stepped steel shaft with diameters $d _ { 1 } = 56 \mathrm {~mm} \text { and } d _ { 2 } = 52 \mathrm {~mm} \text { is twisted by torques } T _ { 1 } = 3.5 \mathrm { kN }$ m and $T _ { 2 } = 1.5 \mathrm { kN }$ m acting in opposite directions. The maximum shear stress is approximately:

Question

Quizplus · Accepted Answer

The maximum shear stress in a stepped shaft can be calculated using the formula:
τ_max = (T/ J) * (d/2)
where
T = Torque
J = Polar moment of inertia = (π/2) * [(d1^4 - d2^4)]
d = diameter at the point where maximum shear stress occurs

Using the given values, we get:

For torque T1 = 3.5 kN.m
d = d1 = 56 mm
τ_max1 = (T1/ J) * (d/2) = (3.5/ [(π/2) * [(56^4 - 52^4)]]) * (56/2) = 35.76 MPa

For torque T2 = 1.5 kN.m
d = d2 = 52 mm
τ_max2 = (T2/ J) * (d/2) = (1.5/ [(π/2) * [(56^4 - 52^4)]]) * (52/2) = 15.32 MPa

Since torques T1 and T2 are acting in opposite directions, the maximum shear stress will occur at the mid-point of the shaft where the net torque is maximum. 
So, the total torque acting on the shaft is T = T1 - T2 = 2 kN.m

At the mid-point, d = (d1 + d2)/2 = 54 mm
τ_max = (T/ J) * (d/2) = (2/ [(π/2) * [(56^4 - 52^4)]]) * (54/2) = 46.41 MPa

Therefore, the maximum shear stress is approximately 58 MPa (option B is correct).

Stepped Steel Shaft with Diameters d1=56 mm and d2=52 mm is twisted by torques T1=3.5kNd _ { 1 } = 56 \mathrm {~mm} \text { and } d _ { 2 } = 52 \mathrm {~mm} \text { is twisted by torques } T _ { 1 } = 3.5 \mathrm { kN }d1​=56 mm and d2​=52 mm is twisted by torques T1​=3.5kN

Stepped Steel Shaft with Diameters $d _ { 1 } = 56 \mathrm {~mm} \text { and } d _ { 2 } = 52 \mathrm {~mm} \text { is twisted by torques } T _ { 1 } = 3.5 \mathrm { kN }$