2 Handouts Ch12

©2005 Pearson Education South Asia Pte Ltd

12. Deflections of Beams and Shafts

1

CHAPTER OBJECTIVES

• Use various methods to determine the deflection and slope at specific pts on beams and shafts:1. Integration method2. Discontinuity functions3. Method of

superposition4. Moment-area method

• Use the various methods to solve for the support reactions on a beam or shaft that is statically indeterminate



2

CHAPTER OUTLINE1. The Elastic Curve2. Slope and Displacement by Integration3. *Discontinuity Functions4. *Slope and Displacement by the Moment-Area

Method5. Method of Superposition6. Statically Indeterminate Beams and Shafts7. Statically Indeterminate Beams and Shafts:

Method of Integration



3

CHAPTER OUTLINE8. *Statically Indeterminate Beams and Shafts:

Moment-Area Method9. Statically Indeterminate Beams and Shafts:

Method of Superposition



4

12.1 THE ELASTIC CURVE

• It is useful to sketch the deflected shape of the loaded beam, to “visualize” computed results and partially check the results.

• The deflection diagram of the longitudinal axis that passes through the centroid of each x-sectional area of the beam is called the elastic curve.



5


• Draw the moment diagram for the beam first before creating the elastic curve.

• Use beam convention as shown and established in chapter 6.1.



6


• For example, due to roller and pin supports at B and D, displacements at B and D is zero.

• For region of -ve moment AC, elastic curve concave downwards.

• Within region of +ve moment CD, elastic curve concave upwards.

• At pt C, there is an inflection pt where curve changes from concave up to concave down (zero moment).



7

Moment-Curvature Relationship• x axis extends +ve to the

right, along longitudinal axis of beam.

• A differential element of undeformed widthdx is located.

• ν axis extends +ve upwards from x axis. It measures the displacement of the centroid on x-sectional area of element.

• A “localized” y coordinate is specified for the position of a fiber in the element.

• It is measured +ve upward from the neutral axis.




8


Moment-Curvature Relationship• Limit analysis to the case of initially straight

beam elastically deformed by loads applied perpendicular to beam’s x axis and lying in the x-ν plane of symmetry for beam’s x-sectional area.

• Internal moment M deforms element such that angle between x-sections is dθ.

• Arc dx is a part of the elastic curve that intersects the neutral axis for each x-section.

• Radius of curvature for this arc defined as the distance ρ, measured from center of curvature O’to dx.



9


Moment-Curvature Relationship• Strain in arc ds, at position y from neutral axis, is

( )( )[ ]

( )1-121

'

'

y

dddy

dydsddxdsds

dsds

s

ερ

θρθρθρε

θρθρ

ε

−=

−−=

−===

−=

or

andBut



10


Moment-Curvature Relationship• If material is homogeneous and shows linear-

elastic behavior, Hooke’s law applies. Since flexure formula also applies, we combing the equations to get

ρ = radius of curvature at a specific pt on elastic curve (1/ρ is referred to as the curvature).

M = internal moment in beam at pt where is to be determined.

E = material’s modulus of elasticity.I = beam’s moment of inertia computed about neutral

axis.

( )2-121EIM

=ρ



11


Moment-Curvature Relationship• EI is the flexural rigidity and is always positive.• Sign for ρ depends on the direction of the

moment.• As shown, when M is +ve, ρ extends above the

beam. When M is –ve, ρ extends below the beam.



12


Moment-Curvature Relationship• Using flexure formula, σ = −My/I, curvature is also

• Eqns 12-2 and 12-3 valid for either small or large radii of curvature.

( )3-121Eyσ

ρ−=



13

12.2 SLOPE AND DISPLACEMENT BY INTEGRATION

• Let’s represent the curvature in terms of ν and x.

• Substitute into Eqn 12-2

( ) 23

2

22

1

1

⎥⎦⎤

⎢⎣⎡ +

=

dxd

dxd

υ

υ

ρ

( )( )4-12

12

32

22

EIM

dxd

dxd

=

⎥⎦⎤

⎢⎣⎡ + υ

υ



14


• Most engineering codes specify limitations on deflections for tolerance or aesthetic purposes.

• Slope of elastic curve determined from dν/dx is very small and its square will be negligible compared with unity.

