# by วิธีของ castigliano assoc. prof. dr. sittichai...

Post on 10-Nov-2018

213 views

Category:

## Documents

Embed Size (px)

TRANSCRIPT

• THEORY OF STRUCTURES

By

Assoc. Prof. Dr. Sittichai SeangatithSCHOOL OF CIVIL ENGINEERING

INSTITUTE OF ENGINEERINGSURANAREE UNIVERSITY OF TECHNOLOGY

7Deflection

1. truss frame statically determinate double integration, moment-area, conjugate-beam, virtual work ( unit load), Castigliano

Deflection statically indeterminate

/deflection v

/slope

• frame

sidesway/

7.1 Deflection Diagram and Elastic CurveElastic curve /Deflection diagram - centroid / deflection diagram elastic curve

moment diagram

+-

elastic curve1.

support (slope) (deflection) ?

• 0y =

0y =0x =

0x

0y =0x =

0 =

0

0

elastic curve

• elastic curve frame moment diagram sign convention

/

MM

/ MM

elastic curve moment diagram

+-

elastic curve moment diagram

+-

• elastic curve moment diagram

+-

elastic curve frame moment diagram

elastic curve frame moment diagram 7.2 Elastic Beam Theory ( 10 )

homogeneous and isotropic material ( ) linear elastic

• Transverse shear stress (shear strain)

Curvature Differential Elementdv/dx = (slope) d2v/dx2 =

(curvature)v =

(deflection)

2

21 M d v

EI dx= =

flexural rigidity

curvature differential element1 d

ds

=

slope neutral axistan dv

dx=

arctan dvdx

=

s x

1 d dxdx ds

= (arctan )d dv dx

dx dx ds=

2 2 + ds dx dv=

1/ 221 ( )ds dv

dx dx = +

1/ 22

1

1 ( )

dxds dv

dx

= +

calculus,

(arctan )d dvdx dx

2

2

21 ( )

d vdx

dvdx

= +

• (curvature) differential element

1 (arctan )d dv dxdx dx ds

=

2

2

2 3/ 2

1

[1 ( ) ]

d vdxdvdx

=

+

2

2

1/ 22 2

1

1 ( ) 1 ( )

d vdx

dv dvdx dx

= + +

Moment-Curvature Relationships:

: dx ds d = =

( - )ds y d =

ds dsds

=

1y

=

( - ) -

y d d

d

=

y

=

isotropic and homogenous material linear elastic

1y Ey

= =

flexural formula,1

( )My

Ey Ey I

= =

flexural rigidity1 MEI

=

y +

Differential Equations Deflection Curve2

2

2 3/ 2

1

[1 ( ) ]

d vMdx

dv EIdx

= =

+

nonlinear second order differential equation elastica deflection L/240 L/360 slope 1

2( ) 0dvdx

2

21 M d v

EI dx= =

flexural rigidity

• 7.3 Double Integration

2

2 ( )d vEI M xdx

=

V = dM/dx 3

3 ( )d vEI V xdx

=

EI v = deflection dv/dx = slope d2v/dx2 = curvature

-w = dV/dx 4

4 ( )d vEI w xdx

=

slope deflection integration integration constant of integration boundary conditions continuity conditions

Sign Convention

Boundary Continuity Conditions (boundary conditions) slope / deflection

(continuity condition)

1 2( ) ( )a a =

1 2( ) ( )v a v a=

1 2( ) ( )a b =

1 2( ) ( )v a v b=

• EXAMPLE slope B C P flexural rigidity EI

( ) ( )M x Px PL P L x= =

FBD

elastic beam2

2d vEI Px PLdx

=

2

12dv xEI P PLx Cdx

= +

3 2

1 26 2x xEIv P PL C x C= + +

2

12dv xEI P PLx Cdx

= +

2

1(0)(0) (0)

2EI P PL C= +

1 0C =

3 2

26 2x xEIv P PL C= +

3 2

2(0) (0)(0)

6 2EI P PL C= +

2 0C =

2

02

dv xEI P PLxdx

= +

21 ( )2xP PLx

EI =

3 2

06 2x xEIv P PL= +

3 21 ( )6 2x xv P PL

EI=

B, x = 2a

C, x = L3

3CPLvEI

=

)(2 aLEIPa

=

x = 2ax = L

slope 2

( )2 2

wLx wxM x =

elastic beam2 2

2 2 2d v wLx wxEIdx

=

2 3

14 6dv wLx wxEI Cdx

= +

3 4

1 212 24wLx wxEIv C x C= + +

EXAMPLE

• boundary condition v = 0 x = 0

3 4

1 2(0) (0)(0) (0)

12 24wL wEI C C= + +

2 0C = = 0 x = L/2

2 3

1( / 2) ( / 2)(0)

4 6wL L w LEI C= +

3

1 24wLC =

2 3

14 6dv wLx wxEI Cdx

= +

3 4

1 212 24wLx wxEIv C x C= + +

x = L/2

2 0C =3

1 24wLC =

3 2 3(4 6 )24

w x Lx LEI

= +

3 4 3

12 24 24wLx wx wLEIv x = +

4 3 3( 2 )

24wv x Lx L xEI

= +

3

max 24wL

EI =

4

max5

384wLv

EI=

3 4

1 212 24wLx wxEIv C x C= + +

2 3

14 6dv wLx wxEI Cdx

= +

2 3 3

4 6 24dv wLx wx wLEIdx

=

slope deflection

• 1. FBD

2. FBD moment x

3. elastic beam integrate slope deflection

4. boundary condition / continuity condition slope deflection integration

5. integration slope deflection

slope EI EXAMPLE

slope A B a = 2b

slope

10 ;x a 1 1PbM xL

=

2 ;a x b 2 2 2

2

( )

