deflection & stiffness of members.pdf

7/27/2019 DEFLECTION & STIFFNESS OF MEMBERS.pdf

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CHAPTER 5

DEFLECTION AND STIFFNESS

Beam deflection can be found by 3 methods: 1) Integration, 2) Superposition and 3)

Castiglianos Theorem. These are detailed as follows:

1) Beam deflection by Integration (Section 5-3, pg. 188 in the textbook):

If the deflection of a beam is mainly due to the bending moment, then the following

formula is applied to find the deflection of the beam:

EI

M

dx

yd

2

2

=

where y is the vertical deflection, x is the horizontal distance, E is the modulus

elasticity and I is the area moment of inertia as before. We also find the slope as:

+== 1cdxEIM

dx

dy

where c1 is a constant found from the end or boundary conditions of the beam.

Example: Find general expressions for the deflection and slope of the following

beam.

y ya FBD:w FF waa/2

B x xC

First we draw a FBD for the system to find the reactions, which are found from

Newtons equations as RA = F+wa and MA = FL+wa2/2. We then cut the beam

between A-B and B-C to find the bending moments ofMAB andMBCas:

We sum the moments at the cut sections and let them to be zero to find:

MAB = RAxwx2/2MA = wx2/2+(F+wa)x(FL+ wa2/2) and2MBC= RAxwa(xa/2)MA = (F+wa)xwa(xa/2)(FL+ wa /2) =F(xL).

ww a

MBC

y

MAMA

MAB

yVAB

RA x

VBC

RA x

A RAMA

LL


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Deflection and Slope between A-B, i.e. 0 x a:

1

22

1 )2

()(2

1cdx

awFLxwaFx

w

EIcdx

EI

M

dx

dy AB +

+++

=+= Slope =AB =

= 1

223 )

2(

2

)(1cx

awFLx

waFx

w+

++

+

6EI

Since AB = 0 atx =0 then c1 = 0. Hence:

AB =

+

++

x

awFLx

waFx

w

EI)

2(

2

)(

6

1 223 .

Deflection=yAB =2

223

2)

2(

2

)(

6

1cdxx

awFLx

waFx

w

EIcdx

AB+

+

++

=+

= 22

234

4

)(2

6

)(

24

1cx

awFLx

waFx

w

EI+

+

++

SinceyAB = 0 atx =0 then c2 = 0. Hence:

yAB =

+

++

22

34

4

)(2

6

)(

24

1x

awFLx

waFx

w

EI.

eflection and Slope between B-C, i.e. ax L:D

lope = BC= 33 )(1

cdxLxFEI

cdxEI

M

dx

dy BC +=+= S

= 3

2

)2

( cLxxF

+EI

.

ince AB = BCatx =a where from the above equations:S

axAB = =

++ aFLaa

EI

)

2

(

26

+

)

2

(

6

1 3 La

Faaw

EI

+ awwaFw )(1 223 =

axBC = = 3

2

)2

( cLaa

EI

F+ and hence c3 =

EI

wa

6

3. Therefore:

BC=

6)

2(

1 2 3waLx

xF

EI.


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Deflection =yBC= 4

2

46

)2

(1

cdxwa

Lxx

FEI

cdx3

BC +

=+ .

= 4

23

6

)

26

(1

cxwax

Lx

F

EI

3

+

SinceyAB =yBCatx =a where from the above equations:

axABy = =

+

++

22

34

4

)(2

6

)(

24a

awFLa

waFa

w

EI

1=

=

+

268

234 FLaFa

EI1 wa

.

axBCy = = 4

23

6

)

26

(1

cawaa

La

F

EI

3

+

and hence c4 =

EI

wa

24

4

. Therefore:

yBC=

+

246)

26(

1 423 wax

waxL

xF

EI

3

.

Note: Review example 5-1, pg. 190 in the textbook.

