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If the rigid rod fitted snugly between the supports as shown in the figure below, is heated, the stress induced in it due to 20℃ rise in temperature will be (α = 12.5 x 10-6/℃ E = 200GPa)
  • a)
    0.07945 MPa
  • b)
    −0.07945 MPa
  • c)
    −0.03972 MPa
  • d)
    0.03972 MPa
Correct answer is option 'B'. Can you explain this answer?

Gate Funda answered
Expansion of rod due to temperature rise δ = α∆tL
= 12.5 x 10−6 x 20 x 0.5
= 1.25 x 10−4m
Force induced by the spring due to compression
F = kδ = 50 x 103 x 1.25 x 10−4 = 6.25 N
This stress is compressive for the bar,
∴ Stress = −0.07945 MPa

A steel rod, 2 m long, is held between two walls and heated from 20℃ to 60℃. Young’s modulus and coefficient of linear expansion of rod material are 200 x 103 MPa and 10 x 10−6/℃ respectively. The stress induced in the rod, if walls yield by 0.2 mm, is
  • a)
    60 MPa tensile
  • b)
    80 MPa tensile
  • c)
    80 MPa compressive
  • d)
    60 MPa compressive
Correct answer is option 'D'. Can you explain this answer?

Neha Joshi answered
L = 2 m, t1 = 20° and t2 = 60°
Temperature change (∆t) = t2 − t1 = 40℃
E = 200 x 103 MPa
α = 10 x 10−6/℃
yielding of the wall (λ) = 0.2 mm
BB1 = δth = α(∆t)L
B1 B2 = Expansion restricted by supports
λ = BB2 = yielding of support
= −100[(2000 x 10 x 10−6 x 40) − 0.2]
= −60 MPa = 60 MPa compressive

A rigid beam of negligible weight is supported in a horizontal position by two rods of steel and aluminum, 2m and 1m long having values of cross-sectional areas 1 cm2 and 2 cm2 and E of 200 GPa and 100 GPa respectively. A load P is applied as shown in the figure.
If the rigid beam is to remain horizontal then
  • a)
    the forces on both sides should be equal
  • b)
    the force on aluminum rod should be twice the force on steel
  • c)
    the force on the steel rod should be twice the force on aluminum
  • d)
    the force p must be applied at the center of the beam
Correct answer is option 'B'. Can you explain this answer?

Sarita Yadav answered
If the beam is to remain horizontal ⇒ δAB = δCD Let force on rod AB be P′
⇒ Force on rod CD = P − P′
⇒ 2P′ = P − P′
⇒ 3P′ = P
⇒ P′ = P/3(force on steel rod) Force on aluminum rod
⇒ Force on aluminum rod
= 2 x force on steel rod.

The stress state at a point in a material under plane stress condition is equi-biaxial tension with a magnitude of 10 MPa. If one unit on the σ−τ plane is 1 MPa, the Mohr's circle representation of the state-of-stress is given by
  • a)
    a circle with a radius equal to principal stress and its center at the origin of the σ−τ plane
  • b)
    a point on the σ axis at a distance of 10 units from the origin
  • c)
    a circle with a radius of 10 units on the σ−τ plane
  • d)
    a point on the τ axis at a distance of 10 units from the origin
Correct answer is option 'B'. Can you explain this answer?

Shalini Deshpande answered
Explanation:

Equi-biaxial tension state:
- In this case, the stress state at a point in the material is equi-biaxial tension with a magnitude of 10 MPa.
- This means that the stress state is characterized by equal tensile stresses in two perpendicular directions.

Mohr's circle representation:
- When representing the state-of-stress on a σ-τ plane using Mohr's circle, the center of the circle corresponds to the average normal stress and the radius of the circle corresponds to the maximum shear stress.
- In this case, the equi-biaxial tension state has a magnitude of 10 MPa, which means the maximum shear stress is also 10 MPa.

Representation on σ-τ plane:
- Since one unit on the σ-τ plane is 1 MPa, a magnitude of 10 MPa for the equi-biaxial tension state corresponds to a distance of 10 units from the origin along the σ axis.
- Therefore, the Mohr's circle representation of the state-of-stress for this equi-biaxial tension state would be a point on the σ axis at a distance of 10 units from the origin.
Therefore, the correct answer is option 'B' - a point on the σ axis at a distance of 10 units from the origin.