(1 )

PbM x P x aL

xPaL

=

=

elastic beam2

112

1

d v PbEI xdx L

=10 ;x a

211 1

1 2dv PbEI x Cdx L

= +

31 1 1 1 2 6

PbEI v x C x CL

= + +

• 2 ;a x b 22 22 (1 )

=

22 2

2 3( )2dv xEI Pa x Cdx L

= +

2 32 2

2 3 2 4 ( )2 6x xEI v Pa C x C

L= + +

integration elastic beam 4 boundary condition

1 10; 0 :x v= =3

1 1 1 1 2 6PbEI v x C x C

L= + + 2 0C =

2 2; 0 :x L v= =2 3

3 40 ( )2 6L LPa C L C

L= + +

2

3 40 3PaL C L C= + +

continuity condition 1 1 2 2( ) ( ) :v x a v x a= = =

2 33

1 3 4( )6 2 6Pb a aa C a Pa C a C

L L+ = + +

2 32 2

2 3 2 4 ( )2 6x xEI v Pa C x C

L= + +

31 1 1 1 6

PbEI v x C xL

= +

1 21 2

1 2

( ) ( ) :dv dvx a x adx dx

= = = 21 1 11 2

dv PbEI x Cdx L

= +

22 2

2 3( )2dv xEI Pa x Cdx L

= +

22

1 3( )2 2Pb aa C Pa a C

L L+ = +

2

3 40 3PaL C L C= + +

2 33

1 3 4( )6 2 6Pb a aa C a Pa C a C

L L+ = + +

22

1 3( )2 2Pb aa C Pa a C

L L+ = +

2 21 ( )6

PbC L bL

= 2 23 (2 )6PaC L a

L= +

3

4 6PaC =

slope 211 1

1 2dv PbEI x Cdx L

= +

31 1 1 1 2 6

PbEI v x C x CL

= + +

22 2

2 3( )2dv xEI Pa x Cdx L

= +

2 32 2

2 3 2 4 ( )2 6x xEI v Pa C x C

L= + +

2 ;a x b

10 ;x a 2 2 211 11

( 3 )6

dv Pb L b xdx EIL

= =

2 2 211 1( )6

Pbxv L b xEIL

=

2 2 222 2 2

2

(3 2 6 )6

dv Pa x L a Lxdx EIL

= = + +

3 2 2 2 22 2 2 2( (2 ) 3 )6

Pav x L a x Lx a LEIL

= + +

• slope A B A, x1 = 0:

B, x2 = L:

2 21 ( ) ( )( ) ( )6 6 6

Pb Pb PabL b L b L b L bEIL EIL EIL

= = + = +

2 22 ( ) ( )( ) )6 6 6

Pa Pa PabL a L a L a L aEIL EIL EIL

= = + = +

a = 2b C slope

2 2 21 1( 3 )6

Pb L b xEIL

= 3b

2 2 21(3 ) 3 0b b x =

1 1.633x b=

3

max 0.48385PbvEI

=

2 2 21 1( 3 )6

Pb L b xEIL

=

7.4 Moment-Area Theorems 1 moment-area method2

2

d v Mdx EI

=

2

2 ( )d v d dv ddx dx dx dx

= =

d Mdx EI=

Md dxEI

=

d slope dx M/EI

/

B

B AA

M dxEI

=

• 2 moment-area methoddt xd=

Md dxEI

= integrate A B

/

B

A BA

Mt x dxEI

=

xd A xdA= centroid

/

B

A BA

Mt x dxEI

=

A elastic curve B M/EI (A B) (A)

slope B C W200x36EXAMPLE

6 434.4(10 ) mmI =1. moment diagram

2. IAB = 1.5IBC M/EI diagram

3. (elastic curve)

4. slope B B A B/A 1

/

6.667 1 13.333 (2) [ ](2)2

B A B

EI EI

=

= +

226.667 kN.mB EI

=

23

6 2 6 4

26.667kN.m 3.88(10 ) rad[200(10 ) kN/m ][34.4(10 )m ]B

= =

• 4. C C A C tC/A 2

/6.667 1 13.333 4 1 10 2[ (2)](2) [ ](2)(1 ) [ ](1)( )

2 3 2 3C A Ct

EI EI EI= = + + +

360.944 kN.mC EI

=

2

6 2 6 4

60.944 kN.m 0.0089 m 8.9 mm[200(10 ) kN/m ][34.4(10 )m ]C

= = =

W360x64 E = 200 GPa I = 176(106) mm4 conjugate beam

EXAMPLE

1. FBD

2. moment diagram

3. moment diagram elastic weight

4. ( C) slope = 0 tan C

/

2A Bt =

//2

A BC C B

t t =

2

/160 8 28 53.33 213.33( ) (6)

3 33520

A Bt EI EI EI

EI

= + + +

=

/53.33 2 213.33 160 10( ) (1) ( )2 3 2 3

657.78

C Bt EI EI EI

EI

= + +

=

3520 657.78 1102.222C EI EI EI

= =

• 3

max 6 2 6 12 4

1102.2 kN.m 0.0308 m 30.8 mm[200(10 )kN/m ][179(10 )(10 )m ]

= = =

12000 389.6 36030.8

L= = >

span linear elastic

40.2 mm12000 298.5 36040.2

L= =