For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is
From the equation of torsion
A stepped steel shaft shown below is subjected to 10 Nm torque. If the modulus of rigidity is 80 GPa, the strain energy in the shaft in Nmm is
The strain energy in the shaft is 1.73 N mm.
Given
From the figure we can say that
T₁ = T₂ = T = 10 Nm
Here
J₁ = π/32 d⁴
By substituting the value we get
J₁ = (π/32) × 25⁴ × 10⁻¹²
G₁ = 80 × 10⁹
l = 0.1 m
θ₁ = 0.33 × 10⁻³
So,
T₁ = J₁G₁θ₁/l₁ ⇒ θ₁ = T₁l₁/J₁G₁
Hence,
θ₁/θ₂ = (J₂/J₁) = l₁/l₂ = G₂/G₁
Now according to the question
θ₁ = 4 θ₂ So,
θ = θ₁ + θ₂ = 5/4 × 0.33 × 10⁻³
Now strain energy can be given by
Strain energy = 1/2 Tθ
By substituting the value we get
Strain energy = 1/2 × 10 × 5/4 × 0.33 × 10⁻³ = 1.73
While transmitting the same power by a shaft, if its speed is doubled, what should be its new diameter if the maximum shear stress induced in the shaft remains same?
A solid shaft of diameter 100 mm, length 1000 mm is subjected to a twisting moment ‘T’, the maximum shear stress developed in the shaft is 60 N/mm^{2}. A hole of 50 mm diameter is now drilled throughout the length of the shaft. To developed a maximum shear stress of 60 N/ mm^{2 }in the hollow shaft, the torque ‘ T must be reduced by
For solid shaft
............... (i)
for hollow shaft
.................. (ii)
From (i) and (ii), we get
Hence reduced torque
Maximum shear stress developed on the surface  of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled then the maximum shear stress developed corresponding to the same torque will be
When shaft diameter is doubled then
= 30 MPa
A circular shaft is subjected to a twisting moment T and bending moment M. The ratio of maximum bending stress to maximum shear stress is given by
Maximum bending stress,
Maximum shear stress,
A section of a solid circular shaft with diameter D is subjected to bending moment M and torque T. The expression for maximum principal stress at the section is
Principal stresses,
Maximum principal stress is obtained by choosing '+' sign and minimum principal stress is be obtained by '' sign.
Two shafts, one of solid section and the other of hollow section, of same material and weight having same length are subjected to equal torsional force. What is the torsional stiffness of hollow shaft?
Let the diameter of solid shaft be d_{s}. Let the external and internal diameter of the hollow shaft be d_{0} and d_{i} respectively.
It is given that both the shafts are made of same material, have same weight and length and are subjected to equal torsional force.
∴ Weight of solid shaft = Weight of hollow shaft
Now, we know that, torsional stiffness is given as
For same value of G and L
Now, the quantity is always greater than 1
∴ K_{h }> K_{s}
A long shaft of diameter d is subjected to twisting moment T at its ends. The maximum normal stress acting at its crosssection is equal to
Maximum shear stress
Normal stress = 0
A circular shaft shown in the figure is subjected to torsion T at two points A and B. The torsional rigidity of portions CA and BD is GJ_{1} and that of portion AB is GJ_{2}. The rotations of shaft at points A and B are q_{1} and q_{2}. The rotation q_{1} is
The symmetry of the shaft shows that there is no torsion on section AB.
∴ Rotation,








