Question

Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for which the factor of safety will be 2.45. Assume that the link will be adequately reinforced around the pins at A and B.

Answer 1

A_ab = 0.206 in^2

Step-by-step explanation:

Given:

- ultimate normal stress T = 65 ksi

- factor of safety FS = 2.45

Find:

- Determine the cross-sectional area of AB

Solution:

- Compute a point load for the distributed force P:

P = 600*4.2 = 2520 lb

- Apply equilibrium equations:

(M)_D = 0

- F_ab *sin (35)*2.8 + 0.7*2520 + 5000*1.4 = 0

F_ab = 5457 lb

- Stress in AB:

stress = F_ab / A_ab

stress_all = T / FS

F_ab / A_ab = T / FS

A_ab = FS*F_ab / T

A_ab = 2.45*5.457/65

A_ab = 0.206 in^2