2001, W. E. Haisler
1
Chapter 13: Beam Bending
Chapter 13 – Beam Bending
The Classical Beam Bending Theory
Classical bending of beams is characterized by the following
assumptions/restrictions:
Geometry - long and slender prismatic beam (symmetric about z
axis) with x axis along its length:
y
distributed load
x
z
y
A
L
A
typical cross sections
z
x-axis is at
centroid of A
2001, W. E. Haisler
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Chapter 13: Beam Bending
Applied Loading - Transverse forces or tractions normal to
the x axis. Moments about the y or z axis.
po
F
po
L
L
x
m
x
L
x
 Kinematics - Predominate deflection is normal to the x
axis. Predominate strain is in the axial (x) direction.
Assume small strain and rotations of beam so that “axial
strains vary linearly over cross-section” or “plane sections
remain plane”. We will explain these assumptions shortly.
In terms of the general elasticity problem, the above can be
stated as follows:
2001, W. E. Haisler
Chapter 13: Beam Bending
3
 Equilibrium. If the cross-sectional dimensions are small
compared to the beam length, then applied transverse tractions
(in y and z directions) will be small compared to the resultant
internal stress in the x direction. --> small transverse loads
produce large axial stresses. Thus we assume that the only
major stress is xx (all other stresses are zero or negligible).
 xx
 0 . Implies that xx = xx (y,z).
Equilibrium reduces to
x
 Stress-Strain. We assume a linear isotropic material so that
 xx  E xx
and
 yy   zz   xx
 Kinematics/Displacements. Includes strain-displacement
 ux
relation  xx 
and the kinematic (displacement)
x
boundary conditions for u x .
2001, W. E. Haisler
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Chapter 13: Beam Bending
We will only consider “bending about the z-axis” in this course.
External transverse loading must therefore be in the z direction
(as distributed loads or point loads), and any applied moments
must be about the z-axis.
y
po
x
y
F
x
L
m
y
po
x
L
L
Such applied loadings (distributed normal loads, point loads and
moments) will produce “bending” of the beam and internal axial
stress  xx and shear stresses  xy on the cross-section.
Consider for example, the cantilevered T-section problem with a
300 lbf transverse force applied at its end as shown below.
2001, W. E. Haisler
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Chapter 13: Beam Bending
y
z
x
y
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Chapter 13: Beam Bending
The idealized internal stresses at x=0 will be given the following.
 xy
 xx
( xx )
( xy )
Note that the axial bending stress  xx (figure a) varies through
out the cross-section and tends to be a maximum at either the top
or bottom and is zero at the “neutral axis” (which is the centroid
2001, W. E. Haisler
Chapter 13: Beam Bending
7
of the cross-section for homogeneous cross-sections). In this
case, it is tensile on top and compressive on the bottom. The
shear stress  xy (figure b) likewise varies over the cross-section,
is zero at the top and bottom of the cross-section, and is a
maximum at the neutral axis (centroid of the cross-section for
homogeneous cross-sections). We will derive the equations with
which you can predict this axial bending and shear stress in this
chapter.
While one could work directly with the internal stresses ( xx and
 xy ) within the Conservation of Linear Momentum equations,
the approach can be simplified using classical beam bending
theory. These stresses can be put in terms of an equivalent
internal axial force, P, shear force, V y , and bending moment, M z ,
resultant that acts at the centroid of the cross section.
2001, W. E. Haisler
y
x
8
Chapter 13: Beam Bending
 xy
 xx
Mz
Vy
y
x
P
z
shear and axial
stress distribution
on cross-section
equivalent axial force, shear
and moment resultants on
cross-section
For axial stress distribution  xx and shear stress distribution
 xy acting on the entire cross-sectional area (left picture above),
we need to determine an equivalent set of forces and moment that
are equivalent to the stress distribution (right picture). Assume
the bar has a cross-sectional area of A as below and the xaxis is along the centroid of the beam cross-section:
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Chapter 13: Beam Bending
y
y
 xy
 xx
z
dA=dydz
x
A
Vy
=
P x
Mz
z
Equating the stress distribution over area A to forces P and V y ,
and moment M z gives:
Internal axial force in x direction = P    xx dA
A
Internal moment about z axis = M z     xx ydA
A
Internal shear force in y direction = V y    xy dA
A
2001, W. E. Haisler
Chapter 13: Beam Bending
10
Kinematic Observations/Assumptions - In order to obtain a
“feel” for the kinematics (deformation) for a beam subjected
to bending loads, it is informative to conduct some
experiments. The following photograph shows a long beam
with a square cross-section. Straight longitudinal lines have
been scribed on the beam’s surface, which are parallel to the
top and bottom surfaces (an thus parallel to a centroidallyplaced x-axis). Lines are also scribed around the
circumference of the beam so that they are perpendicular to
the longitudinals (these circumferential lines form flat planes
as shown). The longitudinal and circumferential lines form
a square grid on the surface. The beam is now bent by
moments at each end as shown in the lower photograph.
2001, W. E. Haisler
Chapter 13: Beam Bending
y
x
z
11
2001, W. E. Haisler
Chapter 13: Beam Bending
12
 After loading, we note that the top line has stretched
(tension) and the bottom line has shortened (compression).
If measured carefully, we see that the longitudinal line at
the center has not changed length ( xx  0 at centroid).
The longitudinal lines now appear to form concentric
circular lines.
 We also note that the vertical lines originally perpendicular
to the longitudinal lines remain straight and perpendicular
to the longitudinal lines. If measured carefully, we will see
that the vertical lines remain the same length ( yy  0).
Each of the vertical lines (as well as the planes they form)
has rotated and, if extended downward, they will pass
through a common point that forms the center of the
concentric longitudinal lines.
2001, W. E. Haisler
Chapter 13: Beam Bending
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 The flat planes originally normal to the longitudinal axis
remain flat planes and remain normal to the deformed
longitudinal lines.
 The squares on the surface are now quadrilaterals and each
appears to have tension (or compression) stress in the
longitudinal direction (since the horizontal lines of a
square have changed length) and perhaps also some shear
stress (since opposite vertical lines of a square have rotated
different amounts).
Thus, to begin the theoretical development, we make
some kinematic assumptions based on the experimental
observation. We assume that the predominate deflection is
normal to the x axis, i.e., u y ( x, y) . Predominate strain is in
2001, W. E. Haisler
Chapter 13: Beam Bending
14
the axial (x) direction, i.e.,  xx . Assume small strain and
rotations of beam so that “axial strains vary linearly over
cross-section” or “plane sections remain plane”.
In terms of the general elasticity problem, we make the
following assumptions for the case of pure bending of a
beam:
1. Conservation of Linear Momentum. If the crosssectional dimensions are small compared to the beam length,
then applied transverse tractions (in y direction) will be
small compared to the resultant internal stress in the x
direction. --> small transverse loads produce large axial
stresses  xx and  yy is small. For pure bending, we
2001, W. E. Haisler
Chapter 13: Beam Bending
15
assume that the shear stress  xy is negligible. Thus we
assume that the only major stress is  xx (all other stresses
are zero or negligible). The stress tensor reduces to
 xx 0 0
[ ]   0
0 0  . Equilibrium (Conservation of Linear
 0 0 0
Momentum) reduces to
 xx
x
 0 . (neglecting body forces)
2. Stress-Strain. We assume a linear elastic isotropic
material so that stress and strain are linearly related to each
other:
E
 xx  E xx and  xy 
 xy  G xy
1 
2001, W. E. Haisler
Chapter 13: Beam Bending
3. Strain-Displacement.  xx
16
u x u y
u x


