Torsion of Shaft MCQ Quiz in मल्याळम - Objective Question with Answer for Torsion of Shaft - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

For the same weight of the hollow shaft and solid shaft, polar section modulus (Zp) of hollow circular cross-section is more than that of solid circular cross-section.

Hence torque transmitting capacity of hollow circular cross-section is more.

Distribution of shear stresses in circular Shafts subjected to torsion :

\(\frac{\tau }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

\(\tau = \frac{{G\theta }}{L}r\)

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft.

Hence the maximum shear stress occurs on the outer surface of the shaft where r = R

For solid circular shaft:

\(\frac{\tau }{r} = \frac{T}{J}\)

\(J=\frac{\pi D^4}{32}\)

\(\tau = \frac{16T}{\pi D^3}\)

From the torsion of solid shafts of circular x – section, it is seen that only the material at the outer surface of the shaft can be stressed to the limit assigned as allowable working stress.

All of the material within the shaft will work at lower stress and is not being used to full capacity. Thus, in these cases where weight reduction is important, it is advantageous to use hollow shafts.

For hollow circular shaft:

\(J=\frac{\pi (D^4-d^4)}{32}\)

\(\tau = \frac{16TD}{\pi (D^4-d^4)}=\frac{16T}{\pi D^3 (1-(\frac{d}{D})^4)}=\frac{16T}{\pi D^3 (1-k^4)}\)

Assume d = D/2 ⇒ k = 0.5

\(\tau =\frac{16T}{\pi D^3 (1-0.5^4)}=1.066 \frac{16T}{\pi D^3 }\)

• It may be seen that the τmax in the case of the hollow shaft is 6.6% larger than in the case of a solid shaft having the same outside diameter
• The stiffness of the hollow shaft is more than the solid shaft with the same weight.
• In the hollow shaft, the material at the center is removed and spread at a large radius. 

Therefore, hollow shafts are stronger and rigid than solid shaft having the same weight.

Concept:

The torque equation is given as,

\(\frac{{\rm{T}}}{{\rm{J}}} = \frac{\tau_{max}}{R}=\frac{{{\rm{G}} \times {\rm{\theta }}}}{{\rm{L}}}\)

Where

T = Torsion, G = modulus of rigidity, θ = Angle of twist, and J = Polar moment of inertia.

The torsion T is nothing but the Maximum Torsional Resisting Moment.

And from the equation, \(\frac{{\rm{T}}}{{\rm{J}}} = \frac{{{\rm{G}} \times {\rm{\theta }}}}{{\rm{L}}}\)

i.e. Torsional resistance is proportional to the angle of twist, modulus of rigidity, and Polar moment of inertia and inversely proportional to the length of the bar.

Mistake Points

Torsional resistance is not directly proportional to the moment of inertia (It is directly proportional to Polar moment of inertia)

Explanation:

We know that Torsion equation is:

\(\frac{{{\tau _{max}}}}{R} = \frac{{G\theta }}{L} = \frac{T}{J}\)

where θ is the angle of twist, T = torque applied, L = length of the shaft, J = polar moment of inertia, d = diameter of the shaft

\(J = \frac{{\pi {d^4}}}{{32}}\)

From torsion equation

\(T \propto G\)

\(T \propto J\)

\(T \propto \frac 1L\)

\(T \propto \frac 1R\)

The twisting moment of a circular shaft increases when the Rigidity modulus & Polar moment of inertia Increase and the Length of the shaft & Radius of the shaft decreases.

Concept:

Torsion equation \(= \frac{T}{J} = \frac{{G\theta }}{\ell } = \frac{\tau }{R}\)

Where,

T is the torque applied, J = polar moment of inertia, τ = shear strength of material, G = modulus of rigidity, θ = angular deflection of shaft (in radians).

\(J = \frac{\pi }{{32}}{d^4}\)

\(\theta = \frac{{TL}}{{GJ}}\)

Calculation:

Given:

d₁ = d, d₂ = 2d, L₁ = L, L₂ = 2L

Explanation:

Different assumptions made in torsion theory are as follows:

1. Shaft must be straight and should have uniform cross-section.
2. The shear stress induced in shaft should not exceed the elastic limit.
3. Twist along the shaft is uniform.
4. Twisting has no effect on circularity of shaft.

Concept:

Power is given by;

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Where P is in watt

N = Angular velocity

T = Torque

Calculation:

1 H.P. = 745.7 watt

Given;

N = 150 rpm

T = 1500 N.m

So from the above equation;

\(P = \frac{{2π \times 150 \times 1500}}{{60}}\)

P = 7500π watt

\(P = \frac{{7500π }}{{745.7}}\) H.P.

P = 10π H.P.

Explanation:

Distribution of shear stresses in circular Shafts subjected to torsion is given by,

\(\frac{\tau }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

\(\tau = \frac{{Tr}}{J}\)

where, T = twisting moment, r = Radius of the shaft and J = polar moment of inertia

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft.

The maximum shear stress occurs on the outer surface of the shaft where r = R.

For the solid shaft, at the centre, the stress is zero.

Stress is never zero in the hollow shaft although it is minimum at the inner surface of the shaft.

The correct answer is "The length and the radius of the shaft remain fixed"

Solution:

Following assumptions are made for torsional force in a circular shaft:

1.Material of the shaft is homogenous throughout the length of the shaft.

2.Shaft is straight and of uniform circular cross section over its length.

3.Radial lines remain radial during torsion.

4.Torsion is constant along the length of the shaft.

5. Cross section of the shaft remains plane.  

Concept:

From the torsional equation of the shaft

\(\mathop \tau \nolimits_{\max } = \frac{{16T}}{{\pi \mathop D\nolimits^3 }}\)

\(T = \frac{{\mathop \tau \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}}\)

Calculation:

Given:

D = 4 cm

\(\mathop \tau \nolimits_{\max } = 20 ~\frac{kN}{cm^2}\)

\(T = \frac{{\mathop \tau \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}}\)

\(T =\frac{20 \;\times\; \pi\; \times \;4^3}{16}=80\pi\) kN-cm

Explanation:

\( \frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{L}\)

Torque per radian twist is known as torsional stiffness (k)

\(k=\frac{T}{\theta}=\frac{GJ}{L}\)

The parameter GJ is called torsional rigidity of the shaft. 

Torsional rigidity is also defined as torque per unit angular twist per unit length

\(GJ=\frac{T}{\theta/L}\)