A rigid plate ABCD, 1 m2 is supported over four legs of equal lengths of same crosssection and same material. A vertical load 1200 N acts at point E such that
AM = LE = 0.5 m
AL = ME = 0.4 m
The reactions RA , RB , RC and RD are
  • a)
    RA = 360N, RB = 360N, RC = 240N, RD = 240N
  • b)
    RA = 240N, RB = 240N, RC = 360N, RD = 360N
  • c)
    RA = 360N, RB = 240N, RC = 360N, RD = 360N
  • d)
    RA = 240N, RB = 360N, RC = 240N, RD = 360N
Correct answer is option 'A'. Can you explain this answer?

Neha Joshi answered
For equilibrium RA + RB + RC + RD = 1200 N ⋯ ①
ABCD is a rigid plate, under load, its plane is changed, it is not deformed. Say under load legs deform by δA , δB , δC , δD respectively.
Diagonal AC and diagonal BD remain straight. Centre of the plate lies on both diagonals AC and BD.
δO = deflection at centre of plates
Moreover within the elastic limit, deflection is proportional to the load (or reaction) so
RA + RC = RB + RD
= 1200 / 2
= 600N ⋯ ③
Now taking moments of the forces about the edge AB
1200 x ME = I x AC + RD x 1
1200 x 0.4 = RC + RD
But RC + RD = 480
Taking moments about edge AD
1200 x 0.5 = RB x 1 + RC x 1
RB + RC = 600
RB − RC = 120
RB = 360N
RC = 240N
RA = 600 − RC
RA = 600 − 240 = 360N
RB + RD = 600
RD = 600 − RB
= 600 − 360 = 240N
Finally the reactions are RA = 360N,
RB = 360N, RC = 240N, RD = 240N

An aluminum bar of 8 m length and a steel bar of 5 mm longer in length are kept at 30℃.If the ambient temperature is raised gradually, at what temperature the aluminum bar will elongate 5 mm longer than the steel bar? (The linear expansion coefficients for steel and aluminum are 12 x 10−6/℃ and 23 x 10−6/℃ respectively)
  • a)
    50.7℃
  • b)
    69.0℃
  • c)
    143.7℃
  • d)
    33.7℃
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
L = 8 m Initially,
αs = 12 x 10−6/℃
αa = 23 x 10−6/℃
t1 = 30℃
Total elongation of aluminum-total elongation of steel
= (AA3) − (A1A2)
= (AA1) + A1A2 + A2A3) − (A1A2)
= AA1 + A2A3 = 10mm
L αal∆t − Lαs∆t = 10
8 x 103 x ∆t[23 x 10−6 − 12 x 10−6] = 10
t2 − 30° = 113.636℃
t2 = 143.636℃

The independent elastic constants for homogeneous and isotropic material are
  • a)
    E, G, K, V
  • b)
    E, G, K
  • c)
    E, G, V
  • d)
    E, G
Correct answer is option 'D'. Can you explain this answer?

Vibhor Goyal answered
Only two independent elastic constant are required to express the stress-strain relation for a linearly elastic, isotropic material. We can find third elastic constant and Poisson’s ratio for the material.

Consider a beam with circular cross-section of diameter d. The ratio of the second moment of area about the neutral axis to the section modulus of the area is.
  • a)
    d/2
  • b)
    πd/2
  • c)
    d
  • d)
    πd
Correct answer is option 'A'. Can you explain this answer?

Mrinalini Sharma answered
Understanding the Beam Properties
To analyze the properties of a circular cross-section beam, we need to define two key terms: the second moment of area (I) and the section modulus (Z).
Second Moment of Area (I)
- For a circular cross-section with diameter d, the second moment of area about the neutral axis is given by:
I = (π/64) * d^4
Section Modulus (Z)
- The section modulus is derived from the moment of inertia and is used to determine the strength of the beam. For a circular section:
Z = I / (d/2) = (π/32) * d^3
Calculating the Ratio
To find the ratio of the second moment of area to the section modulus, we perform the following calculation:
- Ratio = I / Z
Substituting the expressions for I and Z:
- Ratio = [(π/64) * d^4] / [(π/32) * d^3]
This simplifies to:
- Ratio = (1/2) * d = d/2
Conclusion
Thus, the ratio of the second moment of area to the section modulus for a beam with a circular cross-section of diameter d is indeed d/2, confirming that option 'A' is correct. This relationship is crucial for engineers to understand the beam's ability to resist bending and deflection.

