Question

#2) (35p.) Calculate the slopes "OB” and “Oc” in the frame shown in figure using “Slope-Deflection Equations”. Do not solve the system. Do not draw any diagram. (E: Constant). (“A” is the fixed type of support and “C” is the pin). 20 kN/m 20 kN/m D (I) B (1) 2m 30 KN (I) מר | 2m 5m X X X

Answer 1

20 kN/m D С tam B am 30KN an А K M minimo CS Scanned with CamScanner

YOKN 40KNM 20kn/m G с ta sm 2m minimi am mm A CS Scanned with CamScanner

Step-1 Finding fixed and moments MFAB = -wl 8 - 30 x(u) => -15kNm 8 MEBA WJ 15 kNm MEBC = -20(672 a = - 41.67 Krom 12 12 MFOB = +41.67 kNim CS Scanned with CamScanner

Step-a slope deflection equation MARS MFAB & QER (204 +88 + 35 ;) MAB- -15 + Ef [06] QA=0 [since fired 2 MBA = 154 EI QB 1:1) MBC 11 -41-67 + AVS 2 E IL ( 208 +04] [2B(+ OB iv) MCB 41.67 + ZEI Step-3 Joint equilibrium equations E MB 40 ii) EMCE - 0 CS Scanned with CamScanner

j) 19.BA + Moc -40 =0 15+ EI Of + (-41.67) + RED PROB+ oc] -40 5 EI OBT EDOC + 4 ET06 - 66167 5 EL OB + Z EIⓇC = 66.67 0 EMC =0 MeB=0 U EI OC + 2 EI OB = - 41.67 2 5 Solving equation 0 8 0 OB = 544.690 EI Oca 79.433 E indicates Anti Clockwise direction ЕТ Thank you. please give an upvote. Stay safe. CS Scanned with CamScanner