Design of Slab e

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Budapest University of Technology and EconomyFaculty of Civil EgeneeringDepartment of Structural Egineering

DESIGN OF TWO WAY SPANNING FLOOR-SLAB

Manuel v1.11

Koris, Kálmán Dr. Ódor, PéterPéczely, Attila Dr. Strobl, András Dr. Varga, László

Budapest, 2005.

1 It is not a final version. It could be loaded from http:/www.vbt.bme.hu/oktatas/vb2 website



Two way spannig slab - v1.1

2

Content

1. Sizing .....................................................................................................................................3

1.1. Data ..................................................................................................................................31.2. Sizing of beams ................................................................................................................ 4

1.3. Sizing of the thickness of the slabs L1 and L2.................................................................. 4

1.3.1. The effective depth .................................................................................................... 4

1.3.2. Cover ......................................................................................................................... 4

1.3.3. Appropriate cover ...................................................................................................... 4

1.3.4. The final thickness of the slab: ..................................................................................5

1.4. Materials .......................................................................................................................... 5

1.5. The characteristic value of loads...................................................................................... 5

1.6. The design (ultimate) load: .............................................................................................. 5

1.7. Approximate analysis of the thickness of slab.................................................................5

1.8. L1and L2 slabs .................................................................................................................. 61.9. Spans ................................................................................................................................6

2. Calculation of the moments................................................................................................. 8

2.1. The maximum positive and negative moments ............................................................... 8

2.2. Calculating of moment maxm in direction x and y at point ..........................................9

2.3. Calculating of maxm moment in x and y direction in point ...........................................9

2.4. Negative bending moment in point ............................................................................. 9

2.5. Bending moments in practice......................................................................................... 10

2.6. Bending moments

max, xm és

max, ym in point of slab L1 ............................................ 10

2.7. Bending moment maxm in point (above the support)............................................. 11

2.8. Bending moment maxm in point (above the support)............................................. 12

2.9. The bending moment diagram .......................................................................................13

2.10. Redistribution of moments........................................................................................... 15

2.11. The modified bending moment diagram...................................................................... 16

3. Analysing of sections ..........................................................................................................17

3.1. Effective slab thickness and the area of steel required .................................................. 17

3.2. The moment of resistance .............................................................................................. 18

3.3. The minimal reinforcement required ............................................................................. 18

3.4. The anchorage length..................................................................................................... 18

4. Rules of reinforcement....................................................................................................... 19

5. 5. The drawing .................................................................................................................... 19

6. A törteher számítása ........................................................................................................ 20

6.1. Energia módszer.............................................................................................................20

6.2. Egyensúlyi módszer ....................................................................................................... 22

7. APPENDIX ......................................................................................................................... 24

7.1. Bar s’s tables for moments of two way spanning slab .................................................. 24




3

1. Sizing

1.1. Data

The span can be calculated according to the data sheet. The width of the beam G 2 in direction

y approximately is113

2

2

1 0 y

x

lb

. From this value the length of the span can be

determined in x direction:

mml

lcl

x

x y

38028011

05,1

3

2

2

1 b

m2,70,62,1

10x

1020

say: bx =350 mm.

The span (in y direction) of the two way spanning slab can be calculated using the ratio given

on data sheet (in case of l0y/l0 x = 1,2):

80,110,635,045,5

m45,51,1

0,6

1

1020

l

c

ll x

x

210

0

x

y

l

l

h 1

l0 x,1 l0 x,2

x

y

b 1

l 0 y

r

h2

h 2

h

30b2 30

l

h

l x,2 l x,1

h1

b 1

L2 L1

O1

O1

G1 G1

G1 G1

G2




4

1.2. Sizing of beams

In a case of usual load and loaded area for a beam has a hight of (acc. of a rule of thumb):

mmmml

h x

x 5505,54511

00,6

1210

mmmml

h y y 6505,654

11

20,7

1210

mmmmh

b y

x 3504,3827,1

650

25,1

mmmmh

b x y 3005,323

7,1

550

25,1

Remark: Size b x of beam in y direction should be calculated from h y , and b y of beam in x direction should becalculated from h x.