and  xy 
y
x
x
4. Kinematic assumptions: Since  yy  0, this implies that
u y ( x, y)  u y ( x). Let u y  uoy ( x) be the transverse
displacement of the centroidal axis in the y direction (the
zero subscript means the transverse displacement is
measured at y=0). Hence,
u y ( x, y )  uoy ( x)
Draw a sketch of the undeformed and deformed beam and
consider the geometry in light of assuming that a normal to
centroidal axis remains normal and straight after bending:
2001, W. E. Haisler
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Chapter 13: Beam Bending
y
normal
point A at (x,y)
beam centroidal axis
normal
y
 ( x)
ux ( x)   y ( x)
A
y
x
beam centroidal
axis (does not
stretch)
 ( x)
u0 y ( x )
du0 y
 ( x)  rotation of beam 
dx
x
2001, W. E. Haisler
Chapter 13: Beam Bending
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After bending, the transverse displacement of the centroidal
x-axis will be defined by uoy ( x) as shown below. The
subscript “0” means that uy is measured at y=0 (i.e,, at the
centroidal axis position). The rotation of the beam at any
point x is given by the derivative of the transverse
displacement u0 y with respect to x:
duoy
 ( x) 
dx
Since me make the assumption that a normal to the
centroidal axis remains straight and normal, then the normal
will also rotate by this same amount . For a point “A”
located at some position y above the centroidal axis, we note
that point A will have moved to the left as shown on the
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Chapter 13: Beam Bending
sketch. From geometry, the displacement in the x direction
u x ( x, y ) can be written as (note that for +, u x is negative):
u x ( x, y )   y tan  ( x)   y ( x)   y
du0 y ( x)
dx
Above means that the axial displacement u x can be written
entirely in terms of the transverse displacement of the
centroidal axis uoy and that the displacement is linear with
transverse position y. Substituting the above into the axial
strain, we obtain
 xx
 duoy ( x) 
 y
2