For an Oldham coupling used between two shafts, which among the following statements are correct?
I. Torsional load is transferred along shaft axis.
II. A velocity ratio of 1:2 between shafts is obtained without using gears.
III. Bending load is transferred transverse to shaft axis.
IV. Rotation is transferred along shaft axis.
  • a)
    I and III
  • b)
    I and IV
  • c)
    II and III
  • d)
    II and IV
Correct answer is option 'B'. Can you explain this answer?

Vibhor Goyal answered
Oldham coupling is used to connect two shafts which are not on the same axis means they are not aligned to the same axis
So,
(1) Torsional load is transferred along shaft axis as both shafts are rotating member.
So, that statement is correct.
(2) A velocity ratio 1:1 between shafts is obtained using gears.
So, this statement is wrong
(3) Bending load is not transferred transverse to shaft axis as there is no transverse load.
(4) Rotation is transferred along shaft axis
So, this statement is correct

Consider the following statements: The thermal stress is induced in a component in general, when
1. a temperature gradient exists in the component
2. the component is free from any restraint
3. it is restrained to expand or contract freely
Which of the above statements are correct?
  • a)
    1 and 2
  • b)
    2 and 3
  • c)
    3 alone
  • d)
    2 alone
Correct answer is option 'C'. Can you explain this answer?

Sanvi Kapoor answered
1. If a temperature gradient exists in the component, then there will be thermal stresses generated if the component is restrained to expand or contract freely.
2. If the component is free from restraints, no stresses will be generated. However, strain will be there.
3. Thermal stresses will be induced when it is restrained to expand or contract freely.
Ex.: fixed beam with a rise in temperature.

Two circular shafts made of same material, one solid (S) and one hollow (H), have the same length and polar moment of inertia. Both are subjected to same torque. Here, θs​ is the twist and τs​ is the maximum shear stress in the solid shaft, whereas θH​ is the twist and τH​ is the maximum shear stress in the hollow shaft. Which one of the following is TRUE?
  • a)
    θs​=θh​andτs​=τH​
  • b)
    θs​>​θhandτs​>τH​
  • c)
    θs​<θh​andτs​<τH​
  • d)
    θs​=θh​andτs​<τH
Correct answer is option 'D'. Can you explain this answer?

Milan Ghosh answered
Understanding the Problem
Two shafts, one solid (S) and one hollow (H), are subjected to the same torque and have the same polar moment of inertia. The goal is to compare their twists and maximum shear stresses.
Key Concepts
- Twist (θ): The angle of rotation per unit length due to applied torque.
- Maximum Shear Stress (τ): The highest shear stress experienced in the material due to the applied torque.
Formulas for Twisting
- The twist for a circular shaft is given by:
θ = (T * L) / (J * G)
where T is torque, L is length, J is polar moment of inertia, and G is the shear modulus.
Solid vs. Hollow Shaft
- Both shafts have the same polar moment of inertia (J), length (L), and material (same G).
- The solid shaft will have a higher maximum shear stress compared to the hollow shaft, given the same torque.
Comparison of Twists
- Since both shafts have the same J and are subjected to the same torque, they will experience the same twist:
θs = θH.
Comparison of Shear Stresses
- The maximum shear stress in a solid shaft (τs) is higher than that in a hollow shaft (τH) because the solid shaft has more material resisting the torque.
Conclusion
Thus, the correct answer is:
- Option D: θs = θH and τs < τh.="" />
This indicates that while both shafts undergo the same amount of twist, the solid shaft experiences less maximum shear stress than the hollow shaft due to differences in geometry and material distribution.

