Compute depth of yielding in the hollow shaft:
A solid shaft 80 mm diameter is solid for a certain length from one end but hollow for the remaining length along inner diameter of 40 mm. If a pure torsion is applied such that yielding occur at the surface of the solid part of the shaft. Compute:
(1) the depth of yielding in the hollow shaft, and
(2) the ratio of the angles of twist per unit length.
Solution
τ = Shear stress at yield point.
Torsion in the solid shaft =τ × ( π× 80^{3} /16)
T_{1} =τ × (π× 80^{3} /16) ------------- (1)
For Hollow Shaft
Torsion in the unyielded part
T_{2} =τ × (π/16 D) (D ^{4} - 40^{4}) ----------- (2)
where D = diameter of the hollow section at which yielding begins.
Torsion in yielded part
= (πτ /12 )(80^{3} - D^{3} ) -------------- (3)
T_{1} = T_{2} + T_{3}
⇒ 80^{3}/16 = D4 - 40^{4}/16 D + (80^{3} - D^{3}/12)
⇒ 12 × 80^{3} D = ( D^{4} - 40^{4 }) 12 + 16 D (80^{3} - D^{3 })
⇒ 6144 × 10^{3} D = 12D^{4 } - 3072 × 10^{4} + 8192 × 10^{3} D - 16 D^{4}
⇒ 4D^{4} - 2048 × 10^{3} D + 3072 × 10^{4} = 0
⇒ D ^{4} - 512 × 10^{3} D + 768 × 10^{4} = 0 ----------- (4)
By trial and error, D = 74.8 mm
T / J = τ/ R = G θ/ l
⇒ θ= τl /GR ---------- (5)
θ_{S} =2τl / (G × 80) ------ (6)
θ_{H} = 2τl / (G × D) --------- (7)
θ_{H}/ θ_{S} = 80/ D = 80/74.5 = 1.07