UNIFORM AND SYMETRIC LOADING - ppa-europe.eu · 11 Conclusions: Final complete load tables and...

Guideline of Examples

WP2 Guidance - Design by testing of sandwich panels -

Deliverable 2.4 Example n°28

Determination of the load table below a negative loading, PU roof panel, thickness 60 mm on 2 supports with thermal gradient

Method 1 : used results without thermal gradient x k1test

UNIFORM AND SYMETRIC LOADING

Version 1

September 2010

2

Project co-funded under the European Commission Seventh Research and Technology

Development Framework Programme (2007-2013) Theme 4 NMP-Nanotechnologies, Materials and new Production Technologies

Prepared by : D IZABEL Drafting History Draft version 1 (creation of the document) September 2010 Dissemination Level

PU Public x PP Restricted to the other programme participants (including

the Commission Services)

RE Restricted to a group specified by the Consortium (including the Commission Services)

CO Confidential, only for members of the Consortium (including the Commission Services)

Warning : This document is an informative document. The user alone is responsible of the use partially or totally of this document.

3

Summary 1 Definition of the roof panel p4 2 Definition of the tests p6 2.1 Mechanical test alone (without thermal gradient) p7 2.2 Thermal test + mechanical test (with thermal gradient) p7 3 Test results p8

3.1 For the determination of the rigidities and and the design bending SkB 1FB kcc AGmoments strength of fasteners p9 TRdTM Δ,2 RdTF ,2

3.2 shear test results – shear resistance p9 RdTV ,2

3.3 Statistic factor k function of the number of tests p8 3.4 Resistance of the fasteners and assemblies p10 3.5 With thermal gradient p11

4 Exploitation of the test results p12 4.1 Determination of the rigidity without thermal gradient p13 SB 1FB cc AG4.1.1 By calculation p13 4.1.2 By testing p13 4.1.3 Mixed calculation + testing p14 4.2 Determination of the rigidities ; and with thermal gradient p20 TSB Δ 1FB Tcc AG Δ

5 Determination of the load capacities in ULS ; ; p21 TRdTUM Δ,2 TRdTUV Δ,2 TRdTUF Δ,2

6 Determination of the load capacities in SLS ; ; ; p27 TRdTEM Δ,2 TRdTEV Δ,2 TRdTEF Δ,2

7 Synthesis of the different capacities of the panel p35 7.1 Mechanical proprieties p35 7.2 Load strength capacity p35 8 Determination of the load tables below the ULS and SLS combinations p36 8.1 Construction of the load tables– pression (positive) load p37 9 Synthesis of the load tables for the spans tested p52 9.1 General tables of results p53 10 Interpolation and extrapolation of the load table below the ULS and SLS combinations p56 10.1 Expression of the bending and shear rigidities p57 10.2 Expression of the bending and shear capacities p57 11 Conclusions: Final complete load tables and comparison with the Polish and French agreement AT and EN 14509 p58 12 Annex: test results ; curves load deflexion –shear tests p60 13 Annex: Official performances of the PU roof panel technical assessment and French technical agreement p74 14 Annex: Analyse of the tkk Helsinki test results p78

4

1 Definition of the panel

5

1 Definition of the roof panel - Manufacturer : XYZ - Designation of the product : A - Usage : roof product - Nominal thickness of the panel : 60mm =e- Type of core : PU - Nominal value of the shear strength of the core =cvf 0.13 MPa - Characteristic value of the shear strength of the core =cvf 0.1366MPa - Characteristic value of the compression strength of the core =cvf 0.120166 MPa - Nominal value of the compression strength of the core =ccf 0.12 MPa - Nominal density of the core : between 34 and 42 daN/m3 - Facing steel : 320 MPa ribbed facing and =yf =yf 300 MPa flat facing - Facing highly profiled: nominal thickness: =1t 0.50 mm - Facing lightly profiled: nominal thickness: =2t 0.50 mm - Weight of the panel G: 11.46 daN / m² for the panel of 3 m and 11.6 daN / m² for the panel of 4 m - Tolerances: normal - Limit deflection L/200 (excepted if end of elastic behaviour reach before L/200) - Definition of the fasteners

Diameter 5.5 mm Tensile strength unknown Pull out strength (see 3.3) Length 175 mm Head of screw 0.63 mm Number of fasteners by line of support 4 2t

t 1

6

2 Definition of the tests

7

2 Definition of the tests 2.1 Mechanical test alone (without thermal gradient) - Testing laboratory: TKK Helsinki - Bending test report performed in Helsinki University of Technology (tkk) work package 2 (WP2) test series B test report draft 20.10.2009 - shear test report performed in Helsinki University of Technology (tkk) work package 2 (WP2) Test series E test report Draft 02.01.2010 - fastener test following ECCS recommendation Helsinki University of Technology (tkk) work package 2 (WP2) test report (email of the 13 May 2009 - Testing loading in depression: EN 14509 Fig A7 - Panel fixed on the two supports below a negative loading (see after) - Measurements: the deflexion at mid span, the load applied - Tests on panels fixed following - Spans tested: EN 14509 Fig A7 - 1 test by span

- 3m - 4m

2.2 Thermal test + mechanical test (with thermal gradient) - Test laboratory: TKK Helsinki - Test report tests performed in Helsinki University of Technology (tkk/Aalto) work package 2 (WP2) test series D test report Draft 2010 - Test loading in pression: EN 14509 Fig A7 - Panel fixed on the two supports in succion (see after) - Measurements: the deflexion at mid span, the load applied - Spans tested: EN 14509 Fig A7 - 1 test by span for the bending test

- 3m - 4m

8

3 Test Results

9

3. Test results 3.1 For the rigidity and and the design bending moments SB cc AG RdTM ,2

Spans tested L (m)

Load applied

daN

eld QQ /

Load applied cQ

daN

obsyf ,1 (Flat) (MPa)

1yf

obsyf ,2 (ribbed (MPa)

