deflection of beams - university of virginiapeople.virginia.edu/~ttb/deflect.pdf · deflection of...

1

Deflection of Beams:

1. Equations for Beam Deflection

1.1. Model Elements

• Equilibrium: From the statics of beams,

)()(

)(

2

2

xqdx

Md

xqdx

dV

xVdx

dMz

y

yz

=→

=

= (a)

• The Stress Resultants are obtained from the stresses as

∫∫σ−=A

xxz ydAM (b)

∫∫ τ−=A

xyy dAV (c)

• Stress Strain Relationship: For linearly elastic material

xxxx Eε=σ (d)

• From the Notes on Strains, and Stresses,

2

2

)()( dx

vdyxyxy

xx −≈κ−=ρ−=ε (e)

(where v(x)= transverse deflection of the beam.)

CE 206 - Deflection of Beams 10/17/01

2

1.2. Model Assembly

• Substituting (e) into (d) into (b) and integrating

2

2

)()(

)(dx

vdEIxEIx

EIxM zzzzzz

z ≈κ=ρ

= [1]

• Differentiating eq. [1],

( )xVdx

vdEIdxd

dxdM

yzzz =

= 2

2

(f)

)(2

2

2

2

2

2

xqdx

vdEIdxd

dxMd

dxdV

zzzy =

== [2]

Either [1] or [2] may be used to obtain deflections of beams.

If E, and zzI are constant, the above equations reduce to

zzEIxM

dxvd )(2

2

= [1']

zzEIxV

dxvd )(3

3

= (f')

zzEIxq

dxvd )(4

4

= [2']


3

• Equation [1] or [1'] are 2d order o.d.e.s, which requires 2 boundary conditions to yield unique solutions. Equilibrium is imposed separately using eq. (a)

• Equation [2] or [2'] is a 4th order o.d.e., which requires 4 boundary conditions (two at each end) to yield unique solutions.

• Equations (f) or (f') impose only one equilibrium condition. These are typically used for boundary condition formulation only.

1.3. Solution Strategies

• Strategy 1: For a statically determinate beam, we may find )(xM z by statics. Then, the differential equation

zz

zzzz EI

xMdx

vdxMdx

vdEI )()( 2

2

2

2

=→=

may be integrated directly for v', and v(x). (Two boundary conditions needed)

• Strategy 2: The governing differential equation

)(2

2

2

2

xqdx

vdEIdxd

zz =

may be integrated to obtain v, v', v", v''' as functions of x. (4 boundary conditions are needed.)


4

1.4. Boundary Conditions: Depending upon whether 2d order or 4th order equations are to be solved, we have several possible boundary conditions to be considered. • If moment is imposed on the end, note that

zz

z

EIM

dxvd =2

2

• If shear is imposed on the end, note that

yzz Vdx

vdEIdxd =

2

2

These ideas allow a variety of boundary conditions to be handled. Also note that • For 2d order o.d.e's, a total of 2 b.c. are needed. • For 4th order o.d.e.'s, a total or 4 b.c. are needed.


5

Boundary Conditions to be applied:

1. All 2d order BC also are used for 4th order o.d.e. 2. 4th order BC are established by statics for 2d order o.d.e. Condition Appearance 2d Order

BC 4th Order BC

Fixed end 0 ,0 ==dxdvv

same

Simple support 0=v 0''

0=→

=v

M z

Free end See note 3. 0)'''(

0''==

==vEIVvEIM

zzy

zzz

intermediate roller

'' 21

21

vvvv

==

'''' 21 21

21

vEIvEIMM

zzzz

zz

=→

=

Concentrated end load

See note 3

0'')'''(

0

===→

=

vEIMPvEI

PV

zzz

ozz

y

Concentrated end moment

See note 3. 0''

0)'''(MvEIM

vEIV

zzz

zzy

−==

==

Concentrated intermediate load

'' 21

21

vvvv

==

02

21

)'''()'''(

''''21

PvEIvEIvEIvEI

zzzz

zzzz

=−

=

Concentrated intermediate moment

'' 21

21

vvvv

==

012

21

''''

