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Structural Mechanics I
CE 203

KFUPM

023
Summary of lecture 41

45
Beam Deflections:
Computations of Linear and Angular Deformations Using Singularity Functions.
Introduction:
In any structural system beams are mainly subjected to transverse load (normal t
o the
x

axis). The bending action is the main internal effect in these structural members.
Under the effect of internal bending action M(x), a structural beam will deform (i.e.:
deflect by linear deformation and angular deformations:
v(x);
x
). The beam shown
in Fig. 1 below shows the typical deformation of a simple beam.
The beam is in the x

y plane and the deformations are mainly displacement in the y

direction and rotation about the z

axis. The deformations
v(x);
x
at any position
al
ong the x

axis
are both measured from the centroidal axis.
Fig. 1
:
Moment

curvature equation: for small deformations (dy/dx)
2
is neglected, then the
curvature of the deformated shape is mainly
d
2
v/dx
2
, and the
Moment

curvature
equation is:
EI d
2
v/dx
2
= M(x),
… (1)
it is found by differentiations that
EI d
3
v/dx
3
= V(x),
… (2)
EI d
4
v/dx
4
=
q
e
(x)
… (3)
where:
q
e
(x)
is a uniform load ( i.e. force per unit length).
Solution of equation (1) by integration becomes complicated if more than expression is
required for M(x). Example 1 below shows the solution of a typical beam problem using the
method of integration.
The solution shows the use of support and continuity conditions to solve
for all the integration constants. The use of singularity functions (special functions) becomes
valuable to reduce the efforts to solve a problem that requires several beam segments t
o
express bending moment M(x).
Final position
y;
v
(x)
x
v(x);
x
2
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The Method of Singulartiy Functions:
Background for the Method:
The method of singularity functions reduces the efforts to determine the bending
deflections of beams. Since the moment

curvature equation [2
nd
order ordinary
differential equation] requires that the bending moment expression M(x) be
continuous, a beam with N
s
segments where M(x) is continuous only within each
segment, the moment curvature equation has to be written for each segment. The
method o
f singularity functions overcomes the requirements of continuous M(x) and
one expression M
e
(x) is written for the whole span. Therefore only one moment

curvature is written for the beam.
For a general loading that may consist of: a) concentrated loads and
moments; b)
distributed loads and moment (i.e.: uniform, linear, etc….), the load distributed load
function q
e
(x) is used to replace q(x). Then once the equivalent distribute load is
known, equation (3) is integrated twice to obtain the moment curvature eq
uation given
by equation (1). The equivalent loads in terms of singularity functions for different
types of load distributions for beams of length
L
each are as follows:
a)
Uniform load: q(x)=q
0
(x)
x=0
q
e
(x) =

q
0
< x
–
0>
b)
Linear load:
q
e
(x
) =

q
0
( x
–
0)/ L
q
e
(x) =

q
0
< x
–
0> / L
c)
Load not covering the whole span: As the singularity function < x

a>
n
starts
from a and end at +
∞, it is important to make sure the function used is valid for
a given problem when the load is
discontinuous
as shown below. The load 4
kN/m is valid only between x=L
1
and x= L
1
+ L
2
, where x extends from 0 to
L
1
+L
2
+L
3
. For the given load shown below the e
xpression of equivalent
uniform load q
e
<x> is obtained by superposition of two load cases as follows:
q
e
<
x
>
=

4
<
x

L
1
>
+ 4
<
x

(L
1
+ L
2
)
>
.
L
L
1
L
2
L
3
4 kN/m
x
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Expression for Equivalent Uniform Load q
e
(x):
Notes: Upward load is considered positive and equivalen
t load is written using Singularity
functions.
Moment

Curvature Equation
:
)
(
2
2
x
M
dx
d
EI
v
…
(1)
From eqn. (1) we have by (differentiation)
)
(
3
3
x
V
dx
d
EI
v
… (2)
)
(
4
4
x
q
dx
d
EI
e
v
….
(3)
Equation (3) can be used to obtain
q
e
(
x
)
[
force per unit length
] for:
uniform
loads;
linear
loads;
concentrated
loads and
concentrated
moments using the
singularity
function
procedure.
a
.
Uniform load
:
2
/
)
0
(
)
(
2
x
q
x
M
o
q
2
/
0
)
(
2
x
q
x
M
o
q
o
o
e
x
q
x
q
0
)
(
q
0
b
.
Linear loads
:
l
x
q
x
q
o
/
)
(
l
x
q
o
/
)
0
(
3
2
)
0
(
1
)
(
x
x
x
q
l
x
M
o
q
or
l
x
q
x
q
o
e
/
0
)
(
1
q
0
l
c
.
Concen
trated load
:
)
0
(
)
(
x
l
b
P
x
M
o
x
<
a P
0
)
(
)
0
(
)
(
a
x
P
x
l
b
P
x
M
o
o
x
>
a a b
x
1
1
1
)
(
0
)
(
b
a
x
R
a
x
P
x
R
x
q
B
o
A
tot
e
)
(
0
b
a
x
d
.
Concentrated moment
:
M
0
2
a
x
M
q
o
o
M
e
a
b
A
B
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