Chapter Objectives To determine the deflection and slope at specific points on beams and shafts...

Chapter Objectives

To determine the deflection and slope at specific points on beams and shafts using the integration method, discontinuity functions, and the method of superposition.

To use the method of superposition to solve for the support reactions on a beam or shaft that is statically indeterminate.

Copyright © 2011 Pearson Education South Asia Pte Ltd

1. Reading Quiz

2. Applications

3. Elastic Curve

4. Integration Method

5. Use of discontinuity functions

6. Method of superposition

7. Statically indeterminate beams and shafts

8. Use of the method of superposition

9. Concept Quiz

In-class Activities


READING QUIZ

1) The slope angle θ in flexure equations is

a) Measured in degree

b) Measured in radian

c) Exactly equal to dv/dx

d) None of the above


READING QUIZ (cont)

2) The load must be limited to a magnitude so as to not change significantly the original geometry of the beam. This is the assumption for:

a) The method of superposition

b) The moment area method

c) The method of integration

d) All of them


READING QUIZ (cont)

3) A statically indeterminate structure

a) is always a stable structure

b) has more number of unknown reactions than the available number of equilibrium equations

c) is dynamically determinate

d) None of the above


APPLICATIONS


ELASTIC CURVE


• The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve

ELASTIC CURVE (cont)


• Moment-curvature relationship:– Sign convention:

ELASTIC CURVE (cont)


• Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or

• Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/I

y

1

yEEI

M

1

or 1

SLOPE AND DISPLACEMENT BY INTEGRATION


• Kinematic relationship between radius of curvature ρ and location x:

• Then using the moment curvature equation, we have

232

22

1

1

dxdv

dvvd

2

2

2/32

22

1

1

dx

vd

dxdv

dxvd

EI

M

SLOPE AND DISPLACEMENT BY INTEGRATION (cont)


• Sign convention:

SLOPE AND DISPLACEMENT BY INTEGRATION (cont)


• Boundary Conditions:

– The integration constants can be determined by imposing the boundary conditions, or

– Continuity condition at specific locations

EXAMPLE 1


The cantilevered beam shown in Fig. 12–10a is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.

EXAMPLE 1 (cont)


• From the free-body diagram, with M acting in the positive direction, Fig. 12–10b, we have

• Applying Eq. 12–10 and integrating twice yields

Solutions

PxM

(1) 6

(2) C2

(1)

21

3

1

2

2

2

CxCPx

EIv

Px

dx

dvEI

Pxdx

vdEI

EXAMPLE 1 (cont)


• Using the boundary conditions dv/dx = 0 at x = L and v = 0 at x = L, equations 2 and 3 become

• Substituting these results, we get

Solutions

3 and

2

60

20

3

2

2

1

21

3

1

2

PLC

PLC

CLCPL

CPL

(Ans) 236

2

323

22

LxLxEI

Pv

xLEI

P

EXAMPLE 1 (cont)


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

• If this beam was designed without a factor of safety by assuming the allowable normal stress is equal to the yield stress is 250 MPa; then a W310 x 39 would be found to be adequate (I = 84.4(106)mm4)

Solutions

(5) 3

(4) 2

3

2

EI

PLv

EI

PL

A

A

mm 1.74

104.842003

1000530

rad 0222.0104.842002

1000530

6

22

6

22

A

A

v

EXAMPLE 2


The simply supported beam shown in the below figure supports the triangular distributed loading. Determine its maximum deflection. EI is constant.

EXAMPLE 2 (cont)


• Due to symmetry only one x coordinate is needed for the solution,

• The equation for the distributed loading is .

• Hence

Solutions

2/0 Lx

xLw

L

xwM

xLwx

L

xwMM NA

43

043

;0

02

0

02

0

xL

ww 02

EXAMPLE 2 (cont)


• Integrating twice, we have

• For boundary condition,

Solutions

2,0 and 0,0 Lxdxdvxv

213050

12040

0302

2

2460

812

43

CxCxLw

xL

wEIv

CxLw

xL

w

dx

dvEI

xLw

xL

wM

dx

vdEI

0,192

52

30

1 CLw

C

EXAMPLE 2 (cont)


• Hence

• For maximum deflection at x = L/2,

Solutions

(Ans) 120

40

max EI

Lwv

xLw

xLw

xL

wEIv

192

5

2460

303050

USE OF CONTINUOUS FUNCTIONS


• Macaulay functions

USE OF CONTINUOUS FUNCTIONS


• Macaulay functions

• Integration of Macaulay functions:

an

axax

axax

n

n

for

for 0

Cn

axdxax

nn

1

1

USE OF CONTINUOUS FUNCTIONS (cont)


• Singularity Functions:

axP

axaxPw

for

for 01

axM

axaxMw

for

for 0

0

2

0

USE OF CONTINUOUS FUNCTIONS (cont)


• Note: Integration of these two singularity functions yields results that are different from those of Macaulay functions. Specifically,

• Examples of how to use discontinuity functions to describe the loading or internal moment in a beam:

:

2,1,1

naxdxaxnn

EXAMPLE 3


Determine the maximum deflection of the beam shown in Fig. 16–16a. EI is constant.

EXAMPLE 3 (cont)


• The beam deflects as shown in Fig. 16–16a. The boundary conditions require zero displacement at A and B.

