Pure Torsion MCQ Quiz in मराठी - Objective Question with Answer for Pure Torsion - मोफत PDF डाउनलोड करा

Concept:

The torsional equation for the shaft is given by,

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

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

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

Here,

The parameter GJ is called the torsional rigidity of the shaft. 

Torsional rigidity is also defined as torque per unit angular twist over the length of the shaft

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

Explanation:

Twisting moment impart an angular displacement of one end cross-section with respect to the other end and it will setup shear stresses on any cross section of the bar perpendicular to its axis.

Torsion Equation of a shaft is given by,

\(\frac{T}{J} = \frac{τ }{r} = \frac{{Gθ }}{L}\)

Polar section modulus of shaft is given by,

\({Z_p} = \frac{J}{r} = \frac{{\frac{{\pi {D^4}}}{{32}}}}{{\frac{D}{2}}} = \frac{{\pi {D^3}}}{{16}}\)

Where, T = Torque, J = Polar moment of inertia, τ = Shear stress, r = Radius of shaft, G = Shear modulus, θ = Angle of twist and L = Length of shaft

Shear stress distribution across the section of the circular shaft is shown below.

Using Torsinal formula

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

Forgiven T and J; τ ∝ r i.e.

The shear stress distribution is linear.

Shear stress at centre, τcentre = 0

Concept:

Power (P)

P = T × W

Torsion equation

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

Calculation:

Given:

P = 40 kW, N = 500 rpm, D = 40 mm

As, P = T × W

\(40 \times {10^3} = {\rm{T}} \times \frac{{2{\rm{\pi }} \times 500}}{{60}} ⇒ {\rm{T}} = 763.9{\rm{\;Nm}} = 763.9 \times {10^3}{\rm{Nmm}}\)

Now,

\(\frac{T}{J} = \frac{\tau }{R} ⇒ \tau = \frac{{TR}}{J} = \frac{{763.9 \times {{10}^3} \times 20 \times 32}}{{\pi \times {{40}^4}}}\)

Concept:

Let torque developed in brass and steel is T_b & T_s respectively

T_b + T_s = 1000      ----(i)

Since both the bars are firmly joined angle of Twist at the junction will be the same

θ_s = θ_b

\(\frac{{{T_S}L}}{{{G_s}{j_s}}} = \frac{{{T_b}L}}{{{G_b}{j_b}}}\)

\(\frac{{{T_s}}}{{{T_b}}} = \frac{{{G_s}}}{{{G_b}}}\times\frac{{({{50}^4} - {{32}^4})}}{{{{(32)}^4}}}\)

\(\frac{{{T_s}}}{{{T_b}}} = 4.96\times2=9.92\)      ----(ii)

(i) and (ii)

T_s = 908.42

T_b = 91.57

\({({z_{max}})_{steel}} = \frac{{Ts}}{{{{(zp)}_s}}} =16\times \,\,\frac{{908.42 \times {{10}^3} }}{{\pi \times 50^3 ({1} - {{\frac{32^4}{50^4}}})}}\)

z_max = 44.44 N/mm²

Concept:

\({\rm{Shear\;stress\;at\;any\;section\;is\;given\;by}},\;\tau \; = \frac{{V \times A \times \bar y\;}}{{I \times B}}\)

where V = Transverse shear, A = Area above or below the section, \(\bar y\) = distance of CG form NA, I = Moment of inertia

Shear Stress Distribution different section are as follows:

Rectangular Section

The transverse shear stress acting in a beam of rectangular cross section, subjected to a transverse shear load is variable with maximum on the neutral axis.

