Deflection of Beams

Deformation of a Beam Under Transverse LoadingLoading

• Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings.

EIxM )(1

=ρ

• Cantilever beam subjected to concentrated• Cantilever beam subjected to concentrated load at the free end,

EIPx

−=ρ1

EIρ

• Curvature varies linearly with x

1• At the free end A, ∞== A

A ρ

ρ,01

• At the support BEI

B =≠ ρ01

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• At the support B, PLBB

=≠ ρρ

,0

Deformation of a Beam Under Transverse Loading

h i b• Overhanging beam

• Reactions at A and C

B di t di• Bending-moment diagram

• Curvature is zero at points where the bending moment is zero, i.e., at each end and at E.

EIxM )(1

=ρ

• Beam is concave upwards where the bendingBeam is concave upwards where the bending moment is positive and concave downwards where it is negative.

M i t h th t• Maximum curvature occurs where the moment magnitude is a maximum.

• An equation for the beam shape or elastic curve

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is required to determine maximum deflection and slope.

Equation of the Elastic Curve• From elementary calculus, simplified for beam

parameters,2d

2

2

232

2

2

1

1dx

yd

dy

dxyd

≈

⎥⎤

⎢⎡ ⎞

⎜⎛+

=ρ

1dx ⎥

⎥⎦⎢

⎢⎣ ⎠

⎜⎝

+

• Substituting and integrating,

( )2

21

d

xMdx

ydEIEI

x

==ρ

( )

( )

10

CdxxMdxdyEIEI

xx

+=≈

∫∫

∫θ

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( ) 2100

CxCdxxMdxyEI ++= ∫∫

Equation of the Elastic Curve

( ) CCdMdEIxx

∫∫

• Constants are determined from boundary conditions

( ) 2100

CxCdxxMdxyEI ++= ∫∫

• Three cases for statically determinate beams,

– Simply supported beam0,0 == BA yy

– Overhanging beam0,0 == BA yy

– Cantilever beam0,0 == AAy θ

• More complicated loadings require multiple i t l d li ti f i t f

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integrals and application of requirement for continuity of displacement and slope.

Direct Determination of the Elastic Curve From the Load Distribution

• For a beam subjected to a distributed load,dVMddM 2

• Equation for beam displacement becomes

( ) ( )xwdxdV

dxMdxV

dxdM

−=== 2

( )xwdx

ydEIdx

Md−== 4

4

2

2

( ) ( )2131

dxxwdxdxdxxyEI −= ∫∫∫∫• Integrating four times yields

432

2213

161 CxCxCxC ++++

• Constants are determined from boundary conditions

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conditions.

Statically Indeterminate BeamsC id b i h fi d A d ll• Consider beam with fixed support at A and roller support at B.

• From free-body diagram, note that there are four k ti tunknown reaction components.

• Conditions for static equilibrium yield000 =∑=∑=∑ Ayx MFF y

The beam is statically indeterminate.

xx• Also have the beam deflection equation,

( ) 2100

CxCdxxMdxyEIxx

++= ∫∫which introduces two unknowns but provides pthree additional equations from the boundary conditions:

0At000At ===== yLxyx θ

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0,At 00,0At ===== yLxyx θ

Sample Problem 1

SOLUTION:

• Develop an expression for M(x) and derive differential equation for elastic curve.

• Integrate differential equation twice and apply boundary conditions to obtain elastic curve

For portion AB of the overhanging beam, ( ) d i th ti f th l ti

obtain elastic curve.

• Locate point of zero slope or point of maximum deflection.

(a) derive the equation for the elastic curve, (b) determine the maximum deflection, (c) evaluate ymax.

• Evaluate corresponding maximum deflection.

