Question

Problem 3: The 1.00-in-diameter shaft has a 50-ksi yield strength and is in equilibrium when subject to the shown forces and torques. Determine the corresponding design factors $n_{MSST}$ and $n_{DET}$ for the maximum shear stress theory (MSST) and the distortion energy theory (DET). Answer 3: $n_{MSST} = 1.49$ $n_{DET} = 1.53$

          Problem 3: The 1.00-in-diameter shaft has a 50-ksi yield strength and is in equilibrium when subject to the shown forces and torques. Determine the corresponding design factors $n_{MSST}$ and $n_{DET}$ for the maximum shear stress theory (MSST) and the distortion energy theory (DET).
Answer 3:
$n_{MSST} = 1.49$
$n_{DET} = 1.53$

Problem 3: The 1.00-in-diameter shaft has a 50-ksi yield strength and is in equilibrium when subject to the shown forces and torques. Determine the corresponding design factors nMSST and nDET for the maximum shear stress theory (MSST) and the distortion energy theory (DET).
Answer 3:
nMSST = 1.49
nDET = 1.53

Added by Brandy C.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

11/21/2023

Step 1

The maximum shear stress theory states that the maximum shear stress should not exceed the yield strength of the material. Maximum shear stress = Force / Area Area = π*(diameter/2)^2 = π*(1.00/2)^2 = 0.785 in^2 Force = 5000 lb + 3000 lb = 8000 lb Maximum shear Show more…

Show all steps

Thanks for your feedback!

Problem 3: The 1.00-in diameter shaft has a 50-ksi yield strength and is in equilibrium when subject to the shown forces and torques. Determine the corresponding design factors nussr and n_(IET) for the maximum shear stress theory (MSST) and the distortion energy theory (DET). Answer 3: nuss = 1.49 nDBr = 1.53