by,

ÿ _ Transversal displacement

Distance AD

d!

C,

orDD1

=

dt

AD ~

h...(1.4)

T-

;-*

T i tt 9i i/ /

1 / /i /

h t

tii

1 (Si Q'V/

ÿf/ /

i /

ÿ > CDTT7T7777777777777777777

*////////

Fig. 1.4 id)

1.5. EIASTICITY AND EI-ASTIC LIMIT

When an external force acts on a body, the body tends to undergo some deformation. Ifthe external force is removed and the body comes back to its original shape and size (whichmeans the deformation disappears completely), the body is known as elastic body. This property,

5


by virtue of which certain materials return back to their original position after the removal ofthe external force, is called elasticity.

The body will regain its previous shape and size only when the deformation caused bythe external force, is within a certain limit. Thus there is a limiting value of force up to andwithin which, the deformation completely disappears on the removal of the force. The valueof stress corresponding to this limiting force is known as the elastic limit of the material.

If the external force is so large that the stress exceeds the elastic limit, the materialloses to some extent its property of elasticity. If now the force is removed, the material willnot return to its original shape and size and there will be a residual deformation in the material.

1.6

. HOOKE'S IxAW AND ELASTIC MODULI!

Hooke,s Law states that when a material is loaded within elastic limit, the stress is

proportional to the strain produced by the stress. This means the ratio of the stress to thecorresponding strain is a constant within the elastic limit. This constant is known as Modulus

of Elasticity or Modulus of Rigidity or Elastic Modulii.

MODULUS OF ELASTICITY (OR YOUNG,S MODULUS)

The ratio of tensile stress or compressive stress to the corresponding strain is a constant.This ratio is known as Young,s Modulus or Modulus of Elasticity and is denoted by E.

E =Tensile stress

Tensile strainor

Compressive stress

Compressive strain

or ...(1.5)

1.7.1. Modulus of Rigidity or Shear Modulus. The ratio of shear stress to thecorresponding shear strain within the elastic limit, is known as Modulus of Rigidity or ShearModulus. This is denoted by C or G or N.

Shear stress tC (or G or AO =

Shear strain <j)

Let us define factor of safety also.

...(1.6)

1.8. FACTOR OF SAFETY

It is defined as the ratio of ultimate tensile stress to the working (or permissible) stress.Mathematically it is written as

Ultimate stressFactor of safety =

Permissible stress..(1.7)

1.9. CONSTITUTIVE RELATIONSHIP BETWEEN STRESS AND STRAIN

1.9.1

. For One-Dimensional Stress System. The relationship between stress andstrain for a unidirectional stress (i.e., for normal stress in one direction only) is given byHooke*s law, which states that when a material is loaded within its elastic limit, the normal

stress developed is proportional to the strain produced. This means that the ratio of the normal

6


stress to the corresponding strain is a constant within the elastic limit. This constant isrepresented by E and is known as modulus of elasticity or Young

,s modulus of elasticity.Normal stress

= Constant orCorresponding strain

where o = Normal stress, e = Strain and E = Young,

s modulus

e

ora

C~

E...

[1.7 (A))

The above equation gives the stress and strain relation for the normal stress in onedirection.

1.9.2. For Two-Dimensional Stress System. Before knowing the relationship between

stress and strain for two-dimensional stress system, we shall have to define longitudinalstrain, lateral strain, and Poisson,s ratio.

1. Longitudinal strain. When a body is subjected to an axial tensile load, there is anincrease in the length of the body. But at the same time there is a decrease in other dimensionsof the body at right angles to the line of action of the applied load. Thus the body is havingaxial deformation and also deformation at right angles to the line of action of the applied load(i.e., lateral deformation).

The ratio of axial deformation to the original length of the body is known as longitudinal(or linear) strain. The longitudinal strain is also defined as the deformation of the body perunit length in the direction of the applied load.

Let L = Length of the body,

P = Tensile force acting on the body,

8L = Increase in the length of the body in the direction of P.

Then, longitudinal strain = -.Lj

2. Lateral strain. The strain at right angles to the direction of applied load is knownas lateral strain. Let a rectangular bar of length L, breadth b and depth d is subjected to anaxial tensile load P as shown in Fig. 1.5. The length of the bar will increase while the breadthand depth will decrease.Let 8L = Increase in length,

56 = Decrease in breadth, and

6d = Decrease in depth.

Then longitudinal strain = ÿ ...11.7 (fill

and lateral strain = ÿb

or ...[1.7 (C)]

7


Note. (i) If longitudinal strain is tensile, the lateral strains will be compressive.

(ii) If longitudinal strain is compressive then lateral strains will be tensile.(iii) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of

the opposite kind in all directions perpendicular to the load.3. Poisson,s ratio. The ratio of lateral strain to the longitudinal strain is a constant

for a given material, when the material is stressed within the elastic limit. This ratio is calledPoisson's ratio and it is generally denoted by [i. Hence mathematically,

. . Lateral strain ,, _ /r>N.Poisson s ratio, n = --:---:-- ...11.7 (D)]Longitudinal strain

or Lateral strain = ji x Longitudinal strainAs lateral strain is opposite in sign to longitudinal strain, hence algebraically, lateral

strain is written as

Lateral strain = -[ix Longitudinal strain4. Relationship between stress and strain. Consider a

two-dimensional figure ABCD, subjected to two mutuallyperpendicular stresses a, and a2.

Refer to Fig. 1.5 (a).

Let o, = Normal stress in x-directiona2 = Normal stress in y-direction

Consider the strain produced by <j2.The stress a

, will produce strain in the direction ofx andalso in the direction ofy. The strain in the direction ofx will be

C, .... j 5

...11.7 <E)1

I<*1

° JL'%

longitudinal strain and will be equal to -J- whereas the strainE

in the direction of y will be lateral strain and will be equal to - n x

(v Lateral strain. =-[ix longitudinal strain)

Now consider the strain produced by o2.The stress o2 will produce strain in the direction ofy and also in the direction of*. The

strain in the direction of y will be longitudinal strain and will be equal to ~r whereas thehj

strain in the direction of x will be lateral strain and will be equal to - ÿ x ÿ .E

Let e, = Total strain in x-directione2 = Total strain in y-direction

Now total strain in the direction of x due to stresses a, and o., = - n -ÿE

Similarly total strain in the direction ofy due to stresses o, and o2 = - \xE E

1 E M E -11.7 (F)l

Oo Oi

eo = - - n .

E E...(1.7 (G))

8


The above two equations gives the stress and strain relationship for the two-dimensionalstress system. In the above equations, tensile stress is taken to be positive whereas thecompressive stress negative.

1.9.3. For Three-Dimensional Stress System. Fig. 1.5 (6) shows a three-dimensionalbody subjected to three orthogonal normal stresses op o2, o;) acting in the directions of x, yand 2 respectively.

Consider the strains produced by each stressseparately.

The stress a, will produce strain in the direction of

x and also in the directions of y and z. The strain in the

direction of x will be - whereas the strains in the directionE°i

ofy and z will be - nE

Similarly the stress c2 will produce strain -f inE

o2

Y,

f1/

j*

/ X

Fig. 1.5 (b)the direction of y and strain of - p in the direction of x

Eand y each.

Also the stress a3 will produce strain in the direction of 2 and strain of - p x inthe direction of x and y.

