Chapter 8 Deflection

Structural Mechanics 2Dept of Architecture

Outline

• Deflection diagrams and the elastic curve• Elastic-beam theory• The double integration method• Moment-area theorems• Conjugate-beam method

8-1 Deflection diagrams and the elastic curve

• Deflections occur due to– Loads– Temperature– Fabrication errors– Settlement

• Deflections and vibrations must be limited (controlled) for the comfort of occupants → Serviceability requirement

8-1 Deflection diagrams and the elastic curve

• We concern only linear elastic material response at present

• This means a structure will return to its original undeformed position after the load is removed

• It is useful to sketch the deformed shape of the structure when it is loaded

8-1 Deflection diagrams and the elastic curve

Deflection at some typical supports and joints

0

0

8-1 Deflection diagrams and the elastic curve

8-2 Elastic Beam theory

• Consider an initially straight beam• Due to loading, the beam deforms under

shear and bending• If beam L >> d, greatest deformation will

be caused by bending• When moment deforms the element of a

beam, the angle between the cross sections becomes d

d

8-2 Elastic Beam Theory

dsdsds /)'( strain in arc ds located at position y from the neutral axis:

ydddy

dydsddxds

1)(

)(' and

For linear elastic material, Hooke's Law holds true

E/

From elasticityIMy /

initially straight beam

dxEIMd

EIM

1

Substituting, the governing equation is obtained

in which = curvature of the elastic deformation curve1

View in axial direction

Side view

N.A.

※ Neutral axis (N.A.) is the imaginary axis whose length does not change before and after bending.

8-2 Elastic Beam theory

232

21

212

22222

1

2

2

1

1

11

tan

1

1

tan1

tan

tan

v

vv

vvdxd

vdxds

vdxdvdxds

dsdx

dxvd

dsddx

vddxd

dxdv

dxdvv

8-2 Elastic Beam theory

2/32

22

2/32

22

])/(1[/

])/(1[/1

dxdvdxvd

EIM

dxdvdxvd

Mathematically, the curvature is expressed as

For small deflection, the expression for curvature becomes

2

22

01dx

vdEIM

dxdv

dxdv

dxdxdvdxdvdxds 222 1

This implies that points on the elastic curve will only be displaced vertically but not horizontally.

8-3 The Double Integration Method

dxdxEI

xMv

dxEI

xMdxdv

EIxM

dxvd

)(

)(

)(12

2

※ Integral constants are determined by boundary conditions.

Example 8-3 Determine the equation of the elastic curve

oMM

21

2

1

2

2

2 CxCxMvEI

CxMdxdvEI

Mdx

vdEI

o

o

o

ntdisplaceme : 2

0,0

slope :

0,0

22

1

EIxMv

CvxatEI

xM

Cdxdvxat

o

o

x

8-4 Moment-Area theorems• Theorem 1

– The change in slope between any 2 points on the elastic curve equals the area of the M/EI diagram between the 2 points

dxEIMd

dxEIMB

AAB /

Curvature:

Change in slope between two points:

8-4 Moment-Area theorems• Theorem 2

– The vertical deviation tA/B of the tangent at a point A on the elastic curve with respect to the tangent extended from another point B equals the “moment” of the area about point A under the M/EI diagram between the 2 points

dxEIMxt

B

ABA /

dxEIMxt

xdAdAx

B

ABA /

8-4 Moment-Area theorems

• It is important to realize that the moment-area theorems can only be used to determine the angles and deviations between 2 tangents on the beam’s elastic curve

• In general, they do not give a direct solution for the slope or displacement at a point

• However, the moment-area theorems are a powerful tool giving us a significant insight and a shortcut in determination of deformation

Example 8-5 Determine the slope at points B and C of the beam where E = 200 GPa and I = 360106 mm4

ABB /

rad00521.010360200000

1075.3

1075.3

21025.1105.2

25000105050001050

6

11

11

1111

66

EI

EI

EIEIB

Moment-Area theorems

Unit: N, mm

EIM

EI50

EI100

A B5 m 5 m

ACC /

rad00694.010360200000

105

105

100001010021

6

11

11

6

EI

EIC

Moment-Area theorems

Unit: N, mm

EIM

EI50

EI100

A B5 m 5 m

Example 8-6 Deflection at points B and C

mm

mmEI

mN

mmmEI

mNt

mm

mmEI

mNt

BC

AB

ACC

AB

ABB

06.981.225.6

21041020010310500

1081020010510410500

233500

32

44500

5.2210810200

104105002

44500

63

623

63

333

63

623

EI60

EI30

Example 8-7 Slope at point C

0DUse

※ Refer to “animation” in CD Rom attached in the textbook.

Example 8-8 Slope at point C

rad

rad

mEI

mkN

mm

mmmmmEI

mkNt

Lt

C

AC

AB

ACAB

ACAC

0011.0000278.0108

1.11

000278.01036010200

1020

22021

1.111036010200

10800

23222

316660

21

3

63

9

63

12

A

EI20

EI60

Example 8-9 Deflection at point C

mm

mm

mmEI

mkNt

mm

mmEI

mkNt

tt

t

C

BC

BA

BCBA

BCC

1.147.425.37

7.410410200

1075.3

33135.2

21

5.3710410200

1030

63165

21

2

63

12

63

12

EI5 EI

5.2

Example 8-10 Deflection at point C

mm

mm

mmEI

mkNt

mm

dxxxxxEI

mmmEI

mkNt

tt

C

AB

AC

ABACC

3.1434123.225

411025010200

102048

8318192

21

3.22510250102001030728192

83728241

88318192

21

2

63

12

63

12

118

02111

211 372192 xx

x1

Example 8-11 Slope at point B

BCt BC

B

A B C D

A B C D

30 kN-m

20 kN-m

A B C D

20 kN-m30 kN-m

+=

8-5 Conjugate-Beam Method

wdxdV

wdx

Md2

2

EIM

dxd

EIM

dxvd2

2

wdxV wdxdxM

dx

EIM dxdx

EIMv

8-5 Conjugate-Beam Method

• Here the shear V compares with the slope , the moment M compares with the displacement v and the external load w compares with the M/EI diagram

• To make use of this comparison we will now consider a beam having the same length as the real beam but referred to as the “conjugate beam”

8-5 Conjugate-Beam MethodTheorem 1: The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam

Theorem 2: The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam

Sign convention:Upward for negative momentDownward for positive moment

8-5 Conjugate-Beam Method

• A pin or roller support at the end of the real beam provides zero displacement but the beam has a non-zero slope

• Consequently from Theorem 1 and 2, the conjugate beam must be supported by a pin or roller since this support has zero moment but has a shear or end reaction

• When the real beam is fixed supported, the conjugate beam has a free end since at this end there is zero shear and moment

8-5 Conjugate-Beam Method

Example 8-12 Determine the slope & deflection at point B of the steel beam where E = 200 GPa and I = 475106 mm4

rad00263.010475200000

105.2

50001010021

6

11

6

'

EI

VBB

mmM BB

9.21325000500000263.0'

EI100

EI250

Example 8-13 Determine the max deflection of the steel beam with E = 200 GPa and I = 60106 mm4

mmM

EImkNM

xxEI

xxEI

M

mx

xEI

xEI

V

x

x

x

x

8.16106010200

1025.201

25.201312

214571.653

022145

71.6max

63

12

3

71.6

Example 8-14 Deflection at center C

Example 8-15 Displacement at pin B and slope of each beam segment about the pin