Question

Use the direct stiffness method and calculate the forces, deflections and moments in each node.

Mass of UDL = 115 kg

E=200 Gpa

I= 4 x 10^-5 m^4

The downward arrow's force = 1 128,15 N

UDL wa O ह ह 14 Ri Rz

.

Answer 1

F=1128 VOL RI R2 F = Given up = 115bg, 1150N 19.17 N/m. By Equilibrium conditions. F M Efy co. R, & R₂ - 1150 & 11 28.15 RitR2 = 2278015 N EMA=0. M₂ + R2 x6 = 1128.15x3 + 115083 EMB=0 M, & R,x6 = 1128.15x34 1150 x 3 From the above m = M₂ and R = R2 as it is an equilbrium everywhere in the system R = R2 = 1 139. IN 1₂ = M = -0.15 NM &o, moment is clockwise sense

consider section Box V= 1139.1 19.12 x aloexes m(x) = 0.15 + 1139.1 x for section (Be) v= 1 139.1 - 19.178 1128.15 m.r) = 0.15 + 1139-1 30 - 1128-15 (3-3) [3exeo so, we have Node at B at x = 3. 0= dr FI S = de FI consider er os m s=sS 0-15 & 113901x dx EI 1.15102 + 1139-10 2x6 EI E 200x109 I = 4 x 10 0.1503) 1139.4(3) 8= [ + x 2 200x4x10" 2X6