• Therefore, by approximation 1/ρ = d2ν /dx2, Eqn 12-4 rewritten as

• Differentiate each side w.r.t. x and substitute V = dM/dx, we get

( )5-122

2

EIM

dxd =υ

( ) ( )6-122

2xV

dxdEI

dxd

=⎟⎟⎠

⎞⎜⎜⎝

⎛ υ



15


• Differentiating again, using −w = dV/dx yields

• Flexural rigidity is constant along beam, thus

( ) ( )7-122

2

2

2xw

dxdEI

dxd

−=⎟⎟⎠

⎞⎜⎜⎝

⎛ υ

( ) ( )8-124

4xw

dxdEI −=υ

( ) ( )9-123

3xV

dxdEI =υ

( )10-12)(2

2xM

dxdEI =υ



16


• Generally, it is easier to determine the internal moment M as a function of x, integrate twice, and evaluate only two integration constants.

• For convenience in writing each moment expression, the origin for each x coordinate can be selected arbitrarily.

Sign convention and coordinates• Use the proper signs for M, V and w.



17


Boundary and continuity conditions• Possible boundary

conditions are shown here.



18


Boundary and continuity conditions• If a single x coordinate cannot be used to express

the eqn for beam’s slope or elastic curve, then continuity conditions must be used to evaluate some of the integration constants.



19


Procedure for analysisElastic curve• Draw an exaggerated view of the beam’s elastic

curve. • Recall that zero slope and zero displacement

occur at all fixed supports, and zero displacement occurs at all pin and roller supports.

• Establish the x and ν coordinate axes. • The x axis must be parallel to the undeflected

beam and can have an origin at any pt along the beam, with +ve direction either to the right or to the left.



20


Procedure for analysisElastic curve• If several discontinuous loads are present,

establish x coordinates that are valid for each region of the beam between the discontinuties.

• Choose these coordinates so that they will simplify subsequent algrebraic work.



21


Procedure for analysisLoad or moment function• For each region in which there is an x coordinate,

express that loading w or the internal moment Mas a function of x.

• In particular, always assume that M acts in the +ve direction when applying the eqn of moment equilibrium to determine M = f(x).



22


Procedure for analysisSlope and elastic curve• Provided EI is constant, apply either the load eqn

EI d4ν/dx4 = −w(x), which requires four integrations to get ν = ν(x), or the moment eqns EI d2ν /dx2 = M(x), which requires only two integrations. For each integration, we include a constant of integration.

• Constants are evaluated using boundary conditions for the supports and the continuity conditions that apply to slope and displacement at pts where two functions meet.



23

12.2 SLOPE AND DISPLACEMENT BY INTEGRATIONProcedure for analysisSlope and elastic curve• Once constants are evaluated and substituted

back into slope and deflection eqns, slope and displacement at specific pts on elastic curve can be determined.

• The numerical values obtained is checked graphically by comparing them with sketch of the elastic curve.

• Realize that +ve values for slope are counterclockwise if the x axis extends +ve to the right, and clockwise if the x axis extends +ve to the left. For both cases, +ve displacement is upwards.



24

EXAMPLE 12.1Cantilevered beam shown is subjected to a vertical load P at its end. Determine the eqn of the elastic curve. EI is constant.



25

EXAMPLE 12.1 (SOLN)Elastic curve: Load tends to deflect the beam. By inspection, the internal moment can be represented throughout the beam using a single x coordinate.Moment function: From free-body diagram, with Macting in the +ve direction, we have

PxM −=



26

EXAMPLE 12.1 (SOLN)

Slope and elastic curve:Applying Eqn 12-10 and integrating twice yields

( )

( )

( )36

22

1

21

3

1

2

2

2

CxCPxEI

CPxdxdEI

PxdxdEI

++−=

+−=

−=

υ

υ

υ



27

EXAMPLE 12.1 (SOLN)

Slope and elastic curve:Using boundary conditions dν/dx = 0 at x = L, and ν = 0 at x = L, Eqn (2) and (3) becomes

21

3

1

2

60

20

CLCPL

CPL

++−=

+−=



28

EXAMPLE 12.1 (SOLN)Slope and elastic curve:Thus, C1 = PL2/2 and C2 = PL3/3. Substituting these results into Eqns (2) and (3) with θ = dν/dx, we get

Maximum slope and displacement occur at A (x = 0),

( )( )323

22

236

20

LxLxEIP

xLEIP

−+−=

−−=

υ

EIPL

EIPL

AA 32

32−== υθ



29

EXAMPLE 12.1 (SOLN)Slope and elastic curve:Positive result for θA indicates counterclockwise rotation and negative result for A indicates that νA is downward.

Consider beam to have a length of 5 m, support load P = 30 kN and made of A-36 steel having Est = 200 GPa.



