Question 1 |

A cylindrical rod of diameter 10 mm and length 1.0 m is fixed at one end. The other end is twisted by an angle of 10^{\circ} by applying a torque. If the maximum shear strain in the rod is p \times 10^{-3}, then p is equal to ______ (round off to two decimal places).

0.25 | |

0.5 | |

0.8 | |

0.6 |

Question 1 Explanation:

\begin{aligned} &\mathrm{d}=10 \mathrm{mm}\\ &\theta=10^{\circ}=10^{\circ} \times \frac{\pi}{180} \text { radian }\\ &\mathrm{R}=\frac{\mathrm{d}}{2}=5 \mathrm{mm}\\ &\mathrm{L}=1 \mathrm{m}\\ &\phi_{\max }=P \times 10^{-3}\\ &\text { from torsion equation, } \\ \frac{\mathrm{T}}{\mathrm{J}}&=\frac{\tau_{\max }}{\mathrm{R}}=\frac{\mathrm{G} \theta}{\ell}\\ &\Rightarrow \frac{\tau_{\max }}{G}=\frac{R \theta}{\ell}\\ &\phi_{\max }=\frac{R \theta}{\ell} \\ P \times 10^{-3}&=\frac{5 \times 10^{\circ} \times \frac{\pi}{180}}{1000}=0.8726 \times 10^{-3} \\ \therefore \;\;P&=0.8726\end{aligned}

Question 2 |

A bar of circular cross section is clamped at ends P and Q as shown in the figure. A
torsional moment T=150 Nm is applied at a distance of 100 mm from end P. The
torsional reactions \left ( T_{P}, T_{Q} \right ) in Nm at the ends P and Q respectively are

\left (50,100\right) | |

\left (75,75\right) | |

\left (100,50 \right ) | |

\left (120,30 \right ) |

Question 2 Explanation:

\begin{aligned} T_{P}+T_{Q} &=T \ldots(i)\\ Q_{P R}+Q_{R Q} &=0 \\ \frac{T_{P} \cdot L}{G I_{P}}+\frac{\left(T_{P}-T\right) 2 L}{G I_{P}} &=0 \\ T_{P}+2 T_{P} &=2 T \\ T_{P} &=\frac{2 T}{3}=100 \mathrm{Nm}\\ \text{From equation (i)}\\ T_{Q}&=\frac{T}{3}=50 \mathrm{N}-\mathrm{m} \end{aligned}

Question 3 |

A hollow circular shaft of inner radius 10 mm, outer radius 20 mm and length 1 m is to be used as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is ________ (correct to two decimal places).

12.36 | |

18.26 | |

35.34 | |

48.22 |

Question 3 Explanation:

\qquad \begin{aligned} \text { Torsional stiffness }&=\frac{G I_{P}}{L} \\ &=\frac{150 \times 10^{9} \times \frac{\pi}{32}\left[0.04^{4}-0.02^{4}\right]}{1} \\ &=35343 \mathrm{Nm} / \mathrm{rad}=35.343 \mathrm{kNm} / \mathrm{rad} \end{aligned}

Question 4 |

For an Oldham coupling used between two shafts, which among the following statements are correct?

I. Torsional load is transferred along shaft axis.

II. A velocity ratio of 1:2 between shafts is obtained without using gears.

III. Bending load is transferred transverse to shaft axis.

IV. Rotation is transferred along shaft axis.

I. Torsional load is transferred along shaft axis.

II. A velocity ratio of 1:2 between shafts is obtained without using gears.

III. Bending load is transferred transverse to shaft axis.

IV. Rotation is transferred along shaft axis.

I and III | |

I and IV | |

II and III | |

II and IV |

Question 4 Explanation:

Oldham coupling is used to connect two shafts which are not on the same axis means they are not aligned to the same axis

So,

(1) Torsional load is transferred along shaft axis as both shafts are rotating member.

So, that statement is correct.

(2) A velocity ratio 1:1 between shafts is obtained using gears.

So, this statement is wrong

(3) Bending load is not transferred transverse to shaft axis as there is no transverse load.

