Question

An aluminum alloy [E = 73 GPa] cylinder (2) is held snugly between a rigid plate and a foundation by two steel bolts (1) [E = 180 GPa]. The aluminum cylinder has a length of L2 = 275 mm and a cross-sectional area of 2300 mm2. Each steel bolt has a length of L1 = 335 mm and a diameter of d1 = 14 mm. The bolt has a pitch of 1 mm. This means that each time the nut is turned one complete revolution, the nut advances 1 mm. The nut is hand-tightened on the bolt until all slack has been removed from the assembly but no stress has yet been induced. Determine the magnitude of the normal stress produced in the aluminum cylinder when the nut is tightened one complete turn past the snug-tight condition.

Answer 1

`\sigma_A = 58.43 N/mm^2`

Step-by-step explanation:

Given data:

length of Steel bolt
`L_1 = 335 mm`

Length of aluminium cylinder
`L_2 = 275 mm`

Pitch of bolt p = 1mm

Modulus of elasticity of steel E = 215 GPa

Modulus of elasticity of aluminium = 74 GP

Area of bolt
`= (\pi)/(4) 14^2 =  153.93 mm^2`

Area of cylinder = 2300 mm^2

n =1

By equilibrium

`\sum F_y = 0`

`P_A -2P_S = 0`

`P_A =2P_S`

By the compatibility

`\delta _s + \delta_A = nP`

Displacement in steel is
`\delta_s = (P_sL_s)/(E_sA_s)`

Displacement in Aluminium is
`\delta_A = (P_AL_A)/(E_AA_A)`

from compatibility equation we have

`(P_sL_s)/(E_sA_s) +  (P_AL_A)/(E_AA_A) = nP`

`(P_s* 335)/(180* 10^3* 153.93) +  (P_A* 275)/(73* 10^3* 2300) = 1* 1`

`1.20* 10^[-5} P_s  + 1.44* 10^(-6)P_A = 1`

substitute
`P_A =2P_S`

`1.20* 10^[-5} P_s  + 1.44* 10^(-6) (2* P_s) = 1`

`1.488* 10^(-5) P_s = 1`

`P_s = 67204.30 N`

`P_A = 134,408.60 N`

Stress in Aluminium
`\sigma  = (P_A)/(A_A)`

`= (134,408.60)/(2300)`

`\sigma_A = 58.43 N/mm^2`