BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS - Assakkaf

1

• 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.

2



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)



Introduction

x

y

w

y

xwPFigure 9

(a)(b)L

LL/2

L/4

3



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



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



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



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.

5



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.



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−

6



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)



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)

7



Integration and Differentiation of Singularity Functions

0 when C1

1 100 >+−

+=− +

∫ nxxn

dxxx nn

0 when 100 >−=− − nxxnxx

dxd nn

(19)

(20)


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)


8


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




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.

9



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.



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)

10



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



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



Typical Singularity Functions

( )

LxxxwxxMxxPxRxM AL

<<−−

−+−−=

0for 2

23

02

11

1 2 3

4

MAPw

RRRLx2

x3

L

x1

x

y

(22)



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.

12



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.



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.

13



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




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



x

y

x2

L

x

y

Lx2

w0

w0

-w0

x1

x1


22

021

0

220xxwxxwMw −+−−=

1

Figure 14





( ) ( )

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


15



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 −

−=

−

−

−−

−=



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



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



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)

17



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)



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)

18



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)



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

19



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.



Example 5 (cont’d)y

LL

A CxB

M

Figure 17

IE ,

20



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



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



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)



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



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

=

=

+−==

=⇒=+−+−

=′+−+−=′

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BEAMS: DEFORMATION BY SINGULARITY FUNCTIONS - Assakkaf