Mechanical properties of solids - abhyaasclasses.in

Mechanical properties of solids

1) A steel wire of length 4.7 m and cross sectional area 3.0 x 10-5

m2 stretches by

the same amount as a copper wire of length 3.5 m and cross sectional area of 4.0

x 10-5

m2 under a given load. What is the ratio of the young’s modulus of steel to

that of copper?

Solution:

Let the Young’s modulus of the steel wire be Ys and that of copper be Yc.

Young’s modulus of a wire, Y = Fl/Ae

As F, e are same for both the wire

�� =

�� x A/��

�� =

�.�.�×�� x

�.�×��.�

Thus, the ration of the Young’s modulus of steel to that of copper is 1.79:1.

2) Figure 9.11 shows that strain curve for a

given material. What are?

(a) Young’s modulus and

(b) Approximate yield strength for this material?

Solution:

(a) Young’s modulus, Y = Stress/Strain

From the graph, when strain = 0.002,

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Stress = 150 x 106 Nm-2

Y = 150 x 106/0.002

Y = 7.5 x 1010

Nm-2

(b) Yield strength is the maximum stress that the material can withstand before the

wire breaks.

From this graph, the maximum stress is nearly 3 x 108 N m

-2.

3) The stress strain graphs for materials

A and B are shown in Fig. 9.12.

The graphs are drawn to the same scale.

(a) Which of the materials has the

greater Young’s modulus?

(b) Which of the two is the stronger

material?

Solution:

(a) Within elastic limit, strain is directly proportional to stress. So, the slope of the

straight line in the graph represents Young’s modulus of the material. From the given

graphs, the slope of A is more. Hence, Young’s modulus of material A is more.

(b) Material that can withstand more stress before fracture is a stringer material. From

the graphs, the maximum stress is more for A as compared to B. so, material A is

stronger than B.



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4) Read the following two statements below carefully and state, with reasons if

it is true or false.

(a) The Young’s modulus of rubber is greater than that of steel.

(b) The stretching of a coil is determined by its shear modulus.

Solution:

(a) False. When the same stress is applied on both the steel and rubber, strain is more

for rubber. So, rubber is less elastic than steel.

(b) True. When we stretch a coil, its shape changes. So, the stretching of a coil is

determined by its shear modulus.

5) Two wires of diameter 0.25 cm, one made of

steel and the other made of brass are loaded as

shown in Fig.9.13. The unloaded length of steel

wire is 1.5 m and that of brass wire is 1.0 m.

compute the elongations of the steel and the brass

wires.

Solution:

According to the problem

Radius of the wires,

r = 0.25/2 cm

r = 0.125 cm = 1.25 x 10-3

m

Length of steel wire, l1 = 1.5 m

Length of brass wire, l2 = 1.0 m

Elongation, e = Fl/Ay



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(1) For steel wire,

F= 6+4 = 10 kg wt

Y= 2 x 1011

Nm2

e = ��×�.�×�.�

�� ×��.��×��×�×��

(A = πr2)

14.97 x 10-5

m

e = 0.15 mm

so, the elongation in the steel wire is 0.15 mm

(2) For brass wire

F= 6 kg wt

Y= 9.0 x 1010 N m-2

e = �×�.�×�.�

�� ×��.��×��×�.�×��

e = 13.3 x 10-5

m

e = 0.13 mm

6) The edge of an aluminium cube is 10 cm long. One face of the cube is firmly

fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of

the cube. The shear modulus of aluminium is 25 GPa. What is the vertical

deflection of this face?

Solution:

According to the problem,

Side of the cube, a = 10 cm



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a = 0.1 m

Mass attracted, m = 100 kg

Shear modulus of aluminium, G = 25 x 109 Pa

Vertical deflection of the face, ∆x = ?

Shear modulus, G = ��

��

= mga/a2 (∆x) [l = a, A = a x a]

∆x = mg/aG

= ��×�.�

�.�×��×��!

= 392 x 10-9

m

= 3.92 x 10-7

m

= 3.92 x 10-4

mm

7) Four identical hollow cylindrical columns of mild steel support a big structure

of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60

cm, respectively. Assuming the load distribution to be uniform, calculate the

compression strain of each column.

