Question 1 |

A prismatic bar PQRST is subjected to axial loads as shown in the figure. The segments having maximum and minimum axial stresses, respectively, are

QR and PQ | |

ST and PQ | |

QR and RS | |

ST and RS |

Question 1 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 2 |

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

0.01 and 0.01 | |

0.02 and 0.01 | |

0.01 and 0.02 | |

0.02 and 0.02 |

Question 2 Explanation:

Elastic strain : Which can be recovered = 0.03 - 0.01 = 0.02

Plastic strain : Permanent strain = 0.01

Plastic strain : Permanent strain = 0.01

Question 3 |

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).

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).

16.56 | |

56.94 | |

78.25 | |

68.42 |

Question 3 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}

\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}

Question 4 |

Bars of square and circular cross-section with 0.5 m length are made of a material with
shear strength of 20 MPa. The square bar cross-section dimension is 4 cm x 4 cm and
the cylindrical bar cross-section diameter is 4 cm. The specimens are loaded as shown
in the figure.

Which specimen(s) will fail due to the applied load as per maximum shear stress theory?

Which specimen(s) will fail due to the applied load as per maximum shear stress theory?

Tensile and compressive load specimens | |

Torsional load specimen | |

Bending load specimen | |

None of the specimens |

Question 4 Explanation:

\begin{aligned} \sigma &=\frac{80 \times 10^{3}}{40^{2}}=50 \mathrm{N} / \mathrm{mm}^{2} \\ \tau_{\max } &=\frac{\sigma}{2}=25 \mathrm{N} / \mathrm{mm}^{2} \gt 20 \mathrm{MPa} \\ \tau_{\max } &=\frac{16 T}{\pi d^{3}}=\frac{16 \times 64 \pi \times 10^{3}}{\pi(40)^{3}} \\ &=16 \mathrm{N} / \mathrm{mm}^{2} \lt 20 \mathrm{MPa} \\ \sigma &=\frac{320 \times 10^{3}}{40^{3}}=30 \mathrm{N} / \mathrm{mm}^{2} \\ 6 &=\frac{\sigma}{2}=15 \mathrm{N} / \mathrm{mm}^{2} \lt 20 \mathrm{MPa} \\ \tau_{\max } &=\frac{\sigma}{2} \end{aligned}

Question 5 |

Consider a linear elastic rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume plane stress conditions in the plane normal to the thickness. The Young's modulus E=200 MPa and Poisson's ratio v=0.3 are given. The principal strains in the plane of the sheet are

(0.35, -0.15) | |

(0.5, 0.0) | |

(0.5, -0.15) | |

(0.5, -0.5) |

Question 5 Explanation:

\begin{aligned} \sigma_{\mathrm{x}}&=100 \mathrm{MPa} \\ v&=\mu=0.3 \\ \sigma_{\mathrm{y}}&=0, \sigma_{\mathrm{z}}=0, \mathrm{E}=200 \mathrm{MPa} \\ &\text { Principal strain in x-direction } \\ &=\epsilon_{1}=\epsilon_{\mathrm{x}}=\frac{\sigma_{\mathrm{x}}}{\mathrm{E}}-\mu \frac{\sigma_{\mathrm{y}}}{\mathrm{E}} \\ &=\frac{100}{200}-0=0.5\\ &\text { Principal strain in y-direction } \\ &=\epsilon_{2}=\epsilon_{\mathrm{y}}=\frac{\sigma_{\mathrm{y}}}{\mathrm{E}}-\mu \frac{\sigma_{\mathrm{x}}}{\mathrm{E}}\\ &=0-(0.3)\left(\frac{100}{200}\right)=-0.15\\ &\therefore\left(\epsilon_{\mathrm{x}}, \epsilon_{\mathrm{y}}\right)=(0.5-0.15) \end{aligned}

Question 6 |

In a UTM experiment, a sample of length 100 mm, was loaded in tension until failure. The failure load was 40 kN. The displacement, measured using the cross-head motion, at failure, was 15 mm. The compliance of the UTM is constant and is given by 5 \times 10^{-8} m/N. The strain at failure in the sample is _______%

8 | |

16 | |

13 | |

6 |

Question 6 Explanation:

Total Strain =\frac{\delta L}{L}=\frac{15 \times 100}{100}=15

Deformation recoverable at applied failure load =5 \times 10^{-8} \times 40 \times 10^3 \times 10^3 mm

\delta _{recov}=2mm

Recoverable strain =\frac{2 \times 100}{100}=2

Permanent strain = total strain – recoverable strain = 15% - 2% = 13%

Deformation recoverable at applied failure load =5 \times 10^{-8} \times 40 \times 10^3 \times 10^3 mm

