Question

+ Question 2 20 MAR Using virtual work theory. dolermine the reaction to C. A well Figure 2. El 137500 km 10121%-10:56 10 5 O Please assist with structural analysis – Expert Q&A Figure 2 rollers at A and sig

Answer 1

10kN/m 3 ET-137500 4m lom 5m A E Fig! on 2 because no Horizontal force foame, thats why {Ex=0 is ignoned Solution - Ds = re-2 Ds=3-2=1 @IDs=17 Assuming vertical reaction After removing redundant at as redundant. Mest of stoucture is determinade 10kN/m C с B D B D Rc 4m lom Tom ST + Re=2.5 km losky E А TRA= 245kN Fig-2 WRA= 3.5km TIKN Fig. 3 RE Finding EMC=0, support reacth in fig-2 frame. FRA*) + (10*14) IRA 245KN {Fyzo, RA + Rc - (fox 1 ) Rc = los KN

ה Vertical Finding support reacth in fig.3 foame. EMC =0, (- | * 14) +(RA *4) = 0 - TRA= 3.5kn Efyzo, Rc- RA+1=0 Re=2.5KN Using unit load method- Postion AB BC BD DE origin A с D Limit 0-5 0-4 0-10 0-5 -(59e%+105x). (iony 2-5 (0) неве, M = B.M. at any section due to given loading, fig. 2 m = B.M at any section when vertical unit load is applied at Efig 3 NOTE Since there is. onyielding and non-movable support at E Hence net at support E will be Zero AEy =o. → (compatibility condition ) o O o DELF Mm da da EI DEL s moment ET DEL = 0+ [(5x2 +105x)=9.5x) dx *] (193%Ex)) + €5*2.5237 (10542 522)] da 625 ** - 59529]+ (-57"] 1 5-sedie

EIDEL - 800 5600 - 12500 DEL =-18900 (1) ЕТ Vertical displacement of support Edue to vertical reaction de AER= Vertical displacement of E due to vesticeluoittood *PE AER EI DER o S m2 de tre og mismo tie + jcxdu 10) a Re imf jaas ** do of visual (2695077)*. * (T):)" (133-33 + 333.33 ) RE DER= 466-66 RE (1) ET from compatibility condition DEV=0 DER+ AEL=0 466-66 RE ЕТ 18900 EI RE 340.50 KN 1

Now, from fig-1. ΣΜ. το, (Re *14) - [Ra*9) + (10x14°) - 0 640-50814) -(4+4) + ( 10 ) >> A = 03:2s ky {Fy=0, Ret RA+RE-(10x14) Rc + 103.25 + 40.50 140-0 [Re= -3.75KN