Final answer:
The smallest height h required to prevent shear failure in an aluminum bracket supporting a load of 8 kips with a thickness of 0.5 inches, using a factor of safety of 2.5, is 1.74 inches.
Step-by-step explanation:
To determine the smallest height h to prevent shear failure in the aluminum bracket A supporting a centrally applied load of 8 kips (8000 lbs), we first calculate the maximum allowable shear stress using the given failure shear stress (Tfail) and factor of safety (F.S.). The allowable stress (Tallow) can be found through the equation Tallow = Tfail / F.S. Substituting the given values, Tallow = 23 ksi / 2.5 = 9.2 ksi.
To prevent failure, the shear stress (τ) in the bracket must be less than Tallow. Shear stress (τ) is calculated by the formula τ = V / A, where V is the shear force and A is the cross-sectional area. Given the load is 8 kips and the thickness of the bracket is 0.5 inches, we can express the required area (A) to prevent failure as A = 8000 lbs / (9.2 ksi * 1000 lbs/ksi). Solving for A gives us a minimum cross-sectional area of approximately 0.87 in².
The height h, considering the thickness, can be found by rearranging the area formula (A = thickness * height) as h = A / thickness. Inserting our values, h = 0.87 in² / 0.5 in, we find that the minimum height h required to prevent shear failure is 1.74 inches.