Question

30 kN/m [100 kN B A 4= -30mm 9 m 3 m 3 m

draw the shear and moment diagram for the beam

Answer 1

ANSWER

O This structure is cºin determinade, and so would be difficult to selve using the force Method. Aedes Adc, are fined End so do not have any degree of freedom. Node 8' Cannot move horizontaly Since et is restrained by members. As and A. which are both fixed horizontally, node 8. is also restraint from moving vertically due to the Yonies support of the localion overall this structure has only one of which Means is good structure to analyse using the slope deflection method. at rodes. & for the sole rotation the cauilibrium condition will be MBA+MBc zo z 은 08 LAS somm PAR Z Focomm ФАе -0.00333 Yad. 48c A2 LBC 30 6000 48c 2 Y8C 2 +0.oo so you

@ Tort 87 cum 3m amy 100KN 43.30MM element chord 8 200 m . WAB You om 6 m MAB 2 25 (20n+04-30)+feunt 9€! (20A+OB-3948)+ REMAB T (90.4 3(-0.00333/18) १ 951 (081 0.0100 178/ MAB 21 A+CB- MAB 9

на А 5 147 (00+ 0.0100)+a1 241 (20970A-3928) + FBHBA 21 ( 208 + 08-3(0.00333) )- 21.5 MBA 4 Again Node Ais Pined SODA 20 M%CE Ива 4 2013 (268+00-348417 PEMBC RE? (208 +0<-3(0.00500)+45 PCI (208-0.01500)776 6 2 LEI (20(+6B-3YBCIT FEMCB :: HRC McG MCG <DEL (2017 TOR - 30.00 500)-75 6 PEI (708 +0.0100)- 121:57 24 4268-0.01500) +75 20 30 kwim 100 Fixeden moment B min.-SOMM amy 37 1.

[ ub wy 30 PL PL UNIE 2-125 tm isken USt-2 7:30 A B 6. ou my S7.5 nn09 166.9 160-9 (17) 09 119 C EI 0.00278 (67)+46.5 = ( R00600MPa) (12.9 //Ommy) 1000000 N/mm) (12.7 xdoºmasy 17 E Z 7 Suookwmv

11 08(EI) 0.00278/77 1746.5 02 0.00278 1.111 + 46.5 THIEI 0.00278 u 6.5 PIN Nili(suco :: OBZ+0.8041syad HBA 2-18. Anm MBC 718.3 kNm MLB e--1669 KNN