Arch Exam Academy

1. Concurrent Forces 2. Strain 3. Modulus of elasticity. 4. 0.053″ 5. internal forces act in the opposite direction from that assumed. 6. I and II 7. 200 psi 8. unit stress at which the material continues to deform with not increase…

Temperature decrease of 50 degrees

Find required diameter if elongation not to exceed 0.005″ and find required diameter if unit stress not exceed 24,000 psi.

ksi of steel = 20,000,000 psi modulus of elasticitY = hooke’s law or young’s modulus (same thing)

f= P/A

Unit stress (f) is equal to the load (P) divided by the cross sectional area (A).

Divide into rectangles and calculate the area of each rectangle.