Question

a) Draw the bending moment diagram,
b) Determine the maximum normal stress due to bending.

300 N 400 N/m D B 200 N.m с Hinge 2.00 m 2.00 m 2.00 m 300 mm N А. 750 mm x = 292 mm 560 mm

Answer 1

a SFD & B MD 300N YooNIM MA I 0 2m 2m 2 m 200N.m B С. A 300N 400 Nm 2400 Nm 2 m A em 200 Nm 2m 2 FBD YOON 400N Reactione & Fe-0, 1A=0 Ź Mc=0; (for CD) Dy x2-400x2x120 (By = 400/N (1) Ź Fy=0, Ay+by-30 -300-Yox2 O Ay = 1100-400 Ay = 700 N (1) EMATO) 400X6-400x2x5-27-3rox2 -MAro [MA = -2400 Nm (3) Vi V2 4 SED a=im SEN parabolic curve MS My V = 700 N N2 = 1900-300= 400 N VS-410-410 x 2 = -400 N zero shear force location /V3! (M3 BMD M 2 400 = 2-2 no 2-X 1x=1 Im from 'cl (Hinge)

BOMD (Area of SFD) M = -2400 N.m M2=-2400 +V, x2 = -2400 + 700x21000 Nim M3=-1000+200-800 Nom My = -800 + V2x2 -800+ 400x2 0 My = = 0 + V2 xx = you x 200 Nm M6=200 + V3 (2-2) = 200- 400 x (2-1) ²0 Maxemum nosmal stress 3oomm 8t = 756-292 = 458 mm y b=292 mm x 8+ 750 wiu Iman = 458 mm NA Yb 292 mm moment of vnestia (I) K K •130mm 5bomm using parall axis theosam I= IGt Ad² 1 = 2 [tar 130(252) + 5x Box750/212- 4x70)] + [300(752) 3 + 3x 780 x ( 575-292)2] I2 (1523437500+85995000) + (10546875000+15500 15000) I= 15315765000 mm

mon Mman & mans I G Mman = 2400 N·m = 24x16 Nimm Hogging moment man = y += 458mm (Top) 1. Oman - 24x105x458 N/mm 15315765000 max = 0.0717692 mpa At fixed suppost Toman = 71.769/kpa (Tensile) - At top of beam.