2yf

obst ,1 (flat) (mm)

1t

obst ,2 (ribbed)(mm)

2t

Type of failure

Fixed Thickness of the panel

; e obse(mm)

obscG , (MPa)

obscvf , (MPa)

3 550 550 750+65.576 815.576

1511.4+62.7761574.176

402 300

395 320

0.427 0.50

0.435 0.50

Buckling of the upper face

yes 61.3 60

3 0.1366

4 420 420 550+73.576 623.576

1231.9+73.5761305.476

402 300

395 320

0.427 0.50

0.435 0.50


yes 62.16 60

3 0.1366

Self weight of panel of 60 * 3 m = 34.4 daN Self weight of panel of 60 * 4 m = 45.2 daN Wood loading = 4 * 7.094 = 28.376 daN L/200 is considered as the limit deflection excepted if information mentioned in the table 3.2 Shear test results - shear resistance RdTV ,2

Thickness of the panel

Shear load applied Self weight of the panel

Global collapse shear load (Qc,v daN) Failure mode

60 1105 33 1138 shear of core 60 1188 33 1221 shear of core 60 1054 33 1087 shear of core 1148.666 Paavo hassinen test data (following test report 02/01/2010 Thickness of the panel

Global collapse shear load (Qc,v daN)

Failure mode Global SLS load (Qel,v daN)

60 964.8 shear of core 60 1036.6 shear of core 60 921.8 shear of core 974.4 We retain the minimum values. - total length of the specimen 3 m - L1 = 400 mm - L2 = 2400 mm - L3 = 170 mm - Ls = 60 mm = support width - L = 2800 mm

tested

sticcharacteri

meanSSk =3.3 Statistic factor k function of the number of test

Only one test was done we retain for the bending behaviour k=0.85 k = 0.83 for the shear test ULS and 0.8966 SLS (see annexe 12)

10

3.4 Resistance of the fasteners and assemblies Pull out test vakue daN UtestF380 daN

Pull out test vakue daN EtestF350 daN Linear behaviour 150 daN Deformation of the washer (waterproof efficiency?)

The case with 50 mm were retained for the test Fu = 380 daN At SLS the question depend of SLS criteria (linear behaviour/etancheity/deformation of the washer

11

3.5 With thermal gradient

Spans tested L (m)

Load applied

daN

eld QQ /

Load applied cQdaN

obsyf ,1 (ribbed) (MPa)

1yf

obsyf ,2 (flat (MPa)

2yf

obst ,1 (ribbed) (mm)

1t

obst ,2 (flat (mm)

2t

Type of failure

Fixed on support

Thickness of the panel

obsee (mm)

obscG , (MPa)

obscvf , (MPa)

cvf

3 585 585 690+68.56 758.56

1575.3 +68.56 1643.86

395 320

402 300

0.435 0.50

0.427 0.50


No 61.366 60

3 0.1366 0.13

4 420 420 700+80.56 780.56

1231.4 +80.56 1311.96

395 320

402 300

0.435 0.50

0.427 0.50


No 62.13 60

3 0.1366 0.13

Self weight of panel of 60 * 3m = 34.4 daN Self weight of panel of 60 * 4 m =46.4 daN Wood loading = 4 * 8.54 = 34.16 daN

dQ load corresponding at a deflection L/200

elQ elastic load

cQ collapse load

12

4 Exploitation of the test results

13

4 Exploitation of the test results 4.1 Determination of the rigidities ; and without thermal gradient SB 1FB cc AG 4.1.1 First approach by calculation We use the value obtained by calculation by Is mainz

ckc AG = 19690 daN

skB = 22439.3583 daN.m²

1FB = 1478.4 daN.m² 4.1.2 By testing Without thermal gradient 3 spans must be tested to find the 3 rigidities It must be known for this case:

1dQ and 1δ = L1 / 200

2dQ and 2δ = L2 / 200

3dQ and 3δ = L3 / 200 And we look for ; ; skB ckc AG 1FBThe deflection equation is:

( ) ( )ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−= 1

81

307241 3

ccs AGQL

BQL

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

K

BB

B

sF

F

411281

1

1β 2

3LAG

BKcc

s=

( ) ( )111

1

311

1 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

( ) ( )222

2

322

2 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

( ) ( )333

3

333

3 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

1

1

11

411281 K

BB

B

sF

Fβ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

2

1

12

411281 K

BB

B

sF

Fβ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

3

1

13

411281 K

BB

B

sF

Fβ

21

13

LAGB

Kcc

s= 22

23

LAGB

Kcc

s= 23

33

LAGB

Kcc

s=

As we have only two spans (3 and 4 m) tested we don’t use this approach.

14

4.1.3 Mixed calculation + testing 4.1.3.1 Approximate value We neglect the effect of the rib alone by comparison with the sandwich bending rigidity

1dQ = 550 daN and 1δ = L1 / 200 = 1.5 cm L1 = 3 m ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

3

3

2

2

8

1

iid

jjdij

i

jjjd

CC

LQLQ

ff

LL

LQ

AG

( )

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

=

iid

jjdij

ijjjd

s

LQLQ

ff

LLLQ

B3072

41 22

2dQ = 420 daN and 2δ = L2 / 200 = 2.0 cm L2 = 4 m

We obtain without corrections:

ckc AG 22838 daN =B 33201.6226 daN.m² =skB 33201.6226 daN.m² - 1478.4 = 31723.22 daN.m²

Correction of the bending rigidity :

2

,2

2

,1

1, 2

1⎟⎟⎠

⎞⎜⎜⎝

⎛××

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsobstBsadj e

ett

tt

R γ

2

, 062.006.0

21

435.05.0

427.05.0

⎟⎠⎞

⎜⎝⎛××⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=BsadjR

= 1.086548869

=B 33201.6226 daN.m² x 1.086548869 = 36075.185 daN.m² =skB (33201.6226 daN.m² - 1478.4) 1.086548869 = 34468.83164 daN.m²

Note; in case of old test results, it can be taken into account of a γt coefficient following the type of tolerance used.

