Bending
BADI 1
John Errington MSc
Loaded Beams
•
When a beam is supported at both ends
and loaded it will bend.
•
The amount by which it bends depends on
–
the load, and the way it is distributed
–
the distance between supports
–
the elastic modulus for the material, and
–
the distribution of material in the beam
Loaded beam
Here we have a solid square section
beam, centrally loaded with a force
F
and unsupported distance 2x
The beam bends into an arc, so the top
is shorter than the bottom. This means
the top is under compression, and the
bottom under tension.
The material in the middle (the neutral
axis) contributes less to the strength of
the beam than the material in the top and
bottom quarters of the beam.
F
2x
For our design we need to
know how much the beam
will bend, and the
maximum stress in the
beam to determine
whether it will break.
Figure 1.
A side view of a simply
supported beam (top)
bending under a distributed
lateral load (bottom).
Figure 2.
The internal forces and the
cross

sectional stress
distribution of a beam in
bending
.
Note the position of the neutral axis
Joists and floorboards
Floor loading
•
The timber floor of a property obviously weighs something. This weight consists
of the timber joists, the plasterboard ceiling underneath it, and the floorboards.
All of these are known as "the dead load" and the joists themselves must be
able to support this dead load without sagging. For a normal property, with
chipboard or timber floors and plasterboard ceilings, this dead load is generally
taken to be no more than 0.50 kiloNewtons per square metre. (kN/sq.m)
•
The weight we place upon a floor by way of bathroom suites, beds, wardrobes
etc, is known as the "imposed load". It is again accepted that, for normal
household requirements, the imposed load will not exceed 1.5kN/sq.m.
•
The building regulations tables A1 and A2 list the size of joist necessary to
support this weight, over a maximum span. Table A2 uses timbers known as
"SC4" which are high strength timbers containing very few, if any, knots. They
are not common in modern day construction unless specified and we will deal
with the more generally used timbers, dealt with in table A1. These timbers are
known as SC3 and will have C16 stamped on them.
This table is for a dead load of more than 0.25 but not more than 0.50
and allows for an imposed loading of no more than 1.5 kN/sq.m.
This is an abridged representation of Building regulations Table A1
Spacing (distance apart) of joists.
Spacing (distance apart) of joists.
400m
m
450m
m
600m
m
400m
m
450m
m
600m
m
Size of joist
Size of joist
38 x 97
mm
1.72
m
1.56
m
1.21
m
50 x 170
mm
3.61
m
3.47
m
3.08
m
38 x 122
mm
2.37
m
2.22
m
1.76
m
50 x 195
mm
4.13
m
3.97
m
3.5
m
38 x 140
mm
2.72
m
2.59
m
2.17
m
50 x 220
mm
4.64
m
4.47
m
3.91
m
38 x 147
mm
2.85
m
2.71
m
2.33
m
38 x 170
mm
3.28
m
3.1
m
2.69
m
63 x 97
mm
2.19
m
2.08
m
1.82
m
38 x 184
mm
3.53
m
3.33
m
2.9
m
63 x 122
mm
2.81
m
2.7
m
2.45
m
38 x 195
mm
3.72
m
3.52
m
3.06
m
63 x 147
mm
3.37
m
3.24
m
2.95
m
38 x 220
mm
4.16
m
3.93
m
3.42
m
63 x 170
mm
3.89
m
3.74
m
3.4
m
38 x 235
mm
4.43
m
4.18
m
3.64
m
63 x 195
mm
4.44
m
4.28
m
3.9
m
63 x 220
mm
4.91
m
4.77
m
4.37
m
47 x 97
mm
1.92
m
1.82
m
1.46
m
47 x 122
mm
2.55
m
2.45
m
2.09
m
75 x 122
mm
2.97
m
2.86
m
2.6
m
47 x 147
mm
3.06
m
2.95
m
2.61
m
75 x 147
mm
3.56
m
3.43
m
3.13
m
47 x 170
mm
3.53
m
3.4
m
2.99
m
75 x 170
mm
4.11
m
3.96
m
3.61
m
47 x 195
mm
4.04
m
3.89
m
3.39
m
75 x 195
mm
4.68
m
4.52
m
4.13
m
47 x 220
mm
4.55
m
4.35
m
3.79
m
75 x 220
mm
5.11
m
4.97
m
4.64
m
Steel and concrete
•
Steel or concrete are often used to support
loads, in large buildings and for bridges.
•
Steel is good under tension and
compression
•
Concrete is excellent under compression,
and is lighter than steel, but has low
tensile strength
Steel sections
Rounds
Hexagons
Squares
Flats
Equal Leg Angles
Unequal Leg Angles
Channels "H" Beams "I" Beams
Tee Bars Diamond Floor Plate Rectangular Tubing
Seamless and Welded Pipe Seamless and Welded Tubing Half Rounds
Bending moment for a point load
A uniform beam 5m long is supported at its ends
and has a load of 6kN at 3m from one end.
1.
What is the reaction in the supports?
2.
What is the bending moment?
6kN
5m
3m
Calculate reaction in supports
Taking moments about A
we find a CW moment of 3m * 6kN
and a CCW moment of