(4) Rotation is transferred along shaft axis

So, this statement is correct

So,

(1) Torsional load is transferred along shaft axis as both shafts are rotating member.

So, that statement is correct.

(2) A velocity ratio 1:1 between shafts is obtained using gears.

So, this statement is wrong

(3) Bending load is not transferred transverse to shaft axis as there is no transverse load.

(4) Rotation is transferred along shaft axis

So, this statement is correct

Question 5 |

A motor is driving a solid circular steel shaft transmits 40kW of power at 500 rpm. If the diameter of the shaft is 40 mm, The maximum shear stress in the shaft is ____MPa.

60.7 | |

50.5 | |

55.25 | |

51.36 |

Question 5 Explanation:

\begin{aligned} T &=\frac{P \times 60 \times 10^{6}}{2 \pi N}=\frac{40 \times 60 \times 10^{6}}{2 \pi(500)} \\ &=763943.7268 \mathrm{N}-\mathrm{mm} \\ \tau_{\max } &=\frac{16 T}{\pi d^{3}}=\frac{16 \times 763943.7268}{\pi(40)^{3}} \\ &=60.792 \mathrm{MPa} \end{aligned}

Question 6 |

Two circular shafts made of same material, one solid (S) and one hollow (H), have the same length and polar moment of inertia. Both are subjected to same torque. Here, \theta_{s} is the twist and \tau_{s} is the maximum shear stress in the solid shaft, whereas \theta_{H} is the twist and \tau_{H} is the maximum shear stress in the hollow shaft. Which one of the following is TRUE?

\theta_{s}=\theta_{h} and \tau_{s}=\tau_{H} | |

\theta_{s} \gt \theta_{h} and \tau_{s} \gt \tau_{H} | |

\theta_{s}\lt\theta_{h} and \tau_{s}\lt\tau_{H} | |

\theta_{s}=\theta_{h} and \tau_{s}\lt\tau_{H} |

Question 6 Explanation:

\begin{aligned} L_s=L_H, & T_s=T_H=T,\\ I_{ps} &=I_{pH}\\ \therefore \; \frac{\pi}{32}d^4&=\frac{\pi}{32}(d_o^4-d_i^4)\\ \therefore \; d_o & \gt d\\ \therefore \;r_o &\ gt r\\ \text{Angle of twist}\\ \theta &=\frac{TL}{GI_p}\\ \therefore \; \theta _s&=\theta _H\\ \tau _{max_s}&=\frac{I}{I_p}r\\ \tau _{max_H}&=\frac{I}{I_p}r_o\\ \therefore \; \tau _{max_H}& \gt \tau _{max_s} \end{aligned}

Question 7 |

The cross-sections of two solid bars made of the same material are shown in the figure. The square cross-section has flexural (bending) rigidity I1, while the circular cross-section has flexural rigidity I2. Both sections have the same cross-sectional area. The ratio I1/I2 is

1/\pi | |

2/\pi | |

\pi/3 | |

\pi/6 |

Question 7 Explanation:

\begin{aligned} \therefore \quad b^{2}&=\frac{\pi}{4} d^{2} \\ \Rightarrow \quad \frac{b^{2}}{d^{2}}&=\frac{\pi}{4} \\ I_{1} &=\frac{b^{4}}{12} \\ I_{2} &=\frac{\pi d^{4}}{64} \\ \therefore \qquad \frac{I_{1}}{I_{2}} &=\frac{b^{4}}{12} \times \frac{64}{\pi d^{4}}=\frac{16}{3 \pi} \frac{b^{4}}{d^{4}} \\ &=\frac{16}{3 \pi} \cdot \frac{\pi^{2}}{16}=\frac{\pi}{3} \end{aligned}

Question 8 |

A simply supported beam of length 2L is subjected to a moment M at the mid-point x = 0 as shown in the figure. The deflection in the domain o\leq x\leq L is given by

w=\frac{-Mx}{12EIL}(L-x)(x+c)

where E is the Young's modulus, I is the area moment of inertia and c is a constant (to be determined)