E E

Total strain in the direction of x due to stresses a., a, and a, = - - li - - u -.12 3 E E E

Similarly total strains in the direction of y due to stresses Oj, o2 and o3

E M E * E

and total strains in the direction ofz due to stresses olf a2 and a3

O 3 O j G 2

Let e.f e2 and e3 are total strains in the direction of x, y and z respectively. Then

G\ O o Oqp - _L - ii _£ _ || _iE E * E

Oo Oo a i

£ÿ= e-*

E

and e - °3 „ °1 u °2

3 E E E

...[1.7 (H) 1

...(1.7 </)!

...11.7 («/)]

The above three equations give the stress and strain relationship for the three orthogonalnormal stress system.

Problem 1.1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of20 kN. If the modulus of elasticity of the material of the rod is 2 x 105 N/ mm2 ; determine :

(i) the stress,(ii) the strain, and

(iii) the elongation of the rod.

9


Sol. Given : Length of the rod, L = 150 cm

Diameter of the rod, D = 2.0 cm = 20 mm

Area, A = - (20)2 = IOOji mm24

Axial pull, P = 20 kN = 20,000 N

Modulus of elasticity, E = 2.0 x 10& N/mm2(i) The stress (a) is given by equation (1.1) as

a = P = 20000 = 63 662 N/mm2> Ans

A IOOtu

(ii) Using equation (1.5), the strain is obtained as

£=

°.

e

a 63.662Strain, e = - = ~--g- = 0.000318. Ans.

ÿ Z X iu

(iii) Elongation is obtained by using equation (1.2) asdL

e~

L '.

*

. Elongation, dL-ex L= 0

.000318 x 150 = 0.0477 cm. Ans.

Problem 1.2. Find the minimum diameter of a steel wire, which is used to raise a loadof4000 N if the stress in the rod is not to exceed 95 MN/m2

.

Sol. Given : Load, P = 4000 N

Stress, a = 95 MN/m2 = 95 x 106 N/m2 (v M = Mega = 106)= 95 N/mm2 (v 106 N/m2 = 1 N/mm2)

Let D = Diameter of wire in mm

Area, A = - D24

M . Load PNow stress =-= -

Area A

_ 4000 4000x4 _ 4000 x495=-=--ÿ- or DZ =---=53.61

K D2 71 D K x 954

D = 7.32 mm. Ans.

Problem 1.3. Find the Young,s Modulus of a brass rod of diameter 25 mm and oflength 250 mm which is subjected to a tensile load of 50 kN when the extension of the rodis equal to 0.3 mm.

Sol. Given : Dia. of rod, D = 25 mm

/. Area of rod, A = - (25)2 = 490.87 mm24

Tensile load, P = 50 kN = 50 x 1000 = 50,000 N

Extension of rod, dL = 0.3 mm

Length of rod, L = 250 mm

10

SIMPLE STRESSES ANI) STRAINS

Stress (o) is given by equation (1.1), as

o = 4 = = 101-86 N/mm2.A 490.87

Strain (e) is given by equation (1.2), as

e = ÿ = ÿ3 =0.0012.

L 250

Using equation (1.5), the Young,s Modulus (E) is obtained, as

E _ Stress = 101.86 N/mm' = 84883 33 N/mm2Strain 0.0012

= 84883.33 x 10f N/m2. Ans. (v 1 N/mm2 = 106 N/m2)= 84

.883 x 109 N/m2 = 84.883 GN/ma. Ans. (v 109 = G)

Problem 1.4. A tensile test was conducted on a mild steel bar. The following data wasobtained from the test:

(i) Diameter of the steel bar = 3 cm

(ii) Gauge length of the bar = 20 cm(iii) Load at elastic limit = 250 kN

(iv) Extension at a load of 150 kN = 0.21 mm(u) Maximum load = 380 kN

(ui) Total extension = 60 mm

(vii) Diameter of the rod at the failure = 2.25 cm.

Determine : (a) the Young's modulus, (b) the stress at elastic limit,

(c) the percentage elongation, and (d) the percentage decrease in area.

Sol. Area of the rod, A = - D2 = - (3)2 cm24 4

= 7.0685 cm2 = 7.0685 x 10"» m2. 2 | 1

,2

cm = -m100

(a) To find Young,s modulus, first calculate the value of stress and strain within elasticlimit. The load at elastic limit is given but the extension corresponding to the load at elasticlimit is not given. But a load of 150 kN (which is within elastic limit) and correspondingextension of 0.21 mm are given. Hence these values are used for stress and strain withinelastic limit

- Load 150x1000 9 , . XT i/w\xt\Stress = --=-7- N/m2 (v 1 kN = 1000 N)Area 7.0685 xlO"4

and Strain =

= 21220.

9 x 104 N/m2

Increase in length (or Extension)Original length (or Gauge length)

0.21mm

20 x 10 mm= 0.00105

Young,s Modulus,

£ = Stress = 21220.9 x 10 = y 1q4 N/m2

Strain 0.00105

11


= 202.095 x 109 N/m2 (v 109 = Giga = G)

= 202.095 GN/m2. Ans.

(6) The stress at the elastic limit is given by,

Load at elastic limit 250 x 10001reSS * Area

~

7.0685 x 10"4= 35368 x 104 N/m2

= 353.68 x 106 N/m2 (v 106 = Mega = M)

= 353.68 MN/m2. Ans.

(c) The percentage elongation is obtained as,

Percentage elongation

Total increase in length

Original length (or Gauge length)60 mm

=- x 100 = 30%. Ans.20 x 10 mm

(d) The percentage decrease in area is obtained as,

Percentage decrease in area

(Original area - Area at the failure)

x 100

IOriginal area

* X 32 - X 2.252

4 ' x 100

x 100

* x<t2

4

' 32 - 2.2523i

x 100 = ,9 5 0625, x 100 = 43.75%. Ans.9

Problem 1.5. The safe stress, for a hollow steel column which carries an axial load of2.1 x 103 kN is 125 MN/m2. If the external diameter of the column is 30 cm, determine the

internal diameter.

Sol. Given :

Safe stress*, o = 125 MN/m2 = 125 x 106 N/m2

Axial load, P = 2.1 x 103 kN = 2.1 x 106 N

External diameter, D = 30 cm = 0.30 m

Let d = Internal diameter

/. Area of cross-section of the column,

A = - (D2 -d2) = - (,30i - d!) m24 4

PUsing equation (1.1), o = -

A

*Safe stress is a stress which is within elastic limit.

12


or

or

125 x 106 =2

.1 x 1Qf

- (.302 - di)

4

or (.302 - d2) =

4 x 2.1 x 10f

k x 125 x 106

0.09 - d2 = 213.9 or 0.09 - 0.02139 = d2

d = ÿ0.09 - 0.02139 = 0.2619 m = 26.19 cm. Ans.Problem 1.6. The ultimate stress

, for a hollow steel column which carries an axial loadof 1.9 MN is 480 N/mm2. If the external diameter of the column is 200 mm, determine theinternal diameter. Take the factor of safety as 4.

