This document solves for the reaction forces and moments at the supports of a beam with a distributed load w(x). The beam has roller support at x=L and fixed support at x=0. It finds the equations for w(x), the shear S(x), and moment M(x). It then uses the boundary conditions and equations to solve for the specific reaction forces S(0), S(L), M(0), and M(L).

1. :problemH.W (2)
find out a relation for M(x), S(x) and w(x)
and find the reaction forces and moments at each support
Let w(x) = . + . + . +
The boundary conditions are:
at X=L >> it is roller support //// so, w(L) = and EIw''(L)=0
at X=0 >> it is fixed support //// so, w(0) = 0 and w'(0) = 0
w(x) equation, we have that w'(x)= . +2 . +
w''(x) = . +2 and w'''(x) =
B.C's:From
w'(0) =0 ////// . +2 . + =0 >> =0
w(0) =0///// . + . + . + =0 >> =0
EIw''(L)=0/////w''(L)=0 >> . +2 =0 >> = - .
w(L) = ////// . + . = …..
. +- . . = ////// = and =
w(x) = +
M(x) = EIw''(x) = . +2 = EI( + )
S(x) = = M'(x) = ///// constant
The reaction forces and moments at X=0:
S(0) = and M(0) =

2. The reaction forces and moments at X=L:
S(L) = and M(L) =EI( + ) = 0
We get::
S(0) =
S(L) =
M(0) =
M(L) =