I did not ask you to read the following three sections, primarily because I find them
terribly confusing and essentially impossible to understand.
Section 3 – 9, Uniformly Distributed Stress.
This section is merely re-stating what you learned in MAE 3201, strength of materials,
with regard to St. Venant’s principle. In essence, the stress distribution at a location “far from”
an applied load is independent of the details of that load and is essentially the same for any
alternative applied load that is statically equivalent to the actual applied load. “Far from” means
on the order of two times the largest lateral dimension.
Section 3 – 10, Normal Stresses for Beams in Bending
In this course, beams will satisfy the following criteria.
1. The beam is initially straight and horizontal.
2. The cross-section of the beam is constant.
3. The beam is made of only one material, and that material is homogeneous, linear and
isotropic.
4. The cross-section of the beam has a vertical or horizontal axis of symmetry.
5. There is no axial force applied to the beam.
The following cross sections satisfy criterion 4.
The following cross sections do not satisfy criterion 4.
MAE3501 - 4 - 1
In MAE 3201 you learned the following relationship for the normal stress in a beam
along the axis of the beam.
MY
(4.1)
I
Y denotes a vertical coordinate, positive upward, measured from the neutral axis of the
beam. For a homogeneous material, the neutral axis is a horizontal line through the centroid of
the section. I denotes the second-area moment of the cross section, about the Z - axis. In
strength of materials, I was referred to as the “moment of inertia” of the cross section. Shigley is
correct; Mr. Hibbeler is not. This term is more correctly denoted as I Z or I ZZ but the subscript
is frequently dropped.
X  
 MAX , tension, will occur, for positive M, when Y is minimum, at the bottom of the
beam.  MIN , compression, will occur, for positive M, when Y is maximum, at the top of the
beam. Use common sense. If M is positive at some section, the beam is curved upwards and
there is tension below the neutral axis and compression above the neutral axis.
Criterion 4 requires the beam cross section to have either a vertical or horizontal axis of
symmetry. Following is the reasoning behind this criterion.
Vertical loads, parallel to the Y - axis, cannot exert a moment about the Y - axis.
Therefore, the normal stresses acting on a cross section cannot exert a moment about the positive
Y - axis.
(4.2)
MY  0 
   X Z dA
SECTION
The minus sign is required by the sign convention, so that the normal stresses will
contribute positive moments about the positive Y - axis.
0
  MY

 Z dA
I


SECTION

(4.3)
Move the constant bending moment and second-area moment outside the integral.
0
M
I
 Y Z dA
(4.4)
SECTION
The integral is referred to as the second-area moment about the Y - and Z - axes, and is
denoted by I YZ .
Equation (4.4) is satisfied by any cross section with either a vertical or horizontal axis
of symmetry.
MAE3501 - 4 - 2
Example
This very important example demonstrates the method used to determine the location of
the neutral axis and I Z , the second-area moment, for a composite shape composed of rectangles.
10 inch
2
1 inch
8 inch
S
1
1 inch
The first step is to determine the location of the neutral axis. It is a very good idea to use
a coordinate system such that there are no negative values in the problem. Choose S, positive
upwards, with the origin at the bottom of the section. Do not use Y to denote this coordinate,
because Y is a coordinate measured positive upward from the neutral axis. Number each
rectangle in the section.
S
S
S1 A1  S 2 A 2
(E.1)
A1  A 2
4 (8)  8.5 (10 )
 6.50 inch
8  10
(E.2)
For each individual rectangle, I C denotes the second-area moment of that
rectangle about the center of that rectangle. S denotes the vertical distance from the center
of that rectangle to the neutral axis of the section. It does not matter whether S is positive or
negative, because it will be squared. Develop the relationship for I C . Let S  denote a vertical
coordinate with its origin at the center of the rectangle.
H
S
H/2
IC 
 S dA 
2
H / 2
H/2

