PYQs / GATE CE / 2023 / Set 1 / Q39 GATE CE 2023 Set 1 — Question 39 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Hard Unit Load Method Determinate Structures & Energy Methods Structural Engineering Trusses (Joints & Sections)
Structural Engineering → Determinate Structures & Energy Methods → Trusses (Joints & Sections)
Last updated 5 September 2026
Question Consider the pin-jointed truss shown in the figure (not to scale). All members have the same axial rigidity,
A E AE A E . Members
Q R QR QR ,
R S RS R S , and
S T ST S T have the same length
L L L . Angles
Q B T QBT QB T ,
R C T RCT R C T ,
S D T SDT S D T are all
90 ∘ 90^\circ 9 0 ∘ . Angles
B Q T BQT B QT ,
C R T CRT C R T ,
D S T DST D S T are all
30 ∘ 30^\circ 3 0 ∘ . The joint
T T T carries a vertical load
P P P . The vertical deflection of joint
T T T is
k P L A E k \frac{PL}{AE} k A E P L . What is the value of
k k k ?
Solution Using the unit load method, the vertical deflection at
T T T is
δ T = ∑ F i u i L i A E \delta_T = \sum \frac{F_i u_i L_i}{AE} δ T = ∑ A E F i u i L i . Since
u i = F i / P u_i = F_i / P u i = F i / P ,
δ T = 1 P ∑ F i 2 L i A E \delta_T = \frac{1}{P} \sum \frac{F_i^2 L_i}{AE} δ T = P 1 ∑ A E F i 2 L i .
1. Geometry:
Q R = R S = S T = L QR = RS = ST = L QR = R S = S T = L From the given angles, Q B , R C , S D QB, RC, SD QB , R C , S D B T , C T , D T BT, CT, DT B T , C T , D T In △ S D T \triangle SDT △ S D T S D = L cos 30 ∘ = 3 2 L SD = L \cos 30^\circ = \frac{\sqrt{3}}{2}L S D = L cos 3 0 ∘ = 2 3 L D T = L sin 30 ∘ = 0.5 L DT = L \sin 30^\circ = 0.5L D T = L sin 3 0 ∘ = 0.5 L In △ R C T \triangle RCT △ R C T R C = 2 L cos 30 ∘ = 3 L RC = 2L \cos 30^\circ = \sqrt{3}L R C = 2 L cos 3 0 ∘ = 3 L C T = 2 L sin 30 ∘ = L CT = 2L \sin 30^\circ = L C T = 2 L sin 3 0 ∘ = L In △ Q B T \triangle QBT △ QB T Q B = 3 L cos 30 ∘ = 1.5 3 L QB = 3L \cos 30^\circ = 1.5\sqrt{3}L QB = 3 L cos 3 0 ∘ = 1.5 3 L B T = 3 L sin 30 ∘ = 1.5 L BT = 3L \sin 30^\circ = 1.5L B T = 3 L sin 3 0 ∘ = 1.5 L Horizontal members: C D = C T − D T = 0.5 L CD = CT - DT = 0.5L C D = C T − D T = 0.5 L B C = B T − C T = 0.5 L BC = BT - CT = 0.5L B C = B T − C T = 0.5 L
2. Force Analysis (under load P P P T T T
At joint T T T F S T = P cos 30 ∘ = 2 P 3 F_{ST} = \frac{P}{\cos 30^\circ} = \frac{2P}{\sqrt{3}} F S T = c o s 3 0 ∘ P = 3 2 P F D T = P tan 30 ∘ = P 3 F_{DT} = P \tan 30^\circ = \frac{P}{\sqrt{3}} F D T = P tan 3 0 ∘ = 3 P At joint D D D F C D = F D T = P 3 F_{CD} = F_{DT} = \frac{P}{\sqrt{3}} F C D = F D T = 3 P F S D = 0 F_{SD} = 0 F S D = 0 At joint S S S F R S = F S T = 2 P 3 F_{RS} = F_{ST} = \frac{2P}{\sqrt{3}} F R S = F S T = 3 2 P F C S = 0 F_{CS} = 0 F C S = 0 At joint C C C F B C = F C D = P 3 F_{BC} = F_{CD} = \frac{P}{\sqrt{3}} F B C = F C D = 3 P F R C = 0 F_{RC} = 0 F R C = 0 At joint R R R F Q R = F R S = 2 P 3 F_{QR} = F_{RS} = \frac{2P}{\sqrt{3}} F QR = F R S = 3 2 P F B R = 0 F_{BR} = 0 F B R = 0
3. Deflection Calculation: δ T = 1 P [ 3 × ( 2 P / 3 ) 2 L A E + 3 × ( P / 3 ) 2 ( 0.5 L ) A E ] \delta_T = \frac{1}{P} \left[ 3 \times \frac{(2P/\sqrt{3})^2 L}{AE} + 3 \times \frac{(P/\sqrt{3})^2 (0.5L)}{AE} \right] δ T = P 1 [ 3 × A E ( 2 P / 3 ) 2 L + 3 × A E ( P / 3 ) 2 ( 0.5 L ) ] δ T = 1 P [ 3 × 4 P 2 3 L A E + 3 × P 2 3 0.5 L A E ] = 4 P L A E + 0.5 P L A E = 4.5 P L A E \delta_T = \frac{1}{P} \left[ 3 \times \frac{4P^2}{3} \frac{L}{AE} + 3 \times \frac{P^2}{3} \frac{0.5L}{AE} \right] = 4 \frac{PL}{AE} + 0.5 \frac{PL}{AE} = 4.5 \frac{PL}{AE} δ T = P 1 [ 3 × 3 4 P 2 A E L + 3 × 3 P 2 A E 0.5 L ] = 4 A E P L + 0.5 A E P L = 4.5 A E P L Thus,
k = 4.5 k = 4.5 k = 4.5 .
