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

Concept:

A shaft of diameter d and length l undergoes torque T at one end having permissible shear stress τ.

Using equation for maximum shear stress under torque,

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

For constant torque we have,

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

Calculation:

Given

Initial diameter d₁ and final diameter d₂. Similarly, initial shear stress τ₁ and final shear stress τ₂

d₂ = 0.5× d₁

\(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{{d_2^3}}{{d_1^3}}\)

\(\frac{{{\tau _1}}}{{{\tau _2}}} = \frac{1}{8} \Rightarrow {\tau _2} = 8{\tau _1}\)

So, the correct option is 1)

Bending Equation:

\({\frac{M}{I} = \frac{{{σ _b}}}{y} \Rightarrow {σ _b} = \frac{M}{I}y = \frac{{32M}}{{\pi {d^3}}} \Rightarrow {σ _{\max }} = \frac{{32{M_{\max }}}}{{\pi {d^3}}}}\)

The bending moment will be maximum at the fixed end i.e. X so Bending stress will be maximum at X.

Torsion Equation:

\(\frac{T}{J} = \frac{\tau }{r} \Rightarrow \tau = \frac{T}{J}r = \frac{{16T}}{{\pi {d^3}}}\)

At center bending stress and Torsion shear stress are zero.

Because Bending stress and Torsion shear stress directly depends on the radial distance from the centroidal axis. So it will be maximum at the circumference.

Stress due to tensile load P i.e. σ = P/A is constant throughout.

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 = 2 × Di

Ds = 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)

\(\frac{{{M_{t1}}}}{{{M_{t2}}}} = \frac{{\frac{\pi }{{32}} \;\times \;\left( {D_o^4 - D_i^4} \right)}}{{\frac{\pi }{{32}} \;\times \; D_s^4}}\)

Put, Di = (1/2) × Do 

Ds = D0 

\(\frac{{{M_{t1}}}}{{{M_{t2}}}} = \frac{{\left( {D_o^4 - {{\left( {\frac{{{D_0}}}{2}} \right)}^4}} \right)}}{{D_0^4}}\)

\(\frac{{{M_{t1}}}}{{{M_{t2}}}} = \frac{{15}}{{16}}\)

⇒\(\frac{{{M_{t2}}}}{{{M_{t1}}}} = \frac{{16}}{{15}}\)

From the torsion formula, \(\frac{\tau }{r} = \frac{T}{J}\)

\(\Rightarrow \tau = \frac{T}{{\frac{\pi }{{32}}{d^4}}}r\)

\(\Rightarrow {\tau _{max}}\left( {atr = R} \right) = \frac{{16T}}{{\pi {d^3}}}\)

Concept:

Torsion Equation:

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

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

\({I_{zz}} = J = {I_{xx}} + {I_{yy}} = \frac{{\pi {D^4}}}{{32}}\)

\(Z_p=\frac{I_{zz}}{r}=\frac{{\pi {D^3}}}{{16}}\)

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

Calculation:

Given:

d = 40 mm, T = 1200 Nm = 1200 × 10³ N.mm

\(\tau = \frac{{16T}}{{\pi {d^3}}} = \frac{{16 × 1200 × {{10}^3}}}{{\pi × {{\left( {40} \right)}^3}}} = 95.5\;MPa\)

∴ Maximum shear stress is 95.5 MPa

Explanation:

Torsion equation:

\({T\over J}={τ_{max}\over r}={Gθ\over L}\)

Here: 

• T = torque or twisting moment(Nm)
• J = polar moment of inertia or polar second moment of area about shaft axis (m⁴ ) for circular bar J = \({\pi d^4}\over 32\),
• τ_max = shear stress at outer fiber
• r = radius of the shaft,
• G = modulus of rigidity or shear modulus, θ = angle of twist, (rad), L = length of the shaft(m)

so from the above expression, we can say that θ (Angle of twist) is directly proportional to T (Twisting Moment)

Concept:

Using torsion equation

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

\(\Rightarrow L = \frac{{G\theta R}}{\tau }{{\;\;\;\;\;}} \ldots \left( 1 \right)\)

Calculation:

Given:

G = 80000 MPa, R = 25 mm, τ = 45 MPa, θ = 0.02 rad

By using equation (1),

\(\Rightarrow L = \frac{{80000\times0.02\times 25}}{45 } \)

∴ L = 888.88 mm

Concept:

θ = Angle of Twist

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

Calculation:

Since Torque is applied at point B

SO, θ = θ_C = θ_B

\(\Rightarrow \theta = \frac{{TL}}{{GJ}} = \frac{{10 \times 0.5 \times 32}}{{77 \times {{10}^9} \times \pi \times (0.02)^4}}\times \frac{180}\pi = 0.236\)

Explanation:

The simple torsion equation is written as

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

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft and the following is the stress distribution

Hence the maximum shear stress occurs on the outer surface of the shaft where r = R and at the centre the shear stress is zero.

Concept

Torsion equation is given by,

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

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

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

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

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

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