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The thickness of a rectangular steel girder equals h, find the deflection lambda caused by the weight of the girder in two cases: (a) one end of the girder is embedded into a wall with the length of the protruding section being equal to l (figure) (b) the girder of length 2l rests freely on two supports (figure). |
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Answer» Solution :(a) In this case `N(x)=1/2rhogbh(l-x)^2` where b=width of the girder. Also `I=bh^3//12`. Then, `(Ebh^2)/(12)(d^2y)/(dx^2)=(rhogbh)/(2)(l^2-2lx+x^2)`. Integrating, `(dy)/(dx)=(6rhog)/(Eh^2)(l^2x-lx^2+(x^3)/(3))` using `(dy)/(dx)=0` for `x=0`. Again integrating `y=(6rhog)/(Eh^2)((l^2x^2)/(2)-(lx^3)/(3)+(x^4)/(12))` Thus `lambda=(6rhogl^4)/(Eh^2)(1/2-1/3+(1)/(12))` `=(6rhogl^4)/(Eh^2)(3)/(12)=(3rhogl^4)/(2Eh^2)` (b) As before, `EI(d^2y)/(dx^2)=N(x)` where `N(x)` is the bending moment due to section `PB`. This bending moment is clearly `N=underset(x)overset(2l)intwdxi(xi-x)-wl(2l-x)` `=w(2l^2-2xl+x^2/2)-wl(2l-x)=w(x^2/2-xl)` (Here `w=rhogbh` is weight of the beam PER unit length) Now integrating, `EI(dy)/(dx)=w(x^3/6-(x^2l)/(2))+c_0` or since `(dy)/(dx)=0` for `x=l`, `c_0=wl^3//3` Integrating again, `EIy=w((x^4)/(24)-(x^3l)/(6))+(wl^3x)/(3)+c_1` As `y=0` for `x=0`, `c_1=0`. From this we find `lambda=y(x=l)=(5wl^4)/(24)//EI=(5rhogl^4)/(2Eh^2)` 
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