FE Other Disciplines Strength of Materials Practice Problems

The current FE Other Disciplines specification assigns 9-14 of the 110 exam questions to Strength of Materials. Form A contains 10 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A stepped steel anchor rod consists of a 1.2 m length of 40 mm diameter bar butt-welded to a 0.8 m length of 25 mm diameter bar. The rod carries an axial tensile load of 90 kN, and $E = 200$ GPa. The total elongation of the rod is most nearly:

(A)0.43 mm
(B)0.73 mm
(C)1.16 mm
(D)1.83 mm

Show worked solution

Answer: (C)

The slender 25 mm portion supplies most of the movement, and the two series stretches add to about 1.16 mm at the loaded end.

Series segments transmit the identical 90 kN, so each one deforms according to its own area.

A_{1} = \frac{π ( 40 ) ^{2}}{4} = 1257 mm^{2}

A_{2} = \frac{π ( 25 ) ^{2}}{4} = 490.9 mm^{2}

The 40 mm portion is more than two and a half times as stiff per unit length, so its share stays small despite being the longer piece.

δ_{1} = \frac{90 , 000 ( 1200 )}{1257 ( 200 , 000 )} = 0.430 mm

δ_{2} = \frac{90 , 000 ( 800 )}{490.9 ( 200 , 000 )} = 0.733 mm

Compatibility places the free-end movement at the sum of the two segment stretches.

δ = 0.430 + 0.733 = 1.16 mm

FE Reference Handbook — Mechanics of Materials: Uniaxial Loading and Deformation

Why the other choices appear

(A)Counts only the 40 mm segment, 90,000(1200)/[1257(200,000)] = 0.430 mm, ignoring the smaller bar.
(B)Counts only the 25 mm segment, 90,000(800)/[490.9(200,000)] = 0.733 mm.
(D)Applies the full 2.0 m length to the 25 mm area, 90,000(2000)/[490.9(200,000)] = 1.83 mm.