2) Beam Deflection by Superposition (Section 5-5, pg. 192 in the textbook):

Refer to 16 cases given in the Appendix A-9, from pg. 969 to 976 in the textbook. Thesuperposition for linear systems means that we can find the reactions, bending

moments, shear forces, slope and deflection by summing up these entities for each

loading case. So lets solve the above example by the superposition technique. We

have,

y yya wa w

FFB Bx

+C C CA AA L LL

Deflection and Slope between A-B, i.e. 0 x a:

Slope =AB = 1AB + 2AB , where from Table A-9-3, pg. 970 in the textbook,

1AB =dx

dy= )xaxax(

EI

w 232 336

where lis replaced by a in the formula. Also,

from Table A-9-1, pg. 969 in the textbook, 2AB =dx

dy= )Lxx(

EI

F2

2

2 . Hence,


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AB= )xaxax(EI

w 232 336

+ )Lxx(EI

F2

2

2 =

+

++

x

awFLx

waFx

w

EI)

2(

2

)(

6

1 223 ,

which is the same result obtained before using the integration method.

Deflection =yAB =y1AB +y2AB , where from Table A-9-3, pg. 970 in the textbook,

y1AB = )6(424

222 axaxEI

wx where again l is replaced by a in the formula. Also,

from Table A-9-1, pg. 969 in the textbook,y2AB = )Lx(EI

Fx3

6

2

. Hence,

yAB= )6(424

222

axaxEI

wx + )3(

6

2

LxEI

Fx =

+

++

22

34

4

)(2

6

)(

24

1x

awFLx

waFx

w

EI,

which is again the same result obtained before using the integration method.

Deflection and Slope between A-B, i.e. ax L:

Slope =BC= 1BC+ 2BC. Since there is no bending moment between B-C for the case

of distributed load w, we can assume that the slope remains the same as 1AB atx=a.

Hence,1BC= axAB1 = =EI

wa3

6

. Also, from Table A-9-1, pg. 969 in the textbook,

2BC= 2AB = )2(2

2 LxxEI

F . Hence,

BC=EIwa3

6 + )2(

22 Lxx

EIF =

6)

2(1 2 3waLxxF

EI,

which is the same result obtained before using the integration method.

Deflection =yBC=y1BC+y2BC.

For the case of distributed load w, the angle 1BCremains the same between B-C, and

hence the deflected curve between B-C is a linear one.

ywa

Deflected CurveB

C1BA

y1BCx 1BC

From the figure,y1BC = 1BC(xa)+ y1B, wherey1B = axAB1y = =EI

wa

8

4. Therefore,

y1BC=

EI

axwa

6

)(3 +

EI

wa

8

4= )4(

24

3

ax

EI

wa+ .


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For the end load ofF, again referring to Table A-9-1, pg. 969 in the textbook, y2BC=

y2AB = )3(6

2

LxEI

Fx . Hence,

yBC= =y1BC+y2BC= )4(24

3

axEI

wa + + )Lx(EI

Fx 36

2

=

+

246)

26(1

423waxwaxLxF

EI

3

,

which is again the same result obtained before using the integration method.

Note: Review examples 5-2 and 5-3, pgs. 192 and 193 in the textbook.

3) Beam Deflection by Castiglianos Theorem (Sections 5-7 and 5-8, pgs. 198 and

201 in the textbook):

When a machine element is deformed, it stores a potentialorstrain energy U. This

energy can be calculated for different loading cases, as given below:

Loading Strain Energy (U)

Tension or CompressionEA

LF

2

2

Direct ShearGA

LF

2

2

TorsionGJ

LT

2

2

Bending Moment EIdxM

2

2

Bending Shear GAdxcV

2

2

where for bending shearc is a factor that depends on the cross-section of the beam

and is taken from Table 5-1, pg. 200, in the textbook. The Castiglianos theorem then

states that the displacement for any point on a machine element is the partial

derivative of the strain energy with respective to a force which is on the same point

and in the same direction of the displacement. In short:

ii F

U

=

where i is the displacement for an ith point in the machine element andFi is the force

acting at the same point and in the same direction of the displacement asked for. If

there is no force at the point for which the displacement is required, then we have to

assume an imaginary force Q and let it be zero at the end.

