P V M Cantilever Beam Triangular Distributed Load

The cantilever beam has a linearly varying distributed loading, w = w(x).
A. Find the shearing force and bending moment as a function of distance along the beam: V = V(x) and M =
M(x).
B. Draw the shear force and bending moment diagrams.

Solution
First draw the free body diagram and solve for the reaction force and moment. wo L
 Fvertical  2  VA 0 wo L
VA 
2
 w o L  L 
M


M


  0
 A
A
 2  3  w o L2
MA 
6

To find the shear force and bending moment at any point along the wing, cut the beam at an arbitrary point x. Since we have cut the beam, the distributed loading is now in the shape of a trapezoid.

w(x)  w o L2
6
Find w(x) using similar triangles: wo w(x)
L

L-x

A w oL
2

w o (L  x)
L

M xx 33

V x 2

x

 1  w o x 
 
 x 
 2  L 

w(x)  w o L2
6

A w oL
2

w o (L  x)
L

M x 3

V x 2

x

x
3

Summing vertical forces to solve for Shear Force at any point on the beam

w oL  x  w o x  w o (L  x) 
 

x  V
2
L
2 L

 w wL
V  o x 2  w o x  o
2L
2

 FV 

Summing moments about the free end to solve for
Bending Moment at any point on the beam

ΣM cut

end

 w 0 L2 w 0 L
1  w x   2x   w  L  x    x 
 x    M 0

 x   0   x     0

6
2
2 L   3  
L
 2

w 0 L2 w 0 L w w w M

x  0 x3  0 x 2  0 x3
6
2
3L
2
2L
w 0 3 w 0 2 w 0L w 0 L2
M  x  x  x 6L
2
2
6

Notice also:

dM
V
dx