Question 1 |

A circular solid shaft of span L=5 \mathrm{~m} is fixed at one end and free at the other end. A torque T= 100 \mathrm{kN} . \mathrm{m} is applied at the free end. The shear modulus and polar moment of inertia of the section are denoted as \mathrm{G} and \mathrm{J}, respectively. The torsional rigidity \mathrm{GJ} is 50,000 \mathrm{kN} . \mathrm{m}^ 2 / \mathrm{rad}. The following are reported for this shaft:

Statement i) The rotation at the free end is 0.01 \mathrm{rad}

Statement ii) The torsional strain energy is 1.0 kN.m

With reference to the above statements, which of the following is true?

Statement i) The rotation at the free end is 0.01 \mathrm{rad}

Statement ii) The torsional strain energy is 1.0 kN.m

With reference to the above statements, which of the following is true?

Both the statements are correct | |

Statement i) is correct, but Statement ii) is wrong | |

Statement i) is wrong, but Statement ii) is correct | |

Both the statements are wrong |

Question 1 Explanation:

\phi_{\mathrm{BA}}=\frac{\text { T.L. }}{\mathrm{GJ}}=\frac{(100)^{*} 5}{50000}=0.01 \mathrm{rad}

\Rightarrow Torsional strain energy (U)

\begin{aligned} & U=\frac{T^{2} L}{2 G J}=\frac{1}{2} \times T * \phi_{B A} \\ & U=\frac{1}{2} * 100 * 0.01=0.5 \mathrm{kN}-\mathrm{m} \end{aligned}

Hence, statement (i) correct and statement (ii) is incorrect.

Question 2 |

A solid circular torsional member OPQ is subjected to torsional moments as shown in the figure (not to scale). The yield shear strength of the constituent material is 160 MPa.

The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is ________

The absolute maximum shear stress in the member (in MPa, round off to one decimal place) is ________

15.3 | |

12.6 | |

18.2 | |

11.8 |

Question 2 Explanation:

\begin{aligned} \tau_{\max _{O P}}=\frac{16 T_{O P}}{\pi d_{O P}^{3}}=\frac{16 \times 3 \times 10^{3}}{\pi \times 100^{3}}=15.27 \mathrm{~N} / \mathrm{mm}^{2} \\ \tau_{\max \mathrm{PQ}}=\frac{16 T_{P Q}}{\pi d_{P Q}^{3}}=\frac{16 \times 1 \times 10^{6}}{\pi \times 80^{3}}=9.94 \mathrm{~N} / \mathrm{mm}^{2}\\ \tau_{\max }=9.94 \mathrm{~N} / \mathrm{mm}^{2} \end{aligned}

Absolute max shear stress = 15.27 \mathrm{~N} / \mathrm{mm}^{2}

Question 3 |

A plane frame shown in the figure has linear elastic springs at node H. The spring constants are k_x=k_y=5 \times 10^5kN/m and k_\theta =3 \times 10^5kNm/rad.

For the externally applied moment of 30 kNm at node F, the rotation (in degrees, round off to 3 decimals) observed in the rotational spring at node H is_____

For the externally applied moment of 30 kNm at node F, the rotation (in degrees, round off to 3 decimals) observed in the rotational spring at node H is_____

0.001 | |

0.003 | |

0.005 | |

0.009 |

Question 3 Explanation:

No moment can be taken by segment FE.

\therefore \;\;M_{FE}=0

\begin{aligned} M_\theta &=k_\theta \times \theta \\ 30kNm&=3 \times 10^5 \times \theta \\ \theta &=1 \times 10^{-4} \; \text{radians} \\ &=0.0057^{\circ} \end{aligned}

\therefore \;\;M_{FE}=0

\begin{aligned} M_\theta &=k_\theta \times \theta \\ 30kNm&=3 \times 10^5 \times \theta \\ \theta &=1 \times 10^{-4} \; \text{radians} \\ &=0.0057^{\circ} \end{aligned}

Question 4 |

A closed thin-walled tube has thickness, t, mean enclosed area within the boundary of the centerline of tube's thickness, A_m, and shear stress, \tau. Torsional moment of resistance, T, of the section would be

0.5 \tau A_m t | |

\tau A_m t | |

2 \tau A_m t | |

4 \tau A_m t |

Question 4 Explanation:

Shear Stress,

\begin{aligned} \tau &=\frac{T}{J}R=\frac{T}{2 \pi R^3 t}R \\ \tau &=\frac{T}{2 \pi R^2 t}=\frac{T}{2 A_m t} \\ T &=2\tau A_mt \end{aligned}

Question 5 |

A solid circular beam with radius of 0.25 m and length of 2 m is subjected to a twisting moment of 20 kNm about the z-axis at the free end, which is the only load acting as shown in the figure. The shear stress component \tau_{xy} at Point 'M' in the cross-section of the beam at a distance of 1 m from the fixed end is

0.0 MPa | |

0.51 MPa | |

0.815 MPa | |

2.0 MPa |

Question 5 Explanation:

The only non-zero stresses are \tau_{\theta z}=\tau_{z \theta}=\tau

if \theta is 90^{\circ} then \theta=y

\begin{aligned} \text{Hence}\quad\tau_{z y} &=\tau_{y z}=\tau_{\max } \\ &=16 T / \pi \mathrm{d}^{3}=0.815 \mathrm{MPa} \end{aligned}

But in rest of the planes shear stresses are zero, hence, \tau_{x y}=\tau_{y x}=0

There are 5 questions to complete.