1.3. Sizing of the thickness of the slabs L1 and L2.

1.4. The effective depth

4020 short

slab

ld ; but the minimum thickness is d min=50 mm.

mmmmd slab 1800,18035

0,6·05,1

1.5. Cover

minimum cover (1. class ):

mm32hamm][

mm32hamm15min. c

increasing due to inaccuracy: 5 mm c 10 mm

For the lower mesh c = 5 mm could be recomended, for the upper mesh c = 10 mm because of the tread down during assembling.

1.6. Appropriate cover

For the lower layer of reinforcement:

cl = nom. c = min. c + c = 15 + 5 = 20 mm

For the upper layer of reinforcement:

cu = nom. c = min. c + c = 15 + 10 = 25 mm




5

1.7. The final thickness of the slab:

mm210hmm2072

1420180

2

acd h

1.8. Materials

Grade of concrete: C20/25 f ck = 20,0 N/mm2 safety factor c=1,5

E cm = 28,8 kN/mm2

Grade of steelbar: B 60.40 f yk = 400 N/mm2 safety factor s=1,15

1.9. The characteristic value of loads

The masses: (floor layers)

Materials thickness

[mm]

density [kN/m3] dead load (gi)

[kN/m2]

1. tiling 10 23,00 0,23

2. embedding plaster 20 22,00 0,44

3. pure concrete 40 22,00 0,88

4. technological isolation - - -

5. Nikecell foamlayer 60 1,50 0,09

6. r.c. slab 210 25,00 5,25

7. plaster 15 20,00 0,308. partition wall 1,50

dead load: Gk = gi = 8,69 kN/m2

live load: Qk = 5 kN/m2

1.10. The design (ultimate) load:Design load: Gd + Qd = G·Gk + P·Qk , where are the safety factors 1,35 and 1,5 for the

dead and the live loads , respectively.

Gd + Qd = 1,35·8,69 + 1,5·5 = 19,23 kN/m2

1.11. Approximate analysis of the thickness of slabApproximate moment at the support:

m

kNm52,54

14

00,605,1·23,19

14

)05,1·(22

0

max

xd d lQGm

In design state for slab c 0,2 could be recommended

2mm

N

33,135,1

20

ck

cd

f

f

d

cl Ø/2

cu




6

The ultimate moment of resistance for a unit wide strip:

)2

(1000maxc

cc

ck Rd

d d d

f mm

From the equation we can obtain the effective depth of the section:

2

2,01·2,0201000

5,110·52,54

21·1000

6max

ccck

c

f

md

d = 150,1 mm < d slab = 150 mm, so the overall thickness of the slab mmh 210 is suitable.

1.12. L1and L2 slabs

m x = ?

m y = ?

in points

1, 2 and 3

1.13. Spans

The spans l x and l y (for the accurate analysis) can be determined: the clear span l0 x, l0 y should

be increased by 1/3·t 1/2·t at exterior span and by 1/2·t at interior span, where t is the length of the support.

m275,6

2

35,01,000,6

23

cm30101 x

x x

bll

l y

l x2 l x1

12

3




7

m725,52

35,01,045,5

23

cm30202

x x x

bll

m50,730,020,7

2

20 y y y

bll




8

2. Calculation of the moments(Using elastic theory of slabs)

2.1. The maximum positive and negative moments

Generally the maximum moments can be obtained if the arrangement of the loads is:

a.) "Gd " (dead load) is all over slabs

b.) "Qd " (imposed load) is arranged according to the influence line theory on

some slabs.

This is an accurate method but an approximate method can be used if the ratio between the

spans is 25,18,0 b

j

l

l. According to the method the total (ultimate) load should be devided

into two parts:

q’ is UDL load in every fields

q’’ is positive or negative alternating UDL load in every fields, respectively.

