Question

USE METHOD OF JOINTS OR SECTIONS

Problem 4: Find the Reactions at all the pins. The Distributed load at the top will instead be a uniform (rectangular) distributed load of 5kN/m. The load at point C will be 7kN. m C D 1.5 m 1.5 m

Answer 1

FBD of AB xS53.75kN NSLS Ay A AZ FBC FAD 3 m ア+ EMA=0 Point A At 3.75×2 ナFBLX3:o 8.7S X) 1 2 S -3. 75 KN FBC ED C FBD Of 3.75ドN FAD Ey 53.13 C Er polnt ア+ EmE: o At

FADY Sin(53.13)15 t 3.7sx 3 t1X3 o - 32.25 FAD sin 53,13)X FAD 26 87 KN - 26.87 KN FAD 3.75-70 FADX SIn(S3.13) Ey -26.27xsin (S3.13)- 3.75 - EY O74 Ey t N -1074 0.7u N Ey FAD x COS ( 53.13B) :o E FAD COS 53. 13 ) EX

26.81x coS(S3.13 ) EN 16.12 FBD 0f AB FADX SinS3.13) - 3.75 -3.75 Ay t FB C (53 13) 26-8 X SL.s-SL O -3.1S KN 25 2S Ay FAD y COSS3 3) AX o - A 2687x COs ( S3. 13) 16. 12 EN

16.12とN AX 2 S 2S KN AV 16.12 KN EX2 0.74 KN Ey