6. UNIFORM SECTIONS subject to BENDING
Bending stresses
When a uniform section of material is bent one side stretches
and the other compresses. At some point between the two there
is a plane where no stretching or compression occurs - this is
known as the neutral plane
top
Neutral plane
y1
bottom
y2
R
q
115 - Uniform Sections subject to Bending
1
Length of top plane = (R+y1) x q
original length = R x q
Straintop =
(R+y1) x q - R x q
y1
=
R
Rxq
Length of bottom plane = (R-y2) x q
original length = R x q
Strainbottom =
but
(R-y2) x q - R x q
y2
=R
Rxq
s=Ee
y1
st = E R
sb = -E
&
y2
R
Generally, if we measure y as positive above the neutral plane,
and negative when below:
s=y
E
R
E
s
=
y
R
+
neutral
compressive
stresses
115 - Uniform Sections subject to Bending
tensile
stresses
plane
-
2
The net effect of the internal stresses caused by the externally
applied bending moment (M) must be equal and opposite to it.
Top half of the stress
distribution.
w is the width of the
beam at distance y
from the neutral plane
dy
y
neutral
s
plane
Force = stress x area = s x dy x w
Moment = force x distance
dM = s x dy x w x y
To find the total moment we have to integrate from the
top to the bottom of the beam.
but
M =ós w y dy
õ
y
s=E
R
M = ó E y w y dy
õ R
M = E óy² w dy
Rõ
This is a geometric property of the cross-section
known as the
.
It is an important section property given symbol I .
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3
We can summarise the formulae for bending as below:
M
=
I
E
R
=
s
y
This is known as the bending equation.
M = external bending Moment applied (Nm).
I = Second moment of area about the neutral plane. (m4)
E = Elastic modulus of the material under bending.(N/m2)
R = Radius of curvature of the beam at the where the bending
moment (M) exists. (m)
= stress level in the beam (N/m2) at distance y (m) measured
from the neutral plane, above (+) or below (-) it.
Care - when using the Bending Equation make sure you
use standard SI units
M often given in kNm [=103Nm]
I often given in cm4 [= (10-2m)4 = 10-8m4]
E often given in GN/m2 [ = 109 N/m2]
s often given in MN/m2 [ =106 N/m2]
y may be in mm or cm [10-3 or 10-2 m]
For any particular cross-section of beam we must either look
up the I-value in tables or calculate it from first principles.
115 - Uniform Sections subject to Bending
4
BENDING
Section properties:
Location of the Neutral Axis
The neutral axis of a beam’s cross-section always passes
through the centroid of the section. i.e. the point (or line) about
which the shape would balance.
For symmetrical sections the centroid is at the centre.
For unsymmetrical sections it has to be found by
dividing into regular shapes, experimentally, or by
using a computer (CAD).
?
For 3 or more shapes it is
better to draw up a table.
A2
A1
We divide the section into
shapes where we know the
centroid position and find the
total effect.
y2
-y
y1
A1y1+A2y2=(A1+A A1y1+A2y2
y- =
(A1+A2)
115 - Uniform Sections subject to Bending
Shape Area
1
2
3
A
y- =
5
y
(A x y
A
Area x y
(A
x
y
Second Moment of Area
For certain geometrical shapes we can look up a
formula which gives us the I-value about its neutral axis,
otherwise it has to be found by dividing the shape into
recognised smaller shapes, or by using a computer
(CAD).
When we have a composite section (one made up of
different regular shapes) we cannot just add the
I-values to obtain the total I-value.
We need to use the Parallel Axis Theorem
IXX = Ina + Ay²
IXX = I-value of the Area about axis X-X
Ina = I-value of the Area about its own neutral axis n-a
y
= distance between the two axes X-X and n-a
A = Area of the shape
115 - Uniform Sections subject to Bending
6
Example - Find the position of the Neutral axis and the
I-value of the section below: (lengths are in cm)
8
A
2
N
A (for the section as a whole)
6
7
B
y-
3
X
Shape
A
B
Sum(S)
X
3
A
2
(cm )
16.00
18.00
34.00
y
(cm)
7.00
3.00
Ay
3
(cm )
112.00
54.00
166.00
Ina
4
(cm )
5.33
54.00
59.33
formula
3
bd /12
3
bd /13
2
y = SAy/SA
= 166/34
= 4.88
cm
IXX = SAy +SIna
= 946 + 59.33
4
= 1005.33 cm
For the section as a whole
115 - Uniform Sections subject to Bending
2
Ay
4
(cm )
784.00
162.00
946.00
INA = IXX-Ay²
2
= 1005.33 - 34x4.88
4
cm
= 194.86
7
Shear Force and Bending Moment diagrams
In order to design or specify a beam we need to know the forces
and moments acting on it at all points along its length.
We find these by drawing their Shear Force and Bending
Moment Diagrams
Rules:
1. Determine the reactions.
(Use SMreaction=0 and SFy=0)
2. Working from the LH end ‘cut’ the beam between each
applied force. Distance x is measured from the LH end.
3. Apply a downwards SF and anticlockwise BM at the cut.
4. Determine the SF and BM using SMcut=0 and SFy=0
5. Repeat ‘3’ & ‘4’ for each ‘cut’.
Notes:
It may sometimes be easier to work from the RH end in which
case apply an upwards SF and clockwise BM.
Continuous loads act through their mid point - this is true for
finding the Reactions, SF and BM.
a
A
b
w N/m
x
(used to find reactions)
MA = wb(a+b/2)
Mx = w(x-a)(x-a)/2=½w(x-a)²
(at the ‘cut’)
Pure applied moments: have the same effect irrespective of their
point of application. They appear in the moment equations only.
115 - Uniform Sections subject to Bending
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