Past Year Questions: Properties of Metals, Stress & Strain - Notes | Study Civil Engineering SSC JE (Technical) - Civil Engineering (CE)

Q.1. A plate in equilibrium is subjected to uniform stresses along its edges with magnitude σ_xx = 30 MPa and σ_yy = 50 MPa as shown in the figure.
The Young’s modulus of the material is 2 x 10¹¹ N/m² and the Poisson’s ratio is 0.3. If σ_zz is negligibly small and assumed to be zero, then the strain ε_zz is _________. 
[2018 : 2 Marks, Set-I]
(a) -120 x 10^-6 
(b) -60 x 10^-6 
(c) 0.0 
(d) 120 x 10-6
Solution: 
So the answer is (d).

Q.2. A 2 m long, axially loaded mild steel rod of 8 mm diameter exhibits the load-displacement (P-δ) behaviour as shown in the figure.
Assume the yield stress of steel as 250 MPa. The complementary strain energy (in N-mm) stored in the bar up to its linear elastic behaviour will be ________.
[2017 : 2 Marks, Set-II]
Solution:
Elastic strain: 

Note: For linear elastic material both complementary energy and strain energy is the same.

OR

By considering the given graph in question, between Axial Load and Displacement, the solution will be as follows:
Complementary stain energy:

It means there seems some error in the given data.

Q.3. In a material under a state of plane strain, a 10 x 10 mm square centred at a point gets deformed as shown in the figure.
If the shear strain γ_xy at this point is expressed as 0.001 k (in rad), the value of k is _________.  
[2017 : 1 Mark, Set-II]
(a) 0.50 
(b) 0.25 
(c) -0.25 
(d) -0.50
Ans. (d)
Solution:
According to the sign convention:

In question, since angle has been increased, therefore, shear strain should be negative.


Q.4. Consider the stepped bar made with a linear elastic material and subjected to an axial load of 1 kN, as shown in the figure.
Segments 1 and 2 have a cross-sectional area of 100 mm² and 60 mm². Young’s modulus of 2 x 10⁵ MPa and 3 x 10⁵ MPa, and length of 400 mm and 900 mm, respectively. The strain energy (in N-mm up to one decimal place) in the bar due to the axial load is _________.    
[2017 : 2 Marks, Set-I]
Solution:
 

Question for Past Year Questions: Properties of Metals, Stress & Strain

Try yourself:An elastic isotropic body is in a hydrostatic state of stress as shown in the figure. For no change in the volume to occur, what should be its Poisson’s ratio?
[2016 : 2 Marks, Set-II]

• A.

  0.00
• B.

  0.25
• C.

  0.50
• D.

  1.00

Explanation

Volumetric strain:

⇒ 
As ΔV = 0

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Q.7. A tapered circular rod of diameter varying from 20 mm to 10 mm is connected to another uniform circular rod of diameter 10 mm as shown in the following figure. Both bars are made of the same material with the modulus of elasticity, E = 2 x 10⁵ MPa. When subjected to a load P = 30πkN, the deflection at point A is ___ mm.    
[2015 : 2 Marks, Set-I]

Solution: 

Total elongation,
AB is uniform

BC is tapered


= (9 + 6) mm = 15 mm

Q.8. For the state of stresses (in MPa) shown in the figure below, the maximum shear stress (in MPa) is _______.    
[2014 : 2 Marks, Set-II]

Solution:

= 5.0 MPa

Question for Past Year Questions: Properties of Metals, Stress & Strain

Try yourself:The values of axial stress (σ) in kN/m², bending moment (M) in kNm, and shear force (V) in kN acting at point P for the arrangement shown in the figure are respectively.    
[2014 : 2 Marks, Set-II]

• A.

  1000, 75 and 25
• B.

  1250, 150 and 50
• C.

  1500, 225 and 75
• D.

  1750, 300 and 100

Explanation

Loading after removing the cable.


Bending moment = 50 x 3 = 150 kNm
Shear force = 50kN

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Question for Past Year Questions: Properties of Metals, Stress & Strain

Try yourself:A box of weight 100 kN shown in the figure is to be lifted without swinging. If all forces are coplanar, the magnitude and direction (θ) of the force (F) with respect to x-axis should be    [2014 : 2 Marks, Set-I]

• A.

  F = 56.389 kN and θ = 28.28º
• B.

  F = - 56.389 kN and θ = - 28.28º
• C.

  F = 9.055 kN and θ = 1.414º
• D.

  F = - 9.055 kN and θ = - 1.414º

Explanation

For no swinging force should be balanced, i.e.,

or


Dividing eq. (ii) by eq. (i) we get


Substituting


or

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Question for Past Year Questions: Properties of Metals, Stress & Strain

Try yourself:A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of 'K) at points T and ‘2’, and an inclined load acts at ‘2’, as shown.

If the load Pequals 100 kN, which of the following options represents forces P₁ and P₂ in the springs at points ‘1’ and ‘27'? 
[2011 : 2 Marks]

• A.

  R₁ = 20 kN and R₂ = 40 kN
• B.

  R₁ = 50 kN and R₂ = 50 kN
• C.

  R₁ = 30 kN and R₂ = 60 kN
• D.

  R₁ = 40 kN and R₂ = 80 kN

Explanation

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Question for Past Year Questions: Properties of Metals, Stress & Strain

Try yourself: A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of 'K) at points T and ‘2’, and an inclined load acts at ‘2’, as shown.

Which of the following options represents the deflections δ₁ and δ₂ at points ‘Y and ‘2’?  [2011 : 2 Marks]

• A.

  
• B.

  
• C.

  
• D.

  

Explanation

The free diagram of the beam is shown below:

From similar triangles, we get:

⇒  ...(i)
Taking moments about hinge, we get:

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