Warren Truss Member Force by Joints
Hardmethod-of-sectionswarren-truss
A Warren truss has joints A (0, 0), B (4, 0), C (8, 0), D (2, 3) and E (6, 3), in meters, and seven pin-connected members: AB, BC, AD, DB, DE, BE and EC. A is a pin and C is a vertical-reaction roller. Downward loads are 18 kN at D and 30 kN at E. Neglect self-weight. Find the force in diagonal BE, including tension or compression.
- A3.61 kN tension
- B3.00 kN compression
- C5.41 kN compression
- D3.61 kN compression
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Answer: D. 3.61 kN compression
Moments about A: RC(8) = 18(2) + 30(6) = 216 → RC = 27 kN, RA = 21 kN. Cut BC, BE, DE and take the right portion (E and C): BC and DE are horizontal, so ΣFy = 27 − 30 + NBE(−3/√13) = 0 (BE assumed tension pulls toward B) → NBE = −√13 = −3.606 kN, compression. Wrong: 3.61 tension has the sign reversed; 3.00 kN omits the slope factor (uses the vertical component as the force); 5.41 uses the horizontal ratio 2/√13 instead of 3/√13.
kN
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