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Sample Problem 1SOLUTION:

• Develop an expression for M(x) and derive differential equation for elastic curve.q

− Reactions:

↑⎞⎜⎛ +=↓=

aPRPaR 1 ↑⎠

⎜⎝+=↓=

LPR

LR BA 1

− From the free-body diagram for section AD,y g ,

( )LxxLaPM <<−= 0

ayd 2

− The differential equation for the elastic curve,

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xLaP

dxydEI −=2

Sample Problem 1• Integrate differential equation twice and apply

boundary conditions to obtain elastic curve.

121 CxaPdyEI +=

213

1

61

2

CxCxLaPyEI

CxL

Pdx

EI

++−=

+−=

PaLCLCLLaPyLx

Cyx

61

610:0,at

0:0,0at

113

2

=+−===

===x

LaP

dxydEI −=2

2

L 66

xPaLdyady 11 22 ⎥

⎤⎢⎡ ⎞

⎜⎛

Substituting,

⎤⎡ ⎞⎛32PaLxxaPyEI

Lx

EIPaL

dxdyPaLx

LaP

dxdyEI

11

3166

121

3

2

+−=

⎥⎥⎦⎢

⎢⎣ ⎠

⎞⎜⎝⎛−=+−=

9 - 10⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

32

6 Lx

Lx

EIPaLyL

y66

Sample Problem 1• Locate point of zero slope or point

of maximum deflection.

LP Ld 2⎤⎡ ⎞⎛

⎤⎡ 32

LLxL

xEI

PaLdxdy

mm 577.0

331

60 ==

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−==

• Evaluate corresponding maximum

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

32

6 Lx

Lx

EIPaLy

• Evaluate corresponding maximum deflection.

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Sample Problem 2

SOLUTION:

• Develop the differential equation for the elastic curve (will be functionally dependent on the reaction at A).

For the uniform beam, determine the reaction at A, derive the equation for th l ti d d t i th

• Integrate twice and apply boundary conditions to solve for reaction at Aand to obtain the elastic curve.

the elastic curve, and determine the slope at A. (Note that the beam is statically indeterminate to the first degree )

• Evaluate the slope at A.

degree.)

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Sample Problem 2• Consider moment acting at section D,

M D 0

⎞⎛

=∑

MxLxwxRA 0

321

3

20 =−⎟

⎟⎠

⎞⎜⎜⎝

⎛−

LxwxRM A 6

30−=

• The differential equation for the elastic

LxwxRM

dxydEI A 6

30

2

2−==

The differential equation for the elastic curve,

Ldx 6

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Sample Problem 2• Integrate twice

14

02242

1 CLxwxREI

dxdyEI A +−== θ

215

031206

1 CxCL

xwxRyEI A ++−=

LxwxRM

dxydEI A 6

30

2

2−==

• Apply boundary conditions

1

0:0,0at 3

2 ===

L

Cyx

01:0,at

0242

1:0,at

214

03

13

02

=++−==

=+−==

CLCLwLRyLx

CLwLRLx

A

Aθ

01206

:0,at 21 ++ CLCLRyLx A

• Solve for reaction at A

011 40

3 =− LwLRA ↑= LwRA 01

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0303 0 =− LwLRA ↑= LwRA 010

Sample Problem 2

⎞⎛⎞⎛ 5 111

• Substitute for C1, C2, and RA in the elastic curve equation,

xLwL

xwxLwyEI ⎟⎠⎞

⎜⎝⎛−−⎟

⎠⎞

⎜⎝⎛= 3

05

030 120

112010

161

( )xLxLxwy 43250 2+= ( )xLxLxEIL

y 2120

−+−=

Diff ti t t fi d th l

( )42240 65120

LxLxEIL

wdxdy

−+−==θ

• Differentiate once to find the slope,

EILw

A 120

30=θat x = 0,

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Method of Superposition

Principle of superposition:

• Deformations of beams subjected to • Procedure is facilitated by tables of• Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual

• Procedure is facilitated by tables of solutions for common types of loadings and supports.

the deformations from the individual loadings.

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Sample Problem 3

For the beam and loading shown, determine the slope and deflection at

i t Bpoint B.