Circular Section

I Section

Concept:

The equation for Shaft Subjected to Torsion T

\(\frac{τ }{R}\) = \(\frac{T}{J}\) = \(\frac{{Gθ }}{L}\)

Solid shaft J = \(\frac{{\pi D_s^4}}{{32}}\)

Hollow Shaft J = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)

τ = Shear stress induced due to torsion T, G = Modulus of Rigidity, θ = Angular deflection of the shaft, R, L = Shaft radius and length respectively, Ds = Diameter of the solid shaft, D0 = Outer Diameter of Hollow shaft, Di = Inner Diameter of Hollow shaft

Calculation;

Given

D₀ = 3 × D_i

D_s = D₀  (Solid shaft diameter is equal to an outer diameter of hollow shaft)

Material is the same for the hollow and solid shaft (i.e. G is the same for both because G depends upon the material)

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

∴ T  \( \propto \;\) J  (θ, G, L same for both or constant)

\(\frac{{{T_H}}}{{{T_s}}}\) = \(\frac{{{J_H}}}{{{J_s}}}\) = \(\frac{{\frac{\pi }{{32}}\left( {D_0^4 - D_i^4} \right)}}{{\frac{\pi }{{32}}{{\left( {{D_s}} \right)}^4}}}\)

Put, D₀ = 3 × D_i 

D_s = D₀ 

\(J_H\over J_s\) = \(80\over 81\)

Additional Information

Polar moment of inertia

J =  \(\mathop \smallint \limits_0^R 2\pi \)r³dr

For Solid Shaft, J = \(\pi D^4 \over 32\)

For Hollow Shaft, J = \(\mathop \smallint \limits_r^R 2\pi \)r³dr = \(\frac{{\pi \left( {D_0^4 - \;D_i^4} \right)}}{{32}}\)

The assumption for Torsion equation:

\(\tau\over R\) = \(T\over J\) = \(G\theta\over L\)

• The bar is acted upon by pure torque.
• The section under consideration is remote from the point of application of the load and from a change in diameter.
• The material must obey Hooke's Law
•  Cross-sections rotate as if rigid, i.e. every diameter rotates through the same angle.

Concept

Maximum shear stress induced in the shaft under pure torsion is given by,

\({τ _{max}} = \frac{{16{\rm{}}T}}{{\pi {d^3}}}\)

where T = Torque applied on the shaft

Calculation

Given

Maximum shear stress developed, τ_max = 160 MPa

d₁ = d, d₂ = 2d

Since torque is constant, the only diameter is varying. Therefore the maximum shear stress varies as,

\({τ _{max}} \propto \frac{1}{{{d^3}}}\)

\(\frac{{{τ _{max}}_1}}{{{τ _{max}}_2}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^3}\)

\(\frac{{160}}{{{τ _{max}}_2}} = {\left( {\frac{{2d}}{d}} \right)^3}\)

τ_max2 = 20 MPa

Concept:

The strain energy stored in a shaft due to torsion can be derived using the relationship between torque, shear stress, and the angle of twist.

Calculation:

Given:

Torque, T

Shear modulus, G

Polar moment of inertia, J

Length of the shaft, dx

The relationship between torque T, shear stress τ, and the polar moment of inertia J is given by:

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

The angle of twist θ over a length L of the shaft is related to the torque and the material properties by:

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

The strain energy density u per unit volume for shear is given by:

\(u = \frac{1}{2} \tau \times γ \)

Shear strain γ is related to the angle of twist θ and the radial distance r by:

\(γ = \frac{r \times \theta}{L} \)

Substituting τ and γ:

\(\gamma = \frac{r \times \theta}{L} \),  \( \tau = \frac{T \times r}{J} \)  and, we get:  \(\gamma = \frac{r \times T \times L}{G \times J \times L} = \frac{r \times T}{G \times J} \)

\(u = \frac{1}{2} \left( \frac{T \times r}{J} \right) \left( \frac{r \times T}{G \times J} \right) = \frac{1}{2} \frac{T^2 \times r^2}{G \times J^2} \)

To find the total strain energy U, we integrate the strain energy density over the volume of the shaft. For a differential length dx of the shaft:

