Compute depth of yielding in the hollow shaft, Mechanical Engineering

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× 803 /16)

T1 =τ × (π× 803 /16)            ------------- (1)

 For Hollow Shaft

Torsion in the unyielded part

T2 =τ × (π/16 D) (D 4 - 404)                    ----------- (2)

where D = diameter of the hollow section at which yielding begins.

Torsion in yielded part

= (πτ /12 )(803  - D3 ) --------------  (3)

         T1  = T2  + T3

⇒ 803/16 = D4  - 404/16 D +  (803 - D3/12)

⇒         12 × 803 D = ( D4  - 404 ) 12 + 16 D (803  - D3 )

⇒         6144 × 103 D = 12D4  - 3072 × 104  + 8192 × 103 D - 16 D4

⇒         4D4  - 2048 × 103 D + 3072 × 104  = 0

⇒         D 4  - 512 × 103 D + 768 × 104  = 0           ----------- (4)

2411_Compute depth of yielding in the hollow shaft.png

By trial and error, D = 74.8 mm

T / J =  τ/ R = G θ/ l

⇒         θ=  τl /GR                                 ---------- (5)

θS  =2τl / (G × 80)            ------ (6)

θH  = 2τl / (G × D)           --------- (7)

θH/ θS  = 80/ D =   80/74.5 = 1.07

Posted Date: 1/21/2013 5:23:16 AM | Location : United States







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