- UDL substituting load is acting totally and its value is:

2' k

Qk G

QGq

- alternating load:

2" k

Q

Qq

Qd

Gd

= +

Qd /2 -Qd /2 +Qd /2

Q d

1+

If the span condition lr /ll mentioned above is true, the rotations of adjacent spans at thecommon support are almost equel, and it is zero. Thus the ends of the spans may be supposed

as fixed ones for both spans at the adjacent section, so that the slabs can be analysed

separately. The value of rotation of adjacent spans due to alternating loads at the common

support is almost the same but the direction is opposite. That is why we can suppose a hinged

support for the end conditions of the slabs.

The final moment will be the algebraical sum of the moments due to the load q’ and q’’.

"'max qq mmm




9

2.2. Calculating of moment maxm in direction x and y at point

?1

max m

- q' has to be put totally,

- q'' has to be arranged alternally!

maxm = mq' ± mq"

2.3. Calculating of maxm moment in x and y direction in point

?1

max m

The method of calculation is the same as at point 2.2 except instead of using lx1 the lx2 should

be used!

2.4. Negative bending moment in point

?2

max m

On the shaded area q' should be placed totally downward, the remained part should be loaded

by ±q'' . Take care of the edge conditions!

In case of non-equal l xi values both slab should be analysed.

Qd

Gd

=

Qk /2

Q d

- Q d / 2

+ Q d / 2

Q d

G d

2

+

Q d / 2

L1 L1 +

load: q' load: q''

L1 L1 L1 L1 + +ill.

load: q' load: q''




10

In the point the bending moments may be not equal because of the different lx1 and lx2. In

that case the out-of-balance bending moment should be redistributed in the ratio of the

relative stiffness of the slabs (in the ratio 1/lx1 and 1/lx2, respectively)

The bending moment coefficients are given in Tables Bar s (see attached!)

1 2 3 4 5 6

1 2 3 4 5 6

L1: 837,05,7

275,6

y

x

l

l

L2:

763,05,7

725,5

y

x

l

l

2.5. Bending moments in practice(using the Tables)

2.6. Bending moments max, xm és

max, ym in point of slab L1

2m

kN48,15

2

5·5,169,8·35,1

2··' k

Qk G

QGq

2m

kN75,3

2

5·5,1

2·" k

QQq

l y

l x1

L10,837

0,0298 0 , 0

1 8 5

L10.837

0,0571 0 , 0

2 9 7

+

q' q"

Tab. 1.11 Tab. 1.7

"0571,0'0298,02

1

1

max, qqlm x x

m

kNm

59,26)75,3·0571,048,15·0298,0·(275,621

max,

xm




11

"0297,0'0185,021

max, qqlm y y

m

kNm37,22)75,3·0297,048,15·0185,0·(5,7 21

max, ym

2.7. Bending momentmaxm in point

(from L1, above support)

L10,837

-0.0762

l y

l x

+

q' q"

L11.195

Tab. 1.11 Tab. 1.8

-0.0977

As mentioned above that bending moment should be calculated in slab L1 and L2,

respectively. The bending moments should be finally equalised. Pay attention, the Table 1.8

refers to the slab with one short continuous edge. In our project the slab has one long

continuous edge, so the lx/ly ration should be exchanged!

)"·0977,0'·0762,0·(2

1

2

max, qqlm x x

m

kNm87,60)75,3·0977,048,15·0762,0·(275,6 22

max,

xm




12

2.8. Bending moment maxm in point (above the support)

L10.837

l y

l x

+

q' q"

