# Mechanical Metallurgy Questions and Answers – Element of Plasticity Theory – True Stress & True Strain

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This set of Mechanical Metallurgy Multiple Choice Questions & Answers (MCQs) focuses on “Element of Plasticity Theory – True Stress & True Strain”.

1. The engineering stress-strain curve does not give an accurate indication of the deformation characteristic of the material because it’s calculation is based on the original dimension of the specimen. These dimensions continuously change during the test.
a) True
b) False

Explanation: True stress considers the instantaneous area of the sample in the calculation; therefore, it is a real indication of the state of the material. In the metalworking process like rolling, forging where material undergoes a significant change in the cross-sectional area, the true stress-strain curve represents the correct state of raw product.

2. The initial length of sample is Lo and instantaneous length of the sample is L. The true strain of the material will be equal to _________
a) L/Lo
b) Lo/L
c) ln{(L-Lo)/L}
d) ln{L/Lo}

Explanation: True stress is defined physically as infinitesimal change in the dimension of material over the instantaneous length of the material. The conventional form of strain can be used to derive the formula of true stress.

3. The relationship between the true strain and engineering strain is given as _____
Where e=conventional strain, ε= Engineering strain
a) ε=e
b) ε=ln(e)
c) ε=ln(e+1)
d) ε=ln(1/(1+e))

Explanation: Assuming initial length=Lo, final length=L
=> e=(L-Lo)/Lo
=> e = L/Lo-1
=> e+1=L/Lo
taking log on both sides
=> ln(e+1)=ln(L/Lo)
=> By definition, ln(L/Lo) = ε
So finally, ε=ln(e+1).

4. A uniform cylinder of length L is elongated to twice of its original length. Calculate true strain and engineering strain for the cylinder in percentage?
a) Engineering strain=100%, True strain=69.31%
b) Engineering strain=69.31%, True strain=100%
c) Engineering strain=50%, True strain=50%
d) Engineering strain=50%, True strain=100%

Explanation: Initial length of cylinder=L; Final Length of cylinder=2L
Engineering strain = (2L-L)/L= L/L=1
So, in the percent 100%
True strain = ln(2L/L) = ln(2) = 0.6931
In percentage, it is 69.31%.

5. The advantage of using true strain is that the total true strain is equal to the sum of the incremental strains.
a) True
b) False

Explanation: The statement is true, Consider this with this example.
A rod on 50 mm length with initial length is elongated in 3 steps as shown in the diagram: Step 1 = 5mm
Step 2 = 5.5 mm
Step 3 = 6.05 mm
Calculate the engineering strain in each step:
Step 1 = 5/50 = 0.1
step 2 = 5.5/55 = 0.1
Step 3 = 6.05/60.5 = 0.1
So total strain => 0.1+0.1+0.1=0.3
Now calculate the engineering strain in single step=
16.55/50=0.331.
So, the strain value calculated in steps and all together for the same specimen is different.
But in the case of true strain calculation, it will be the same in both cases.

6. For perfectly plastic material, Poisson’s ratio is equal to ________
a) 1
b) 0.5
c) 0.33
d) 0

Explanation: For the perfectly plastic material, volume change on deformation is zero. And the slope of the stress-strain curve is zero (as it is parallel to the x-axis)
=> E=3K(1-2ν)
=> 0=3K(1-2ν)
=> (1-2ν) = 0
=> ν=1/2=0.5.

7. What is the relationship between engineering stress and true stress?
Given that; s = engineering stress, σ = true stress, e = engineering strain, ε = true strain.
a) σ = s(e+1)
b) σ = ln(s)
c) σ = ln[s(e+1)]
d) σ = se+1

Explanation: True stress is defined as load divided by the instantaneous area.
=> σ=P/A
Engineering stress = P/Ao
=> σ=(P/A0)(Ao/A) : (Ao/A) = (L/Lo) = e+1
=> σ=(P/A0) e+1
=> σ=s(e+1).

8. A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the engineering stress at maximum load (Ultimate tensile strength).
a) 796 MPa
b) 398 MPa
c) 512 MPa
d) 52 MPa

Explanation: Engineering stress at maximum load = Pmax/Amax
=> Amax = π (diameter max.)2
=> 45000/ π (6)2 *10-6
=> 398 MPa.

9. A tensile specimen of 6 mm diameter and gauge length 25 mm reached a maximum load of 45 kN and fractured at 35 kN. The maximum diameter at fracture is 5mm. Determine the true stress at the fracture point.
a) 445 MPa
b) 371.13 MPa
c) 398 MPa
d) 518.87 MPa

Explanation: Engineering stress at maximum load = Pfracture/Afracture
=> Amax = π (fracture max.)2
=> 35000 / π (5)2 *10-6
=> 445 MPa
So true stress is σ=s(e+1)
Engineering strain= the Change in the diameter / original diameter
=> (6-5)/6 = 1/6 = 0.166
true stress is σ=445(1+0.166) = 518.87 MPa.

10. A uniform cylinder of length L is compressed to half of its original length. Calculate true strain and engineering strain for the cylinder.
a) Engineering strain = 1, True strain = 0.69
b) Engineering strain = 0.5, True strain = 0.69
c) Engineering strain = -0.5, True strain = -0.69
d) Engineering strain = -1, True strain = -0.69

Explanation: Initial length = L
Final length = L/2
Engineering strain = [(L/2)-L]/L = -1/2 = -0.5
True Strain = ln[(L/2)/L] = ln(1/2)=-0.693.

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