Chapter 7 Bending Deformation -...

2008 Architecture LJM 1

7.1 Introduction

7.2 Approximately Differential equation of deflection

curve

7.3 Integration method of determining the beam

deflections

7.4 Superposition method of determining the beam

deflections

7.5 Statically indeterminate beams

7.6 Stiffness criteria of beams; Optimum design of

beams for stiffness

Chapter 7 Bending DeformationChapter 7 Bending Deformation


7.1 Introduction


curve

7.3 Integration methof of determining the beam

deflections

7.4 Superposition methof of determining the beam

deflections

7.5 Statically inderminate beams


beams for stiffness



Highlights：：：：Feflecion calculation of beams and plane frames

Objectives：：：：①①①①Stiffness check；；；；

②②②②Solution of Statically Inderminate Problems

7.1 Introduction


⑴⑴⑴⑴ Deflection, w：：：：vertical displacement of centriod of the

cross-section.

⑵⑵⑵⑵ Rotational angle, θθθθ, ：：：：Angle rotated about neutral axis

2. Basic amounts measuring the beam deflection

3. Relation between w and θθθθ：：：：

1. Deflection curve：：：： axis of the beam after deformation,

smooth ever-curve

dtan w'

d

w

xθ θ= = =

F

xθθθθ

C

w

C1

w

w =w (x)

7.1 Introduction


7.1 Introduction


curve


deflections


deflections



beams for stiffness



z

z

EI

)x(M=

ρρρρ1

1. Differential equation of deflection curve

EI

)x(M)x("w ±=∴

挠曲线近似微分方程挠曲线近似微分方程挠曲线近似微分方程挠曲线近似微分方程。。。。

)x("w 'w

)x("w±≈

+±=

23

2 )1(

1

ρρρρ

小变形小变形小变形小变形

w

x

M>0

0>)x(''w

w

x

M<0

0<)x("w EI

)x(M)x("w =∴

7.2 Differential equation of deflection curve


7.1 Introduction


curve


deflections


deflections



beams for stiffness



CxdEI

xMx'w +== ∫

)()(θθθθ

DCxxdxdEI

Mw ++= ∫ ∫ )(

1. Integrate the equation:

2.Boundary conditions FA BC

F

D

For uniform straight beams:

)()( xMx"EIw =

Equation of

rotational angle

Deflection equation

0=Aw

0=Bw

0=Dw0=Dθθθθ

7.3 Integration methof of determining

the beam deflections


Discussions：：：：

①①①①Suitable to small deformation members, linear-elasticity

materials.

②②②②Advantage: Find deflection and rotational angle of

arbitrary sections; Weakness: Troublesome

Continuity condition：

Smooth condition：：：：

+− =CC

ww

+− =CC

θθθθθθθθP

A BC

wc




FD

a a a

BA

C

MFa/2

Fa/2x

Fa/2Ex.7.1 Draw the deflection curve of the cantilever beam，，，，EI.

Solution：：：：1. Basic foundation

EI

xM"w

)(=

Determine curve，，，，Obey B.C and C.C

2. Draw sketch of the

deflection curve

DCGA

w

7.3 Integration method of determining



Ex.7.2 Determine the deflection curve |w|max |θ|max of the

cantilever beam .

�Bending equation

)()( lxFxM −=

�Differential equation and integration

)()( xlPxM"EIw −−==

CFlxxF

'EIw +−= 2

2

DCxxFl

xF

EIw ++−= 23

26

Solution:

F

l

xw

x

BA

Bθθθθ




��Find constant from B.C.

X = 0: θA=0, C=0

wA=0, D=0

FlxxF

'EIw −= 2

223

26x

Flx

FEIw −=

Substitution B.C. into the above equations, gets:

④④④④Maximum deflection and rotational angle at B.

