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Truss: force method, EA and 2EA members

Short answer

Using the force method with diagonal CE as redundant, the member forces (kN, + tension) are: AB 66.67, BC 45.06, DE -200, EF -21.61, BE -66.21, CF -16.21, AE 166.67, BD 0, BF 27.01, CE -56.32 (compression).

Find out the member forces in the truss shown in figure below using the force method. The axial rigidity of all vertical and horizontal members is EA and that for all inclined members is 2EA. [Figure: two-panel tall truss ABCDEF, 3 m wide, two panels of 4 m height each, with X-bracing; 50 kN horizontal at C and 50 kN horizontal at B; A hinged, D roller.]

Assumptions (figure not fully clear): joints A(0,0), B(0,4), C(0,8) on the left and D(3,0), E(3,4), F(3,8) on the right (3 m wide, two panels of 4 m). Members: verticals AB, BC, DE, EF and horizontals BE, CF (axial rigidity EA); diagonals AE, BD, BF, CE (axial rigidity 2EA, lengths 5 m). Both 50 kN loads act horizontally to the right at B and C. A is a hinge and D a roller (vertical reaction only). There is no member joining A and D (the supports hold them).

Degree of indeterminacy

m=10m=10, r=3r=3, j=6j=6: Ds=m+r−2j=10+3−12=1D_s=m+r-2j=10+3-12=1. Take the force XX in the diagonal CE as the redundant.

Reactions

∑MA=0\sum M_A=0: RD×3=50×8+50×4⇒RD=200R_D\times3=50\times8+50\times4\Rightarrow R_D=200 kN (up), RAy=200R_{Ay}=200 kN (down), HA=100H_A=100 kN (to the left).

Forces S0S_0 (CE removed, loads only) and S1S_1 (unit tension in CE)

Solved joint by joint. Compatibility: ∑S0S1LAE+X∑S12LAE=0\sum S_0S_1\dfrac{L}{AE}+X\sum S_1^2\dfrac{L}{AE}=0 (the L/AEL/AE values are in units of 1/EA1/EA, with the 2EA2EA of the diagonals already taken into account).

MemberL (m)AES0S_0 (kN)S1S_1S0S1L/AES_0S_1L/AES12L/AES_1^2L/AEFinal SS (kN)
AB4.000EA66.670.0000.0000.000066.67
BC4.000EA0.00-0.8000.0002.560045.06
DE4.000EA-200.000.0000.0000.0000-200.00
EF4.000EA-66.67-0.800213.3332.5600-21.61
BE3.000EA-100.00-0.600180.0001.0800-66.21
CF3.000EA-50.00-0.60090.0001.0800-16.21
AE5.0002EA166.670.0000.0000.0000166.67
BD5.0002EA0.000.0000.0000.00000.00
BF5.0002EA83.331.000208.3332.500027.01
CE5.0002EA0.001.0000.0002.5000-56.32
Sum691.66712.2800
X=−691.66712.2800=−56.32 kNX=-\frac{691.667}{12.2800}=-56.32\ \text{kN}

Final member forces (S=S0+XS1S=S_0+XS_1; + tension, - compression)

Verticals: AB = 66.67, BC = 45.06, DE = -200.00, EF = -21.61 kN. Horizontals: BE = -66.21, CF = -16.21 kN. Diagonals: AE = 166.67, BD = 0.00, BF = 27.01, CE = -56.32 kN.

As a check, the whole truss was also analysed by the direct stiffness method and gives identical forces.

Answer: CE = -56.32 kN (compression), AE = 166.67 kN, BF = 27.01 kN, BD = 0.00 kN, verticals and horizontals as tabulated.

✓ Answer checked by Claude Opus 5.5 (high effort)

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