Question

Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions in order to gain confidence in your plots. 2 kN 11 kN 5 kNm +3m +4m + 6m — +3m 8 KN Questions: When x = 2.1 m, V = KN.m When x = 5.7 m, V = KN.m When x = 8.3 m, V = KN.m When x = 13.7 m, V = KN-m
Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions in order to gain confidence in your plots.

Answer 1

Qut, RA Reaction at Support A Ro Reaction at support by Sagging will e positive A loging will be negah -ve ZZZZ 57925 G-25 165737 AM aka 6M on Haucing moment of all forces about support 'B'! and applying equilibrium II condition, le EMB=0 RAX16-2213-5+8x9 H G3W 165 - 11 X3 = 0 Clock wise ve an HClOceWZ -ve 1 1% fe46-5472 -33 = 0 k IGRAT G4772= 0 8h, 3M, M24 -0-5x3=-1.5lening RA =-9/16=-0.5kN Between Dandic 3 042 Ra = 0,5kA 041 | V = – R 2 = = 0,5-2 = -25 F| durchon gt will be constant from Dto e on applying equilibrium at, te-Siam, vs2=-251N4 Bending moment between and Condition in vertical directions Ma = RA2 -2(X2-3) Epyo, Rb+8-11-2-0.5=0 at, 2 = 5.7MMx = -05x57-2(575 T Ro= 5.5 KN7 - 25 wrt at = 7 MX-0-5x7-20-3) Take a section between A ando MX2 - - 11,5 KM at x, clistance from 'A' Between c and E, shear force Sheen force at this section MR-RA-216 = -05-1+8=5PN) Vy=-RA => N=-0.5 KNY This will be constant upto e OX3 = 83M, V = 5. 5 l atx 21M, V = -0.5kn! Bending moment My=- Racx.)-243-5 at, 26=8m. Nau=-0.5 KN +8 ( -) at, 3= 7M, T S + MX -0.5*4 -2 (1-3)-5 +0 Bending moment for suchen 0-0 MX -165KNM May = - RAX [Hogging -e] at, x 83M at, 34= May= 0 Mog -0.5x8-3-2 (83-3)-5 t. 2021M, MY-0.5x2! + (6:3-7) MS-1.65 KNMI MM -9.35 KM Between £ and B Shea Horce VX = -RA-2 +0-11 = -05-2+841 = -55KW vxy = -5.5 kN] It will be constant up to 'B'. and at Bending moment, nhu 24 = 13.7M V -5.5 km Mxq = - RAXX4 - 20-3) L + 8024-1)-5-11684-13) at, 24 = 13 M, MX4= -1.5x13-2013-3)+8 (13-7)-5-11 (13-13) M = -65- 20 + 43-5-0 = 16.5 KNM MxA = 16.5 KNM at, xq = 1817M, Max = -0.5X13:7-2013-7-3)+60137-7)-5-(13-7-13) Mx4 = .12.65 KNM ot, x4 = 16, My -6,5x16-2(16-3)+6016--5-(16-13) -- Mxy =o) Il Vosed 5.5 N M = 16:5 NM/