Beam Deflection 4th Order Method

8/13/2019 Beam Deflection 4th Order Method

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11Beam Deflections:4th Order Method

& Adl Topics

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Lecture 11: BEAM DEFLECTIONS: 4TH ORDER METHOD & ADL TOPICS 112

TABLE OF CONTENTS

Page

11.1. Fourth Order Method For Beam Deflections 113

11.1.1. Example 1: Cantilever under Triangular Distributed Load . . . 113

11.2. Superposition 114

11.2.1. Example 2: Cantilever Under Two Load Cases . . . . . . 114

11.2.2. Example 3: A Statically Indeterminate Beam . . . . . . . 115

11.3. Continuity Conditions 116

11.3.1. Example 4: Simply Supported Beam Under Midspan Point Load 116

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113 11.1 FOURTH ORDER METHOD FOR BEAM DEFLECTIONS

11.1. Fourth Order Method For Beam Deflections

The fourth-order method to find beam deflections gets its name from the order of the ODE to be

integrated: E Izz vI V(x) = p(x)is a fourth order ODE. The procedure can be broken down into the

following steps.

1. Express the applied load p(x)as function ofx , using positive-upward convention.23. Integrate p(x)twice to get Vy(x)andMz(x)

4. Pause. Determineintegrationconstants from static BCs, and replace inMz(x). (If theconstants

are too complicated in terms of the data, they might be kept in symbolic form until later.)

58. From here on, same as the second order method.

An example of this technique follows.

11.1.1. Example 1: Cantilever under Triangular Distributed Load

This example has been worked out in the previous lecture using the second order method. It is

defined in Figure 11.1, which is reproduced for convenience.

w

w(x) = w x /L

x

yB

BConstantEIzz

A B

L

Figure11.1. Beam problem for Example 1.

From inspection the applied load is

p(x) = wBxL

(11.1)

Integrating p(x)twice yields

Vy(x) =

p(x) dx= wBx2

2L+ C1,

Mz(x) =

Vy(x) dx= wBx

3

6L C1x + C2.

(11.2)

Apply now the static BCs at the free end A: Vy A=

Vy(0)=

C1=

0 andMz A=

Mz(0)=

C2=

0.

Hence

Mz(x) = 12 w(x)x( 13x) = wBx

3

6L(11.3)

From here on the steps are the same as in the second order method worked out in Lecture 10. The

deflection curve is

v(x) = wB120E IzzL

(x5 5L4x + 4L5) (11.4)

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P

w

w(x) = w x /L

x

yB

BConstantEIzz

A B

(a) Original problem

L

P

x

y

A B

(b) Decomposition into two load cases and superposition

L

w

w(x) = w x /L

x

yB

B

A B

L

= +


The maximum deflection occurs at the tip A, and is given by

vA= v(0) = wBL4

30E Izz (11.5)

The negative sign indicates that the beam deflects downward ifwB >0.

11.2. Superposition

All equations of the beam theory we are using are linear. This makes possible to treat complicated

load cases by superposition of the solutions of simpler ones. Simple beam configurations and load

cases may be compiled in textbooks and handbooks; for example Appendix D of Beer-Johnston-DeWolf. The following example illustrates the procedure.

11.2.1. Example 2: Cantilever Under Two Load Cases

Consider the problem shown in Figure 11.2(a). The cantilever beam is subject to a tip point force

as well as a triangular distributed load. This combination can be decomposed into the two load

cases shown in Figure 11.2(b). Both of these have been separately solved previously as Examples

1 and 2 of Lecture 10 (the latter also as the example in the previous section). The deflection curves

for these cases will be distinguished as v P (x)andvw(x), respectively.

We had obtained

v P (x) = P6E Izz

(L x)2(2L +x), vw(x) = wB120E Izz L

(x5 5L4x + 4L5) (11.6)

The deflection under the combined loading is obtained by adding the foregoing solutions:

v(x) = v P (x) + vw(x) = P6E Izz

(L x)2(2L +x) wB120E Izz L

(x5 5L4x + 4L5).(11.7)

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115 11.2 SUPERPOSITION

R

w

w(x) = w x /L

x

yB

BConstantEIzz

A B

(a) Original (statically indeterminate) problem

L

x

y

A B

(c) Decomposition into two load cases and superposition

L

w

w(x) = w x /L

x

yB

B

A B

L

=+

A B

(b) Support reactions

M

A

RA

RB

B


The tip deflection is

vA= v(0) = P L3

3E Izz wBL

4

30E Izz= L

3

30E Izz(10P + wBL) (11.8)

Superposition can be also used for any other quantity of interest, for example transverse shear

forces, bending moments and deflection curve slopes. An application to statically indeterminate

beam analysis is given next.

11.2.2. Example 3: A Statically Indeterminate Beam

The problem is defined in Figure 11.3(a). The beam is simply supported at A and clamped at B. Ifthe supports are removed 3 reactions are activated: RA,RBand MB , as pictured in Figure 11.3(b).

But there are only two nontrivial static equilibrium equations:

Fy= 0 and

Many poin t =0because

Fx= 0 is trivially satisfied. Consequently the beam is statically indeterminatebecause

the reactions cannot be determined by statics alone. One additional kinematic equation is required

to complete the analysis.