)'''()'''(21

MvEIvEIvEIvEI

zzzz

zzzz

−=−

=

Intermediate hinge

21 vv = (no slope continuity)

0'' ,0''

)'''()'''(

21

012 12

==

=−

vvPvEIvEI zzzz


6

2. Slope and Deflection by Integration

2.1. Statically Determinate Beams

For these beams, it's often convenient to do the statics separately and use the 2d order o.d.e.'s... S.P. 1: Uniformly loaded cantilever beam

( ) 2/2)(

2/))(()(022

0

0

xLxLwxM

xLxLwxMM

z

zz

−−−=→

−−−−==∑

Therefore,

2/)2('' 220 xLxLwvEI zz +−−=

Subject to the B.C.

v(0)=0, v'(0)=0. We may solve by either direct or indirect integration, since there are no lower degree derivatives in v in the o.d.e.


7

Direct Integration Solution:

(i) First integration:

)

3(

2)('

)3

(2

)0(')('

3220

3220

xLxxLEIwxv

xLxxLEIwvxv

zz

zz

+−−

=→

+−−

=−

(ii) Second integration:

+−−

=→

+−−

=−

12322)(

12322)0()(

43220

43220

xLxxLEIwxv

xLxxLEIwvxv

zz

zz

Indirect integration solution: (i) Integrate twice, picking up constants of integration along the way.

21

43220

1

3220

12322)(

32'

CxCxLxxLEIwxv

CxLxxLEIwv

zz

zz

++

+−−=

+

+−−

=


8

(ii) Substitute the boundary conditions at x=0:

0)0('0)0(

1

2

====

CvCv

Therefore,

+−−=12322

)(4322

0 xLxxLEIwxv

zz

the same as was obtained by direct integration.

It is informative to write this solution as

+

−

−

=4324

0

31

342

8)(

Lx

Lx

Lx

EILwxvzz

Note that the displacement caused by the uniform load is proportional to the ratio zzEILw /4

0 , which has units of displacement. at x=L. the tip deflection is

zzEILwLv

8)(

40−

=


9

S.P. 2: Solution of the 4th order o.d.e.

(This solution incorporates the statics into the d.e. )

The equation is

oiv

zz wvEI −=

subject to the boundary conditions

shear) (zero 0)('''moment) (zero 0)(''

0)0('0)0(

==

==

LvLv

vv

Indirect integration:

432231

40

3221

30

21

20

10

2624

26'

2''

'''

CxCxCxCEI

xwv

CxCxCEI

xwv

CxCEI

xwv

CEI

xwv

zz

zz

zz

zz

++++−

=

+++−

=

++−

=

+−

=


10

The boundary conditions determine the four constants. At x=0:

0)0('0)0(

3

4

====

CvCv

So, the last two constants vanish because of the fixed end at x=0, leaving the solution in the form .

22314

0

2624)( xCxC

EIxwxv

zz

++−

=

At x=L:

zzzz

zz

EILwCC

EILwLv

CLCEI

LwLv

011

0

21

20

0)('''

02

)(''

=→=+−

=

=++−=

Backsubstituting,

zzEI

LwC2

20

2 −=

Inserting these constants into the solution gives

zzzzzz EIxLw

EILxw

EIxwxv

4624)(

20

30

40 −+

−=

which is the same as the previous solution.