\( dU = u \times dV = \left( \frac{1}{2} \frac{T^2 \times r^2}{G \times J^2} \right) \times (2\pi r \times dr \times dx) \)

where \( dV = 2\pi r \times dr \times dx \) is the differential volume element in cylindrical coordinates

Integrating r from 0 to R (the radius of the shaft):

\( U = \int_0^L \int_0^R \frac{T^2 \times r^2}{2 G J^2} \times 2\pi r \times dr \times dx \)

Simplifying the integrand:

\( U = \int_0^L \int_0^R \frac{T^2 \times 2\pi r^3}{2 G J^2} \times dr \times dx = \int_0^L \frac{T^2 \times \pi}{G \times J^2} \left[ \frac{r^4}{4} \right]_0^R \times dx \)

Evaluating the integral:

\( U = \int_0^L \frac{T^2 \times \pi}{G \times J^2} \times \frac{R^4}{4} \times dx = \int_0^L \frac{T^2 \times \pi \times R^4}{4 G \times J^2} \times dx \)

Since :  \( J = \frac{\pi \times R^4}{2} \)

\( U = \int_0^L \frac{T^2}{2GJ} \times dx \)

The strain energy due to torsion in the shaft is:

\( U = \int_0^L \frac{T^2}{2GJ} \times dx \)

Additional InformationThe elastic strain energy stored in a member of length s (it may be curved or straight) due to axial force, bending moment, shear force and torsion is summarized below:

T₀ = T₁ + T₂                            _________________(i)

but from torsional equation

\(\frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{l} \Rightarrow T = \frac{{GJ\theta }}{l}\)

or \(T \propto \frac{1}{l} \Rightarrow {T_1}{l_1} = {T_2}{l_2}\)

\(\Rightarrow {T_1} \times \frac{L}{4} = {T_2} \times \frac{{3L}}{4} \Rightarrow {T_1} = 3{T_2}\)

from equation (i) \({T_0} = 3{T_2} + {T_2} = 4{T_2} \Rightarrow {T_2} = \frac{{{T_0}}}{4}\)

and then T₁ = 3T₀/4

hence T₁ > T₂

So, maximum shear stress is developed due to T1

\(\frac{{{T_1}}}{J} = \frac{\tau _{max}}{r} \Rightarrow {\tau _{max}} = \frac{{{T_1}}}{J}r\)

\(\Rightarrow {\tau _{max}} = \frac{{\frac{{3{T_0}}}{4}}}{{\frac{\pi }{{32}}{d^4}}} \times \frac{d}{2} = \frac{{32 \times 3{T_0}}}{{8\pi \times {d^3}}} = \frac{{12{T_0}}}{{\pi {d^3}}}\)

Concept:

Equation of Pure torsion

\(\frac{T}{{{I_P}}} = \frac{τ }{r} = \frac{{Gθ }}{L}\)  where,

T = Maximum torque, IP = Polar moment of inertia, τ = shear stress at any point at a distance r from center, G = Modulus of rigidity, θ = angle of twist in radians, R = radius of shaft

Considering the rigidity of shaft, The maximum shear stress induced in the solid shaft of diameter d, to transmit twisting moment T is given by,

\(\tau_{max}\;=\frac{{16T}}{{\pi d^3 }}\)

The maximum shear stress induced in the hollow shaft of outer diameter do and inner diameter di to transmit the twisting moment T.

\(\tau_{max}\;=\frac{{16T}}{{\pi d_o^3 (1-K^4)}}\), where K is the ratio of outer and inner diameter, \(K =\frac{{d_i}}{{d_o}}\)

Calculation:

Given:

d = 50 mm, L = 0.7 M, T = 1200 N-m

Maximum shear stress-induced, \(\tau_{max}\;=\frac{{16T}}{{\pi d^3 }}\)

⇒ \(\tau_{max}\;=\frac{{16~\times~1200~\times~1000}}{{\pi ~\times~50^3 }}~=~48.9~MPa\)