Consider the following statements:
X. Two-dimensional stresses applied to a thin plater in its own plane represent the plane stress condition.
Y. Normal and shear stresses may occur simultaneously on a plane.
Z. Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
Which of the above statements are correct?
  • a)
    2 only
  • b)
    1 and 2
  • c)
    2 and 3
  • d)
    1 and 3
Correct answer is option 'D'. Can you explain this answer?

Dishani Desai answered
Understanding Plane Stress Condition
In the context of mechanical engineering, particularly in materials and structural analysis, the following explanations address the correctness of the statements provided.
Statement X: Two-dimensional stresses applied to a thin plate in its own plane represent the plane stress condition.
- This statement is incorrect.
- Plane stress conditions are typically relevant for thin plates subjected to loads in their own plane, but this does not mean that all two-dimensional stresses necessarily imply a plane stress condition.
Statement Y: Normal and shear stresses may occur simultaneously on a plane.
- This statement is correct.
- In mechanics, it is common for both normal and shear stresses to act on a plane simultaneously. This is a fundamental aspect of stress analysis in materials.
Statement Z: Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
- This statement is correct.
- In a plane stress condition, the thickness direction (perpendicular to the plane) experiences negligible stress, which results in zero strain in that direction.
Conclusion
- Based on the analysis above:
- Statements Y and Z are correct, while statement X is incorrect.
- Therefore, the correct answer to the question is option D (1 and 3), as it confirms that Y and Z are valid statements regarding stresses in a plane stress scenario.

For a loaded cantilever beam of uniform cross-section, the bending moment (in N.mm) along the length is M(x)=5x2+10x , where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x=10 mm is
  • a)
    100
  • b)
    105
  • c)
    110
  • d)
    115
Correct answer is option 'C'. Can you explain this answer?

Nitin Joshi answered
Understanding the Problem
For a cantilever beam with a given bending moment function M(x) = 5x² + 10x, we need to find the shear force at a specific point (x = 10 mm) along the beam.
Key Concepts
- Bending Moment (M): Represents the internal moment that causes the beam to bend.
- Shear Force (V): Represents the internal force that acts perpendicular to the beam's axis.
Finding Shear Force
To find the shear force (V) at a position x, we can use the relationship between bending moment and shear force:
V(x) = dM/dx
This means we need to differentiate the bending moment function M(x).
Calculate the Derivative
1. Given M(x) = 5x² + 10x
2. Differentiate M with respect to x:
- dM/dx = 10x + 10
Evaluate Shear Force at x = 10 mm
1. Substitute x = 10 mm into the derivative:
- V(10) = 10(10) + 10
- V(10) = 100 + 10
- V(10) = 110 N
Conclusion
Thus, the magnitude of the shear force at x = 10 mm is 110 N, which corresponds to option 'C'. This calculation confirms the internal force acting on the beam at that point, which is crucial for understanding the beam's structural integrity.

Two steel rods of 20 mm diameter are joined end to end by means of a turn buckle as shown in figure, other end of each rod is rigidly fixed. Initially, there is no tension in each rod. Effective length of each rod is 5 m. Threads on each rod has a pitch of 2.4 mm. Calculate the increase in tension when the turn buckle is tightened by one third of revolution. E = 208 GPa. If the temperature of the rods is increased, then at what temperature, tension in each rod is reduced to half of its magnitude. αs = 11 x 10−6/℃.
  • a)
    9.62℃
  • b)
    7.27℃
  • c)
    11.63℃
  • d)
    3.18℃
Correct answer is option 'B'. Can you explain this answer?

Sarita Yadav answered
A turn buckle is a type of nut having internal threads on two ends but in opposite directions. Ends of tie rods are inserted from both the sides as shown. By rotating the turn buckle, both the tie rods are tightened.
Length of each rod = 5 m Pitch of thread = 2.4 mm
Axial movement of rods on both the sides = 2.4/ 3 = 0.8 mm
Strain in each rod = 0.8 / 5000 = 0.16 x 10−3
E = 208000 N/mm2
σ, stress in each rod
= 0.16 x 10−3 x 208 x 1000 = 33.28 N/mm2
Area of cross–section of each rod,
A =π / 4 x 202 = 100 π mm2
Tension in each rod, T = 100π x 33.28
= 10455 N
= 10.455 kN
Tension in rod is to be reduced to half by increasing the temperature of the bar

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