)(EI

FlwB ↓−=

3

3

EI

Fl'wBB

2

2

−==θθθθ （）

F

l

w

wBx

BA

Bθθθθ

x




FBC

Ex 7.3 Discuss the deformation of the simple-supported

beam shown.

x1

l

w

ASolution:

FA=Fb/l, FB=Fa/l

(2) List bending-moment equations

Portion AC :

)ax(l/FbxM ≤≤= 111 0

)()( 2222 lxaaxFl/FbxM ≤≤−−=

x2

Protion BC :

(1) Find constraint reactions

a bx



FA FB


(3) List differential equations and then integrate

)lxa()ax(Fxl

FbM"EIw ≤≤−−== 22222

For portion CB:

)ax(xl

Fb)x(M"EIw ≤≤== 1111 0

111

3

11

1

2

1

6

2 1

DxCxl

Fb'EIw

Cxl

Fb'EIw

++=

+=

222

3

2

3

22

2

2

2

2

22

66

22

DxC)ax(l

Fx

l

FbEIw

C)ax(P

xl

Pb'EIf

++−−=

+−−=

For portion AC :




(4) Determine integration constants by B.C and C.C

)ax(D:)(w ≤≤== 111 0000

:)a('w)a('w 21 =

2

22

1

2

222C)aa(

Fa

l

FbCa

l

Fb+−−=+

066

22

33

22 =++−= DlCbl

Fl

l

Fb)l(w

Continuity conditions:

22

33

11

3

21

666DaC)aa(

Fba

l

FbDaCa

l

Fb

:)a(w)a(w

++−−=++

=

(a)

(b)

(c)

(d)

Boundary conditions:




)bl(l

FbCC

DD

22

21

21

6

0

−−==

==

Upon solving the four equations simultaneously, find

)xbl(xl

FbEIw

)xbl(l

Fb'EIw

2

1

22

11

2

1

22

1

6

36

−−−=

−−−=

])()[(6

])(3

)3[(6

3

22

2

2

22

2

2

2

2

2

22

2

axb

lxxbl

l

FbEIw

axb

lxbl

l

Fb'EIw

−+−−−=

−+−−−=

AC:

CB:

FBCA




(5) Discussion

2 2

A

B

Fb Fabθ = - (l - b ) = - (l + b)

6EIl 6EIl

Fabθ = (l + a)

6EIl

When a>b, θB > θA

Maximum deflection:

When a>b, θA<0, θC >0. Thus, the point of

θ1(w1' )=0 occurs in the longer segment of the beam.

wmax

Maximum rotational angle:

2 2

0x = (l - b )/3

x0

EI

Flw,

la max

482

3

==



FBCA

)bl(EIl

Fab)bl(

EIl

Fbw

max+−=−−=

639

322


7.1 Introduction


curve


deflections


deflections



beams for stiffness



1. Superposition by loads：：：：

∑=⋅⋅⋅ )()( 21 iin FF,F,F θθθθθθθθ

2. Superposiyion using analysis of

portion-by-portion

∑=⋅⋅⋅ )()( 21 iin FwF,F,Fw

Where F I is generalized force, including force and couple

7.4 Superposition method of determining



Ex.7.4 Find deflection at C and

rotational angle at A by

superposition.

Solution：：：：�Exerting load alone

�Deformation caused by one load

EI

Faw F,C

6

3

=EI

FaF,A

4

2

=θθθθ

EI

qaw q,C

24

54

=EI

qaq,A

3

3

=θθθθ

qF

A B

Ca a

�Superposition

EI

qaFq,AF,AA

12

43 +=+= θθθθθθθθθθθθ

3 4

C

F a 5 q aw = +

6 E I 2 4 E I

F =

A B

q

+

A B

7.4 Superposition method of finding deflections


F

lBA

qMe

BA

Me

q

BA

F

BA

+

+

w = wMe+wq+wF

EI

Fl

EI

ql

EI

lMw e

B382

342

−−=

EI

Fl

EI

ql

EI

lM eB

262

232

−−=θθθθ

Ex.7.4 Find deflection and

rotational angle at B by

superposition.

( )

( )

7.4 Superposition method of determining


Solution：：：：


l

BA

Principle of portion-by-portion analysis

=+

Fl a

A

BC

CB

Fa

w2

EI

alFawwwC

3

)(2

21

+=+=

FM=Fa

)(3

2

1⇓==

EI

lFaaw

Bθθθθ

w1

Bθθθθ

Basic consideration：：：：Deflection equalization

Basic theory：：：：Force

transition

Basic results：：：：Application

directly

)(3

3

2⇓=

EI

Faw



A Cq

B

F

a a/2 a/2

Aq

FAy

FBy

(a)

B

F

a/2 a/2

(b)

A CB

q,Aθθθθ

wB

Ex.7.5 Combined beam AC。。。。EI，，，，F=qa。。。。Find wB and θB.

