Question

Problem 1 A simply supported beam of length L = 5m is subjected to a point load P= 20 kN at the mid span. Draw the shear force and bending moment diagram for the beam. If the beam is 300mm x 500mm, calculate the deflection at the midspan for the following orientations where the dashed line shows the bending axis. Explain the difference in results. Which orientation is better for beam performance and why? Take E = 30,000 MPa 300 mm 500 mm - 500 mm 300 mm Note: Deflection 8 = PL3/48EI Ix = bh3/12 ly = hb3/12

Answer 1

• Draw the Shear Force (SED) Sol wiren data. Simply supported beam. • Length =L=5m 20K N =P Load at mid span (P) = 20KN + 2. 5m the Shear Force & Bendinamoment A4 2.5m3 c d k L=5m Diagram (BMD): HAT - Support reactions use equations of equidibrium: Ev=o Section shape & stze VATVB = 200 Taking moment about Et=0 point Al HA=0 1 → VRX5=20 X 215 By equation 0 - Vp = 10 KN (1) tsooms & VA= 20-10 > VA = 10KN EM=0 300mm mm mm 120KN 2.5m ie 2.5m TORN sign convention & Left to right Clockwise ) " tres Left Right (104 - SED) • Bending moment calculation a (10) 20KN A4 2.5ml MORN 25KN.m (1 2.5m C log RM D ALL) lokly Mac = 1oxx Xx(4/2) Maxm. at point ci Me=25KNM at xco, MA=0 KNM at x=40 (M = 10x215 (X=255 m. = 25 KN.my (-25), at x=L=5m MX Flux- 20x(x-25) at a=2= mool

• Calculate the deflection for the at Midspan given osientation : - a 300 mm Section 300 x 500 a San mm ly=250 Ty=150 300 mm um mm we know & siren, p=20kN. bending axis. L orientation orientation. h = 30 mm - T - brez 12 300 x 500 » Ix - Se = geflection at midspan sc = PL? I assume п, 1м = 103 мм T R =constant = 30x103 m Pq = 30 x 103 x 1062 - 48 EI E=com 30x103 X 166 N (103) 2 noma None Calculate E=30kN/mm2] I For given orientation • For orientation : ommh 500mm Tromm • For orientation For ② b = 500mm Ix = 6h Ix = bxh3 - Ex = 500 X 3003 12 ( [x1 = 3125 x 106 mm Ix2 = 1125x166 mm 4 since P, L&E is same for both (5), = 20 KIV X(5x103) 3 mm3 1 (orientations 48 X 304 %om 2 * 3125x106mma (Sc), = 20x (5x103)3 se = 0.556 mm 49x30 X 11 25 X 106 o which orientation is better. Alwhy) 2 = 1.543 mm a oriention ② is better, because of C'we can see (scu) < (Sc,) · 6 section Modulus of beam orientation is more than that of (2) orientition 2 (73) *= 3125X105, 12:5X10*mme (2x), = I x = 1125X186 – 7 Tmax 150 = 75X106 y = G =/ Hence Zx > Zx2 Section modules is more then gives strength of beame Strength of beam is more. ex) Tmax 250 2 SO - mm3