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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.

Respuesta :

Answer:

A_ab = 0.206 in^2

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

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