Bending a Cantilevered Beam
Bram Sadlik
2 Feb. 2005
1
Objective
We look at the idealized case of a mass-less cantilever beam with a point force
acting on its far end, and derive an equation for the deflection of such a beam.
Although this calculation has been done many times, here I describe it a 1st year
level without any assumptions of previous knowledge of mechanics (Calculus is
inevitable...).
2
The Bending Moment
A cantilever beam can be imagined as a series of “springy” hinges, connected
tip to tail, with the first hinge fastened to a rigid wall (Fig. 1).
Top view of single
hinge
Wall
~τ
~r
Direction of bending
F~
Figure 1: Top view of a single spring loaded hinge. A series of
hinges connected head to tail can be used to visualize a beam.
The twisting force acting on a given hinge (otherwise known as ‘the torque’)
is given by,
~τ = ~r × F~ ,
(1)
where ~τ stands for torque, ~r is the vector pointing from the hinge to the point
where F~ , the force, acts. The ‘×’-symbol stands for ‘cross product’, which may
be familiar to you from Linear algebra. It means that ~τ points in a direction
normal to the plane created by ~r and F~ and has the magnitude,
|~τ | = |~r| · |F~ | sin(φ),
where φ is the angle subtended between ~r and F~ (Fig. 2).
1
(2)
~r × F~ = ~τ
~r
φ
F~
Figure 2: The cross product.
We make two approximations that simplify the equations. First, we say that
a bent beam is really only a two dimensional object, so that we can just assume
that the torque acts in a direction normal to the page. From now on we will call
this the ‘ẑ-direction’ (since it is k to the z-axis). Next, we say that since the
beam isn’t deflected much, the angle between its tip and the horizontal is small,
and therefore the angle between the force vector, F~ and ~r is approximately 90o
(Fig. 3).
y
z
x
Horizontal
Cantilevered beam
Wall
Angle
aproximately 90o for
small
deflcections
F~
Figure 3: The orientation of the coordinate system: x along the
beam, y in the vertical direction, and z into the page. The angle
between a cantilevered beam and vertical force is approximately
90o for small displacements.
These approximations lead to,
~τ = |~r||F~ |ẑ.
(3)
Or, if we just look at magnitudes, asserting that ~τ points along ẑ we get,
|τ | = |r||F |.
(4)
Here we note that a large ~r with a small F~ can produce the same toqrue as a
small ~r with a large F~ .
When torque refers to bending of rigid objects, we refer to it as the bending
~ ). The magnitude of the bending moment at a given point
moment (symbol M
along a rigid cantilevered beam is given by,
|M | = |r||F |.
(5)
The bending moment is exactly like a torque, except that we are dealing with
continuous material– it doesn’t bend at specific points (or hinges). The tendency
to twist is replaced with a tendency to bend at the point of interest. So, rather
~ - to make it clear that this is bending and not twisting.
than using ~τ we use M
2
3
Mechanical Equilibrium
After “loading” the beam we see that it bends, at first we observe small oscillations, but eventually these ‘die-out’ and the beam is stationary. This means
that the beam has reached mechanical equilibrium. The torque from the external bending moment is balanced by torques from the internal stresses in the
beam.
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→
~ +−
−M
{All internal stresses leading to torques} = 0
(6)
To understand how the internal stress works, imagine that the beam only
bends at one point along its length (Fig. 4). From Eq. 5, the magnitude of the
bending moment at this point is given by
M = F (L − x),
(7)
where L is the length of the beam, x is the position of the “kink” along the beam,
and F is the force. This moment is balanced by internal stress shown in the
diagram as small springs. Springs near the top of the beam are extended while
those near the bottom are compressed. Between these two types of springs there
is a neutral surface which is not extended or contracted. Each spring applies a
torque around the neutral axis. The further the spring from this axis, the larger
the torque it applies. This torque tends to bend the beam back to its horizontal
position. In this way the horizontal forces applied by the springs are translated
into vertical forces (opposing the hanging mass).
Kink
x
l
F~
Enlarged view of
kink
Neutral surface
Figure 4: A beam which has been bent at single point (kink),
illustrating the internal stresses at this kink and the bending
moment M = F (L − x).
A similar thing happens when you try to pull a nail out with a crow bar.
The force you apply on the crow bar is perpendicular to the force applied on
the nail, and the further the nail from the axis of rotation the harder it is to
pull out. In this analogy the nail is the spring, and the bending moment is the
torque on the crow-bars end. The axis of rotation is around the “neck” of the
crow bar.(Fig. 5)
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Figure 5: A crow-bar translates a force in the vertical direction
into a horizontal force.
4
The Bending Shape and Deflection
Usually beams bend uniformly along their entire length, rather than at a single
point. The deflection of the beam is actually quite complicated. However if we
assume small deflections, the local shape of the beam can be approximated by a
circle. That is, at any point along the beam we can draw a tangent circle with
radius R. (Fig. 6)
We treat the beam like a multilayered sandwich of springy surfaces (Fig. 7).
Each layer behaves like a wide ribbon of fabric. You can bend individual layers
very easily but it is harder to stretch them. Next we will show how the sandwiching of these layers, couples forces between neighboring layers and makes
the beam rigid to bending.
Fig. 3 shows a loaded cantilever beam. Fig. 6 shows an enlarged segment
of this beam with an exaggerated bend. The line NN’ is a neutral surface, its
length doesn’t change when the beam is bent. Surfaces above NN’ are stretched
during bending, while surfaces below it are contracted, just like we saw with the
springs 1 .
