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Truss joint L2 by virtual work and temperature

Short answer

The vertical deflection of joint L2 due to the three 10 kN loads is 4.69 mm downward. A 20°C temperature rise in the top boom causes an upward deflection of 5.40 mm. The net deflection is 0.71 mm upward.

Determine, using virtual work method, the vertical deflection of joint L2L_2. The L/A values for diagonal and vertical members are 12 mm⁻¹ and for horizontal members are 6 mm⁻¹. Take E=200×103 N/mm2E = 200\times10^3\ \text{N/mm}^2 for all members. (i) Find vertical deflection due to loads as shown in figure. (ii) Find the additional deflection if the top boom is subjected to a temperature rise of 20°C. Take α=10.8×10−6/∘\alpha = 10.8\times10^{-6}/^\circC. [Figure: truss, height 12 m, 4 panels of 10 m; bottom joints L1L_1 to L4L_4 and top joints U1U_1 to U5U_5 as drawn; three 10 kN loads on the bottom chord; L1L_1 hinge, L4L_4 roller.]

Given data and assumed geometry

Truss of height 12 m, 4 panels of 10 m. The joint labels in the figure could not be read exactly, so the support joints are taken as the two ends of the bottom chord, labelled here L0 (hinge) and L4 (roller), with bottom joints L1, L2, L3 between them; top joints U1 to U5 (U1 above L0). Verticals Ui+1LiU_{i+1}L_i, diagonals U1L1, U2L2, U4L2, U5L3. 10 kN down at L1, L2, L3. L/AL/A = 12 mm−1^{-1} for diagonals and verticals, 6 mm−1^{-1} for horizontal members. E=200×103E = 200\times10^3 N/mm2=200^2 = 200 kN/mm2^2.

By symmetry the reactions at L0 and L4 are 15 kN. A unit downward load is applied at L2L_2.

(i) Deflection due to loads

Δ=∑F k LAE\Delta = \sum F\,k\,\frac{L}{AE}
MemberFF (kN)kkL/AL/A (mm−1^{-1})F k (L/A)F\,k\,(L/A) (kN/mm)
U1U2-12.50-0.417631.25
U2U3-16.67-0.833683.33
U3U4-16.67-0.833683.33
U4U5-12.50-0.417631.25
L0L10.000.00060.00
L1L212.500.417631.25
L2L312.500.417631.25
L3L40.000.00060.00
U1L0-15.00-0.5001290.00
U2L1-5.00-0.5001230.00
U3L2-0.00-0.000120.00
U4L3-5.00-0.5001230.00
U5L4-15.00-0.5001290.00
U1L119.530.65112152.50
U2L26.510.6511250.83
U4L26.510.6511250.83
U5L319.530.65112152.50
Δ1=∑Fk(L/A)E=938.33200=4.692 mm (↓)\Delta_1 = \frac{\sum Fk(L/A)}{E} = \frac{938.33}{200} = 4.692\ \text{mm}\ (\downarrow)

(ii) Additional deflection due to 20∘^\circC rise in the top boom

The top boom has 4 members, each 10 m long: αΔTL=10.8×10−6×20×10 000=2.16\alpha\Delta T L = 10.8\times10^{-6}\times20\times10\,000 = 2.16 mm each.

Δ2=∑k α ΔT L=2.16 (−0.417−0.833−0.833−0.417)=2.16×(−2.5)=−5.40 mm\Delta_2 = \sum k\,\alpha\,\Delta T\,L = 2.16\,(-0.417 - 0.833 - 0.833 - 0.417) = 2.16\times(-2.5) = -5.40\ \text{mm}

(upward, because the top boom lengthens and its kk values are negative: the top boom is in compression under a downward unit load).

Answer: (i) vertical deflection of L2 due to loads =4.69= 4.69 mm downward; (ii) additional deflection due to temperature =5.40= 5.40 mm upward; net =0.71= 0.71 mm upward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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