30

EXAMPLE 12.1 (SOLN)Slope and elastic curve:Using methods in chapter 11.3, assuming allowable normal stress is equal to yield stress σallow = 250 MPa, then a W310×39 would be adequate (I = 84.8(106) mm4). From Eqns (4) and (5),

EIPL

EIPL

AA 32

32−== υθ



31

EXAMPLE 12.1 (SOLN)Slope and elastic curve:From Eqns (4) and (5),

( ) ( )( )[ ] ( )( )

( ) ( )( )[ ] ( )( ) mm7.73

mm108.84N/mm102003

mm/m10m5N/kN10kN30

rad0221.0mm108.84N/mm102002

mm/m10m5N/kN10kN30

4623

3233

4623

2233

−=⎥⎦⎤

⎢⎣⎡×

−=

=⎥⎦⎤

⎢⎣⎡×

=

A

A

υ

θ



32

EXAMPLE 12.1 (SOLN)Slope and elastic curve:Since 2θA = (dν/dx)2 = 0.000488 << 1, this justifies the use of Eqn 12-10 than the more exact 12-4. Also, since it is for a cantilevered beam, we’ve obtained larger values for θ and ν than would be obtained otherwise.



33

EXAMPLE 12.1 (SOLN)

SOLUTION 2Using Eqn 12-8 to solve the problem. Here w(x) = 0 for 0 ≤ x ≤ L, so that upon integrating once, we get the form of Eqn 12-19

VCdxdEI

dxdEI

==

=

13

3

4

4

'

0

υ

υ



34

EXAMPLE 12.1 (SOLN)Solution IIShear constant C’1 can be evaluated at x = 0, since VA = −P. Thus, C’1 = −P. Integrating again yields the form of Eqn 12-10,

Here, M = 0 at x = 0, so C’2 = 0, and as a result, we obtain Eqn 1 and solution proceeds as before.

MCPxdxdEI

PdxdEI

=+−=

−=

22

2

3

3

'υ

υ



35

EXAMPLE 12.4Beam is subjected to load P at its end. Determine the displacement at C. EI is a constant.



36

EXAMPLE 12.4 (SOLN)Elastic curveBeam deflects into shape shown. Due to loading, two x coordinates will be considered, 0 ≤ x1 ≤ 2a and 0 ≤ x2 ≤ a, where x2 is directed to the left from C since internal moment is easy to formulate.



37

EXAMPLE 12.4 (SOLN)

Moment functionsUsing free-body diagrams, we have

Slope and Elastic curve: Applying Eqn 10-12,

2211 2PxMxPM −=−=

( )

( )212

14

220

2113

11

12

11

1

121

12

1

CxCxPEI

CxPdxdEI

xPdxdEIax

++−=

+−=

−==≤≤

υ

υ

υfor



38

EXAMPLE 12.4 (SOLN)

Slope and Elastic curve: Applying Eqn 10-12,

( )

( )46

32

0

4233

22

32

22

2

222

22

2

CxCxPEI

CxPdxdEI

PxdxdEIax

++−=

+−=

−==≤≤

υ

υ

υfor



39

EXAMPLE 12.4 (SOLN)

Slope and Elastic curve: The four constants of integration determined using three boundary conditions, ν1 = 0 at x1 = 0, ν1 = 0 at x1 = 2a, and ν2 =0 at x2 = a and a discontinuity eqn.Here, continuity of slope at roller requires dν1/dx1 = −dν2/dx2 at x1 = 2a and x2 = a.

( ) ( ) 212

11

11

2212

0;20

2000;00

CaCaPaxat

Cxat

++−===

++===

υ

υ



40

EXAMPLE 12.4 (SOLN)

Slope and Elastic curve:

Solving, we obtain

( ) ( ) ( ) ( ) ⎟⎠⎞

⎜⎝⎛ +−−=+−−=

++−===

32

12

2

2

1

1

433

22

22

4;2

60;0

CaPCaPdx

addx

ad

CaCaPaxat

υυ

υ

34232

2

1 670

3PaCPaCCPaC −====



41

EXAMPLE 12.4 (SOLN)

Slope and Elastic curve: Substituting C3 and C4 into Eqn (4) gives

Displacement at C is determined by setting x2 = 0,

EIPax

EIPax

EIP 3

2

23

22 67

6−+−=υ

EIPa

C

3−=υ



42

*12.3 DISCONTINUITY FUNCTIONS

• A simplified method for finding the eqn of the elastic curve for a multiply loaded beam using a single expression, formulated from the loading on the beam , w = w(x), or the beam’s internal moment, M = M(x) is discussed below.

Discontinuity functionsMacaulay functions• Such functions can be used to describe distributed

loadings, written generally as

( ) ( )0

11-12{

0{

≥≥−

<=−

naxforax

axforaxn

n



43


Discontinuity functionsMacaulay functions• x represents the coordinate position of a pt along

the beam• a is the location on the beam where a

“discontinuity” occurs, or the pt where a distributed loading begins.

