Strength of Materials Questionbank – SARKARI LIBRARY

Strength of Materials

For an axial member, the formula for normal stress (\(\sigma\)) is given by:

\(\sigma = F \cdot A\)
\(\sigma = \frac{F}{A}\)
\(\sigma = \frac{M \cdot c}{I}\)
\(\sigma = \frac{V \cdot Q}{I \cdot t}\)

Explanation:

Stress is force per unit area, making \(\sigma = F \cdot A\) incorrect as it is force times area.
\(\sigma = \frac{F}{A}\) is the correct formula for normal (axial) stress.
\(\sigma = \frac{M \cdot c}{I}\) is the flexure formula for bending stress, not axial stress.
\(\sigma = \frac{V \cdot Q}{I \cdot t}\) is the formula for shear stress in beams.

The bending stress at a point in a beam is zero. This point is most likely:

At the top surface of the beam
At the bottom surface of the beam
At the neutral axis
Where the shear force is maximum

Explanation:

Top and bottom surfaces experience the maximum bending stress (tension or compression).
The neutral axis is the layer within the beam where the stress transitions from tension to compression; the longitudinal strain and stress are zero here.
The location of maximum shear force is unrelated to where bending stress is zero.

The state of stress at a point is defined by the components \(\sigma_x = 100 MPa\), \(\sigma_y = -40 MPa\), and \(\tau_{xy} = 50 MPa\). What is the maximum in-plane shear stress (\(\tau_{max}\)) at this point?

50 MPa
70 MPa
86 MPa
140 MPa

Explanation:

The formula for maximum in-plane shear stress is: \(\tau_{max} = \sqrt{ \left( \frac{\sigma_x – \sigma_y}{2} \right)^2 + \tau_{xy}^2 }\)
Plugging in the values: \(\tau_{max} = \sqrt{ \left( \frac{100 – (-40)}{2} \right)^2 + (50)^2 } = \sqrt{ (70)^2 + (50)^2 } = \sqrt{4900 + 2500} = \sqrt{7400} \approx 86.02 MPa\)
Therefore, the correct answer is approximately 86 MPa.

Which stress transformation equation is used to find the principal stresses?

The equation for maximum shear stress
The equation for normal stress on a plane defined by angle \(\theta\)
The equation for the center of Mohr’s circle
The von Mises yield criterion equation

Explanation:

Principal stresses are the maximum and minimum normal stresses at a point.
They are found by maximizing/minimizing the normal stress transformation equation: \(\sigma_{n} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x – \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta\).
The other options are related to different concepts in stress analysis.

According to the Maximum Distortion Energy (von Mises) Theory, yielding begins when:

The maximum normal stress reaches the yield strength
The maximum shear stress reaches the shear yield strength
The distortion energy per unit volume equals the distortion energy per unit volume in a tensile test at yield
The strain energy per unit volume reaches a critical value

Explanation:

The first option describes the Maximum Normal Stress theory.
The second option describes the Tresca (Maximum Shear Stress) theory.
The von Mises criterion is based on the energy associated with the change in shape (distortion) of the material, not the volume change.
The final option describes the Total Strain Energy theory, which is different.

Within the elastic limit, the slope of the stress-strain curve for a material represents its:

Proof Resilience
Factor of Safety
Modulus of Elasticity
Thermal Stress

Explanation:

Proof Resilience is the maximum strain energy stored without permanent deformation.
Factor of Safety is the ratio of ultimate stress to working stress.
According to Hooke’s Law, \( \sigma = E \epsilon \), the modulus of elasticity (E) is the constant of proportionality, which is the slope of the linear stress-strain curve.
Thermal stress is the stress induced due to restriction of thermal expansion or contraction.

The total area under the engineering stress-strain curve up to the point of fracture represents:

Modulus of Resilience
Proof Resilience
Toughness
Factor of Safety

Explanation:

Modulus of Resilience is the area under the curve only up to the elastic limit.
Proof Resilience is the total strain energy up to the elastic limit.
Toughness is the ability of a material to absorb energy and plastically deform without fracturing. It is measured by the total area under the stress-strain curve.
Factor of Safety is a design parameter, not a material property derived from the stress-strain diagram.

A steel rod is heated from 20°C to 80°C but is completely constrained from expanding. The stress developed in the rod is:

Tensile
Zero
Compressive
Shear

Explanation:

The rod tries to expand due to heating (\( \delta_{thermal} = \alpha L \Delta T \)).
Since the ends are fixed, this expansion is fully restricted.
The supports exert a force on the rod to keep it at its original length, effectively compressing it and generating compressive stress (\( \sigma_{thermal} = E \alpha \Delta T \)).

The Factor of Safety is defined as the ratio of:

Working stress to Ultimate stress
Strain energy to Proof resilience
Yield stress (or Ultimate stress) to Working stress
Thermal stress to Strain energy

Explanation:

It is a factor that provides a safety margin against failure.
It is calculated as \( \text{Factor of Safety} = \frac{\text{Yield Stress}}{\text{Working Stress}} \) or \( \frac{\text{Ultimate Stress}}{\text{Working Stress}} \), ensuring the working stress is well below the failure stress.
The other options incorrectly define the ratio.