This extension and contraction of surfaces above and below NN’ is what
gives the beam its vertical “springiness”. Consider an isolated layer of thickness
dξ and a distance ξ from the line NN’, marking the neutral surface (Fig. 6).
When at rest, the length of this layer is equal to the length of NN’, but when
the beam is bent, the layers length becomes l + ∆l. From geometry in Fig. 6
we see that,
ξθ = dl,
(8)
Rθ = l.
Dividing through we get
∆l
ξ
=
.
(9)
R
l
The layer is stretched by ∆l from its rest length, l. The general form of
Hooke’s Law (HL) relates the stress to the strain,
F
∆l
=E ,
A
l
(10)
1 If you want to do an experiment, take a rubber eraser and place it on the table with its
large face down, mark a bunch of parallel lines on its side perpendicular to the table. Now
bend it up. The lines will no longer be parallel to each other. They get closer to each other
along the top surface and farther from each other along the bottom.
4
A
t/2
B
dξ
t
C
∆l
l
ξ
N’
−t/2
R
θ
N
D
θ
O
Figure 6: A small segment of the beam, with exaggerated bending. The radius of curvature is R, and is taken to the arc N N 0 , the
neutral surface, which is halfway down the beam. The line OA
k BD, and represents the length of the segment before bending.
The small arrows between BD and CO represent the extension
and contraction of the beam. Above N N 0 the beam is extended,
below N N 0 it is contracted.
y
t
dξ
ξ
x
Neutral surface N N 0
w
z
Figure 7: A ‘cut’ through the beam shows the layered structure
used for solving. Note the orietation shown with the coordinate
system.
∆l
where F
A is the applied stress, l is the strain, and E is Young’s modulus. A is
the cross section, l is the length, and ∆l is the extension of the material. We
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can derive the more common form by multiplying Eq. 10 by A,
F =
EA
∆l.[2]
l
(11)
Therefore the magnitude of the force exerted by the layer on each one of its
ends due to this stretch is given by Eq. 11,
dAE∆l
,
l
dF =
(12)
where dA is the cross sectional area of a single layer and E is Young’s Modulus.
Substituting Eq. 9 into Eq. 12 we get,
dF =
dAEξ
.
R
(13)
The cross section area dA of a single layer is given by the product of its width
with its height (Fig. 7),
dA = wdξ.
(14)
Substituting this into Eq. 13 we get,
dF =
Ewξ
dξ.
R
(15)
Now that we have the force exerted by a single layer we can calculate the
torque due to such a force. The magnitude of the torque is given by |~r||F~ | since
φ ≈ 90o . Substituting this into the mechanical equilibrium condition we get,
X
F ((L − x) =
dF ξ =
cross−section
Ewξ 2
dξ.
R
cross−section
X
(16)
Remember, a torque is balanced by a torque, that’s why we must multiply dF
by ξ, the distance between the ξ th layer and the neutral layer. We can turn this
sum into a definite integral with limits given by the top and bottom surface of
the beam, t/2, −t/2,
Zt/2
F (L − x) =
Ewξ 2
Ew ξ 3 t/2
Ewt3
dξ =
=
.
R
R 3 −t/2
12R
(17)
−t/2
From Eq. 17 we can get the curvature (Defined as 1/R),
1
12F (L − x)
=
.
R
Ewt3
(18)
Note that this equation makes intuitive sense, the curvature decreases as the
beam becomes thicker, wider or stiffer (Young’s Modulus) and increases as the
force increases and as the wall is approached (the last statement is less obvious).
We can get the shape and deflection of the beam by recalling that the curvature of a function is given by its second derivative. The slope is the first deriv.
2 From
Eq. 11 we see that the larger the cross section A, the more force we need to keep
∆l constant. We also see that a longer piece of material stretches by ∆l more easily than a
short piece.
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and the slope of the slope (or curvature) is given by the deriv. of the slope or
second deriv.
1
d2
(19)
= 2 (f unction)
R
dx
So that we can write:
dy 2
12F (L − x)
=
,
(20)
dx2
Ewt3
where we have replaced 1/R with the second derivative of y, the deflection at
point x along the beam. If you don’t get this, you can think of the beam as
defining some line on the plane. This line can be treated like a function where
deflections (y-coordinates) are related to position (x-coordinates). (Fig. 8) In
y
x
y = f (x)
Cantilever
Figure 8: The shape of the beam can be described as a function on the xy-plane.
order to get the deflection of the beam at a point x we multiply through by dx2
and integrate twice w.r.t. x,
2
ZZ
ZZ
12F (L − x) 2
x3
12F
Lx
2
y(x) =
d y=
−
(21)
dx =
Ewt3
Ewt3
2
6
The deflection at the beam’s end is given by substituting x = L,
y=4
F L3
.
Ewt3
(22)
In our case the force F was supplied by a hanging mass so,
y=4
(−mg)L3
Ewt3
(23)
where mg is the weight of the mass. (The minus sign accounts for the fact that
F is pointing down.)
5
Conclusion
We found an approximate equation describing the deflection of a cantilever beam
with a rectangular cross section when a point mass is hung from its end. The
equation holds in the limit of small deflections. Although the vector calculus and
l=knowledge of mechanics is minimal a solid background in calculus is required
to follow the derivation.
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