LECTURE
Third Edition
BEAMS: DEFORMATION BY
SINGULARITY FUNCTIONS
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
17
Chapter
9.5 – 9.6
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 1
ENES 220 ©Assakkaf
Introduction
– The integration method discussed earlier
becomes tedious and time-consuming
when several intervals and several sets of
matching conditions are needed.
– We noticed from solving deflection
problems by the integration method that
the shear and moment could only rarely
described by a single analytical function.
1
Slide No. 2
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Introduction
– For example the cantilever beam of Figure
9a is a special case where the shear V and
bending moment M can be represented by
a single analytical function, that is
V ( x ) = w(L − x )
and
(15a)
(
M ( x ) = w − L2 + 2 Lx − x 2
)
(15b)
Slide No. 3
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Introduction
Figure 9
y
ENES 220 ©Assakkaf
P
w
x
y
L/4
L/2
w
L
x
L
(a)
(b)
2
Slide No. 4
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Introduction
– While for the beam of Figure 9b, the shear
V or moment M cannot be expressed in a
single analytical function. In fact, they
should be represented for the three
intervals, namely
0 ≤ x ≤ L/4,
L/4 ≤ x ≤ L/2, and
L/2 ≤ x ≤ L
Slide No. 5
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Introduction
– For the three intervals, the shear V and the
bending moment M can are given,
respectively, by
y
P
w

wL
P+
for 0 ≤ x ≤ L / 4

2

wL

V ( x) = 
for L/4 ≤ x ≤ L / 2
2

 wL − w x − L  for L / 2 ≤ x ≤ L
 2
2

x
L/4
L/2
L
3
Slide No. 6
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Introduction
y
and
ENES 220 ©Assakkaf
P
w
x
L/4
L/2
L
 PL 3wL2
wL
−
+ Px +
x
 −
4
8
2

3wL2 wL

M ( x) = 
−
+
x
8
2

2
2
w
L
 3wL wL
−
+
x
−
x
−



8
2
2
2

for 0 ≤ x ≤ L / 4
for L/4 ≤ x ≤ L / 2
for L / 2 ≤ x ≤ L
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 7
ENES 220 ©Assakkaf
Introduction
– We see that even with a cantilever beam
subjected to two simple loads, the
expressions for the shear and bending
moment become complex and more
involved.
– Singularity functions can help reduce this
labor by making V or M represented by a
single analytical function for the entire
length of the beam.
4
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 8
ENES 220 ©Assakkaf
Basis for Singularity Functions
– Singularity functions are closely related to
he unit step function used to analyze the
transient response of electrical circuits.
– They will be used herein for writing one
bending moment equation (expression)
that applies in all intervals along the beam,
thus eliminating the need for matching
equations, and reduce the work involved.
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 9
ENES 220 ©Assakkaf
Definition
A singularity function is an expression for x
n
written as x − x0 , where n is any integer
(positive or negative) including zero, and x0
is a constant equal to the value of x at the
initial boundary of a specific interval along
the beam.
5
Slide No. 10
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Properties of Singularity Functions
– By definition, for n ≥ 0,
x − x0
n
( x − x0 )n
=
0
when x ≥ x0
when x < x0
(16)
– Selected properties of singularity functions
that are useful and required for beamdeflection problems are listed in the next
slides for emphasis and ready reference.
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 11
ENES 220 ©Assakkaf
Selected Properties
x − x0
n
x − x0
0
( x − x0 )n
=
0
when n > 0 and x ≥ x0
1
=
0
when n > 0 and x ≥ x0
when n > 0 and x < x0
when n > 0 and x < x0
(17)
(18)
6
Slide No. 12
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Integration and Differentiation of
Singularity Functions
∫
n
x − x0 dx =
d
x − x0
dx
n
1
x − x0
n +1
= n x − x0
n +1
+ C when n > 0
(19)
when n > 0
(20)
n −1
Slide No. 13
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Examples: Singularity Functions
y
y = x −1
y
1
y=
2
1
x
4
1
2
2
1
x
-1
1
(a)
2
3
x
-1
Figure 10
1
2
3
(b)
7
Slide No. 14
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Examples: Singularity Functions
y
y = 2 x−2
y
1
y = x +1 − x −1
0
−2 x−2
2
1
1
0
2
1
x
-1
1
(c)
2
3
x
-1
Figure 10
1
2
(e)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
3
Slide No. 15
ENES 220 ©Assakkaf
Application of Singularity Functions in
Developing a Single Equation to
Describe the Bending Moment
– By making use of singularity functions
properties, a single equation (expression)
for the bending moment for a beam can be
obtained.
– Also, the corresponding value of M in any
interval can be computed.
8
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 16
ENES 220 ©Assakkaf
Application of Singularity Functions in
Developing a Single Equation to
Describe the Bending Moment
– To illustrate this, consider the beam of the
following figure (Fig.11).
– The moment equations at the four
designated sections are written as shown
in the following slide.
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 17
ENES 220 ©Assakkaf
Applications of Singularity Functions in
Developing a Single Equation to
Describe the Bending Moment
M 1 = RL x
M 2 = RL x − P(x − x1 )
M 3 = RL x − P(x − x1 ) + M A
M 4 = RL x − P(x − x1 ) + M A −
for 0 < x < x1
for x1 < x < x2
for x2 < x < x3
(21)
w( x − x3 )
for x2 < x < xL
2
9
Slide No. 18
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
y
ENES 220 ©Assakkaf
Application of Singularity Functions in
Developing a Single Equation to
Describe the Bending Moment
P
1
MA
2
Figure 11
w
3
x
x1
4
x2
x3
RL
RR
L
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 19
ENES 220 ©Assakkaf
Application of Singularity Functions in
Developing a Single Equation to
Describe the Bending Moment
– These for moment equations can be
combined into a single equations by means
of singularity functions to give
M (x ) = RL x − P x − x1 + M A x − x2
1
0
−
w
x − x3
2
2
for 0 < x < L
(22)
Where M(x) indicates that the moment is a function of x.
10
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 20
ENES 220 ©Assakkaf
Typical Singularity Functions
y
P
1
MA
2
w
3
x
x1
RL
4
x2
RR
x3
L
M ( x ) = RL x − P x − x1 + M A x − x2
1
−
w
x − x3
2
2
0
for 0 < x < L
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
(22)
Slide No. 21
ENES 220 ©Assakkaf
Notes on Distributed Loads
– When using singularity functions to
describe bending moment along the beam
length, special considerations must be
taken when representing distributed loads,
such as those shown in Figure 12.
– The distributed load cannot be represented
by a single function of x for all values of x.
11
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 22
ENES 220 ©Assakkaf
Notes on Distributed Loads
– The distributed loading should be an openended to the right (Figure 11) of the beam
when we apply the singularity function.
– A distributed loading which does not
extend to the right end of the beam or
which is discontinuous should be replaced
as shown in Figures by an equivalent
combination of open-ended loadings.
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 23
ENES 220 ©Assakkaf
Notes on Distributed Loads
– Examples of open-ended-to-right
distributed loads, which are ready for
singularity function use are shown in
Figure 12.
– Examples of non open-ended-to-right or
discontinuous distributed loads are shown
in Figures 13, 14, and 15.
12
Slide No. 24
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
y
Moment due to Distributed Loads
w0
w0
y
x
x1
2
M w0 = −
x
x1
L
w0
x − x1
2
w0
y
x
x1
L
M w0 = −
ENES 220 ©Assakkaf
L
w0
x − x1
6(L − x1 )
3
M w0 = −k x − x1
n+2
Figure 12. Open-ended-to-right distributed loads
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 25
ENES 220 ©Assakkaf
Moment due to Distributed Loads
y
w0
1
x
x1
The moment at section 1
due to distributed load
alone is
L
Figure 13