Sol. Given :

= 480 N/mm2

P = 1.9 MN = 1.9 x 106 N

= 1900000 N

D = 200 mm

= 4

d = Internal diameter in mm

or

Ultimate stress

Axial load, (v M = 10f)

External dia.,

Factor of safetyLet

Area of cross-section of the column,

A = - (D2 - d2) = - (2002 - d2) mm24 4

Using equation (1.7), we get

Factor of safetyUltimate stress

4 =

Working stress or Permissible stress480

or Working stress

Working stress

-= 120 N/mm2

o = 120 N/mm2

Now using equation (1.1), we getP

a = - or 120 =1900000 1900000 x 4

- (2002 - di) *<40000 " d >4

40000 - d? = 19°°°°? X 4 = 20159.6

7cx 120

or d2 = 40000 - 20159.6 = 19840.4

d = 140.85 mm. Ans.

Problem 1.7 A stepped bar shown in Fig. 1.6 is subjected to anaxially applied compressive load of 35 kN. Find the maximum andminimum stresses produced.

Sol. Given :

Axial load, P = 35 kN = 35 x 103 N

Dia. of upper part, D, = 2 cm = 20 mm

35 kN

\2 cm

DIA

3 em

"

777

DIA///////////7*r7~

Fig. 1.6


Area of upper part, A, = (202) = 100 7t mm2

Area of lower part, A2 = ÿ D2i = (302) = 225 Jt mmiThe stress is equal to load divided by area. Hence stress will be maximum where area

is minimum. Hence stress will be maximum in upper part and minimum in lower part.

Load 35 x 103/. Maximum stress

Minimum stress

A, 100 X 71

Load 35 x 10

A2 225 x k

= 111.408 N/mma. Ans.

= 49.5146 N/mm2. Ans.

1.10. ANALYSIS OF BARS OF VARYING SECTIONS

A bar of different lengths and of different diameters (and hence of different crosssectional areas) is shown in Fig. 1.6 (a). Let this bar is subjected to an axial load P.

Section 3Section 2

Fig. 1.6(a)

Though each section is subjected to the same axial load P, yet the stresses, strains andchange in lengths will be different. The total change in length will be obtained by adding thechanges in length of individual section.

Let P

1ÿ2»A2ÿ3« A3

E

Axial load acting on the bar,

Length of section 1,

Cross-sectional area of section 1,

Length and cross-sectional area of section 2,Length and cross-sectional area of section 3, and

Young,s modulus for the bar.Then stress for the section 1,

a.=

Load

Area of section 1 A,

Similarly stresses for the section 2 and section 3 are given as,

P A Pa. = - ana a.. = -

Using equation (1.5), the strains in different sections are obtained.

Strain of section 1, e, = ÿ

2 A,Ea. =

Ai,

14


Similarly the strains of section 2 and of section 3 are,

and e- = - =

or

e2 =

But strain in section 1 =

=

E A,EE A2E "3Change in length of section 1

Length of section 1dL,

L,

where dLx = change in length of section 1./. Change in length of section 1, c/L, = e,L,

PL iAXE

Similarly changes in length of section 2 and of section 3 are obtained as

Change in length of section 2, dL2 = e2 L,,

-

a2e

and change in length of section 3, rfL3 = eaLr,

AaE

6i =A,E

e2 =A.E

*3 =AaE

Total change in the length of the bar,

dL = dL.+ dL2 + dL, =PL, Pin PL-- + -- + -

AXE A2E A:,E

P L, L9 Lo-e[a;+

a;+

a-

3 .a8,

Equation (1.8) is used when the Young,s modulus of different sections is same. If theYoung,s modulus of different sections is different, then total change in length of the bar isgiven by,

dL = P*

»» , L2 ÿ A,

...(1-9 >ÿ1ÿ1 2 "2 ÿ-3 As

Problem 1.8. An axial pull of35000 N is acting on a bar consisting of three lengths asshown in Fig. 1.6 (b). If the Young,s modulus =2.1 x 105 N/mm

'

~\ determine :(i) stresses in each section and

(ii) total extension of the bar.

Section 3

20 cm 25 cm 22 cm

Fie. 1.6 «/.>

15


Sol. Given :

Axial pull,

Length of section 1,Dia. of section 1,

.

*

. Area of section 1,


Area of section 2,


.

*

. Area of section 3,

Young,s modulus,

(i) Stresses in each section.

Stress in section 1,

p = 35000 N

= 20 cm = 200 mm

= 2 cm = 20 mm

A, - (202) = 100 it mm4

*ÿ2= 25 cm = 250 mm

d2 ZZ 3 cm = 30 mm

= -7 (302) = 225 71 mm4

LZ =

*

22 cm = 220 mm

d3 s 5 cm = 50 mm

ÿ3 = - (502) = 625 7i mm4

E - 2.1 x 10& N/mm2.

Axial loado. = -

Area of section 1

P 35000= -- = --- = 111.408 N/mm2. Ans.

Stress in section 2,

Stress in section 3.

A, 100 7iP 35000

A2 225 x it

35000=

3 A3 625 x k

(ii) Total extension of the bar


Total extension

= 49.5146 N/mm2. Ans.

= 17.825 N/mm2. Ans.

P

E

L

Al +h.+h1 "2 "3 .

35000

2.1xl0&

35000

2.1x 10

200 250+-+

220

v 100 k 225 X7U 625X71,

r (6.366 + 3.536 + 1.120) = 0.183 mm. Ans.

Problem 1.9. A member formed by connecting a steel bar to an aluminium bar is shownin Fig. 1.7. Assuming that the bars are prevented from buckling sideways, calculate themagnitude of force P that will cause the total length of the member to decrease 0.25 mm. Thevalues of elastic modulus for steel and aluminium are 2.1 x 105 N/mm2 and 7 x 104 N/mm2respectively.

Sol. Given :

Length of steel bar, L. = 30 cm = 300 mm

16


Area of steel bar, A, = 5 x 5 = 25 cm2 = 250 mm2Elastic modulus for steel bar,

E, = 2.1 x 10® N/mm2Length of aluminium bar,

5 cm x 5 cm

Steel bar

L2 = 38 cm = 380 mm

Area of aluminium bar,

A2 =

Elastic modulus for aluminium bar,A2 = 10 x 10 = 100 cm2 = 10000 mm2

10 cm x 10 cmAluminium bar

Fig. 1.7

E2 = 7 x 104 N/mm2Total decrease in length, dL = 0.25 mm

Let P = Required force.As both the bars are made of different materials

, hence total change in the lengths ofthe bar is given by equation (1.9).

"

L, ,

L2 .

or

dL = P

0.25 = P

E\A\+

E2A2

300 380

2.1 x 10S x 2500 7 x 104 x 10000

X *

= P (5.714 X 10"7 + 5.428 x 10~7) = P x 11.142 x 10"7

P =0

.25

v-7

0.25 x 10

11.14211.142x10

= 2.2437 x 10& = 224.37 kN. Ans.

Problem 1.10. The bar shown in Fig. 1.8 is subjected to a tensile load of 160 kN. Ifthe stress in the middle portion is limited to 150 N/mm2, determine the diameter of themiddle portion. Find also the length of the middle portion if the total elongation of the baris to be 0.2 mm. Young,s modulus is given as equal to 2.1 x 105 N/mm2.

Sol. Given :

P = 160 kN = 160 x 103 NTensile load,

Stress in middle portion,Total elongation,

Total length of the bar,Young's modulus,

Diameter of both end portions, D, = 6 cm = 60 mmArea of cross-section of both end portions,

o2 = 150 N/mm2dL = 0.2 mm

L = 40 cm = 400 mm

E = 2.1 x 10f N/mm2

. = 7X60- 900 7t mm2.

