A uniformly distributed load of density \'w\' over a span \'a\' at one end of a

Question

A uniformly distributed load of density 'w' over a span 'a' at one end of a simply supported beam of length 'L' generates the following deflection 'y' as a function of 'x', where 'x' measures the distance from the end of the beam where the load, where 'E' and 'I' are constants depending on the characteristics of the geometry of the cross section of the beam and the material the beam is made of. What is the maximum value of the slope of the beam? Where does this maximum occur??

A uniformly distributed load of density 'w' over a span 'a' at one end of a simply supported beam of length 'L' generates the following deflection 'y' as a function of 'x', where 'x' measures the distance from the end of the beam where the load, where 'E' and 'I' are constants depending on the characteristics of the geometry of the cross section of the beam and the material the beam is made of. What is the maximum value of the slope of the beam? Where does this maximum occur??

Explanation / Answer

For maximum deflection,

d^2y/dx^2=0 and d^3y/dx^2<0

dy/dx=-w/(24EIL)*(4Lx^3-6a*(2L-a)x^2+a^2(2L-a)^2) 0<x<a

=-wa^2/(24EIL)*(-(-2x^2+4Lx-a^2)+(L-x)*(-4x+4L))=-wa^2/(24EIL)*(2x^2+4Lx-a^2+4L^2-8Lx+4x^2)

=-wa^2/(24EIL)*(6x^2-4Lx+4L^2) a<x<L

d^2y/dx^2=-w/(24EIL)*(12Lx^2-12a*(2L-a)x) 0<x<a

=-wa^2/(24EIL)*(12x-4L) a<x<L

d^y/dx^2=0 gives

x=0,a*(2L-a)/L, L/3

d^3y/dx^3=-w/(24EIL)*(24Lx-12a*(2L-a)) 0<x<a

=-wa^2/(2EIL)

x=a*(2L-a)/L, L/3 are where maxima occurs.

Now substitute these values of x in the equation for dy/dx to find the larger one.

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