FE Civil Structural Engineering Practice Problems

The current FE Civil specification assigns 10-15 of the 110 exam questions to Structural Engineering. Form A contains 12 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A determinate portal frame consists of vertical columns AB and DC, each 4 m tall, and horizontal beam BC spanning 6 m between the column tops B and C. The frame is supported by a pin at base A and by a roller at base D that resists vertical force only. A horizontal load of 30 kN is applied at joint B, directed toward C. The maximum bending moment in the frame is most nearly:

(A)60 kN·m
(B)80 kN·m
(C)120 kN·m
(D)180 kN·m

Show worked solution

Answer: (C)

Only the pin at A can resist horizontal force, so column AB carries the full 30 kN and the moment peaks at its top: $30 \times 4 = 120$ kN·m.

With the roller at D unable to take horizontal force, the pin at A resists the full 30 kN; the vertical reactions form a couple balancing the overturning.

A_{x} = 30 kN

Σ M_{A} = 0 : D_{y} (6) = 30 (4) \Rightarrow D_{y} = 20 kN (↑), A_{y} = 20 kN (↓)

Cut column AB just below B: the 30-kN base shear acting through 4 m sets the joint moment. Along the beam it falls linearly to zero at C, since column DC carries axial force only.

M_{B} = 30 kN \times 4 m = 120 kN \cdot m

FE Reference Handbook — Statics: Equations of Equilibrium (frames and free-body diagrams)

Why the other choices appear

(A)Splits the 30-kN lateral load equally between the two columns as in the portal method, giving 15 x 4 = 60 kN·m, although the roller cannot resist horizontal force.
(B)Multiplies the 20-kN vertical reaction at D by the 4-m column height instead of tracking the lateral load path.
(D)Multiplies the 30-kN load by the 6-m beam span instead of the 4-m column height.