y
w0
M w0 = −
-w0
x1
x
w0
w
2
x − 0 + 0 x − x1
2
2
2
L
13
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Moment due to Distributed Loads
y
w0
x1
1
x
x2
L
Figure 14

y
x1
w0
The moment at section 1
due to distributed load
alone is
w
w
2
M w0 = − 0 x − x1 + 0 x − x2
2
2
x2
L
Singularity Functions
„
2
x
-w0
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
y
Slide No. 26
ENES 220 ©Assakkaf
Slide No. 27
ENES 220 ©Assakkaf
Moment due to Distributed Loads
w0
x1
w1 L − x1
w (L − x1 )
=
⇒ w1 = 0
w0 x2 − x1
x2 − x1
x
The moment at section 1
due to distributed load
alone is
1
x2
L

y
w0
w1
M w0 = −
x
x1
-w0
-w1
x2
L
w0
w0
3
x − x1 +
x − x2
6(x2 − x1 )
6(x2 − x1 )
+
w0 x − x2
3
2
2
Figure 15
14
Slide No. 28
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Moment due to Distributed Loads
Note that in Fig. 14, the linearly varying
load at any point x ≥ x1 is
w0
x1
x2
w
w=
w0 ( x − x1 )
x2 − x1
The moment of this load for
Any point x ≥ x1 is

