5-Flexural Analysis Service Slides 2015.pdf

8/9/2019 5-Flexural Analysis Service Slides 2015.pdf

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Flexural Analysis of

Prestressed Concrete underService Loads

CEE 453University of Washington

Prof. Marc Eberhard


2/67

Lecture Topics

Beam Behavior

Goals of Service Load Analysis

Prestress Losses

Stresses by Direct Calculation

Stresses by Equivalent Loads

Stresses by Pressure Line (C line)

Kerns

Composite Construction (SKIP FOR NOW)


3/67

Self-weight

Decompression

Balanced

Cracking

Yield

Failure

Concrete becomes non-linear

fr

fpy fps

Curvature

Moment

Typical Beam Behavior


4/67

Tendon

stress

Time

Jacking

Service

Relaxation + thermal

Release PS (initial)

Elastic shortening

Live Load

Shrinkage, creep, relaxation

Typical Beam Behavior

fpj

fpi

fpe


5/67

Goals of Service Load Analysis

Find stresses in the concrete and steel

various times during life of girdervarious x-sections along beam

various (critical) locations within x-section

Compare with (Code) allowable values


6/67

Check stresses at various times:

Jacking (tendon only, fpj)

Initial (directly after release, fpi)

Service (after all PS losses, fpe)

Need to know how prestress is lost between

stages to know how much remains.

Key Times


7/67

Prestress Losses

Calculation Method Pre- Post-tensioned

Lump-Sum Methods

AASHTO 45 ksi 23 - 33 ksi + friction

ASCE/ACI 423 35 ksi 25 ksi + friction

PTI (-) 15 - 35 ksi + friction

Simplified Methods

AASHTO AASHTO LRFD Specifications

PCI Simplified PCI Jo. July 1975

Zia et al. Conc. Int. June 1979

Time-Step Methods

PCI general PCI Jo. July 1975

Ghali et al. PCI Jo. March 1977

NCHRP 446


8/67

Prestress Losses

In this class we do not have time to address

prestress losses in detail. We will:

Use a given value of fpi(will be given in the problem.

Use a given value of h= fpe/fpi(given in the problem).

Typically

fpi 0.9 fpjh 0.85


9/67

Allowable Stresses

ACI and AASHTO specify allowable stresses (as functions offciand fc) to ensure good behavior at service loads.

In our notation:

fci= allowable compression initial (-ve)

fti = allowable tension initial (+ve)

fcs= allowable compression service (-ve)fts = allowable tension service (+ve)


10/67

Allowable Stresses

Design for service loads:Make sure all stresses lie within allowable values:

fci< f1i< fti

Initial conditions: Actual stress

Allowable stress

fci< f2i< fti

Bottom Top

Service conditions: Similar equations


11/67

STRESSES BY DIRECTCALCULATION


12/67

Linear Elastic Material (superposition)

Uncracked

nAp


13/67

PS Alone (M=0)

Prestressing

Moment ppccps FeFeM

Prestressing

Axial Force

c pF F pc

axial

c c

FF

f A A

bMc M

f

I S

FcFp

ec= ep

cgc

cgp

PS alone


14/67

PS Alone (M=0)

FcFp

ec= ep

cgc

cgp

PS alone

S

Ae

A

F

S

Fe

A

Ff

ppppp1


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PS + Applied Moment

FcFp

ec= ep

cgc

cgp

PS alone PS + Mex t

ecFc

ep

Fp

1p p p p pext ext

F e F F e AM Mf

A S S A S S


16/67

Notation and Sign Convention

cgc = origin

y = positive down

Section properties are signed

even powers are always positive (A, I, Cw)

odd powers are signed according to y.

Tension is positive (force, stress, strain deformation)

Beam sign convention Locations numbered from bottom up (but cgc = 0)


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Example: Cross Section30

4

20

cgc

8

6

2

0

1

Ac= 240 in.2

c1= +16.0 in.

c2= -8.0 in.

I= 12,800 in.4

r2

= I/A= +12,800/(+240) = 53.333 in.4

S1= I/c1= +12,800/(+16) = 800 in.3

S2= I/c2=+12,800/(-8) = -1600 in.3

Note the signs on c and S


18/67

Notation and Sign Convention

Stresses due to prestressing alone

S

Ae

A

F

S

Fe

A

Ff

ppppp1

11

1 1

S

Ae

A

F

S

Fe

A

Ff

ppppp

22

2 1

S

Ae

A

F

S

Fe

A

Ff

ppppp


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Example: Prestressing Alone

ksiS

Fe

A

F

f ppp

475.0975.05.01600

120*13

240

120

2

2

ksiS

Fe

A

Ff

ppp450.295.15.0

800

120*13

240

120

1

1 Bot:

Top:

The T-section used in the notation example is prestressed with atendon force of 120 kips, located 3 in. up from the bottom face. Find

the stresses due to prestress alone.