11

S.P. 3: Simply supported beam under end moment

(a) 2d Order Eqn:

By statics, the moment can be shown to be

LxMxM /)( 0=

Therefore:

21

30

1

20

0

6)(

2)('

)(''

CxCLEI

xMxv

CLEI

xMxv

LEIxMxv

zz

zz

zz

++=

+=

=

B.C: 0 0)0( 22 =→== CCv

zzzz EILMCLC

EILMLv

660)( 0

11

20 −=→+==

Solution:

−

=−=

Lx

Lx

EILM

EILxM

LEIxMxv

zzzzzz

3200

30

666)(


12

(b) 4th order eqn: No load on the beam, so

43

2

2

3

1

32

2

1

21

1

26

2'

'''''

00

CxCxCxCv

CxCxCv

CxCvCv

vvEI ivivzz

+++=

++=

+==

=→=

B.C:

zz

zzzz

EILMC

LEIMCLC

EIMLv

LCCLCLCLv

CvCv

6

)(''

6

60)(

0C 0)0(''0C 0)0(

03

011

0

2

133

3

1

22

44

−=

=→==

−=→+==

=→===→==

−

=−=

Lx

Lx

EILM

EILxM

LEIxMxv

zzzzzz

3200

30

666)(

as before.


13

S.P. 4: Simply supported beam under sinusoidal load

D.E.: Use

π−=

LxwvEI zz sin')'''( 0

B.C.: 0)( 0)0( == Lvv 0)('' 0)0('' == Lvv

Indirect Integration:

43

23

14

40

32

2

13

30

212

20

10

26sin

2cos'

sin''

cos)'''(

CxCxCxCLx

EILwv

CxCxCLx

EILwv

CxCLxLwvEI

CLxLwvEI

xzz

zz

zz

zz

++++

π

π−=

+++

π

π−=

++

π

π=

+

π

π=

B.C. at x=0:

0)0( 4 == Cv 04 =C

0)0('' 2 == Cv 02 =C


14

So the solution reduces to

xCxCLx

EILwv

zz3

3

14

40

6sin ++

π

π−= \

B.C. at x=L:

LCLCLv 3

3

1 60)( +== (Since sin(π)=0)

0)('' 1 == CEI

LLvzz

→ 0 ,0 31 == CC

So, in this case, all constants are zero, and

π

π=

Lx

EILwv sin4

40

The moments and shears are

π

π==

π

π==

LxLwvEIV

LxLwvEIM

zzz

zzz

cos'''

sin''

0

2

20

Note: It isn't obvious here, but this example illustrates an important building block of beam and plate analysis.


15

If the loading is described piecewise on the member or structure, the solution must be constructed by piecewise solutions, with • Boundary conditions at supports or ends • continuity conditions at the points of discontinuous

load description Solution Strategy: (1) Subdivide the beam into sections at the points of

load discontinuity. (2) Solve the o.d.e (either 2d order or 4th order) for

each section. (3) Match continuity conditions between portions and

boundary conditions at the ends. Side Note: In later courses, you'll study much more

efficient ways of doing this. This semester we'll just learn the basic concepts.


16

S.P. 5: Solve using the second order o.d.e.'s (Do the statics separately.)

(a) Beam reactions:

4

42210 00

0LwVLwVLwVF yyyy =→−=−==∑

The resultant of the distributed load acts at 5L/6, so

245

06

54

200 LwMLLwMM zAz −=→=⋅−−=∑

(b) For x < L/2: The beam moment is

424

5

0424

5

02

0

02

0)(

LxwLwM

LxwLwMM

z

zx

z

+−=→

=−+=∑


17

(c) Integrate the moment for 0 < x < L/2:

21302

20

1

120

20

1

02

01

24485

8245'

4245)(''

CxCxEI

LwxEI

Lwv

CxEI

LwxEI

Lwv

EILxw

EILw

EIxMv

zzzz

zzzz

zzzzzz

+++−=

++−=

+−==

From the cantilever beam boundary conditions at x=0:

0)0('0)0(

11

21

====

CvCv

So, the solution for 0 < x < L/2 reduces to

3022

01 2448

5 xEI

LwxEI

Lwvzzzz

+−=

This solution only applies over the first half of the beam. For the second half, the distributed load must be properly accounted for.