解解解解2

qaFF ByAy ==

)(48

13

)2

3(6

)2(

324

33

↓=

−+⋅=

+=

EI

qa

aa

EI

/aF

EI

aqa

wwwF,BF,BB By

EI

qa

EI

qa

EI

qa

EI

/aF

EI

aF

BBB

q,BB

By

B

12

5

24

8

3

2

)2(

2

3

3

322

−=−−=

==

=+=

−+

−

+

ππππθθθθθθθθππππθθθθ∆∆∆∆

θθθθθθθθ

θθθθ

+Bθθθθq,B

θθθθ



Ex.7.6 Determine wC .

��For infinitesimal portion dx

��From Table of deflection of beams.

��Superposition

EI

)bL(b)Fd(w dF,C

48

43 32 −−−−====

bdL

bqxd)x(qFd 02========

bdEI

)bL(qb

24

43 322 −−−−====

EI

qLbd

EIL

)bL(qbww

L.

dF,Cq,C24024

43 450

0

322

====−−−−

======== ∫∫∫∫∫∫∫∫

q0

0.5L 0.5L

x dx

b

x

f

C




7.1 Introduction


curve


deflections


deflections



beams for stiffness



1. Take equalization system

2. List compatibility equation

3. Introduce physical law to get supplementary

4. Solve the equation to get redundant reaction

A

FB

A

FB

q q

Redundant reactions(多余支反力多余支反力多余支反力多余支反力)

FBy

Redundant restraints(多于约束多于约束多于约束多于约束)




A

F

l/2 l/2

A

FByF

B

MA

A

F


相当系统相当系统相当系统相当系统E.E

①①①①Take equalization equation

②②②②Compability condition

③③③③Fond redundant reaction

Change S.ID.P to S.D.P！！！！

0=Bw

Bw = 0

3 3By

B

F l 5Flw = - = 0

3EI 48I

)(16

5↓=

FFBy

Ex.7.7 Find reactions of the beam

④④④④Find other reactions A

FByFMA

FAy

16

3)(

16

11 FlM

FF AAy =↓= ( )



BC

A

F

D

E

Example7.8 Two cantilever beams of AD and BE are joined

by a steel rod CD. Determine the deflection of the cantilever

beam AD, at D due to a force F applied at E.

EI

EA

l

l

l

l



Solution:

(1) Set up a equivalent system

(2) Compatible condition

wD=wC -△△△△rod (a)

wC = (wC)p- (wC)FD

WhereEI

lFw D

D3

3

=

EI

Fl

EI

Flw FD

23)(

33

+=

EI

lFw D

FD D 3)(

3

=

wc

△△△△rod

B CF

E

Fig.b

FD

A

D

FD

wD

Fig.a



BC

A

F

D

E

FD

wc

△△△△rod

wD

Substitute wD, wC and △△△△rod

into Eq.(a):

(3) Find the deflection of D, wD

EA

lFD

rod=∆∆∆∆

A/Il

FlFD

64

52

2

+=

)()64(3

52

5

↓+

=A/IlEI

FlwD



=

q0

lA

B

l

MA

BA

Ex.7.9：：：： Shown is the beam AB

of length l, EI, subjected to

uniform load q. Draw M-

diagram.

MA, FB--多余约束反力多余约束反力多余约束反力多余约束反力。。。。

q0

EI

q0

L FB

AB



(1) Set up a equivalent system

Different Equivalent systems!


�Compatible equation

0=+=BF,Bq,BB www

+

q0

L FB

AB

=

FB

AB

q0

AB

�physical relations

�Supplementary equation

EI

LFw

EI

qLw B

F,Bq,B B 3;

8

34

−==

038

34

=−EI

LF

EI

qL B

8

3qLF

B=∴

�M-diagram


M

12823 2 /ql

82 /ql


Solution：：：：��Set up E.SMA

Ex.7.9：：：：Plot Bending-moment

diagram of simple-supported

beam AB shown.