Eccentricity: ep= +16 -3 = +13 in.


20/67

Example: Applied Moment

Positive Applied Moment M = 960 in-kips:

ksi

S

Mf 200.1

800

960

1

1

ksiS

Mf 600.0

1600

960

2

2

Bottom (tension)

Top (compression)

The sign of the stress is automatically correct.


21/67

Example: Superposition

-Fp/A-epFp/S +Mext/S

total

Superposition of stresses

Includes stress due to Msw


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STRESSES BYEQUIVALENT LOADS


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Equivalent LoadsChanges in tendon slope induce transverse loads. The prestressing tendon

can be replaced by the equivalent loads that it causes.

cgc

cgp

Compute the loads caused by all the changes in slope

Apply them to the concrete beam.

Find moments.

Find stresses in the concrete.


24/67

q+dq

Fp+dFp

Fp

q

w

dz

qqq sinsin ppp FddFFwdz

qq sintan dz

deslope

p ''" pppp eFeFw

"''" pppppp eFeFeFw

Equivalent Loads


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Straight tendon: No load

Parabolic, p2(z) : Uniform load

Cubic, p3(z) : Linearly varying load

Sharp angle: Point load.

Equivalent Loads


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Fp= 120 kips

cgc

cgp

12 1216

3

10.8 k 10.8 k

12 16 12

c1= 16

13

144 120 k10.8 k

Tendon is so flat that q= sinq= tanq

Treat horizontal component = diagonal component.

Equivalent Loads

130 ft-k = 1560 k-in.


27/67

cgc

cgp

12 1216

3

10.8 k 10.8 k

12 16 12

130 ft-k = 1560 k-in.

Equivalent Loads

M

2

120 15600.5 0.975 0.475

240 1600f ksi

1

120 1560

0.5 1.95 2.450240 800f ksi

Stresses Between

Harping Points

Same as calculated

directly earlier!. But

now, easier to see

variation over span


28/67

Fp= 120 kipsweq= -8Fp*sag/L2

= -8*120*13/4802

= 0.05517 k/in

= 0.65 k/ft

cgc

cgp

12 1216

13

=

sag

0.60 k/ft

120 ft-k

c1= 16

epmax= 13

ep= 4*sag*(x/L)(1-x/L)

ep= -8*sag/L2

Mpmax= weqL2/8

=0.65*402/8

= 130 ft-kips = 1560 k-in.

Equivalent Loads

Same at midspan,

but varies differently

over span


29/67

STRESSES BY PRESSURELINE (C LINE)


30/67

Effect of Mext on Fc

FcFp

ec= ep

cgc

cgp

PS alone PS + Mex t

ecFc

ep

Fp

Mext= Fp(ep-ec)

We know Mext(x), Fpand ep(x)

We can calculate and plot ec(x)

extc P

P

Me e

F


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Example: Pressure Line

Same T-Beam Fp= 120k

ep_max= 13 in.

W = 0.4 k/ft

Mext_max= 1/8*0.4 * 402

Mext_max= 80 k-ft = 960 k-in.

ec= ep- Mext/ Fp

ec= ep- 8 in. = 5 in.


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Why Should We Care About ec?

1. We want the stress at top and bottom to be within allowable

limits

2. We know Fp, ep

3. We can change (2) to keep location of Fc(ec) within certain

limits

i. Fp too close to bottom: compression on bottom too high

or tension on top too high

ii. Fp too close to top: compression on top too high or

tension on bottom too high


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Kerns

Kern = location of resultant force on cross-section that

causes zero stress on the opposite face

cgc

Kern distance

b

h

k = h/6 for a rectangular section. Middle third rule


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22

2 1S

Ae

A

F

S

Fe

A

Ff c

ppcp

Express the top and bottom stresses in terms of ep/k

2

1 1k

e

A

Ff c

p

1

2 1k

e

A

Ff c

p

ec> k1

cgck1

ec=k1 ec< k1

Kerns


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Keep the resultant force inside the kerns and the

section will never experience tension

cgc

k1

k2OK

zone

Kerns


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Location of compressive resultant if stresses at top and bottomjust equal the allowable stress.

cgc

k1

ec=k1 ec= k1ti

f2= fti

ec

Kern Modified kern

f2= 0.0

Extended Kerns


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If top stress is equal to the initial allowable stress, fti,

1

0

1

2 11

k

ef

k

e

A

Fff ci

cpi

iti

i

ticti

ffkek0

11 1'

Solve for ecto get k1ti

Kern

Modifier

Extended Kerns


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Note:

The (basic ) kern is a property of the section geometry alone.

