# Stress-strain Relationship and Elastic Constants

 Question 1
Which one of the following is the definition of ultimate tensile strength (UTS) obtained from a stress-strain test on a metal specimen?
 A Stress value where the stress-strain curve transitions from elastic to plastic behavior B The maximum load attained divided by the original cross-sectional area C The maximum load attained divided by the corresponding instantaneous crosssectional area D Stress where the specimen fractures
GATE ME 2022 SET-2   Strength of Materials
Question 1 Explanation:
Tensile Strength: The tensile strength, or ultimate tensile strength (UTS), is the maximum load obtained in a tensile test, divided by the original cross-sectional area of the specimen.
$\sigma _u=\frac{P_{max}}{A_o}$
where, $\sigma _u=$ Ultimate tensile strength, $kg/mm^2$
$P_{max}=$ Maximum load obtained in a tensile test, kg
$A_o=$ Original cross-sectional area of gauge length of the test piece, $mm^2$
The tensile strength is a very familiar property and widely used for identification of a material. It is very easy to determine and is a quite reproducible property. It is used for the purposes of specifications and for quality control of a product. Tensile strength can be empirically correlated to other properties such as hardness and fatigue strength. For brittle materials, the tensile strength is a valid criterion for design.
 Question 2
Assuming the material considered in each statement is homogeneous, isotropic, linear elastic, and the deformations are in the elastic range, which one or more of the following statement(s) is/are TRUE?
MSQ
 A A body subjected to hydrostatic pressure has no shear stress B If a long solid steel rod is subjected to tensile load, then its volume increases. C Maximum shear stress theory is suitable for failure analysis of brittle materials. D If a portion of a beam has zero shear force, then the corresponding portion of the elastic curve of the beam is always straight.
GATE ME 2022 SET-1   Strength of Materials
Question 2 Explanation:
(c) Wrong : Maximum shear stress theory good for ductile material.
(d) If shear force = 0, M = C, But elastic curve is always non-linear.
 Question 3
A steel cubic block of side 200 mm is subjected to hydrostatic pressure of $250 N/mm^2$. The elastic modulus is $2 \times 10^5 \; N/mm^2$ and Poisson ratio is 0.3 for steel. The side of the block is reduced by _____mm (round off to two decimal places).
 A 0.18 B 0.22 C 0.1 D 0.05
GATE ME 2021 SET-2   Strength of Materials
Question 3 Explanation:

\begin{aligned} E &=200 \mathrm{GPa} \\ \sigma &=250 \mathrm{MPa} \\ \mu &=0.3 \\ \epsilon_{x} &=\epsilon_{y}=\epsilon_{z}=\frac{\delta a}{a} \\ \frac{1}{E}\left[\sigma_{x}-\mu\left(\sigma_{y}+\sigma_{z}\right)\right] &=\frac{(\delta a)}{a} \\ \left(\frac{-\sigma}{E}\right)(1-2 \mu)&=\frac{\delta a}{a}\\ \delta a &=-\frac{(250)(1-0.6)(200)}{200 \times 10^{3}} \\ \delta a &=(-) 0.10 \mathrm{~mm} \end{aligned}
Reduction in side of cube is 0.10mm.
 Question 4
A wire of circular cross-section ofdiameter 1.0 mm is bent into a circular arc of radius 1.0 m by application of pure bending moments at its ends. The Young's modulus of the material of the wire is 100 GPa. The maximum tensile stress developed in the wire is ______MPa.
 A 25 B 50 C 65 D 85
GATE ME 2019 SET-2   Strength of Materials
Question 4 Explanation:
\begin{aligned} &\mathrm{d}=1.0 \mathrm{mm}\\ &\mathrm{y}_{\max }=\frac{\mathrm{d}}{2}=\frac{1.0}{2}=0.5 \mathrm{mm}\\ &\mathrm{R}=1.0 \mathrm{m}=1000 \mathrm{mm}\\ &\mathrm{E}=100 \mathrm{GPa}=100 \times 10^{3} \mathrm{MPa}\\ &\text { From bending equation, } \frac{\mathrm{f}_{\max }}{\mathrm{y}_{\max }}=\frac{\mathrm{E}}{\mathrm{R}}\\ &\mathrm{f}_{\max }=\mathrm{y}_{\max } \cdot \frac{\mathrm{E}}{\mathrm{R}}\\ &=0.5 \times \frac{100 \times 10^{3}}{1000}=50 \mathrm{MPa}\\ &\therefore \mathrm{f}_{\max }=50 \mathrm{MPa} \end{aligned}
 Question 5
At a critical point in a component, the state of stress is given as $\sigma _{xx}=100 MPa$, $\sigma _{yy}=220 MPa$, $\sigma _{xy}=\sigma _{yx}=80 MPa$ and all other stress components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is ______ (round off to one decimal place).
 A 1.7 B 0.8 C 1.2 D 2.6
GATE ME 2019 SET-1   Strength of Materials
Question 5 Explanation:
$\begin{array}{l} \sigma_{x}=100 \mathrm{MPa}, \sigma_{y y}=220 \mathrm{MPa}, \tau_{x y}=80 \mathrm{MPa} \\ S_{y t}=468 \mathrm{MPa}, \mathrm{F.S.}=?(\mathrm{MSST}) \\ \sigma_{1, 2}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\left(\tau_{x y}\right)^{2}} \\ =\frac{100+220}{2} \pm \sqrt{\left(\frac{100-220}{2}\right)^{2}+80^{2}} \\ \sigma_{1}=257.08 \mathrm{MPa}, \sigma_{2}=62.9 \mathrm{MPa}, \sigma_{3}=0 \end{array}$
According to maximum shear stress theory,
$\begin{array}{l} \tau_{\max }=\frac{S_{y t}}{2 \times F S}=\operatorname{Max}\left[\left|\frac{\sigma_{1}-\sigma_{2}}{2}\right|,\left|\frac{\sigma_{1}}{2}\right|,\left|\frac{\sigma_{2}}{2}\right|\right]\\ \tau_{\max }=\frac{257.08}{2}=128.54 \mathrm{MPa} \\ 128.54=\frac{468}{2 \times \mathrm{FS}} \\ \mathrm{FS}=1.78 \end{array}$
 Question 6
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
GATE ME 2018 SET-1   Strength of Materials
Question 6 Explanation:

