Question 1 |

Ignoring the small elastic region, the true stress (\sigma) - true strain ( \varepsilon) variation of a
material beyond yielding follows the equation \sigma =400 \varepsilon ^{0.3} MPa. The engineering
ultimate tensile strength value of this material is ________ MPa.
(Rounded off to one decimal place)

145.6 | |

325.6 | |

206.6 | |

125.2 |

Question 1 Explanation:

Given

\begin{aligned} \sigma_{\mathrm{T}} & =400 \varepsilon^{0.3}=400 \varepsilon_{\mathrm{T}}^{0.3} \;\;\;...(i)\\ \varepsilon & =\text { true strain }=\varepsilon_{\mathrm{T}} \\ \mathrm{n} & =0.3=\text { strain hardening exponent } \end{aligned}

The true strain \left(\varepsilon_{T}\right) at the onset of necking equal to the strain hardening exponent i.e. at ultimate tensile point.

So, \quad \varepsilon_{\mathrm{T}}=\mathrm{n}=0.3 at ultimate point

So, from equation (i)

\sigma_{\mathrm{T}} =400 \times(0.3)^{0.3} =278.74 \mathrm{MPa} \text { at ultimate point }

As \varepsilon_{\mathrm{T}}=\ln (1+\varepsilon)

So, \varepsilon=\mathrm{e}^{\varepsilon} \mathrm{T}-1=\mathrm{e}^{0.3}-1=0.35

and \sigma_{\mathrm{T}}=\sigma(1+\varepsilon)

So \sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}

\begin{aligned} \sigma_{\mathrm{T}} & =400 \varepsilon^{0.3}=400 \varepsilon_{\mathrm{T}}^{0.3} \;\;\;...(i)\\ \varepsilon & =\text { true strain }=\varepsilon_{\mathrm{T}} \\ \mathrm{n} & =0.3=\text { strain hardening exponent } \end{aligned}

The true strain \left(\varepsilon_{T}\right) at the onset of necking equal to the strain hardening exponent i.e. at ultimate tensile point.

So, \quad \varepsilon_{\mathrm{T}}=\mathrm{n}=0.3 at ultimate point

So, from equation (i)

\sigma_{\mathrm{T}} =400 \times(0.3)^{0.3} =278.74 \mathrm{MPa} \text { at ultimate point }

As \varepsilon_{\mathrm{T}}=\ln (1+\varepsilon)

So, \varepsilon=\mathrm{e}^{\varepsilon} \mathrm{T}-1=\mathrm{e}^{0.3}-1=0.35

and \sigma_{\mathrm{T}}=\sigma(1+\varepsilon)

So \sigma_{\text {ultimate }}=\frac{\sigma_{T}}{1+\varepsilon}+\frac{278.74}{1+0.35}=206.55 \mathrm{MPa}

Question 2 |

The principal stresses at a point P in a solid are 70 MPa, -70 MPa and 0. The
yield stress of the material is 100 MPa. Which prediction(s) about material failure
at P is/are CORRECT?

Maximum normal stress theory predicts that the material fails | |

Maximum shear stress theory predicts that the material fails | |

Maximum normal stress theory predicts that the material does not fail | |

Maximum shear stress theory predicts that the material does not fail |

Question 2 Explanation:

\sigma _1=70MPa, \sigma _2= -70MPa, \sigma _3=0 , S_{yt}=100MPa

For maximum shear stress theory:

\frac{\sigma _1-\sigma _2 }{2}=\frac{70-(-70)}{2}=70

and \frac{S_{yt} }{2}=\frac{100}{2}=50

i.e. \frac{\sigma _1-\sigma _2 }{2} \gt \frac{S_{yt} }{2}

So material will fail As per maximum normal stress theory:

\sigma _1 \; and \; sigma _2 \gt sigma _{yt}

then material will fail

Here 70 & -70 < 100

So material is safe.

For maximum shear stress theory:

\frac{\sigma _1-\sigma _2 }{2}=\frac{70-(-70)}{2}=70

and \frac{S_{yt} }{2}=\frac{100}{2}=50

i.e. \frac{\sigma _1-\sigma _2 }{2} \gt \frac{S_{yt} }{2}

So material will fail As per maximum normal stress theory:

\sigma _1 \; and \; sigma _2 \gt sigma _{yt}

then material will fail

Here 70 & -70 < 100

So material is safe.

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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}

There are 5 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