This is true because, when the (top) stress is zero, the only

relevant characteristic of the compression force is its location.Its magnitude does not affect the (top) concrete stress, which

is zero.

The modif ied kern is a property of the section geometry. The

allowable stresses, and the applied stress, f0i.

Extended Kerns


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Similarly, if the bottom stress is equal to the initial allowablecompressive stress, fci,

2

01 1

k

efff ciici

i

cicci

f

fkek

0

21 1'

Solve for ecto get k1ci

Kern

Modifier

Extended Kerns


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0

21 1'

f

fkek ccc

At any loading stage, four modified kerns exist:

0

11 1'

f

fkek tct

ct kkk 111 ','min'

0

12 1'

f

fkek ccc

0

22 1'

f

fkek tct

ct kkk 222 ','max'

To

prevent

excess

tension

To

prevent

excess

compn.

To avoid

all

trouble

Extended Kerns


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0

21 1'

f

fkek ccc

Note the pattern:

0

11 1'

f

fkek tct

0

12 1'

f

fkek ccc

0

22 1'

f

fkek tct

Note the pattern:

Subscript number stays the same if tension limit.

Subscript number switches if compression limit.

Same subscript:

Different subscript:

Extended Kerns


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At any loading stage, four modified kerns:

k2t

k2c

k1ck1t

k2

k1

cgc

C-linemay be

located

anywhere in this

range

Extended Kerns


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Geometric way of thinking about stress limits:

Stress approach:Keep all stresses within all limiting values.

Geometric approach:Keep stress resultant (C-line) within modified kerns.

Extended Kerns


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Analysis Example

8DT24+2.5

dc,min=3

24 12 24

dc,min=3 dc,min=3

24

5

48 24

8

22

4.5


46/67

Analysis Example

A precast 8DT24+2 1/2 spans 60 ft simply supported.

It is reinforced with ten -in. diameter, low-relaxation

strands stressed to 185 ksi initial and 165 ksi effective.

The strands are harped at 3/8 and 5/8 of the span. The

dcdistance is 10 in. at the ends, 3 in. in the central 25%of the span and varies linearly in between those points.

Under initial conditions, the beam carries only its self-

weight. Under effective (i.e. service) conditions, it carries

10 psf permanent DL plus 50 psf LL.

The material properties are: f'ci= 3500 psi, f'c= 6000 psi

and the concrete density is 150 pcf.

The member is to satisfy ACI 318 allowable stresses


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Analysis Example

The member properties are:

h = 26.5 in. A= 718 in.2 I= 38535 in.4

c2= - 7.191 in. S2= -5359 in.3 k2= -2.779 in.

c1= 19.309 in. S1= 1996 in.3 k1= 7.464 in.

wsw= 748 lbs/ft

Ap= 10*0.153 = 1.530 in.2

dc,min= 3 at midspan, 10 at the ends.

ep= c1dcmin= 16.309 (mid-span), 9.309 (ends).

Check that the stresses satisfy ACI 318 at mid-span


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Analysis Example

Allowable Stresses

From ACI 318, Section 18.4

fci= -0.6 f'ci = -0.6*3.5 = -2.100 ksi

fti= 3.0 f'ci = (3*3500)/1000 = +0.177 ksi

fcs= -0.60 f'c = 0.6*6.0 = -3.600 ksi (for total load)

fts= 7.5 f'c = 7.5*6000)/1000 = +0.581 ksi

A l i E l S iti


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Elastic Analysis Using Stress Superposition

At midspan, initial conditions:

Fpi= 1.530 in.2*185 ksi = 283.1 kips

Mi= Msw= wsw L2/8 = 12*(0.748*602/8) = 4039 in.-kips

bot: f1 = - Fpi/A - epFpi/S1+ Mi/S1

= -(283.1/718) - (16.309*283.1/1996) + (4039/1996)

= -0.684 ksi > - 2.1 ksi = fci OK

top: f2 = - Fpi/A - epFpi/S2+ Mi/S2

= -(283.1/718) - (16.309*283.1/-5359) + (4039/-5359)

= -0.286 ksi < 0.177 ksi = fti OK

Analysis ExampleSuperposition

A l i E l S iti


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At midspan, effective (i.e., service) conditions:

Fpe= 1.530 in.2*165 ksi = 252.45 kips

Ms= Msw + MDL+ MLL = 4039 + 12*(0.080 + 0.400)* 602/8 = 6631 in.-kips

bot: f1= -Fpe/A - epFpe/S1 + Ms/S1

= -(252.45/718) - (16.309*252.45/1996) + (6631/1996)

= 0.908 ksi > 0.581 ksi = fts No Good

top: f1= -Fpe/A - epFpe/S2+ Ms/S2

= -(252.45/718) - (16.309*252.45/-5359) + (6631/-5359)

= -0.821 ksi > -3.600 ksi = fcs OK

The conditions at other locations could be found similarly. Two things are clear:

1. Design is controlled by bottom tension under service conditions. This

condition is very common and can be cured by increasing the prestress force.

