2001, W. E. Haisler
Chapter 13: Beam Bending
108
Chapter 13 – Beam Bending - Continued
Shear Stress in Beam Bending
 For pure bending, we considered only the axial stress  xx
under the assumption that all shear stresses were zero.
 The conservation of linear momentum equation reduced to
 xx
 0 which gives a solution of  xx  constant .
x
 However, when transverse loads p y or transverse shear
forces are applied to the structure, the moment distribution
is no longer a constant. Hence the axial stress  xx is now a
function of x. Because the axial stress is not constant, a
shear stress  xy must exist on the cross-section in order to
satisfy global equilibrium.
2001, W. E. Haisler
Chapter 13: Beam Bending
109
 We can show this by constructing a free-body of the beam
at any point x by passing a horizontal cutting a plane at a
distance y=h and two vertical cutting planes at point x and
x+x, respectively, as shown below.
y
We consider a beam of arbitrary cross section as
shown on the right. Note that the y-z axis is
located at the centroid of the cross-section (i.e.,
you must find this location first). The thickness
at any point y is given by t(y). Note also, that
loads (i.e., distributed or point load) will be in y
direction. Moments are about the z-axis.
Cutting through the beam with a horizontal
cutting plane gives:
z
end view
t(y)
c
2001, W. E. Haisler
Chapter 13: Beam Bending
110
x-axis at centroid of
cross-section
z
end view
z
x
y=h
y
 xy ( y)
t(y)
 xx ( x)
y
c
c
 xx ( x  x)
 xy
x
x
h
t(y)
dy
dA = t dy
x+x
y
Note that the figure is drawn with the y-axis pointing
downwards for visualization purposes only and that no sign
conventions have been changed. The cross section may be
2001, W. E. Haisler
Chapter 13: Beam Bending
111
any shape but the x-axis is at the centroid of the crosssection. Apply conservation of linear momentum in the x
direction to obtain the following equilibrium equation:
0   Fx   
x x
x
c
 yx ( x, y  h)t ( y )dx
c
   xx ( x  x, y )t ( y )dy    xx ( x, y )t ( y )dy
h
h
Note that the bending stress and width of the beam (t) are
functions of y. Divide the above equation by x. Note from
calculus that
1 x x
 yx ( x, y  h)t ( y )dx   yx ( x, h)t ( y )
lim

x o x x
2001, W. E. Haisler
Chapter 13: Beam Bending
Hence, we have
 yx ( x, h)t ( y )  
112
c  xx ( x  x, y )   xx ( x, y )
t ( y )dy
x
h
Take the limit as x0 for the stress term to obtain
lim
x o
 xx ( x  x, y )   xx ( x, y )
x
d xx ( x, y )

dx
The shear stress equation becomes
 yx ( x, h)t ( y )  
c d xx
h
dx
t ( y )dy
2001, W. E. Haisler
Chapter 13: Beam Bending
The axial stress is given by  xx
Mzy

. Hence
I zz
 Mzy
d

I
d xx
y dM z


zz


dx
dx
I zz dx
dM z
 V y . Thus
Bending moment is related to shear by
dx
d xx
y
  (V y )
dx
I zz
With the above result, the shear stress equation becomes
113
2001, W. E. Haisler
Chapter 13: Beam Bending
 yx ( x, h)t ( y )  
c
y
h
I zz
114
V y t ( y ) dy
The shear and moment of inertia terms may be taken outside
the integral since they are functions of x only. Hence, the
last result may be written
 yx ( x, h)t ( y ) 
V y ( x)
I zz
c
 h t ( y) y dy
Dividing by the width t(y) gives
 yx ( x, h) 
V y ( x)
c
t ( y ) ydy

