Final answer:
The cross-sectional area required for member AE to experience a normal stress of 15 ksi under a load of 3600 lb is 0.24 in².
Step-by-step explanation:
To determine the cross-sectional area of member AE for which the normal stress in the member is 15 ksi, given that P = 3600 lb, we first convert the stress to pounds per square inch (psi) since 1 ksi equals 1000 psi. Therefore, 15 ksi is 15,000 psi. Normal stress is defined as the force divided by the area over which it acts (normal stress = force / area). Hence, to find the cross-sectional area (A), we rearrange the equation to A = force / stress.
Now we have P = 3600 lb and the normal stress, which is 15,000 psi. So, the cross-sectional area needed is:
A = P / stress = 3600 lb / 15,000 psi = 0.24 in².
Thus, the required cross-sectional area for member AE to have a normal stress of 15 ksi with a force of 3600 lb applied is 0.24 in².