Joint deflection of a truss by virtual work: 16-Civ-A1 Elementary Structural Analysis, December 2013, Q6

Question 6: Joint deflection of a truss by 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.

Truss 6: pin at U₂ and horizontal roller at L₃ (both to a vertical wall); 36 kN down at L₁ and at L₂. Deflection sought at L₂.

Given. Nodes $L_1(0,0),L_2(4,0),U_1(4,3),L_3(8,3),U_2(8,6)$ (metres); pin at $U_2$, horizontal roller at $L_3$ (both anchored to a right-hand wall). Loads 36 kN downward at $L_1$ and $L_2$. Axial rigidity $EA = 4.0\times10^{4}$ kN for all seven bars.

Find. Vertical deflection at $L_2$.

Member	$L$ (m)	$N$ real (kN)	$n$ unit	$NnL$
L₁L₂	4	−48	−0.667	128.0
L₁U₁	5	+60	+0.833	250.0
U₁L₂	3	+72	+1.000	216.0
U₁U₂	5	+180	+1.667	1500.0
U₁L₃	4	−96	−0.889	341.3
L₂L₃	5	−60	−0.833	250.0
L₃U₂	3	−36	−0.333	36.0

Real forces. With vertical loads only, the pin at $U_2$ carries all 72 kN vertically; the horizontal roller at $L_3$ balances the horizontal thrust. Method-of-joints gives the bar forces $N$ tabulated above (T positive).
Unit system. Remove the real loads, apply a 1 kN downward dummy at $L_2$, and re-solve for the $n$ forces.
Virtual work. $\displaystyle \delta_{L_2} = \sum \frac{N\,n\,L}{EA} = \frac{\sum NnL}{EA} = \frac{2721.3}{4.0\times10^{4}}$.
Result. $\delta_{L_2} = \boxed{0.0557\ \text{m} = 55.7\ \text{mm}\ \downarrow}$.

Quantity	Value
Vertical deflection at L₂	55.7 mm downward

Question 6 of 8: Joint deflection of a truss by virtual work

Question 6: Joint deflection of a truss by virtual work (20 marks)