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Truss: bar forces due to load and temperature

Short answer

Using consistent deformation with the C roller as redundant: load case gives AB 7.10, BC 2.20, CD 2.20, EF -7.10, AE -8.87, BE -3.68, CF -20.03, DF -2.75, BF 6.13 kN; temperature case gives AB 7.49, BC 14.99, CD 14.99, EF -7.49, AE -9.37, BE 5.62, CF 16.86, DF -18.74, BF -9.37 kN.

Compute the bar forces in all members due to: (i) given load and (ii) temperature rise by 30°C in the upper chord. Take E=2×105 N/mm2E = 2\times10^{5}\ \text{N/mm}^2, α=10.8×10−6/°C\alpha = 10.8\times10^{-6}/°C. Take area of all members to be 30 cm². [Figure: truss with top joints under 9 kN and 18 kN downward loads; three panels of 4 m (total 12 m), height 3 m, diagonals; left support hinged, two roller supports at the lower chord.]

Assumptions (the figure is not clear): bottom joints A, B, C, D at 0, 4, 8, 12 m; top joints E and F, 3 m above B and C. Members: bottom chord AB, BC, CD; top chord EF; verticals BE, CF; end diagonals AE, DF and one middle diagonal BF (9 members). 9 kN acts downward at E and 18 kN downward at F. A is a hinge; the two rollers are at C and D (vertical reactions only). EA=2×105 N/mm2×3000 mm2=6×108EA=2\times10^5\ \text{N/mm}^2\times3000\ \text{mm}^2=6\times10^8 N =6×105=6\times10^5 kN for every bar.

Degree of indeterminacy

m=9m=9, r=4r=4, j=6j=6: Ds=9+4−12=1D_s=9+4-12=1 (external). Take the roller reaction at C, XX (upward), as the redundant. Primary structure: remove the roller at C, leaving a truss hinged at A and on a roller at D.

(i) Bar forces due to the loads

S0S_0 = forces in the primary truss under the loads; S1S_1 = forces for a unit upward force at C.

BarL (m)S0S_0 (kN)S1S_1S0S1LS_0S_1LS12LS_1^2LFinal SS (kN)
AB4.00016.00-0.444-28.440.7907.10
BC4.00020.00-0.889-71.113.1602.20
CD4.00020.00-0.889-71.113.1602.20
EF4.000-16.000.444-28.440.790-7.10
AE5.000-20.000.556-55.561.543-8.87
BE3.0003.00-0.333-3.000.333-3.68
CF3.0000.00-1.0000.003.000-20.03
DF5.000-25.001.111-138.896.173-2.75
BF5.000-5.000.556-13.891.5436.13
Sum-410.4420.494

Vertical displacement of C in the primary truss (positive upward) ΔC=∑S0S1LEA=−410.446×105=−0.684\Delta_C=\dfrac{\sum S_0S_1L}{EA}=\dfrac{-410.44}{6\times10^5}=-0.684 mm, so C would move 0.684 mm downward without the roller. Compatibility (C cannot move):

X=−∑S0S1L∑S12L=−−410.4420.494=20.03 kN (up)X=-\frac{\sum S_0S_1L}{\sum S_1^2L}=-\frac{-410.44}{20.494}=20.03\ \text{kN (up)}

Final bar forces: AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN (+ tension, - compression).

(ii) Temperature rise of 30°C in the upper chord (EF)

Free elongation of EF: e=α ΔT L=10.8×10−6×30×4=1.296e=\alpha\,\Delta T\,L=10.8\times10^{-6}\times30\times4=1.296 mm. The primary truss expands freely and no forces develop in it (S0=0S_0=0); only the displacement at C changes. The compatibility equation with the temperature term is

X∑S12LEA+S1EF e=0⇒X=−EA S1EF e∑S12L=−16.86 kN\frac{X\sum S_1^2L}{EA}+S_1^{EF}\,e=0\Rightarrow X=-\frac{EA\,S_1^{EF}\,e}{\sum S_1^2L}=-16.86\ \text{kN}

(with S1EF=0.444S_1^{EF}=0.444). Bar forces S=XS1S=XS_1:

AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN (+ tension, - compression).

Answer: (i) AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN. (ii) AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN. Both were checked with a full stiffness analysis.

✓ Answer checked by Claude Opus 5.5 (high effort)

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