Question

Determine the reactions at A and B for the beam subjected to the distributed and concentrated loads. w w = wo + kx2 4.0 KN 5.6 kN/m 1.6 kN/m TI 1.6 kN/m T 0.4 m -----*- x 0.6 m 1.3 m 0.4 m 0.6 m Answers: KN 13.13 Br = 7113.18 KN |-7.56 KN

Answer 1

3+ W = Works ² 4KN 0.4 m KNU Solne first of all we have to calculate total area da 5.6kN total load of distributed so load given w swot kx² AT 163m 0.49 By Gm which is equal to area of the distribution above the span. mR dA= Wola

given distribution of load W=Wotka² At x = 0, w = 1.6 ; 1.6= Wot ko)2 Wo=1.6 KN. At x=1.9, W = 5.6' 5.6 = endo+k(1.9)? K = 1.108 1.9 P= Loact = Total area - Jora - Swolx = 5(wo+ ka) olx I Wola - = (Wox + Kyun) = wox1.9 + K (19° = 16x1.9 + lyg (1033 P = load = 5.573 KNI And the distributed load through centroid of the Now calculating centroid load. can be assumed acting area. of the area of distributed lxda x (Wotka) dx Worx

X = (ew. z oka do 5.573 u Jo 6.5 = - 1.166 m. 1.166 m. 5.573 Now considering equilibrium and taking moment about B RAX 17 - Px1-134-440.4=0 Bx-B RA = 4.658 KN. . Ay = 4.658 KNI 1.134 By 1.7m Now equating forces in verticle direction: RA + By - p =0 By= P - RA= 5.573 - 4.658 By = 0.914 KN Now equating forces in horizontal alirection. Bar4 =0 B x = 4 kn ]