18

(d) Determine the moment for x < L/2:

LxwxwLwM

LxL

wxLwLwM

LxLxLxw

xLwLwMM

z

z

zx

z

326

0224424

5

231

212

21

4245

30

20

20

300

20

0

02

0)(

−+−=→

=

−+−+=

−

−

−+

−+=∑

(e) Integrate the moment for L/2 < x < L.

43

50

402

20

2

3

40

30

20

2

30

20

20

2

602412

1266'

326''

CxCEIxw

EIxwx

EILwv

CEIxw

EIxwx

EILwv

LEIxw

EIxw

EILw

EIMv

zzzzzz

zzzzzz

zzzzzzzz

z

++−+−=

+−+−=

−+−==


19

(f) To determine 43 and CC , the continuity conditions at L/2 are applied...

)2/(')2/(' )2/()2/( 2121 LvLvLvLv ==

where

zz

zz

EILwLv

EILwLv

967)2/('

48)2/(

30

1

40

1

−=

−=

1

30

2

21

40

2

19213

)2/(

21603

)2/(

CEI

LwLv

CLCEILw

Lv

zz

zz

+−=

++−=

Setting )2/(')2/(' 21 LvLv = :

zzzzzz EILwCC

EILw

EILw

192

19213

967 3

011

30

30 −=→+−=−

Setting )2/()2/( 21 LvLv = : (and substituting 1C from above)

zzzzzz EILwCC

EILw

EILw

1920

192041

48

40

22

40

40 =→+−=−

Therefore, for x > L/2:

zzzzzzzz EILw

xEILw

LEIxw

EIxw

EIxLw

xv1920192602412

)(4

03

05

04

022

02 +−−+−=


20

S.P. 6: We may also use the 4th order o.d.e.'s

Here there's no load except at L/2. The reactions at each of the two ends are P/2. (But we don't need that for the solution.) Assume that zzIE , are constant

To the left of the concentrated load,

432231

1

3221

1

211

11

1

26

2'

'''''

00')'''(

CxCxCxCv

CxCxCv

CxCvCv

vvEI ivzz

+++=

++=

+==

=→=

B.C.: At x=0,

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

221

441

=→===→==

CCvCCv

so,

xCxCv 331

1 6+=


21

To the right of the concentrated load:

87

26352

7625

2

652

52

22

26

2'

'''''

00')'''(

CxC

xCxCv

CxCxCv

CxCvCv

vvEI ivzz

++

+=

++=

+==

=→=

B.C. At x=L,

LCLCCLC

CLCLv

CLCLCLCLv

7

2

5856

652

87

2

6

3

52

2 ,C

0)(''26

0)(

−=−=→

+==

+++==

so

)(326 7

323

52 LxCLxLxCv −+

+−=

We still have 4 undetermined constants, and haven't included the load effect. To take care of this, we need the continuity conditions at the load point.


22

At the load point,

(i) Continuity of displacements:

75

3

31

3

21 24811

24822CLCLCLCLLvLv −=+→

=

(ii) Continuity of slopes:

75

2

31

2

21 83

82'

2' CCLCCLLvLv +−=+→

=

(iii) Continuity of moments:

)2/('')2/('')2/('')2/(''

)2/()2/(

21

21

21

LvLvLEIvLvEI

LMLM

zz

=→=→

=

since zzIE , are constant. Therefore,

515121 22)2/('')2/('' CCLCLCLvLv −=→−=→=


23

(iii) Shear equilibrium:

