How is the total deflection of E found?

Add the effects of loading, lack of fit in CE and DE, and cooling of CD

Which expression gives the loading effect?

sum kFL / AE = 366.667 / (10^-3 x 2 x 10^8) = 1.833 mm

Why does cooling CD by 15°C move E downward by 0.208 mm?

CD shortens by 0.36 mm and k_CD is negative, so the product is positive

Temperature has no effect on deflection

Asked 2/24
2068 Bhadra · 12 marks

Truss joint E: loading, lack of fit, temperature

Short answer

The vertical deflection at joint E is the sum of three effects: 1.83 mm from the 100 kN load, 9.24 mm from members CE and DE being 8 mm too long, and 0.21 mm from a 15°C temperature drop in CD. Total downward deflection = 11.28 mm.

Determine the vertical deflection of joint E of the truss due to (i) loading shown, (ii) members CE and DE being 8 mm too long and (iii) temperature of member CD alone is decreased by 15°C. Given: cross-sectional area of all members = 1000 mm², Young's modulus = $2\times10^5\ \text{N/mm}^2$ , and coefficient of thermal expansion = $12\times10^{-6}/^\circ$ C. [Figure: truss with A (hinge) and B (roller), E at the bottom middle, AE = EB = 2 m; top joints C and D; angles 60°; 100 kN downward at E.]

Given data

A(0,0) hinge, E(2,0), B(4,0) roller; top joints C(1, 1.732) and D(3, 1.732); all members 2 m (60 $^\circ$ ). 100 kN down at E. $A = 1000$ mm $^2 = 10^{-3}$ m $^2$ , $E = 2\times10^5$ N/mm $^2 = 2\times10^8$ kN/m $^2$ , $\alpha = 12\times10^{-6}$ / $^\circ$ C. Unit vertical load (down) at E.

Member forces (tension +)

Member	L (m)	F (kN)	k	kFL (kN m)	Temp. term (mm)	Fit term (mm)
AE	2.000	28.87	0.289	16.667	0.0000	0.0000
EB	2.000	28.87	0.289	16.667	0.0000	0.0000
AC	2.000	-57.74	-0.577	66.667	-0.0000	-0.0000
CE	2.000	57.74	0.577	66.667	0.0000	4.6188
CD	2.000	-57.74	-0.577	66.667	0.2078	-0.0000
ED	2.000	57.74	0.577	66.667	0.0000	4.6188
DB	2.000	-57.74	-0.577	66.667	-0.0000	-0.0000

Reactions: $A_y = B_y = 50$ kN.

(i) Loading

\Delta_1 = \frac{\sum kFL}{AE} = \frac{366.667}{10^{-3}\times2\times10^8} = 1.833\ \text{mm}\ (\downarrow)

(ii) CE and DE 8 mm too long

\Delta_2 = (0.577)(8) + (0.577)(8) = 9.238\ \text{mm}\ (\downarrow)

(iii) Temperature of CD decreased by 15 $^\circ$ C

\Delta L_{CD} = 12\times10^{-6}\times(-15)\times2000 = -0.36\ \text{mm},\qquad \Delta_3 = k_{CD}\Delta L = (-0.577)(-0.36) = 0.208\ \text{mm}\ (\downarrow)

Total

\Delta_E = 1.833 + 9.238 + 0.208 = 11.279\ \text{mm}

Answer: vertical deflection of E = 11.28 mm downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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