160 kN t f6 cm DIA

1

1

6 cm DIA1

160 kN

40 cm

Fig. 1.8

17


or

or

Let D2 = Diameter of the middle portionL2 = Length of middle portion in mm.

.\ Length of both end portions of the bar,Lj = (400 - L2) mm

Using equation (1.1), we have

LoadStress = --.

Area

For the middle portion, we haveP

- K

or 150 =

On = - where A., = - D 2

2

160000

2 -

4-2

- Do2

4x160000 loco o

D'1 =--- = 1358 mm2

2 ji x 150

or D2 = Vl358 = 36.85 mm = 3.685 cm. Ans.Area of cross-section of middle portion,

A„= - x 36

.85 = 1066 mm2

® 4Now using equation (1.8), we get

PL LTotal extension, dL = -

E A i A2

160000 ["

(400-L2) | L2 I" = 2.1 x 10® L 900tt + 1066 J

[ v L. = (400 - L.,) and A2 = 1066]0.2 x 2.1x 10& (400-L2) | L2

160000 900 tc 1066

1066(400 - L2) + 9007C L2900 71 x 1066

or 0.2625 x 900ti x 1066 = 1066 x 400 - 1066 L2 + 900;: x L2

or 791186 = 426400 - 1066 L2 + 2827 L2or 791186 - 426400 = L2 (2827 - 1066)or 364786 = 1761 L

2

L, = = 207.14 mm = 20.714 cm. Ans.

1761

1.10.1. Principle of Superposition. When a number of loads are acting on a body,the resulting strain, according to principle of superposition, will be the algebraic sum of strainscaused by individual loads.

While using this principle for an elastic body which is subjected to a number of directforces (tensile or compressive) at different sections along the length of the body, first the freebody diagram of individual section is drawn. Then the deformation of the each section isobtained. The total deformation of the body will be then equal to the algebraic sum ofdeformations of the individual sections.

18


Problem 1.11. A brass bar, having cross-sectional area of 1000 mm2, is subjected toaxial forces as shown in Fig. 1.9.

50 kN

600 mm1 m 1

.20 m

_

F"8 1®Find the total elongation of the bar. Take E = 1.05 x 105 N/mm2.Sol. Given :

Area, A = 1000 mm2Value of E = 1.05 x 105 N/mm2

Let dL = Total elongation of the bar.The force of 80 kN acting at B is split up into three forces of 50 kN, 20 kN and 10 kN.

Then the part AB of the bar will be subjected to a tensile load of 50 kN, part BC is subjectedto a compressive load of 20 kN and part BD is subjected to a compressive load of 10 kN asshown in Fig. 1.10.

B

Fig. 1.10

Part AB. This part is subjected to a tensile load of 50 kN. Hence there will be increasein length of this part.

Increase in the length of AB

.AxL-

«

50 x1000x 600 (v P, = 50,000 N, L, = 600 mm)1000 x 1.05 x 10f

= 0.2857.

Part BC. This part is subjected to a compressive load of 20 kN or 20,000 N. Hencethere will be decrease in length of this part.

Decrease in the length of BC

20,000_ 2 x T __

AE 2 1000 x 1.05 xlO&

= 0.1904.

x 1000 (v L2 = 1 m = 1000 mm)

19


Part BD. This part is subjected to a compressive load of 10 kN or 10,000 N. Hencethere will be decrease in length of this part.

Decrease in the length of BD

Po . 100003 x I __

AE 2 1000 x 1.05 xlO5= 2200

(v L3 = 1.2 + 1 = 2.2 m or 2200 mm)= 0.2095

.

.

*

. Total elongation of bar = 0.2857 - 0.1904 - 0.2095

(Taking +ve sign for increase in length and-ve sign for decrease in length)

= - 0.1142 mm. Ans.

Negative sign shows, that there will be decrease in length of the bar.

Problem 1.12. A member ABCD is subjected to point loads Pv P# P3 and P4 as shown

in Fig. 1.11.

B C;

OJ

;

A T ,1250 mm

1* 2

625 mmÿ

CM

*

h 120 cm 60 cm

Fig. 1.11

90 cm -ÿ]

Calculate the force P2 necessary for equilibrium, if P, = 45 kN, P3 = 450 kN andP4 = 130 kN. Determine the total elongation of the member, assuming the modulus ofelasticity to be 2.1 x 105 N/mm2

.

Sol. Given :

Part AB :

Part BC

Part CD

Value of

Value of P2 necessary for equilibrium

Area, Al - 625 mm2 andLength, Lj = 120 cm = 1200 mmArea, A2 = 2500 mm2 andLength, L2 = 60 cm = 600 mmArea, A3 = 1250 mm2 andLength, L3 = 90 cm = 900 mm

E = 2.1 x 105 N/mm2.

Resolving the forces on the rod along its axis (i.e., equating the forces acting towardsright to those acting towards left), we get

P1 + P3 = P2 + P4

20


But P. = 45 kN,

P3 = 450 kN and P, = 130 kN

45 + 450 = P2 + 130 or P2 = 495 - 130 = 365 kNThe force of 365 kN acting at B is split into two forces of 45 kN and 320 kN {i.e., 365 -

45 = 320 kN).

The force of 450 kN acting at C is split into two forces of 320 kN and 130 kN (i.e., 450 -320 = 130 kN) as shown in Fig. 1.12.

From Fig. 1.12, it is clear that part AB is subjected to a tensile load of 45 kN, part BCis subjected to a compressive load of 320 kN and part CD is subjected to a tensile load 130 kN.

45 kN jV E

45 kN

4 1 1

320 kN 320 kN

130 kN

C D

;_

fÿ. i.i2 _ _JHence for part AB, there will be increase in length ; for part BC there will be decrease

in length and for part CD there will be increase in length.

Increase in length of AB

A,Ex L. =

45000

625 x 2.1 x 10'*x 1200 (v P = 45 kN = 45000 N)

= 0.4114 mm

Decrease in length of BCP

AoEx Lo =

320,000

2500 x 2.1 x 10&

= 0.3657 mm

x 600 (v P = 320 kN = 320000)

Increase in length of CDP 130.000

AZEx 900 < v P = 130 kN = 130000)

1250 x 2.1 x 10&

= 0.4457 mm

Total change in the length of member= 0.4114 - 0.3657 + 0.4457

(Taking +ve sign for increase in length and-ve sign for decrease in length)

= 0.4914 mm (extension). Ans.

Problem 1.13. A tensile load of 40 kN is acting on a rod of diameter 40 mm and oflength 4 m. A bore of diameter 20 mm is made centrally on the rod. To what length the rod

21


should be bored so that the total extension will increase 30% under the same tensile load. Take

E = 2 x 105 N/mm2.

Sol. Given :--

40 kN 40 kN

N-4 m-H

Fig. 1.12 (a)

Tensile load,

Dia. of rod,

P = 40 kN = 40,000 N

D = 40 mm

nArea of rod, A = - (403) = 40071 mm-

4

E

4 m ÿ

Length of rod,Dia. of bore,

Area of bore.

Fig. 1.12(6)

L = 4 m = 4 x 1000 = 4000 mm

d = 20 mm

a = - x 202 = 100 7i mm24

Total extension after bore = 1.3 x Extension before bore

Value of E = 2 x 105 N/mm2

Let the rod be bored to a length of x metre or x x 1000 mm. Then length of unboredportion = (4 - x) m = (4 - x) x 1000 mm. First calculate the extension before the bore is made.