Solution:


Mass supported by each column, m = 50000/4 kg

m = 12500 kg

Inner radius of the column, Ri = 30 cm = 0.3 m

Outer radius of the column, Ro = 60 cm = 0.6 m



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Young’s modulus of steel, Y = 2 x 1011 N m-2

Y = ��"

Strain, "� = F/YA =

#$%&'��(')�*� [ F = mg, A = +&,�� − ,.�*]

"� =

��×�.�� ×��.��(��.��×�×��

= ��×��

�.��

"� = 7.22 x 10

-7

Thus, the compression strain of each column is 7.22 x 10-7

8) A piece of copper having a rectangular cross section of 15.2 mm x 19.1 mm is

pulled in tension with 44,500 N force, producing only elastic deformation.

Calculate the resulting strain.

Solution:


Force, F = 44,500 N

Cross sectional area, A = 15.2 x 19.1 mm2

= 290.32 mm2

A = 2.9 x 10-4

m2

Y = Stress/ Strain

Strain = F/AY (Stress = F/A)

= ��,��

�.�×��1×�.�×��



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= 139.5 x 10-5

= 1.39 x 10-3

9) A car weighs 1800 kg. The distance between its front and back axles is 1.8 m.

its centre of gravity is 1.05m behind the front axle. Determine the force exerted

by the level ground on each front wheel and each back wheel.

Solution:


Weight of the car, W = 1800 x 9.8

W = 17640 N

Distance between the front (F) and back

(B) axles, d = 1.8 m

Distance between the front axle (F) and centre of gravity (G), x =1.05 m

Consider RF and RB as the forces exerted by the level ground on each front wheel and

each back wheel, respectively.

For translational equilibrium,

RF +RB = W

RF + RB = 1800 x9.8 = 17640 N -------------(1)

For rotational equilibrium, on taking the torque about the centre of gravity, about F,

we have,

RF x 1.0 = RB x (0.75)

RF/RB = 0.75/1.05 = 5/7

RF/RB = 5/7



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RB = 1.4 RF --------------(2)

Solving (1) and (2) we get,

1.4 RF + RF 17640

RF = 17640/2.4

Therefore, RB = 10290 N

Hence, the force exerted on each front wheel is 3675 N and on each back wheel is

5145 N.

10) A rigid bar of mass 15 kg is supported symmetrically by three wires, each

2.0 m long. Those at each end are of copper, and the middle one is of iron.

Determine the ratios of their diameters if each is to have the same tension.

Solution:

Young’s modulus of a material, Y = Stress/Strain

Y = ��"

As the three wires are of the same length and the tension in them is also the same,

their elongation will also be the same.

Since, l, e and F are same.

A is directly proportional to 1/Y

Diameters of copper wire/diameter of iron wire == dc/dFe = 2�34�5

dc/dFe = 2�.��×��!��×��!

dc/dFe = 1.31



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11) A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0

m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom

of the circle. The cross sectional area of the wire is 0.065 cm2. Calculate the

elongation of the wire when the mass is at the lowest point of its path.

Solution:


Load on the steel wire, m=14.5 kg

Length of the steel wire, l = 1.0 m

Angular velocity, n = 2 rps

ω = 4π rad s-1

(ω= 2πn)

Cross sectional area, A = 0.065 cm2

A = 65 x 10-7

m2

At the lowest point, the tension in the string, T = mg + mv2/T

= mg + mrω2

= m (g +r ω2)

= 14.5 x [9.8 +1 x (4π)2]

= 14.5 x 167.7

T = 2432 Newton

Elongation of wire, e = Tl/Ay

e = ��×�

��×��×�×��

e = 18.7 x 10-4

m

e = 1.87 mm



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12) Compute the bulk modulus of water from the following data: Initial volume

= 100.0 litre, pressure increase = 100.0 atm (1 atm = 1.013 x 105 Pa), final

volume = 99.5 litre. Compare the bulk modulus of water with that of air (at

constant temperature). Explain in simple terms why the ratio is so large. [When

pressure is increased volume decreases, so the volume given the problem is

taken as 99.5 litres in place of 100.5 litre]

Solution:

Initial volume, Vi = 100.0 litre = 100 x 10-3

m3

Increase in pressure, ∆p = 100 x 1.013 x 105 Pa

Final volume, Vf = 99.5 litre = 99.5 x 10-3

m3

Bulk modulus of air, Bair = 1x 10-4

Pa

Bulk modulus of water, Kw = - �67�868

For water, ∆v = Vf – Vi

∆v = (99.5 – 100) x 10-3

m3

∆v = - 0.5 x 10-3

m3

K = - ��×�.��×��×��×��

(�.�×��

K = 2.02 x 109 Pa

9:�;<=>?@"A9:�;#<B:�:C<=?.A = Bw/Bair =

�.��×��!�×��1 = 2 x 10

13

The change in volume of air is very large as compared with that of water for the same

change in pressure. This is due to the weak bonding of the molecules in gases.



