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Truss Joint Deflection by Virtual Work

Hardvirtual-worktruss-deflection
A planar pin-jointed truss has joints A (0, 0), B (4, 3) and C (8, 0), coordinates in meters, with members AB, BC and AC. A is a pin and C is a roller restraining vertical displacement. A 24 kN downward load acts at B. Every member has EA = 40,000 kN. Consider axial deformation only. Find the vertical displacement of joint B, including direction.
  1. A6.30 mm upward
  2. B4.22 mm downward
  3. C2.03 mm downward
  4. D6.30 mm downward
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Answer: D. 6.30 mm downward

Unit-load method: δ = ΣNnL/EA. Reactions are 12 kN at A and C. AB = BC = 5 m (sin 3/5, cos 4/5), AC = 8 m. Joint A: 12 + NAB(3/5) = 0 → NAB = NBC = −20 kN; NAC = +16 kN. Dividing by 24 gives nAB = nBC = −5/6, nAC = +2/3. ΣNnL = 2(−20)(−5/6)(5) + 16(2/3)(8) = 166.67 + 85.33 = 252 kN·m. δ = 252/40,000 = 6.30 mm, positive so in the direction of the downward unit load. Wrong: 6.30 upward has the direction reversed; 4.22 mm counts only one inclined member (83.33 + 85.33 = 168.67); 2.03 mm gives the chord the wrong sign (166.67 − 85.33 = 81.33).
m mm
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