80 mm P=40kN w-20kN/m 15 mm 120 mm 12 3m 6m 15 mm x3 Figure1 (for problem 2) Fig

Question

80 mm P=40kN w-20kN/m 15 mm 120 mm 12 3m 6m 15 mm x3 Figure1 (for problem 2) Figure 2 (for problem 3) Consider the 3 m long simply supported beam whose cross section is shown in Figure 2The beam is made from an isotropic linearly elastic ideally plastic, material with modulus of elasticity E- 200GPa and yield stress in uniaxial tension 250 MPa. The beam is subjected to a concentrated force of P at the middle point of its span. (35 points) a. Compute the elastic failure force PY and the plastic collapse force P b. Establish the distribution of the normal component of stress acting on the cross section of the 3, beam when P = (p" + P") / 2. Draw the stress diagram.

Explanation / Answer

let us determine the elastic moment capapcity and plastic moment capapcity of the section

elastic moment capapcity:

moment of inertia of section about neutral axis = 2*{80*153/12+[80*15*(60-7.5)2]}+15*903/12=7571250 mm4

elastic section modulus of section = 7571250/60=126187.5 mm3

elastic moment capacity of section = 250*126187.5*10-6 = 31.55 kNm

plastic section modulus of section = 2*[(80*15*52.5)+(15*45*22.5)]=156375 mm3

plastic moment capapcity of section = 250*156375*10-6=39.09 kNm

a)Elastic failure force PY = 4*31.55/3=42.07 kN

plastic collapse force = PP = 4*39.09/3=52.12 kN

b)When load =(PY+PP)/2=(42.07+52.12)/2=47.1 kN, the stress distribution will be such that some part of the section would have completely yielded whereas the other part would still be exibiting linear variation in normal stress.

moment in beam = 47.1*3/4=35.32 kNm

amount of load exceeding the elastic failure force=47.1-42.07=5.03 kips

The problem can be solved by trial and error:

Let us determine the moment the section would be resisting if only the flanges have completely yielded:

moment resisted = 2*[(80*15*250*52.5)+(0.5*15*45*250*2*45/3)]*10-6=36.56kNm

Since this moment is less than kNm, the stress distriution is such that the flanges have partially yielded and the remianing flange and the entire web is within linear stress distribution

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