GATE CE 2018 Set 1 — Question 53
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Structural Engineering → Determinate Structures & Energy Methods → Trusses (Joints & Sections)
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Question
Consider the deformable pin-jointed truss with loading, geometry and section properties as shown in the figure.
Given that
E=2×1011 N/m2,
A=10 mm2,
L=1 m and
P=1 kN. The horizontal displacement of Joint C (in mm, up to one decimal place) is
______Solution
1.Identify member properties and geometry:
- Member AB: length L, area A, modulus E, stiffness AE/L.
- Member BC: length L, area A, modulus E, stiffness AE/L.
- Member AC: length 2L, area 2A, modulus E, stiffness 2AE/(2L)=2AE/L.
2.
Calculate member forces (S) due to external loads (P horizontal, 2P vertical at C):- From joint C equilibrium:
∑Fx=0⇒P−SACcos45∘=0⇒SAC=2P (Tension)
∑Fy=0⇒−2P−SBC−SACsin45∘=0⇒SBC=−2P−P=−3P (Compression)
- From joint A (roller) equilibrium:
∑Fx=0⇒SAB+SACcos45∘=0⇒SAB=−P (Compression)
3.Apply unit horizontal load at C to find displacement (k forces): - kAC=2
- kBC=−1
- kAB=−1
4.
Calculate horizontal displacement ΔC using ∑AESkL:- ΔC,AC=2AE(2P)(2)(2L)=AE2PL
- ΔC,BC=AE(−3P)(−1)L=AE3PL
- ΔC,AB=AE(−P)(−1)L=AEPL
- Total ΔC=(4+2)AEPL
5.
Substitute values (P=1000 N,L=1 m,AE=2×106 N):- ΔC=(4+1.414)×2×1061000×1=5.414×0.0005=0.002707 m=2.707 mm.
Rounding to one decimal place, the displacement is 2.7 mm.
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