The extension (6L) is given by,

5L = -f- x L =AE 400ti x 2 x 10"

40000 x 4000 2= - mm

71

...(i)

Now extension after the bore is made

= 1.3 x Extension before bore

1 o 2 26= 1

.3 x - =-mm

71 71

The extension after the bore is made, is also obtained by finding the extensions of theunbored length and bored length.

For this, find the stresses in the bored and unbored portions.Stress in unbored portion

= Load P = 40000 = 100

"

A "N/mm2

Area A 40071

Extension of unbored portionStress

-

Ex Length of unbored portion

22


JQ0 x(4-*)x 1000 = mm71 x 2 x 10® 2ti

Stress in bored portion

Load P 40000 = 40000

Area (A-a) (400ti - IOOti) 300tt

/. Extension of bored portion

Stress

Ex Length of bored portion

4QQ0Q 1AA, 4*x 1000a: = - mm

300tc x 2 x 10® 6tt

Total extension after the bore is made(4-*) 4x ....

+ - ...(«)2tt 67t

Equating equations (i) and (ii),2

.6 _

4 - x 4x71 271 671

or 2.6 = -ÿ- + ÿ- or 2.6 x 6 = 3 x (4-x) + 4x2 6

or 15.6 = 12 - 3* + 4x or 15.6 - 12 = x or 3.6 = x

.

,

. Rod should be bored upto a length of 3.6 m. Ans.

Problem 1.14. A rigid bar ACDB is hinged at A and supported in a horizontal positionby two identical steel wires as shown in Fig. 1.12 (c). A vertical load of 30 kN is applied at B.Find the tensile forces T. and T., induced in these wires by the vertical load.

E F

1 T,

I,T! '

lm

(IBlA C D

W- 1 m ÿ 4 1 m 1 m ->

30 KN

Fig. 1.12 (c)

AC D B

Fig. 1.12(d)

Sol. Given :

Rigid bar means a bar which will remain straight.Two identical steel wires mean the area of cross-sections, lengths and value of E for

both wires is same.

Ay = A2, Ex = E2 and L1 = L2Load at B = 30 kN = 30,000 N

Fig. 1.12 (c) shows the position of the rigid bar before load is applied at B. Fig. 1.12 (d)shows the position of the rigid bar after load is applied.

23


as

Let Tx = Tension in the first wireT2 = Tension in the second wire51 = Extension of first wire82 = Extension of second wire

Since the rigid bar remains straight, hence the extensions 8, and 82 are given by6

_

l AC = 1

82 AD ~

2

28, = 82 ...(i)But 8, is the extension in wire EC

(ry x LiStress in EC x Lx U

,

" r,xz,,

1 Ej Ex Aj x E,

Sjmi]ar1J

Substituting the values of 8, and 82 is equation (i),

2 x ÿ _

ÿ2 X ÿ2A, x i?j A2 x E2

But A, = A2, = E2 and L, = Lr Hence above equation becomes

2 T1 = T2 ...(H)Now taking the moments of all the forces on the rigid bar about A, we get

7\ x 1 + T2 x 2 = 30 x 3

or T, + 2 T2 = 90 ...(m)

Substituting the value of T2 from equation (zi), into equation (m), we getT

, + 2(2T,) = 90 or 5T, = 90

T1 = - = 18 kN. Ans.

From equation (ii),

T2 = 2 x 18 = 36 kN. Ans.\nt««. After calculating the values of T, and Tv the stresses in the two wires can also be obtained

Stress in wire EC = ÿQa{' ÿ

T*

and Stress in wire FD = -r~.

Area A,

sA>l

1.11. ANALYSIS OF UNIFORMLY TAPERING CIRCULAR ROD

A bar uniformly tapering from a diameter D, at one end to a diameter D2 at the otherend is shown in Fig. 1.13.

Let P = Axial tensile load on the bar

L = Total length of the bar

E = Young,s modulus.

24


Fig. 1.13

Consider a small element of length dx of the bar at a distance x from the left end. Letthe diameter of the bar be D at a distance x from the left end,

ThenD, -D

= D} - kx where k =

Dl-D

Area of cross-section of the bar at a distance x from the left end,

4. = f = f (D, - k

.x)\

Now the stress at a distance x from the left end is given by,

Load°x =

4 P

liDÿk.x)*

4

The strain er in the small element of length dx is obtained by using equation (1.5).

Stress

E E

4 P_

1 _

4 P

n{Dx-k.x)2 E n E(Dl - k.x)

Extension of the small elemental length dx= Strain

, dx = e

x. dx

4 P

nE(Dt-k.x)2 .dx Ui)

Total extension of the bar is obtained by integrating the above equation between thelimits 0 and L

.

25


Total extension,

dL = r *P dX

Jo 7T E(D, - k.

4 P cL

x) nE j (Dl-k.xrZ.dx

4r r-

7t£ Jo-2

4P f/- (D,x<-A)(-*>

. [Multiplying and dividing by (- A)J

4P

nE

4P

tiE*

4P

-lWx-k.x)(- 1) x (- k)

4 P

nEk (D,-k

.x)

1_

1

Dl- k.L D{- kx 0

1 1

nEk

Substituting the value of £ =

Total extension,

Dt-k.L Dy

D\- Do .

in the above equation, we get

dL =4 P

nE,'Oi-D2j Dl- ajau *4PL

nE .{Dx- D2)

4PL

nEAD1-D2)APL

Dx - Z)j + D2 Dx

1 1

d2 d,

(D, - D2) 4 PL...(1.10*

nEADx-D2) D\D2 nEDlD2If the rod is of uniform diameter, then D

, - D2 = D

Total extension, dL = -Q ...(1.11)

nE .D

Problem !. 15. A rod, which tapers uniformly from 40 mm diameter to 20 mm diameter

in a length of400 mm is subjected to an axial load of5000 N. IfE = 2.1 x 105 N/mm2, find theextension of the rod.

Sol. Given :

Larger diameter,Smaller diameter,

Length of rod, L = 400 mmAxial load, P = 5000 N

Young's modulus, E = 2.1 x 10& N/mm'-'Let dL = Total extension of the rod

D, = 40 mm

D.. = 20 mm

26


Using equation (1.10),

4 PLdL =

4 x 5000 x 400

n E DxD2 7i x 2.1 x 10& x 40 x 20

= 0.01515 mm. Ans.

Problem 1.1 tt. Find the modulus of elasticity for a rod, which tapers uniformly from30 mm to 15 mm diameter in a length of 350 mm. The rod is subjected to an axial load of5

.5 kN and extension of the rod is 0.025 mm.Sol. Given :

Larger diameter, Dl =30 mmSmaller diameter, D., = 15 mm

Length of rod, L = 350 mm

Axial load, P = 5.5 kN = 5500 N

Extension, dL = 0.025 mm

Using equation (1.10), we get4 PL

or

dL =

E =

tc E Z)j D24 PL 4 x 5500 x 350

7i D,D2 dL 7i x 30 x 15 x 0.025= 217865 N/mm2 or 2.17865 x 10S N/mm

2. Ans.

1.12. ANALYSIS OF UNIFORMLY TAPERING RECTANGULAR BAR

A bar of constant thickness and uniformly tapering in width from one end to the otherend is shown in Fig. 1.14.

