Use discontinuity functions to find a general expression for

Use discontinuity functions to find a general expression for the deflection curve for the beam shown. Use your result to evaluate the deflection at (a) the right end of the beam (b) mid-way between the two supports

Solution

solution:

here we consider point load wL is acting at -L/2 and UDL is acting between L to 3L/2,hence load equation using delta and doublet is s follows

P(x)=-wl<x+L/2>^-1-w<x-L>^0+w<x-3L/2>^0

on integrating once we get shear force equation,again integrating we get moment equation,again twice integrating we get deflection of beam as follows

V(x)=wL<x+L/2>^0+w<x-L>^1-w<x-3L/2>^1-c1

M(x)=-wL<x+L/2>^1-w/2<x-L>^2+w<x-3L/2>^2+c1x-c2

y=(1/EI)(-wL/6<x+L/2>^3-w/24<x-L>^4+w/24<x-3L/2>^4+c1x^3/6-c2^2x^2/2+c3x+c4)

where boundary conditionare

x=0,y=0,v=0

x=L,y=0,v=0

x=-L/2,M=0

x=3L/2,M=0

so we get

c1=3wL/2

c2=wl^2/8

c3=.3802wL^3

c4=-wL^4/127.83

so equation becomes

y=(1/EI)(-wL/6<x+L/2>^3-w/24<x-L>^4+w/24<x-3L/2>^4+wLx^3/4-wL^2x^2/16+.3802wL^3x-wL^4/127.83)

here deflection at

x=3L/2

y=53.08wL^4/EI

at x=L/2

y=.070wL^4/EI