Question

Use the graphical method to construct the shear-force and bending-moment diagrams for the beam shown. Label all significant points on each diagram and identify the maximum moments along with their respective locations. Additionally: (a) Determine V and M in the beam at a point located 1.50 m to the right of B. (b) Determine Vand M in the beam at a point located 1.25 m to the left of D. Leta - 3.0m, b = 6.1 m,w = 38 kN/m, P-55 kN. Q - 34 kN, and MC-152 kNm. Construct the shear-force and bending- moment diagrams on paper and use the results to answer the questions in the subsequent parts of this GO exercise. Р Q 4 B смс D E a b . a a Part 1 Calculate the reaction forces By and D y acting on the beam. Positive values for the reactions are indicated by the directions of the red arrows shown on the free body diagram below. (Note: Since Bx0, it has been omitted from the free-body diagram.)

Answer 1

* Given ba 6.) m, a = 30 p=55kn W=38 kN/m 9:34KN A A с. E ->X B Mc =152kNm * X a=3m bobim X X T 41=3 q=3 since 'B' is at hinge B a hinge point, so is Zero. net moment EMB -0 a 55x3 + RoX 9.1 = 34 X12.1 + 38 x (6.1+3)? +152 2 ญ) Ro 216.68 KN 8 RB - 55+34 +38X 9.1 - 216.68 RB 218.12 KN AS we know that, S.F, (ut of ) VB==-55, VB+ = -55+ 218.12 VB t = 163-12 Gm portion BD CS Scanned with CamScanner

So Shear force diagram (in KN) 163.12 KN 34 C B lo 0 55 68.68 182.68 KN and we know that, sm=gu da OM-V = from= fudx an 275.04 parabolic parabolic 123.04 B E D с 102 165 (in kom) csl

1.5 m to right of B. @ shear force, u at Fe Swee) de v V = 163.12 • 38 8 1.5 106:12 KN M MB + Svenda Ź ( 106-12 +163.12) *1-5 -165 t М. 36.93 kNm Shear force, u 1.25 m to left of D. VE fo + Swin) da 182.68 + 38 X1.25. ☺ - 135:18 KN M -102 + 0.51 135-18+182.68)x1.25 = 96.66 kNm a at x= 3.0m v =-55 KN at a = 3.0 + m from graph V = 163.12 KN at x = 12.1- m v =-182.68 KW d at x = 12:1 tm v - 34 KN CS

a > >Bending moment at x = 3.0m Cat Support B) m =-165 KNM D 9.1 - m. Cat Just left of a =123.04 KNM 9.1 tm. ( just right of e) = 275.04 km d X 21 ) C x M M =-102 KNM Zoro since Shear force is 163.12 - 38 x(x-3) = 0 Sif = = a = 7:29 m As we zero all know that if Sif és B.m. (m) will be maximum. 223-165 + 0.5X163-12)x 4.29 185.10 KNM m In whole beam ชชาaxi ชาน ชร) positive moment = 275.04 KNM at 7 = 9 © So maximuon negative bending moment, M-165 kNm at location, n = 3m CS