What unit load is applied to find the deflection of E?

A unit downward load at E (the bottom middle joint Q2)

A unit load at every bottom joint

For the middle vertical P2Q2, F = 0 and k = 0. What follows?

It contributes nothing to the sum of kFL

Which formula gives the deflection of E?

Delta = sum kFL/AE, with AE = 80,000 kN

2071 Magh (old course) · 16 marks

Truss deflection at E by virtual work

Short answer

Virtual unit load at E. Sum member kFL = 9550.4 kNm. With AE = 80,000 kN, deflection = 9550.4/80000 = 119.4 mm downward.

Determine the deflection at E of the pin-jointed truss shown in figure below by using virtual work method. Take area for all members $A = 400\ \text{mm}^2$ , $E = 200\times10^3\ \text{N/mm}^2$ . [Figure: truss 8 m high with four bottom panels of 10 m each; three 50 kN downward loads on the bottom chord joints, E being the bottom joint at the middle load; supports at the two ends.]

Given data and assumed geometry

Pin-jointed truss, 4 panels of 10 m, height 8 m. Top joints P0 to P4 and bottom joints Q0 to Q4 (E is the bottom middle joint Q2). Members: top chord (4), bottom chord (4), five verticals, and four diagonals P0Q1, P1Q2, P3Q2, P4Q3. 50 kN down at Q1, Q2 and Q3; Q0 hinge, Q4 roller. $A = 400$ mm $^2 = 4\times10^{-4}$ m $^2$ , $E = 200\times10^3$ N/mm $^2 = 2\times10^8$ kN/m $^2$ ( $AE = 80\,000$ kN).

Step 1: Reactions and real forces

By symmetry $R_{Q0} = R_{Q4} = 75$ kN. Forces by the method of joints (tension +, compression -), e.g. at Q0: $F_{P0Q0} = -75$ kN, then joint P0 gives $F_{P0Q1} = +120.06$ kN and $F_{P0P1} = -93.75$ kN, and so on.

Step 2: Virtual forces

Apply a unit downward load at E (Q2). By symmetry the reactions are 0.5 each, and the $k$ forces follow in the same way.

Step 3: Table of $F$ , $k$ , $L$

Member	L (m)	F (kN)	k	kFL (kN m)
P0P1	10.000	-93.75	-0.625	585.938
P1P2	10.000	-125.00	-1.250	1562.500
P2P3	10.000	-125.00	-1.250	1562.500
P3P4	10.000	-93.75	-0.625	585.938
Q0Q1	10.000	-0.00	-0.000	0.000
Q1Q2	10.000	93.75	0.625	585.938
Q2Q3	10.000	93.75	0.625	585.938
Q3Q4	10.000	0.00	0.000	0.000
P0Q0	8.000	-75.00	-0.500	300.000
P1Q1	8.000	-25.00	-0.500	100.000
P2Q2	8.000	-0.00	-0.000	0.000
P3Q3	8.000	-25.00	-0.500	100.000
P4Q4	8.000	-75.00	-0.500	300.000
P0Q1	12.806	120.06	0.800	1230.600
P1Q2	12.806	40.02	0.800	410.200
P3Q2	12.806	40.02	0.800	410.200
P4Q3	12.806	120.06	0.800	1230.600

Step 4: Deflection

\Delta_E = \sum\frac{kFL}{AE} = \frac{9550.4}{80\,000} = 119.38\ \text{mm}

Answer: vertical deflection at E = 119.4 mm downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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