γt =1.00 if the decaled tolerances are used and 0.93 if the normal tolerances are used.

Correction of the shear rigidity :

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsc

cAGadj e

eGG

Rcc

,,

⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛=

062.006.0

33

, ccAGadjR = 0.9677

ckc AG = 22838 daN x 0.9677 =22100 daN very near of the calculated value 19690 daN

15

4.1.3.2 More exact value a) We apply the exact formula of δ with the previous values

ckc AG 19690 daN =skB 22439.3583 daN.m² =1FB 1478.4 daN.m²

1dQ = 550 daN L1 = 3 m

2dQ = 420 daN L2 = 4 m

( ) ( )111

1

311

1 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

We obtain:

1δ =0.01687 m for 0.015m β1 =0.1258895778

2δ =0.024015127 m for 0.02 m β2 =0.098965753 c) We use the approximate values and and we use the exact formula of δ B ckc AG

ckc AG 22100 daN =B 36075.185 daN.m² =skB 34468.83164 daN.m² =1FB 1478.4 daN.m²

1dQ = 550 daN L1 = 3 m

2dQ = 420 daN L2 = 4 m

( ) ( )111

1

311

1 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

We obtain:

1δ =0.013557 m for 0.015m

2δ =0.1840 m for 0.02 m With the values without correction

ckc AG 22838 daN =B 33201.6226 daN.m² =skB 31723.22 daN.m² =1FB 1478.4 daN.m²

1dQ = 550 daN L1 = 3 m

2dQ = 420 daN L2 = 4 m

( ) ( )111

1

311

1 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

Conclusion So there is a good correlation between the design by testing values and the calculated values

16

b) Resolution of the equation system We fix one value by calculation and we find the two others by testing The bending sandwich rigidity is fixed : skB

( ) ( )111

1

311

1 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

( ) ( )222

2

322

2 18

1307241

ββδ −⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

cc

d

s

d

AGLQ

BLQ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

1

1

11

411281 K

BB

B

sF

Fβ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ +

+

=

2

1

12

411281 K

BB

B

sF

Fβ

21

13

LAGB

Kcc

s= 22

23

LAGB

Kcc

s=

Numerical application :

1dQ = 550 daN and 1δ = L1 / 200 = 1.5 cm L1 = 3 m

2dQ = 420 daN and 2δ = L2 / 200 = 2.0 cm L2 = 4 m For we propose to test 3 values: skB

skB = 22439.3583 daN.m² (obtained by calculation Is Mainz)

skB = 34468.83164 daN.m² (obtained with a simplify approach)

skB = 34123.6 daN.m² (obtained after several iteration with the aim to have = 1478.4 daN.m²) 1FB And we look for and G 1FB ckc A

17

Synthesis of the procedure the case of 2 tests without thermal gradient with the Paavo Hassinen loading The tests are the following: Test 1 4 x 4/diQ Li/8 Li/4 Li/4 Li/4 Li/8 miniL Test 2

Panel on 2 supports with load to have no deflection at mid span maxjL 4 x 4/djQ maxjL The data are: The geometric and mechanical data are:

sFFFFji BBAAEEEeLL ,,,,;;; 2121 == 3; 4; 0.06m E=210 000 MPa The tests results are:

jidjdi ffQQ ;;; 550 daN; 420 daN; 0.015m; 0.020 m We look for by testing:

1FB ; cc AG Step1 : Calculus of the bending rigidity : sB

( )BAEAEeAEAEB

FFFF

FFFFs

2211

22211

+=

Step 2 : Calculus of the different constant :

[ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

dii

ii QL

fA 130721 [ ] ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

djj

jj QL

fA 130721

18

[ ] ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

dii

ii QL

fA 141

117964832 [ ]

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

djj

jj QL

fA 1

411179648

32

[ ] [ ]1313072 idii

ii A

QLfA =⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛= [ ] [ ]13

13072 jdjj

jj A

QLf

A =⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

[ ] 3844 =iA [ ] 3844 =jA [ ] 2

5 41 ii LA = [ ] 25 41 jj LA =

{ [ ][ ] [ ][ ] }1221

2ijijs AAAABA −=

{ [ ][ ] [ ][ ] [ ][ ] [ ][ ] }++−+−= 14255241 ijijijijs AAAAAAAABB { [ ][ ] [ ][ ] }2332

2ijijs AAAAB +−

{ [ ][ ] [ ][ ] }++−= 3443 ijijs AAAABC { [ ][ ] [ ][ ] }5445 ijij AAAA −

Step 3: To solve the equation:

ACB 42 −=Δ

ABX

21Δ−−

= A

BX22

Δ+−=

Step 4 : Calculus of with 1FB

s

F

BBX 1= :

sF XBB =1 Step 5: Calculus of Y :

[ ] [ ] [ ][ ] [ ] ⎟⎟

⎠

⎞⎜⎜⎝

⎛−−−

=42

2135

isi

sisii

ABAXBXAXBAXAY

Step 6: Calculus of with cc AG

cc

F

AGBY 1=

YBAG F

cc1=

Numerical application:

19

With Is mainz value skB = 22439.3583 daN.m²

We obtain : Error in the resolution With

=skB 34468.83164 daN.m² We obtain X =5.151682347E-2 Y =9.968989671E-2 = 1775.724 daN.m² =17812.484 daN 1FB cc AG With

skB = 34123.6 daN.m² X =4.332573209E-2 Y =7.943562499E-2 = 1478.429952 daN.m² =18611.67394 daN 1FB cc AG skB (daN.m²) 1FB (daN.m²) cc AG (daN) Design by calculation 22439.3583 1478.4 19690 Design by testing 34123.60 1478.429952 18611.67394 Deviation of the design by calculation in %

52 0 -5.4765

Comment The augmentation of rigidity seen with the design by testing is probably due to the fastener effect – small bending moment at each end of the panel we can also see this augmentation of the rigidity in the load applied Qi (daN) Qj (daN) Not fixed in pression 500 375 Fixed in succion (4 fasteners by line of support)

550 420

Conclusion Design by testing shows a benefit in the global bending rigidity of the system created by the fastener (around 50%) This increase of the bending rigidity is present in the reality and could by taken into account in the calculation of the deflection.