We select reactionRAas redundant forceto be carried along as afictitious applied load. Removing

the support at A and includingRAmakes the beam statically determinate. See Figure 11.3(b). This

beam may be viewed as being loaded by a combination of two load cases: (1) the actual triangular

loadw(x), and (2) a point load RAat A. But this is exactly the problem solved in Example 2, if we

replace Pby RA. The deflection curve of this beam isv(x) = RA

6E Izz(L x)2(2L +x) wB

120E Izz L(x5 5L4x + 4L5). (11.9)

Now the tip deflection must be zero because there is a simple support at A. Setting vA= v(0) = 0provides the value ofRA:

vA= v(0) = RAL3

3E Izz wBL

4

30E Izz= 0 RA= wBL

10(11.10)

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This reaction value can be substituted to complete the solution. For example, the bending moment

is

Mz(x) = RAx wBx

3

6L= wBL x

10 wBx

3

6L= wBx

30L(3L2 5x2) (11.11)

The moment is zero at A (x= 0), becomes positive for 0 < x < L3/5 .7746L , crosses zeroatx= 0.7746Land reaches MzB= wBL2/15 at thefixed end. The deflection is

v(x) = wBL60E Izz

(L x)2(2L +x) wB120E Izz L

(x5 5L4x + 4L5), (11.12)

which may be simplified to

v(x) = wBx120E IzzL

x(L2 x2)2 (11.13)

11.3. Continuity ConditionsIf the applied load is discontinuous, i.e., not a smooth function ofx , it is necessary to divide the

beam into segments separated by the discontinuity points. The ODEs are integrated over each

segment. These solutions are patchedby continuityconditions expressing that the slope v(x)and the deflectionv(x)are continuous between segments. This matching results in extra relations

between integration constants, which permits elimination of all integration constants except those

that can be determined by the standard BCs. The procedure is illustrated with the next example.

11.3.1. Example 4: Simply Supported Beam Under Midspan Point Load

The problem is defined in Figure 11.4(a). The calculation of the deflection curve will be done by the

second order method. Divide the beam into two segments: AC, which extends over 0 x 12L ,and CB, which extends over 1

2L x L . For brevity, these are identified as segments 1 and 2,

respectively, in the equations below.

The expression of the bending moment over each segment is easily obtained from statics. From

symmetry, the support reactions are obviously RA= RB= 12 Pas shown in Figure 11.4(b). Byinspection one obtains that the bending moment Mz (x), diagrammed in Figure 11.4(c), is

Mz (x) =

Mz1(x) = Px2 over segment 1 (AC),Mz2(x) = P(L x)2 over segment 2 (CB).

(11.14)

IntegrateMz /E Izz over each segment:

E Izz v(x) =

E Izz v1(x) = P x

2

4 + C1 over segment 1 (AC),

E Izz v2(x) = P x(2L x)4 +C1 over segment 2 (CB).

(11.15)

It is convenient to stop here and get rid ofC1to avoid proliferation of integration constants. Todo that, note that the midspan slopev Cmust be the same from both expressions: v

C= v1( 12L) =

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117 11.3 CONTINUITY CONDITIONS

(a) Problem definition

A x

y

C B

L/2 L/2

L

ConstantEIzz

(b) Support reactions and division

into two segments

A B

PP

(c) Bending moment diagram

P

2R =

A

P

2R =

B

P2

R =AP2

R =B

C

A B

P

C

Px2

M (x)=z1P(L-x)

2M (x)=z2

+ +

segment 1

segment 1

segment 2

segment 2


v2(12L). Else the beam would have a kinkat C. This is called a continuity condition. Equating

P L2/16 + C1= (3/16)P L2 +C1yields C1= C1 P L2/8, which is replaced in the secondexpression above:

E Izz v(x)

=

E Izz v

1(x) = P x

2

4 + C1 over segment 1 (AC),

E Izz v 2(x) = P x(2L x)4 P L2

8 + C1 over segment 2 (CB).(11.16)

NowC1is gone. Integrate again both segments:

E Izz v(x) =

E Izz v1(x) = P x3

12 + C1x + C2 over segment 1 (AC),

E Izz v2(x) = P x2 (3L x)

12 P L2x

8 + C1x +C2 over segment 2 (CB).

(11.17)

To get rid ofC2we say that the midspan deflection vCmust be the same from both expressions:vC= v1( 12L)= v2( 12L). This continuity condition gives C2= C2+ P L3/48, which replacedyields

E Izz v(x) =

E Izz v1(x) = P x3

12 + C1x + C2 over segment 1 (AC),

E Izz v2(x) = P x2 (3Lx)

12 P L2x

8 + P L3

48 + C1x + C2 over segment 2 (CB).

(11.18)

We have now only two integration constants. To determine C1and C2use the kinematic BCs at A

and B. vA= v1(0)= C2=0 and vB= v2(L)=0 C1= P L2/16. Substitution gives, after

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some simplifications,

E Izz v(x) =

E Izz v1(x) = P x48 (4x2 3L2) over segment 1 (AC),

E Izz v2(x) = P48 (4x3 12Lx2 + 9L2x L3) over segment 2 (CB).

(11.19)

The midspan deflection, obtainable from either segment, is

vC= v1( 12L) = v2( 12L) = P L3

48E Izz(11.20)

As can be seen the procedure is elaborate and error prone, even for this very simple problem. It can

be streamlined by using Discontinuity Functions (DFs), which are covered in Lecture 12.

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Beam Deflection 4th Order Method

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