2001, W. E. Haisler
Chapter 13: Beam Bending
38
Beam Bending (Chapter 13)
Shear and Moment Diagrams/Equations
The internal forces and moments must be in equilibrium on a
“free-body diagram.” Consider a segment x (free-body)
from a beam such as:
p y(x)
y
x
x
2001, W. E. Haisler
Chapter 13: Beam Bending
39
p y(x)
M z(x)
M z(x+x)
P(x)
P(x+x)
y
x
Vy (x+x)
Vy(x)
x
x
x+x
Sum forces in x direction (COLM): P( x  x)  P( x)  0 .
P
0.
Divide by x to obtain (as x  0):
x
2001, W. E. Haisler
Chapter 13: Beam Bending
40
Sum forces in y direction (COLM):
0   Fy Vy ( x  x)  Vy ( x)  
x x
x
p y dx
Divide by x to obtain
Vy ( x  x)  V y ( x)
x

1 x x
x x

p y dx  0
Take limit for x. The integral term becomes 1x p y x  p y
(since p y is a constant for x  0 ). Hence:
 Vy
  py
x
2001, W. E. Haisler
Chapter 13: Beam Bending
41
Sum moments about the center of the differential element (COAM):
0   M M z ( x  x)  M z ( x)  V y ( x  x)(x / 2)
 V y ( x)(x / 2)  
x x
x x / 2
xp y dx  
x x / 2
x
xp y dx
Divide by x and take limit. Note that p y is a constant as
x  0 and hence the integral terms cancel. Thus
 Mz
 V y
x
Note:
V y ( x)
zero
constant
linear in x
M z ( x)
constant
linear in x
quadratic in x
2001, W. E. Haisler
Chapter 13: Beam Bending
42
 Vy
 2M z
You can combine last two equations to get:

 py
2
x
x
The differential equations for shear and moment can be
integrated. Consider the moment equation and integrate
from some point x0 to x1. We assume that Vy ( x) has already
been determined.
x
1 dM
z
x
0

x
1V dx
x y
0
 
or
x
1V dx
x y
0
M z ( x1)  M z ( x0 )  
Note: Last equation says that Bending Moment is related to
the “area under the shear diagram.” If you know the
bending moment at xo , then you can obtain the bending
2001, W. E. Haisler
Chapter 13: Beam Bending
43
moment at x1 by integrating the shear diagram from xo to x1.
The area under the shear diagram from xo to x1 represents
the change in moment from xo to x1.
Shear and Moment Diagrams are important because:
 they provide information on internal shear forces and
bending moments due to the applied loads,
 the internal shear and bending moment can be related to
internal axial stress  xx and shear stress  xy ,
 they identify the location of maximum internal shear and
moment (and stress), and
 they will be used to determine the transverse deflection,
u y ( x) .
2001, W. E. Haisler
Chapter 13: Beam Bending
44
Shear and Moment Diagrams can be obtained in two
ways:
1. Using the differential equations of equilibrium
 Vy
 Vy
 Mz
 2M z
 V y 

 py
  py ,
x
x
x
 x2
x
V y ( x1)  V y ( x0 )  1 p y dx
x
0
x
M z ( x1)  M z ( x0 )  1V y dx
x
0


