Question

Q3: Determine the vertical and horizontal displacement of joint A for the truss shown in Fig. (3), cach bar is made of steel and has the cross-sectional area of 400mm? Take E = 200 GPa. Use the method of virtual work. 25% E 2 m А. cool B 1.5 m -1.5 m 20 KN 40 KN

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Answer 1

We can use Method of Joints to solve the Member Forces FAE 53.13 2 Joint a Fy = 0 FAE sin = 40 40 2/2-5 FAE = 50 KN 6 FAB 2 FAE = 0 1.5 40 KN g=53.13 E For = Sin - 2 2-5 COS - 1.5 2.5 FAE Cos 0 + FAB -0 3 5 FAB 50 x 3 FAB 30 KN ii) Joint B FBE FAB FBC Fy =o FBE 20 kn B E Fx =0 20 kN FBC 2 FAB FBC = - 30 kn

iii) Joint E FDE E Efyo е 0 50 FAE FCE FBE FAE Sin 0 + FBE + Fee sine 50x ? + 20+ 2 2.5 FCE 2-5 FCE = 75kN I Fx = 0 FDE + Fce cos = FAE Cosa FDE = FAE COSO - FCe cose 50*2/3 = 75* 3 / 5 FDE = 75 kn

a undt load Now we have to remove all loads and place in downward direction at A. Them solue the bress. E 4 A B I unit FAE ;) Joint Fy=0 PAE SIND = 1 fAE LAB 2/25 FAE 1.25 E Fx = 0 PAR + LAE Cose do FAB fag - 0-75 -1.25 x 3

17 Joint B PRE O Efy = fee =0 FAR B FBC FAB = fee fBc -0.75 in Joint E E FDE e e IFy zo FAE Sin e + fee + fee shimo FAE FBE E fce 1.25 x 2 0 + flex 2 =O 2-5 02.5 fCE -1.25 Σκ το fce coso + for fAE cos @ = FDE - 1-25 x 2 2.5 -1.250? 2.5 ADE 2

As per Virtual Work Method, Equation for de flection in buss is " F fi Li Δε Σ A; E , where F ial Membor Force clue to actual load fi - Member force due to unit load 11 Length Ai Area of cross section Here 2 А = 400 mm = 200 G Pa = 200: kN/mm2 AE - 80000 KN

Member Length (min) Fi (KN) fi FIL AE AB 1500 -30 -0-75 0.421875 Bc 1500 - 30 -0.75 0.421875 AE 2500 50 1.25 1.953125 BE 2000 20 0 2500 -75 -1-25 2.9296875 DE 75 1500 2 2.8125 Total - 8.54 Vertical Displaerent at Fifi Li AE Vertical Displacement, 4 = 8.54 mm Downwards at 4

Now we have to find the horizontal de flection at A. We have do lemone all loads and place a unot horizontal load at E A as gruens below, D B Consider ; 1 Joint A APAE Fy=0 fAE line =0 A Fag FAE 20 E Fx = 0 FAB

11 Joint B IBE ΣFs SBE B FAB foc 2 Fx = 0 far = fgc fec 111) Joint E E toe ΣF : fee Sino+ FAE Sino+ fre fie g AE=0 "fee fresc I Fx=0 & fcE cose LAE cose 11 IDE fore o

© Member Length (mm) Li Fi fi Fi fi Li AE AB 1500 -30 -0.5625 Вс 1500 - 30 -0.5625 AE 2500 50 0 BE 2000 20 0 CE 2500 -75 DE 1500 75 0 Total = - 1.125 Horizontal Displacement of Joint = 1.125mm towards right A

The given truss is solved using method of joints and the Displacement of joint A is found using Virtual Work Method.