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?

Which of the option(s) given represent(s) the bending stress distribution at crosssection AA after plastic yielding?

A | |

B | |

C | |

D |

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.

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?

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 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.

\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 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 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.

(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).

0.18 | |

0.22 | |

0.1 | |

0.05 |

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.

25 | |

50 | |

65 | |

85 |

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.

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))

q:24 is wrong

ans is c