Two bars, one composed of aluminum and one of steel are place end to end as show
ID: 1973704 • Letter: T
Question
Two bars, one composed of aluminum and one of steel are place end to end as shown in Figure P11.60. The bars are both initially of length L = 32 cm, and both have a square cross section with h = 2.8 cm. If a compressive force of F = 8160 N is applied to each end, find the change in length of the(a) aluminum bar
(b) steel bar
I think that I am suppose to use the equation F/A=Y(Young's modulus)*change in L/L0 but, I am unsure on the width to use to find the area?
The Young modulus for Aluminum is 7.0x10^10. For steel 2.0x10^11
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Explanation / Answer
In the question it is given that it is a square cross section and the height is 2.8cm hence the width will also be 2.8cm. hence area of cross section is 7.84* 10-4 m2 .
(F/A)/(L/L)= Y
from this you can solve the change length