S  2 B dS 
H / 2
B
MAE3501 - 4 - 3
B H3
12
(E.3)
The parallel-axis theorem.
I  I C  A ( S) 2
(E.4)
The parallel-axis theorem relates the second-area moment of an area about some
horizontal axis to the second-area moment of the area about a horizontal axis through its own
centroid. This theorem can never be used to relate the second-area moments about two arbitrary
horizontal axes.
1 ( 8) 3
I1 
 (8) (1) (6.5  4) 2  92.7 inch4
12
I2 
(E.5)
10 (1) 3
 (10) (1) (8.5  6.5) 2  40.8 inch4
12
(E.6)
I = 92.7 + 40.8 = 134 inch 4
(E.7)
Section 3 – 11, Shear Stresses for Beams in Bending
It is necessary to determine the shear-stress distribution,  XY as a function of position on
the cross section of a beam in bending. Determine V, the shear force acting on the cross section.
V
It is not possible to determine  XY directly. It is necessary to determine  YX , and use the
relationship that the two shear stresses are identical everywhere. The derivation will utilize a
rectangular cross section for ease of drawing, but the derivation is general. Despite the drawing,
assume the height of the beam is at least twice the width of the beam.
Y
Y
N.A.
X
H
Z
X
b
MAE3501 - 4 - 4
Examine all the forces acting in the X - direction on the shaded block of material at the
top of the beam. Consider a FBD of that block, looking inward along the positive Z - axis.
X
 X   X
 YX
Y
X
N.A.
All stresses are drawn positive.  YX could vary in both the X - and Z - directions. The
derivation will take the limit as X  0 , so the possible variation of  YX in the X - direction is
of no concern. Assume  YX does not vary in the Z - direction. Exact solutions from the theory
of elasticity demonstrate that this assumption is within a few percent of being correct for
“ordinary” beams with H  2 t .
The axial normal stress,  X , at any value of Y, is proportional to M Z , the bending
moment.
During the second lesson, we developed the following relationship.
V
dM Z
dX
(4.5)
Therefore, if there is a shear force present, the bending moment and the axial normal
stresses will be different at the two ends of the block of material.
Static equilibrium of the shaded block of material in the X – direction yields the
following relationship.
 XY   YX
V
   
 Ib 
TOP
 Y dA
(4.6)
Y
The integral in equation (4.33) is denoted by Q. It is conventional to drop the minus
sign. The shear formula.
 XY   YX 
VQ
Ib
(4.7)
In MAE 3201, the thickness of the section was denoted by t rather than b.
MAE3501 - 4 - 5
Examine Q more closely. The analysis will again use a rectangle for simplicity. Again
assume the height of the rectangle is greater than twice the width.
H
2
Y
N.A.
b
Q (Y) 
 Y dA
(4.8)
SHADED
Q (Y)  A SHADED YSHADED
(4.9)
For any cross section, Q = 0 at the top and bottom. Therefore,  XY = 0 at the top and
bottom. For any cross section, Q is maximum at the neutral axis. Therefore,  XY will be
maximum at the neutral axis for any “reasonable” cross section.
For more complicated cross sections, such as a wide-flanged beam, it will be necessary to
determine Q for an area composed of more than one smaller area.
Q   A J YJ
(4.10)
J
MAE3501 - 4 - 6
Example
Determine the maximum shear stress acting on the following section subjected to a
vertical shear force, V  20 kN . Dimensions are in mm.
200
S
300
20
20
20
175
S
20 (175) (10)  20 ( 300) (170)  20 ( 200) ( 330)
 175.9 mm
20 (175)  20 ( 300)  20 ( 200)
IZ 
(E.1)
175 ( 20) 3
 175 ( 20) (175.9  10.0) 2
12


20 ( 300) 3
 300 ( 20) (175.9  170.0) 2
12
(E.2)
3
200 ( 20)
 200 ( 20) ( 330.0  175.9) 2
12
 237 (10) 6 mm 4  237 (10)  6 m 4
The preceding calculations are reasonably straightforward. However, when determining
Q, I believe it is helpful to use a more-detailed schematic diagram of the cross section, one in
which important values of S are labeled. The schematic diagram is not to scale.
200
340
320
20
N.A. 175.9
S
20
0
175
MAE3501 - 4 - 7
 320.0  175.9 
Q MAX  200 ( 20) ( 330.0  175.9)  20 ( 320.0  175.9) 

2


(E.3)
 824 (10) 3 mm 3  824 (10)  6 m 3
 MAX 
Q MAX V
 3,480 kPa 
IZ b
(E.4)
The calculation of Q MAX was reasonably straightforward because the bottom of the
vertical rectangle was located at the neutral axis. However, it may be necessary to determine  ,
the shear stress, at some other value of S, e.g., S  120 . I strongly suggest adding another
horizontal line to the previous schematic diagram.
200
340
320
20
S
N.A. 175.9
120
20
0
The contribution of the top, horizontal rectangle remains the same. Neither its area nor
the Y - value of its centroid has changed. However, it is necessary to be careful in determining
the Y - value of the centroid of the bottom, vertical rectangle. That Y - value is the S - value of
the centroid of the rectangle minus the S - value of the centroid of the section.
 320.0  120.0

