Truss Member Forces by the Principle of Virtual Work: 22-Mec-A7 Advanced Strength of Materials, December 2013, Q7

Question 7: Truss Member Forces by the Principle of Virtual Work (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.

Given. A determinate planar truss (top chord E–F–G, bottom chord A–B–C–D) with 1 m panels; pin at A, roller at C.

Given data
Node coordinates (m)	E(0,1) F(1,1) G(2,1); A(0,0) B(1,0) C(2,0) D(3,0)
Loads	20 kN → at E; 15 kN ↓ at D
Supports	pin at A (A_x, A_y), roller at C (C_y)
Members sought	FG (top chord), GD (diagonal), CD (bottom chord)

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

Determinate truss (7 joints, 11 members, 3 reactions). Diagonals AF, FC, GD; verticals EA, FB, GC.

Approach. First find the reactions from overall equilibrium. The principle of virtual work for a member force isolates that member: with the real loads in place, imposing a compatible unit virtual displacement (equivalently, cutting the member and enforcing joint equilibrium) yields the bar force. Here the target members are reached directly through joint equilibrium at D, G and F, which is the virtual-work result member-by-member.

Support reactions. Overall equilibrium (moments about A, with the 20 kN acting 1 m above A and the 15 kN acting 3 m to its right): $$C_y=\frac{20(1)+15(3)}{2}=32.5\ \text{kN}\uparrow,\quad A_y=15-C_y=-17.5\ \text{kN},\quad A_x=-20\ \text{kN}.$$
Joint D — members CD and GD. At D only CD (horizontal, toward C) and GD (diagonal up-left at 45°) meet, with the 15 kN load down. Vertical equilibrium: $$\tfrac{1}{\sqrt2}F_{GD}-15=0\ \Rightarrow\ \boxed{F_{GD}=15\sqrt2=21.2\ \text{kN (tension)}.}$$ Horizontal equilibrium then gives $$\boxed{F_{CD}=-\tfrac{1}{\sqrt2}F_{GD}=-15\ \text{kN (compression)}.}$$
Joint G — carry into member FG. At G the members are FG (horizontal), GC (vertical) and GD. Horizontal equilibrium (GD pulls right-down toward D): $$F_{FG}=\tfrac{1}{\sqrt2}F_{GD}=15\ \text{kN},$$ so before even reaching joint F the top chord force is fixed.
Confirm at joint F — member FG. Solving the joints in sequence (E→A→B, giving F_EF = −20 kN, F_AF = +24.7 kN, F_FB = 0) and applying equilibrium at F reproduces $$\boxed{F_{FG}=+15\ \text{kN (tension)},}$$ in agreement with joint G — the virtual-work consistency check.
Interpretation. FG and GD carry tension (they are being stretched by the load path from D up through the right panel), while the bottom chord CD is in compression, thrusting back against the roller at C. The diagonal GD alone carries the 15 kN into the truss, so it sees the largest force of the three.

Results — Question 7
Member	Force	Sense
FG	15.0 kN	Tension
GD	21.2 kN (15√2)	Tension
CD	15.0 kN	Compression

Topic: Member forces in a statically determinate truss by the principle of virtual work (unit-load / method-of-joints equivalent).

Key relations: global equilibrium fixes the reactions; at each joint ΣF_x = ΣF_y = 0. Virtual work states Σ N·δe = external virtual work; for a determinate truss this returns exactly the equilibrium bar forces.

Why this works: a determinate truss has as many equilibrium equations as unknowns (2j = m + r: here 14 = 11 + 3), so the virtual-work and joint methods coincide. Starting at a joint with only two unknown members (D) and marching through the structure resolves every force without simultaneous equations.

Common pitfall: mis-reading the geometry — the verticals are EA, FB, GC and the diagonals AF, FC, GD (all at 45°); assuming a member such as B–G that is not present will corrupt the joint equations. Also track tension (+) vs compression (−) consistently.

Source: Hibbeler, Structural Analysis, method of joints and virtual work for trusses (Ch. 3 & 8); Gere & Goodno, energy methods.

Question 7 of 7: Truss Member Forces by the Principle of Virtual Work

Question 7: Truss Member Forces by the Principle of Virtual Work (20 marks)