L10.837

Tab. 1.8Tab. 1.11

- 0 , 0 5

0 2

- 0 , 0 6

8 9

)"·0689,0'·0502,0·(23

max, qqlm y y

m

kNm25,58)75,3·0689,048,15·0502,0·(9,7 23

max,

ym

2.9. Bending moments max, xm és

max, ym in point of slab L2

"0654,0'0350,022

4max, qqlm x x

m

kNm80,25)75,3·0654,048,15·0350,0·(725,5 24

max, xm

"0240,0'0151,024

max, qqlm y y

m

kNm21,18)75,3·0240,048,15·0151,0·(5,7 24

max, ym

2.10. Bending moment maxm in point (from L2, above support)

)"·1036,0'·0854,0·(22

2max, qqlm x x

m

kNm06,56)75,3·1036,048,15·0854,0·(275,6 22

max, xm

2.11. Bending moment maxm in point (above the support)

)"·0609,0'·0438,0·(23

max, qqlm y y

m

kNm98,50)75,3·0609,048,15·0438,0·(9,7 23

max, ym




13

2.12. Balancing of moment in point

There is an out-of-balance bending moment above the support.

m

kNm

87,602

1max,,

L xm

m

kNm06,562

2max,, L xm

m

kNm81,406,5687,602

xm

on the slab L1

m

kNm29,281,4

725,5

1

275,6

1275,6

1

11

1

2

21

12

1

x

x x

x x m

ll

lm

on the slab L2

m

kNm52,281,4

725,5

1

275,6

1

725,5

1

11

1

2

21

22

2

x

x x

x x m

ll

lm

The balanced moment in point

m

kNm58,5852,206,562

, new xm

and the positive midfield moments

m

kNm74,27

2

29,259,26

2

2

11

max,

1

max,,

x xnew x

mmm

m

kNm54,24

2

52,280,25

2

2

24

max,

4

max,,

x

xnew x

mmm




14

2.13. The bending moment diagram




15

2.14. Redistribution of moments

As the ratio of

m

m is disadvantageous in reinforcing, a redistribution of bending moments

may be made. The redistribution may be appplied, if

the resultant moment is in equalibrium with the loads applied,

for the ratio of the spans it is true that 5,02 b

j

l

l;

d

x 25,144,0 , if the grade of concrete is not greater than C35/40;

7,0 using high ductility steel, i.e 5ud % ( ud fracture strain);

85,0 using normal ductility steel, i.e. 5,2ud %;

where.

.

nondistr

distr

m

m ;

x: depth of compression zone after redistribution;

d : effective depth of the section.

An economical reinforcement can be designed, if 5,1

m

m, taking the conditions into

consideration given above.

As the sum of the bending moments must not be changed therefore

mmmm ecec

Moment redistribution for m x (in slab L1):

m

kNm53,34

5,2

58,5874,27

5,11

1

,

mmm xec

m

kNm80,5153,345,15,12

, ec xec mm

85,088,058,58

80,512

,

m

m xec OK.

Moment redistribution for m y (in slab L1):

m

kNm25,32

5,11

25,5837,221,

yecm

m

kNm38,4825,325,13,

yecm

85,083,025,58

38,48 , it is not OK.

As the condition of 85,0 should be fulfilled:

m

kNm51,4925,5885,085,03, mm yec




16

Of course the sum of the positive and negative moments after the redistribution must not be

changed (equalibrium condition!), from where we get:

m

kNm11,3151,4925,5837,222,

ec yec mmmm

2.15. The modified bending moment diagram

Because of the partial fixing of the outer support, the 25 % of the midfield moment should betaken into calculation on this support.

According to the EUROCODE:

m

kNm98,693,2725,0

25,0 4

2,

xecleft mm

m

kNm63,853,3425,0

25,0 1

1,

xecright mm




17

3. Analysing of sections

As the bending moment is bigger in the shorter direction, the lower (layer) reinforcement is

running in the shorter direction.