• Integrating Macaulay functions, we get

• The functions describe both uniform load and triangular load.

( )∫ ++

−=−

+12-12

1

1C

naxdxax

nn



44


Discontinuity functionsMacaulay functions



45


Discontinuity functionsSingularity functions• Used to describe the pt location of concentrated

forces or couple moments acting on a beam.• A concentrated force P can be considered as a

special case of distributed loading, where w = P/esuch that its width is ε, ε→0.

( )13-12{0{1

axforPaxforaxPw

==

≠=−= −



46


Discontinuity functionsSingularity functions• Similarly, couple moment M0, considered +ve

counterclockwise, is a limitation as ε→0 of two distributed loadings. Hence,

• Integration of the two functions yields

( )14-12{0{

0

20

axforMaxforaxMw

==

≠=−= −

( )∫ −−=−=− + 15-122,1,1 naxdxax nn



47


Discontinuity functionsSingularity functions



48


Procedure for analysisElastic curve• Sketch the beam’s elastic curve and identify the

boundary conditions at the supports.• Zero displacement occurs at all pin and roller

supports, and zero slope and zero displacement occurs at fixed supports.

• Establish the x axis so that it extends to the right and has its origin at the beam’s left end.

Load or moment function• Calculate the support reactions and then use the

discontinuity functions in Table 12-2 to express either the loading w or the internal moment M as a function of x.



49


Procedure for analysisLoad or moment function• Calculate the support reactions and then use the

discontinuity functions in Table 12-2 to express either the loading w or the internal moment M as a function of x.

• Make sure to follow the sign convention for each loading as it applies for this equation.

• Note that the distributed loadings must extend all the way to the beam’s right end to be valid. If this does not occur, use the method of superposition.



50


Procedure for analysisSlope and elastic curve• Substitute w into EI d4ν/dx4 = −w(x) or M into the

moment curvature relation EI d2ν/dx2 = M, and integrate to obtain the eqns for the beam’s slope and deflection.

• Evaluate the constants of integration using the boundary conditions, and substitute these constants into the slope and deflection eqns to obtain the final results.

• When the slope and deflection eqns are evaluated at any pt on the beam, a +ve slope is counterclockwise, and a +ve displacement is upward.



51

EXAMPLE 12.5Determine the eqn of the elastic curve for the cantilevered beam shown. EI is constant.



52

EXAMPLE 12.5 (SOLN)Elastic curveThe loads cause the beam to deflect as shown. The boundary conditions require zero slope and displacement at A.



53

EXAMPLE 12.5 (SOLN)Loading functionsSupport reactions shown on free-body diagram. Since distributed loading does not extend to C as required, use superposition of loadings to represent same effect. By sign convention, the 50-kNm couple moment, the 52-kN force at A, and portion of distributed loading from B to Con the bottom of the beam are all –ve.



54

EXAMPLE 12.5 (SOLN)Loading functionsTherefore,

The 12-kN load is not included, since x cannot be greater than 9 m. Because dV/dx = −w(x), then by integrating, neglect constant of integration since reactions are included in load function, we have

02

01

m5m/kN8m5mkN50

0mkN2580kN52

−−−⋅−

−⋅+−−=−

−

xx

xxw

11

110

08050

080258052

−+−+

−−−−−=−

−

xx

xxxV



55

EXAMPLE 12.5 (SOLN)

Loading functionsFurthermore, dM/dx = V, so integrating again yields

The same result can be obtained directly from Table 12-2.

( ) ( )

( ) mkN55054452258

582155008

210258

022

2020

⋅−+−+−+−=

−+−+−−−−=

xxxx

xxxxM



56

EXAMPLE 12.5 (SOLN)Slope and elastic curveApplying Eqn 12-10 and integrating twice, we have

214

2432

1312

2022

2

531

52531

326129

534550

3426258

54550452258

3

CxCx

xxxxEI

CxxxxxdxdEI

xxxxdxdEI

++−+

−+−+−=

+−+−+−+−=

−+−+−+−=

υ

υ

υ



57

EXAMPLE 12.5 (SOLN)Slope and elastic curveSince dν/dx = 0 at x = 0, C1 = 0; and ν = 0 at x = 0, so C2 = 0. Thus

m)531525

31

326129(1

42

432

−+−+

−+−=

xx

xxxEI

υ



58

*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD

Theorem 2• The vertical deviation of the tangent at a pt (A) on

the elastic curve w.r.t. the tangent extended from another pt (B) equals the moment of the area under the ME/I diagram between these two pts (A and B).

