Question

C2 Done session.masteringengine AA <HW10 - Attempt 1 Analysis of Frames - No Sidesway < 4 of 7 > The frame ABCD is subject to the distributed W B es Je, H L load w = 5.2 kN/m as shown. The dimensions are H = 4 m and L = 20 m. EI is constant in each span. The supports do not move. Part A - Write the slope-deflection equations All three spans are fixed supported end spans, SO

Answer 1

Part-A 3-5-? fined end moments Mãe Ar: 0 H: 4 MESA=0 ME BC = -5.2x20? 1? DAFO Mfoc-173-33 MF C B = 173.33 2-२०m ME co= mfocco Slope deflection Sovations IMAGE O + 381 (20A +8s) MAG= € I OB 2 in Maa- 392 (A +206) MBA = EIOS → (i) cli) MBLE -173-33 + ZEL (204+280) Moc = = -173.33 + § ET OBTIBET OC si (iv) Mcg= 173.33 + 2 EI 20 (B3 3+20) Mg: 173.357 , E10s + ar 0. Lor five

(v) M co- Ot ? EI 201 202+0) McD = EIO را ) (vi) Mocs 0+ O + RET REF (Oct (oc +200) Μοι- και Ετθς (vi) Past-B We know, ² MB = MBA + MBL=0 0 - EI OB-173-33 + ŠEL OB* to EI Oe. EL OR + to EI Oc = + 173.33 (vii) 3) Me= McBt McD = 0 = -173-33+ to EI Os+ Í EI OC + EI Oc. SET OL it to EI OB = -H73-33 -Sini) Solve Gii) & (Vid OB=157.573 Ocz-157. 573 EL EL clockerise *B=0CP 157.573x109 200x103x R-5 x108 9 -0.003151 rad c = -0.003151 rad to counterclockwir

Part-c Substitute OB & Oc in Soucation (is MAB= 32.78.785 IN-M 1) MBA = 157.87 kn-m 157.57 59.01 Rx 4 + (78785+15757) رام 89.08 R=-236.355 4 R=-59.08 KNI > MB2B-78-785 moment at mid spant of AB M= 78-73-59.08*4 x 4 M=-39.397 KN-m dos

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