5m * Rb
As the system is in equilibrium
CW = CCW so
3m * 6kN =

5m * Rb
Rb =

18/5 kN =

3.6kN
Also Ra + Rb =

6kN
Ra =

6kN
–
(

3.6kN) =

2.4kN
Ra =

2.4kN Rb =

3.6kN
6kN
5m
3m
A
B
Ra
Rb
CW = clockwise
CCW = counter clockwise
Bending moment for beam with point load
To find the bending moment at a point p, x metres
from a we take moments about P.
To the left of P:
Reaction Ra causes a CW moment of 2.4x
To the right of P:
The load causes a CW moment of 6 * (3

x)
Reaction Rb causes a CCW moment of 3.6*(5

x)
Total moment = (18

6x)
–
(18

3.6x) =

2.4x
i.e. 2.4x CCW
So the bending moment for this loaded beam at a
distance x from A is
M
x
= 2.4x
.
6kN
5m
3m
A
B
Ra
Rb
3.6kN
2.4kN
p
X m
Bending moment and shear force for point load
Plotting BM vs x we get this graph,
showing BM is a maximum of 7.2 at x =3.
We can also plot the shear force acting
on the beam. Imagine the beam cut at
point p and work out the force acting on
the left side of the cut.
6kN
5m
3m
A
B
Ra
3.6kN
2.4kN
p
X m
Bending moment and shear force for point load
4
3
2
1
0
1
2
3
4
5
6
7
8
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
distance from A
Bending moment
Shear force
Bending moment for a distributed load
A beam 3m long is supported at its ends and
has a distributed load of 2 tonnes/m.
1.
What is the reaction in the supports?
2.
What is the bending moment?
3m
2 tonnes/m
A
B
Calculate reaction in supports
Distributed load of 2 tonnes / m
= 2000kg * 9.81 = 19620N / m
The total weight of the load is
19620N * 3m = 58860N
By symmetry this is evenly split
between the ends, so
Ra = Rb = 588600/2 = 29430N
Ra
Rb
3m
2 tonnes/m
A
B
Bending moment for beam with distributed load
To find the bending moment at a point p, x metres
from a we take moments about P.
To the left of P:
Reaction Ra causes a CW moment of 29430x
The load causes a CCW moment of 19620 * x *x/2
= 9810 x
2
Bending moment M
x
= 29430x

9810 x
2
We can show (by plotting, or by differentiating this
expression and looking for stationary turning points,
then differentiating again) that the maximum bending
moment M! for this loaded beam occurs at the mid
point, where
M! = 29430* 1.5

