Question

Question 3 Draw the AFD, SFD and BMD for the frame shown below. 10 kNm 20 KN

Answer 1

무튼 10 2OkN 10 kN/m moment about F. + EME=o, YAX10 – 10*3*(+7) – 20%2 –104550 7) v = 34,5 KN forces Summation of all vertical it I fj = 0, 34.5+ Ve = 1043 +20 5 15,5 KM

Summation of all horizontal forces at Fx=0, H = 10 kN. Analyze the members of the Isolating the members and frame separately finding AF, SF, BM.
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☆ To LOKN Member AB Axial Force - locarlon Ankar we have to balance T3m the other han end. I 34.5 kn 405 kn essing IFy 20 und 2 fx=0. Axial force = 10 un (compressive) ty tension Lover r o lown of ARD for Ab. E-comp.. Shear force At A, SPA = 34.5 kn. At B, SF = 3415-104 3-4.5 = 0 kwa - 3405 ner. . SED for AB.

Bending moment AT A, BMA=0 At B, BM= 34.5 % -103* = 58.5 km .. 58. Som z BMD for AB. Member Be 1.4.5kn it lour, 58.5 unm 15m GF 10 KN 415 UN 4.5 en (compressive). 58.5 resim B. Af = 4.5kn 4.5UN (AFD) och

sf at B = -10 kn Sf at c = +10+10=okr. A 3 (SFO)BC loten BM at B = 58.5 urm BM at ca 58.5-10x5 28.5 krm 8.5km 21 (BMD) Best X 58.5 kNm Member ce . 4.5und 8.5 kom. cm D 2 10KN 10 UN 18.5 urm doc 4.5kN Axial 10KN 20kN 15.5 kN. force = 10 un (compressive) 10 kn Jioun 4. (AFP)

SF at c = 4.5 kn St Just left of D = 4.5 UN sf just night of D=4.5-20 = -15.5 un Sf at E= -15.5415.5=0 kara 405 KN € (SAD) CA 15.5kn

BM at c = 8.5 kerm BM at D = 8.5 44.545=gl korm BM at E= 8.5 44.547 -20x2 = 0 krim sl urm go5 + (Bano Dec

Member Ef 15.5 不Ek1buy 15.5kn Axial Force = 15.5 kr (compressives 1515 kn (AFD) EF 18.5kn sf at Ero Sf at fao I (SED) EF BM at E=0 BM at F=o (BMD)