Truss Member Forces by the Principle of Virtual Work: 22-Mec-A7 Advanced Strength of Materials, May 2013, Q8

Question 8: Truss Member Forces by the Principle of Virtual Work (20 marks)

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Given. A panelized truss, all panels 1 m × 1 m; nodes A(0,0), B(1,0), C(2,0), D(3,0) on the bottom chord and E(0,1), F(1,1), G(2,1) on the top chord.

Given data
Panel size	1 m (all horizontal and vertical members)
Loads	12 kN horizontal at E; 18 kN vertical (down) at D
Supports	pin at A, roller at C
Diagonals	AF, CF, GD (each length √2 m)

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

Determinate truss; member forces FG, GD, CD isolated at joints D and G.

Approach. The principle of virtual work for a truss states $\sum F\,\delta = 0$ for any admissible virtual displacement. Cutting a member and giving the released mechanism a unit virtual displacement isolates that member’s force; here joints D and G supply the three requested forces directly, and global equilibrium confirms the support reactions.

Reactions (virtual work / equilibrium). A rigid-body virtual rotation about A and a horizontal sway give $$C_y=\frac{12(1)+18(3)}{2}=33\ \text{kN}\ (\uparrow),\quad A_y=18-33=-15\ \text{kN},\quad A_x=-12\ \text{kN}.$$
Joint D — member GD. Give D a unit vertical virtual displacement; only GD (at 45°) and the 18 kN load do work. Vertical work balance $F_{GD}\tfrac{1}{\sqrt2}-18=0$: $$\boxed{F_{GD}=18\sqrt2=25.5\ \text{kN (tension)}.}$$
Joint D — member CD. A unit horizontal virtual displacement of D balances CD against the horizontal component of GD, $F_{CD}+F_{GD}/\sqrt2=0$: $$\boxed{F_{CD}=-18\ \text{kN}\ \Rightarrow\ 18\ \text{kN (compression)}.}$$
Joint G — member FG. A unit horizontal virtual displacement of G balances FG against GD’s horizontal pull, $-F_{FG}+F_{GD}/\sqrt2=0$: $$\boxed{F_{FG}=18\ \text{kN (tension)}.}$$ (The vertical balance at G gives $F_{CG}=-18$ kN, a consistency check.)

Question 8 — results
Member	Force
FG	18 kN (tension)
GD	25.5 kN (tension)
CD	18 kN (compression)

Topic: Truss member forces via the principle of virtual work.

Key relations: virtual work $\sum F_i\,\delta_i=0$; for a determinate truss the released-member mechanism gives one equation per unknown; equivalently method-of-joints equilibrium $\sum F_x=\sum F_y=0$.

Why this works: Virtual work replaces force balance with an energy statement: for any kinematically admissible virtual displacement the total work of all forces vanishes. Choosing a mechanism that displaces only the member of interest (and the applied loads) isolates its force in a single scalar equation, avoiding a full simultaneous solution. Members FG, GD and CD are reachable from the lightly-connected joints D and G, so their forces come out immediately. The right diagonal GD carries the vertical load to support C in tension, and CD is pushed into compression by GD’s horizontal thrust.

Common pitfall: mismatching the sign of the diagonal’s components (each √2-length diagonal has equal horizontal and vertical components of magnitude F/√2), or forgetting that a virtual displacement must respect the support constraints.

Source: Hibbeler, Structural Analysis, Ch. 9 (virtual work); Boresi & Schmidt, Advanced Mechanics of Materials, Ch. 5.

Question 8 of 8: Truss Member Forces by the Principle of Virtual Work

Question 8: Truss Member Forces by the Principle of Virtual Work (20 marks)