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13) What is the density of water at a depth where pressure is 80.0 atm, given that

its density at the surface is 1.03 x 103 kg m

-3?

Solution:

Let us consider density of water on the surface as d1 and at depth where pressure is 80

atm as d2.

Volume occupied by water of mass ‘M’.

On the surface, V1 = M/d1

At the depth, V2 = M/d2

Increase in pressure, ∆p = 80 – 1 = 79 atm

∆p = 79 x 1.013 x 105

Pa

Change in volume, ∆v = V2 – V1

= (M/d2) – (M/d1)

Bulk modulus of water at STP

B= 2.2 x 109 Pa

B= - ��D�E��F�

2.2 x 109 = -

��×�.��×��×GB�H

I�(HI�

�.�×��!�×�.��×�� = -

B�B�(B�

275 = - B�

��(B�

J� = 1033.76



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J� = 1.034 x 103 kg m

-3

Hence, the density of water at a depth where pressure is 80.0 atm is 1.034 x 103 kgm

-3.

14) Compute the fractional change in volume of a glass slab, when subjected to

a hydraulic pressure of 10 atm.

Solution:

Bulk modulus of glass, B = 37 x 109 Pa

Pressure applied, ∆P = 10 atm

Fractional change in volume, 688 =?

688 =

6K9 =

��×�.��×��×��!

688 = 2.74 x 10

-5

15) Determine the volume contraction of a solid copper cube, 10 cm on an edge,

when subjected to a hydraulic pressure of 7.0 x 106 Pa.

Solution:


Side of the cube, a= 10 cm = 0.1 m

Pressure applied, LM = 7 x 106 Pa

Volume contraction, - LN =?

Bulk modulus of copper, B = 140 x 109 Pa

B= �67�8(68



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- LN = �67�89

= ×��O×��.��

��×��!

(Volume of cube is a3)

- LN = 5 x 10-8

m3

- LN = 0.05 cm3

Thus, the volume of the solid copper cube contracts by 0.05 cm3.

16) How much should the pressure on a litre of water be changed to compress it

by 0.10%?

Solution:

Given, volume of water = 1 litre = 10-3

m3

Change in volume,

688 = -

(�.�� = - 10

-3

Bulk modulus of water, K= 2.2 x 109

LP = - �68�98

LP = 2.2 x 109 x 10

-3 =2.2 x 10

6 Pa

Hence, the pressure should be increased by 2.2 x 106 Pa.



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17) Anvils made of single crystals of diamond,

with the shape as Shown in figure, are used to

investigate behaviour of materials under very

high pressures. Flat faces at the narrow end of

anvil have a diameter of 0.50 mm, and the wide

ends are subjected to a compression force of

50,000 N. what is the pressure at the tip of the

anvil?

Solution:

Here, D = 0.5 mm = 0.5 x 10-3

m = 5 x 10-4

m

F = 50,000 N = 5 x 104 N

Pressure at the tip of anvil is the force per unit area.

P = �

QR�1

= ��%S�

P = �×��×��1�

T�� U×��×��1�� = 2.5 x 10

11 Pa

18) A rod of length 1.05 m having negligible mass

is supported at its ends by two wires of steel (wire

A) and aluminium (wire B) of equal lengths as

shown in Figure. The cross sectional areas of wires

A and B are 1.0 mm2 and 2.0 mm

2, respectively. At

what point along the rod should a mass m be

suspended in order to produce equal stresses in

both steel and aluminium wires.

Solution:

For steel wire A, l1 = l; A1 = 1 mm2;



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Y1 = 2 x 1011

N m-2

For aluminium wire B, l2 = l; A2 =

2 mm2;

Y2 = 7 x 1010

N m-2

Let mass ‘m’ be suspended from the rod at distance x from the end where wire A is

connected. Let F1 and F2 be the tensions in two wires and there is equal stress in two

wires, then

�� =

�� or

�� =

�� = 1/2 ---------(1)

Taking the moment of forces about the point of suspension of mass from the rod,

We have

F1x = F2 (1.05 – x)

�.��(VV =

�� = ½

2.10 – 2x = x

x = 0.70 m = 70 cm

19) A rod of length 1.05 m having negligible

mass is supported at its ends by two wires of steel

(wire A) and aluminium (wire B) of equal lengths

as shown in Figure. The cross sectional areas of

wires A and B are 1.0 mm2 and 2.0 mm

2,

respectively. At what point along the rod should a

mass m be suspended in order to produce equal

strains in both steel and aluminium wires.



