Pin-jointed steel structure:

Find out the forces in the members of the pin-jointed steel structure.

Solution

Joint D

 Suppose f_CD and f_DE both as tensile forces; and also suppose the nature of other members forces as illustrated.

f_CD = 2 t            ⇒         ∴ f_CD = 2 t (tension)

and      f_DE = 0

Joint C

∠ BCE = ∠ DCE = 45^o  (obviously)

f_CD = f_CE  cos 45^o ⇒  ∴ f_CE  = 2 √2

= 2.8284 (comp.)

i.e. CE is a strut

f _CB = f _CE cos 45^o  = 2 × √2 × (½) = 2 t  (tension)

i.e. CB is a tie.

Joint E

f _EF = f _CE cos 45^o  = 2 × √2 ×  (1/√2) = 2 t  (comp.)

 f _EB = f _CE cos 45^o  = 2 × √2 ×  (1/√2) = 2 t   (tension)

Joint B

∴          f _BE  = f_BF cos 45^o    (comp.)

               f _BE  =2 √2 t(comp.)

f _AB - f _FB cos 45^o  = f_BC  

f _AB = 2 × √2 ×  (1/√2) = 4t  (comp.)