2. Drawing a free-body diagram and using equations of
equilibrium ( Forces = 0,  Moments = 0).
2001, W. E. Haisler
Chapter 13: Beam Bending
45
Example 1. Cantilever beam with end moment, m
m
+
L
x
m
x
cut beam and find
internal forces and
moment
Sum forces in x and y direction to
obtain P(x)=0 and Vy ( x)  0. Sum
moments at any point x to obtain
M z ( x)  m.
P( x)
M z ( x)
m
V y ( x)
x
x’=L-x
Mz
M z ( x)  m
x
L
2001, W. E. Haisler
Chapter 13: Beam Bending
46
Example 2. Cantilever beam with end load, F.
F
+
L
x
F
P( x)
x
cut beam and find
internal forces and
moment
M z ( x)
F
V y ( x)
x
x’=L-x
Sum forces in y direction to find Vy ( x)  F . Sum moments
at any point x to obtain 0   M M z ( x)  F ( L  x) or
M z ( x)  F ( L  x) .
2001, W. E. Haisler
Chapter 13: Beam Bending
47
The internal shear and bending moment diagrams are:
Vy
F
Mz
V y ( x)  F
FL
M z ( x)  F ( L  x)
x
x
L
L
2001, W. E. Haisler
Chapter 13: Beam Bending
48
Example 3. Cantilever beam with uniform distributed load,
po
p0
+
L
x
p0
M z ( x)
p0
P( x)
x
cut beam and find
internal forces and
moment
x
V y ( x)
x’=L-x
2001, W. E. Haisler
Chapter 13: Beam Bending
49
Sum forces in y direction to obtain Vy ( x)  p0 ( L  x) .
Sum moments at any point x to obtain
0   M  M z ( x)  [ p0 ( L  x)]( L  x) / 2
or
M z ( x)  p0 ( L  x)2 / 2
Internal Shear and bending moment diagrams are given by:
Vy
Mz
p0 L
2
p
L
V y ( x)  p0 ( L  x) 0 / 2
M z ( x)  p0 ( L  x) 2 / 2
x
x
L
L
2001, W. E. Haisler
Chapter 13: Beam Bending
p0
Example 3a. Cantilever beam with
uniform distributed load, po. Use the
integration method to obtain V&M
diagrams.
 Vy
  p y   p0
x
50
L
x
We know the shear at x=L is 0, so lets integrate from x=L to
any point x:
or
or
x
x
L dVy   L p0dx
V y ( x )  V y ( L)
0
x
  p0 L
  p0 ( x  L)
Vy ( x)  p0 ( L  x)
2001, W. E. Haisler
Chapter 13: Beam Bending
51
Next, integrate the shear equation to obtain the moment
equation.
 Mz
 V y   p0 ( L  x)
x
We know the moment is zero at the free end, x=L, so
integrate from x=L to any point x:
x
x
L dM z   L p0 ( L  x)dx
or
0
M z ( x)  M z ( L)   p0 ( Lx  x / 2)
2
x
L
  p0 ( Lx  x 2 / 2  L2 / 2)  p0 ( L  x) 2 / 2
or
M z ( x)  p0 ( L  x)2 / 2
2001, W. E. Haisler
Chapter 13: Beam Bending
52
Example 4. Cantilever beam with 2 shear forces
75 lbf
M z ( x)
100 lbf
100 lbf
P( x)
x
x
8 in
20 in
V y ( x)
20-x
20 in
Make a free-body by cutting at some point x to right of the
75 lbf load (sketch on right above). Sum forces and
moments about left end of free-body to obtain:
8"  x  20"
0   Fx  P( x)  P( x)  0
0   Fy  Vy ( x)  100  Vy ( x)  100 lbf
2001, W. E. Haisler
Chapter 13: Beam Bending
53
0   M  M z ( x)  100(20  x)
 M z ( x)  100(20  x)  2,000  100 x ft  lbf
Make a free-body by cutting at some point x to the left of the
75 lbf load. Sum forces and moments about left end of freebody to obtain:
M z ( x) 75 lbf
100 lbf
P( x)
V y ( x ) 8-x
x
20-x
20 in
0  x  8"
0   Fx  P( x)  P( x)  0
0   Fy  Vy ( x)  75  100  Vy ( x)  25 lbf
2001, W. E. Haisler
Chapter 13: Beam Bending
54
0   M  M z ( x)  75(8  x)  100(20  x)
 M z ( x)  1, 400  25 x ft  lbf
Now draw the internal shear and moment diagrams.
M z (ft-lbf)
Vy (lbf)
100
1,400
1,200
25
x
x
8”
20”
8”
20”
2001, W. E. Haisler
Chapter 13: Beam Bending
55
Example 5. Simply Supported Beam with Point Load
P
NO MOMENTS AT ENDS
WITH SIMPLE SUPPORTS
8”
20”
x
P
R1
M(x)
R2
8”
R2
V(x)
20”
x'=L- x
From equilibrium: R1 + R2 = P and R1(8) - R2(12) = 0. Thus
R1 = P (12/20) and R2 = P (8/20).
V(x)
M(x)
(8/20)P
P(8/20)(12)
L=20”
x
x
-(12/20)P
L
2001, W. E. Haisler
Chapter 13: Beam Bending
56
Example 6. Simply Supported Beam with Distributed Load
p
o
NO MOMENTS AT ENDS
WITH SIMPLE SUPPORTS
L
x
po
M(x)
R1
po
V(x)
R2
R2
x'=L-x
From equilibrium: R1 = R2 = poL/2.
V(x)
M(x)
M(x)=(po /2) x (x-L)
p L/2
o
x
x
L
L
2001, W. E. Haisler
Chapter 13: Beam Bending
57
Recitation Exercise #1:
For each case determine the shear and moment boundary
conditions.
75 lbf
100 lbf
100 lbf
x
x
8 in
8 in
20 in
20 in
75 lbf
75 lbf
100 in-lbf
x
x
8 in
8 in
20 in
75 lbf
6 in
20 in
100 in-lbf
2001, W. E. Haisler
Chapter 13: Beam Bending
58
50 lb/in
x
20 in
10 in
50 lb/in
1,000 lb
500 in-lb
x
10 in
10 in
10 in
50 lb/in
500 in-lb
x
20 in
10 in
2001, W. E. Haisler
Chapter 13: Beam Bending
59
Recitation Exercise #2:
50 lb/in
x
20 in
10 in
Determine the shear (V y ) and bending moment ( M z )
equations as a function of x. Draw the V & M diagrams.
Repeat for the following two cases:
2001, W. E. Haisler
Chapter 13: Beam Bending
60
50 lb/in
x
20 in
10 in
50 lb/in
x
10 in
10 in
10 in
2001, W. E. Haisler
Chapter 13: Beam Bending
61
Example 7. Simply Supported Beam of length L=10 ft, with
distributed normal load po = 50 lb/ft (up) and force P = 200
lb (down) at x=6 ft.
p =50 lb/ft
o
P=200 lb
a=6 ft
po =50 lb/ft
R1
P=200 lb
a=6 ft
R2
L=10 ft
x
L=10 ft
For equilibrium: R1 = 170 lb, R2 = 130 lb
Obtain the shear diagram by drawing free-body diagrams for
the segment to left and right of the force P:
2001, W. E. Haisler
Chapter 13: Beam Bending
62
0  x  6' . Free-body of beam to left of 200 lb force:
50lb / ft
x
M z ( x)
V y ( x)
+
P( x)
170 lb
0   Fy  Vy ( x)  170  50 x  Vy ( x)  170  50 x lb
0   M  M z ( x)  170lf ( x)  [50 lbft ( x)]( x / 2)
 M z ( x)  170 x  25 x 2
ft-lb
2001, W. E. Haisler
Chapter 13: Beam Bending
63
6  x  10'. Do a free-body to right of 200 lb force:
M z ( x) 50lb / ft
P( x)
x
V y ( x)
+
10-x
130 lb
0   Fy  V y ( x)  370  50 x
 V y ( x)  370  50 x (lbf )
0   M  M z ( x)  130(10  x)  50(10  x)[(10  x) / 2]

