Now we are going to use our free body analysis to look at
Beam Bending (W3L1)
Problems 17, F2002Q1, F2003Q1c
One of the most useful applications of the free body analysis
method is to be able to derive equations for stresses in beams –
no matter what the geometry or material. The procedure is no
more complicated than what we just did for pressure vessels,
although it has a lot more steps.
When you bend something, like the cantilevered beam shown
below, one surface will be in tension, the other in compression
(indicated by arrows below). As you move through the thickness
the stress gets smaller until it is zero at the (centre) neutral
surface. As long as the beam is fairly thin there is no vertical
stress in the beam (even though the applied force is vertical) –
all of the stress is in the longitudinal direction.
In this and the next couple of lectures we will use the free body
analysis method to develop equations to calculate longitudinal
stresses in beams of different shapes and material types.
neutral surface (zero stress)
tension
Try bending
your pen to
visualize this
compressio
n
Side view of a cantilevered
beam with a force W at the end.
w
1
Try bending
your pen to
visualize this
neutral surface (zero stress)
tension
compressio
n
Points to note:
→ Vertical Forces: (not usually of interest)
Vertical external
force W (down)
=
w
Vertical reaction
force at the wall (up)
→ Assume large radius of curvature (R) for the beam
(in other words it isn’t bent too much)
→ Beam has only longitudinal stresses if it is thin
with respect to its length
→ Top surface in tension
The further from
the neutral surface
→ Bottom surface in compression
the higher the
→ neutral surface in between
stress
(no longitudinal stress at neutral surface)
→ Since F (external) is vertical then there is no
horizontal external force, so the net internal force
across a vertical cut = 0
→ Assume initially that the beam is “light”
2 in
The following page outlines the specific steps we will work through
order to obtain an equation for longitudinal stress in a bent beam. We
will be going through each step separately in this and the next lecture.
Free Body Analysis Steps – Bending
Step 1: We want to analyze the longitudinal stresses so make the
cut perpendicular to the length.
NS
cut
Step 2: Reaction force is not constant over the cross section, w
therefore we must determine the reaction force as a function of
distance from the neutral surface.
Yz
F(on dA) = dA
R
Step 3: Static equilibrium condition. Longitudinal forces = 0 across
the cut, since no longitudinal external force exists. However the
torques must balance as well (all beam bending problems are
based on free body analysis of the torque equilibrium).
Y
Izz=2nd moment of cross sectional area: it
W (l − x ) = ∑ z 2dA
R
defines dimensional characteristics
Step 4: Solve for R (the radius of curvature) in terms of the x and y
co-ordinates of the beam. 1
d2y
R
=
dx 2
Step 5: Sub this into the equation from Step 3 to give a 2nd order
d2y
differential equation.
YI zz 2 = W (l − x )
general internal torque
dx
eqn for bent beams
Step 6: Solve the differential equation and apply boundary
conditions to give (in the simplest case):
(the deflection y at any
W ⎛ x2 x3 ⎞
⎜⎜ l − ⎟⎟
y=
point x along the beam)
YI zz ⎝ 2 6 ⎠
Step 7: Determine an equation relating the position on the beam
surface to stress and strain (by combining above equations).
surface strain at x
surface stress at x
3a (l − x ) yend
2l 3
3aY (l − x) yend
τs =
2l 3
εs =
3
So now lets go through each step individually:
→ Step 1: Make a cut perpendicular to the
direction of interest
→ We want to analyze the longitudinal stress, so…
Make a vertical cut
NS
cut
w
(load
force)
Now, unlike the case for the pressurized cylinders, the
internal reaction force (represented by the arrows above) is not
constant throughout the cross section. So next we have to
4
develop an expression for how the internal force varies as a
function of distance away from the neutral surface (NS).
Step 2 – determine reaction force as a function of distance
Note this is not drawn
from the neutral surface (NS)
z
A
A’
z
δθ
R
to scale – the
curvature is much
exaggerated but
really should be very
small (i.e. large R)
neutral
surface (NS)
Defining terms:
→ z is co-ordinate perpendicular to the NS with z = 0 at NS
→ R = radius of curvature for the NS
→ R ± z is the radius of curvature for any fibre in the bent rod
So now...
A. consider the small fibre A-A’, subtending δθ and a
distance z from the NS
B.
the unstretched length of A-A’ (if beam is unbent) is the
same as it would be at the NS
= Rδθ
C.
stretched length of A-A’ = (R + z)δθ
D.
