Pin-jointed steel structure, Mechanical Engineering

Pin-jointed steel structure:

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


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.)

Posted Date: 1/29/2013 7:35:15 AM | Location : United States

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