Strength of Material Sheet 20

1) Question.
A steel bar of $2$ $m$ length is fixed at both ends at $20 ° C$ . The coefficient of thermal expansion is $11 \times 10^{- 6} / ° C$
and the modulus of elasticity is
2 × 106
Answer.

1
$22 k g / c m^{2}$ ( tensile)

2

$22 k g / c m^{2}$ (compressive)

3

$44 k g / c m^{2}$ (compressive)

4

$44 k g / c m^{2}$ (tensile)

Answer: 4

Explanation:

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2) Question.

A mild steel bar is in three parts, each $20 c m$ long. The diameters of parts $A B$ , $B C$ and $C D$ are $2 c m$ , $1 c m$ and

$3 c m$
Answer.
- 1
  
  $9$
- 2
  
  $4$
- 3
  
  $3$
- 4
  
  $2$
Answer: 1

Explanation:
3) Question.

A steel cable of $2 c m$ diameter is used to lift load of $500 π kg$ . Given that, $E = 2 x 10^{6} k g / c m^{2}$ and the

length of the cable is $10 m$ , the elongation of the cable due to the load
Answer.
- 1
  
  $0.5 c m$
- 2
  
  $0.25 c m$
- 3
  
  $1 c m$
- 4
  
  $1 / π cm$
Answer: 2

Explanation:
4) Question.

Consider the following statements :

In a uni-dimensional stress system, the principal plane is defined as one on which the

1. shear stress is zero

2. normal stress is zero

3. shear stress is maximum

4. normal stress is maximum

of these statements
Answer.
- 1
  
  1 and 2 are correct
- 2
  
  2 and 3 are correct
- 3
  
  1 and 4 are correct
- 4
  
  3 and 4 are correct
Answer: 3

Explanation:
5) Question.

For the design of a cast iron member, the most appropriate theory of failure is
Answer.
- 1
  
  Mohr's theory
- 2
  
  Rankine's theory
- 3
  
  Maximum strain theory
- 4
  
  Maximum shear energy theory
Answer: 2

Explanation:
6) Question.

At a point in a steel member, the major principal stress is 2000 kg/cm² and the minor principal stress is compressive. If the uni-axial tensile yield stress is 2500 kg/cm², then the magnitude of the minor principal stress at which yielding will commence, according to the maximum shearing stress theory, is
Answer.
- 1
  
  1000 kg/cm²
- 2
  
  2000 kg/cm²
- 3
  
  2500 kg/cm²
- 4
  
  500 kg/cm²
Answer: 4

Explanation:
7) Question.

For ductile materials, the most appropriate failure theory is
Answer.
- 1
  
  maximum shear stress theory
- 2
  
  maximum principal stress theory
- 3
  
  maximum principal strain theory
- 4
  
  shear strain energy theory
Answer: 1

Explanation:
8) Question.

A three hinged parabolic arch rid is acted upon by a single load at the left quarter point. If the central rise is increased and the shape of arch altered to segmental without changing the other details, the horizontal thrust will
Answer.
- 1
  
  increase definitely
- 2
  
  decrease definitely
- 3
  
  be difficult to predict
- 4
  
  increase or decrease depending upon the radius of the segmental arch
Answer: 2

Explanation:
9) Question.

The stress below which a material has a high probability of not failing under reversal of stress is known as
Answer.
- 1
  
  tolerance limit
- 2
  
  elastic limit
- 3
  
  proportional limit
- 4
  
  endurance limit
Answer: 4

Explanation:
10) Question.

The property of metal which allows it to deform continuously at slow rate without any further increase in stress is known as
Answer.
- 1
  
  fatigue
- 2
  
  creep
- 3
  
  plasticity
- 4
  
  resilience
Answer: 2

Explanation:
11) Question.

The phenomenon of decreased resistance of a material to reversal of stress is called
Answer.
- 1
  
  creep
- 2
  
  fatigue
- 3
  
  resilience
- 4
  
  plasticity
Answer: 2

Explanation:
12) Question.

Impact test enables one to estimate the property of
Answer.
- 1
  
  hardness
- 2
  
  toughness
- 3
  
  strength
- 4
  
  creep
Answer: 2

Explanation:
13) Question.

The specimen in a Charpy impact test is supported as a
Answer.
- 1
  
  cantilever beam
- 2
  
  simply supported beam
- 3
  
  fixed beam
- 4
  
  continuous beam
Answer: 2

Explanation:
14) Question.

If the depth of a beam of rectangular section is reduced to half, strain energy stored in the beam becomes
Answer.
- 1
  
  1/4 times
- 2
  
  1/8 times
- 3
  
  4 times
- 4
  
  8 times
Answer: 4

Explanation:
15) Question.

Two identical bars, one simply supported and other fixed at ends, are acted upon by equal loads applied at the midpoints.

The ratio of strain energy stored in the simply supported beam and the fixed ended beam is
Answer.
- 1
  
  1
- 2
  
  2
- 3
  
  3
- 4
  
  4
Answer: 4

Explanation:
16) Question.

Two circular bars $A$ and $B$ of same material and same length are of diameters $D_{A}$ and $D_{B}$ respectively. The bars are

subjected to the same torque T. The ratio of strain energies stored in the bars $A$ and $B$ is proportional to
Answer.
- 1
  
  $\frac{D_{A}}{D_{B}}$
- 2
  
  $\frac{D_{B}}{D_{A}}$
- 3
  
  ${(\frac{D_{A}}{D_{B}})}^{2}$
- 4
  
  ${(\frac{D_{B}}{D_{A}})}^{4}$
Answer: 4

Explanation:
17) Question.

A steel rod of cross sectional area equal to 1000 mm² is 5 m long. If a pull of 100 kN is suddenly applied to it, then the

maximum stress intensity will be
Answer.
- 1
  
  50 N/mm²
- 2
  
  100 N/mm²
- 3
  
  200 N/mm²
- 4
  
  400 N/mm²
Answer: 3

Explanation:
18) Question.

A rectangular block of size $200 m m x 100 m m x 50 m m$ is subjected to a shear stress of $100 N / m m^{2}$ . If

modulus of rigidity of material is $1 x 10^{5} N / m m^{2}$ , strain energy stored will be
Answer.
- 1
  
  $10 N . m$
- 2
  
  $25 N . m$
- 3
  
  $50 N . m$
- 4
  
  $100 N . m$
Answer: 3

Explanation:
19) Question.

Strain energy stored in a member is given by
Answer.
- 1
  
  0.5 x stress x volume
- 2
  
  0.5 x strain x volume
- 3
  
  0.5 x stress x strain x volume
- 4
  
  0.5 x stress x strain
Answer: 3

Explanation:
20) Question.

Proof resilience is the maximum energy stored at
Answer.
- 1
  
  limit of proportionality
- 2
  
  elastic limit
- 3
  
  plastic limit
- 4
  
  none of the above
Answer: 2

Explanation:
Resilience is the total energy stored in a body within its elastic limit. Proof resilience is the maximum strain energy stored in a body. Modulus of resilience is the greatest amount of strain energy per unit volume that a material can absorb without exceeding the elastic limit.

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