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?

Stress value where the stress-strain curve transitions from elastic to plastic behavior | |

The maximum load attained divided by the original cross-sectional area | |

The maximum load attained divided by the corresponding instantaneous crosssectional area | |

Stress where the specimen fractures |

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.

\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 body subjected to hydrostatic pressure has
no shear stress | |

If a long solid steel rod is subjected to tensile
load, then its volume increases. | |

Maximum shear stress theory is suitable for
failure analysis of brittle materials. | |

If a portion of a beam has zero shear force,
then the corresponding portion of the elastic
curve of the beam is always straight. |

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.

(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).

0.18 | |

0.22 | |

0.1 | |

0.05 |

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.

25 | |

50 | |

65 | |

85 |

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).

1.7 | |

0.8 | |

1.2 | |

2.6 |

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}

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

increases linearly with the true strain | |

decreases linearly with the true strain | |

first increases linearly and then decreases linearly with the true strain
| |

remains constant |

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.

4.547 | |

4.845 | |

5.015 | |

6.348 |

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}

\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

Yield stress | |

Proportional limit | |

Maximum stress | |

Fracture stress. |

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.

245.76 | |

240.12 | |

248.57 | |

251.98 |

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}

\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

1 | |

0.5 | |

0 | |

infinity |

Question 10 Explanation:

For perfectly incompressible material, poisson's ratio =0.5

(if change in vol. should be zero, \mu=0.5)

(if change in vol. should be zero, \mu=0.5)

There are 10 questions to complete.

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

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