I zz t ( y ) h
2001, W. E. Haisler
Chapter 13: Beam Bending
115
The integral term is a geometrical property (like moment of
inertia) so that the last result may be written
 yx ( x, h) 
V y ( x)
I zzt ( y )
Q ( h)
where Q is called the first moment of the area (area from h
to c) and is given by
c
Q(h)   y t ( y ) dy
h
Note that the integral could be written as Q(h) 
c
 h y dA and
we are integrating over the area between y=h and y=c.
2001, W. E. Haisler
Chapter 13: Beam Bending
116
 The shear stress equation provides the magnitude of the
shear stress at any distance y=h from the centroidal axis.
 Since Q is a function of position (y=h), the shear stress
varies over the cross-section (from top to bottom).
 Note that at y=c (top or bottom surface of the beam), Q=0
and hence the shear stress is zero at the top and bottom
locations of the cross-section.
 For a rectangular cross section, we will show in a later
example that the shear stress varies quadraticly over the
cross section and is a maximum at the centroid of the
cross-section (y=0).
2001, W. E. Haisler
Chapter 13: Beam Bending
117
Determining shear stress at a cross-section: In order to
determine the shear stress at any point of a cross-section, we
must know the shear, V y , moment of inertia, I zz , thickness, t,
and the value of Q at the y coordinate (really, y  h ) where
we want to determine the shear stress  xy .
c
c
h
h
The value of Q as given by Q(h)   y t ( y ) dy   y dA is
a geometrical property just like moment of inertia
I zz   y 2dA. However, its value depends upon the value of
A
h that you select, i.e., you don’t integrate over the entire
area; instead only that portion from y=h to y=c (where c is
location of either the upper or lower surface). Lets look at
several examples.
2001, W. E. Haisler
Chapter 13: Beam Bending
118
2001, W. E. Haisler
Chapter 13: Beam Bending
y 119
Determining Q. Consider a rectangular
cross-section of width t and height 2c
Now calculate Q(h):
c
c
Q( y  h)   ytdy  t  ydy  t (c 2  h 2 )
h
h
2
Note: at top or bottom, Q( y  c)  0
at center, Q( y  0)  t c 2  c (tc)
2
2
Thus, Q is 0 at the top and bottom, max at
the center, and varies quadraticly from top
c
h
z
c
t
2001, W. E. Haisler
Chapter 13: Beam Bending
120
to bottom. For the rectangular cross section, we see that
 xy is a maximum at the centroid and is given by:
V y ( x)
3
.
 xy ( x, y  0) 
4 tc
y
80
Consider a beam with the T
cross-section shown.
34
 The moment of inertia
24
14
about the z axis (centroid) z
centroid
16
is determined to be
of T 46
6
4
I zz  2.3110 mm .
 Assume that at some point all
dimensions
along the length of the
in mm
40
beam, the shear is given
20
60
2001, W. E. Haisler
Chapter 13: Beam Bending
121
by V y  104 N .
 Determine the shear stress at y=+14 mm and y=-14 mm.
The shear stress at any point
is given by:
V y ( x) Q( y )
 xy ( x, y ) 
I zz t ( y )
Substituting the values of
V y and I zz gives:
 xy 
10  103[ N ] Q( y )
2.31 10 [mm ] t ( y )
6
4
80
y
34
z
14
centroid
46
of T
all
dimensions
in mm
40
2001, W. E. Haisler
Chapter 13: Beam Bending
122
At y  14mm :
Q(14mm)  
34
14
yt ( y )dy  
 xy ( y  14mm) 
34
14
2
y
y (80)dy  80
2
10 x103[ N ]38.4 x103[mm3 ]
2.31x106[mm4 ]80[mm]
34
 38.4  103[mm3 ]
14
 2.08
N
mm2
2001, W. E. Haisler
Chapter 13: Beam Bending
Similarly, at y  14mm,
Q (-14) is the first moment
of the area from y = -14 mm
to y = -46 mm (i.e., beyond
the point y where the shear
stress is desired). Note that
the thickness is now 40 mm.
Q(14mm)  
46
14
yt ( y )dy  
 xy ( y  14mm) 
46
14
80
123
34
z
14
-14
centroid
of T
46
all
dimensions
in mm
y (40)dy 
2
y
40
2
10 x103[ N ]38.4 x103[mm3 ]
6
y
4
2.31x10 [mm ]40[mm]
40
46
 38.4  103[mm3 ]
14
 4.16MPa
2001, W. E. Haisler
Chapter 13: Beam Bending
124
Note: at the upper surface, Q(34 mm)=0 and at the lower
surface, Q(-46 mm)=0 which means that the shear stress is
zero at the upper and lower surface. For example,
Q(46mm)  Q(14mm)  
14
46
y 40dy
2 14
y
 38.4  10  40
2
3
0
46
Exercise: Show that at the centroid of the cross-section
(y=0): Q(0)  42,300mm3 and Vy (0)  4.582MPa .
MDSolids gives the shear stress distribution from top to
bottom. Note that it varies quadraticly with position, y; and
is zero at top and bottom, and maximum at the centroid:
2001, W. E. Haisler
Chapter 13: Beam Bending
125
2001, W. E. Haisler
Chapter 13: Beam Bending
126
Consider another cross-section.
Assume the same shear stress
(V y  104 N ) and evaluate the shear
stress at the centroid of the rectangular
cross-section. Note centroid is at
center of rectangle.
Q(0)  
10
0
2 10
y
y5dy  5
2
5cm
y
z
10cm
 50  5  250cm3
0
I zz  5cm(20cm)3 /12  13104 cm 4
 xy
y 0

10  103[ N ]250[cm3 ]
1 10
3
4
4
[cm ]20[cm]
 37.5
N
cm
2
 0.375MPa
20cm
2001, W. E. Haisler
Chapter 13: Beam Bending
For a simple composite
cross-section of rectangles
as shown, Q can be
determined simply in terms
of centroids of each area:
127
y
80
y1  24
z
A1
20
A2
centroid
60
of T
14
16
c
Q(h)   yt ( y )dy
h


c
ydA   yi Ai
for area
from
y  h to c
For example:
h
all
dimensions
in mm
40
Q(0)  y1 A1  y2 A2
 24(80 x 20)  7(40 x14)  42,300mm3
2001, W. E. Haisler
Chapter 13: Beam Bending
y
Notice that you could take
the bottom portion of the T
from y = 0 to y = -46 mm to
y1  24
obtain to obtain the same
z
result:
Q(0)  y3 A3
 23(40 x 46)
 42,300mm3
128
80
20
14
23
A3
all
dimensions
in mm
centroid
60
of T
40
The area is negative because it is below the z-axis.
See the handout “Brief Tutorial on MDSolids for Beam Analysis”
on the web page for information on how to determine section
properties, bending and shear stress, and deflection results.