From equilibrium, at a point load, -P,

PLVLVV yyy −=−=∆ )2/()2/(12

But,

52

11

)'''(

)'''(

2

1

CEIvEIVCEIvEIV

zzzzy

zzzzy

==

==

so

PCEICEI zzzz −=− 15 Using these results together, obtain

zzzz EI

PCEIPC

2 ,

2 51 −==

Substituting 51 and CC into the displacement and slope continuity conditions,

7

3

3

3

224811

2248CL

EIPLCL

EIPL

zzzz

−

−=+

7

2

3

2

283

28C

EIPLC

EIPL

zzzz

+

−−=+

zzEI

PLDPLC163

,16

2

3

2

3 −=−=→


24

Substituting all of these constants into the solutions,

)(

163

3262

16122323

2

23

1

LxEIPLLxLx

EIPv

xEI

PLxEIPv

zzzz

zzzz

−−

+−−=

−=

The displacement at any point of the beam is given by

≤<≤≤

=LxLxv

Lxxvxv

2/ )(2/0 )(

)(2

1

Maximum displacement occurs at L/2, where

zzEI

PLLv48

)2/(3

−=

The - sign means that the displacement is downward, consistent with the downward loading.


25

2.2. Statically Indeterminate Beams

If a beam has more reactions than can be determined by statics, the moments can't be found ahead of time.

The deflected shape of the beam, taking into account the boundary conditions determines the moments. Three strategies are possible:

(a) Perform the statics separately, leaving the excess reactions as unknowns. Then integrate the 2d order o.d.e. and impose the displacement and rotation b.c. to solve for the reactions.

(b) Integrate the 4th order o.d.e. directly, and impose all boundary conditions.

(c) Superimpose solutions of simpler problems to satisfy the boundary conditions.


26

S.P. 7: Analyze the statically indeterminate beam shown.

(a) Conduct the static analysis separately, and incorporate the moments as unknowns:

)(12

2)(1

2/

02)(

0

zBzAL

LowyB

LwBAL

A

LwBALAM

LwBAF

ozzy

ozzyB

Z

yyy

−+=

+−−=→

++−−=

=−+=

∑∑

Note that yy BA and have two components: (i) Effect of load if zz BA , are zero. (ii) Effect of zz BA , .


27

Next, cut a FBD at x and obtain the moment.

LxB

LxAxwLxwxM

xwxBAL

LwAxMM

zzoo

z

ozz

ozz

xZ

+

−+−=→

=

+

−−−−=∑

122

)(

0 2

)(12

)(

2

2)(

Therefore, the curvature is given by

+

−+−=

=

LxB

LxAxwLxw

EI

EIxM

dxvd

zzo

zz

zz

z

122

1

)(

20

2

2

Integrate twice. Before doing this, note that the boundary conditions are

0)(')()0(')0( ==== LvLvvv Here, let's use indirect integration.


28

The first integration yields

1

2320

2641 C

LxB

LxxAxwLxw

EIdxdv

zzo

zz

+

+

−+−=

But, v'(0)=0 implies that ,01 =C so this reduces to

+

−+−=LxB

LxxAxwLxw

EIdxdv

zzo

zz 2641 232

0

The second integration yields

2

232430

26224121)( C

LxB

LxxAxwLxw

EIxv zz

o

zz

+

+

−+−=

But, v(0)=0 implies that 02 =C . So, the displacement is

+

−+−=L

xBL

xxAxwLxw

EIxv zz

o

zz 26224121)(

232430

At this point the boundary conditions at x=0 have been satisfied.


29

We still don't know zz BA and , but we still have the conditions .0)(' ,0)( == LvLv Hence,

02212

)('

06324

)(

3

224

=++=

=++=

zz

z

zz

z

zz

o

zz

z

zz

z

zz

o

EILB

EILA

EILwLv

EILB

EILA

EILwLv

Solving these two equations simultaneously yields

12

20 LwBA zz −==

Therefore, the moments are

1222)(

22 LwxwLxwxM oooz −−=

and the displacements are

zz

o

zz

o

zz EIxLw

EIxw

EILxwxv

242412)(

22430 −−=

These functions are plotted below.


30

Moments as a function of x:

Displacements as a function of x:

Note that • the displacements are zero at the ends. • the slopes are also zero at the ends (as they must be

for the fixed boundary conditions to be satisfied).


