Consider a simply supported beam of length L. The distributed load acting on the
ID: 1854171 • Letter: C
Question
Consider a simply supported beam of length L. The distributed load acting on the beam is nonuniform and given by the function w(x) (with units of force per unit length). The net force clue to the distributed load is given by Fw = L 0 w(x)dx and the moment due to the distributed load about the left end is given by Mw = L 0 xw(x)dx. Determine the quantities Fw and Mw for the following special cases: w(x) = w0, a constant value. w(x) = w0 (x/L), a triangular load. w(x) = w0 (x/L)2, a quadratically varying load. w(x) = w0 sin2pix/L, a sinusoidally varying load.Explanation / Answer
Fw=w(x)dx
Fw=w0 dx=w0(L-0)
=w0L
Fw=w0x/L dx
=w0(L^2-0)/2L
=w0L/2
Fw=w0(x/L)^2 dx
=w0(L^3-0)/3L^2
=woL/3
Fw=w0sin2x/L dx
=woL/2*(1-(-1))
=woL/
Mw=xw(x)dx
Mw=xwodx
=woL^2 /2
Mw=x^2w0/Ldx
=woL^2/3
Mw=xw0(x/L)^2dx
=woL^2/4
Mw=xw0sin2x/Ldx
=w0L^2/