9810 *1.5*1.5
M! = 44145
–
22072 = 22073 Nm
Ra
Rb
3m
19620N/m
A
B
29430N
p
x
CW = clockwise
CCW = counter clockwise
29430N
Bending moment and shear force
for distributed load
Here you can see that
the bending moment
is a maximum at the
centre, while the
shear force is a
maximum at the ends.
Bending moment and shear force
for distributed load
30000
20000
10000
0
10000
20000
30000
0
0.5
1
1.5
2
2.5
3
Distance from A
BM
SF
TO BE CONTINUED
•
Enough for one week.
•
Next week we will use this information to
evaluate how much stress we will get in a
loaded beam, and how much it will sag.
Calculate maximum stress in beam and
radius of curvature of neutral axis
Suppose our beam from the
previous example was a steel box
girder with dimensions as shown
here.
•
How much would it bend?
•
Would it break?
To answer these questions we
need to calculate the radius of
curvature , and the maximum
stress in the beam.
HOW?
0.2m
0.1m
Thickness
t = 0.02m
/ y = M / I = E / R
Where:
is the strain at y metres from the neutral axis
M is the bending moment
I is the Moment of Inertia for the beam
E is Young’s Modulus of elasticity for the material of the
beam, and
R is the radius of curvature at the neutral axis
Maximum stress
The stress is a maximum at the outer faces (top and bottom) of the
beam, at a distance D/2 from the neutral axis, where D is the depth of
the beam.
Using
/ y = M / I = E / R
!
= M! D / 2 I
Remember we are using ! to signify a maximum value
For our beam
M!
is
22073 Nm
and
I
= 0.1 * 0.2
3
/12
–
0.06 *0.16
3
/ 12
I
= 6.667*10

5

2.048*10

5
I
= 4.619*10

5
m
4
(Formula for calculation of I shown later)
!
=
22073 * 0.2 / 2 * 4.619*10

5
! = 47.8 * 10
6
Pa
(1Pa = 1N/m
2
)
Steel has a tensile strength of 1.5*10
9
N/m
2
so no problem
Radius of curvature
For our steel beam
M!
is
22073 Nm
(calculated)
E = 2.1 * 10
11
N/m2
(tables)
I
= 4.619*10

5
m
4
calculated)
Using
/ y = M / I = E / R
R = I E / M
The radius of curvature is given by
R = (4.619*10

5
)* (2.1 * 10
11
) / 22073
R = 439m
So how much will our beam sag?
How far does the beam sag?
The amount by which the beam sags, x is
found using Pythagoras theorem.
R
2
= a
2
+ (L/2)
2
a
2
= R
2