• This moment is computed about pt (A) where the vertical deviation (tA/B) is to be determined.



59


Theorem 2



60


Procedure for analysisM/EI Diagram• Determine the support reactions and draw the

beam’s M/EI diagram.• If the beam is loaded with concentrated forces, the

M/EI diagram will consist of a series of straight line segments, and the areas and their moments required for the moment-area theorems will be relatively easy to compute.

• If the loading consists of a series of distributed loads, the M/EI diagram will consist of parabolic or perhaps higher-order curves, and we use the table on the inside front cover to locate the area and centroid under each curve.



61


Procedure for analysisElastic curve• Draw an exaggerated view of the beam’s elastic

curve.• Recall that pts of zero slope and zero

displacement always occur at a fixed support, and zero displacement occurs at all pin and roller supports.

• If it is difficult to draw the general shape of the elastic curve, use the moment (M/EI) diagram.

• Realize that when the beam is subjected to a +ve moment, the beam bends concave up, whereas -ve moment bends the beam concave down.



62


Procedure for analysisElastic curve• An inflection pt or change in curvature occurs

when the moment if the beam (or M/EI) is zero.• The unknown displacement and slope to be

determined should be indicated on the curve.• Since moment-area theorems apply only between

two tangents, attention should be given as to which tangents should be constructed so that the angles or deviations between them will lead to the solution of the problem.

• The tangents at the supports should be considered, since the beam usually has zero displacement and/or zero slope at the supports.



63


Procedure for analysisMoment-area theorems• Apply Theorem 1 to determine the angle between

any two tangents on the elastic curve and Theorem 2 to determine the tangential deviation.

• The algebraic sign of the answer can be checked from the angle or deviation indicated on the elastic curve.

• A positive θB/A represents a counterclockwise rotation of the tangent at B w.r.t. tangent at A, and a +ve tB/A indicates that pt B on the elastic curve lies above the extended tangent from pt A.



64

EXAMPLE 12.7Determine the slope of the beam shown at pts B and C. EI is constant.



65

EXAMPLE 12.7 (SOLN)M/EI diagram: See below.Elastic curve:The force P causes the beam to deflect as shown.



66

EXAMPLE 12.7 (SOLN)Elastic curve:The tangents at B and C are indicated since we are required to find B and C. Also, the tangent at the support (A) is shown. This tangent has a known zero slope. By construction, the angle between tan A and tan B, θB/A, is equivalent to θB, or

ACCABB θθθθ == and



67

EXAMPLE 12.7 (SOLN)Moment-area theorem:Applying Theorem 1, θB/A is equal to the area under the M/EI diagram between pts A and B, that is,

EIPL

LEI

PLLEI

PLABB

83

2221

222

−=

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−+⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−==θθ



68

EXAMPLE 12.7 (SOLN)

Moment-area theorem:The negative sign indicates that angle measured from tangent at A to tangent at B is clockwise. This checks, since beam slopes downward at B.Similarly, area under the M/EI diagram between pts Aand C equals θC/A. We have

EIPL

LEIPL

ACC

2

21

2−=

⎟⎠⎞

⎜⎝⎛−==θθ



69

EXAMPLE 12.8Determine the displacement of pts B and C of beam shown. EI is constant.



70

EXAMPLE 12.8 (SOLN)M/EI diagram: See below.Elastic curve:The couple moment at C cause the beam to deflect as shown.



71

EXAMPLE 12.8 (SOLN)Elastic curve:The required displacements can be related directly to deviations between the tangents at B and A and Cand A. Specifically, ΔB is equal to deviation of tan Afrom tan B,

ACCABB tt =Δ=Δ



72

EXAMPLE 12.8 (SOLN)Moment-area theorem:Applying Theorem 2, tB/A is equal to the moment of the shaded area under the M/EI diagram between A and B computed about pt B, since this is the pt where tangential deviation is to be determined. Hence,

EILML

EIMLt ABB 824

200 −=⎥⎦

⎤⎢⎣⎡

⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛==Δ



73

EXAMPLE 12.8 (SOLN)Moment-area theorem:Likewise, for tC/A we must determine the moment of the area under the entire M/EI diagram from A to Cabout pt C. We have

Since both answers are –ve, they indicate that pts Band C lie below the tangent at A. This checks with the figure.

( )EILML

EIMLt ACC 22

200 −=⎥⎦

⎤⎢⎣⎡

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛==Δ



74

12.5 METHOD OF SUPERPOSITION

• The differential eqn EI d4ν/dx4 = −w(x) satisfies the two necessary requirements for applying the principle of superposition

• The load w(x) is linearly related to the deflection ν(x)

• The load is assumed not to change significantly the original geometry of the beam or shaft.



