Question

2. Determine the vertical displacement of joint D. Use the method of virtual work. AE is constant. Assume the members are pin connected at their ends. 0 0 ooo 0 0 Oo 3 m ВГ –4 m t 15 KN 20 KN

Answer 1

we can find the vertical displacement at joint D by virtual work method

for all members cross sectional area and E are same

we know the virtual work poinciple Ix4= Ś no Ni Li i detrection 4 - " ni Nili Ter AE meroberi due to unit where we need to find load at deflection Ni - Force in the location N; – Force in opembes due to applyed loads.

Step 1

Calculate all member force Ni in truss due to applyed load by method of joints.

{fj=0 -> YA+= 15 +20 - VATY = 35 KN Efs=0 => n =0 B VISKN 120kn EMD2=0 sum of moment about E =o). YAx8 = 15x4 VA- 7.5 kn =) VE= 27.5 kn

7.5kN FAB Take joind A tan ocean => SIDO COON = Ś Apo {fy =O - FADSIDO 17.500 Fro = 7.5 = 12.5kw (caspression EF=o = FABTFAO COSID =0 FAB =-(-12.5% FAB = 10 kn (ension) Take Joint B AF BD FBs = 15kN FBC = FAB FRO = 10KN = 10KN E tec E B BC FAB diskn Tare Joint D FBD Efy o -) feptFAD SIDO +FCD Sino-0 Fcd= - FBO FADSIDO - Sino = -15 +12.5X375) 315 FCD = -12.5 km Compression

EFx=0 => IDE + Eco coso = FADcoso FDE <fod Coso - Foo coso = -12.5X -(-12.5% =D IN Tare Joint E a DE I "27.5 Efyzo =) FCE=27.5 kn Ucce 12.5kn 15kn \ 12.5kn 27.5kn 10KN VoB VISKN X 20KN

step 2

apply a unit load at D and find all member force (ni)

EFxzo - HE=0 EFg zo = VATVE al EMERO =) Vx8 = 1x4 Yaya => VB=Y2 A

783 Take Joint A {Fy=0 = FADSIDO + Y=0 FAD = -1/2/SIDO - 12/315 = 56 Stoc zo = Face + Fad coBoco FAB = - FAD caso = 3 x 4 = % Taxe Joint B . TAB Ty A FED FBD =0 FBC = FAB = 3% B AB * GE foot ode Take Joint a {Ex=0 a) FB C + Fco cosazo FCD =-FBc/coso =623)| = 5/6 Efy =0 =) FLE+ FOD SIDO =0 FCE = -FCOSMO - 757375 = 1/2 Take Joine E EDE=0 - FLE=Y₂ difce

O EN Length of AD = DC = 5.42732 =5w, so we can make table. mecobeo length Di Lim I Force are to applied load I force aneto I Unit load Di Na L; AD -12.5 -576 52.083 AB 10 2/3 26.667 BD 15 o BC Alwlat Allau 10 26,667 2/3 -516 CD -12.5 52.083 41.250 27.5 1/2 CE. DE EDIN; L; = 198.750

EDIN; L; = 198.750 AE is constant for all member Vertical de fiection at D 4- D;N; L? AE vertical deflection = at D 198.75 AE _ no,

here the value of AE is not given so the numerical value of displacement = 198.75/AE

If we take AreaA in mm2 and E in kN/mm2 the this expression will give displacement at D in Meter.

the direction of displacement is in the direction of applied unit load.(downwards) - since the value is positive.

All the necessary calculations are given above.