Question

Part A Consider the truss shown in (Figure 1). Assume the members are pin connected at their end points. Take A = 2.6 in? and E = 29(103) ksi for each member. Use the method of virtual work to solve this problem. Determine the vertical displacement of joint A measured downward. Express your answer using three significant figures. Enter positive value if the displacement is downward and negative value if the displacement is upward. Figure < 1 of 1 V AC vec O O ? 1.00170 Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining 8 ft - 8 ft 500 lb 1000 lb

Answer 1

Solution - 82 € loost tomo=4 841862 or, O=tantly 509ь ol, o = 26.57 EB DC 1000lb AB - AC → EB= 8 > EB= 4 ft At first we will calculate the member forces using method of Joints. Initially we will assume that all the forces are in tension, therefore

the direction of those forces will be away from joints. Therefore; positive results stands for our assumption is okay and negative results stands for the opposite of our initial assumption (compression). Joint Ai Ar sine – – Farnost To FAB & 1000lb Ifyo 0 FAT sino - 1000 = 0 FAR = 1000 = 2235. 682 lb (t) sin26.57° Ifx=0; FAT COSO & FAB=0 os, FAB - - 2235-682 Cos 26-57° oe, FAB= -1999. 568 lb (c) Joint B: BE FAB {fg=0; FBE- 500 16 (T) {fx=0 FAB = FBC o, FBC - 1999. 568 lb (c) 50026

Joint E A for sine For For cost ) CO FOR COSO FRE & FBE AFE ? FCE Sine FAG sino I fx=0; - FAE Cost & FoE Cose & ECE cose=0 og - FAE +FOR+ FLE=0 or FDE TELE=FAE og FDE = 2235.682 - FCE – – (1) Efro=0; For sino - FBE - FCE sino - FAE sino - o or EE - FLE - FAE) sino - FBE of., FOE - FCE - 2235482 - 500 sina 6.57° or. 2235.682 - FLE - FCE - 2235.682 = 500 sim 26.570 09, 2Fer= 500 sin2657° on. FLE= – 558.920 lb (c) 2235.682 - (-558.920)=2794.c. so FDF = 2235.682 -

Since we know the bar forces bar areas – bor length modulers of elasticity can easily find out the deflection of each member Hook's Law. therefore we individual by wsing oo, S- PL AE AB=8ft = 8x12 in AE = 8742= 8.944 ft = 8.544x12 in. BE = 4ft - 4x12 in BC= 8 ft = 8+12 in. CE= 874,2= 8.944 ft = 8.944 x 12 in DE= 8+42= 8.944 ft = 8.944712 in. Is-PL Bar IPL Chips) in IA LEO I Cime Chipslin2) Emesling) I cm) AB -1.ggg 8x12 -0.00254 AF 12.236 8.044 x12 0:00318 0.5 4x12 2.6 Iran 29YID 0.000319 - 1.999 8x12 -0.00254 CE -0.559 18.944812 1-0-000796 DE 12.795 8.944x12. 0.00398

Now we will apply virtual load on joint A to calculate member forces due to that whit load. Vertical displacement Calculation. .. Da A soo- & 8ft t 8ft & Joint A: st FARSing A {fo=0 FAL sin 26.57 = 1 Fár cost or FAE inze. or, FAÉ= 2.236 hips CD) 1k P { x=0; FART Arcose - o r, FAB= - 2.236 cos 26.57 = -1.99g lips Joint B AFBE {fo=0; FBE = zero force member) Efn=0; FAB FBC on FBc = -1.9gg tips. B. FBC (c)

Joint Ea For Simo te cose For cose do Frecast & BE FAR FCE Fecsino Frosino Efx=0; – FAC Cost + the cose it for cose=0 or, FA & FE & Foro or, FÉ & For = FAE or, for = 2.236 - FCE {fy=0; For Sino-FBE - FC Esino - Ar sind so or, (For - FCE - FAE) sino = FRE om, 2.236-For-F6-2.236 0 or – 2FLE=0 on Fee = 0 (zero force member) . FOR = 2.236-0= 2.236 kips (T) Sim26.579

2 Now we can find the vertical displacement of joint A; AA by using this equation. -(1) - It's where t'a member forces due to mit load. Sa member deflections or displacements in the original ston moal structure there fore we can writer. (1) x AAN = FAB SAB & FAE SAE & F'BE SBE + F'BC SBC + FCE SCE+ FÓR SDE 091, A - [-1.999% (0.00254] + (2.2368 0.00318) + (ox 0.000318) +(-1.999x -0.00254) +(ox -0.000796) +( 2.2367 0.00398) or, Age=0.02616 in So, upto three significant figure; the value will be; AA ² 0.0262 in (positive sign indicales displacement is downwards ) (Answer)