1  w ( x − x1 )
(x − x1 ) x − x1  = w0 (x − x1 )3
M =−  0
2  x2 − x1
 3  6( x2 − x1 )
x
From similar triangles :
w
x − x1
=
w0 x2 − x1
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 29
ENES 220 ©Assakkaf
Example 4
Use singularity functions to write the
moment equation for the beam shown in
Figure 16. Employ this equation to obtain
the elastic curve, and find the deflection at
x = 10 ft. y
2000 lb/ft
Figure 16
E = 29 × 106 psi
I = 464 in
x
2
4 ft
11 ft
5 ft
15
Slide No. 30
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
ENES 220 ©Assakkaf
Example 4 (cont’d)
„
y
2000 lb/ft
x
11 ft
4 ft
5 ft
R1
+
∑M
R2
2
Find the reactions:
= 0; R1 (16 ) − 2000(15)(5 +
15
)=0
2
∴ R1 = 23,437.5 lb
+ ↑ ∑ Fy = 0; R1 + R2 − 2000(15) = 30,000
∴ R2 = 6,562.5 lb
Slide No. 31
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 4 (cont’d)
A single expression for the bending
moment can be obtained using the
singularity functions:
y
2000 lb/ft
x
4 ft
R1
11 ft
5 ft
M (x ) = −
R2
2
2000( x ) 2000 x − 15
+
2
2
2
+ 23,437.5 x − 4
1
(23a)
16
Slide No. 32
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 4 (cont’d)
The elastic curve is found by integrating
Eq. 23 twice
2000 x 2 2000 x − 15
EIy′′ = M (x ) = −
+
2
2
2
+ 23,437.5 x − 4
3
EIy′ = EIθ = −2000
23,437.5 x − 4
x 3 2000 x − 15
+
+
6
6
2
x 4 2000 x − 15
EIy = −2000 +
24
24
4
+
1
2
+ C1 (23b)
3
23,437.5 x − 4
+ C1 x + C2
6
Slide No. 33
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
(23c)
ENES 220 ©Assakkaf
Example 4 (cont’d)
Boundary conditions:
y = 0 at x = 4 ft and at x = 20 ft
EIy (4) = 0 = −2000
(4) 4 2000 4 − 15
+
24
24
+
23,437.5 4 − 4
6
2000 20 − 15
( 20)
+
24
24
4
∴ 20C1 + C2 = −2,718,750
3
+ C1 (4) + C2
0
0
∴ 4C1 + C2 = 21,333.3
EIy (20) = 0 = −2000
4
4
+
23,437.5 20 − 4
6
(23d)
3
+ C1 (20) + C2
(23e)
17
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 34
ENES 220 ©Assakkaf
Example 4 (cont’d)
From Eqs 23d and 23e, the constants of
integrations are found to
C1 = 168,588.54
and
C2 = −653,020.88
Therefore, the elastic curve is given by
y ( x) =
1 
( x) 4 2000 x − 15
+
− 2000
24
24
EI 

4
+
3
23,437.5 x − 4
6

+ 168,588.5 x − 653,020.9

(23f)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 35
ENES 220 ©Assakkaf
Example 4 (cont’d)
The deflection at any point along the beam
can be calculated using Eq. 23f (elastic
curve equation). Therefore, the deflection
0
y at x = 10 ft is

(10) 4 2000 10 − 15
+
− 2000
24
24

4
23,437.5 10 − 4
3

+ 168,588.5(10) − 653,020.9

y (10) =
1
EI
y (10) =
(12)2 [− 833,333.3 + 843,750 + 1,685,885 − 653,020.9]
29 × 106 (464 )
+
6
y10′′ = 0.011165 ft = 0.1339 in
18
Slide No. 36
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 5
A beam is loaded and supported as
shown in Figure 17. Use singularity
functions to determine, in terms of M, L,
E, and I,
a) The deflection at the middle of the span.
b) The maximum deflection of the beam.
Slide No. 37
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 5 (cont’d)
y
M
E, I
A
x
C
B
L
L
Figure 17
19
Slide No. 38
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
ENES 220 ©Assakkaf
Example 5 (cont’d)
„
First find the reactions RA and RB:
y
M
x
A
E, I
L
RA
+
C
B
L
RB
∑M
C
= 0; RA (2 L ) + M = 0
M
2L
+ ↑ ∑ Fy = 0; RA + RB = 0
∴ RA = −
∴ RB =
M
2L
Slide No. 39
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 5 (cont’d)
Singularity functions to describe the
bending moment:
y
M
A
R
A
x
L
Mx
+M x−L
2L
or
M ( x) = −
E, I
C
B
L
R
B
EIy′′ = −
Mx
+M x−L
2L
0
0
(24a)
(24b)
20
Slide No. 40
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 5 (cont’d)
Integrating Eq. 24.b twice, we get
Mx
0
EIy′′ = −
+M x−L
2L
Mx 2
1
(24c)
EIy′ = −
+ M x − L + C1
4L
2
Mx 3 M x − L
(24d)
EIy = −
+
+ C1 x + C2
12 L
2
Slide No. 41
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
ENES 220 ©Assakkaf
Example 5 (cont’d)
Boundary conditions:
At x = 0, y = 0
At x = 2L, y = 0
Thus,
Therefore: C2 = 0
Therefore: C1= ML/12
[
M
2
− x 3 + 6 L x − L + L2 x
48EIL
M
y ( L) =
− L3 + L3 = 0
48EIL
y=
[
]
]
(24e)
(24f)
21
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6)
Singularity Functions
„
Slide No. 42
ENES 220 ©Assakkaf
Example 5 (cont’d)
(b) Finding the maximum deflection:
y′ =
[
M
1
− 3 x 2 + 12 L x − L + L2
12 EIL
]
ymax when y′ = 0
1
Therefore : − 3 x 3 + 12 x − L + L2 = 0 ⇒ x = L / 3
(
)
ymax = y L / 3 =
=
M  L3
L3 
2 ML2
+
−
=
12 EIL  3 3
3  36 3EI
3ML2
54 EI
22