Maximum stress in each material, Mechanical Engineering

The maximum stress in each material:

A copper tube of external diameter 60 mm and internal diameter 40 mm is closely fitted to a steel rod of 40 mm diameter to compose a composite shaft. If a torque of 6 kN-m is to be resisted by the shaft, discover the maximum stress in each material and the angle of twist in 2 m length.

Take  G = 80 kN/mm for steel, and

G = 40 kN/mm2 for copper.

Solution

Copper Tube

G = 40 kN/mm2  = 0.4 × 1011 N/m2 , l = 2 m

d1 = 60 mm = 0.06 m

d2 = 40 mm = 0.04 m

2451_Maximum stress in each material.png

TC = torque resisted by copper tube

θ= Tl/ GJ    =  (TC ) (2) /((0.4 × 1011 ) (1 × 10- 6 )) = 2 TC × 10- 5 radians       ---------- (1)

τmax =  (T/J )× R = (1 × 10- 6 ) × (0.03) = 0.03 TC  × 10  N/m

         = 0.03 TC N/mm2                             ------------ (2)

1689_Maximum stress in each material1.png

Figure

G = 80 kN/mm2  = 0.8 × 1011 N/m2 ,   l = 2 m

d = 40 mm

J =  (π/32) d 4 = (π /32)× 0.044  = 2.5 × 10- 7 m4

R = 20 mm = 0.02 m

θ= Tl  / GJ  = (TS ) (2) / ((0.8 × 1011 ) (2.5 × 10- 7 )) = TS × 10- 4 radians  --------- (3)

τmax =(T/ J) × R =  Ts / ((2.5 × 10- 7 ) (0.02)) = 0.008 TS  × 107  N/m

= 0.08 TS N/mm2                                    --------- (4)

In composite shafts, θC = θS

 ⇒        5 TC  × 10- 5  = TS  × 10- 4

∴          TC  = 2TS                                  -------- (5)

TS  + TC  = T

⇒         TS = 2TS  = 6

∴          TS = 2 kN-m             TC  = 4 kN-m

∴ Angle of twist, θ= (2 × 103 ) × 10- 4  = 0.2 radians

Maximum Shear Stress

For copper shaft : For steel shaft :

τmax  = 0.03 × 4 × 103  = 120 N/m2

τmax  = 0.08 × 2 × 103  = 160 N/m2

Posted Date: 1/21/2013 5:07:19 AM | Location : United States







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