BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS - Assakkaf
Transcript of BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS - Assakkaf
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• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
Third EditionLECTURE
179.5 – 9.6
Chapter
BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS
byDr. Ibrahim A. Assakkaf
SPRING 2003ENES 220 – Mechanics of Materials
Department of Civil and Environmental EngineeringUniversity of Maryland, College Park
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 1ENES 220 ©Assakkaf
Singularity Functions
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.
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 2ENES 220 ©Assakkaf
Singularity Functions
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
( ) ( )
( ) ( )22 2 and
xLxLwxM
xLwxV
−+−=
−= (15a)
(15b)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 3ENES 220 ©Assakkaf
Singularity Functions
Introduction
x
y
w
y
xwPFigure 9
(a)(b)L
LL/2
L/4
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 4ENES 220 ©Assakkaf
Singularity Functions
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, andL/2 ≤ x ≤ L
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 5ENES 220 ©Assakkaf
Singularity Functions
Introduction– For the three intervals, the shear V and the
bending moment M can are given, respectively, by
≤≤
−−
≤≤
≤≤+
=
LxLLxwwL
LxLwL
LxwLP
xV
2/for 22
2//4for 2
4/0for 2
)(
y
x
wP
LL/2L/4
4
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 6ENES 220 ©Assakkaf
Singularity Functions
Introductionand
≤≤
−−+−
≤≤+−
≤≤++−−
=
LxLLxwxwLwL
LxLxwLwL
LxxwLPxwLPL
xM
2/for 2228
3
2//4for 28
3
4/0for 28
34
)(22
2
2
y
x
wP
LL/2L/4
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 7ENES 220 ©Assakkaf
Singularity Functions
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.
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 8ENES 220 ©Assakkaf
Singularity Functions
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) Slide No. 9ENES 220 ©Assakkaf
Singularity Functions
DefinitionA singularity function is an expression for xwritten as , where n is any integer (positive or negative) including zero, and x0is a constant equal to the value of x at the initial boundary of a specific interval along the beam.
nxx 0−
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 10ENES 220 ©Assakkaf
Singularity Functions
Properties of Singularity Functions– By definition, for n ≥ 0,
– Selected properties of singularity functions that are useful and required for beam-deflection problems are listed in the next slides for emphasis and ready reference.
( )
<≥−
=−0
000 when 0
when xxxxxxxx
nn
(16)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 11ENES 220 ©Assakkaf
Singularity Functions
Selected Properties
( )
<>≥>−
=−0
000 and 0 when 0
and 0 when xxnxxnxxxx
nn
<>≥>
=−0
000 and 0 when 0
and 0 when 1xxnxxn
xx
(17)
(18)
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 12ENES 220 ©Assakkaf
Singularity Functions
Integration and Differentiation of Singularity Functions
0 when C1
1 100 >+−
+=− +
∫ nxxn
dxxx nn
0 when 100 >−=− − nxxnxx
dxd nn
(19)
(20)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 13ENES 220 ©Assakkaf
Examples: Singularity Functions
x
y
1 2 3-1
2
41 xy =
2
1
x
y
1 2 3-1
11−= xy
2
1
Figure 10(a) (b)
Singularity Functions
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 14ENES 220 ©Assakkaf
Examples: Singularity Functions
x
y
1 2 3-1
0
11
22
11
−−
−−+=
x
xxy
2
1
Figure 10(c) (e)
x
y
1 2 3-1
022 −= xy
2
1
Singularity Functions
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 15ENES 220 ©Assakkaf
Singularity Functions
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.
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 16ENES 220 ©Assakkaf
Singularity Functions
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) Slide No. 17ENES 220 ©Assakkaf
Singularity Functions
Applications of Singularity Functions in Developing a Single Equation to Describe the Bending Moment
( )( )
( ) ( )LAL
AL
L
L
xxxxxwMxxPxRM
xxxMxxPxRMxxxxxPxRMxxxRM
<<−
−+−−=
<<+−−=<<−−=<<=
23
14
3213
2112
11
for 2
for for
0for
(21)
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 18ENES 220 ©Assakkaf
Singularity Functions
Application of Singularity Functions in Developing a Single Equation to Describe the Bending Moment
1 2 3
4
MAP
w
RRRL x2
x3L
x1
x
y
Figure 11
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 19ENES 220 ©Assakkaf
Singularity Functions
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
Where M(x) indicates that the moment is a function of x.
( ) LxxxwxxMxxPxRxM AL <<−−−+−−= 0for 2
23
02
11 (22)
11
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 20ENES 220 ©Assakkaf
Singularity Functions
Typical Singularity Functions
( )
LxxxwxxMxxPxRxM AL
<<−−
−+−−=
0for 2
23
02
11
1 2 3
4
MAPw
RRRLx2
x3
L
x1
x
y
(22)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 21ENES 220 ©Assakkaf
Singularity Functions
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.
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 22ENES 220 ©Assakkaf
Singularity Functions
Notes on Distributed Loads– The distributed loading should be an open-
ended 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) Slide No. 23ENES 220 ©Assakkaf
Singularity Functions
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.
