# Stress-strain Relationship and Elastic Constants

 Question 1
A beam is undergoing pure bending as shown in the figure. The stress $(\sigma )$-strain $(\varepsilon )$ curve for the material is also given. The yield stress of the material is $(\sigma _Y)$.
Which of the option(s) given represent(s) the bending stress distribution at crosssection $AA$ after plastic yielding?

 A A B B C C D D
GATE ME 2023   Strength of Materials
Question 1 Explanation:
If bending stress at every fibre is equal to yield strength of the material, then entire cross-section undergoes plastic bending. Then, option ( C ) is correct.
If bending stress at inner fibre is less than yield strength and bending stress at extreme fibres is equal to yield strngth, Then, option ( D ) is correct.
 Question 2
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 2 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 3
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 3 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 4
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 4 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 5
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 5 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}

There are 5 questions to complete.

### 3 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