# Strength of Materials (SOM) GATE Solved Practice Questions | Part 1

GATE/ESE/PSU Solved Practice Questions | Competitive Exam Questions | Strength of Materials (SOM) | Part 1 | By Akhand Dutta

# GATE/ESE/PSU Solved Practice Questions (SOM) - Part 1

## Q1. A spherical ball of volume 20 m³ is placed under a certain liquid wherein the ball is subjected to a uniform hydrostatic pressure of 200 MPa. If the material of the ball has a bulk modulus of elasticity of 2.5 ×10^5 MPa and a Poisson's ratio of 0.30, then due to the hydrostatic pressure, the volume of the ball will change by (in m³).

### Solution:

V= 20 m^3
σ= 200 Mpa
K= 2.5 * 10^5 Mpa
μ= 0.3

dV/V= σ/K

dV= 20 * (200 * 10^3) / (2.5 * 10^5 * 10^6)

dV= 0.0160 (B)

## Q2. A Steel rod of circular section tapers from 2 cm diameter to 1 cm diameter over a length of 50 cm. If the modulus of elasticity of the material is 2 x 10^6 kg/cm² then the increase in length under a pull of 3000 kg will be (in cm)

### Solution:

D1= 2 cm
D2= 1 cm
L= 50 cm
E= 2 * 10^6 kg/cm^3
P= 3000 kg

dl= 4PL/πED1D2

= (4 * 3000 * 50) / (π * 2 * 10^6 * 2 * 1)

= 0.15/π

dl= 0.3/2π (A)

## Q3. In a particular material, if the modulus of rigidity is equal to the bulk modulus, then the Poisson's ratio will be

### Solution:

here, K=G

We know that,

E = 2G(1+u)

Also, E = 3K(1-2u)

So, 2G(1+u) = 3K(1-2u)

=> (1+u)/(1-2u) = 3/2 ................................. (As, E=K)

2+2u=3-6u

u=1/8 (A)

## Q4. In a plane strain problem, the tensile stresses along two mutually perpendicular rectangular coordinate axes x and y are σx and σy, respectively with sigma σx > σy and there are no shearing stresses. The Poisson's ratio is u. The stress along the third rectangular coordinate axis z will be.....

### Solution:

Strain along z-axis, 𝞊z = -uσx/E -uσy/E

E.𝞊z = -u(σx+σy)

σz = -u(σx+σy) [B]

## Q6. A rod of material with E= 200 x 10^3 MPa and a = 10^-3 mm/mm °C is fixed at both ends. It is uniformly heated such that the increase in temperature is 30°C. The stress developed in the rod is

### Solution:

E= 200 x 10^3 MPa
a= 10^-3 mm/mm °C
dt= 30°C

σth = Ea.dt

200 x 10^3 * 10^-3 * 30

σth = 6000 N/mm2 (Compressive as it is rise in temperature case) [C]

## Q7. A gradually applied load W is suspended by wire ropes AB and CD as shown in the figure. The wires AB and CD, made of the same material and of the same cross-section are connected to a rigid block from which the load W is suspended in such a way that both the ropes stretch by the same amount. If the load in AB and CD are P1 and P2 respectively, then the value of P1 and P2 will be

### Solution:

A1=A2
E1=E2

ΣFv=0;

P1+P2=W...........................................................(1)

Using Compatibility Equation:

P1L1/A1E1 = P2L2/A2E2

P1=P2(L2/L1)                  {As, A1=A2 And E1=E2}

P1=3P2/2..................................................................(2)

Using (1) and (2):

P2+ 3P2\2 = W

=> P2=2W/5

Substituting P2 in (1)

P1= 3W/5

## Q8. Below Fig. shows a rigid bar hinged at A and supported in a horizontal position by two vertical identical steel wires. Neglect the weight of the beam. The tension T1 and T₂ induced in these wires by a vertical load P applied as shown are

### Solution:

ΣFv=0;

T1+T2+Ra=P.............................................................(1)

Using, 𝛿1/a= 𝛿2/b= 𝛿/l

T1L1/A1E1a= T2L2/A2E2b

or, T1/a=T2/b

=> T1=T2(a/b)..............................................................(2)

ΣMa=0;

T2 * b + T1 * a = Pl

Using (2) and (3);

T2 * b + T2(a/b)a = Pl

T2(a^2 + b^2)/b = Pl

T2= Pbl/(a^2 + b^2)

Similarly, From (1)

T1= Pal/(a^2 + b^2)

## Q9. The reactions at the rigid supports at A and B for the bar loaded as shown in the figure are respectively.

### Solution:

Ra= P*BC/AB

Ra= 10*2/3 = 20/3 kN

Rb= P*AC/AB

Rb= 10*1/3 = 10/3 kN

## Q10. The axial movement of surface C of the stepped column as shown in the figure is

### Solution:

Total Axial Moment or Total Elongation= P1L1/A1E1 + P2L2/A2E2

= PL/2A.2E + PL/AE

= PL/4AE + PL/AE

=5PL/4AE (a)

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