boietz' slope & deflection diagram by parts

8/13/2019 Boietz' Slope & Deflection Diagram by Parts

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BOIETZ Slope & Deflection Diagram by Parts

BOIETZ S CALUAG Page 2

DEFLECTION OF STRUCTURESEngineering structures are constructed from materials that deform slightly when subjected tostress or a change in temperature. As a result of this deformation, points on the structureundergo certain movements called DEFLECTIONS . Provided that the elastic limit of thematerial is not exceeded, this deformation and the resulting deflection disappear when thestress is removed and the temperature returns to its original value. This type of deformation ordeflection is called ELASTIC and can be caused either by loads acting on the structure or by achange in temperature.

CAUSES OF DEFLECTION1. Stress2. Change in temperature3. Support settlement4. play in pin joints5. shrinkage of concrete, creep6. etc.

IMPORTANCE OF DEFLECTION COMPUTATIONS1. Mode of failure in design2. to analyze the vibration and dynamic response characteristics of the structures3. Stress analysis of statically indeterminate structures are based largely on an evaluation

of their deflection under load4. etc.

METHODS OF COMPUTING DEFLECTION1. Virtual Work Method (Applicable to any type of structure beam, truss, frames, etc.)2. Castiglianos Second Theorem (Applicable to any type of structure)3. Williot-Mohr Method (Applicable to trusses only)4. Bar-chain Method (Applicable to trusses only)5. Double Integration Method (Applicable to beams only)6. Moment-Area Method (Applicable to beams and frames only)7. Elastic Load Method (Applicable to beams and frames only)8. Conjugate Beam Method (Applicable to beams only)9. Slope and Deflection Diagram by Parts (Applicable to beams only)


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Relationship between Load, Shear, Moment, Slope and Deflection Diagram by Parts

Maximum Ordinate:

dwL)!dn(

!n VMSD

+= m

where: VMSD = shear, moment, slope or deflection ordinate at the fixed end.

d = type of diagram= 1, for shear= 2, for moment= 3, for slope= 4, for deflection

w = maximum intensity of the load, KN/mL = Length of the cantilever beam, mn = Degree of the load

= 0, for uniform load= 1, for triangular load 2, for spandrel load

Slope and Deflection Diagram Ordinate from M/EI Diagram

Slope:EIML

)!1m(!m

+=

Deflection: 2LEIM

)!2m(!m

+=

where:

EIM

= moment/EI ordinate at fixed support (maximum)

m = degree of the M/EI Diagram= n + 2


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Derivation:

For Concentrated Load: For Concentrated Moment:

For Spandrel Load (n 0):

Shear:

wL)!1n(

!n V

+=

Moment:2wL

)!2n(!n

M+

=

Slope:

3wL)!3n(

!n+

=

Deflection:4wL

)!4n(!n

+=

L

ShearDiagramby Parts

n+1

n+2

n+3

V

M

n w

LoadDiagram

MomentDiagramby Parts

SlopeDiagramby Parts

n+4 DeflectionDiagramby Parts

P

L

V - Diag

M - Diag

- Diag

- Diag

0 1

2

3

V = -P

M = PL

= -2

PL2

=6

PL3

Fixed ML

V - Diag

M - Diag

- Diag

- Diag

0

1

2

V = 0

M = M

= -ML

=2

ML2

Fixed


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Slope and Deflection Diagram by Parts From M/EI Diagram:

Slope:

EI

ML

)!1m(

!m+

=

Deflection:2L

EIM

)!2m(!m

+=


L


m+1

m+2

m =n+2 M/EI

M/EIDiagram

DeflectionDiagramby Parts

Fixed

P

L

- Diag

- Diag

0 1

= - P

= PL

FixedM

L

- Diag

- Diag

0

= 0

= M

Fixed


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HLx

y'm

=

ORDINATE OF SLOPE AND DEFLECTION DIAGRAM

Ordinate: where:y = deflection or slope ordinate at any point.x = distance from vertex/point of tangency to the point.H = highest ordinate or the ordinate at the fixed support.L = Length of the cantilever beam.m = slope of the deflection or slope diagram.

STEPS:1. Compute the reactions of the real beam.2. Choose an arbitrary point on the beam to be held fixed and draw the M/EI Diagram by

parts.


For Spandrel Load (n 0):

Moment Ordinate:

2LEIW

)!2m(!mM

+=

m

H

y

x

L

vertex

P

L

M/EI - Diag1 M = PL

FixedM

L

M/EI - Diag

0 M = M

Fixed

L

m+2M

m =n+2 W

LoadDiagram

M/EIDiagramby Parts


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HLx

y'm

=

3. For the conjugate beam loaded by M/EI Diagram by parts, compute the reactions.

4. Draw the Deflection and Slope Diagram by Parts using the principles in drawing theshear and moment diagram by parts.

Slope and Deflection Diagram by Parts From M/EI Diagram:

Slope:

EIML

)!1m(!m

+=

Deflection:2L

EIM

)!2m(!m

+=

5. Using the curve property, compute the slope and deflection at any point.

Ordinate:

Deflection or Slope at any point,

= y where:= y,y deflection and slope ordinates, resp.

= y

L


m+1

m+2

m =n+2 M/EI

M/EIDiagram

DeflectionDiagramby Parts

Fixed

m

H

y

x

L

vertex


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SAMPLE PROBLEMCompute the slope of the elastic curve at point A, slope and deflection at midspan.

Solution:1. Compute the reactions

0Mc = Ra(18) -30(6) = 0

Ra = 10 KN0Ma =

Rc(18) -30(12) = 0 Rc = 20 KN

2. Draw the M/EI Diagram by parts

Fixed @ a,

3. Compute the reactions of the conjugate beam

0Ma = Rc(18) -1/2(360/Ei)(18)(18/3)

+1/2(360/EI)(12)(12/3) = 0 Rc = 600/EI

M/EI - Diag

30KN

12m 6ma c

EI = constant

30KN

12m 6ma c

Ra Rc

30KN

12m 6ma c

Ra = 10 Rc = 20

20(18)=360

-30(12)=-360

M/EI - Diag

360

-360

RaRc

6m12m

1

1


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4. Draw the Slope & Deflection Diagram by Parts

Fixed @ a, Slope Ordinates:

EIML

)!1m(!m

+=

EI600

A =

EI2160

EI)18(360

)!11(!1

B =

+=

EI3240

EI)18(360

)!11(!1

C =+

=

Deflection Ordinates:2L

EIM

)!2m(!m

+=

EI19440

18EI

360)!21(

!1D 2 =

+=

EI864012

EI360

)!21(!1E 2 =

+=

EI10800

)18(EI600

F

=

=

M/EI - Diag

360

-360Rc =600/EI

6m12m

1

1

SlopeDiag by Parts

2

2

A =EI

600

B=EI

2160

C=EI

3240

DeflectionDiag by Parts

3

3 D=EI

2160

E=EI

3240

F=EI

10800 1


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HL

xy

'm

=

5. Compute the Slope and deflection

Slope @ left support:

= ya ans

EI480

EI3240

EI2160

EI600

=+=

Slope @ midspan:

= ym

ansEI75

EI3240

189

EI2160

123

EI600

22 =

+=

Deflection @ midspan:

= ym

ansEI3105

EI10800

189

EI8640

123

EI19440

189

223 =

=

m

Hy

x

L

vertex

boietz' slope & deflection diagram by parts

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