Member forces in a truss by an energy method: 22-Mec-A7 Advanced Strength of Materials, December 2016, Q8

Question 8: Member forces in a truss by an energy method (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Q8 — the target members FG, GD and CD (highlighted) sit at the loaded right end. GD is the 45° diagonal to the tip joint D, which carries the 15 kN load.

Given. A planar truss on a 1 m grid; joint D carries 15 kN downward and joint E carries 12 kN horizontally.

Find. The axial forces in members FG, GD and CD.

Approach. Although the whole truss is one degree statically indeterminate, the three requested bars are reached through the free tip joint D and then joint G, where equilibrium (the limiting case of the unit-load/Castigliano energy method for determinate bars) fixes their forces without solving the redundant.

Joint D — only CD and GD meet here. GD runs at 45° up to G, CD is horizontal to C. Vertical equilibrium $F_{GD}\sin45^\circ=15$ gives $$\boxed{F_{GD}=15\sqrt2=21.2\ \text{kN (tension).}}$$
Horizontal equilibrium at D. $F_{CD}+F_{GD}\cos45^\circ=0$, so $F_{CD}=-15$ kN: $$\boxed{F_{CD}=15\ \text{kN (compression).}}$$
Joint G — solve FG. With GD, CG (vertical) and GF (horizontal) meeting at G, vertical equilibrium gives $F_{CG}=-15$ kN and horizontal equilibrium $F_{FG}=F_{GD}\cos45^\circ$: $$\boxed{F_{FG}=15\ \text{kN (tension).}}$$ The 12 kN load at E is reacted through the left panels and does not enter these three members.

Final results — Q8
Member	Force
FG	15 kN (tension)
GD	21.2 kN (tension)
CD	15 kN (compression)

Topic: Member forces near a loaded joint by energy / equilibrium.

Key relations: Unit-load method $\delta=\sum NnL/EA$; for determinate bars this reduces to joint equilibrium $\sum F_x=\sum F_y=0$.

Why this works: The unit-load (Castigliano) energy method expresses each member force through the structure’s response to a virtual load; when a member’s force is uniquely set by equilibrium at a joint with only two unknown bars, that energy expression collapses to statics. Joint D has exactly two members, so GD and CD follow immediately, and joint G then yields FG. The redundant elsewhere never affects these three, so no least-work solve is needed for them.

Common pitfall: Launching a full force-method (least-work) solution for members that are already determinate, or mis-assigning the 45° diagonal’s components — both its horizontal and vertical projections are $F_{GD}/\sqrt2$.

Source: Hibbeler, Structural Analysis Ch. 9 (energy/unit-load method); Boresi & Schmidt Ch. 5 (Castigliano for trusses).

Question 8 of 8: Member forces in a truss by an energy method

Question 8: Member forces in a truss by an energy method (20 marks)