The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] has an outside diameter of 39 mm and an inside diameter of 32 mm. The ceramic core [E = 310 GPa, α= 3.2 × 10−6/°C] has a diameter of 32 mm. At a temperature of 8°C, the assembly is unstressed. Assume L = 160 mm. Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

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 Δ⁢t 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

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