\delta _{recov}=2mm

Recoverable strain =\frac{2 \times 100}{100}=2

Permanent strain = total strain – recoverable strain = 15% - 2% = 13%

Question 7 |

A plane-strain compression (forging) of a block is shown in the figure. The strain in the z-direction is zero. The yield strength (S_y) in uniaxial tension/compression of the material of the block is 300 MPa and it follows the Tresca (maximum shear stress) criterion. Assume that the entire block has started yielding. At a point where \sigma _x=40 MPa (compressive) and \tau _{xy}=0, the stress component \sigma _y is

340 MPa (compressive) | |

340 MPa (tensile) | |

260 MPa (compressive) | |

260 MPa (tensile) |

Question 7 Explanation:

\begin{array}{l} \tau_{x y}=0 \\ \text { Principal Stress, } \sigma_{z}=\frac{\sigma_{x}+\sigma_{y}}{2} \\ \therefore \sigma_{x}=\sigma_{\max }, \sigma_{y}=\sigma_{\min } \\ \therefore \sigma_{x}-\sigma_{y}=\sigma_{0} \\ -40-\sigma_{y}=300 \\ \Rightarrow \sigma_{y}=-340 \mathrm{MPa}=340 \mathrm{MPa}(\text {Compression}) \end{array}

Question 8 |

Consider the stress-strain curve for an ideal elastic-plastic strain hardening metal as shown in the figure. The metal was loaded in uniaxial tension starting from O. Upon loading, the stress-strain curve passes through initial yield point at P, and then strain hardens to point Q, where the loading was stopped. From point Q, the specimen was unloaded to point R, where the stress is zero. If the same specimen is reloaded in tension from point R, the value of stress at which the material yields again is _________ MPa.

210 | |

70 | |

420 | |

480 |

Question 8 Explanation:

Yield strength will increase to 210 MPa due to strain hardening.

Question 9 |

A solid cube of side 1 m is kept at a room temperature of 32 ^{\circ}C. The coefficient of linear thermal expansion of the cube material is 1 \times 10^{-5}/^{\circ}C and the bulk modulus is 200 GPa. If the cube is constrained all around and heated uniformly to 42 ^{\circ}C, then the magnitude of volumetric (mean) stress (in MPa) induced due to heating is____

20 | |

60 | |

80 | |

100 |

Question 9 Explanation:

\begin{aligned} &\therefore \quad \text { Volumetric strain } \epsilon_{\mathrm{v}}=\frac{\sigma}{\mathrm{k}}=3 \alpha(\mathrm{t})\\ &\begin{aligned} \therefore \quad \sigma &=3 \alpha \mathrm{t}(\mathrm{k}) \\ &=3 \times 1 \times 10^{-5}(42-32) \times 200 \times 10^{3} \\ \therefore \quad \sigma &=60 \mathrm{MPa} \end{aligned} \end{aligned}

Question 10 |

The true stress (in MPa) versus true strain relationship for a metal is given by

\sigma =1020\, \varepsilon^{0.4}

The cross-sectional area at the start of a test (when the stress and strain values are equal to zero) is 100 mm^{2}. The cross-sectional area at the time of necking (in mm^{2}) is ________ (correct to two decimal places)

\sigma =1020\, \varepsilon^{0.4}

The cross-sectional area at the start of a test (when the stress and strain values are equal to zero) is 100 mm^{2}. The cross-sectional area at the time of necking (in mm^{2}) is ________ (correct to two decimal places)

25.25 | |

67.03 | |

90.21 | |

110.36 |

Question 10 Explanation:

Given,\qquad \sigma=1020 \varepsilon^{0.4}

at UTS, \qquad\varepsilon_{T}=n=0.4

at UTS, neak formation starts,

\begin{aligned} \varepsilon_{T} &=0.4=\ln \left(\frac{A_{0}}{A_{f}}\right) \\ 0.4 &=\ln \left(\frac{100}{A_{f}}\right) \\ \frac{A_{0}}{A_{f}} &=e^{0.4} \\ A_{f} &=\frac{100}{e^{0.4}} \end{aligned}

Cross-section area at the time of necking,

A_{f}=67.032 \mathrm{mm}^{2}

at UTS, \qquad\varepsilon_{T}=n=0.4

at UTS, neak formation starts,

\begin{aligned} \varepsilon_{T} &=0.4=\ln \left(\frac{A_{0}}{A_{f}}\right) \\ 0.4 &=\ln \left(\frac{100}{A_{f}}\right) \\ \frac{A_{0}}{A_{f}} &=e^{0.4} \\ A_{f} &=\frac{100}{e^{0.4}} \end{aligned}

Cross-section area at the time of necking,

A_{f}=67.032 \mathrm{mm}^{2}

There are 10 questions to complete.

Explaination of Q6 is wrong.

Can you please specify with more details..

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%

compliance error = 2mm