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

Answer:

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.

$\delta_1=\delta_2$

$\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}$

$\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}$

$P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})$

$P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}$

$P_2 = 6.6212 \ P_1$

Replace $P_2$ into equation (1)

$P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN$

Finally, to determine the normal stress in aluminum rod:

$\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}$

$\sigma_1 = - 23.523 \ MPa}$

Thus, the normal stress = 23.523 MPa in compression.