• The loading function for the beam can be written as

Solutions

1110608

xxw

EXAMPLE 3 (cont)


• Integrating, we have

• In a similar manner,

• Integrating twice yields

Solutions

0010608 xxV

mkN 1068

106081

11

xx

xxM

(1) 103

4

1034

108

21

33

1

22

1

2

2

CxCxxEIv

Cxxdx

dvEI

xxdx

vdEI

EXAMPLE 3 (cont)


• From Eq. 1, the boundary condition v = 0 at x = 10 m and at x = 30 m gives

• Thus,

Solutions

12000 and 1333

301030360000

10101013330

21

213

213

CC

CC

CC

(3) 120001333103

4

(2) 13331034

33

22

xxxEIv

xxdx

dvEI

EXAMPLE 3 (cont)


• To obtain the displacement of C, set x = 0 in Eq. 3.

• The negative sign indicates that the displacement is downward as shown in Fig. 12–18a

• To locate point D, use Eq. 2 with x > 10 and dv/dx = 0,

Solutions

(Ans) mkN 12000 3EI

vC

m 320 root, positive for the Solving

0163360

133310302

22

.x

xx

xx

D

DD

DD

EXAMPLE 3 (cont)


• Hence, from Eq. 3,

• Comparing this value with vC, we see that vmax = vC.

Solutions

3

33

mkN 5006

120003.201333103.203.203

4

EIv

EIv

D

D

EXAMPLE 4


Determine the equation of the elastic curve for the cantilevered beam shown in Fig. 16-17a. EI is constant.

EXAMPLE 4 (cont)


• The boundary conditions require zero slope and displacement at A.

• The support diagram reactions at A have been calculated by statics and are shown on the free-body,

Solution

020215855000258052

xxxxxw

EXAMPLE 4 (cont)


• Since

• Integrating twice, we have

Solution VdxdMxwdxdV and

1111058550080258052

xxxxxV

mkN 54550452258

582

155008

2

10520258

202

20210

xxxx

xxxxxM

21

42432

1

3132

2022

2

53

1525

3

1

3

26129

53

4550

3

426258

54550452258

CxCxxxxxEIv

Cxxxxxdx

dvEI

xxxxdx

vdEI

EXAMPLE 4 (cont)


• Since dv/dx = 0, x = 0, C1 = 0; and v = 0, C2 = 0. Thus

Solution

(Ans) m 53

1525

3

1

3

26129

1 42432

xxxxx

EIv

METHOD OF SUPERPOSITION


• Necessary conditions to be satisfied:

1. The load w(x) is linearly related to the deflection v(x),

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

• Then, it is possible to find the slope and displacement at a point on a beam subjected to several different loadings by algebraically adding the effects of its various component parts.

STATICALLY INDETERMINATE BEAMS AND SHAFTS


• Definition:

A member of any type is classified statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations, e.g. a continuous beam having 4 supports

STATICALLY INDETERMINATE BEAMS AND SHAFTS (cont)


Strategy:

• The additional support reactions on the beam or shaft that are not needed to keep it in stable equilibrium are called redundants. It is first necessary to specify those redundant from conditions of geometry known as compatibility conditions.

• Once determined, the redundants are then applied to the beam, and the remaining reactions are determined from the equations of equilibrium.

USE OF THE METHOD OF SUPERPOSITION


Procedures:

Elastic Curve

• Specify the unknown redundant forces or moments that must be removed from the beam in order to make it statically determinate and stable.

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

USE OF THE METHOD OF SUPERPOSITION (cont)


Procedures:

Elastic Curve (cont)

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

• Sketch the deflection curve for each beam and indicate the symbolically the displacement or slope at the point of each redundant force or moment.



Procedures:

Compatibility Equations

• Write a compatibility equation for the displacement or slope at each point where there is a redundant force or moment.

• Determine all the displacements or slopes using an appropriate method as explained in Secs. 12.2 through 12.5.



Procedures:

Compatibility Equations (cont)

• Substitute the results into the compatibility equations and solve for the unknown redundant.

• If the numerical value for a redundant is positive, it has the same sense of direction as originally assumed. Similarly, a negative numerical value indicates the redundant acts opposite to its assumed sense of direction.



Procedures:

Equilibrium Equations

• Once the redundant forces and/or moments have been determined, the remaining unknown reactions can be found from the equations of equilibrium applied to the loadings shown on the beam’s free body diagram.

EXAMPLE 5


Determine the reactions at the roller support B of the beam shown in Fig. 16-27a, then draw the shear and moment diagrams. EI is constant.

EXAMPLE 5 (cont)


• By inspection, the beam is statically indeterminate to the first degree.

• Taking positive displacement as downward, the compatibility equation at B is

• Displacements can be obtained from Appendix C.

Solutions

(1) '0 BB vv

EI

B

EI

PLv

EI

PL

EI

wLv

yB

B

33

334

m 9

3'

EI

mkN 25.83

48

5

8

EXAMPLE 5 (cont)


• Substituting into Eq. 1 and solving yields

Solutions

kN 25.9

925.830

y

y

BEI

B

EI

CONCEPT QUIZ

1) The moment-curvature equation 1/ρ = M/EI is applicable to

a) Statically determined member only

b) Beams having uniform cross-sections only

c) Beams having constant Young’s Modulus E only

d) Beams having varying moment of inertia I.


CONCEPT QUIZ

2) The flexure equations imply that

a) Slope and deflection at a point of a beam are independent

b) Moment and shear at a point of a beam are independent

c) Maximum moment occurs at the locations where the shear is zero

d) Maximum moment occurs at the inflection point.


Chapter Objectives To determine the deflection and slope at specific points on beams and shafts...

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