Question

2.4 kN/m Problem No. 1 (25%): 2 kN/m y = kx 3.0 KN 300 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load (b) the reactions at the beam supports (Fig. 1). B A 0.6 m 1.2 m - 0.8 m 2.4 m —- - 10 ml Fig. 1

Answer 1

ANSWER:

Given that

According to given information, 24 INIM yalet 2 kN/m kayo 3.0 KN K30 ITAB (teby] se l te AQ I aon la 2.4 M babx] 1018mf Tom' | - 1.2m set reactions at A&B are Rad & by , f be forca): Y y=kx² 2.4m - parabolic load (y) = kx^ - Centroid from origin, >> X = a- an+) here n=2; a=214 (n+2) 5) a = sou-l pe 23) 2 qull-] 2 Qruly a 2 0r6 m. >) b=214 kaum - Area of this chave = ablnti - (2.4) (2«u))/(2+1) z 1192 kn

NOW, 1 (A2,72) 8.4kn O CABjte) (Alto) Ilid 1.2M ! 0.6M ya 2.0m - Area (AI) 1.8 (2) = 3.6 kW centroid (T) = 118/2 2019m ) Area (A2) = 1/2 (1:87(014) = 0136 KN centroid (r = 2/2 (1118) =112m Area (A3) = 1.98 kn Centroid () - 06+118 = 2.4m. -) combine center of gravity (centroid): - AIR + A212 + Agra AltA 2 +A3 = 3.6(019) + (0.36) (112) + 1.92(2.4) - 8.6 +0,36 +4192 aban - 1.4081M (from hett end) - Resultant Load values; > 3.6 +0136 +1.92 >) 5.88 KN.

tor(6): reactions at the beam supports. -> resultont diagram, 5.88 kW gsingo" Brf Roy ) slosti Gron (ra 225m 49 2) Rbx = 3605(30°) $363 fbx = 21598 kal - su take moment about B. e Ra (4.8) -(5-85*2401.) - 3*$*120 b) Ra = 95/16 25.9375 kW =) Ra +R b = 5.88 +3/ = 7158 EN 2) Rb = 7,38-5-9675 = 1.4uar en