1
Chapter 4
Shear Forces and Bending Moments
In chapter 4
Statically determinate
structures only.
Types of Beams, Loads and Reactions
Fig. 4

1
Beams subjected to
lateral
loads
(concentrated load, moment and
uniform
ly
distributed load).
Beam deflectio
ns
Shown by the
dashed lines.
2
Fig. 4

1
Examples of beams subjected to lateral loads.
Types of beams
Fig. 4

2
(a)
Simply supported beam
or simple beam.
Reactions
Indicated by
slashes
across
the arrows.
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Fig. 4

2
Types of be
ams: (a) simple beam.
Pin support at
A
:
Rotation
0
Vertical movement = 0
R
A
Horizontal movement = 0
H
A
Roller support at
B
:
Rotation
0
Vertical movement = 0
R
B
Horizontal movement
0
4
Fig. 4

2
(b)
Cantilever beam.
Fig. 4

2
Types of beams
: (b) cantilever beam.
Fixed support at
A
:
Rotation = 0
M
A
Vertical movement = 0
R
A
Horizontal movement = 0
H
A
Free end at
B
:
Rotation
0
Vertical movement
0
Horizontal movement
0
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5
Fig. 4

2
(c)
Beam with an overhang.
Fig. 4

2
Types of be
ams: (c) beam with an overhang.
Overhanging segment
BC
Similar to
a cantilever beam, except it may rotate
at point
B
.
Pin support at
A
and roller support at
B
.
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6
Actual construction
Fig. 4

3
(a)

(b)
Beam supported on
concrete wall (roller support).
Fig. 4

3
Slotted hole
Horizontal movement
0
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7
Fig. 4

3
(c)

(d)
Beam

to

column
connection (pin support).
Fig. 4

3
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Fig. 4

3
(e)

(f)
Pole anchored to
concrete pier (fixed support).
Fig. 4

3
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9
Photographs of actual construction
10
11
Types of loads
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Concentrated load
Distributed load
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Moment
Concentrated load
13
Shear Forces and Bending Moments
Sign conventions
Fig. 4

5
Sign conventions for shear
force
V
and bending moment
M
.
Fig. 4

5
Sign conventions for shear force
V
and bending moment
M
.
Sign for shear
V
:
Sign for moment
M
:
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+
+
14
Relationships between Loads
(q)
,
Shear Forces
(V)
and Bending
Moments
(M)
V
dx
dM
q
dx
dV
This equation shows
The rate of
change of the bending moment at any
p
oint on the axis of a beam is equal to
the shear force at that same point.
Shear

Force and Bending Moment
Diagrams
Generally, the maximum and minimum
values of the shear forces and bending
moments are needed when designing a
structure.
15
Principle of sup
erposition
When
several loads act on a structure, the
shear

force and bending

moment
diagrams can be obtained by
summation of the diagrams obtained
for each of the loads acting separately.
Equal,
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B
A
q
P
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q
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B
A
q
P
P
+
16
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M
V
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B
A
q
P
P
4
PL
M
2
P
2
P
V
+
+
17
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B
A
q
P
M
V
18
Ex
ample
(P
rob
. 4.3

9)
A curved bar
ABC
is subjected to loads
in the form of two equal and opposite
forces
P
, as shown in the figure. The axis
of the bar forms a semicircle of radius
r
.
Determine the axial force
N
, shear force
V
and bending moment
M
acting
at a
cross section defined by the angle
.
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19
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O
r
Solution
Free