20

4.2 Determination of the rigidities ; and with thermal gradient TSB Δ 1FB Tcc AG Δ

We use the same value that without thermal gradient (same test valuses)

21

5 Determination of the load capacities in ULS

22

5 Determination of the load capacities in ULS(without thermal gradient) 5.1 Bending capacity 5.1.1 elastic TUM 2

⎟⎟⎠

⎞⎜⎜⎝

⎛××

×=m

cTMUadTU

LQRMγ82,2

This formula corresponds at the maximum bending moment in span. With for a bending test:

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×=

obsobsy

y

obstTMUad e

eff

ttR

,,1

1

,1

12,

αβ

γ

- if no wrinkling in span

0=α if yobsy ff ≤,

1=α if yobsy ff >,

- If wrinkling of the face in compression:

5.0=α if and yobsy ff >, 601.33300

21000027.127.1 =×=>⎟⎠⎞

⎜⎝⎛

y

F

fE

tb so, it is necessary that

that is satisfied with this profile. mm800.601b .33> 1650.0 =× So, 5.0=α because = 402 > =300 MPa and second condition on b/t satisfied obsyf , yf - If no wrinkling in span

1=β - If wrinkling in span In compression:

1=β If ttobs <

2=β If and ttobs > 601.33300

21000027.127.1 =×=>⎟⎠⎞

⎜⎝⎛

y

F

fE


mmb 800.1650.0601.33 =×> That is satisfied with this profile. With for tγ

tγ = 1 if decaled tolerance

tγ = 0.93 if the normal tolerance are used

23

Numerical application Spans considered 3 m 4 m Type of collapse Wrinkling in span Wrinkling in span

tγ 1 1

obst ,1 (mm) 0.427 0.427

1t (mm) 0.5 0.5

obsyf ,1 (MPa) 402 402

1yf (MPa) 300 300 α 0.5 0.5 β 1 1

obse, (m) 0.0613 0.06216 e (m) 0.06 0.06

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×=

obsmobsy

y

obstTMUad e

eff

ttR

,,1

1

,1

12,

αβ

γ 0.9901032954 0.9764049551

Spans tested L (m)

Load applied

(daN) cQ

mγ TMUadR 2, TUM 2 daN.m

Type of collapse

3 1574.176 1.10 0.9901032954 531.3398 Wrinkling in span 4 1305.476 1.10 0.9764049551 579.3969 Wrinkling in span

5.1.2 elastic RdTUM ,2

Spans tested L (m)

k (*) TUM 2 daN.m

RdTUM ,2 daN.m

Type of collapse

3 0.85 531.3398 451.638 Wrinkling in span 4 0.85 579.3969 492.487 Wrinkling in span k (*) forfeiture value when only one test done (value based on experience) We retain by security and in first approach the min value:

mdaNM RdTU .638.451,2 = Or if we done an optimize calculation we will retain:

)3(,2 mofspanRdTUM = 451.638 daN.m

)4(,2 mofspanRdTUM = 492.487 daN.m

24

5.2 Shear strength capacity 5.2.1 Calculation of TUV 2

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=m

cTVUadTU

QRVγ22,2 if the bending test on 2 supports is used

⎟⎠⎞

⎜⎝⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛×=

LLLQ

RVm

vcTVUadTU

1,2,2 γ

if the shear load test given in A.15.2 of EN 14509 is used

This formula corresponds at the maximum shear force on the end support. We will use the shear load tests (formula 2) L1 = max 1.5 * e = 1.5x 60 = 90 mm and 400 mm With for a bending test:

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobscv

cvTVUad e

effR

,,2,

We can also use directly the results that are in the tkk test report (serie E) Spans considered 3 m Type of collapse Shear failure on support

cvf (MPa) 0.13 (hypothesis)

obscvf , (mm) 0.1366

obse, (m) (0.0617+0.0619+0.0617)/3 =0.0617666

e (m) 0.06

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobscv

cvTVUad e

effR

,,2,

0.9244644337

Spans tested L (m)

Load applied (daN) )3(, testsofmeanQ vc

mγ TVUadR 2, TUV 2 daN.m

Type of collapse

3 974.4 1.25 0.9244644337 720.638 Shear failure on support

5.2.2 elastic RdTUV ,2

Spans tested L (m)

k (*) TUV 2 daN.m

RdTUV ,2 daN.m

Type of collapse

3 0.829 720.638 597.409 Shear failure on support k (*) see statistic study in annex 12.2 We retain by security and in first approach the min value:

daNV RdTU 409.597,2 =

25

5.3 Strength of the assembly below a tensile action on the support [ ]RdfastenerUeRdanchoragUughtRdpullthroURdU FFFF .;min= in ULS

During the tensile test, it was seen a pulltrought failure:

)(,

1

ULSmvadjUtestURd RFknF

γ××××=

This formula corresponds at the maximum strength capacity of the fastener on the end support.

[ ]321, ;;min fffR vad =

1,1

1

,1 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsm

u

tt

Rf

f 12/1

,2 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCc

Cck

ff

f 12/1

,3 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCv

Cvk

ff

f

As the test were done, we have directly the value for one fixation Pull out test value daN UtestF380 daN

Type of collapse Fastener failure + deformation of the rib and washer

uf (MPa) Unknown,

obsmR , (MPa) Unknown

obst ,1 (mm) 0.48 (data AMC Polska)

1t (mm) 0.5

1,1

1

,1 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsm

u

tt

Rf

f 1

Cckf (MPa) 0.12

obsCcf , (MPa) 0.120168

12/1

,2 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCc

Cck

ff

f 0.999

Cvkf (MPa) 0.13

obsCf , (MPa) 0.1366

12/1

,3 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCv

Cvk

ff

f 0.9755427967

[ 321, ;;min fffR vad = ] 0.9755427967

Pull out test

value daN UtestFk )(ULSMγ vadjR , Number of

fasteners n /support Assembly

strength daN URdF380 0.85 1.70 0.9755427967 4 741.412

Note : 0.85/1.70x0.9755 = 0.48775=>2.05 to be in link with safety factor of the polish agreement.