d
uoy ( x)
dx 
u x



 y
x
x
dx 2
(1)
2001, W. E. Haisler
Chapter 13: Beam Bending
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The above satisfies the assumption that the axial strain at
y=0 (the centroidal axis) is zero. Because the strain is zero
along the x-axis passing through the centroid, it is
sometimes referred to as the neutral axis.
We can now rewrite the internal bending moment in terms of
displacements by substituting the strain-displacement
equation into the stress-strain equation and that result into
the moment equation to obtain
 d 2uoy ( x) 
M z ( x)     xx ydA    E xx ydA    E   y
ydA

2
A
A
A 

dx


Note that we must integrate over the cross-section A which
lies in the y-z plane. Assume that Young’s modulus E is a
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Chapter 13: Beam Bending
constant over the cross-section, i.e., E=E(x). Hence, we
write:
M z ( x)  E
d 2uoy ( x)
dx
2
A
y 2dA
The integral term is a geometrical property of the crosssection and can be easily integrated:
I zz   y dA
2
A
where I zz is called the moment of inertia of the crosssection about the z axis.
2001, W. E. Haisler
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Chapter 13: Beam Bending
Note: We will assume that the cross-section is symmetric
about the y axis. If the cross-section is not symmetric about
the y axis, a transverse load may produce twisting of the
cross-section which we have not considered here. Note that
bending is occurring about the z axis since bending moments
are about the z axis.
With this definition of the moment of inertia, the bending
moment equation becomes
d 2uoy ( x)
M z  EI zz
dx 2
or
d 2uoy ( x)
dx
2
Mz

EI zz
(2)
2001, W. E. Haisler
Chapter 13: Beam Bending
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The last equation is an ordinary, second order differential
equation that defines the transverse displacement in terms of
the bending moment. Note that the bending moment M z
will in general be a function of x. In addition, E and I zz may
be functions of x.
The stress may now be written in terms of the bending
moment. Substitute (2) into the strain equation (1) to obtain:
 xx   y
d 2uoy ( x)
dx 2
Mz
 y
EI zz
Substitute strain into Hooke's Law to obtain:
2001, W. E. Haisler
Chapter 13: Beam Bending
 xx  E xx
24

Mz 
 Ey

 EI zz 
Thus,
 xx
Mzy

I zz
Note that  xx varies linearly with y (i.e., linearly from top to
bottom surface of the beam) and is zero at the centroidal
axis. Since the bending moment M z  M z ( x) , then the
stress also varies with position along the length of the beam.
For a particularly location x, the bending stress  xx will
obviously be a maximum at the maximum value of y (i.e.,
either the top or bottom of the beam).
2001, W. E. Haisler
Chapter 13: Beam Bending
Mz
Internal Bending Moment, M z
 xx
Resultant Bending Stress,  xx
25
Mz
 xx
2001, W. E. Haisler
Chapter 13: Beam Bending
26
The sketch above shows the centroid as if it were halfway
between the top and bottom surface. In this case, the
bending stress on the top and bottom surface would be equal
in magnitude but opposite in sign (i.e., one compression, the
other tension).
Important: For cross-sections like the inverted T section
shown earlier wherein the centroid is not an equal distance
from the top and bottom surface, the bending stress will
have different magnitudes at the top and bottom. For these
types of problems, the surface with maximum y value will
produce the largest magnitude bending stress. We will work
some problems shortly, which show this type of behavior.