SOLUTIONSOLUTION:

Superpose the deformations due to Loading I and Loading II as shown.

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Sample Problem 3

Loading I

( ) wLIB

3−=θ ( ) wLy IB

4−=( )

EIIB 6θ ( )

EIy IB 8

Loading II

( )EI

wLIIC 48

3=θ ( )

EIwLy IIC 128

4=

In beam segment CB, the bending moment is zero and the elastic curve is a straight line.

( ) ( ) wL3( ) ( )

EIwL

IICIIB 48== θθ

( ) wLLwLwLyB7 434

=⎞⎜⎛+=

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( )EIEIEI

y IIB 384248128=

⎠⎜⎝

+=

Sample Problem 3

( ) ( ) wLwL 33 wL7 3

Combine the two solutions,

( ) ( )EI

wLEI

wLIIBIBB 486

+−=+= θθθ

( ) ( ) wLwL 7 44

EIwL

B 487

−=θ

wL41 4

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( ) ( )EI

wLEI

wLyyy IIBIBB 3847

8+−=+=

EIwLyB 384

41−=

Application of Superposition to Statically Indeterminate BeamsBeams

• Method of superposition may be applied to determine the reactions at

• Determine the beam deformation without the redundant support.applied to determine the reactions at

the supports of statically indeterminate beams.

D i t f th ti

without the redundant support.

• Treat the redundant reaction as an unknown load which, together with

• Designate one of the reactions as redundant and eliminate or modify the support.

the other loads, must produce deformations compatible with the original supports.

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Sample Problem 4

For the uniform beam and loading shown, determine the reaction at each support and the slope at end A.

SOLUTION:

• Release the “redundant” support at B, and find deformation.A l ti t B k l d t f di l t t B• Apply reaction at B as an unknown load to force zero displacement at B.

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Sample Problem 4• Distributed loading:

( ) [ ]xLLxxEI

wy wB334 2

24+−−=

2

( ) LLLLLwyB3

34 2222⎥⎤

⎢⎡ ⎞

⎜⎛+⎞

⎜⎛−⎞

⎜⎛−=

At point B, Lx 32=

( )

EIwL

LLLLLEI

y wB

401132.0

332

324

−=

⎥⎥⎦⎢

⎢⎣ ⎠

⎜⎝

+⎠

⎜⎝⎠

⎜⎝

=

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EI

Sample Problem 4• Redundant reaction loading:

bP 22

EILbPayax

3,At

22−==

12

( ) LLRy BRB

222⎠⎞

⎜⎝⎛

⎠⎞

⎜⎝⎛=

LbLa 31

32 and For ==

( )

EILR

EILy

B

RB

301646.0

333

=

⎠⎜⎝⎠

⎜⎝

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Sample Problem 4

• For compatibility with original supports yB = 0For compatibility with original supports, yB 0

( ) ( )EI

LREI

wLyy BRBwB

3401646.001132.00 +−=+=

↑LR 6880

• From statics,

↑= wLRB 688.0

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↑=↑= wLRwLR CA 0413.0271.0

Sample Problem 4

( ) wLwL 33

Slope at end A,

( )EI

wLEI

wLwA 04167.0

24−=−=θ

( ) ( ) wLLLLwLbLPb 322

220339800688.0

⎥⎤

⎢⎡ ⎞

⎜⎛⎞

⎜⎛−θ( ) ( )

EIL

EILEILRA2 03398.0

3366=

⎥⎥⎦⎢

⎢⎣ ⎠

⎞⎜⎝⎛−

⎠⎞

⎜⎝⎛=−=θ

( ) ( ) wLwL 33033980041670 ++ θθθ wL3

007690θ

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( ) ( )EIEIRAwAA 03398.004167.0 +−=+= θθθ

EIA 00769.0−=θ

Moment-Area Theorems• Geometric properties of the elastic curve can

be used to determine deflection and slope.