Q (S  120)  200 ( 20) ( 330.0  175.9)  20 ( 320.0  120.0) 
 175.9 
2


 793 (10) 3 mm 3  793 (10)  6 m 3
Note that the value of Q at any location other than the neutral axis is always less than the
value of Q at the neutral axis.
MAE3501 - 4 - 8
Shear Flow
The shear flow, denoted by q, is the shear force in the X - direction, per unit length in
the X - direction, on the bottom of the horizontal surface exposed by the horizontal section.
QV
(4.11)
I
q also denotes the shear force in the Y - direction, per unit length in the Y - direction, on
the cross section of the beam.
q
Thin-Walled Sections
Determine the local shear stress on the cross section of a thin-walled section.
Make a section parallel to the local thickness of the cross section, at the location where
the value of the shear stress is to be determined. The analysis proceeds in the same manner as
the previous analysis, except the average shear stress determined from the shear formula is now
the average shear stress, in the X - direction, on the surface exposed by the section. That average
shear stress could be either  ZX ,AVG or  YX ,AVG . Because the cross section is thin, the shear
stress is essentially constant across the face of the exposed surface.
That average shear stress is also the average shear stress, on the cross section of the
beam, perpendicular to the surface exposed by the section. This average shear stress could be
either  XZ ,AVG or  XY ,AVG . Again, essentially constant across the local thickness.
The shear stress on the cross section of a beam with a thin-walled section “flows” along
the cross section, essentially perpendicular to the local thickness
In the shear formula, V and I are equal to VY and I Z , regardless of the direction of the
shear stress on the cross section. Q is the first moment, about the neutral axis, of the area
between the section and an outside surface. b is the local thickness of the cross section.
 “flows” along the face of the cross section. q is “conserved” at intersections, in the
same manner as electric current or volumetric flow of an incompressible fluid.
MAE3501 - 4 - 9
Shear Center
Consider a beam with a cross section that does not have a vertical plane of symmetry.
P
The location of the load, P, in the Z - direction, is not yet determined. Make a section as
shown and sketch q, the shear flow, on the exposed section facing the downward force.
This shear flow, q, will create a moment, about an X - axis through the centroid of the
cross section, or about an X - axis through almost any other point selected. This shear flow was
determined using the results of elementary beam theory. Therefore, elementary beam theory will
apply to the bending of this beam only if the load, P, exerts the same moment about an X - axis
through the selected point that the shear flow, q, exerts about that X - axis.
MAE3501 - 4 - 10
Example
Determine the necessary location of the load for the following geometry. The thickness,
b, much smaller than either B or H, is the same for all portions of the section. Z  and Y  are local
coordinates. The shear force is equal to the downward load, P.
B
H
I
Y
Z
Bb3
 H 2 
b H3

2
 (B b ) 


12
12
4



(E.1)
Neglect the term containing b 3 .
b H3 b B H2
I

12
2
(E.2)
Consider the top flange.
 H
Q  ( Z b) 

 2
(E.3)
H
P ( Z b)  
VQ
 2   P b H Z
q

(to the left)
I
I
2I
(E.4)
We will not need q in the shear web or the bottom flange.
Determine the total moment, positive counter-clockwise, of all the shear flows about an
X - axis through a cleverly chosen point. Choose an X - axis through the lower, left corner of the
cross section, eliminating the moments of the shear flows in both the web and the lower flange.
P b H Z
P b H2 B2

d
Z

 2I
4I
Z  0
B
MX  H
Substitute equation (E.2) into equation (E.5).
MAE3501 - 4 - 11
(E.5)
MX




2
B




P
H


b H3
  2B 
 2b B H2
 3

3
P b H2 B2
(E.6)
In order for elementary beam theory to yield the correct result, the force P must also exert
the same M X about the lower, left corner of the cross section. The load P must act through a
point a distance e to the left of the shear web.
e
B2
(E.10)
H
 2B
3
If P does not pass through the shear center, also referred to as the center of twist, the
beam deflects and twists. The twisting causes additional shear stresses, and the shear stresses
determined from elementary beam theory are not correct.
MAE3501 - 4 - 12
Example
A thin-walled circular section with a longitudinal slit.
V
R

b
In evaluating trigonometric integrals, the following identity is very useful.
n
2
 sin
2
 d 
m
2
n
2
 cos
m
2
2
 1   n    m   
 d     


 2   2   2  
2
I   Y dA   ( R sin  ) 2 R b d   R 3 b
2
(E.1)
0


Q (  )   Y dA   R sin  R b d  R 2 b (1  cos  )
0
q() 
(E.2)
0
Q V V R 2 b ( 1  cos  )  V 
 ( 1  cos  )

 
I
 R3 b
 R
(E.3)
Determine the moment of the shear flow about the center of the section,
2
MX 
 VR
 q ( ) (R d ) R    
0
2
 ( 1  cos  ) d  2 V R
(E.4)
0
The shear center is located a distance 2 R to the left of the center of the cross section.
MAE3501 - 4 - 13
MAE3501 - 4 - 14
Example
Determine the location of the shear center for the following cross sections.
The shear flow at a location on the section is “along” the section at that location.
Therefore, the moment of the shear flows about the “intersection” point is equal to zero. The
intersection point for any of these sections is therefore the shear center for that section.
Homework
Problems: 3 - 34 (d) Determine the second moment of the area about the neutral axis.
3 - 44 (a), (b). It is necessary to determine the magnitudes of the largest positive and negative
bending moments.
MAE3501 - 4 - 15