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Pin-jointed steel structure:
Find out the forces in the members of the pin-jointed steel structure.
Suppose fCD and fDE both as tensile forces; and also suppose the nature of other members forces as illustrated.
fCD = 2 t ⇒ ∴ fCD = 2 t (tension)
and fDE = 0
∠ BCE = ∠ DCE = 45o (obviously)
fCD = fCE cos 45o ⇒ ∴ fCE = 2 √2
= 2.8284 (comp.)
i.e. CE is a strut
f CB = f CE cos 45o = 2 × √2 × (½) = 2 t (tension)
i.e. CB is a tie.
f EF = f CE cos 45o = 2 × √2 × (1/√2) = 2 t (comp.)
f EB = f CE cos 45o = 2 × √2 × (1/√2) = 2 t (tension)
∴ f BE = fBF cos 45o (comp.)
f BE =2 √2 t(comp.)
f AB - f FB cos 45o = fBC
f AB = 2 × √2 × (1/√2) = 4t (comp.)
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