Fig. 1.14


Let P = Axial load on the bar

L = Length of bara = Width at bigger endb = Width at smaller end

E = Young,s modulust = Thickness of bar

Consider any section X-X at a distance x from the bigger end.Width of the bar at the section X-X

(a - b)x=a-~r

~

= a - kx

Thickness of bar at section X-X = t

Area of the section X-X

= Width x thickness

= (a - kx)tStress on the section X-X

Load _

PArea

~

(a - kx)t

Extension of the small elemental length dx

= Strain x Length dx

Stress

where k =a-b

Exdx Strain =

Stress \

E J

(a - kx)t

E

P

x dx Stress =(a - kx)t J

axE(a - kx)t

Total extension of the bar is obtained by integrating the above equation between thelimits 0 and L.

Total extension,

P P ri dxdL -j:_,i

_

f1E(a - kx)t aX~ Et fJo (a - kx)

P .= YtA0g< (a - kx)

o

X "" i)=" ~£k1,0gf ,f ÿkL) ÿ'°g? f1p

Etkllogt. a - log{1 (a - kL)J = Etk

loge a - kL

Et ¥)log,

a

a-lÿl L ; *=¥)

mm


PL . a

Problem 1.17. A rectangular bar made of steel is 2.8 m long and 15 mm thick. The rodis subjected to an axial tensile load of 40 kN. The width of the rod varies from 75 mm at oneend to 30 mm at the other. Find the extension of the rod if E = 2 x 105 N/mm2.

Sol. Given :

Length, L = 2.8 m = 2800 mm

Thickness, t =15 mm

Axial load, P = 40 kN = 40,000 N

Width at bigger end, a = 75 mm

Width at smaller end, b =30 mmValue of E = 2 x 105 N/mm2

Let dL = Extension of the rod.


,, PL , adL = EtUÿb) gf

~

b

40000 x 2800 . ( 75log,2 x 105 x 15(75 - 30) V30

= 0.8296 x 0.9163 = 0.76 mm. Ans.

Problem 1.18. The extension in a rectangular steel bar of length 400 mm and thickness10 mm, is found to be 0.21 mm. The bar tapers uniformly in width from 100 mm to 50 mm. IfE for the bar is 2 x 10& N/mm2, determine the axial load on the bar.

Sol. Given :

Extension, dL = 0.21 mm

Length, L = 400 mm

Thickness, t = 10 mm

Width at bigger end, a = 100 mm

Width at smaller end, b = 50 mm

Value of E = 2 x 105 N/mm2

Let P = axial load.


PL . (adL . EtuTÿb) ®" U

0.21= --log, (-2x 10f x 10(100 - 50) K 50

= 0.000004 P x 0.6931

P = -- = 75746 N0

.000004 x 0.6931

= 75.746 kN. Ans.

29


1.13. ANALYSIS OF BARS OF COMPOSITE SECTIONS

mi/m w//////

Fig. 1.15

A bar, made up of two or more bars of equal lengths butof different materials rigidly fixed with each other and behavingas one unit for extension or compression when subjected to anaxial tensile or compressive loads, is called a composite bar.For the composite bar the following two points are important:

1. The extension or compression in each bar is equal.Hence deformation per unit length i.e., strain in each bar isequal.

2. The total external load on the composite bar is equal

to the sum of the loads carried by each different material.

Fig. 1.15 shows a composite bar made up of two differentmaterials.

Let P = Total load on the composite bar,

L = Length of composite bar and also length of bars of different materials,

A, = Area of cross-section of bar 1,A2 = Area of cross-section of bar 2,

Ex = Young,s Modulus of bar 1,E2 = Young,s Modulus of bar 2,P

, = Load shared by bar 1,

P2 = Load shared by bar 2,<Tj = Stress induced in bar 1

, and

ct2 = Stress induced in bar 2.Now the total load on the composite bar is equal to the sum of the load carried by the

two bars.

...«)

Load carried by bar 1

p=pi+p

2

The stress in bar 1,Area of cross-section of bar 1

a, =AA,

or P, = a, A, ...(»)

Similarly stress in bar 2, ÿ A,

or P2 = a,A2 ...(iii)

Substituting the values of P, and P2 in equation (i), we getP = o,A1 + a2A2 ...(iv)

Since the ends of the two bars are rigidly connected, each bar will change in length bythe same amount. Also the length of each bar is same and hence the ratio of change in lengthto the original length (i.e., strain) will be same for each bar.

_ . . Stress in bar 1 a,

But strain in bar 1, = --:--7Z-i = "F~-Young s modulus of bar 1

Similarly strain in bar 2,

30

E,


But strain in bar 1 = Strain in bar 2

= ÿ- = ÿ- Uv)Ei E2

From equations (iv) and (i>), the stresses a, and a2 can be determined. By substitutingthe values of c, and a2 in equations (ii) and (iii), the load carried by different materials maybe computed.

E,

Modular Ratio. The ratio of - is called the modular ratio of the first material to the

second.

Problem 1.19 .A steel rod of 3 cm diameter is enclosed centrally in a hollow copper tubeof external diameter 5 cm and internal diameter of 4 cm. The composite bar is then subjectedto an axial pull of45000 N. If the length of each bar is equal to 15 cm, determ ine :

(i) The stresses in the rod and tube, and

(ii) Load carried by each bar.Take E for steel =2.1 x 105 N/mm2 and for copper = 1.1 x 105 N/mm2.Sol. Given :

Dia. of steel rod = 3 cm = 30 mm

.

*

. Area of steel rod,

or

A=- (30)2 = 706.86 mm2

s 4

External dia. of copper tube= 5 cm = 50 mm

Internal dia. of copper tube= 4 cm = 40 mm

/. Area of copper tube,

Ae = ÿ [502 - 402] mm2 = 706.86 mm2

Axial pull on composite bar, P = 45000 N

Length of each bar, L = 15 cmYoung,s modulus for steel, Eg = 2

.1 x 105 N/mm2

Young,s modulus for copper, Ec = 1.1 x 10f N/mm2

Ci) The stress in the rod and tubeLet os = Stress in steel

,

Ps = Load carried by steel rod,

o(. = Stress in copper, and

Pc = Load carried by copper tube.

Now strain in steel = Strain in copper

fk = ÿE

. E.

15 cm

Wn*Coppertube

P = 45000 N

Fig. 1.16

- = strainE

°f E

2.1 x 10

x a = -

C 1.1x10

5* x oc = 1

.909 o Ui)

31


LoadNow stress = -- , .*. Load = Stress x Area

Area

Load on steel + Load on copper = Total loadag x As + ar x Ac = P (v Total load = P)

or 1.909 af x 706.86 + ar x 706.86 = 45000

or ac (1.909 x 706.86 + 706.86) = 45000

or 2056.25 ac = 45000

CC< = = *1'88 NW- AnS-

Substituting the value of ct(

. in equation (i), we get

a8 = 1.909 x 21.88 N/mm2

= 41.77 N/mm2. Ans.

(ii) Load carried by each barAs load = Stress x Area

/. Load carried by steel rod,P, =

= 41.77 x 706.86 = 29525.5 N. Ans.

Load carried by copper tube,P = 45000 - 29525.5

= 15474.5 N. Ans.

