Question

The type of stress and the maximum stress value in the link are respectively, b=5 in., t=0.25 in., d=2in. 200 lb d=2 inch in Pin

a. Normal,160 psi
b. Shear, 63.7 psi
c. Normal,63.7 ksi
d. Normal, 266.7 psi

Answer 1

The type of stress in the link is normal stress and the maximum stress value is 387 psi.

Step-by-step explanation:

The type of stress and maximum stress value in the link can be determined by calculating the normal and shear stresses.

To calculate the normal stress, we use the formula σ = F/A, where σ is the normal stress, F is the force applied, and A is the cross-sectional area of the link. The force applied can be calculated by multiplying the weight (200 lb) by the acceleration due to gravity (32.2 ft/s^2), and the cross-sectional area can be calculated using the formula A = b*t, where b is the width and t is the thickness of the link.

For the given values b = 5 in. and t = 0.25 in., the cross-sectional area is A = 5*0.25 = 1.25 in^2. Substituting the values into the formula σ = F/A, we get σ = (200 lb * 32.2 ft/s^2) / (1.25 in^2 * 0.00694 ft^2/in^2) = 387 psi.

Therefore, the type of stress is normal stress and the maximum stress value in the link is 387 psi. The correct answer is option d.