Question

The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod is d = 25 mm and the outside diameter of the steel tube is D= 45 mm. The length of the composite column is L = 761 mm. A force P = 88 kN is applied at the top surface, distributed across both the rod and tube.

Required:
Determine the normal stress σ in the steel tube.

Answer 1

Step-by-step explanation:

From the information given:

`E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN`

The total load is distributed across both the rod and tube:

`P = P_1+P_2 --- (1)`

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

`\delta_1=\delta_2`

`(P_1L)/(A_1E_1)= (P_2L)/(A_2E_2)`

`(P_1 * 0.761)/(((\pi)/(4)* .0025^2 ) * 68* 10^4)= (P_2* 0.761)/(((\pi)/(4)* (0.045^2-0.025^2))* 201 * 10^9)`

`P_1(2.27984775* 10^(-8)) = P_2(3.44326686* 10^(-9))`

`P_2 = ( (2.27984775* 10^(-8)) P_1)/((3.44326686* 10^(-9)))`

`P_2 = 6.6212 \ P_1`

Replace
`P_2` into equation (1)

`P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = (-88)/(7.6212) \\ \\ P_1 = -11.547 \ kN`

Finally, to determine the normal stress in aluminum rod:

`\sigma _1 = (P_1)/(A_1) \\ \\ \sigma _1 = (-11.547 * 10^3)/((\pi)/(4) * 25^2)`

`\sigma_1 = - 23.523 \ MPa}`

Thus, the normal stress = 23.523 MPa in compression.