Problem 1.20. A compound tube consists of a steel tube 140 mm internal diameterand 160 mm external diameter and an outer brass tube 160 mm internal diameter and

180 mm external diameter. The two tubes are of the same length. The compound tube carriesan axial load of 900 kN. Find the stresses and the load carried by each tube and the amountit shortens. Length of each tube is 140 mm. Take E for steel as 2 x 105 N/mm2 and for brassas 1 x 105 N/mm2

.

Sol. Given :Internal dia. of steel tube = 140 mm

External dia. of steel tube = 160 mm

Area of steel tube, As = (1602 - 1402) = 4712.4 mm2

Internal dia. of brass tube = 160 mm

External dia. of brass tube = 180 mm

.

-

. Area of brass tube, Ab = (1802 - 1602) = 5340.7 mm2Axial load carried by compound tube,

P = 900 kN = 900 x 1000 = 900000 N

Length of each tube, L = 140 mm

E for steel, Es = 2 x 105 N/mm2

E for brass, Eh = 1 x 105 N/mm2

Let os = Stress in steel in N/mm2 and

a. = Stress in brass in N/mm2

32


Now strain in steel = Strain in brass

ÿ6E

s Eb

E,

a = - x a8

Eh

Strain = Stress ÿE J

_

2 x 10&'

=

lxlO5Now load on steel + Load on brass = Total load

or

or

or

°s*As + °

b*Ah =2o6x4712.4 + Gh x 5340.7 =

14765.5 nL =

a, =

900000

900000

900000

900000

14765.5

(v

".(i)

Load = Stress x Area)

(v a = 2a.)

= 60.95 N/mm2. Ans.

Substituting the value of pb in equation (i), we getOs = 2 x 60.95 = 121.9 N/mm2. Ans

.

Load carried by brass tube= Stress x Area

= ohxAb = 60.95 x 5340.7 N= 325515 N = 325.515 kN. Ans

.

Load carried by steel tube= 900-325.515 = 574.485 kN. Ans

.

Decrease in the length of the compound tube= Decrease in length of either of the tubes= Decrease in length of brass tube= Strain in brass tube x Original length

ob 60.95* "

lxio'x 140 = 0.0853 mm. Ans.

Problem 1.21. Two vertical rods one of steel and the other of copper are each rigidlyfixed at the top and 50 cm apart. Diameters and lengths of each rod are 2 cm and 4 m respectively.A cross bar fixed to the rods at the lower ends carries a load of5000 N such that the cross barremains horizontal even after loading. Find the stress in each rod and the position of the loadon the bar. Take E for steel = 2 x 105 N/mm2 andE for copper = 1 x 105 N/mm2.

Sol. Given :

Distance between the rods

= 50 cm = 500 mm

Dia. of steel rod

= Dia. of copper rod

= 2 cm = 20 mm

.

"

. Area of steel rod

= Area of copper rod

= £ X (20)2 = 100 it mm24

Steel2 cm din

50 cm

Copper2 cm dia

4 0

cm

Cross bar

33


.

*

. As = Ac = 100 it mm2

Length of each rod = 4 m = 4000 mmTotal load carried by rods,

P = 5000 N

E for steel, Es = 2 x 10& N/mm2

E for copper, Ec = 1 x 105 N/mm2

Let as = Stress in steel rod and

ac = Stress in copper rod.Since the cross bar remains horizontal, the extensions of the steel and copper rods are

equal. Also these rods have the same original length, hence the strains of these rods areequal.

Strain in steel = Strain in copper

ÿ = ÿ f.. Strain =E

, Ec { E

a = tt X , ift5 X °c = 2 x ac ...(i)E„ 2 x 10&TT* X CCc =-Ec 1x10

Now load on steel rod + Load on copper rod = Total load= 5000 N

or as x As + ac x Ac = 5000 (v Load = Stress x Area)

or cts x lOOir + ac x IOOti = 5000 (v As = Ac = 100 mm2)or 2a

. x lOOrc + ac x IOOti = 5000 (v oN = 2a

.

)

or 3ac x IOOti = 5000

= 5.305 N/mm2. Ans.

f 30071

Substituting this value of a. in equation (i), we get

a.

= 2 x 5.305 = 10.61 N/mm2. Ans.

Position of the load of5000 N on cross barLet x = The distance of the 5000 N load from the copper rod (i.e., from

the right hand rod).Now first calculate the load carried by each rod.Load carried by steel rod,

PB = og x As (v Load = Stress x Area)

= 10.61 x IOOti = 3333 N

Load carried by copper rod,Pc = Total load - P

s

= 5000 - 3333 = 1667 N

Now taking the moments about the copper rod and equating the same, we get5000 x x = P8 x 50

= 3333 x 50 (v Pb = 3333)

3333 x 50x = = 33.33 cm. Ans.

5000

Problem 1.22. A load of 2 MN is applied on a short concrete column 500 mm x 500 mm.

The column is reinforced with four steel bars of 10 mm diameter, one in each corner. Find the

34


stresses in the concrete and steel bars. Take E for steel as 2.1 x 10S N/mm2 and for concrete as1.4 x 104 N/mm2.

Sol. Given :

Total load applied, P = 2 MN = 2 x 106 NArea of column - 500 x 500 = 250000 mm2

Area of 4 steel bars, As = 4 x -(10)2 = 314.159 mm2

or

or

Area of concrete,

1 Steel bars ÿ

E for steel,

E for concrete,

Let

Ac = Area of column

- Area of steel bars

= 250000-314.159

= 249685.841 mm2 ,

E8 = 2.1 x 10& N/mm2

Ec = 1.4 x 104 N/mm2

ag = Stress in steel bar in N/mm2

o„ = Stress in concrete in N/mm2

500

mm

500 mm *1

Fig. 1.18

Now strain in steel = Strain in concrete

ftE

.

=

ft

E.

Strain =

2.1xl05xar =

1.4x10

Toc = 15or

Stress 1E )

Now load on steel + Lx>ad on concrete = Total load

<VA, + °c-Ac = P15oc x 314.159 + oc x 249685.841 = 2000000

254398ac = 2000000

(v Load = Stress x Area)

(v os=15o

c)

2000000Cc = = 7.

86 N/mma. Ans.254398

Substituting this value in equation (t), we getas = 15 x 7.

86 = 117.92 N/mm2. Ans.

Problem 1.23. A reinforced short concrete column 250 mm x 250 mm in section isreinforced with 8 steel bars. The total area of steel bars is 2500 mm

'2. The column carries a

load of 390 kN. If the modulus of elasticity for steel is 15 times that of concrete, find thestresses in concrete and steel.

or

Sol. Given :

Area of concrete column

Area of steel bars,

/. Area of concrete,

Total load on column,

E for steel

Let

= 250 x 250 = 62500 mm2

As = 2500 mm2

Ac = 62500 - 2500 = 60000 mm2

P = 390 kN= 390,000 N

= 15 x E for concrete

Et = 15 Ec

= Stress in steel in N/mm2

o(

.

= Stress in concrete in N/mm2

35


Now strain in steel = Strain in concrete

or

or

_

Ec

E= TT* Gn = 15(T

E. C

Strain = Stress ÿE )

...(*)

or

or

Also, we know thatTotal load

P

390000

Load on steel + Load on concrete

(y Load = Stress x Area)= aa.A

t-a

e.A

e

= 15ac x 2500 + ac x 60000

= 97500a

(v a = 15a )

390000a = = 4.0 N/mm2. Ans.

97500

Substituting this value in equation (i), we getas = 15 x 4.0 = 60.0 N/mm2. Ans.