 Question 7
A rectangular region in a solid is in a state of plane strain. The (x,y) coordinates of the corners of the under deformed rectangle are given by P(0,0), Q (4,0), S (0,3). The rectangle is subjected to uniform strains,$\varepsilon _{xx}=0.001$ , $\varepsilon _{yy}=0.002$,$\gamma _{xy}=0.003$.The deformed length of the elongated diagonal, up to three decimal places, is _________ units.
 A 4.547 B 4.845 C 5.015 D 6.348
GATE ME 2017 SET-1   Strength of Materials
Question 7 Explanation:
$\tan \theta=\frac{4}{3}$

\begin{aligned} \theta &=53.13^{\circ} \\ 2 \theta &=106.26^{\circ} \\ \cos 2 \theta &=-0.27999 \\ \sin 2 \theta &=0.96 \end{aligned}
Normal strain on a oblique plane (i.e. RP) inclined
at an angle $(\theta)$ is given by
\begin{aligned} \left(\varepsilon_{n}\right)_{\theta}=& \frac{1}{2}\left[\varepsilon_{x}+\varepsilon_{y}\right]+\frac{1}{2}\left[\varepsilon_{x}-\varepsilon_{y}\right] \cos 2 \theta \\ & +\left(\frac{\gamma x y}{2}\right) \sin 2 \theta \\ =& \frac{1}{2}[0.001+0.002]+\frac{1}{2}[0.001-0.002] \\ &[-0.27999]+\left(\frac{0.003}{2}\right)(0.96) \\ &\left(\varepsilon_{n}\right)_{\theta=53.13^{\circ}}=0.00308 & \\ =& \frac{\text { Change in length of diagonal }}{\text { Original length of diagonal }}=0.00308 \end{aligned}
Hence, change in length of diagonal
$=(0.00308) 5=0.0154 \mathrm{mm}$
deformed length of diagonal
$=5+0.0154=5.0154 \mathrm{mm}$
 Question 8
In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to
 A Yield stress B Proportional limit C Maximum stress D Fracture stress.
GATE ME 2017 SET-1   Strength of Materials
Question 8 Explanation:
Ultimate tensile strength represents the max. stress that a material can withstand without fracture.
 Question 9
In a metal forming operation when the material has just started yielding, the principal stresses are $\sigma _{1}=+180$ Mpa,$\sigma _{2}=-100$ Mpa,$\sigma _{3}=0$. Following Von Mises criterion, the yield stress is ________ MPa.
 A 245.76 B 240.12 C 248.57 D 251.98
GATE ME 2017 SET-1   Strength of Materials
Question 9 Explanation:
Safe condition for design as per von-Mises criterion,
$\sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2} \leq\left(\frac{S_{y t}}{N}\right)^{2}$
When yielding occurs, N=1
$\begin{array}{r} \sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2}=\left(S_{y t}\right)^{2} \\ (180)^{2}+(-100)^{2}-(180)(-100)=\left(S_{y t}\right)^{2} \\ S_{t y}=245.7641 \mathrm{MPa} \end{array}$
 Question 10
The poisson's ratio for a perfectly incompressible linear material is
 A 1 B 0.5 C 0 D infinity
GATE ME 2017 SET-1   Strength of Materials
Question 10 Explanation:
For perfectly incompressible material, poisson's ratio =0.5
(if change in vol. should be zero,$\mu=0.5$)
There are 10 questions to complete.

### 2 thoughts on “Stress-strain Relationship and Elastic Constants”

1. Calculation mistake in question 18,
Sigma x is negative, and sigma y is positive, and in under root how there is 25+64 …there should be 1+64 so root 65

Thanks

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2. Check out question 24, it should be root((150²)+(100²))

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