Question

For the beam shown, a) Draw the bending moment diagram, b) Determine the maximum normal stress due to bending. 300 N 400 N/m b B 200 N.m с Hinge 2.00 m 2.00 m 2.00 m 300 mm 750 mm x 292 mm 560 NA: Neutral Axls

Answer 1

Given beam: 300N 400N-m А с B DOON-m) Ro Rc RAL 2m 2m 6m Taking moment at point "n"; RoX6 + Rc = 400 x 2x [4 + + 200 + 300 x 2 X4 GR +4R, 4800 - 0 Considering i EV = 0 Equilibrium condition RAAR CARD 300 +(400 x2) m 1100 RA+RctRo Solving 0 x 0; We get ;Eq0 - Eq@ GRp +4 Rc -RA-RD - Rc 4800 מ По о 3700 –B +5Rp+ 3R.

Solving +0.③ equations; We get RA= 140 N Rc = 480N Ro - 480N Bending Moments :- At point o; Mo = 0 = 480 X2 -400 x2x3 => 160N-m. At point B; M3 = 480 x4 - 400 x2x (2+3) +480 x2 +200 MB 680 N-m At point A; MA=0 aBending moment diagram. At point c; Mc 2 = 300 400 Nm A B 2pory-m RD. - 480N RA - 40N Rq=480N 2m IMB =GBon-mos - Më 1600- tve parabolic Curve MAO Mo=0 'B

b] Maximum Nom al stress due to bendingi Vь A lexy Here; M = Max - Bending moment = 680 I-M Distance of the layer under consideration from neutral axis. y = 292 mm = 0.292 m. I. Moment of inertia

300mm I: I, + I + I3 From figure ; I- Is then; I-I, +2 I Ii - bd + A, CY,-5) 750mm 12 300mm 130mm In - Iz - bh? 560 mm - 5* + Ax[ys- ) Here; y = Any + A₂Y+A 3Y₃ A, HA at Az 300 X 75 0 A, = bxd 225 x103 mm? A2 = A3 - Ź bh : { X130 X750 => 48.750 X10 mm 750 : 375 mm 750 From 0 - 4,- From @-- 75 250mm theo; B = (225 xinºx 315 + 48 750 (250)+48 750C259) (225X103 + 48750 + 48750) 9 : 337,20 ) 3 then I, 300 (750) +225 12 2 X103 [375 -337.20] 10 I, - 1.0868 X10" mmt.

5 - Iz- 130x750 + 48750(250-337.2012 36 I, - Iz 1.894 x109 mma I =I, +2 Iz 10 = 1.0868 X10'° +2[1.894 X109] I = 1.4656 x 100 mmt 4 I- 0.014656 m then; Maximum bending stress :- Vь = М xy I = 680 X0.292 0.014656 Vo = 13548.03 N/m cor) V = 13.548 kN/m?