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Solution Using the unit load method, the vertical deflection at
T T T is
δ T = ∑ F i u i L i A E \delta_T = \sum \frac{F_i u_i L_i}{AE} δ T = ∑ A E F i u i L i . Since
u i = F i / P u_i = F_i / P u i = F i / P ,
δ T = 1 P ∑ F i 2 L i A E \delta_T = \frac{1}{P} \sum \frac{F_i^2 L_i}{AE} δ T = P 1 ∑ A E F i 2 L i .
1. Geometry:
Q R = R S = S T = L QR = RS = ST = L QR = R S = S T = L From the given angles, Q B , R C , S D QB, RC, SD QB , R C , S D B T , C T , D T BT, CT, DT B T , C T , D T In △ S D T \triangle SDT △ S D T S D = L cos 30 ∘ = 3 2 L SD = L \cos 30^\circ = \frac{\sqrt{3}}{2}L S D = L cos 3 0 ∘ = 2 3 L D T = L sin 30 ∘ = 0.5 L DT = L \sin 30^\circ = 0.5L D T = L sin 3 0 ∘ = 0.5 L In △ R C T \triangle RCT △ R C T R C = 2 L cos 30 ∘ = 3 L RC = 2L \cos 30^\circ = \sqrt{3}L R C = 2 L cos 3 0 ∘ = 3 L C T = 2 L sin 30 ∘ = L CT = 2L \sin 30^\circ = L C T = 2 L sin 3 0 ∘ = L In △ Q B T \triangle QBT △ QB T Q B = 3 L cos 30 ∘ = 1.5 3 L QB = 3L \cos 30^\circ = 1.5\sqrt{3}L QB = 3 L cos 3 0 ∘ = 1.5 3 L B T = 3 L sin 30 ∘ = 1.5 L BT = 3L \sin 30^\circ = 1.5L B T = 3 L sin 3 0 ∘ = 1.5 L Horizontal members: C D = C T − D T = 0.5 L CD = CT - DT = 0.5L C D = C T − D T = 0.5 L B C = B T − C T = 0.5 L BC = BT - CT = 0.5L B C = B T − C T = 0.5 L
2. Force Analysis (under load P P P T T T
At joint T T T F S T = P cos 30 ∘ = 2 P 3 F_{ST} = \frac{P}{\cos 30^\circ} = \frac{2P}{\sqrt{3}} F S T = c o s 3 0 ∘ P = 3 2 P F D T = P tan 30 ∘ = P 3 F_{DT} = P \tan 30^\circ = \frac{P}{\sqrt{3}} F D T = P tan 3 0 ∘ = 3 P At joint D D D F C D = F D T = P 3 F_{CD} = F_{DT} = \frac{P}{\sqrt{3}} F C D = F D T = 3 P F S D = 0 F_{SD} = 0 F S D = 0 At joint S S S F R S = F S T = 2 P 3 F_{RS} = F_{ST} = \frac{2P}{\sqrt{3}} F R S = F S T = 3 2 P F C S = 0 F_{CS} = 0 F C S = 0 At joint C C C F B C = F C D = P 3 F_{BC} = F_{CD} = \frac{P}{\sqrt{3}} F B C = F C D = 3 P F R C = 0 F_{RC} = 0 F R C = 0 At joint R R R F Q R = F R S = 2 P 3 F_{QR} = F_{RS} = \frac{2P}{\sqrt{3}} F QR = F R S = 3 2 P F B R = 0 F_{BR} = 0 F B R = 0
3. Deflection Calculation: δ T = 1 P [ 3 × ( 2 P / 3 ) 2 L A E + 3 × ( P / 3 ) 2 ( 0.5 L ) A E ] \delta_T = \frac{1}{P} \left[ 3 \times \frac{(2P/\sqrt{3})^2 L}{AE} + 3 \times \frac{(P/\sqrt{3})^2 (0.5L)}{AE} \right] δ T = P 1 [ 3 × A E ( 2 P / 3 ) 2 L + 3 × A E ( P / 3 ) 2 ( 0.5 L ) ] δ T = 1 P [ 3 × 4 P 2 3 L A E + 3 × P 2 3 0.5 L A E ] = 4 P L A E + 0.5 P L A E = 4.5 P L A E \delta_T = \frac{1}{P} \left[ 3 \times \frac{4P^2}{3} \frac{L}{AE} + 3 \times \frac{P^2}{3} \frac{0.5L}{AE} \right] = 4 \frac{PL}{AE} + 0.5 \frac{PL}{AE} = 4.5 \frac{PL}{AE} δ T = P 1 [ 3 × 3 4 P 2 A E L + 3 × 3 P 2 A E 0.5 L ] = 4 A E P L + 0.5 A E P L = 4.5 A E P L Thus,
k = 4.5 k = 4.5 k = 4.5 .
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