3.1. Effective thickness of slab and area of steel required

The effective slab thickness would be different in x és y direction for the positive and negative

bending moments xm ;

xm ; ym ;

ym , respectively.

mm1832

1420210

2:

a x x chd m for

mm1782

1425210

2:

f x x chd m for

mm16914183: x y y d d m for

mm16414178: x y y d d m for

secti

onm x( y)

[kNm/m]d x( y)

[mm] xc

[mm]as,cal

[mm2/m]as,provided as

[mm2/m]

0’x -8,63 178

1x +34,53 183

2x -51,80 178

4x +27,93 183

0’’x -6,98 178

1y +31,11 169

3y -49,51 164

4y 169

5y 164

Calculate xc from this equation

2

ccd c

xd f xbm

mmb 1000

Then

yd calscd c f a f x ,1000

yd

cd ccals f

f xa

1000,




18

3.2. The moment of resistance

km as

[mm2/m]

xc [mm]

m Rd [kNm/m]

m Rd > mSd m Rd

mSd >1

0’x

1x

2x

4x

0’’x

1y

3y

4y

5y

cd c yd s f xb f a 0ccd

yd sc x

f b

f a x

2

ccd c Rd

xd f xbm

3.3. The minimal reinforcement required

d b f

d b

a yk s

0015,0

6,0

maxmin, ( f yk in [N/mm2

])

cs Aa 04,0max,

3.4. The anchorage length

The basic value:bd

yd b f

f l

4 (where f bd is taken from Table below)

f ck 12 16 20 25 30 35 40 45 50Smooth

surface

0,9 1,0 1,1 1,2 1,3 1,4 1,5 1,6 1,7

ribbed

1,6 2,0 2,4 2,8 3,2 3,6 3,9 4,2 4,5

Anchorage length required: min,

,

,

b provs

recsbsb l

A

All

s = 1 in case of smooth surface steel

As,rec = area of steel required

As,prov = area of steel providedThe minimal anchorage length:




19

103,0min, bb ll tension steeel bars

mm1006,0min, bb ll compression steel bars

The anchorage length of bend-up bars for shear resistance:

In tension zone.: 1,3

lb,net In compression zone: 0,7lb,net

4. Rules of reinforcement

Maximal distance between bars:

main bars: mm3505,1 h (h: depth of slab)

distributors: mm4005,2 h

At least the half of midfield reinforcement should be let to the support and fixed there

At pinned support a reinforcement should be provided

5. The drawing Signing of bars

The place of the bar should be given from the moulding

Bottom view !!! (the section is taken below the slab, seeing in a mirror)

Do not use many types of bars, or diameters next to each other

Use stays (supporting reinforcement)!

In notes indicate!:

grades of materials (concrete, steelbars);concrete cover;

characteristic value of imposed load;

and any other data, if there are Schedule of bars




20

6. A törteher számítása

6.1. Energia módszer

A kinematikailag lehetséges tör terhet a küls és bels munkák egyenlsége alapján lehet

meghatározni.

Lk = Lb

Vegyünk fel egy lehetséges törésképet a lemezek törésvonal elmélete alapján. Nyomatéki paraméterként a hosszabbik oldalhoz tartozó, m pozitív nyomatékot választjuk. A rövidebbik

irányban fellép nyomatékot -val való szorzással, a támasznyomatékokat 1 4 szorzóksegítségével számíthatjuk az m nyomatékból. Geometriai paraméterként a törésvonalak

metszéspontját meghatározó 1, 2 és tényezk vehetk fel.

2 = 4 (a szimmetriából adódóan)

1 = 1 - 2

l x = 5,6 m l y = 7,9 m 710 ,l

l

y

x




21

m

kNm8336 ,m (lásd 3.2.)

m

kNm31,93 m 87,0

m

kNm66181 ,m

51,01

m

kNm614742 ,m m

30,142

m

kNm46553 ,m

51,13

A p teher által az elmozduláson végzett küls munka a töréskép által meghatározott térfogatalapján számítható.

12

212

3

11

22

3

11

22

3

11

2

21 y x y x y x y xk k

llllllll p L

236

y xk k

ll p L

A bels munka a nyomatéknak a törésvonalak menti elforduláson végzett munkájával

egyenl.