body diagram
20
Efficient Modeling of Loads: the
Dirac Delta Function
As
x
0,
q
behaves as follows:
1.
highly concentrated (
q
) at
x
= 9;
2.
q
= zero
elsewhere;
3.
9
9
1
x
x
qdx
qdx
x = 9
P = 1 k
q = 100 k/ft (
q
x =
k
8.95
9.05
x = 0.01
q = 500 k/ft for
x
within
(8.999,9.001)
,
keeping
qdx
q
x =
9.001
8.999
x =
0.002
equivalent load density:
(point load)
or
21
Such a localized load density
is called
a Delta
function, denoted as:
q
(
x
) =
(
x
–
9)
In general,
if
a function
= 0
everywhere
except at
x
=
x
0
,
but its integral is 1, it is
a
(Dirac) Delta function:
(
x
–
x
0
) = 0 if
x
x
0
, while
1
)
(
0
dx
x
x
(
x
)
can
also be thought of as the derivative of the step
function
H
(
x
)
:
H
(
x
) = 1 if
x
> 0;
H
(
x
) = 0 if
x
< 0
(x) =
d
H
(x) / dx
22
Picture:
Check
: integral property is satisfied:
0
0
0
0
1
0
1
x
x
x
x
H
H
dH
dx
dx
dH
1
H(x)
x
0
H =
1
as long
as
x
= 0 is
contained in
x
ut
x
,
hence
H /
x
blows up in here
H /
x
= 0
H /
x
= 0
23
Important
integrals
for
and
H
Property 1:
x
x
x
H
dx
x
x
0
0
0
)
(
'
)
'
(
Picture:
Property 2:
x
x
0
H(xx
0
)
1
x
(xx
0
)
x
0
“
The narrow
but
high
area of 1 is
picked up
by
integration
only if
we
integrate past
x
0
”
24
x
x
x
H
x
x
dx
x
x
H
0
0
0
0
)
(
)
(
'
)
'
(
Picture:
“
shaded area is picked up only if
x
>
x
0
“
Now, to model the total e
ffect of
Point loads P
1
, P
2
, … applied at x
1
, x
2
,
…, respectively,
and
Distributed load(s)
w
(x),
We just need one “total load density”:
w
total
=
w
(x) +
P
i
(x
–
x
i
)
x
0
x
H(xx
0
)
1
xx
0
H(x’
–
x
0
)dx’
25
Then,
simply integrate w
total
twice to get
M(x).
** no need to cut sections, even fo
r
many point loads!! **
Don’t forget
(
from
CIVL 113)
:
Reaction R
A
(upward = +ve) at x = 0
contributes V
0
= R
A
(add it to V(x) after
integrating
–
w).
E
ach concentrated moment
M
i
(clockwise = +ve) at
x
=
x
i
will
add
M
i
H
(
x
–
x
i
) to
M
(
x
)
in addition to
x
dx
x
V
0
'
)
'
(
26
Example 1:
R
A
=
23.59k
R
B
=
2
1
.
41
k
27
You can also use
CAS (e.g.
Mathematica
) for Example 1
. In
either case, you can efficiently obtain V and M diagrams:
28
29
Using delta function to get
BM
diagram
:
Example
2
:
Only 3 simple steps needed; no
need to
cut sections!
(1)
Define the loading (
㴠⭶攩=
w(x)
= 3.2
–
10.8*
(
x
–
3) (kN/m)
This
can
also
be done by Mathematica as follows:
w=3.2

10.8*DiracDelta[x

3]
(2) Integrate
–
w(x) to get V(X),
remembering constants of
integration:
Recall: R
A
= 3.6kN
at left end constitutes a
+ve shear, V
0
= 3.6kN, to be added to
integral of
–
w(x), i.e.
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R
A
=
3.6KN,
R
B
=
10.8kN
R
A
R
B
30
X
dx
w
V
X
V
0
0
)
(
)
(
X
X
dx
x
dx
0
0
)
3
(
8
.
10
2
.
3
6
.
3
V
(
X
) = 3.6
–
3.2
X
+ 10.8
H
(
X
–
3)
Y
ou can
also
use
CAS
to integrate:
e.g. in Mathematica
syntax:
Integrate[Fn[x],{
x,x1,x2}]:
v=3.6+Integrate[

w,{x,0,X}]
(3) Integrate V(X) to get M(x),
remembering constants of integration
(if any):
x
dX
X
V
x
M
0
)
(
)
(
(with M
0
= 0 at pinned end)
=
x
0
3.6
–
3.2
X
+ 10.8
H
(
X
–
3) d
X
M
(x) = 3.6x

1.6x
2
+ 10.8(x

3)H(x

3)
This
can
be
plot
ted for
the BM diagram.
31
T
he integration of V can
also
be done by Mathematica
using the following syntax
:
m=Integrate[v,{X,0,x}]
Note: to get M
max
or M
min
, check at the x
values where
1.
dM/dx = 0, i.e.V = 0 (by solving
algebraic equation for V
x = 3.6/3.2 = 1.125),
and
2.
V is discontinuous (x = 3)
Maximum values of M are:
M
x = 1.125
= 2.025 kN
m
= M
pos
M
x = 3
=

3.6 kN
m = M
neg
Mathematica output
:
32
33
Property 3 (Step function):
x
0
x
0
0
0
)
(
)
x
(
)
(
)
(
x
dx
x
f
x
H
dx
x
x
H
x
f
Picture:
f(x)
H(x  x
0
)
x
0
x
0
f(x)H(x  x
0
)
=
x
times
34
Step function
s
:
also useful for
modeling
“piecewise” loads
:
a
distributed load f(x)
that
only exists
for
x
1
< x < x
2
is given by
f(x)
[
H(x
–
x
1
)
–
H(x
–
x
2
)
]
Picture:
x
1
minus
H(x  x
1
)
x
2
H(x  x
2
)
x
2
x
1
=
=
times
f(x)
x
1
x
2
x
1
x
2
35
Example 3
:
From CIVL 113: effect of the c
oncentrated
ccw
mm
’
t
M
0
=
–
12 at x =
12
: adds

12 H(x
–
12) to M(x) expression
36
37
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