75

EXAMPLE 12.16Steel bar shown is supported by two springs at its ends A and B. Each spring has a stiffness k = 45 kN/m and is originally unstretched. If the bar is loaded with a force of 3 kN at pt C, determine the vertical displacement of the force. Neglect the weight of the bar and take Est = 200 GPa, I = 4.6875×10-6 m.



76

EXAMPLE 12.16 (SOLN)

End reactions at A and B are computed and shown. Each spring deflects by an amount

( )

( ) m0222.0kN/m45kN1

m0444.0kN/m45kN2

1

1

==

==

B

A

υ

υ



77

EXAMPLE 12.16 (SOLN)If bar is considered rigid, these displacements cause it to move into positions shown. For this case, the vertical displacement at C is

( ) ( ) ( ) ( )[ ]

[ ]

m0370.0

m0282.0m0444.032m0222.0

m3m2

1111

=

−+=

−+= BABC υυυυ



78

EXAMPLE 12.16 (SOLN)We can find the displacement at C caused by the deformation of the bar, by using the table in Appendix C. We have

( ) ( )( )( )( ) ( ) ( ) ( )[ ]( )( ) ( )( ) ( )

mm422.1m3m106875.4kN/m102006

m1m2m3m2m1kN36

4626

222

2222

=

−−=

−−=

−

abLEIL

PabCυ



79

EXAMPLE 12.16 (SOLN)Adding the two displacement components, we get

( )mm4.38m0384.0

m001422.0m0370.0==+=+ Cυ



80

12.6 STATICALLY INDETERMINATE BEAMS AND SHAFTS

• A member of any type is classified as statically indeterminate if the no. of unknown reactions exceeds the available no. of equilibrium eqns.

• Additional support reactions on beam that are not needed to keep it in stable equilibrium are called redundants.

• No. of these redundants is referred to as the degree of indeterminacy.



81

12.7 STATICALLY INDETERMINATE BEAMS AND SHAFTS: METHOD OF INTEGRATION• For a statically indeterminate beam, the internal

moment M can be expressed in terms of the unknown redundants.

• After integrating this eqn twice, there will be two constants of integration and the redundants to be found.

• The unknowns can be found from the boundary and/or continuity conditions for the problem.



82

EXAMPLE 12.17Beam is subjected to the distributed loading shown. Determine the reactions at A. EI is a constant.



83

EXAMPLE 12.17 (SOLN)Elastic curve:Beam deflects as shown. Only one coordinate x is needed. For convenience, we will take it directed to the right, since internal moment is easy to formulate.



84

EXAMPLE 12.17 (SOLN)Moment function:Beam is indeterminate to first degree as indicated from the free-body diagram. We can express the internal moment M in terms of the redundant force at A using segment shown below.



85

EXAMPLE 12.17 (SOLN)

Moment function:

Slope and elastic curve:Applying Eqn 12-10,

LxwxAM y

3

061

−=

21

5

03

1

4

02

3

02

2

1201

61

241

21

61

CxCLxwxAEI

CLxwxA

dxdEI

LxwxA

dxdEI

y

y

y

++−=

+−=

−=

υ

υ

υ



86

EXAMPLE 12.17 (SOLN)Slope and elastic curve:The three unknowns Ay, C1 and C2 are determined from the boundary conditions x = 0, ν = 0; x = L, dν/dx = 0; and x = L, ν = 0. Applying these conditions yields

214

03

13

02

2

1201

610;0,

241

210;0,

0000;0,0

CLCLwLALx

CLwLAdxdLx

Cx

y

y

++−===

+−===

++−===

υ

υυ



87

EXAMPLE 12.17 (SOLN)Slope and elastic curve:Solving,

Using the result for Ay, the reactions at B can be determined from the equations of equilibrium. Show that Bx = 0. By = 2w0L/5 and MB= w0L2/15

0120

1101

23

01

0

=−=

=

CLwC

LwAy



88

12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS: MOMENT-AREA METHOD• Draw the ME/I diagrams such that the redundants

are represented as unknowns.• Apply the 2 moment-area theorems to get the

relationships between the tangents on elastic curve to meet conditions of displacement and/or slope at supports of beam.

• For all cases, no. of compatibility conditions is equivalent to no. of redundants.



89

12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS: MOMENT-AREA METHODMoment diagrams constructed by method of

superposition• Since moment-area theorems needs calculation of

both the area under the ME/I diagram and centroidal location of this area, the method of superposition can be used to combine separate ME/I diagrams for each of the known loads.

