Design of Two Way Spanning Floor-slab

Budapest University of Technology and Economy Faculty of Civil Egeneering Department of Structural Egineering

DESIGN OF TWO WAY SPANNING FLOOR-SLAB

Manuel v1.11

Koris, Kálmán Dr. Ódor, Péter Péczely, 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 .................................................................................................................................. 3 1.2. 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 .................................................................................................................. 6 1.9. 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 törőteher számítása ........................................................................................................ 20 6.1. Energia módszer............................................................................................................. 20 6.2. Egyensúlyi módszer ....................................................................................................... 22

7. APPENDIX ......................................................................................................................... 24 7.1. Baręs’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 G2 in direction

y approximately is 113

221 0 y

x

lb ⎟

⎠⎞

⎜⎝⎛ ≈≅ . From this value the length of the span can be

determined in x direction:

mml

lcl

x

xy

38028011

05,132

21b

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/l0x = 1,2):

80,110,635,045,5

m45,51,10,6

1

1020

=++=

===

l

cll x

x

210

0 ≤<x

y

ll

h 1

l0x,1 l0x,2

x

y

b 1

l 0y

r

h2

h 2

h

30 b2 30 l

h

lx,2 lx,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 xx 5505,545

1100,6

1210⇒=≅

÷≅

mmmml

h yy 6505,654

1120,7

1210⇒=≅

÷≅

mmmmh

b yx 3504,382

7,1650

25,1⇒=≅

÷≅

mmmmh

b xy 3005,323

7,1550

25,1⇒=≅

÷≅

Remark: Size bx of beam in y direction should be calculated from hy, and by of beam in x direction should be

calculated from hx. 1.3. Sizing of the thickness of the slabs L1 and L2. 1.4. The effective depth

4020 ÷≅ short

slabl

d ; but the minimum thickness is dmin=50 mm.

mmmmd slab 1800,18035

0,6·05,1⇒=≅

1.5. Cover minimum cover (1. class ):

⎩⎨⎧

≥∅⇒∅<∅⇒

≥mm32hamm][

mm32 hamm15min. c

increasing due to inaccuracy: 5 mm ≤ Δc ≤ 10 mm For the lower mesh Δc = 5 mm could be recomended, for the upper mesh Δc = 10 mm because of the tread down during assembling. 1.6. Appropriate cover For the lower layer of reinforcement: cl = nom. c = min. c + Δc = 15 + 5 = 20 mm For the upper layer of reinforcement: cu = nom. c = min. c + Δc = 15 + 10 = 25 mm


5

1.7. The final thickness of the slab:

mm210hmm2072

14201802

=⇒=++=∅

++= acdh

1.8. Materials Grade of concrete: C20/25 fck = 20,0 N/mm2 safety factor γc=1,5 Ecm = 28,8 kN/mm2 Grade of steelbar: B 60.40 fyk = 400 N/mm2 safety factor γs=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 0,30 8. partition wall 1,50

dead load: Gk = Σgi = 8,69 kN/m2 live load: Qk = 5 kN/m2 1.10. The design (ultimate) load: Design load: Gd + Qd = γG·Gk + γP·Qk, where γι 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 slab Approximate moment at the support:

( ) ( )m

kNm52,5414

00,605,1·23,1914

)05,1·( 220

max =⋅

=+

≅− xdd lQGm

In design state for slab ξc ≈ 0,2 could be recommended

2mmN33,13

5,120

==γ

= ckcd

ff

d

cl Ø/2

cu


6

The ultimate moment of resistance for a unit wide strip:

)2

(1000maxc

cc

ckRd

dddfmm ξξγ

−==−

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

⎟⎠⎞

⎜⎝⎛ −⋅⋅

⋅=

⎟⎠

⎞⎜⎝

⎛ −⋅=

−

22,01·2,0201000

5,110·52,54

21·1000

6max

ccck

c

f

md

ξξ

γ

d = 150,1 mm < dslab = 150 mm, so the overall thickness of the slab mmh 210= is suitable. 1.12. L1and L2 slabs mx = ? my = ? in points 1, 2 and 3 1.13. Spans The spans lx and ly (for the accurate analysis) can be determined: the clear span l0x, l0y 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.

m275,6235,01,000,6

23cm30

101 =++=++= xxx

bll

l y

lx2 lx1

12

3


7

m725,5235,01,045,5

23cm30

202 =++=++= xxx

bll

m50,730,020,72

20 =+=⋅+= yyy

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

ll

. 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

QkGQGq ⋅γ+⋅γ=

- alternating load:

2" k

QQq ⋅γ±=

Qd

Gd

= +

Qd/2 -Qd/2 +Qd/2Q

d

1+

If the span condition lr/ll mentioned above is true, the rotations of adjacent spans at the common 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

?1max =+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

?2max =−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 lxi values both slab should be analysed.