Example: Lets solve the above problem for its vertical displacement at the free end,

i.e. at point C, using the Castiglianos theorem. So the question is to find C in the

vertical (y) direction. We have,

FUC=


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where the strain energy U is contributed by the bending shear force Vand bending

momentM, but well only consider the effect ofMin U. Thus, as given above,

U= EI

dxM

2

2

= +L

a

BCa

AB

EI

dxM

EI

dxM

22

2

0

2

and

+

=L

a

BCBCa

ABAB

CEI

dxF

MM

EI

dxF

MM

0

where as found before: MAB = wx2/2+(F+wa)x (FL+ wa2/2) and MBC = F(xL).

Therefore: LxF

MAB =

and Lx

F

MBC =

. Substituting these in the formula

above yields:[ ]dx

EI

LxFdx

EI

LxwaFLxwaFwxL

a

a

C

++++

=2

0

22 )()(/2)()(/2 .

which upon calculation results in

EI

FLaL

EI

waC

3)(4

24

33

+= .

This result is in agreement with the result that can be found from the two methods

above by substitutingx=L inyBC. But remember that LxBCy = = C. Why?

Note: Review examples 5-10, 5-11 and 5-12, pgs. 202 to 206 in the textbook.

5.10 Statically Indeterminate Problems

When a static problem has more unknowns than the Newtons static equations, then

this problem is known as the statically indeterminate problem. In these cases, we need

extra displacement equations to solve the problem. There are two procedures to

handle such problems, among which well follow the Procedure 1 in the textbook,

whose steps are given as follows:

1) Choose the redundant (extra) reaction(s). This redundant reaction could bea force or a moment.

2) Write the Newtons equations of static equilibrium in terms of the appliedloads and the redundant reaction(s) of step 1.

3) Write the deflection or slope equation(s) for the point(s) of the redundantreaction(s) of step 1. Normally this deflection or slope will be zero.

4) Solve the equations in steps 2 and 3 for the reactions.Note: Review example 5-14, pgs. 212 and 213 in the textbook.


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5.12 Long Columns with Central Loading

The long columns with central compressive loading may exhibit a phenomenon

called buckling, which needs to be checked for the safe use of these columns. Such

columns with 4 different end conditions are shown in the textbook in Figure 5-18, pg.

217. The formula used for the long columns is namedEuler formula, given as:

2cr L

EIcP

2

= or2

2

)/( kL

Ec

A

Pcr =

where Pcr is the critical or maximum central load for the column, c is the end

condition constant that should be taken from Table 5-2, pg. 220, in the textbook, and k

is the radius of gyration found fromI=Ak2, which is equal to d/4 for a column having

a solid round cross-section with a diameter ofd. L/k in the formula is known as the

slenderness ratio. There are 2 scenarios here:

1) If the critical load Pcr is known, then one should use the above equations tofind an adequate cross-section for the column.

2) If the cross-section of the column is known, one can use the above equationsto solve for thePcr.

How do we know a column is long enough or it is of intermediate length? First we

define:1/2

2

1

2

=

yS

cE

k

L

where Sy is the yield strength of the column material. We follow the following

criterion:

1) If

k

L>

1

k

Lthen use the aboveEuler formula.

2) If

k

L

1

k

Lthen use theJohnson formula given below in section 5-13 for

the intermediate-length columns.

5.13 Intermediate-Length Columns with Central Loading

Use theJohnson formula:

cEk

LSS

A

P yy

cr 1

2

2

=

Note: Review examples 5-16, 5-17, 5-18 and 5-19, pgs. 223 to 225 in the textbook.

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