1. The true strain for a low carbon steel bar which is doubled in length by forging is
(a) 0.307
(b) 0.5
(c) 0.693
(d) 1.0
(1 Mark, 1992)

Ans: c
True strain \left( \varepsilon \right) =ln\frac { l }{ { l }_{ 0 } }
=> \varepsilon =ln\frac { 2{ l }_{ 0 } }{ { l }_{ 0 } }
=> \varepsilon =ln2 = 0.693
2. The ultimate tensile strength of a material is 400 MPa and the elongation up to maximum load is 35%. If the material obeys power law of hardening, then the true stresstrue strain relation (stress in MPa) in the plastic deformation range is:
(a) σ = 540ε^{0.30}
(b) σ = 775ε^{0.30}
(c) σ = 540ε^{0.35}
(d) σ = 775ε^{0.35}
(2 Mark, 2006)

Ans: b
Given, Engineering strain \left( { \varepsilon }_{ E } \right) = 0.35
True strain \left( { \varepsilon } \right) =ln\left( 1+{ \varepsilon }_{ E } \right)
=> \varepsilon =ln\left( 1+0.35 \right) = 0.3
True stress \left( \sigma \right) ={ \sigma }_{ E }\left( 1+{ \varepsilon }_{ E } \right)
=> \sigma =400\left( 1+0.35 \right) = 540 MPa
General form of true stress and true strain relation, \sigma =K{ \varepsilon }^{ n }
At maximum load, \varepsilon =n
To find K, 540=K{ \times 0.3 }^{ 0.3 }
=> K = 775 MPa
After putting the values of n and K in general form, \sigma =775{ \varepsilon }^{ 0.3 }
3. In opendie forging, disc of diameter 200 mm and height 60 mm is compressed without any barreling effect. The final diameter of the disc is 400 mm. The true strain is
(a) 1.986
(b) 1.686
(c) 1.386
(d) 0.602
(2 Mark, 2007)

Ans: c
True strain \left( \varepsilon \right) =2\times ln\frac { { d }_{ 0 } }{ d }
=> \varepsilon =2\times ln\frac { 200 }{ 400 }
=> \varepsilon =1.386
4. A metal rod of initial length L_{0} is subjected to a drawing process. The length of the rod at any instant is given by the expression, L(t) = L_{0}(1 + t^{2}), where t is the time in minutes. The true strain rate (in min^{−1}) at the end of one minute is _______.
(2 Mark, 2014[1])

Ans: 1
Given, L(t) = L_{0}(1 + t^{2})
=> dL = L_{0}(2t) dt
Now, d{ \varepsilon }_{ T }=\frac { dL }{ L }
=> d{ \varepsilon }_{ T }=\frac { { L }_{ 0 }\times 2tdt }{ { L }_{ 0 }\left( 1+{ t }^{ 2 } \right) }
=> \frac { d{ \varepsilon }_{ T } }{ dt } =\frac { 2t }{ 1+{ t }^{ 2 } }
At t = 1 min.,
\frac { d{ \varepsilon }_{ T } }{ dt } = \frac { 2 }{ 2 } =1
5. The relationship between true strain (ε_{T}) and engineering strain (ε_{E}) in a uniaxial tension test is given as
(a) ε_{E} = ln (1 + ε_{T})
(b) ε_{E} = ln (1 – ε_{T})
(c) ε_{T} = ln (1 + ε_{E})
(d) ε_{T} = ln (1 – ε_{E})
(1 Mark, 2014[2])

Ans: c
6. The flow stress (in MPa) of a material is given by σ = 500ε^{0.1}, where ε is true strain. The Young’s modulus of elasticity of the material is 200 GPa. A block of thickness 100 mm made of this material is compressed to 95 mm thickness and then the load is removed. The final dimension of the block (in mm) is _____.
(2 Mark, 2015[2])

Ans: 95.176
7. The value of true strain produced in compressing a cylinder to half its original length is
(a) 0.69
(b) − 0.69
(c) 0.5
(d) − 0.5
(1 Mark, 2016[2])

Ans: b
8. Engineering strain of a mild steel sample is recorded as 0.100%. The true strain is
(a) 0.010%
(b) 0.055%
(c) 0.099%
(d) 0.101%
(1 Mark, 2016[3])

Ans: c
9. A rod of length 20 mm is stretched to make a rod of length 40 mm. Subsequently, it is compressed to make a rod of final length 10 mm. Consider the longitudinal tensile strain as positive and compressive strain as negative. The total true longitudinal strain in the rod is _______
(a) 0.5
(b) 0.69
(c) 0.75
(d) 1.0
(2 Mark, 2017[2])

Ans: b
Explanation:
10. A bar is compressed to half of its original length. The magnitude of true strain produced in the deformed bar is _________. (correct to two decimal places).
(2 Mark, 2018[1])

Ans: 0.69
11. The true stress (σ) – true strain (ε) diagram of a strain hardening material is shown in figure. First, there is loading up to point A, i.e., up to stress of 500 MPa and strain of 0.5. Then from point A, there is unloading up to point B, i.e., to stress of 100 MPa. Given that the Young’s modulus E = 200 GPa, the natural strain at point B (ε_{B}) is ____ (correct to three decimal places).
(2 Mark, 2018[1])

Ans: 0.498
Explanation:
12. In a linearly hardening plastic material, the true stress beyond initial yielding
(a) increases linearly with the true strain
(b) decreases linearly with the true strain
(c) first increases linearly and then decreases linearly with the true strain
(d) remains constant
(1 Mark, 2018[1])

Ans: a
Explanation:
13. The true stress (in MPa) versus true strain relationship for a metal is given by σ = 1020ε^{0.4}. The crosssectional area at the start of a test (when the stress and strain values are equal to zero) is 100 mm^{2}. The crosssectional area at the time of necking (in mm^{2}) is ________ (correct to two decimal places).
(2 Mark, 2018[2])

Ans: 67.032
Explanation: