Determine the final stresses in the bolt:
A steel bolt of diameter 12 mm and length 175 mm is used to clamp a brass sleeve of length 150 mm to a rigid base plate as in Figure. The sleeve has an internal diameter of 25 mm and a wall thickness of 3 mm. The thickness of the base plate is 25 mm. Initially, the nut is tightened until there is tensile force of 5 kN in the bolt. The temperature is now increased by 100°C. Determine the final stresses in the bolt and the sleeve.
For computation purposes, take following values:
E_{b} = 105 GNm^{-2} ; E_{s} = 210 GNm^{-2}
α_{b} = 20 × 10 ^{-6} K-1 ; α_{s} = 12 × 10 ^{-6} K^{-1}
Figure
Solution
Let the free thermal expansions of steel and brass be Δs and Δb and Δ be the common expansion. Then
Δ_{s} = (175) (12 × 10^{-6}) (100) = 0.21 mm
Δ_{b} = (150) (22 × 10^{-6}) (100) = 0.3 mm
If the initial stresses in steel and brass due to 5 kN load are σ_{s1} and σ_{b1}, then
σ_{s1} = +( 5 × 10^{3} × 4)/π (12)^{2} = + 44.21 N/mm^{2 }(Tensile)
σ_{b1} = +( 5 × 10^{3} × 4)/π (31^{2} - 25^{2}) = + 44.21 N/mm^{2 }(Compressive)
Equilibrium of thermal stresses σs2 and σb2 requires that
σ_{s2} × π (12)^{2}/4 +σ_{b2} × π (31^{2} - 25^{2} )/4 = 0
or 3 σ_{s2} + 7 σ_{b2} = 0
The thermal strains are given by
ε_{s} = (Δ- Δ_{s} )/175 and ε_{b} = (Δ- Δ_{b} )/150
Thus, the thermal stresses are as follows:
σs_{2} = 210/175 (Δ- Δ_{s}) × 10^{3 }N/mm^{2}
and σb_{2} = 105/150 (Δ- Δ_{b}) × 10^{3 }N/mm^{2}
Substituting these in the equilibrium equation,
3 × 210/175 (Δ - 0.21) × 10^{3} + 7 × 105/150 (Δ - 0.30) × 10^{3} = 0
Hence, Δ = 0.262 mm
Thus,
σ_{s2} = 210/175 (0.262- 0.21) × 10^{3} = + 62.26 N/mm^{2 }(Tensile)
σ_{b2} = 105/150 (0.262- 0.3) × 10^{3} = - 26.28 N/mm^{2 }(Compressive)
Total stresses are therefore as follows:
σ_{s} = σ_{s1} + σ_{s2} = - 106.5 N/mm^{2} (Tensile)
σ_{b} = σ_{b1} + σ_{b2} = - 4.56 N/mm^{2} (Compressive)