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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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3

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Learning

is a trademark used herein under license.
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Learning

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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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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
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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Learning

is a trademark used herein under license.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
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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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
7




Fig. 4
-
3
(c)
-
(d)


Beam
-
to
-
column
connection (pin support).



Fig. 4
-
3




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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
8



Fig. 4
-
3
(e)
-
(f)


Pole anchored to
concrete pier (fixed support).




Fig. 4
-
3



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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
9

Photographs of actual construction










10





















11

Types of loads













©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
Concentrated load

Distributed load

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Learning

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Learning

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12





















©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
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
:





©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
+

+

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,








©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.

B


A


q


P

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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.

q


©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.

B


A


q


P


P

+

16





















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Learning

is a trademark used herein under license.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.
M

V



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Learning

is a trademark used herein under license.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson
Learning

is a trademark used herein under license.

B


A


q


P


P

4
PL

M

2
P

2
P


V

+

+

17






















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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.

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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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
19

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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.
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(x-x
0
)
1
x

(x-x
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(x-x
0
)
1
x-x
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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Learning

is a trademark used herein under license.
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Learning

is a trademark used herein under license.

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