Question

50 kN 40 kN/m Q1: For the overhanging beam shown below, draw the shear force and bending moment diagrams. Write the equations for the shear force and the bending diagrams as needed. For bonus points sketch the deformation shape (elastic shape of the beam under the given loads). Hint: A is pin support, and B is roller. 4 m 2 m

Answer 1

Solution y -Oy i 50 kn . 40 kN/m # 4m - am * I figure 1 © Calculation of end reactions → E fac = 0 THA zo Efy = 0 +YA-(40X4) + V2 -50=0 YA+YB = (4074) +50 VA + VB = 160 +50 YA & YB =210 KN IMA=O (YAKO) + (Haro) +(40X4x9_) -(VBX4) . +(50^6 ) = 0

+(40X4*2)+ (5076) = (V8x4) 3204 300 = YBX4 620 = VBX4 YB = 620 YB = 155 kN put value of VB in equation @ YA+YB = 210 VA+ 155 = 210 YA = 210 -155 YA: 55 kN] - @ calculation of sheave force → Ls for calculation of sheque force, · Considering a section between AB and Bc. • Taking either deft or right side of section and equating veutical forces.

(1) Taking AB . from figivee 1, taking side of the section, XX left XXIX - 40 kulm i 150KN Cc x 4m x Sfxe = t VA - 40700 Is fx = 55-4088 linear. - at a=0 ; at x = 4m SFA= 55-4000 SFB = 55-40x4) ISTA = 55 kN] SfB= 5.5.-160 SFB. -105 KN

(12) Taking BC → from figure it, taking right sicle of the section x'x! x'4x4 50KN Sfx = 50kN constant at x'=o Sfe=50kn at x'- 2 m ISI B = 50KN 40 kN/m SOKN k Ilm - am to constant linear Shear force diaquam +55 450 L +50 Z d750 dia

o calculation of Bending moment → ů Taking AB porom figurie į teleing left side of section x-x. 40kplm It At a L ltidlo x 4m x 1188 = + VA * 20 – (40 %*) ML=55% – Home Ma = 551-2002 parabolic at 0 = 0 . M = (55x0) – (0X02) . MA=0

at x = 4m 118=5584 - P0x42 = 220 - 320 MB = -100 kNm docation where bending moment become maximum, CM2 от 1С - о ax da (5532012) = 0 55 - 40n O 55 = 40x H€ 55 40 x= 1,375 m NE value of max. bending moment BMman = 55*1.375 – pox (-375) = 75-625 - 37.8125 B 97 mar = 37.8125 knºm

location at Bending moment become zero, 55- 2012-0 (55 – 20 m)x=0 [a=o 55-20 x 20 x = 2.75m

(1) Taking BC from figevel, taking Section 1x! left side of 50KN to k Band Kamat I Mx = -50xx lineam, at x=0 at X = 2m 19,= -5070 MB = -5042 M. O kpm 17132-100 km

30 KN 40 kulm - um *am- * - - 37.8125 kPm barabolic tosz baraberlic A e 1.3750 t 2.75m - f -liner - -00 iram Bending moment diagram,