Question

2. Derive a formula for the vertical deformation at “c” in the truss shown. 2 ft 10 kips 2 ft All bars have: h E = 10000 k/in2 A = 2 in2 2 ft 12 ft 12 ft

Answer 1

Given, K- 1 Kips Virtual load for all bars- E = 10000 Klin? A = 2 in2 12Ft a 0 & 12Ft 1) force acting at point 'a ? 10 kips 12ft 0 Pab 12Ft 3 ОО. goo K 12 Et tano = 12 o = tan - 1 (1) 0 = 45° by applying Sine rule.at 'o'. Pab 10 Pai Sin 90° Sino Sin 135° Pab 10x Singo I 34.1 kips sin 45° Pai 108 sin 135 = 10 kips. Sin 450 2) force acting at point 'b' 135 Poc Pab to by applying Sine rule at 'b' al Poc. Poi Paba Sino sin 185° - Singooi Pbc » 14. 2. X Sin 45 = 10 kips a Singo Phi → * 14.1 X Sin195 - 10 kips singol up bi 0-45° (Same as above)

Poc a - = - 25 Pci Pod o = 45° Poi 7 Pci Pai por 0=45 O= 45° 37 Force acting at point 'c! by applying sine rule. Poc Pci pod Sin 45° Sin 270º Sin 45° Pai Pbox sin 270 = -14 1 Kips, Sin 45° Pcd Pboysin 45 = 10 Kips. Sin 45 47 Force acting at point i n here Sine sue is not applicable. $o. {fx=0 Oz Pai - Pai - Pcicoso O= 10 - Pai - (-14 1 Cos 45°) 0= 10-Pai + 10 Pdi = 20 kips, Efy=0 O = Pbi + Peisino -Phi 0= 10 - 14. 1 Sin 45° - Phi o = 10-10-Phi Phi=0 5) force acting at point d'. Efx=0 o Pdit Pahcoso Poi - Pan cos 45° Pan + -20/C0545º =-28.28 Kips. I fy=0 o= Pdet (- Pde - Pdh sino) o=10 - Pde - Pansin 45° Pde + Pansin 45º = 10 - (a) XL Oud

oni 1 Phd Phe V 67 force acting at point' h. Efx=0 0 = Phet Phd Coso o= Phe- 28.28 Cos 45° The = -20 kips. Efyzo o Phi + Phd sino - Png O= 0+ Phd sino - Png Ping = Phd Sino (6) 77 force acting at point é Png Peg => 20% = 2828 Kips, o=Phet Pea Coso Pheone I cos 45o - 28128 Kips. Efyeo o = Ped - Peg sino - Pet O = Ped - 28.28 8in45° - Pet 20 = Ped - Pef -(C). By force acting at point g' and 'p? Ping Pge Solving as above.

U ( Here we have to derive formu or Vertical deformation at point & so we are assuming Vartual load at point ¿ SK = 1 kips} After applying this load we check force in all memberis. force in all members are Zero Excepting, at member cd, ed. Δ. Σ ΣΚΑ - Ked (PL) + Kde ( 1 ) + Kef When ,K= 1 (Vertual load) K = Ked Kde Kef = 1 A c. = 3K (PL) = 0.072 inch L=12 Ft A = ín? E = 10000 klin² P = 10 kipe. Ke 1 Rips.