21

11

2

111

2

2

13

21

21

y x

x y x y y x x x yb

llm

llm

llm

llm

lllm L

mmmmmm

Lb 21

2

3

1

22111

49,1131

33,781

20,1301

21

b L

A küls és bels munkák egyenlsége alapján ( Lk = Lb):

49,11333,7820,130

2337,721

k p

12 1

A tör teher tehát két paraméter függvény, amikbl a szélsérték parciális deriválással

kapható.

, f p 1




22

37,7·3·2··1·

·49,113··20,13033,78·20,130·49,113 2

k p 211 ,

0

0

p

p

A deriválást elvégezve: 563,01 , 436,02 , 424,0

A kapott értékeket behelyettesítve a fenti egyenletbe megkapható pk értéke:

2m

kN74,42k p

6.2. Egyensúlyi módszer




23

Ez is törési határállapot-vizsgálat. A kinematikai tételen alapszik. (A feladatban taláncélravezet bb!) A küls terhek nyomatékának és a bels nyomatékok egyensúlyának

felírásából számítható a határer .

és lemezdarab azonos

3

2

2

22

2

y x x

ll pmml

lemezdarabra:

32

22

2122

22

2

1 x y x x y

y

lllll pmml

lemezdarabra:

322

2

2111

22

13 x y x x y y

lllll pmml

ismeretlenek: p, 1 ,

adott: 3 egyenlet

56,01 , 44,02 , 42,0 ,2m

kN74,42 p

22 m

kN45,17

m

kN74,42 d p p

145.2

45,17

74,42

d p

p Megfelel!