Qd

Gd

=

Qk/2

Qd

-Qd/2

+Q

d/2

Qd

Gd

2 ×

+

Qd/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 Baręs (see attached!)

1 2 3 4 5 6

1 2 3 4 5 6

L1: 837,05,7

275,6==

y

x

ll

L2:

763,0

5,7725,5

==y

x

ll

2.5. Bending moments in practice (using the Tables) 2.6. Bending moments +

max,xm és +max,ym in point of slab L1

2mkN48,15

25·5,169,8·35,1

2··' =+=+= k

QkGQ

Gq γγ

2mkN75,3

25·5,1

2·" ±=±=γ±= k

QQq

l y

lx1

L1 0,837

0,0298

0,01

85

L10.837

0,0571

0,02

97

+

q' q"

Tab. 1.11 Tab. 1.7

( )"0571,0'0298,021

1max, qqlm xx ⋅±⋅⋅=+

mkNm59,26)75,3·0571,048,15·0298,0·(275,6 21

max, =±=+xm


11

( )"0297,0'0185,021max, qqlm yy ⋅±⋅⋅=+

mkNm37,22)75,3·0297,048,15·0185,0·(5,7 21

max, =±=+ym

2.7. Bending moment −

maxm in point (from L1, above support)

L1 0,837

-0.0762 l y

lx

+

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·(21

2max, qqlm xx ±=−

mkNm87,60)75,3·0977,048,15·0762,0·(275,6 22

max, −=±=−xm


12

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

L1 0.837

l y

lx

+

q' q"

L10.837

Tab. 1.8Tab. 1.11

-0,0

502

-0,0

689

)"·0689,0'·0502,0·(23max, qqlm yy ±=−

mkNm25,58)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

( )"0654,0'0350,02

24max, qqlm xx ⋅±⋅⋅=+

mkNm80,25)75,3·0654,048,15·0350,0·(725,5 24

max, =±=+xm

( )"0240,0'0151,024max, qqlm yy ⋅±⋅⋅=+

mkNm21,18)75,3·0240,048,15·0151,0·(5,7 24

max, =±=+ym


maxm in point (from L2, above support)

)"·1036,0'·0854,0·(22

2max, qqlm xx ±=−

mkNm06,56)75,3·1036,048,15·0854,0·(275,6 22

max, −=±=−xm


maxm in point (above the support)

)"·0609,0'·0438,0·(23max, qqlm yy ±=−

mkNm98,50)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.

mkNm87,602

1max,, −=−Lxm

mkNm06,562

2max,, −=−Lxm

mkNm81,406,5687,602 =−=Δ −

xm

on the slab L1

mkNm29,281,4

725,51

275,61

275,61

11

12

21

121 =⋅

+=Δ⋅

+=Δ −−

x

xx

xx m

ll

lm

on the slab L2

mkNm52,281,4

725,51

275,61

725,51

11

12

21

222 =⋅

+=Δ⋅

+=Δ −−

x

xx

xx m

ll

lm

The balanced moment in point

mkNm58,5852,206,562

, =+=−newxm

and the positive midfield moments

mkNm74,27

229,259,26

2

211

max,1max,, =+=

Δ+=

−++ xxnewx

mmm

mkNm54,24

252,280,25

2

224

max,4max,, =−=

Δ−=

−++ xxnewx

mmm


14

2.13. The bending moment diagram


15

2.14. Redistribution of moments

As the ratio of +

−

mm 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

ll

;

• dx

⋅+≥ 25,144,0δ , if the grade of concrete is not greater than C35/40;

• 7,0≥δ using high ductility steel, i.e 5>udε % ( udε fracture strain); • 85,0≥δ using normal ductility steel, i.e. 5,2>udε %;

where .

.

nondistr

distr

mm

=δ ;

x: depth of compression zone after redistribution; d: effective depth of the section.

An economical reinforcement can be designed, if 5,1≅+

−

mm , taking the conditions into

consideration given above. As the sum of the bending moments must not be changed therefore

−+−+ +=+ mmmm ecec Moment redistribution for mx (in slab L1):

mkNm53,34

5,258,5874,27

5,111, =

+=

++

=−+

+ mmm xec

mkNm80,5153,345,15,12

, =⋅=⋅= +−ecxec mm

85,088,058,5880,512

, >===−

−

mm xecδ OK.