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Solution:

For steel wire A, l1 = l; A1 = 1 mm2;

Y1 = 2 x 1011

N m-2

For aluminium wire B, l2 = l; A2 = 2 mm2;

Y2 = 7 x 1010

N m-2

Let mass ‘m’ be suspended from the rod at distance x from the end where wire A is

connected. Let F1 and F2 be the tensions in two wires and there is equal stain in two

wires, then

��W� =

��W�

�� =

��W��W� =

�� x

�×��×�� =

��

As the rod is stationary, X�Y = X� (1.05 – x)

�.��(VV =

�� =

��

10x = 7.35 – 7x

x = 0.4324 m = 43.2 cm

20) A mild steel wire of length 1.0 m and cross sectional area 0.50 x 10-2 cm2 is

stretched, well within its elastic limit, horizontally between two pillars. A mass

of 100 g is suspended from the midpoint of the wire. Calculate the depression at

the midpoint.

Solution:

It is given that length, 2l = 1m

Cross sectional area, A = 0.50 x 10-2

cm2



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Let ‘x’ be the depression at the midpoint, i.e.

CD = x in the figure, AC = CB = l =0.5 m;

m = 100 g = 0.1 kg

AD = BD = (l2 + x

2)1/2

Increase in length, ∆l = AD +DB – AB = 2AD – AB

∆l = 2(l2 +x

2)

1/2 -2l

∆l = 2l(1 + V�� )

1/2 – 2l

Using binomial theorem and neglecting higher order terms, we get

∆l = 2l (1 + V��) – 2l =

V��

Strain = �Z�Z =

V��

If ‘T’ is the tension in the wire, then

2T cos⍬ = mg or T = mg/2cos⍬

Here,

Cos ⍬ = V

��[V��/� = V

�]�[^�_�`�/�

Using binomial theorem and neglecting higher order terms, we get

Cos ⍬ = V

�]�[��^�_�`

As x<<1, so 1>>��V�� and 1 + �

�V�� ≈ 1

Cos ⍬ = x/l

Hence,



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T = #$��^_�

= mgl/2x

Stress = T/A =#$��V

Y= Stress/Strain = #$��V x

� ��V� =

#$��V�

x = l [ #$�� ]1/3

= 0.5 [�.�×��

�×��×�.�×��O]1/3

= 1.074 x 10-2

m = 1.074 cm

Hence, the depression at the centre of the wire will be 1.074 cm.

21) Two strips of metal are riveted together at their ends by four reverts, each of

diameter 6.0 mm. what is the maximum tension that can be exerted by the

reverted strip if the shearing stress on the rivet is not to exceed 6.9 x 107 Pa?

Assume that each rivet is to carry one quarter of the load.

Solution:

Here, r = 6/2 = 3 mm = 3 x 10-3

m

Maximum stress = 6.9 x 107 Pa

Maximum tension on a rivet = maximum stress x area of cross section

Maximum tension on a rivet= 6.9 x 107 x 22/7 x (3 x 10

-3)

2

Maximum tension that can be exerted by the strip on a rivet

= 4[6.9 x 107 x (22/7) x 9 x 10

-6] = 7.8 x 10

3 N



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22) The Marina trench is located in the Pacific Ocean, and at one place it is

nearly eleven km beneath the surface of water. The water pressure at the bottom

of the trench is about 1.1 x 108 Pa. a steel ball of initial volume 0.32 m

3 is

dropped into the ocean and falls to the bottom of the trench. What is the change

in the volume of the ball when it reaches to the bottom?

Solution:


Pressure at the bottom, P= 1.1 x 108 Pa

Volume of the ball on the surface of the ocean, V =0.32 m3

Bulk modulus of steel, B = 1.6 x 1011

Pa

We know that

B= - PV/∆V

The change in the volume of the ball when it reaches the bottom,

- ∆V = PV/B

∆V = ��.�×��d�×�.��

�.�×�� = -2.2 x 10-4

m3

Hence, the volume of the ball decreases by 2.2 x 10-4

m3

When it reaches the bottom.



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