Answer:
ΔT = 62.11°C
Explanation:
Given:
- Brass Shell:
Inner Diameter d_i = 32 mm
Outer Diameter d_o = 39 mm
E_b = 93 GPa
α_b = 15.1*10^-6 / °C
- Ceramic Core:
Outer Diameter d_o = 32 mm
E_c = 310 GPa
α_c = 3.2*10^-6 / °C
- Unstressed @ T = 8°C
- Total Length of the cylinder L = 160 mm
Find:
Determine the largest temperature increase Δt that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.
Solution:
- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:
б_b = - 60 MPa
- The force equilibrium can be written as:
б_b*A_b + б_c*A_c = 0
Where, б_b is the stress in core
A_b is the cross sectional area of the shell
A_c is the cross sectional area of the core
б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0
б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0
б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)
Plug in the values:
б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)
б_c = 29.121 MPa , б_b = - 60 MPa
- The total strains in both brass shell and ceramic core is given by:
ξ_b = α_b*ΔT + б_b / E_b
ξ_c = α_c*ΔT + б_c / E_c
- The compatibility relation is:
ξ_b = ξ_c
α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c
ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b
ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )
Plug in values and solve:
ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )
ΔT = 62.11°C
