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 D B 200 N.m с Hinge 2.00 m 2.00 m 2.00 m 300 mm N A 750 mm x = 292 mm 560 mm NA: Neutral Axis

Answer 1

Part a Free Bely Diegren is 300 N 200 m 400 N/m A MA AD am & amt am RA RD Efy co RA + Rp = 300+ 400X2 RAIRD = 1100 N ΣΜ το at C to the sight; 2 Ro 400x2x1 Ro=400N . 700 N RA ale to the left; 4RA + 200- 30012 - ES MA MA =-600+200+ 2800 => MA = 2400 Nm

Now we can find the Bending Moment Equation for each segment. o consider AB 2400 Nim M CA b) M- 7002 – 2400 2 tv 700 N ii) Consider BC 300N M 2400 km t M = 700x+ 200 – 2400 – 30064-2) 200NM 700N M = 4001 - 1600 X iii) consider CD Mtx- X 400N/m 2 M = 4008 4008 e possono 2 400N M2-200x'+ + 4008

Bending Moment Diagram 200 Nm A B c D -800 Ning * -1000NM -2400 Nm

Bending chess, on what Mr Bending Moment 1. y - Distarme from Neached ana's I - Menyent of Inertia. So we have to find the Moment of Inertia of the Cross section. h=750 mm * 560mm n3 (a + hab+ 62) Moment of Inentia of trapezium Is 36 (a+b) 750' (30 I = 300*+ 4x 300x550 5603) 36x (300+ 560) I 1.4657 xiolo mom a

Stress Distribution is, 458 mm NA 29LM location of maxinoum Maximum Bending Sbess occurs al Bending Moment.d M Mere 2400 Nm - 2400 X 1000 X 458 oman =0.075 N/mm2 1.4657 81010 Therefore, Marimum Normal Shress 20.075 N/mm2 der to bending 75 kN/m²

Bending moment diagram is drawn

Maximum normal stress due to bending = 0.075 N/mm²