BA

0====++++==== q,AM,AA Aθθθθθθθθθθθθ

EI

lMA

M,A A 3====θθθθ

EI

qlq,A

24

3

====θθθθ

�Geometry equation

8

2qlM A −−−−====

�Reactions：：：：8

3

8

5 qlF,

qlF

BA========

�Bending-moment DiagramM 29 /128ql

82 /ql

㈠㈠㈠㈠

㈩㈩㈩㈩

lA

BEI

q0

FBFA

MA

BA


Have：：：：



相当相当相当相当系统系统系统系统E.S.

①①①①Equivalent system

②②②②Compatible equation

21 CC ww =

EI

lF

EI

lF

w RR

C243

)2

( 33

2 ========

Ex.7.10 Determine maximum deflection of the cantilever beam

AB shown. Suppose that EI for the two beams are equal.

④④④④Maximum deflection

A FR C

F

B

FR

A C

C BA

l/2 l/2

EI

lFFw R

C48

)25(3

1

−−−−====

EI

lFF

EI

lF RR

48

)25(

24

33 −=

4

5FFR =

33 3

RB

5F lFl 13Flw = - =

3EI 48EI 64EI

B

B'

w%

w= 61

F



BCF,Bq,BB LwwwB

∆∆∆∆=−=

=

Ex.7.11 Determine axial load in rod BC for the structure shown.

q0

FB

FB

A

+

q0

A

LBCEA

q0

LA B

C

EI


①①①①Equivalent system

②②②②Compatible equation


E.S.


=

LBCEA

x

f

q0

L FB

A B

C

FB

AB

+

q0

AB

�Physical relations

�Supplementary

�Others（（（（Reactions, stresses,

deflections, and so on.））））

EI

LFw;

EI

qLw B

R,Bq,B B 38

34

−==

EA

LF

EI

LF

EI

qL BCBB =−38

34

)3(8 3

4

I/LA/LI

qLF

BC

B +=∴

EA

LFL BCB

BC =∆∆∆∆



7.1 Introduction


curve


deflections


deflections



beams for stiffness



[ ] wmax

δδδδ≤ [ ]θθθθθθθθ ≤max

[w]: Allowable deflection，，，，

[θθθθ]: Allowable rotational angle

� Check the stiffness;

��、、、、Determine allowable loads.

1. Stiffness conditions of beams

Three types of stiffness calculation

��、、、、Design sections;

7.6 Stiffness criteria of beams;

Optimum design of beams for stiffness


Strength：：：： [ ] W

M

z

max σσσσσσσσ ≤=

zEI

XM''w

)(=Stiffness：

Methods：：：：

→3S are related to internal forces and

properties of the cross-section

Reducing Bending moment M

Enhancing Inertia moment I or Modulus of section W

Select materials rationally

Stability：：：： l

EImimcr 2

2ππππσσσσ =

2. Methods of enhancing deformations of beams




Ex.7.12 The hollow circular rod AC of in-diameter d=40mm

and out-diameter D = 80mm. E = 210GPa，，，，[δ]= 10-5 m at C.

[θθθθ] = 0.001. at C, F1 = 1kN, F2 = 2kN. Check the stiffness of

the overhanging beam.

F2

B

l=400

P2

A

Ca=100

200 D

F1

B

+

F2 M

=

F1

+F2



Analysis of deformations


2

1C1 B1

F l aw = θ a =

16EI

2

1B1

F lθ =

16EI

2B3

F laθ = -

3EI2

2

C 3 B 3

F law = θ a = -

3E I

��From Table of deformations

02 =Bθθθθ

C

F aw

EI= −

3

22

3

F2

B C

++

=

( )

( )

( )

CF1

A BD

The overhang beam

C

F2

B

D

A

M





C

F l a F a F law

EI EI EI= − −

2 3 2

1 2 2

16 3 3

2

1 2B

F l F laθ = -

16EI 3EI

��Deformations by superposition

481244441018810)4080(

64

143)(

64m

. dDI

−− ×=×−=−=ππππ

[ ] 00101042304

..max =<×= − θθθθθθθθ

[ ] mm.w max

56 1010195 −− =<×= δδδδ

��Check the stiffness

)(104230)3

200

16

400(

1880210

40 4 弧度−×−=−×

= ..

Bθθθθ

m.wC

610195

−×−=

( )



Chapter 7 Bending Deformation -...

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