A timber beam, with rectangular cross section (h Times b) is reinforced with add

Question

A timber beam, with rectangular cross section (h Times b) is reinforced with additional full width (b) steel plates. There is a plate of thickness t securely connected on the bottom and a plate of thickness 2t on the top of the timber to ensure composite action in bending. The section is then used as a simply supported beam of length L. The beam is oriented such that the minor principal axis is vertical. It should be taken that rho steel = 7850 kg / m3, rho timber = 1000 kg / m3, Etimber = 11000 MPa and Esteel = 200000 MPa, and that both materials exhibit linear elastic behaviour. The design engineer needs to ensure that the normal bending stress (tension or compression] does not exceed 8 MPa (for the timber) or 250 MPa (for the steel). The beam experiences a vertically downwards point load of magnitude P acting mimdspan. It can be assumed that the self-weight of the steel-timber beam is negligible compared to the value of P. Before the steel plates are connected (ie based on the timber beam only) what is the maximum value of P that can be applied before the timber reaches its maximum stress. For the value of P calculated in (a), what is the maximum deflection of the beam. The steel plates are then added. What new value of P will induce the critical bending stress (8 MPa (for the timber) or 250 MPa (for the steel))? Draw the stress and strain distributions (values required) for both the steel and the timber on the critical cross-section when this occurs. h (mm) b (mm) t (mm) L (m) 250 210 7 3.1

Explanation / Answer

a)

v=0.250*0.210*3.1=0.1627 m^3
m=pV=0.1627*1000=162.7 Kg
W=mg=162.7*9.81=1596.08 N;
since the weight is uniformly distributed along the entire length, the uniformly distributed load, w, is then w= 1596.08/3.1=514.86 N/m
Max moment=w*L^2/8=514.86*3.1^2/8=618.4755Nm=618475.5 Nmm
So M=618475.5 Nmm, y=250/2=125 mm, Ix=210*250^3/12=273.4E6;
therefore, f=(618475.5 *150)/273.4E6=0.33932 MPa

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