26

5.4 Corrected coefficient due to the thermal gradient k1uls This coefficient is obtained directy by the ratio of the load with and without thermal gradient Span (m) Qc without thermal

gradient (mechanical loading only) (daN)

Qcth with thermal gradient and mechanical loading (daN)

k1uls = Qcth / Qc

3 1574.176 1643.86 1.044266969 4 1305.476 1311.96 1.004966771 5.5 Final values of the strength capacities at ULS Span (m) ULSk ,1 RdTUM ,2

(daN.m) TRdTUM Δ,2

(daN.m) RdTUV ,2

(daN) TRdTUV Δ,2

(daN) 3 1.044266969 451.638 471.6306453 597.409 623.854 4 1.004966771 492.487 494.9330701 597.409 600.376

27

6 Determination of the load capacity in SLS

28

6 Determination of the load capacities in SLS 6.1 Bending capacity 6.1.1 elastic TEM 2

⎟⎟⎠

⎞⎜⎜⎝

⎛××

×=m

elTMEadTE

LQRMγ82,2

This formula corresponds at the maximum bending moment in span. With for a bending test:

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×=

obsobsy

y

obstTMEad e

eff

ttR

,,1

1

,1

12,

αβ

γ

- if no wrinkling in span

0=α if yobsy ff ≤,

1=α if yobsy ff >,

- If wrinkling of the face in compression:

5.0=α if and yobsy ff >, 601.33300

21000027.127.1 =×=>⎟⎠⎞

⎜⎝⎛

y

F

fE


that is satisfied with this profile. mm800.601b .33> 1650.0 =× So, 5.0=α because = 402 > =300 MPa and second condition on b/t satisfied obsyf , yf - If no wrinkling in span

1=β - If wrinkling in span In compression:

1=β If ttobs <

2=β If and ttobs > 601.33300

21000027.127.1 =×=>⎟⎠⎞

⎜⎝⎛

y

F

fE


mmb 800.1650.0601.33 =×> That is satisfied with this profile. With for tγ

tγ = 1 if decaled tolerance

tγ = 0.93 if the normal tolerance are used

29

Numerical application Spans considered 3 m 4 m Type of collapse Wrinkling in span Wrinkling in span

tγ 1 1

obst ,1 (mm) 0.427 0.427

1t (mm) 0.5 0.5

obsyf ,1 (MPa) 402 402

1yf (MPa) 300 300 α 0.5 0.5 β 1 1

obse, (m) 0.0613 0.06216 e (m) 0.06 0.06

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×⎟

⎟⎠

⎞⎜⎜⎝

⎛×=

obsmobsy

y

obstTMEad e

eff

ttR

,,1

1

,1

12,

αβ

γ 0.9901032954 0.9764049551

Spans tested L (m)

Load applied

(daN) elQ

mγ TMEadR 2, TE

daN.m M 2 Type of collapse

3 815.576 1.0 0.9901032954 302.814 Wrinkling in span 4 623.576 1.0 0.9764049551 304.431 Wrinkling in span

6.1.2 elastic RdTEM ,2

Spans tested L (m)

k (*) TE

daN.m M 2 RdTEM ,2

daN.m Type of collapse

3 0.85 302.814 257.391 Wrinkling in span 4 0.85 304.431 258.766 Wrinkling in span k (*) forfeiture value when only one test done (value based on experience) We retain by security and in first approach the min value:

mdaNM RdTE .391.257,2 = Or if we done an optimize calculation we will retain:

)3(,2 mofspanRdTEM = 257.391 daN.m

)4(,2 mofspanRdTEM = 258.766 daN.m

30

6.2 Shear strength capacity 6.2.1 Calculation of TEV 2

⎟⎟⎠

⎞⎜⎜⎝

⎛×

×=m

elTVEadTE

QRV

γ22,2 if the bending test on 2 supports is used

This formula corresponds at the maximum shear force on the end support

⎟⎠⎞

⎜⎝⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛×=

LLLQ

RVm

velTVEadTE

1,2,2 γ

if the shear load test given in A.15.2 of EN 14509 is used

This formula corresponds at the maximum shear force on the end support. We will use the shear load tests (formula 2) L1 = max 1.5 * e = 1.5x 60 = 90 mm and 400 mm With for a bending test: With for a bending test:

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobscv

cvTVEad e

effR

,,2,

Spans considered 3 m Type of collapse Shear failure on support

cvf (MPa) 0.13 (hypothesis)

obscvf , (mm) 0.1366

obse, (m) (0.0617+0.0619+0.0617)/3 =0.0617666

e (m) 0.06

⎟⎟⎠

⎞⎜⎜⎝

⎛×⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobscv

cvTVUad e

effR

,,2,

0.9244644337

Spans tested L (m)

Load applied

(daN) )3(, testsofmeanQ vel

mγ TVEadR 2, TEV 2 daN.m (formula 2)

Type of collapse

3 870 1.1 0.9244644337 676.715 shear 6.2.2 elastic RdTEV ,2

Spans tested L (m)

k (*) TEV 2 daN.m

RdTEV ,2 daN.m

Type of collapse

3 0.89667 676.715 561.912 shear k (*) see statistic study in annex 12.2 We retain by security and in first approach the min value:

daNV RdTE 912.561,2 =

31

6.3 Strength of the assembly below a tensile action on the support [ ]RdfastenerEeRdanchoragEghtRpullthrouERdE FFFF .;min= in SLS

During the tensile test, it was seen a pulltrought failure:

)(,

1

SLSmvadjEtestERd RFknF

γ××××=

This formula corresponds at the maximum strength capacity of the fastener on the end support.

