Truss member forces by virtual work (FG, GD, CD): 22-Mec-A7 Advanced Strength of Materials, December 2018, Q8

Question 8: Truss member forces by virtual work (FG, GD, CD) (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 Pratt-type truss with bottom chord A–B–C–D and top chord E–F–G; all horizontal and vertical members 0.75 m.

Given data (from the figure)
Bottom joints	A(0,0), B(0.75,0), C(1.5,0), D(2.25,0)
Top joints	E(0,0.75), F(0.75,0.75), G(1.5,0.75)
Supports	A pinned, C roller (vertical reaction)
Load at E	12 kN horizontal, to the right
Load at D	18 kN vertical, downward

Truss with pin at A, roller at C, and a right-hand overhang to D. Horizontal 12 kN at E; vertical 18 kN at the overhang tip D.

Find. Axial forces in FG (top chord), GD (diagonal) and CD (bottom chord), with sense (tension/compression).

Approach. The truss is statically determinate ($m+r=11+3=14=2j$), so the virtual-work (unit-load) method reduces to equilibrium of the joints at the overhang end. Isolate joint D (only CD and GD), then joint G (FG follows). The 12 kN horizontal load is carried straight to the pin at A and never enters the right-hand panel.

Joint D (members CD horizontal, GD diagonal at 45°). With the 18 kN load down: $$\sum F_y:\ \tfrac{1}{\sqrt2}F_{GD}-18=0\;\Rightarrow\;F_{GD}=18\sqrt2=\boxed{25.5\text{ kN (T)}},$$ $$\sum F_x:\ -F_{CD}-\tfrac{1}{\sqrt2}F_{GD}=0\;\Rightarrow\;F_{CD}=-18\Rightarrow\boxed{18\text{ kN (C)}}.$$
Joint G (members FG horizontal, GC vertical, GD diagonal). Horizontal balance with $F_{GD}=25.5$ kN pulling toward D: $$\sum F_x:\ -F_{FG}+\tfrac{1}{\sqrt2}F_{GD}=0\;\Rightarrow\;F_{FG}=18\Rightarrow\boxed{18\text{ kN (T)}}.$$ (The vertical balance gives $F_{GC}=18$ kN compression, not requested.)
Role of the 12 kN load. The horizontal force at E is equilibrated by the pin reaction at A ($A_x=-12$ kN); it produces no force in the right-panel members FG, GD, CD — confirmed by the full determinate solution.

Final results — Question 8
Member	Force	Sense
GD (diagonal)	25.5 kN ($18\sqrt2$)	Tension
CD (bottom chord)	18 kN	Compression
FG (top chord)	18 kN	Tension

Topic. Member forces in a determinate truss via the principle of virtual work (unit-load method).

Key relations. Virtual work $1\cdot\Delta=\sum n\,\delta$; for a determinate truss the required forces follow directly from joint equilibrium $\sum F_x=\sum F_y=0$.

Why this works. When the number of members plus reactions equals twice the number of joints, the truss is statically determinate and internal forces are found from equilibrium alone; the virtual-work statement for member forces collapses to the method of joints/sections. Starting at the free overhang tip (joint D) exposes only two unknowns, which cascade to the neighbouring joint.

Common pitfall. Assuming the horizontal load at E must load every member; here it is reacted entirely by the pin at A. Also, reading the diagonal GD as horizontal — it spans one bay across and one bay down, hence the $1/\sqrt2$ direction cosines.

Source. Hibbeler, Structural Analysis, §3 (trusses) & §9 (virtual work); Boresi & Schmidt, Ch. 5.

Question 8 of 8: Truss member forces by virtual work (FG, GD, CD)

Question 8: Truss member forces by virtual work (FG, GD, CD) (20 marks)