The strain energy per unit volume stored in a material when strained within its elastic limit is known as:

Toughness
Proof Resilience
Modulus of Resilience
Thermal Stress

Explanation:

Toughness is the total energy per unit volume up to fracture.
Proof Resilience is the total strain energy stored up to the elastic limit.
Modulus of Resilience is the strain energy per unit volume, which is the area under the stress-strain curve up to the proportional limit. It is given by \( U = \frac{\sigma_y^2}{2E} \).

On a stress-strain diagram, the point beyond which the material will not return to its original shape upon unloading is called the:

Proportional Limit
Ultimate Tensile Strength
Yield Point
Fracture Point

Explanation:

The Proportional Limit is the point where stress-strain linearity ends, but the material may still be elastic.
The Ultimate Tensile Strength is the maximum stress on the curve.
The Yield Point is the stress level at which a material begins to deform plastically. Deformation beyond this point is permanent.
The Fracture Point is where the material breaks.

For a bar of cross-sectional area A, length L, and modulus of elasticity E, subjected to an axial load P, the expression for strain energy (U) stored is:

\( U = \frac{P L}{A E} \)
\( U = \frac{P^2 L}{2 A E} \)
\( U = \frac{\sigma \epsilon}{2} \)
\( U = \frac{A E \delta}{L} \)

Explanation:

The first option \( \frac{P L}{A E} \) is the formula for deflection (δ), not energy.
The formula for strain energy stored in a linearly elastic body is \( U = \frac{P^2 L}{2 A E} \) or equivalently \( U = \frac{A E \delta^2}{2 L} \).
The third option \( \frac{\sigma \epsilon}{2} \) is for strain energy per unit volume, not total energy.
The fourth option is incorrect.

For a simply supported beam with a uniformly distributed load (UDL) of intensity \( w \) over its entire length \( L \), what is the maximum bending moment?

\( \frac{wL}{2} \)
\( \frac{wL^2}{8} \)
\( \frac{wL^2}{2} \)
\( wL \)

Explanation:

\( \frac{wL}{2} \) is the value of the maximum shear force, not the bending moment.
\( \frac{wL^2}{8} \) is the correct formula for the maximum bending moment at the mid-span of a simply supported beam carrying a UDL.
\( \frac{wL^2}{2} \) is the maximum bending moment for a cantilever beam with a UDL.
\( wL \) is the total load on the beam.

The point on a beam’s bending moment diagram where the bending moment changes sign (from positive to negative or vice versa) is called a:

Maximum moment point
Inflection point
Point of contraflexure
Fixed point

Explanation:

A maximum moment point is where the bending moment is at its highest value.
An inflection point is a concept in mathematics where the curvature changes.
A point of contraflexure is the specific term in structural engineering for the point in a beam (like a continuous or fixed beam) where the bending moment is zero and changes sign.
A fixed point refers to a support condition, not a point on the diagram.

The torsion formula for a circular shaft is given by \( \frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L} \). What does the term \( J \) represent?

Shear Modulus
Polar Radius
Polar Moment of Inertia
Torque

Explanation:

Shear Modulus is represented by \( G \) in the formula.
Polar Radius is represented by \( r \), the distance from the center.
Polar Moment of Inertia (\( J \)) is a geometric property that measures a shaft’s resistance to twisting. For a solid shaft of radius \( R \), \( J = \frac{\pi R^4}{2} \).
Torque is represented by \( T \).

For a thin cylindrical shell subjected to an internal pressure \( p \), with diameter \( D \) and thickness \( t \), the circumferential (hoop) stress is given by:

\( \frac{pD}{t} \)
\( \frac{pD}{2t} \)
\( \frac{pD}{4t} \)
\( \frac{2pD}{t} \)

Explanation:

\( \frac{pD}{t} \) is incorrect and would overestimate the stress.
\( \frac{pD}{2t} \) is the correct formula for the hoop stress, which is the largest stress and acts circumferentially.
\( \frac{pD}{4t} \) is the formula for the longitudinal stress in a thin cylinder.
\( \frac{2pD}{t} \) is incorrect and would greatly overestimate the stress.

In a fixed beam with a point load at the center, how many points of contraflexure are typically present?

0
1
2
4

Explanation:

0 is the case for a cantilever or a simply supported beam without a moment change.
1 is atypical for a fixed beam under a central point load.
2 points of contraflexure exist in a fixed beam with a central load. These are the points where the bending moment diagram crosses the zero axis, located between the supports and the center.
4 is an incorrect number for this loading condition.

The stiffness of a helical spring is given by \( k = \frac{Gd^4}{8D^3N} \). If the wire diameter \( d \) is doubled, the stiffness becomes:

Half
Double
Four times
Sixteen times

Explanation:

Half would be if stiffness was inversely proportional to \( d \).
Double would be if stiffness was proportional to \( d \).
Four times would be if stiffness was proportional to \( d^2 \).
The stiffness \( k \) is proportional to \( d^4 \). If \( d \) is doubled (i.e., becomes \( 2d \)), the new stiffness is \( k_{new} \propto (2d)^4 = 16d^4 \). Therefore, it becomes 16 times the original stiffness.

Strength of Materials Questionbank

Post published:September 3, 2025

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