FE Mechanical Mechanical Design and Analysis Practice Problems

The current FE Mechanical specification assigns 10-15 of the 110 exam questions to Mechanical Design and Analysis. Form A contains 11 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A ductile steel bracket has a yield strength of 46 ksi. At the critical point the plane stress state is $σ_{x} = 18$ ksi, $σ_{y} = - 6$ ksi, and $τ_{x y} = 8$ ksi. Using the maximum distortion energy theory, the factor of safety against yielding is most nearly:

(A)1.79
(B)1.99
(C)2.18
(D)2.56

Show worked solution

Answer: (A)

A von Mises stress of 25.7 ksi against the 46 ksi yield strength gives a factor of safety of roughly 1.79.

Distortion energy collapses the biaxial state into one equivalent uniaxial stress, and for plane stress all three in-plane components enter directly.

σ^{'} = σ_{x}^{2} - σ_{x} σ_{y} + σ_{y}^{2} + 3 τ_{x y}^{2}

σ^{'} = 324 + 108 + 36 + 192 = 660 = 25.7 ksi

Yielding becomes a single scalar comparison; the compressive $σ_{y}$ increases $σ^{'}$ rather than relieving it, which holds the margin below two.

n = \frac{S _{y}}{σ ^{'}} = \frac{46}{25.7} = 1.79

FE Reference Handbook — Mechanical Design and Analysis: Static Loading Failure Theories, Distortion Energy Theory

Why the other choices appear

(B)Factor of 3 omitted from the shear term: $324 + 108 + 36 + 64 = 23.1$ ksi gives $46/23.1 = 1.99$ .
(C)Sign of $σ_{y}$ taken as positive: $324 - 108 + 36 + 192 = 21.1$ ksi gives $46/21.1 = 2.18$ .
(D)Shear and transverse components ignored, checking $σ_{x}$ alone: $46/18 = 2.56$ .