M z ( x)  1,200  370 x  25 x 2
ft-lb
Internal shear and moment diagrams are given by:
2001, W. E. Haisler
Vy (lbf)
170
Chapter 13: Beam Bending
V y ( x)  170  50 x
V y ( x )  370  50 x
70
10 ft
6 ft
-130
-130
M z (ft-lbf)
10 ft
6 ft
-120
-169
-289
x
x
M z ( x )  170 x  25 x 2
M z ( x)  1, 200  370 x  25 x 2
64
2001, W. E. Haisler
Chapter 13: Beam Bending
65
Alternately, we could determine the moment diagram using
differential equation of equilibrium
x
x
 Mz
1
 V y   dM z    1V y dx
x
x
x
0
0
x
1V dx
x y
0
M z ( x1)  M z ( x0 )  
We can determine the moment diagram by integrating the
shear diagram. Consider the left segment of the beam and
integrate from x=0 to any point x6 (note: M z (0)  0 ):
x
x
0
0
M z ( x)  M z (0)   Vy dx  0   (170  50 x)dx  25 x 2  170 x
The above equation is valid only from x=0 to 6. At x=6 ft, the
2001, W. E. Haisler
Chapter 13: Beam Bending
66
above moment equation gives M z (6 ft )  120 ft  lb .
We can now integrate the shear diagram for the right portion of
the beam (from x=6 to 10) knowing that M(6) = -120.
x
x
6
6
M z ( x)  M z (6)   V y dx  120   (370  50 x)dx
 25 x 2  370 x  1200
2001, W. E. Haisler
Chapter 13: Beam Bending
67
Determining Shear & Moment Diagrams Graphically
Consider the last example (example 7):
p =50 lb/ft
o
P=200 lb
a=6 ft
po =50 lb/ft
R 1=170 lb
P=200 lb
a=6 ft
R 2 =130 lb
L=10 ft
x
L=10 ft
We use the following relations:
 Vy
 Mz
 V y . Integrate each from x0 to x1
  p y and
x
x
x
1 p dx
y
x
0
V y ( x1)  V y ( x0 )  
x
1V dx
x y
0
and M z ( x1)  M z ( x0 )  
Hence the shear at any point is related to the area under the
distributed load ( p y ) diagram, and the moment is related to the area
2001, W. E. Haisler
Chapter 13: Beam Bending
68
under the shear diagram. Draw the distributed load diagram.

