Question

X w long beam 1 subjected to a site concentrated force 1.450 kn, a concentrated couple, Mo- 100 kN, and a triangularly distributed load starting at q = 80 kN/m as shown. a) Find the shear, Von and moment Mas as fumetons of 6) Construct ecorate shear and moreel dagstany, calulate the location Year of the maximum manent and its absolute magnitude b) 9 Mmas q= 80 kn/m basoka "Mo= looks in 1 1m elm ,

Answer 1

go Po Po=SO KN 1 м Mo 100 KN.m Tw Мо AX IM Boo im A 90 = 80 kN/m By Reactons SFn=0, An=of EMAO, - gexix/3 + 3x1) + Mo + Byx 2 - Pox1=0 -50x1=0 - 80 xx 1 +100+2By ( By = 125 KN (1) 2 Fy=0; Ay +By-Po - 90 x1=0 Ay + 125 - 50- 89x1=0 Ay = 145 KN (1) a sheas & moment equations for Regon ocakim & Min) 1 KN Vn= 145 Viw 145 kn Ma = 1458 KN.m 3 A4 >= im M = 145 x 48.333 KNM

for Region Imen<2m 50KN Vx= 145 - 50 Wa-- KN My > I 2 Vn 145 kn Ma = 1452 - 50(2-) Mn = 50-54 At x=2m; M = 50 - 5x2 = 4148 m = +46.667 K Nom for Region 2m<2<im SON My V2 = 14s - 50+ 125 im 11 Vn 145 KN 1V=40 KN INSKN Mn = 1451 - 50(x-1) + 125 (x-2) Mn = SA + 12s4 - 200 56- Mn = 40x-100 Al x=2m; M = 40x2-19 = 46.667k Nur At no3m, M-40x3- - 100 = 86.66760 Nm (Iust left)

for Regnon 3m (alum go 9 * * x-3 Mo -Mu Im im "x-3 44- tex(4-0) 125 3 qu=80(4-2) Va = 145-50 + 128 - 2xx (x-3) - 1x [ 9. 2x} x (X-3) 40- 90(4-2)(x-3) - (x-2)[7-90(4-»)] Va-40- 90 (4-6) (2-3) - 90 (2-3) (21-3) Vn- 40- 20 (1-3) 5-2) - 40-80 (5x-x215+34) 2 [Vn = 40(2²8x+16)| KN At x = 3m; v = 40 (9-24+16) - YOUN At n.4m; - 40 (16-32+16) = 0 Puth 3 Mr = 1451 - 502 X-1) + 125 (2-2) -M.- 90 + 2 x 2 *x (4-3) 9n+290, (n-3) Mn = 40x- 100 100 - (1-312 [vo (4-2) + 296] -400 80 (x²79-64) (6-2) Mar 400-

Mn = 40 (x² 12 8² +484-64) | At 2-3m) M= 4 (27-1249 +48 X3-64) M = -13.333 KN (Just right) Ata = ym; M = 40 (64-12 x 16 +48x4-64) = 0 O SFD & BMP SOKNim 50KN G A 12 man = 3m To im Im IM Thoknu 48.333 KN 41.667KN 48.333KN HOGN YOKN Mman = 86.667 KNM (Absolute) parabolec X (2° cerore) 1267 KN SED 8 6.667 Kalm 48.833kN.M 46.66 km ibid BMD (3 carne) -13.333 KNim