• This will be relevant if the resultant moment diagram is of a complicated shape.



90

12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS: MOMENT-AREA METHODMoment diagrams constructed by method of

superposition• Most loadings on beams are a

combination of the four loadings as shown.



91

12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS: MOMENT-AREA METHOD



92

12.8 STATICALLY INDETERMINATE BEAMS AND SHAFTS: MOMENT-AREA METHOD



93

EXAMPLE 12.20Beam is subjected to couple moment at its end C as shown. Determine the reaction at B. EI is constant.



94

EXAMPLE 12.20 (SOLN)M/EI Diagram:Free-body diagram as shown. By inspection, beam is indeterminate to first degree. To get a direct solution, choose By as the redundant.Using superposition, the M/EI diagrams forBy and M0, each applied to the simply supported beam are shown.



95

EXAMPLE 12.20 (SOLN)Elastic curve:Elastic curve as shown. Tangents at A, B and C has been established. Since ΔA = ΔB = ΔC = 0, then tangential deviations shown must be proportional,

( )121

CACB tt =



96

EXAMPLE 12.20 (SOLN)Elastic curve:From ME/I diagram, we have

( ) ( )

( )

( ) ( ) ( ) ( )⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛+⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛+

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛+⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

LEIMLL

EILB

Lt

LEIML

LEIMLL

EILB

Lt

yCA

yCB

2212

322

221

22

221

32

221

31

0

0

0



97

EXAMPLE 12.20 (SOLN)Elastic curve:Substituting into Eqn (1), we have

Equations of equilibrium:Reactions at A and C can now be determined from the eqns of equilibrium. Show that Ax = 0, Cy = 5M0/4L, and Ay = M0/4L.

LMBy 2

3 0=



98

EXAMPLE 12.20 (SOLN)Equations of equilibrium:From figure shown, this problem can also be worked out in terms of the tangential deviations,

ACAB tt21

=



99

12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS: METHOD OF SUPERPOSITION• First, identify the redundant support reactions on

the beam.• Remove these reactions from the beam to get a

primary beam that is statically determinate and stable and subjected to external load only.

• Add to this beam with a series of similarly supported beams, each with a separate redundant, then by principle of superposition, the final loaded beam is obtained.

• After computing the redundants, the other reactions on the beam determined from the eqns of equilibrium.

• This method of analysis is sometimes called the force method.



100

12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS: METHOD OF SUPERPOSITIONProcedure for analysisElastic curve• Specify unknown redundant forces or moments that

must be removed from the beam in order to make it statically determinate and stable.

• Use principle of superposition, draw the statically indeterminate beam and show it to be equal to a sequence of corresponding statically determinate beams.

• The first beam (primary) supports the same external loads as the statically indeterminate beam, and each of the other beams “added” to the primary beam shows the beam loaded with a separate single redundant force or moment.



101

12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS:METHOD OF SUPERPOSITIONProcedure for analysisElastic curve• Sketch the deflection curve for each beam and

indicate symbolically the displacement or slope at the pt of each redundant force or moment.

Compatibility equations• Write a compatibility eqn for the displacement or

slope at each pt where there is a redundant force or moment.

• Determine all the displacements or slopes using an appropriate method explained in chapter 12.2−12.5.

• Substitute the results into the compatibility eqns and solve for the unknown redundants.



102

12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS: METHOD OF SUPERPOSITIONProcedure for analysisCompatibility equations• If a numerical value for a redundant is +ve, it has

the same sense of direction as originally assumed. • Similarly, a –ve numerical value indicates the

redundant acts opposite to its assumed sense of direction.

Equilibrium equations• Once the redundant forces and/or moments have

been determined, the remaining unknown reactions can be found from the eqns of equilibrium applied to the loadings shown on the beam’s free-body diagram.



103

EXAMPLE 12.22Determine the reactions on the beam shown. Due to loading and poor construction, the roller support at Bsettles 12 mm. Take E = 200 GPa and I = 80(106) mm4.



104

EXAMPLE 12.22 (SOLN)Principle of superpositionBy inspection, beam is indeterminate to the first degree. Roller support at B is chosen as the redundant. Principle of superposition is shown. Here, By is assumed to act upwards on the beam.