Design of two way spanning slab

7. APPENDIX

7.1. Bars’s tables for moments of two way spanning slab

Tab. 1.7 =0,15

ws M xs M ys

0,50 0,1189 0,0991 0,0079

0,55 0,1101 0,0923 0,0103

0,60 0,1015 0,0857 0,0131

0,65 0,0931 0,0792 0,01620,70 0,0851 0,0730 0,0194

0,75 0.0777 0,0669 0,0230

0,80 0,0708 0,0611 0,0269

0,85 0,0644 0,0557 0,0307

0,90 0,0584 0,0507 0,0344

0,95 0,0529 0,0462 0,0383

1,00 0,0476 0,0423 0,0423

1,10 0,0390 0,0353 0,0500

1,20 0,0320 0,0293 0,0575

1,30 0,0262 0,0244 0,0644

1,40 0,0216 0,0204 0,0710

1,50 0,0179 0,0173 0,0772

1,60 0,0149 0,0146 0,0826

1,70 0,0124 0,0124 0,0874

1;80 0,0105 0,0107 0,0916

1,90 0,0088 0,0091 0,0954

2,00 0,0074 0,0079 0,0991

q· a4

E· h3 q· a2 q· b2

Tab. 1.8 =0,15

ws M xs M ys M yvs

0,50 0,1087 0,0908 0,0084 -0,0305

0,55 0,0981 0,0826 0,0109 -0,0362

0,60 0.0881 0,0747 0.0135 -0,0421

0,65 0,0786 0,0670 0,0162 -0,04790,70 0,0698 0,0599 0,0192 -0,0537

0,75 0,0618 0,0533 0,0221 -0,0594

0,80 0,0544 0,0472 0,0249 -0,0650

0,85 0,0479 0,0417 0,0277 -0,0703

0,90 0,0421 0,0369 0,0304 -0,0750

0,95 0,0370 0,0327 0,0330 -0,0797

1,00 0,0326 0,0291 0,0354 -0,0840

1,10 0,0253 0,0228 0,0399 -0,0917

1,20 0,0197 0,0180 0,0438 -0,0980

1,30 0,0155 0,0143 0,0471 -0,1032

1,40 0,0123 0,0115 0,0500 -0,1075

1,50 0,0099 0,0094 0,0524 -0,1109

1,60 0,0079 0,0076 0,0544 -0,1136

1,70 0,0063 0,0062 0,0561 -0,1160

1,80 0,0052 0,0052 0,0575 -0,1184

1,90 0,0043 0,0044 0,0586 -0,1203

2,00 0,0036 0,0037 0,0594 -0,1213

q· a4

E· h3 q· a2 q· b2 q· b2

a y

M x b M y s

xq

q

x=0 x=a

y = 0

y = b

=ab =

ab

M xbs = · M yvs

a y

q M x b M y s

xq y = 0

y = b

x=0 x=a

M xb M y v




25

Tab. 1.9 =0,15

ws M xs M ys M yvs

0,50 0,0990 0,0835 0,0088 -0,0297

0,55 0,0872 0,0738 0,0113 -0,0350

0,60 0,0759 0,0647 0,0137 -0,0400

0,65 0,0657 0,0563 0,0166 -0,0450

0,70 0,0565 0,0489 0,0187 -0,0497

0,75 0,0484 0,0423 0,0212 -0,0540

0,80 0,0414 0,0363 0,0233 -0,0578

0,85 0,0355 0,0313 0,0254 -0,0612

0,90 0,0305 0,0270 0,0274 -0,0644

0,95 0,0262 0,0232 0,0292 -0,0677

1,00 0,0225 0,0201 0,0309 -0,0699

1,10 0,0167 0,0151 0,0335 -0,0741

1,20 0,0126 0,0113 0,0357 -0,0770

1,30 0,0096 0,0088 0,0374 -0,0793

1,40 0,0073 0,0068 0,0386 -0,0811

1,50 0,0057 0,0053 0,0396 -0,0815

1,60 0,0045 0,0042 0,0404 -0,0825

1,70 0,0036 0,0034 0,0410 -0,0830

1;80 0,0029 0,0028 0,0414 -0,0832

1,90 0,0023 0,0023 0,0416 -0,0833

2,00 0,0018 0,0019 0,0417 -0,0833

q· a4

E· h3 q· a2

q· a2

q· b2

Tab. 1.10

=0,15

ws M xs M xvmin M ys M yvmin

0,50 0,0549 0,0570 -0,1189 0,0040 -0,0205

0,55 0,0520 0,0543 -0,1148 0,0054 -0,0249

0,60 0,0490 0,0514 -0,1104 0,0072 -0,0294

0,65 0,0458 0,0483 -0,1057 0,0092 -0,0341

0,70 0,0425 0,0451 -0,1008 0,0114 -0,0390

0,75 0,0393 0,0418 -0,0957 0,0139 -0,0442

0,80 0,0361 0,0385 -0,0905 0,0164 -0,0496

0,85 0,0330 0,0354 -0,0852 0,0191 -0,0548

0,90 0,0301 0,0324 -0,0798 0,0217 -0,0598

0,95 0,0273 0,0295 -0,0745 0,0243 -0,0648

1,00 0,0246 0,0269 -0,0699 0,0269 -0,0699

