Pin-jointed steel structure:
Find out the forces in the members of the pin-jointed steel structure.
Solution
Joint D
 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
Joint C
∠ 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.
Joint E
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)
Joint B
∴          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.)