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 24ENES 220 ©Assakkaf
Singularity Functions
Moment due to Distributed Loads
x
y
x1
L
w0
x
y
x1
L
w0
x
y
x1
L
w0
Figure 12. Open-ended-to-right distributed loads
21
0
20xxwMw −−=
( )3
11
0
60xx
xLwMw −−
−= 210
+−−= nw xxkM
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 25ENES 220 ©Assakkaf
Singularity Functions
Moment due to Distributed Loads
x
y
x1
L
x
y
Lx1
w0
w0
-w0
1
The moment at section 1due to distributed loadalone is
21
020
20
20xxwxwMw −+−−=
Figure 13
14
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 26ENES 220 ©Assakkaf
Moment due to Distributed Loads
x
y
x2
L
x
y
Lx2
w0
w0
-w0
x1
x1
The moment at section 1due to distributed loadalone is
22
021
0
220xxwxxwMw −+−−=
1
Figure 14
Singularity Functions
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 27ENES 220 ©Assakkaf
Moment due to Distributed Loads
The moment at section 1due to distributed loadalone is
( ) ( )
2
662
20
32
12
031
12
00
xxw
xxxx
wxxxx
wMw
−+
−−
+−−
−=
( )12
101
12
1
0
1
xxxLww
xxxL
ww
−−
=⇒−−
=x
y
x2
L
x
y
Lx2
w0 w1
-w0
x1
x1
1
-w1
w0
Figure 15
Singularity Functions
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 28ENES 220 ©Assakkaf
Singularity Functions
Moment due to Distributed LoadsNote that in Fig. 14, the linearly varying load at any point x ≥ x1 is
12
10 )(xxxxww
−−
=
12
1
0
:ianglessimilar tr From
xxxx
ww
−−
=
w0 wx1
x2
x The moment of this load forAny point x ≥ x1 is
( ) ( ) ( ) ( )3112
011
12
10
6321 xx
xxwxxxx
xxxxwM −
−=
−
−
−−
−=
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 29ENES 220 ©Assakkaf
Singularity Functions
Example 4Use 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.
x
y 2000 lb/ft
4 ft 11 ft 5 ft
Figure 16
2
6
in 464psi1029
=
×=
IE
16
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 30ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)
x
y 2000 lb/ft
4 ft 11 ft 5 ft
R1 R2
Find the reactions:
( ) ( )
( )lb 5.562,6
000,30152000 ;0lb 5.437,23
0)2
155(15200016 ;0
2
21
1
12
=∴
=−+=↑+
=∴
=+−=+
∑
∑
RRRF
R
RM
y
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 31ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)A single expression for the bending moment can be obtained using the singularity functions:
x
y 2000 lb/ft
4 ft 11 ft 5 ftR1 R2
( ) ( ) 122
45.437,232
1520002
2000−+
−+−= x
xxxM (23a)
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 32ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)The elastic curve is found by integrating Eq. 23 twice
( )
21
344
1
233
122
645.437,23
24152000
242000
245.437,23
6152000
62000
45.437,232
1520002
2000
CxCxxxEIy
CxxxEIyEI
xxxxMyEI
++−
+−
+−=
+−
+−
+−==′
−+−
+−==′′
θ (23b)
(23c)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 33ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)Boundary conditions:
y = 0 at x = 4 ft and at x = 20 ft
750,718,220
)20(6
4205.437,2324
1520200024
)20(20000)20(
3.333,214
)4(6
445.437,2324
154200024
)4(20000)4(
21
21
344
21
21
344
−=+∴
++−
+−
+−==
=+∴
++−
+−
+−==
CC
CCEIy
CC
CCEIy
0 0(23d)
(23e)
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 34ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)From Eqs 23d and 23e, the constants of integrations are found to
Therefore, the elastic curve is given by
88.020,653 and 54.588,168 21 −== CC
−+
−+
−+−= 9.020,6535.588,168
645.437,23
24152000
24)(20001)(
344
xxxx
EIxy
(23f)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 35ENES 220 ©Assakkaf
Singularity Functions
Example 4 (cont’d)The deflection at any point along the beam can be calculated using Eq. 23f (elastic curve equation). Therefore, the deflection y at x = 10 ft is
( )( ) [ ]
in 0.1339ft 111650.0y
9.020,653885,685,1750,8433.333,8334641029
12)10(
9.020,653)10(5.588,1686
4105.437,2324
1510200024
)10(20001)10(
01
6
2
344
==
−++−×
=
−+
−+
−+−=
′′
y
EIy
0
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LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 36ENES 220 ©Assakkaf
Singularity Functions
Example 5A 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.
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 37ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)y
LL
A CxB
M
Figure 17
IE ,
20
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 38ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)First find the reactions RA and RB:
( )
LMR
RRFLMR
MLRM
B
BAy
A
AC
2
0 ;02
02 ;0
=∴
=+=↑+
−=∴
=+=+
∑
∑
y
LL
A CxB
MIE ,
RARB
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 39ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)Singularity functions to describe the bending moment:
0
0
2
or 2
)(
LxMLMxyEI
LxMLMxxM
−+−=′′
−+−= (24a)
(24b)
y
LL
A CxB
M IE ,
RA
RB
21
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 40ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)Integrating Eq. 24.b twice, we get
21
23
11
2
0
212
4
2
CxCLxM
LMxEIy
CLxML
MxyEI
LxMLMxyEI
++−
+−=
+−+−=′
−+−=′′
(24c)
(24d)
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 41ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)Boundary conditions:
At x = 0, y = 0 Therefore: C2 = 0At x = 2L, y = 0 Therefore: C1= ML/12
Thus,
[ ][ ] 0
48)(
648
33
223
=+−=
+−+−=
LLEILMLy
xLLxLxEILMy (24e)
(24f)
22
LECTURE 16. BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS (9.5 – 9.6) Slide No. 42ENES 220 ©Assakkaf
Singularity Functions
Example 5 (cont’d)(b) Finding the maximum deflection:
[ ]
( )
EIML
EIMLLL
EILMLyy
LxLLxx
yyLLxLxEILMy
543
3362
333123/
3/ 0123 :Therefore
0 when 12312
2
233
max
213
max212
=
=
+−==
=⇒=+−+−
=′+−+−=′