∴ stretch of A-A’
= (R + z)δθ - Rδθ
= zδθ
5
Step 2 cont…
So strain of any fibre A-A’
a distance z away from NS
ε=
∆l zδθ
z
=
=
l
Rδθ R
(this strain is positive [+z] above the NS, negative [-z] below
the NS)
longitudinal strain ε =
at any z
z
R
1
longitudinal stress τ = Y
at any z
z
2
R
If the beam thickness is “a” then at the surface z = a/2
a
strain
ε=
at surface:
2R
stress
at surface:
1a
τ=
Ya
2R
2a
Eq2 is τ as a function of z – but we need F as a function of z:
To get this: Look at the Beam End-On (in cross section)
Total force on dA will be:
dA
FdA = τdA =
So
Yz
dA
R
Yz
FdA = dA
R
z
NS
eq2
3
(so this tells us how the reaction
force varies through the cross
section of the beam at any
position z away from the NS)
[reaction
forces are in at
top and out at
bottom in this
diagram]
6
Step 3 – Static Equilibrium Condition:
(summing the forces and torques)
need co-ordinate system:
x
y
(deflection)
(note we didn’t need to
sum torques for the
pressure vessel – since
axisymmetric all cancel
Our cantilever beam is
of length, l, with a
load, W. We define
the cut to be a
distance, x, away from
the wall.
cut
l (overall length)
W (load)
The cut segment is in
static equilibrium under
the influence of W and the
internal reaction forces at
the cut.
Q
Cut segment
l-x
W
→ Summing Forces: no horizontal W component
(since bending is small), so internal longitudinal
reaction forces=0
→ Summing Torques: (sum torques about point Q)
external torque
about Q
=
total internal reaction
torque (about Q)
Yz
W (l − x ) = ∑ ⎛⎜ dA ⎞⎟ z
⎝R ⎠
reaction force
for each dA
A little
rearranging...
W (l − x ) =
distance dA is
from Q
Y
2
∑ z dA
R
Izz : 2nd moment of cross
Y
W (l − x) = I zz
R
sectional area (next page)
4
where
I zz = ∫ z 2 dA
7
Detour!!!!
What is Izz?
→ “The second moment of cross sectional area
about an axis which is in the neutral surface and at
right angles to the external force”. It defines how the
area is distributed about the neutral surface (sort of
analogous to the moment of inertia in rotation, which
defines how mass is distributed about a rotational
axis). Units are m4!!
Calculating Izz for a bar of rectangular cross section
looking at cross section:
I zz = ∫ z 2 dA
dA = bdz
dA
integrate from z = -a/2 to +a/2
dz
a
z
NS
+a 2
I zz = b ∫ z 2 dz =
−a 2
1 3
ba
12
(units m4)
b
See following page for Izz tables for common cross
sections
→ Further note: for composite objects you can add or
subtract Izz for different components – provided they
have the same NS (further note… this is just like adding
moments of inertia but there is no parallel axis theorem …
bummer)
8
Izz values for common beam cross sections
9
Ex. A “light” metre stick clamped to the
bench with the following parameters:
(Cross section) a = 7.8mm
b = 25.4mm
l = 0.5m
Birch wood: Y = 16x109 Pa
mass on end = 1kg
(Don’t actually need these yet)
Calculate Izz for each orientation:
1 3
I zz = ba = 25.4 x 7.83 / 12 = 1x10−9 m 4
12
Note
correction
I zz =
1 3
ba = 7.8 x 25.43 / 12 = 10.6 x10−9 m 4
12
The second orientation makes a much stiffer
beam than the first, (almost 11x) even though the
material is the same!!
Two critical points…
→ Izz is essentially the stiffness of the geometry, just as
Y is the inherent stiffness of the material!
→ since
I zz = ∫ z 2 dA - the further z (the
area) is distributed away from the NS, the
geometrically stiffer it is – eg. I beams are
designed for maximum stiffness – lots of
area a long way from the NS
e.g. an I-beam
10
Ex. Calculate Izz for a c-bar of
the following cross
section and dimensions.
6cm
2cm
4cm
Note there are two
methods you can
use to do this –
both are shown
below
2cm
neutral surface
2cm
Method 1:
Use the previously derived formula for rectangular beam Izz and
subtract the middle from the outside (since both have the same NS)
I zz =
6cm
1 3
ba
12
4cm
8cm
(6 x 83)/12
256cm4
4cm
(4 x 43)/12
21.3 cm4
= 234.7cm4
Note that the middle doesn’t have a high Izz value – why? Because it
is close to the NS. So removing it doesn’t change the value of Izz by
11
very much (only about 9%). BUT you reduce the weight by quite
a
bit – about 33% if you work out the area – so you have increased the
stiffness/weight ratio!!!
Method 2:
Break the piece into two sections (A and B) above the NS (then
multiply the answer by 2).
I zz = ∫ z dA
2
A
NS
B
z
(Look at how we did this for
the rectangle earlier)
Section B
2
2
Section A
I zz = 2 ⎡ ∫ z ( 2dz ) + ∫ z 2 (6dz )⎤
⎢⎣ 0
⎥⎦
2
4
Two halves
=234.7 cm4
12