105

EXAMPLE 12.22 (SOLN)Compatibility equationWith reference to pt B, we require

Using table in Appendix C, displacements are

( ) ( )1'm012.0 BB υυ −=+

( )( )EIEIEI

wLB

344 mkN640768

m8kN/m2457685 ⋅

===υ

( )EI

BEI

BEI

PL yyB

333 m67.1048

m848

' ===υ



106

EXAMPLE 12.22 (SOLN)Compatibility equationThus Eqn (1) becomes

Expressing E and I in units of kN/m2 and m4, we have

yBEI 67.10640012.0 −=

( )( ) ( )[ ]kN0.42

67.10640108010200012.0 66

=

−=−

y

y

B

B



107

EXAMPLE 12.22 (SOLN)Equilibrium equations:Applying this result to the beam, we then calculate the reactions at A and C using eqns of equilibrium.

( ) ( ) ( )

kN51

0kN00.3kN0.42kN96;0

kN00.3

0m8m4kN0.42m2kN96;0

=

=++−=+

=

=++−=+

∑

∑

y

y

y

yA

A

AyF

C

CM



108

EXAMPLE 12.24

Determine the moment at B for beam shown. EI is constant. Neglect the effects of axial load.



109

EXAMPLE 12.24Principle of superposition:Since axial load if neglected, a there is a vertical force and moment at Aand B. Since only two eqns of equilibrium are available, problem is indeterminate to the second degree.Assume that By and MB are redundant, so that by principle of superposition, beam is represented as a cantilever, loaded separately by distributed load and reactions By and MB, as shown.



110

EXAMPLE 12.24

Compatibility equations:Referring to displacement and slope at B, we require

Using table in Appendix C to compute slopes and displacements, we have

( ) ( )( ) ( )2'''0

1'''0

BBB

BBBυυυθθθ++=+++=+

( )

( )( )EIEIEI

wLEIEIEI

wL

B

B

42384

m4kN/m973847

1248

m4kN/m948

44

33

===

===

υ

θ



111

EXAMPLE 12.24

Compatibility equations:

( )

( )

( )

( )EIM

EIM

EIML

EIM

EIM

EIML

EIB

EIB

EIPL

EIB

EIB

EIPL

BBB

BBB

yyB

yyB

82

m42

''

4m4''

33.213

m43

'

82

m42

'

22

33

22

===

===

===

===

υ

θ

υ

θ



112

EXAMPLE 12.24

Compatibility equations:Substituting these values into Eqns (1) and (2) and canceling out the common factor EI, we have

Solving simultaneously, we get

( )( ) By

By

MB

MB

833.21420

48120

++=+

++=+

mkN75.3

kN375.3

⋅=

−=

B

y

M

B



113

CHAPTER REVIEW• The elastic curve represents the centerline

deflection of a beam or shaft.• Its shape can be determined using the moment

diagram.• Positive moments cause the elastic curve to

concave upwards and negative moments cause it to concave downwards.

• The radius of curvature at any pt is determined from 1/ρ = M/EI.

• Eqn of elastic curve and its slope can be obtained by first finding the internal moment in the member as a function of x.



114

CHAPTER REVIEW• If several loadings act on the member, then

separate moment functions must be determined between each of the loadings.

• Integrating these functions once usingEI(d2ν/dx2) = M(x) gives the eqn for the slope of the elastic curve, and integrating again gives the eqn for the deflection.

• The constants of integration are determined from the boundary conditions at the supports, or in cases where several moment functions are involved, continuity of slope and deflection at pts where these functions join must be satisfied.



115

CHAPTER REVIEW• Discontinuity functions allow us to express the eqn

of elastic curve as a continuous function, regardless of the no. of loadings on the member.

• This method eliminates the need to use continuity conditions, since the two constants of integration can be determined solely from the two boundary conditions.

• The moment-area method if a semi-graphical technique for finding the slope of tangents or the vertical deviation of tangents at specific pts on the elastic curve.



116

CHAPTER REVIEW• The moment-area method requires finding area

segments under the M/EI diagram, or the moment of these segments about pts on the elastic curve.

• The method works well for M/EI diagrams composed of simple shapes, such as those produced by concentrated forces and couple moments.

• The deflection or slope at a pt on a member subjected to various types of loadings can be determined by using the principle of superposition. The table in the back of the book can be used for this purpose.



117

CHAPTER REVIEW• Statically indeterminate beams and shafts have

more unknown support reactions than available eqns of equilibrium.

• To solve them such problems, we first identify the redundant reactions, and the other unknown reactions are written in terms of these redundants.

• The method of integration or moment-area theorems can be used to solve for the unknown redundants.

• We can also determine the redundants by using the method of superposition, where we consider the continuity of displacement at the redundant.



118

CHAPTER REVIEW• The displacement due to the external loading is

determined with the redundant removed, and again with the redundant applied and external loading removed.

• The tables in Appendix C of this book can be used to determine these necessary displacements.

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