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The force in the member BD and CD respectively are
  • a)
    6 kN (compressive)
  • b)
    6 kN (tensile)
  • c)
    4 kN (compressive)
  • d)
    4 kN (tensile)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The force in the member BD and CD respectively area)6 kN (compressive)...
Suppose R1, R2 and R3 are the reactions at A and F as shown in figure.

R3 = 4 kN
R1 + R2 = 12
R1 × 8 = 12 × 4 + 4 × 6
R1 = 9 kN
R2 = 3 kN
cos θ = 6/7.21
sin θ = 4/7.21
At joint A
Resolving vertically,
TAB = −9 kN (∵ considered tensile as positive)
At joint B
Resolve vertically,

TBC cos θ + TAB = 0

Resolve horizontally,
TBC sin θ + TBD = 0
TBD = −10.815 × 4/7.21 = −6 kN = 6 kN (compressive)
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The force in the member BD and CD respectively area)6 kN (compressive)b)6 kN (tensile)c)4 kN (compressive)d)4 kN (tensile)Correct answer is option 'A'. Can you explain this answer?
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