(L/2)
2
a = sqrt(R
2

(L/2)
2
)
x = R

sqrt(R
2

(L/2)
2
)
Substituting values for R=439m and L=3m
x = 439

sqrt(439
2

(3/2)
2
)
x = 439

sqrt(192721

2.25)
x = 439
–
438.9974 = 2.56 *10

3
m
The beam sags by 2.56mm in the centre
x
R
L/2
a
Moment of Inertia Defined
•
The moment of inertia measures the resistance
to a change in rotation.
–
Change in rotation from torque
–
Moment of inertia
I
=
m * r
2
for a point mass
•
The total moment of inertia
I
is due to the sum of
masses at a distance from the axis of rotation.
N
i
i
i
r
m
I
1
2
Moment of Inertia
More importantly for our purposes the
moment of inertia describes the way in
which the mass of the body is distributed.
As we have already seen we can use this
parameter to evaluate how a regular
isotropic beam will respond to a load.
Mass at a Radius
•
Extended objects can be
treated as a sum of small
masses.
•
A straight rod (
M
) is a set of
identical masses
D
m
.
•
The total moment of inertia is
•
Each mass element
contributes
•
The sum becomes an
integral
axis
length
L
distance
r
to
r+
D
r
2
)
(
D
r
m
I
D
D
D
r
r
L
M
I
r
L
M
m
2
)
/
(
)
/
(
2
3
0
2
)
3
/
1
(
)
3
/
)(
/
(
)
/
(
ML
L
L
M
I
dr
r
L
M
I
L
Rigid Body Rotation
•
The moments of inertia for many shapes
can found by integration.
–
Ring or hollow cylinder:
I
=
MR
2
–
Solid cylinder:
I
= (1/2)
MR
2
–
Hollow sphere:
I
= (2/3)
MR
2
–
Solid sphere:
I
= (2/5)
MR
2
Point and Ring
•
The point mass, ring and
hollow cylinder all have
the same moment of
inertia.
I
=
MR
2
•
All the mass is equally far
away from the axis.
•
The rod and rectangular
plate also have the same
moment of inertia.
I
= (1/3)
MR
2
•
The distribution of mass
from the axis is the same.
M
R
M
R
axis
length
R
length
R
M
M
Moment of Inertia of common shapes
y is the distance of the NA from the top of the shape. ** indicates raised to power e.g. D**3 = D cubed
b
h
b
2
b
1
h
2
h
1
Rectangle
Box
3
12
1
h
b
I
rect
3
2
2
3
1
1
12
1
12
1
h
b
h
b
I
box
Moment of Inertia of a Cross Section
A box girder of dimension 0.2m * 0.1m and 0.02m thickness has a csa of 0.012m
This amount of material would make a solid bar of 0.12m * 0.1m
I(bar) = 1.44 *10

5
m
4
I(box) = 6.66 *10

5
–
1.36 *10

5
= 5.3 *10

5
m
4
Other shapes
•
We can calculate the M of I for other shapes just
by adding or subtracting, as we did for the box
girder. See also next slide.
•
We can evaluate the M of I at a different point
within the shape, using the parallel axis theorem
•
We can evaluate the M of I about an axis in
another plane, using the perpendicular axis
theorem
Comparison of moments of inertia for different shapes
all having the same cross

sectional area
a: Ia =bh**3 / 12 = 1*(1*1*1)/12 = 1/12
Ia = 1/12
b: Ib = I(b outer)

I(a) = (1.414 * 1.414**3) /12
–
1/12
=
4/12
–
1/12 = 3/12
Ib = 3/12
c: Ic = I

I

I +I
Ic =
{
(2*(2**3))

(1*(2**3))

(2*(1**3)) + (1*(1**3))
}
/ 12
Ic =
{
(2*(8))

(1*(8))

(2*(1)) + (1*(1))
}
/ 12
Ic =
{
16

8

2 + 1
}
/ 12 = 7 / 12
Ic = 7/12
a
1m
1m
b
1.41m
c
2m
Parallel Axis Theorem
•
Some objects don’t rotate
about an end, but rather
about the axis at the
center of mass.
•
The moment of inertia
depends on the distance
between axes.
•
The moment of inertia
for a rod about its
center of mass:
2
Mh
I
I
CM
axis
M
h
=
R
/2
2
2
2
2
2
)
12
/
1
(
)
4
/
1
(
)
3
/
1
(
)
2
/
(
)
3
/
1
(
MR
I
MR
MR
I
R
M
I
MR
CM
CM
CM
Perpendicular Axis Theorem
For flat objects the
rotational moment of
inertia of the axes in the
plane is related to the
moment of inertia
perpendicular to the
plane.
y
x
z
I
I
I
M
I
x
= (1/12)
Mb
2
I
y
= (1/12)
Ma
2
a
b
I
z
= (1/12)
M(a
2
+
b
2
)
Top view
Side view
To Increase the Moment of Inertia
•
Increase the size:
–
But as you increase the size, you increase the weight
and cost
•
Use a denser material
–
But it may not have the properties you need and it will
be heavier
•
Change the cross

sectional shape:
–
A hollow cross

section is stronger for the amount of
material used
Summary
We can now work out the stress in a loaded beam, such as a
table top or beam bridge. We can find out whether the
structure will be strong enough, and how much it will deform
under load.
Question.
A bridge is made of reinforced
concrete using pre

stressed
reinforcing rods. Should the rods be
placed at the top, middle or bottom of
the section?
WHY?
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