Question

Draw the shear and moment diagrams. 2.1 kN/m A C B - 5 m - 2 m

Answer 1

Fr
Whe! CD 2.1 kN/m A .C sm As 2ml FB RA Efy=0 RA+RB – **7 2.1 RATRB = 7:35. KN 0 ग EMA=0 IX 7 X 2.1 x RB X5 =0 2 [R8 = 6.86 KN from o RA=0.49 KN >Shear force + (i) portion AB sfre= RA (osassm) I x x x Woc wse = 2.1 7 2 2 Wa=0a3x Sfx= 0.49-1 xx x 0.32 2 Sfx=0.ug-0.1522

SFB SFA = 0.4g. On 15 (0)2 = 0.49 kn 0.49-0.15 (5)2 = -3.26 kN cil portion BCA (5<x< 7m) SFxa - 1 x x x W x + RB s fra 0.49 -! xx x0.3 + 6.86 RA 2 Sfea 7.35 0.15 x² SFB = 7035-0015(52) 3.6kN SF ca 7.35- 0.15 (73) OKN *Bending moment + c) portion AB e COSU35m Me= RAR I 2 xxx шx x 3 3 0.05x3 = 0.494-1/2 x R x 0.3** 2 = 0.49% - 0.05 MAD 0.4910) -0.05 (03) = 0 Mga 0.49 (5) - 0.05(53) = -3.8 kNm KN-m for BM max, dmx da 2 0.49 - 3x0.05% X x=10807 m

BMmap, = 0.49(1.807) 0.49(1.807) -0.05 (1.80733 0:59 KN-m 1 3 + 34.3 (ii) portions BC + (56257 m) Mx = RAx- 1 x x x W x 2 + RB (X-5) = 0.49a-0.95 x 60862 7.354-00543 - 34.3 MBP 7035(5)- 0.95153) -34-3 -3.8 kNm Me=7.35 (7)-0.05(73) - 34.3 = 0 kN-m 3.6kN 0.49 KN + 2 cuore x 2 cuore Shear Force Diagram a -3.26kN 0.59 kNm 3 cuore sending moment diaz.com А x=1.807m Les curne • 3.8 kNm