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 × 10³ × 4)/π (12)² = + 44.21 N/mm²                       (Tensile)

σ_b1 = +( 5 × 10³ × 4)/π (31² - 25²) = + 44.21 N/mm²           (Compressive)

Equilibrium of thermal stresses σs2 and σb2 requires that

 σ_s2 × π (12)²/4 +σ_b2 × π (31²  - 25² )/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:

σs₂ = 210/175 (Δ- Δ_s) × 10³ N/mm²

and σb₂ = 105/150 (Δ- Δ_b) × 10³ N/mm²

Substituting these in the equilibrium equation,

3 × 210/175 (Δ - 0.21) × 10³ + 7 × 105/150 (Δ - 0.30) × 10³ = 0

Hence,            Δ = 0.262 mm

Thus,

σ_s2 = 210/175 (0.262- 0.21) × 10³ = + 62.26 N/mm²                            (Tensile)

σ_b2 = 105/150 (0.262- 0.3) × 10³ = - 26.28 N/mm²                (Compressive)

Total stresses are therefore as follows:

σ_s = σ_s1 + σ_s2 = - 106.5 N/mm²       (Tensile)

σ_b = σ_b1 + σ_b2 = - 4.56 N/mm²       (Compressive)