[ ]321, ;;min fffR vad =

1,1

1

,1 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsm

u

tt

Rf

f 12/1

,2 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCc

Cck

ff

f 12/1

,3 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCv

Cvk

ff

f

As the test were done, we have directly the value Pull out test vakue daN EtestF350 daN Linear behaviour 150 daN Deformation of the washer (waterproof efficiency?)

Type of collapse Fastener falure + deformation of the rib and washer

uf (MPa) Unknow,

obsmR , (MPa) Unknown

obst ,1 (mm) 0.48 (data AMC Polska)

1t (mm) 0.5

1,1

1

,1 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

obsobsm

u

tt

Rf

f 1

Cckf (MPa) 0.12

obsCcf , (MPa) 0.120168

12/1

,2 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCc

Cck

ff

f 0.999

Cvkf (MPa) 0.13

obsCf , (MPa) 0.1366

12/1

,3 ≤⎟

⎟⎠

⎞⎜⎜⎝

⎛=

obsCv

Cvk

ff

f 0.9755427967

[ 321, ;;min fffR vad = ] 0.9755427967

32

Hypothesis 1 (linear behaviour value, (waterproof assembly?)

Pull out test value daN EtestF

k )(SLSMγ vadjR , Number of fasteners n

Pull out strength daN

ERdF

350 0.85 1.70 0.97554 4 682.878 Hypothesis 2 (limit of deformation of the washer)

Pull out test value daN EtestF

k )(SLSMγ vadjR , Number of fasteners n

Pull out strength daN

ERdF

150 0.85 1.10 0.97554 4 452.295

33

6.4 Corrected coefficient due to the thermal gradient k1sls This coefficient is obtained directy by the ratio of the load with and without thermal gradient Span (m) Qel without thermal

gradient (mechanical loading only) (daN)

Qelth with thermal gradient and mechanical loading (daN)

k1sls = Qelth / Qel

3 815.576 758.56 0.9300911 4 623.576 780.56 1.251747983 6.5 Final values of the strength capacities at SLS Span (m) SLSk ,1 RdTEM ,2

(daN.m) TRdTEM Δ,2

(daN.m) RdTEV ,2

(daN) TRdTEV Δ,2

(daN) 3 0.9300911 257.391 239.397 561.912 522.629 4 1.251747983 258.766 323.909 561.912 703.372

34

7 Synthesis of the different capacities of the panel

35

7 Synthesis of the different capacities of the panel 7.1 Mechanic properties

Span (m) Span (m) Mechanic properties All (without thermal; gradient) All (with thermal;

gradient) )(calculatedskB (daN.m²) 22439.3583

)(1 calculatedFB (daN.m²) 1478.4

)(calculatedckc AG (daN) 19690

)( testingbyfixedskB (daN.m²) 34123.60

)(1 testingbyFB (daN.m²) 1478.429952

)( testingbyckc AG (daN) 18611.67394 7.2 Load strength capacity a) Characteristic value with k = 0.85

Span (m) Strength characteristic value 3 4

RdTUM ,2 (daN.m) (design by testing) 451.638 492.487

TRdTUM Δ,2 (daN.m) (design by testing) 471.630 494.933

RdTUV ,2 (daN) (design by testing) 597.409 597.409

TRdTUV Δ,2 (daN) design by testing) 623.854 600.376

URdF (daN) (design by testing) 741.412 741.412

URdF (daN) (design by testing) 741.412 741.412

RdTEM ,2 (daN.m) (design by testing) 257.391 258.766

TRdTEM Δ,2 (daN.m) (design by testing) 239.397 323.909

RdTEV ,2 (daN) (design by testing) 561.912 561.912

TRdTEV Δ,2 (daN) (design by testing) 522.629 703.372

ERdF (daN) (linear) (design by testing) 682.878 682.878

ERdF (daN) (linear) (design by testing) 682.878 682.878

ERdF (daN) (washer) (design by testing) 452.295 452.295

ERdF (daN) (washer) (design by testing) 452.295 452.295

36

8 Determination of the load tables below the following combinations:

G - 1.50 Q - 1.5 x 0.6 ΔT G - 1.50 ΔT - 1.5 x 0.6 Q

At SLS

1.00 G - 1.00 Q - 0.6 ΔT 1.00 G - ΔT - 0.6Q

G - 1.50 Q at ULS 1.00 G - 1.00 Q at SLS

Roof sandwich panel on 2 supports

37

8.1 Construction of the load tables case of Elastic approach –succion load

38

8.1.1 ULS Criteria 8.1.1.1 Criteria 1 : ULS - no bending collapse at middle span We must verify:

TRdTUTSdTU MM ΔΔ ≤ ,2,2 for the global value of the bending moment

Span (m) iLStrength characteristic value 3 4

TRdTUM Δ,2 (daN.m) 471.630 494.933 Combination 1: TQG MMM Δ×−− int6.05.15.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 850.1

8

211

2

,2iii

TSdTULqLG

M

We must verify that:

2

2

,2

11 5.1

88

i

iTRdTU

i L

LGM

q⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ

Numerical application with

G =11.46 daN/m² for the panel of 3m G =11.6 daN/m² for the panel of 4m

Application 1 with the value affected at each span

Span in m iL Value of (daN/m²) and ΔTmax= 40°C

11iq

3.00 287.124 4.00 172.711

Application 2 with the min Value of the strength bending moment daN.m

Span in m iL Value of (daN/m²) 11iqand ΔTmax = 40°C

3.00 4.00

39

Combination 2: TQG MMM Δ−×− int5.16.05.1

⎭⎬⎫

⎩⎨⎧ ×

−⎭⎬⎫

⎩⎨⎧

=− Δ 86.050.1

8

212

2

,2iii

TSdTULqLG

M


2

2

,2

12 6.05.1

88

i

iTRdTU

i L

LGM

q×

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ




Span in m iL Value of (daN/m²) and ΔTmax= 40°C

12iq

3.00 478.541 4.00 287.851



3.00 4.00

40

8.1.1.2 Criteria 2 :ULS - no shear collapse on the end support We must verify:

TRdTUTSdTU VV ΔΔ ≤ ,2,2 for the global force apply on the assembly.