Area of steel required so that column may support a load of 480 kN and maximumstress in concrete is 4.5 N/mm2

.

Area of steel required62500 - A

s

*

Stress in steel in N/mm2

Stress in concrete = 4.5 N/mm2

Total load = 480 kN = 480000 N

15ac

*

15 x 4.5

67.5 N/mm2

Let

Area of concrete

We know that

A * =

=

Cc

* =

P* =

a * =

(y a * = 4.5)

or

or

or

or

Also, we know thatTotal load = Load on steel + Load on concrete

P* =

480000 =

480000 - 4.5 x 62500

198750

ÿ _

a* x A* + a.

* x A*

67.5 x A* + 4.5 x (62500 -A*)

67.5 x A* + 4.5 x 62500 - 4.5 A*

63 A*

63 A*

198750

63= 3154.76 mm2. An*.

Problem 1.24. A steel rod and two copper rods together support a load of 370 kN asshown in Fig. 1.19. The cross-sectional area of steel rod is 2500 mm2 and of each copper rodis 1600 mm2. Find the stresses in the rods. Take E for steel = 2 x 105 N/mm2 and for copper= 1 x 105 N/mm

2.

Sol. Given :

Load, P = 370 kN = 370,000 N

Area of steel rod, Ar = 2500 mm2

36

SIMPLE STRESSES ANI) STRAINS

Area of two copper rods, Ac = 2 x 1600

= 3200 mm2

E for steel,

E for copper,

Length of steel rod,

E. = 2 x 105 N/mm2

Ec

= 1 x 105 N/mm2

Length of copper rods,

La =15+10

= 25 cm = 250 mm

L = 15 cm = 150 mm

Let ag = Stress in steel rod in N/mm2

o(. = Stress in copper rods in N/mm2

We know that decrease in the length of steel rodis equal to the decrease in the length of copper rods.

But decrease in length of steel rod

= Strain in steel rod x Length of steel rod

Stress in steel

E.

x L.

370 kN

15

cm

1I

1

Copper

10cm

,

I§5

i

ii

77ÿÿ77

Steel

T7

i

I

i

Copper

7T

Fig. 1.19

Strain =Stress

E

= . X 250

2 x 10

Similarly decrease in length of copper rods

= Strain in copper roods x Length of copper rods

Stress in copper=---x L

Ec

...(»)

lx 10,x 150

.Mi)

Equating the decrease in length of steel rod to the decrease in the length of copperrods, we get

x 250 = x 150

or

or

or

or

2 x 10f 1 x 10&

2 x 10® 150= * ÿ50 = L2°

c

Also, we know that

Load on steel + Load on copper = Total load appliedas x Aa + ac x Ac = P

1.2o x 2500 + a x 3200 = 370,000

.Mil)

(v Load = Stress x Area)

(v os = 1.2o

c)

6200ar = 370000

370000=

6200= 59.67 N/mm2

. Ans.

Substituting this value in equation (iii), we getos = 1

.2 x o

c

= 1.2 x 59.67 = 71.604 N/mm2. Ans.

37


Problem 1.25. Two brass rods and one steel rod

together support a load as shown in Fig. 1.20. If the stressesin brass and steel are not to exceed 60 N/mm2 and 120 N/mm

2, find the safe load that can be supported. Take E for

steel = 2 x 105 N/mm2 and for brass = 1 x 105 N/mm2

. The

cross-sectional area of steel rod is 1500 mm? and of eachbrass rod is 1000 mm2.

Sol. Given :

Stress in brass,

Stress in steel,

E for steel,

E for brass,

Area of steel rod,

Area of two brass rods,

ah = 60 N/mm2os = 120 N/mm2

Es = 2 x 10® N/mm2

Eh= 1 x 10& N/mm2A = 1500 mm2

Ah = 2x 1000

L. = 170 mm

o

Lh = 100 mm

= 2000 mm2

Length of steel rod,

Length of brass rods,

We know that decrease in the length of steel rod should be equal to the decrease inlength of brass rods.

But decrease in length of steel rod

= Strain in steel rod x Length of steel rod

= es x Ls where es is strain in steel

Similarly decrease in length of brass rods

= Strain in brass rods x Length of brass rods

= eb x Lb where eb is strain in brass rodEquating the decrease in length of steel rod to the decrease in length of brass rods, we get

or

e. £ÿ_100

.A = L> -

170

But stress in steel = Strain in steel x E_

= <ÿ» * Es

Similarly stress in brass is given by,n/> = e. x E

b -b

Dividing equation (i) by equation (ii), we get

(v Stress = Strain x E)

...(£)

Es e.x£ 100 2 x 10-x- = 1.176

ehxEh 170 1x10°

Suppose steel is permitted to reach its safe stress of 2 x 10& N/mm2, the correspondingstress in brass will be

= 2x_10ÿ_= sN/ 21.176 1.176


1.7 x 10R N/mm2 which exceeds the safe stress of 1 x 10S N/mm2 for brass. Therefore let brass

be allowed to reach its safe stress of 1 x 10" N/mm2. Then corresponding stress in steel will be1.176 x 105 N/mm2 which is less than 2 x 10® N/mm2.

Total load = P = Load on steel + Load on copper= Gs x As + oh* Ab= 1

.176 x 10® x 1500 + 1 x 105 x 2000

= 3764 x 10r' N or 376.4 x 106 N

= 376.4 MN. Ans. (v M = 106)

Problem 1.26. Three bars made of copper, zinc and aluminium are of equal length andhave cross-section 500, 750 and 1000 square mm respectively. They are rigidly connected attheir ends. If this compound member is subjected to a longitudinal pull of250 kN, estimate theproportional of the load carried on each rod and the induced stresses. Take the value of E forcopper = 1.3 x 105 N/mm

2, for zinc = 1.0 x 105 N/mm2 and for aluminium = 0.8 x 105 N/mm2.

Sol. Given : j _

-

P = 250 kN = 250 x 103 NTotal load,

For copper bar,

Area, Ac = 500mm2 and Ec = 1.3 xlO5N/mm2 5For zinc bar, §ÿ

Area, A2 = 750 mm2 and E; = 1.

0 x 10& N/mm2 °

For aluminium bar,

Let

w/)W//i/i////m/i/ii//i/i////)///\

\

| Figr P = 250 kN

1.21 |

c(. = Stress induced in copper bar,

<5Z = Stress induced in zinc bar,

aa = Stress induced in aluminium bar,

Pc = Load shared by copper rod,

P, = Load shared by zinc rod,

Po = Load shared by aluminium rod, andL = Length of each bar.

Now, we know that the increase in length of each bar should be same, as length of eachbar is equal hence strain in each bar will be same.

Strain in copper = Strain in zinc = Strain in aluminium

Stress in copper Stress in zinc Stress in aluminiumor

or

E

Also

Now

E,

Ea

<*c _ O,

Ec Et

c r

a

D

KJ

I

I

I

I

cP3

!

ÿXoa =

1.3 xlO°

0.8 x 10'o„ = 1

.625o

a...(«)

xo„ =1.

0 x 105

0.8 x 10®

x a„ = 1.25a

0...(«)

or

total load = Load on copper + Load on zinc + Load on aluminium250 x 10;< = Stress in copper x Ac + Stress in zinc x A

z

+ Stress in aluminium x A

39

Unit1 From Som by Bansal

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Transcript of Unit1 From Som by Bansal