Final answer:
The factor of safety (FS) for the steel link under the given tensile load of 10 kips is approximately 2.03, which is calculated by comparing the ultimate stress the steel can handle to the actual stress experienced by the link.
Step-by-step explanation:
The given question requires calculating the factor of safety (FS) for a steel link subjected to a tension load. To find the FS, we have to compare the ultimate stress the material can withstand with the actual stress experienced by the link due to the given load. We first compute the actual stress using the formula σ = P/A, where P is the load in pounds (10 kip) and A is the cross-sectional area of the link (width × thickness). Once the actual stress is computed, the factor of safety is found using FS = σ_ultimate / σ_actual. The ultimate stress for steel is given as 65 ksi (65,000 psi), and the link's dimensions are provided for the calculation.
To solve this problem, we'll assume the cross-sectional area is uniform and the force is evenly distributed across it, which is a common condition in statics and strength of materials. With the given dimensions (1/4 in. thick and 1.25 in. high), the cross-sectional area A is 1/4 in x 1.25 in = 0.3125 in². Converting the load P from kips to pounds (1 kip = 1000 pounds), we get 10,000 pounds.
Now, we can calculate the actual stress (σ_actual): σ_actual = P / A = 10,000 lb / 0.3125 in² = 32,000 psi. Finally, calculating the factor of safety: FS = σ_ultimate / σ_actual = 65,000 psi / 32,000 psi = 2.03125. Therefore, the factor of safety for the steel link under the given load is approximately 2.03.