2. The calculations are tedious and repetitive and are therefore ideal candidates

for automation.

Analysis ExampleSuperposition

A l i E l E i l t L d


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Analysis ExampleEquivalent Loads

kipsinFM pend 309.9

kipsFFW ppeq2887

12*24"309.9"309.16

Fp

9.309Fp

Fp

7Fp

/288

12 2424

9.309Fp

7Fp

/288

7Fp

/288

7Fp

/288

Elastic Analysis Using Equivalent Loads

The equivalent loads are:

End moment:

Transverse point load:

= 6.881 kips (initial)

= 6.137 kips (effective)


52/67

Initial conditions:-Fpi/A = -283.1/718 in2 = -0.394 ksi

At mid-span:

Mp = -(283.1k*9.309) - (6.881k*288) = -4617 in.-kips

Msw = 0.748 k/ft *602/8*12 = 4039 in.-kipsMi = -4617 + 4039 = -578 in.-kips

f1i = -Fpi/A + Mi/S1 = -0.394 + (-578)/(+1996) =-0.684 ksi

bottom

f2i

= -Fp

i/A + Mi/S2 = -0.394 + (-578)/(-5359) = -0.236 ksi top


A l i E l E i l t L d


53/67

Effective conditions:

-Fpi/A = -252.5/718 in2 = -0.352 ksi

At mid-span

Mp = -(252.5k*9.309) - (6.137k*288) = -4118 in.-kips

Msw = 0.748 k/ft *602/8*12 = 4039 in.-kips

MD+L= (0.080+0.400)*602/8*12 = 2592 in.-kips

Me = -4118 + 4039 + 2592 = 2513 in.-kips

f1i = -Fpi/A + Mi/S1 = -0.352 + (2513)/(+1996) = -0.908 ksi

bottomf2i= -Fpi/A + Mi/S2 = -0.352 + (2513)/(-5359) = -0.821 ksi top

These results are the same as those from the stress superposition

calculations.


A l i E l P Li


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Elastic Analysis Using Pressure Line and Modified Kerns

Initial conditions:Fpi =283.1 kips

Mext(L/2) = 4039

ep-ec= Mext/Fp= 4039/283.1 = 14.267 in.

ec= ep- Mext/Fp = 16.30914.267 = 2.042 in.

This value must be compared to the range of locations in which the PressureLine may lie, given by the modified kerns. The limits of that range are:

inkki

ti

f

f

ti 027.41779.21' 39 4.017 7.0

220

inkk ici

f

f

ci 319.321464.71' 39 4.010 0.2

120

inkki

ti

f

f

ti 817.101464.71' 39 4.017 7.0

11 0

inkki

ci

f

f

ci 033.121779.21' 39 4.010 0.2

210

ink i 027.4'2

ink i 817.10'1

Analysis ExamplePressure Line

A l i E l P Li


55/67

The C-line lies at y = -2.042 in. This is within the modified kern boundaries (-4.027,

+10.817) under initial conditions, so the results are acceptable.

k1ti

k1ci

k2ci

k2ti

cgc

C-line

OK zone

k2i

k1i

2.042


A l i E l P Li


56/67

Under effective conditions:

Fpi=252.5 kipsMext(L/2) = Msw+ MD+ ML= 6631 in.-kips

ep-ec= Mext/Fp= 6631/252.5 = 26.261 in.

ec= ep- Mext/Fp = 16.30926.261 = -9.952 in.

This value must be compared to the range of locations in which the Pressure

Line may lie, given by the modified kerns. The limits of that range are:

inkks

ts

f

f

ts 366.71779.21' 35 2.058 1.0

220

inkks

cs

f

f

cs 872.681464.71' 35 2.060 0.3

120

inkks

ts

f

f

ts 784.191464.71' 35 2.058 1.0

11 0

inkks

cs

f

f

cs 642.251779.21' 352.0600.3

210

ink s 366.7'2

ink s 784.19'1


A l i E l P Li


57/67

Under effective conditions,

ec= -9.952 in., but the highest it may be without violating a

stress limit isk2=- 7.366 in.