Moment redistribution for my (in slab L1):

mkNm25,32

5,1125,5837,22

1, =++

=+yecm

mkNm38,4825,325,13, =⋅=−

yecm

85,083,025,5838,48

<==δ , it is not OK.

As the condition of 85,0≥δ should be fulfilled:

mkNm51,4925,5885,085,03, −=⋅=⋅= −− 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:

mkNm11,3151,4925,5837,222, =−+=−+= −−++

ecyec mmmm

2.15. The modified bending moment diagram

Because of the partial fixing of the outer support, the 25 % of the midfield moment should be taken into calculation on this support. According to the EUROCODE:

mkNm98,693,2725,0

25,0 42,

−=⋅−=

⋅−= +−xecleft mm

mkNm63,853,3425,0

25,0 11,

−=⋅−=

⋅−= +−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

14202102

: =−−=∅

−−=++axx chdmfor

mm1782

14252102

: =−−=∅

−−=−−fxx chdmfor

mm16914183: =−=∅−= +++xyy ddmfor

mm16414178: =−=∅−= −−−xyy ddmfor

section

mx(y) [kNm/m]

dx(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

⎟⎠

⎞⎜⎝

⎛ −⋅⋅⋅=±

2c

cdcx

dfxbm

mmb 1000=

Then

ydcalscdc fafx ⋅=⋅ ,1000

yd

cdccals f

fxa

⋅=

1000,


18

3.2. The moment of resistance

km as [mm2/m]

xc [mm]

mRd [kNm/m]

mRd > mSd mRdmSd

>1

0’x 1x 2x 4x

0’’x 1y 3y 4y 5y

cdcyds fxbfa ⋅α⋅⋅=⋅ 0ccd

ydsc x

fbfa

x <⋅α⋅

⋅=

⎟⎠⎞

⎜⎝⎛ −⋅⋅α⋅⋅=

2c

cdcRdxdfxbm

3.3. The minimal reinforcement required

⎪⎩

⎪⎨⎧

⋅⋅

⋅⋅=

dbf

dba yks

0015,0

6,0maxmin, (fyk in [N/mm2])

cs Aa ⋅= 04,0max, 3.4. The anchorage length

The basic value: bd

ydb f

fl ⋅

∅=

4 (where fbd is taken from Table below)

fck 12 16 20 25 30 35 40 45 50

Smooth surface

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

∅ ≤ 32 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

AA

ll ≥⋅⋅α=

αs = 1 in case of smooth surface steel As,rec = area of steel required As,prov = area of steel provided The 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: 0,7 lb,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 törőteher számítása 6.1. Energia módszer A kinematikailag lehetséges törőterhet a külső és belső munkák egyenlősége 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 fellépő nyomatékot κ-val való szorzással, a támasznyomatékokat μ1 ÷ μ4 szorzók segítségével számíthatjuk az m nyomatékból. Geometriai paraméterként a törésvonalak metszéspontját meghatározó α1, α2 és β tényezők vehetők fel. μ2 = μ4 (a szimmetriából adódóan) α1 = 1 - α2

lx = 5,6 m ly = 7,9 m 710,llγ

y

x ≅=


21

mkNm8336, m = (lásd 3.2.)

mkNm31,93 =⋅κ m 87,0=κ

mkNm66181 , m =⋅μ

51,01 =μ

mkNm614742 ,mμm μ =⋅=⋅

30,142 =μ=μ

mkNm46553 , m μ =⋅

51,13 =μ

A p teher által az elmozduláson végzett külső munka a töréskép által meghatározott térfogat alapján számítható.

( )⎭⎬⎫

⋅β⋅−⋅⋅

+⋅⋅⋅β⋅⋅⋅α

+⋅⋅⋅β⋅⋅⋅α

⎩⎨⎧

+⋅⋅⋅β⋅⋅

⋅= 12

212

311

22

311

22

311

221 yxyxyxyx

kk

llllllllpL

( )β⋅−⋅⋅⋅

= 236

yxkk

llpL

A belső munka a nyomatéknak a törésvonalak menti elforduláson végzett munkájával egyenlő.

