1. Problem 7.159
w = w0 cos px 2L y
For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment. A x B L

2. Solving Problems on Your Own
w = w0 cos px 2L y
For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment. A B x L
For beams supporting a distributed load expressed as a function w(x), the shear V can be obtained by integrating the function -w(x) , and the moment M can be obtained by
integrating V (x).

3. ( ) ( ) ( ) ò ò dV dx = -w = -w0 cos V = - wdx = - w0 sin + C1 dMy
w = w0 cos px 2L
Problem Solution
The shear V can be obtained by integrating the function -w(x)
and the moment M can be obtained by integrating V (x). A x B x L dV dx px 2L
= -w = -w0 cos
( ) 2L p px 2L
V = - wdx = - w sin C1 ò dM dx
( ) 2L p px 2L
= V = - w sin C1
( ) 2L p 2 px 2L
M = Vdx = w cos C1x + C2 ò

4. ( ) ( ) V = - w0 sin + C1 2 M = w0 cos + C1x + C2 w = w0 cosy
w = w0 cos px 2L
Problem Solution
The shear V can be obtained by integrating the function -w(x)
and the moment M can be obtained by integrating V (x). A x B x L
V = - w sin C1
( ) 2L p px
M = w cos C1x + C2
( ) 2L p 2 px
Boundary conditions
At x = 0 : V = C1 = C1 = 0
At x = 0 : M = w0 (2L/p)2 cos (0) + C2 = 0
C2 = -w0 (2L/p)2

5. ( ) ( ) ( ) ( ) ( ) V = - w0 sin 2 2 M = w0 cos - w0 2y
w = w0 cos px 2L
Problem Solution
The shear V can be obtained by integrating the function -w(x)
and the moment M can be obtained by integrating V (x). A B x x L
V = - w sin
( ) 2L p px
( ) 2L p 2 px 2L
( ) 2L p 2
M = w cos w0
( ) 2L p 2 px 2L
M = w ( -1 + cos )
( ) 2L p 2
Mmax = w [-1 + 0]
Mmax at x = L: 4 p2
Mmax = w0 L2