Question

Po = sokN %=80kN/m Problem 3. A 4 m long beam is subjected to a concentrated force Po = 50 kN, a concentrated couple, Mo = 100 kN, and a triangularly distributed load starting at qo = 80 kN/m as shown. х А B Mo = 100km (a) Find the shear, V(x) and moment M(x) as functions of x. (b) Construct accurate shear and moment diagrams. (c) Calculate the location Xmax of the maximum moment and its absolute magnitude Mmax. im im 1m T Im

Answer 1

i) Ay & By so ( x1 * 80 ) = 0 (1 E Fyso] By x 2)-(50x1)-+100 TEMA 0 => + - { *1x80) x6 x0)+ 37 50-100+ (4ox boy 3) By Ay=[50+40) – 41-67] 니 = 41.67 KM 2 48-33 KN scanned with CamScanner

Po=50KN 9.= 80 SokN/m. 9 ( 0 ) tel a 31 *-2 E А c Y'D Im MoflookNim ay=80 (1-2) 1 m. TIM X Ay=48.33 KN By = 1 41.67 m 4-20 ye am 2 For seen * gmi 1 © 148.33-V, 20 Ay = 48.33 @ 'M, -64 8.33 ** 1m 50 UN For sech X st ag 92 a 48-23 48-33-50 tra =>/V2 = 1.6fum (ü) M2 -(18633x) + 50(x-1) co * M2 = -1067x + 50 CS Scanned with CamScanner

For sec ☺ im im Ma Ilmaz 95 im 48c33 41.67 (2) 48.33-50+ 41.67 +z=0 ind. Vz=40 KN I downwards 1V3 = -40 KN Cu? (48-335) + M3 + 50x(x-1) 41.67(3-2) o Mz= 40 X – 83. 34 T 80 un/m Forsec @: 50 80 - 50 (4->) loo 80 (4-2) Koln Whh. Im im (4-2) ↑ Oma 1m 4 48.33 41.67 _46- 33 - 50+41.67 – **(4-2)* [80+ 80 (9-x)] -40+ (4-2)(40+40 (4-2)) => /va In Antent 14-3 2 ü) M4-(48.33 ) + 50(2-1) +100 - 41.67 (3-2) + [8064-2] *(4-x) * & *(**) *[*0- 80(9-x)] (40 x- +100) + (827) 7 Imp = [402 +16,654 [40 (4-1) +(4-1) [402 – 120] 3 27 Ma X-83.34 +100) + X +(4-2)[70- 4064-22 CSScanned with CamScanner

from, VA :) V=0)40+ (4->) (200- 40) 40 +[(4-») x 40(5-2)] -40 + 40 [ 20- 40 [ 20-43-5x+22] = - 40+ 6292 +20 40 40 [a²g x 120-17 V@= 40 CaZ9X+19) [oseed Amor From M4 3 M= t 40x+161666 +40 (4-») + (9-x)(40x=120) =4032 + 16166 +4064-2003 4-X)* 40(2-3) = 40x+16+66 +40 4^)+ 40x (4x–12-222733) = 404+16166 +40 (4-2)2 + 40C-x772-12) = 40 [(4-2)*+ *-*77*-12] +16.66 • [[4_~)*— x?78X-12] [(4® – 3:47. X+3.4.2?-«?)-2°78x42] 40 -12 +16+66 - 40 & 16.66 [osasa =40[ 64 - 484 +122*-x2x+8*412] +16.66 40x476] +16.66 MA) 40 [x² + 11x? CS Scanned with CamScanner Ah

for, M max =) dM da 0 dm dy 40 [- 3. 6% +11.20 – 40] © -) 307 ai 7 2231-40 322_22at 40 22 622)=6477 38 40) . ma 2x3 - 4 /3.33, . - 274 to as, at free end, Moment is sero il.at, n=4, Mso them gine mayn moment, then, it should be at x=3.33 m. to give may N. 11 M max 40 [-(3:23) +11 02, 3372 40X43.33)+ 6] 76 f16.66 @ amax=3.33m. c 1130,73 kn, m, Amri CSScanned with CamScanner

Im In F A c 2 D Aura 48.33 48.33 el Torbole 167 T SED(UN) 16 eb 40 parabolic 1130.43 1130136 Copbi EERD 46.46 (64) 36.66 (t) BMD (unim) 3:34 48.13 X 13.33 carried with Camsdanner