==EIM

dxyd

dxdθ

2

2

• Consider a beam subjected to arbitrary loading,

∫∫ =D

C

D

C

x

xdx

EIMd

EIdxdx

θθ

θ

∫=−D

C

CC

x

xCD

x

dxEIMθθ

θ

Cx

• First moment-area theorem:

area under (M/EI) diagram between=CDθ

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C and D.CD

Moment-Area Theorems

• Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C.

M

∫=

==

Dx

DC dxMxt

dxEIMxdxdt

1

11 θ

= tangential deviation of C∫Cx

DC dxEI

xt 1 tangential deviation of Cwith respect to D

• Second moment-area theorem:The tangential deviation of C with respect to Dis equal to the first moment with respect to ais equal to the first moment with respect to a vertical axis through C of the area under the (M/EI) diagram between C and D.

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Application to Cantilever Beams and Beams with Symmetric LoadingsSymmetric Loadings

• Cantilever beam − Select tangent at A as the referencereference.

ADD

Aθ

=

=

θθ

,0with

ADD ty =

• Simply supported, symmetrically loadedSimply supported, symmetrically loaded beam − select tangent at C as the reference.

Cθ = ,0with

CBB

CBB

ty =

=θθ

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Bending Moment Diagrams by Parts

• Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately.each load is evaluated separately.

• Construct a separate (M/EI) diagram for each load.load.

− The change of slope, θD/C, is obtained by adding the areas under the diagrams.

− The tangential deviation, tD/C, is obtained by adding the first moments of the areas with respect to a vertical axis through D.

• Bending moment diagram constructed from individual loads is said to be drawn by parts.

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Sample Problem 5

SOLUTION:

Determine the reactions at s pports• Determine the reactions at supports.

• Construct shear, bending-moment and (M/EI) diagrams.

For the prismatic beam shown, determine the slope and deflection at E.

(M/EI) diagrams.

• Taking the tangent at C as the reference, evaluate the slope and tangential deviations at E.

9 - 30

Sample Problem 5

SOLUTION:

• Determine the reactions at supportsDetermine the reactions at supports.waRR DB ==

C h b di d• Construct shear, bending-moment and (M/EI) diagrams.

LwaLwaA22 ⎞

⎜⎛

( )EI

waaEI

waA

EIEIA

6231

42232

2

1

−=⎟⎠

⎞⎜⎜⎝

⎛−=

−=⎠

⎜⎝

−=

EIEI 623 ⎠⎜⎝

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Sample Problem 5• Slope at E:

L

CECECE

32

=+= θθθθ

EIwa

EILwaAA

64

3221 −−=+=

( )aLEI

waE 23

12

2+−=θ ( )

EIE 12

• Deflection at E:

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛−⎥⎦

⎤⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛ +=

−=

LAaALaA

tty CDCEE

443

4 121

⎥⎦

⎤⎢⎣

⎡−−⎥

⎦

⎤⎢⎣

⎡−−−=

EILwa

EIwa

EILwa

EILwa

168164

224223

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( )aLEI

wayE +−= 28

3

Application of Moment-Area Theorems to Beams with Unsymmetrical LoadingsLoadings

• Define reference tangent at support A. Evaluate θAby determining the tangential deviation at B withby determining the tangential deviation at B with respect to A.

Lt AB

A −=θL

• The slope at other points is found with respect to reference tangent.

ADAD θθθ +=

• The deflection at D is found from the tangential d i ti t Ddeviation at D.

ABtLxEF

LHB

xEF

==

9 - 33

ABADD tLxtEFEDy −=−=

Maximum Deflection

• Maximum deflection occurs at point Kwhere the tangent is horizontal.

t

AKAK

ABA L

t

θθθ

θ

+==

−=

0

AAK θθ −=

• Point K may be determined by measuringPoint K may be determined by measuring an area under the (M/EI) diagram equal to −θA .

• Obtain ymax by computing the first moment with respect to the vertical axis through A of the area between A and K.

9 - 34