And has a diameter of 60 mm at A and 40 mm at B. The length of the shaft is 800 mm. The material of the shaft has a shear modulus of 80 GPa. A torque of 10 kN·m is applied at end B. Determine:

(a) the maximum shear stress in the shaft,

(b) the angle of twist of end B with respect to end A.

Solution:

(a) To find the maximum shear stress, we need to calculate the shear stress at the point of maximum diameter change, which is at point C. We can assume that the shaft is in pure torsion, so we can use the torsion formula:

T/J = Gθ/L

where T is the torque, J is the polar moment of inertia, G is the shear modulus, θ is the angle of twist, and L is the length of the shaft.

To find the polar moment of inertia, we can use the formula for a solid circular shaft:

J = π/32 (D^4 - d^4)

where D is the diameter at point A and d is the diameter at point B.

J = π/32 (60^4 - 40^4) = 7.19×10^6 mm^4

Substituting the given values into the torsion formula, we get:

10×10^3 N·m / 7.19×10^6 mm^4 = 80×10^9 N/mm^2 × θ / 800 mm

θ = 0.00014 rad

Now we can find the shear stress at point C using the formula:

τ = Tc/J

where Tc is the torque at point C. Since the shaft is solid, we can assume that the torque is constant along the length of the shaft. Therefore, Tc is equal to the applied torque at end B:

Tc = 10×10^3 N·m

τ = 10×10^3 N·m / 7.19×10^6 mm^4 × 30 mm/2 = 0.92 N/mm^2

(b) To find the angle of twist of end B with respect to end A, we can use the same torsion formula:

θ = T/J × L/G

Substituting the given values, we get:

θ = 10×10^3 N·m / 7.19×10^6 mm^4 × 800 mm / 80×10^9 N/mm^2 = 0.00025 rad

Therefore, the angle of twist of end B with respect to end A is 0.00025 rad.

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