Because member areas differ, the loading deflection is computed as:

(1/E) sum of kFL/A with A for each member

For a 20 C fall in CD of 5 m, the length change and its effect are:

-1.2 mm giving 0.693 mm downward at E, with k = -0.577

Asked 2/24
2075 Bhadra · 12 marks

Truss joint E: temperature decrease and lack of fit

Short answer

Vertical deflection of joint E is 6.98 mm downward. Load contributes 1.67 mm down, 20°C temperature drop in member CD adds 0.69 mm down, and member CE fabricated 8 mm too long adds 4.62 mm down.

Determine the vertical deflection of joint E due to the decrease in temperature by 20°C in member CD and member CE being 8 mm too long. Take $E = 200\times10^3\ \text{N/mm}^2$ , $\alpha = 12\times10^{-6}/^\circ$ C. Horizontal member area = 2000 mm², vertical/inclined member area = 1000 mm². [Figure: truss with A (hinge), E (bottom middle) and B (roller), AE = EB = 5 m; top joints D and C; all inclined members at 60°; 80 kN downward load at C.]

Given data

A(0,0) hinge, E(5,0), B(10,0) roller; top joints D(2.5, 4.330) and C(7.5, 4.330); all members are 5 m long (equilateral triangles). 80 kN acts downward at C. $E = 200\times10^3$ N/mm $^2 = 2\times10^8$ kN/m $^2$ , $\alpha = 12\times10^{-6}$ / $^\circ$ C. Areas: horizontal members AE, EB, DC = 2000 mm $^2$ ; inclined members = 1000 mm $^2$ . Apply a unit load (down) at E.

Reactions (moments about A): $B_y\times10 = 80\times7.5 \Rightarrow B_y = 60$ kN, $A_y = 20$ kN. For the unit load at E: $A_y = B_y = 0.5$ .

Member forces (tension +)

Member	L (m)	F (kN)	k	kFL (kN m)	Temp. term (mm)	Fit term (mm)
AE	5.000	11.55	0.289	16.667	0	0
EB	5.000	34.64	0.289	50.000	0	0
AD	5.000	-23.09	-0.577	66.667	0	0
DE	5.000	23.09	0.577	66.667	0	0
DC	5.000	-23.09	-0.577	66.667	0.6928	0
EC	5.000	-23.09	0.577	-66.667	0	4.6188
CB	5.000	-69.28	-0.577	200.000	0	0

(i) Loading

Because the areas differ, use $\sum kFL/A$ with the area of each member:

\Delta_1 = \frac{1}{E}\sum\frac{kFL}{A} = \frac{1}{2\times10^8}\left[\frac{133.33}{0.002} + \frac{266.67}{0.001}\right] = 1.667\ \text{mm}\ (\downarrow)

(ii) Temperature fall of 20 $^\circ$ C in CD

\Delta L_{CD} = 12\times10^{-6}\times(-20)\times5000 = -1.2\ \text{mm},\qquad \Delta_2 = k\,\Delta L = (-0.577)(-1.2) = 0.693\ \text{mm}\ (\downarrow)

(iii) CE 8 mm too long

\Delta_3 = k_{CE}\,e = (0.577)(8) = 4.619\ \text{mm}\ (\downarrow)

Total

\Delta_E = 1.667 + 0.693 + 4.619 = 6.978\ \text{mm}

Answer: vertical deflection of joint E = 6.98 mm downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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