1,10 0,0201 0,0221 -0,0608 0,0319 -0,0787

1,20 0,0164 0,0182 -0,0530 0,0365 -0,0869

1,30 0,0133 0,0148 -0,0462 0,0406 -0,0937

1,40 0,0108 0,0122 -0,0405 0,0442 -0,0993

1,50 0,0089 0,0100 -0,0358 0,0473 -0,1041

1,60 0,0072 0,0081 -0,0317 0,0499 -0,1082

1,70 0,0059 0,0066 -0,0282 0,0521 -0,1116

1,80 0,0048 0,0055 -0,0252 0,0540 -0,1143

1,90 0,0040 0,0046 -0,0226 0,0556 -0,1167

2,00 0,0034 0,0040 -0,0205 0,0570 -0,1189

q· a4

E· h3 q· a2

q· a2

q· b2

q· b2

=ab

M x0s = M xbs

M xbs = · M yvs

a y

q M x b M y s

xq y =

0

y = b

x=0 x=a

M xb M y v

M x0 M y v

=ab

M xbmin = · M yvmin

M y0min = · M xvmin

a y

q M x b M y s

xq y

= 0

y = b

x=0 x=a

M xbmi M y v m i

M xvmin

M y 0 m i




Tab. 1.11 =0,15

ws M xs M xvs M ys M yvmin

0,50 0,0528 0,0550 0,1135 0,0045 0,0203

0,55 0,0489 0,0514 0,1078 0,0062 0,0247

0,60 0,0450 0,0476 0,1021 0,0081 0,0291

0,65 0,0411 0,0436 0,0964 0,0101 0,0336

0,70 0,0373 0,0398 0,0906 0,0122 0,0381

0,75 0,0336 0,0359 0,0845 0,0145 0,0427

0,80 0,0300 0,0323 0,0881 0,0169 0.0471

0,85 0,0266 0,0289 0,0720 0,0191 0,0513

0,90 0,0236 0,0257 0,0661 0,0211 0,0551

0,95 0,0209 0,0228 0,0603 0,0232 0,0586

1,00 0,0184 0,0202 0,0546 0,0252 0,0617

1,10 0,0142 0,0158 0,0467 0,0287 0,0676

1,20 0,0110 0,0123 0,0399 0,0316 0,0722

1,30 0,0086 0,0096 0,0341 0,0340 0,0757

1,40 0,0068 0,0075 0,0293 0,0359 0,0782

1,50 0,0054 0,0060 0,0254 0,0374 0,0800

1,60 0,0043 0,0048 0,0221 0,0386 0,0814

1,70 0,0034 0,0039 0,0193 0,0395 0,0825

1;80 0,0027 0,0031 0,0171 0.0402 0,0834

1,90 0,0022 0,0026 0,0154 0,0408 0,0342

2,00 0,0018 0,0022 0,0141 0,0412 0,0847

q· a4

E· h3 q· a

2

q· a

2

q· b

2

q· b

2

Tab. 1.12

=0,15

ws M xs M xvs M ys M yvmin

0,50 0,0296 0,0405 0,0833 0,0024 0,0143

0,55 0,0286 0,0394 0,0817 0,0033 0,0172

0,60 0,0275 0,0378 0,0794 0,0046 0,0206

0,65 0,0261 0,0360 0,0767 0,0061 0,0242

0,70 0,0246 0,0339 0,0737 0,0079 0,0280

0,75 0,0231 0,0315 0,0704 0,0098 0,0320

0,80 0,0214 0,0293 0,0668 0,0103 0,0360

0,85 0,0196 0,0269 0,0631 0,0139 0,0400

0,90 0,0180 0,0247 0,0593 0,0160 0,0440

0,95 0,0164 0,0224 0,0554 0,0181 0,0480

1,00 0,0149 0,0202 0,0515 0,0202 0,0515

1,10 0,0121 0,0164 0,0449 0,0242 0,0585

1,20 0,0098 0,0131 0,0388 0,0287 0,0643

1,30 0,0078 0,0105 0,0336 0,0306 0,0690

1,40 0,0063 0,0084 0,0291 0,0332 0,0728

1,50 0,0051 0,0066 0,0254 0,0353 0,0757

1,60 0,0041 0,0053 0,0223 0,0369 0,0779

1,70 0,0033 0,0042 0,0198 0,0383 0,0797

1,80 0,0027 0,0035 0,0176 0,0392 0,0812

1,90 0,0022 0,0028 0,0158 0,0399 0,0824

2,00 0,0018 0,0024 0,0143 0,0405 0,0833

q· a4

E· h3 q· a

2

q· a

2

q· b

2

q· b

2

=a

b

M xbs = M x0s M yas = M y0s

M x0s = · M yvs

M y0s = · M xvs

a y

q M x b M y s

xq y =

0

y = b

x=0 x=a

M xbmi M y v m i

M x0m M y v m i

M xvs

M y a s

=a

b

M x0min = M xbmin

M x0min = · M yvmin

M yas = · M xvs

q

a y

q M x b M y s

x y = 0

y = b

x=0 x=a

M xbs M y v s

M x0s M y v s

M xvs

M y a s

M xvs

M y 0 s

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