Question

A С 1.5 Using virtual work determine the vertical displacement at join E. Assume D is a roller and F is a pin A = 2 in? E = 29000 ksi IE 6k 2'0" 2'0

Answer 1

Solution А. B C les" O2 03 E T 2 11 2 СК C 1 F (Pin) (Rollery's HE Ro RE 2 in² E = 29,000 ksi, e cae) = I (BF) Support Reactions 16)+ (2) 2.5" • Eme = 0 2 tre Gore - O Ro 3 Rox 4 - 682 4 3 Due to Symmeboy, RF = 3 12 .. = = RO ) 10 Ro 3K RE = 3 K 01 Не - 0 .E Fx = 0 @ using method of Joints, find forces in each member subjected to External load.i.eek a) Joint D - Efy og ptve tove FpA 34 FDA (0) F DE 3 fon = -3 * :[on=3](c) ( Assume, All unknown fortes qae Tensile ie quay form Joint)

E Fx = 0, → tve FoE = 0 & -ve, b) Joint A tan o,= 1.5 :. :: 0, = 36.87 2 А * FAB FAE Efy = 0, =o 3 - Fae sin 36.87 +3 3 FAE - 5 K sin 36.87 FAE - SK (T) o Ef =0, FAB + FAE Cos 36.87= fac cos 36.87 = -5 COS 36.87 FAB = -4 K FAB = 4 KCC .. FAB - : FAB 11 c) Joint E l.s tan oz - O2 = 36.87° 2 5 FEB VE Efy = 0, FEB - 6+5 sin 36.87 = 0 FEF : 6 FEB = 3K (T) Efx=0 FEF 5 cos 36.87 = 0 FEF 4 K (T) d) Joint C - AS No External forces and Both Forces are non-coliness. FCB :. FCB FCF = 0 2 C + CF

E fy 37fFB FB 4 Ffe - - 5 C 3 6) using method of section, find member forces. e) Joint F 03 - 36.37° 0, sin 36.87 = 0 -3 sin 36.87 FFB =5K (c) * Now, Removing External loads and Apply Unit load in direction of Ek at'E. Ró IK RE a) suppoot Reactions Due to symmetry of figure, Ro' = RK = 0.5k and HA B A 1,5" 2 E O3 H F 01 2 a i) Joint D Ery = 0, 0.5 +FO A = 0 FDA FO A = -0.5 FDA = 0.5 K O D Fot 0.5 Etx=0, FDE = 0

ii) Joint A = A *FAB o 0, = 36.870 Same as befoe Calculations, Efy = 0, 0.5- FAE sin 36.87=0 0.5 FAE = sin 36.87 0.833 K (1) 0.5 FAE Efx = 0, FAB + FAE Cos 36.87 = 0 FAB = FAE COS 36 87 = -0.833 COS 3 5.37 :. FAB =-0.67 K :. FAB 0.67 K (C) U iii) Joint E A FEB 0.833 Efy = 0, 0, = 02 = 36.87 ° FER - 1 + 0.833 sin 36 37 = 0 1 -0.5 E FEF : FEB FEB = 0.5 K. (T) Efx=0, FEF 0.833 COS 36.87 = 0 FEF = 0.67 K (T) il iv) Joint c FCB FCF = = 0 C FCB FCF

v ) Joint F - Efy = 0, 03 = 36.87 °=0,=02 FFB sin 36-87 +0.5=0 FFB 0.67 0.5 FFB - -0.833 sin 36.87 0.5 FFB = 0.833 Kro PKL member Length Forces due to Actual Virtual load load (P) (K) (K) (K) (inch) -0.5 DA -3 1.5" 2.25 2" O 0 DE 11 -0.67 -4 2" 5-36 AB +5 AE 2.5" +0.833 10:412 +3 to.5 1.5" 2.25 EB EF +4 2 to-67 5-36 O 2" CB 1.5" CF -5 -0.833 FB 2.5" 110.412 (T) = Tension Positive (C) = compression = negative EPKL 36.044

. Actual load - 6K - = virtual load : IK Area = 2 in 2 E = 29,000 KS; Sample calculations © For DA = PX KXL = (-3)*(-0.5)X(115) 2.25 ③ EPKL (2.25 +5.367 10.412 +2.25 +5.36 + 10.412) = 36.044 Using virtual Work method, vestical Displacement at E E PKL AE 36.044 DE 2x 29,000 -4 AE 6.214 x 10 = inches = o Answer Vertical displacement at Joint E. DE = 6.214 x 10-4 inches