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,

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

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

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,

Polar section modulus of shaft is given by,

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

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

Calculation:

Given:

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

As, P = T × W

Now,

τ_max. = 60.79 N/mm² = 60.79 MPa

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

      ----(ii)

(i) and (ii)

T_s = 908.42

T_b = 91.57

z_max = 44.44 N/mm²

Concept:

where V = Transverse shear, A = Area above or below the section,  = 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

 =  = 

Solid shaft J = 

Hollow Shaft J = 

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)

 = 

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

 =  = 

Put, D₀ = 3 × D_i 

D_s = D₀ 

 = 

Additional Information

Polar moment of inertia

J =  r³dr

For Solid Shaft, J = 

For Hollow Shaft, J = r³dr = 

The assumption for Torsion equation:

 =  = 

• 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,

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_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,

Shear modulus,

Polar moment of inertia,

Length of the shaft, dx

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

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

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

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

Substituting and :

,   and, we get: 

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

where is the differential volume element in cylindrical coordinates

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

Simplifying the integrand:

Evaluating the integral:

Since : 

The strain energy due to torsion in the shaft is:

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

or 

from equation (i) 

and then T₁ = 3T₀/4

hence T₁ > T₂

So, maximum shear stress is developed due to T1

Concept:

Equation of Pure torsion

  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,

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

, where K is the ratio of outer and inner diameter, 

Calculation:

Given:

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

Maximum shear stress-induced, 

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