# Stress and Strain

 Question 1
Ignoring the small elastic region, the true stress $(\sigma) -$ true strain $( \varepsilon)$ variation of a material beyond yielding follows the equation $\sigma =400 \varepsilon ^{0.3}$ MPa. The engineering ultimate tensile strength value of this material is ________ MPa. (Rounded off to one decimal place)
 A 145.6 B 325.6 C 206.6 D 125.2
GATE ME 2023   Strength of Materials
Question 1 Explanation:
Given
\begin{aligned} \sigma_{\mathrm{T}} & =400 \varepsilon^{0.3}=400 \varepsilon_{\mathrm{T}}^{0.3} \;\;\;...(i)\\ \varepsilon & =\text { true strain }=\varepsilon_{\mathrm{T}} \\ \mathrm{n} & =0.3=\text { strain hardening exponent } \end{aligned}
The true strain $\left(\varepsilon_{T}\right)$ at the onset of necking equal to the strain hardening exponent i.e. at ultimate tensile point.
So, $\quad \varepsilon_{\mathrm{T}}=\mathrm{n}=0.3$ at ultimate point
So, from equation (i)
$\sigma_{\mathrm{T}} =400 \times(0.3)^{0.3} =278.74 \mathrm{MPa} \text { at ultimate point }$
As $\varepsilon_{\mathrm{T}}=\ln (1+\varepsilon)$
So, $\varepsilon=\mathrm{e}^{\varepsilon} \mathrm{T}-1=\mathrm{e}^{0.3}-1=0.35$
and $\sigma_{\mathrm{T}}=\sigma(1+\varepsilon)$
So $\sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}$
 Question 2
The principal stresses at a point P in a solid are 70 MPa, -70 MPa and 0. The yield stress of the material is 100 MPa. Which prediction(s) about material failure at P is/are CORRECT?
 A Maximum normal stress theory predicts that the material fails B Maximum shear stress theory predicts that the material fails C Maximum normal stress theory predicts that the material does not fail D Maximum shear stress theory predicts that the material does not fail
GATE ME 2023   Strength of Materials
Question 2 Explanation:
$\sigma _1=70MPa, \sigma _2= -70MPa, \sigma _3=0 , S_{yt}=100MPa$

For maximum shear stress theory:
$\frac{\sigma _1-\sigma _2 }{2}=\frac{70-(-70)}{2}=70$
and $\frac{S_{yt} }{2}=\frac{100}{2}=50$
i.e. $\frac{\sigma _1-\sigma _2 }{2} \gt \frac{S_{yt} }{2}$
So material will fail As per maximum normal stress theory:
$\sigma _1 \; and \; sigma _2 \gt sigma _{yt}$
then material will fail
Here 70 & -70 < 100
So material is safe.

 Question 3
A prismatic bar PQRST is subjected to axial loads as shown in the figure. The segments having maximum and minimum axial stresses, respectively, are A QR and PQ B ST and PQ C QR and RS D ST and RS
GATE ME 2021 SET-1   Strength of Materials
Question 3 Explanation: \begin{aligned} P_{\max }&=P_{S T}=25 \mathrm{kN} \\ P_{\min }&=P_{R S}=5 \mathrm{kN} \end{aligned}
Hence, maximum and minimum axial stresses are in ST and RS portions because of prismatic bar.
 Question 4
The loading and unloading response of a metal is shown in the figure. The elastic and plastic strains corresponding to 200 MPa stress, respectively, are A 0.01 and 0.01 B 0.02 and 0.01 C 0.01 and 0.02 D 0.02 and 0.02
GATE ME 2021 SET-1   Strength of Materials
Question 4 Explanation:
Elastic strain : Which can be recovered $= 0.03 - 0.01 = 0.02$
Plastic strain : Permanent strain $= 0.01$
 Question 5
Uniaxial compression test data for a solid metal bar of length 1 m is shown in the figure. The bar material has a linear elastic response from O to P followed by a non-linear response. The point P represents the yield point of the material. The rod is pinned at both the ends. The minimum diameter of the bar so that it does not buckle under axial loading before reaching the yield point is _______ mm (round off to one decimal place).
 A 16.56 B 56.94 C 78.25 D 68.42
GATE ME 2020 SET-2   Strength of Materials
Question 5 Explanation:
For both end pin,
\begin{aligned} P &=\frac{\pi^{2} E I}{L^{2}}=\frac{\pi^{2} E A \cdot \frac{d^{2}}{16}}{L^{2}} \\ \frac{P}{A E} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ \varepsilon_{y} &=\frac{\pi^{2} d^{2}}{16 L^{2}} \\ d &=\sqrt{\frac{16 L^{2} \varepsilon_{y}}{\pi^{2}}} \\ &=\sqrt{\frac{16 \times 1000^{2} \times 0.002}{\pi^{2}}} \\ &=56.94 \mathrm{mm} \end{aligned}

There are 5 questions to complete.

### 4 thoughts on “Stress and Strain”

1. Explaination of Q6 is wrong.

• Can you please specify with more details..

2. The answer for q6 is 13%
As the compliance error elongation = 0.05*40 = 2mm/N
Elongation shown = 15mm
Therefore,
Actual elongation = 15-2 =13
Actual strain % = ((13/100)*100) = 13%

3. 