12.8 use of the moment-area method …contents.kocw.net/kocw/document/2015/hanyang/hanseog...2.5...

12.8 USE OF THE MOMENT-AREA METHOD

Copyright © 2011 Pearson Education South Asia Pte Ltd

Procedures:

• Construct separately the M/EI diagrams for each applied force or moment, and each redundant as well.

• Then use the method of superposition and apply the two moment area theorems to obtain the proper relationship between the tangents on the elastic curve in order to meet the conditions of displacement and/or slope at the supports of the beam or shaft.

USE OF THE MOMENT-AREA METHOD (cont)


Procedures:

EXAMPLE 12.19


The beam is subjected to the concentrated force shown in Fig. 12–39a. Determine the reactions at the supports. EI is constant.

EXAMPLE 12.19 (cont)


• The free-body diagram is shown in Fig. 12–39b.

• Using the method of superposition, the separate M/EI diagrams for the redundant reaction By and the load P are shown in Fig. 12–39c.

• The elastic curve for the beam is shown in Fig. 12–39d.

Solutions



• Applying Theorem 2, we have

• Using this result, the reactions at A on the free-body diagram, Fig. 12–39b, are

Solutions

Ans)( 5.2

021

32

221

32

/

PB

LEIPLLL

EIPLLL

EILB

Lt

y

yAB

(Ans) 5.0025.2 ;0

(Ans) 5105.2 ;0

(Ans) 0 ;0

PLMLPLPMM

P.APPAF

AF

AAA

yyy

xx

EXAMPLE 13.1

12.9 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 statistically determinate and stable.

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

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 12.21


Determine the reactions at the roller support B of the beam shown in Fig. 12–44a, then draw the shear and moment diagrams. EI is constant.



• 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

EIB

EIPLv

EIPL

EIwLv

yB

B

33

334

m 93

'

EImkN 25.83

485

8



• Substituting into Eq. 1 and solving yieldsSolutions

kN 25.9

925.830

y

y

BEIB

EI

EXAMPLE 13.1

EXAMPLE 12.23


Determine the moment at B for the beam shown in Fig. 12–46a. EI is constant. Neglect the effects of axial load.



• Since the axial load on the beam is neglected, there will be a vertical force and moment at A and B.

• Referring to the displacement and slope at B, we require

Solutions

0 ' '' (1)

0 ' '' (2)B B B

B B Bv v v



• Use Appx C to calculate slopes and displacements,Solutions

EIM

EIMLv

EIM

EIML

EIB

EIPLv

EIB

EIPL

EIEIwLv

EIwL

BB

BB

yB

yB

B

B

82

''

4''

33.213

'

82

'

mkN 423847

EImkN 21

48

2

3

2

34

33



• Substituting these values into Eqs. 1 and 2 and cancelling out the common factor EI, we get

• Solving these equations simultaneously gives

Solutions

By

By

MB

MB

833.21420

48120

3.375 kN

3.75 kN m (Ans)y

B

B

M

Chapter Objectives

Understand the behavior of columns and concept of critical load and buckling

Determine the axial load needed to buckle a so-called ‘ideal’ column

Determine the ‘effective length’ of a column with various end-conditions

Design a realistic column against bucklingCopyright © 2011 Pearson Education South Asia Pte Ltd

APPLICATIONS


APPLICATIONS (cont)


13.1 CRITICAL LOAD


• Long slender members subjected to an axial compressive force are called columns, and the lateral deflection that occurs is called buckling.

• The maximum axial load that a column can support when it is on the verge of buckling is called the critical load.

CRITICAL LOAD (cont)


• From the free-body diagram:

• For small θ, tan θ ≈ θ,

• Note: This loading (Pcr = kL/4) represents a case of the mechanism being in neutral equilibrium. Since Pcr is independent of θ, any slight disturbance given to the mechanism will not cause it to move further out of equilibrium, nor will it be restored to its original position. Instead, the bars will remain in the deflected position.

2/tan2 LkkFP

) ofnt (independe 4/

2/2

kLP

LkP

13.2 IDEAL COLUMN


• Ideal column– It is perfectly straight before loading.– Both ends are pin-supported.– Loads are applied throughout the centroid of the cross section.

• Behavior– When P < Pcr, the column remains straight.– When P = Pcr,

xEIPCx

EIPCv

vEIP

dxvd

PvMdx

vdEI

cossin

0

21

2

2

2

2

IDEAL COLUMN (cont)


• Since v = 0 at x = 0, then C2 = 0

• Since v = 0 at x = L, then

• Therefore,

• Which is satisfied if

• Or

0sin1

L

EIPC

0sin

L

EIP

nLEIP

2 2

2 where 1, 2, 3,...n EIP nL

IDEAL COLUMN (cont)


• Smallest value at P is when n = 1, thus

• Corresponding stress is

• Where r = √ (I/A) is called ‘radius of gyration’

• (L/r) is called the ‘slenderness ratio’.

• The critical-stress curves are hyperbolic, valid only for σcris below yield stress

2

2

LEIPcr

22

/ rKLE

cr

EXAMPLE 13.1


The A-36 steel W200×46 member shown in Fig. 13–8 is to be used as a pin-connected column. Determine the largest axial load it can support before it either begins to buckle or the steel yields.

EXAMPLE 13.1 (cont)


• From Appendix B,

• By inspection, buckling will occur about the y–y axis.

• When fully loaded, the average compressive stress in the column is

• Since this stress exceeds the yield stress,

Solutions

21887.6 1000 320.5 N/mm (MPa)5890

crcr

PA

46462 mm 103.15,mm 105.45,mm 5890 yx IIA

2 62

22 2

200 15.3 101887.6 kN

4 1/1000crEIP

L

(Ans) MN 47.1kN 5.14725890

250 PP

13.3 COLUMNS HAVING VARIOUS TYPES OF SUPPORTS


• one fixed end and one pined end

VxPvM

xEIVv

EIP

dxvd

VxPvEIEI

Mdx

vd

2

2

2

2

)(1

0,0:'. vxatsCB

0, vLx

0, dxdvLx

P

V y

x



'' BxAv p

sin coshP Pv A x B xEI EI

h pv v v

xEIVBxA

EIP

)''(

0',' BPVA

xPVx

EIPBx

EIPAv cossin

0

P

V y

x



0;0,0:'. BvxatsCB

0sin;0, LPVL

EIPAvLxat

, ' 0; cos 0P P Vat x L v A LEI EI P

PVLA

LPVLA

cos

sinLL

tan1

LL tan

P

V y

x



----------------------------------• The smallest critical load

occurs when n = 1, thus

• K is called the ‘effective-length factor’

2 2

22 or with 24cr cr

EI EIP P KL KL

error & by trial Solving

LEIPL 4934.4

2

2

219.20

ecr L

EIL

EIP

Then, 0.7eL L


• K for various end conditions:

12.8 use of the moment-area method …contents.kocw.net/kocw/document/2015/hanyang/hanseog...2.5...

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