31

S.P. 8: Solve the problem of S.P. 7 by integrating the 4th order differential equation

zz

oivo

ivzz EI

wvwvEI −=→−=

First Integration:

1''' CEI

xwvzz

o +−=

Second Integration:

21

2

2'' CxC

EIxwv

zz

o ++−=

Third Integration:

32

2

1

3

26' CxCxC

EIxwv

zz

o +++−=

Fourth Integration:

43

2

2

3

1

4

2624CxCxCxC

EIxwv

zz

o ++++−=

4 constants of integration must be determined from the boundary conditions.


32

BC at x=0:

0 0)0('0 0)0(

33

44

=→===→==

CCvCCv

So, the solution reduces (at this point to

2624

2

2

3

1

4 xCxCEIxwv

zz

o ++−=

BC at x=L:

LCLCEI

LwLv

LCLCEILwLv

zz

o

zz

o

2

2

1

3

2

2

3

1

4

260)('

26240)(

++−==

++−==

Solving these two equations for 21 and CC ,

zz

o

zz

o

EILwC

EILwC

12 ,

2

2

21 −==

The solution then becomes

zz

o

zz

o

zz EIxLw

EIxw

EILxwxv

242412)(

22430 −−=

in agreement with the solution of S.P. 7.


33

This approach can also be used for somewhat more general loads and boundary conditions, although the amount of effort to obtain solutions increases. S.P. 9:

We'll solve this one by integrating the two ends separately and matching boundary conditions. The integrals are the same as in S.P. 6, but the boundary conditions are different. To the left of the load, To the left of the concentrated load,

432231

1

3221

1

211

11

1

26

2'

'''''

00')'''(

CxCxCxCv

CxCxCv

CxCvCv

vvEI ivzz

+++=

++=

+==

=→=


34

B.C.: At x=0,

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

221

441

=→===→==

CCvCCv

so,

xCxCv 331

1 6+=

To the right of the concentrated load:

872635

2

7625

2

652

52

22

26

2'

'''''

00')'''(

CxCxCxCv

CxCxCv

CxCvCv

vvEI ivzz

+++=

++=

+==

=→=

B.C. At x=L,

76

2

52

87

2

6

3

52

20)('

260)(

CLCLCLv

CLCLCLCLv

++==

+++==

(These won't be quite as easy to solve as in S.P. 6, so we'll try a more formal approach.)


35

The continuity conditions at the load point are the same as in S.P. 6. Namely, (i) Continuity of displacements:

87

2

6

3

53

3

1

21

2848248

)2/()2/(

CLCLCLCLCLC

LvLv

+++=+→

=

(ii) Continuity of rotations:

76

2

53

2

1

21

288

)2/(')2/('

CLCLCCLC

LvLv

++=+→

=

(iii) Continuity of moments:

651

21

22

)2/('')2/(''

CLCLC

LvLv

+=→

=

(iv) Shear equilibrium:

zzzz EIPCCPvvEI /)''''''( 5121 =−→=− To solve, let's organize these algebraic boundary and continuity condition equations into a matrix format..


36

=

−

−−

−−−

−−−−

zzEIP

CCCCCC

LL

LLL

LLLLL

LL

LLL

00000

000101

0012

02

0128

18

12848248

012

00

126

00

8

7

6

5

2

1

22

233

2

23

These may either be solved by hand (with considerable effort) or symbolically by computer. (Some possibilities include Mathematica, Mathcad, or Maple.) Using the latter approach,

zzEIP

LL

L

L

CCCCCC

−

−−

=

48/32/5

2/16/1132/

16/5

3

2

2

8

7

6

5

2

1


37

Then, the solution is

LxLEI

PLxLLxxv

LxEI

PxLxv

zz

zz

≤<

+−+−=

≤≤

−=

2

48325

49611

2/0 3296

5

3223

2

23

1


38

3. Analysis of Beam Deflections and Statically Indeterminate Beams by Superposition Key Observation: Because the differential equation for the model is linear, we may superposition (add) solutions to obtain combined solutions!

deflection of beams - university of virginiapeople.virginia.edu/~ttb/deflect.pdf · deflection of...

Documents