21

112111

22

13

21

21

⋅⋅β

⋅⋅⋅μ+

+⋅α

⋅⋅⋅μ+⋅α

⋅⋅⋅μ+⋅⋅β

⋅⋅⋅κ+⎟⎟⎠

⎞⎜⎜⎝

⎛⋅α

+⋅α

⋅⋅=

yx

xy

xy

yx

xxyb

llm

llm

llm

llm

lllmL

( )γ⋅⋅μ⋅+γ⋅⋅κ⋅⋅β

+⎟⎟⎠

⎞⎜⎜⎝

⎛γ⋅μ

+γ

⋅α

+⎟⎟⎠

⎞⎜⎜⎝

⎛γ⋅μ

+γ

⋅α

= mmmmmmLb 21

2

3

1

22111

49,113133,78120,1301

21

⋅β

+⋅α

+⋅α

=bL

A külső és belső munkák egyenlősége alapján (Lk = Lb):

( )β

+α

+α

=β⋅−⋅⋅49,11333,7820,1302337,7

21kp

12 1 αα −= A törőteher tehát két paraméter függvény, amikből a szélsőérték parciális deriválással kapható.

( ),βαfp 1=


22

( )( ) ( ) 37,7·3·2··1·

·49,113··20,13033,78·20,130·49,113 2

−ββ−ααα−βα−+β+α

=kp 211 ,

0

0α⇒βα⇒

⎪⎪⎭

⎪⎪⎬

⎫

=β∂

∂

=α∂∂

p

p

A deriválást elvégezve: 563,01 =α , 436,02 =α , 424,0=β A kapott értékeket behelyettesítve a fenti egyenletbe megkapható pk értéke:

2mkN74,42=kp

6.2. Egyensúlyi módszer


23

Ez is törési határállapot-vizsgálat. A kinematikai tételen alapszik. (A feladatban talán célravezetőbb!) A külső terhek nyomatékának és a belső nyomatékok egyensúlyának felírásából számítható a határerő.

és lemezdarab azonos

( )32

2

22

2 ⋅⋅β⋅⋅

=⋅μ+⋅κ⋅ yxx

llpmml

lemezdarabra:

( ) ( )⎪⎩

⎪⎨⎧

⎪⎭

⎪⎬⎫⋅α

⋅⋅β⋅⋅α

⋅+⋅α⋅⋅β⋅−

⋅=μ⋅+⋅32

22

21 2222

21

xyxxyy

lllllpmml

lemezdarabra:

( ) ( )⎪⎩

⎪⎨⎧

⎪⎭

⎪⎬⎫⋅α

⋅⋅β⋅⋅α

⋅+⋅α⋅⋅β⋅−

⋅=μ⋅+⋅32

22

21 1122

13

xyxxyy

lllllpmml

ismeretlenek: p, 1α , β adott: 3 egyenlet

56,01 =α , 44,02 =α , 42,0=β , 2mkN74,42=p

22 mkN45,17

mkN74,42 =>= dpp ⎟⎟

⎠

⎞⎜⎜⎝

⎛>== 145.2

45,1774,42

dpp Megfelel!

Design of two way spanning slab

7. APPENDIX 7.1. Baręs’s tables for moments of two way spanning slab

Tab. 1.7 μ=0,15 γ ws Mxs Mys

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 0,0162 0,70 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 μ=0,15 γ ws Mxs Mys Myvs

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 -0,0479 0,70 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

ay

Mxb Mys

x q

q

x=0 x=a

y=0

y=b

γ = ab γ =

ab

Mxbs = μ·Myvs

a y

q Mxb Mys

x q y=

0 y=

b

x=0 x=a

Mxb

Myv


25

Tab. 1.9 μ=0,15 γ ws Mxs Mys Myvs

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 μ=0,15 γ ws Mxs Mxvmin Mys Myvmin

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

γ = ab

Mx0s = Mxbs

Mxbs = μ·Myvs

a y

q Mxb Mys

x q y=

0 y=

b

x=0 x=a

Mxb

Myv

Mx0 Myv

γ = ab

Mxbmin = μ·Myvmin

My0min = μ·Mxvmin

a y

q Mxb

Mys

x q y=

0 y=

b

x=0 x=a

Mxbmi

Myv

mi

Mxvmin

My0

mi


26

Tab. 1.11 μ=0,15 γ ws Mxs Mxvs Mys Myvmin

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


Tab. 1.12 μ=0,15 γ ws Mxs Mxvs Mys Myvmin

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


γ = ab

Mxbs = Mx0s Myas = My0s Mx0s = μ·Myvs My0s = μ·Mxvs

a y

q Mxb

Mys

x

q y=0

y=b

x=0 x=a

Mxbmi

Myv

mi

Mx0m Myv

mi

Mxvs

Mya

s

γ = ab

Mx0min = Mxbmin

Mx0min = μ·Myvmin

Myas = μ·Mxvs

q

a y

q Mxb

Mys

x

y=0

y=b

x=0 x=a

Mxbs Myv

s

Mx0s Myv

s

Mxvs

Mya

s

Mxvs

My0

s

Documents

Design of Two Way Spanning Floor-slab