Thus the C-line lies outside the modified kern boundaries andthe result is unacceptable.

Not surprisingly, k2sis controlled by k2ts, which means that the

bottom tension stress under service conditions controls the

design. This is the same results as we obtained with the othertwo methods.


Composite Sections


58/67

Precast beam, cast-in-place slab.

Erect precast without formwork. Use it as form for cast-in-place deck.

CIP deck joins the pieces to work as one structure.

Composite Sections

Behavior of Composite Sections


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Composite action:

Connect CIP slab to PC girderno slip.

Entire section acts as one.

Need transformed section analysis if different Ecvalues. Compute stresses at different load stages, then add.




60/67

Load stages:

1. PC beam, self-weight only, initial prestress.

2. PC beam, self weight + slab weight, reduced prestress.

3. Composite beam, self weight + live load, effective prestress.


For stage 3 check, note that the stresses due to PS and self

weight of beam + slab are already locked in, using the PCbeam section alone. The LL acts on the composite section,

so causes stress in the slab and beam.



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Stage 1: at release


12

1 1S

M

k

e

A

Ff sw

ppi

i

21

2 1S

M

k

e

A

Ff sw

ppi

i

Check these stresses againstfci,fti.



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Stage 2: when slab is cast.Include PC self weight, slab self-weight, construction loading.


Check these stresses againstfcs,fts.

(Depends on how much time-dependent loss has occurred).

112

1 1

S

M

S

M

k

e

A

Ff constrslabsw

ppe

222

2 1S

M

S

M

k

e

A

Ff constrsl absw

ppe



63/67

Stage 3: Service conditions

Include only LL in the composite section analysis.

Add those stresses to the previous locked-in stresses.


Check these stresses againstfcs,fts.

112

1 1S

M

S

M

k

e

A

Ff constrsl absw

ppe

222

2 1S

M

S

M

k

e

A

Ff constrsl absw

ppe

comp

LD

S

Mf

,4

4

comp

LD

S

Mf

,3

3



64/67


Include only LL in the composite section analysis.Add those stresses to the previous locked-in stresses.


PS +

PC sw

Slab

sw constr LL Total



65/67



An 18 x 36 deep rectangular precast beam spans 75 ft simply supported. A 96 x

9 slab is cast on top of it. The composite system carries a LL of 100 psf. Find the

stresses under initial conditions, when the slab has been cast, and under full LL.

Allow for 20 psf construction load when the slab is cast.

Precast concrete: fci= 5.0 ksi, fc= 8.0 ksi, Ec= 5000 ksi, g= 0.150 kcf

c.i.p. concrete: fci= 3.0 ksi, fc= 5.0 ksi, Ec= 4000 ksi, g= 0.150 kcf

Strands: 20 strands, dia., fpi= 185 ksi, h= 0.85, epmax= 15.

Behavior of Composite SectionsThe member properties, all transformed to the precast concrete, are:


66/67


Property pc used comp Property pc used comp pc used comp

h = 36 42 c4 cip = -15.272727b min = 18 18 c3 cip = -9.2727273

b max = 18 96 c2 pc = -18 -9.2727273

A = 648 1108.8 c1 pc = 18 26.7272727 n 1 0.8

Ixx = 6.998E+04 190127.127

rxx^2 = 1.080E+02 171.471074 S4 = -12448.800 k4 =

Iyy = 1.750E+04 371390.4 S3 = -20503.906 k3 =

w = 0.056 S2 -3888 -20503.906 k2 = -6.000

ep.max 15.000 S1 3888 7113.600 k1 = 6.000

Member properties

Behavior of Composite SectionsThe member properties, all transformed to the precast concrete, are:


67/67


Summary of stresses

PS pc beam slab LL total

Initial

2 1.638 -1.465 0.000 0.000 0.173 OK

1 -3.822 1.465 0.000 0.000 -2.357 OK

0 -1.092 0.000 0.000 0.000 -1.092 OK

slab cast

2 1.392 -1.465 -1.302 -0.347 -1.722 OK

1 -3.249 1.465 1.302 0.347 -0.135 OK

0 -0.928 0.000 0.000 0.000 -0.928 OKservice

4 0.000 0.000 0.000 -0.434 -0.434 OK

3 0.000 0.000 0.000 -0.263 -0.263 OK

2 1.392 -1.465 -1.302 -0.329 -1.704 OK

1 -3.249 1.465 1.302 0.949 0.467 OK

0 -0 928 0 000 0 000 0 000 -0 928 OK

5-Flexural Analysis Service Slides 2015.pdf

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