10'
Note: Vy (10')  130lb . Then, Vy (10')  Vy (6') 
p y dx or
6'
Vy (6')  Vy (10')  
10'
6'
p y dx . The area
py
under the p y curve is (50lb/ft)(10-6)ft=200
lb. Hence Vy (6')  130  200  70lb .
At x=6 ft, the shear drops by 200 lbs due to V y
170 lb
the point load. Hence, shear just to left of 6'
50 lb/ft
x
6 ft
10 ft

is V y (6 )  70  200  130lb . Area
under p y from 6 to 0 is 50(6-0)=300 lb. So
Vy (0)  130  300  170lb . Since p y is
constant, Vy ( x) is linear. Note that slope of
shear curve is equal to  p y  50lb / ft .
70
6 ft
-50 lb/ft
10 ft
x
-50 lb/ft
-130
-130
2001, W. E. Haisler
Chapter 13: Beam Bending
69
Another approach to obtaining the shear diagram: if we know the
shear is -130 lb at x=10, we can construct a line from that point
with a slope of -50 lb/ft. up to x=6. This gives a value of 70 for
shear at x=6. Due to -200 lb force at x=6, the shear drops to a
value of -130lb. From that point, construct a line with slope
equal to -50 until x=0. This gives a value of shear equal to 170
lb at x=0. Note from the shear diagram that the shear is zero at
x=3.4 ft [by similar triangles 170/x=130/(6-x) or x=3.4 ft].
Now do the moment diagram from the shear diagram. We know
from B.C. that the moment is zero at x=0 and 10 ft. We'll start
from the left this time where M z (0)  0 . We can
write M z (3.4')  M z (0) 
3.4'
0'
Vy dx . The area under the shear
curve from x=0 to 3.4 is 0.5(3.4')(+170 lb) = 289 ft-lb. Thus
2001, W. E. Haisler
Chapter 13: Beam Bending
70
M z (3.4')  289 ft  lb . Area under shear curve from 3.4' to
6' is 0.5(2.6')(-130 lb) = -169 ft-lb. So we have
M z (6')  M z (3.4')  
6'
Vy dx  289  (169)  120 ft  lb
3.4'
Since the shear diagram is linear is a linear function from x=0 to
6, we know the moment diagram is a quadratic. Similarly, for
the region from x=6 to 10, the shear is zero at x=7.4'. Area under
shear curve from x=6 to x=7.4' is 0.5(7.4'-6')(+70lb)=+49 ft-lb.
Thus
M z (7.4')  M z (6')  
7.4'
6'
Vy dx  120  (49)  169 ft  lb .
Similarly, area under shear curve from 7.4 to 10 is
0.5(10'-7.4')(-130 lb) = -169 ft-lb. Thus
M z (10')  M z (7.4')  
10'
Vy dx  169  (169)  0 ft  lb .
7.4'
2001, W. E. Haisler
Chapter 13: Beam Bending
The complete moment diagram is
shown on the next page.
Notice that at x=10', the shear is a
negative value. Hence from
Vy (lbf)
170
 Mz
 V y , the slope of the moment
x
curve at x=10 is a positive slope.
At x=0, the shear is positive, therefore
Mz
the slope of the moment curve is
negative. This is an easy way to check (ft-lbf)
your work.
Notice that drawing the diagrams in
this way does not give us equations
for Vy ( x) and M z ( x) . However the
equations are needed to determine
the deflection equation!
71
70
10 ft
6 ft
-130
-120
-169
-289
-130
10 ft
6 ft
x
x
2001, W. E. Haisler
Chapter 13: Beam Bending
72
Determining location of maximum moment
For example 7, f0r the left segment (x=0 to 6) we have
Vy ( x)  170  50 x
and
M z ( x)  25 x 2  170 x
The point of the maximum moment (for 0 to 6 ft) can be obtained
dM z
 0  50 x  170 . Solving for x,
from calculus, i.e., where
dx
we obtain x=170/50=3.4 ft. Hence, M z (3.4 ft )  289 ft  lb .
Note that from the differential equation relating moment and
 Mz
 V y , so a location of maximum moment is
shear,
x
where shear Vy  0. From the shear equation above for 0-6',
Vy ( x)  0 is zero at x=170/50=3.4 ft. Same result!!
2001, W. E. Haisler
Chapter 13: Beam Bending
73
Some books may use a different sign convention for Shear.
M z(x)
p y(x)
M z(x+x)
P(x)
P(x+x)
y
x
Vy(x)
Vy (x+x)
From equilibrium, the change in direction of the assumed positive
shear changes both the shear and moment equilibrium equations:
 Vy
 py
x
 Mz
 Vy
x
Be careful with this! The sign change in V will flip the shear and
moment diagram vertically. But also changes the sign in stress eq.!