The slope at the center x = 0 is

w=\frac{-Mx}{12EIL}(L-x)(x+c)

where E is the Young's modulus, I is the area moment of inertia and c is a constant (to be determined)

The slope at the center x = 0 is

ML/(2EI) | |

ML/(3EI) | |

ML/(6EI) | |

ML/(12EI) |

Question 8 Explanation:

Equation of deflection

\begin{aligned} W &=\frac{-M}{12 E I L}\left[L x^{2}-x^{3}+L c x-C x^{2}\right] \\ \frac{d W}{d x} &=\frac{-M}{12 E I L}\left[2 L x-3 x^{2}+L C-2 C x\right] \\ \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 E I L}[2 L-6 x-2 C] \\ \therefore \quad E I \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 L}[2 L-6 x-2 C] \\ \text { now } \quad E I \frac{d^{2} W}{d x^{2}} &=\frac{-M}{12 L}[2 L-6 x-2 C]\\ \text{at }\quad x&=L, M_{x}=0 \\ \therefore \quad C&=2 L\\ Slope\\ \frac{d W}{d x} &=\frac{-M}{12 E I L}\left[2 L x-3 x^{2}+2 L^{2}-4 L x\right] \\ \therefore \quad \left|\frac{d W}{d x}\right|_{x=0} &=\frac{-M L}{6 E I} \end{aligned}

Question 9 |

A rigid horizontal rod of length 2L is fixed to a circular cylinder of radius R as shown in the figure. Vertical forces of magnitude P are applied at the two ends as shown in the figure. The shear modulus for the cylinder is G and the Young's modulus is E.

The vertical deflection at point A is

The vertical deflection at point A is

PL^{3}/(\pi R^{4}G) | |

PL^{3}/(\pi R^{4}E) | |

2PL^{3}/(\pi R^{4}E) | |

4PL^{3}/(\pi R^{4}G) |

Question 9 Explanation:

Angle of twist \theta

\theta =\frac{(2PL)L}{GI_p}=\frac{2PL^2}{G\frac{\pi}{2}R^4}=\frac{4PL^2}{\pi GR^4}

\therefore \; Vertical deflection of A

\Delta _A=\theta L=\frac{4PL^3}{\pi G R^4}

Question 10 |

A machine element XY, fixed at end X, is subjected to an axial load P, transverse load F, and a twisting moment T at its free end Y. The most critical point from the strength point of view is

a point on the circumference at location Y | |

a point at the center at location Y | |

a point on the circumference at location X | |

a point at the center at location X |

Question 10 Explanation:

Both bending stress and shear stress (due to torque) will be max at the circumference. The bending stress will be maximum at x. Hence, critical section will be at a point on the circumference at location x.

There are 10 questions to complete.

in question 24 torque is 1.0 Nm not 10 Nm. Please correct it

In this topic only total 23 questions are there. So can you please verify your question number

I think he means question no. 21. The value of Torque is 1 instead of 10. Or the answer is wrong (1.0 radians instead of 0.1 radians)

Question 11, in explanation factors should be ((r4⁴-r3⁴)(r4⁴ + r3⁴)) in both numerator and denominator.

Question 21, torque is 1 Nm

Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=0 at x=-L or assuming Mz=M at x=0. But we actually get different values of ‘c’ in those different assumptions. Hence, it is proved that the question is wrong.

Correction*

Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=M at x=0 since the expression is given to be valid at 0≤x≤L . But we actually get different value of ‘c’ in this assumption. Hence, it is proved that the question is wrong.

I am really sorry.. the question is absolutely correct.. but correct your solution where it is written that c=2L but it would be c=-2L..

Atttt…Kamalasan

Question no 8 is actually wrong.. Let me explain.. In the explanation of this question as shown in this page, we get c=2L assuming Mz=0 at x=L… Now, if the question would be correct, then we would also get c=2L assuming Mz=M at x=0 since the expression is given to be valid at 0≤x≤L . But we actually get c=7L in this assumption. Even if we assume, Mz=-M at x=0, we get c=5L in case you have doubt in sign of M. Hence, it is proved that the question is wrong.