TRdTUV Δ,2 (daN) 623.854 600.376 Combination 1: TQG VVV Δ×−− int6.05.15.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 250.1

221

,2iii

TSdTULqLG

V


i

iTRdTU

i L

LGV

q5.1

22

,2

21⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 284.908 4.00 207.858

Application 2 with the min Value of the strength shear load daN


3.00 4.00

41

Combination 2: TQG VVV Δ−×− int5.16.05.1

⎭⎬⎫

⎩⎨⎧ ×

−⎭⎬⎫

⎩⎨⎧

=− Δ 26.050.1

222

,2iii

TSdTULqLG

V


i

iTRdTU

i L

LGV

q6.05.1

22

,2

22 ×⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 474.846 4.00 346.431



3.00 4.00

42

8.1.1.3 Criteria 3 :ULS - no assembly failure on the end support We must verify:

TURdTSdTU FV ΔΔ ≤,2 for the global value of the shear force


TURdF Δ (daN) 741.412 741.412 Combination 1: TQG VVV Δ×−− int6.05.15.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 250.1

231

,2iii

TSdTULqLG

V


i

iTURd

i L

LGF

q5.1

2231

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 337.456 4.00 254.870



3.00 4.00

43

Combination 2: TQG VVV Δ−×− int5.16.05.1

⎭⎬⎫

⎩⎨⎧ ×

−⎭⎬⎫

⎩⎨⎧

=− Δ 26.050.1

232

,2iii

TSdTULqLG

V


i

iTURd

i L

LGF

q6.05.1

2232 ×

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 562.426 4.00 424.784



3.00 4.00

44

8.1.2 SLS Criteria 8.1.2.1 Criteria 4 : SLS - no wrinkling at middle span We must verify:

TRdTETSdTE MM ΔΔ ≤ ,2,2 for the global value of the bending moment


TRdTEM Δ,2 (daN.m) 239.397 323.909 Combination 1: TQG MMM Δ×−− int6.010.100.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 800.1

800.1 2

412

,2iii

TSdTELqLG

M


2

2

,2

41 0.1

800.1

8i

iTRdTE

i L

LGM

q ⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 224.257 4.00 173.554



3.00 4.00

45

Combination 2: TQG MMM Δ−×− int6.01

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 86.0

800.1 2

422

,2iii

TSdTELqLG

M


2

2

,2

42 6.0

800.1

8i

iTRdTE

i L

LGM

q ⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 289.258 4.00 373.761



3.00 4.00

46

8.1.2.2 Criteria 5 :SLS - no shear collapse on the end support We must verify:

TRdTETSdTE VV ΔΔ ≤ ,2,2 for the global value of the shear force


TRdTEV Δ,2 (daN) 522.629 703.372 Combination 1: TQG VVV Δ×−− int6.010.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 200.1

200.1 51

,2iii

TSdTELqLG

V


i

iTRdTE

i L

LGV

q0.1

200.1

2,2

51⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 359.879 4.00 363.286



3.00 4.00

47

Combination 2: TQG VVV Δ−− int6.000.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 26.0

200.1 52

,2iii

TSdTELqLG

V


i

iTRdTE

i L

LGV

q6.0

200.1

2,2

52⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 599.798 4.00 605.476



3.00 4.00

48

8.1.2.3 Criteria 6 :SLS - no assembly failure on the end support

TERdTSdTE FV ΔΔ ≤,2 for the global force apply on the assembly.


TERdF Δ (daN) 682.878/452.295(*) 682.878/452.295(*) (*) washer criteria Combination 1: TQG VVV Δ×−− int6.010.100.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 200.1

200.1 61

,2iii

TSdTELqLG

V


i

iTERd

i L

LGF

q0.1

200.1

261⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 466.712/312.99 4.00 353.039/237.7475



3.00 4.00

49

Combination 2: TQG VVV Δ−− int6.000.1

⎭⎬⎫

⎩⎨⎧

−⎭⎬⎫

⎩⎨⎧

=− Δ 26.0

200.1 62

,2iii

TSdTELqLG

V


i

iTERd

i L

LGF

q6.0

200.1

262⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎭⎬⎫

⎩⎨⎧

+

=Δ





3.00 777.853/521.65 4.00 588.398/396.245



3.00 4.00

50

8.1.2.4 Criteria 7 :ELS – deflexion criteria a) Whitout creep

We must verify:200lim,,

iESdE

Lww =≤

( ) ( ) ( )2

2

10

27

47

111

24

1, 188384

51

83845

1 βα

ββψβ −Δ

ΨΨ−⎥⎦

⎤⎢⎣

⎡+−−⎥

⎦

⎤⎢⎣

⎡+−=− Δ

ΔΔΔΔ eLT

AGLq

BLq

AGLG

BLG

w isT

Tckc

ii

Tsk

ii

Tckc

i

Tsk

iSdE

With : TΔβltheoritica

measured

ltheoriticaT

measuredT

ww

RR

==Δ

Δ

,

,

k

sF

F

KB

B

B

2.311

11

++

=β

k

TsTF

TF

KB

B

B

67.211

12

++

=Δ

Δ

Δβ

( ) ( )

( ) ⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−−

ΔΨΨ+⎟

⎠⎞

⎜⎝⎛−

−=

ΔΔ

ΔΔΔ

1

24

11

1

24

2

2

10

7

18384

5

18384

51

8200

βψ

ββα

β

Tckc

i

Tsk

i

Tckc

i

Tsk

iisT

i

si

AGL

BL

AGLG

BLG

eLTL

q

( ) ( ) ( )2

2

10

27

47

111

24

1, 188384

51

83845

1 βα

βψψβψβ −Δ

+⎥⎦

⎤⎢⎣

⎡+−−⎥

⎦

⎤⎢⎣

⎡+−= Δ

ΔΔΔΔ eLT

AGLq

BLq

AGLG

BLG

w iwT

Tckc

ii

Tsk

ii

Tckc

i

Tsk

iSdE

( ) ( )

( ) ⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛+−−

Δ−⎟

⎠⎞

⎜⎝⎛

−=

ΔΔ

ΔΔΔ

1

24

11

1

24

2

2

10

5

18384

5

18384

51

8200

βψ

ββα

βψψ

Tckc

i

Tsk

i

Tckc

i

Tsk

iiwT

i

wi

AGL

BL

AGLG

BLG

eLTL

q


G = 11.46 daN/m² for the panel of 3m =0ψ 0.6 =1ψ 1

G = 11.6 daN/m² for the panel of 4m =0ψ 0.6 =1ψ 1

Span (m) iLcharacteristic value (if we suppose = ( ) )( fixedTskB Δ )( fixedskB

and = )( fixedTckc AG Δ )( fixedckc AG

(min)iL = 3 )(int eriL = 4 (max)iL

1β 0.127379249 0.099856598

2β 0.117095223 0.093731474

e 0.06 0.06 ΔT 40 40

)( fixedTskB Δ = ( daN.m²) )( fixedskB 22439.5 22439.5

)( fixedTckc AG Δ = (daN) )( fixedckc AG 19690 19690

)(1 fixedTkFB Δ(daN.m²) 1478.4 1478.4

TΔβ 1 no information 1 no information

wiq 5 = (daN/m²) siq 5

=0ψ 0.6 =1ψ 1 124.06 61.788

wiq 5 = (daN/m²)siq 5

=0ψ 1.0 =1ψ 0.6 207.0021 102.981

51

b) With creep + thermal gradient

ϕϕ +=

1cGG

We must verify:200lim,,

iESdE

Lww =≤

In summer:

( ) ( ) ( ) ( )2

2

10

27

47

111

24

1, 188384

51

81

3845

1 βα

βψψβψϕβ −Δ

−⎥⎦

⎤⎢⎣

⎡+−−⎥

⎦

⎤⎢⎣

⎡++−=− Δ

ΔΔΔΔ eLT

AGLq

BLq

AGLG

BLG

w isT

Tckc

ii

Tsk

ii

Tckc

i

Tsk

iSdE

( ) ( ) ( )

( ) ⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛++−−

Δ+⎟

⎠⎞

⎜⎝⎛−

−=

ΔΔ

ΔΔΔ

1

24

11

1

24

2

2

10

7

18384

5

18

1384

51

8200

βψ

βϕβα

βψψ

Tckc

i

Tsk

i

Tckc

i

Tsk

iisT

i

si

AGL

BL

AGLG

BLG

eLTL

q

In winter:

( ) ( ) ( ) ( )2

2

10

27

47

111

24

1, 188384

51

81

3845

1 βα

βψψβψϕβ −Δ

+⎥⎦

⎤⎢⎣

⎡+−−⎥

⎦

⎤⎢⎣

⎡++−= Δ

ΔΔΔΔ eLT

AGLq

BLq

AGLG

BLG

w iwT

Tckc

ii

Tsk

ii

Tckc

i

Tsk

iSdE

( ) ( ) ( )

( ) ⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

−⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛++−−

Δ−⎟

⎠⎞

⎜⎝⎛

−=

ΔΔ

ΔΔΔ

1

24

11

1

24

2

2

10

7

18384

5

18

1384

51

8200

βψ

βϕβα

βψψ

Tckc

i

Tsk

i

Tckc

i

Tsk

iiwT

i

wi

AGL

BL

AGLG

BLG

eLTL

q



Strength characteristic

value 3 4

)( fixedTskB Δ(daN.m²)

)( fixedTckc AG Δ(daN)

)(1 fixedTkFB Δ(daN.m²)

siq 7(daN/m²)

wiq 7(daN/m²)

52

9 Synthesis of the load table below the following combinations: At ULS :

1.50 Q + 1.5 x 0.6 ΔT 1.50 ΔT + 1.5 x 0.6 Q

At SLS :

1.00 Q + 0.6 ΔT 1.00 ΔT + 0.6Q


53

9.1 Synthesis load tables for the spans tested 9.1.1 General tables of results Criteria: We retain the minimal value of q obtains with:

54321 ;;;; qqqqq k = 0.85 fcv = 0.13 MPa

Elastic ( ) daN/m² 1qNo bending

collapse in span (ULS)

(wrinkling ) Combination 1 Combination 2

3.00 287.124

478.541

4.00 172.711

287.851

Elastic ( ) daN/m² 2qNo shear failure

on (ULS) Combination 1 Combination 2

3.00 284.908

474.846

4.00 207.858

346.431

Elastic ( ) daN/m² 2qNo assembly

failure on (ULS) Combination 1 Combination 2

3.00 337.456

562.426

4.00 254.870

424.784

Elastic ( ) daN/m² 3qNo wrinkling of the sheet in span

(SLS) Combination 1 Combination 2 3.00 224.257

289.258

4.00 173.554

373.761

Elastic ( ) daN/m² 4qNo shear failure

on the end support (SLS)

Combination 1 Combination 2

3.00 359.879 599.798 4.00 363.286 605.476

No Assembly failure on the end Elastic ( ) daN/m² 4q

54

support (SLS) Combination 1 Combination 2

3.00 466.712/312.99

777.853/521.65

4.00 353.039/237.7475

588.398/396.245

Elastic ( ) daN/m² 5qΔT = 40°C ΔT = 40°C

Deflexion at mid span

< L/200 (SLS) Combination 1 Combination 2 Combination 1 Combination 2

3.00 124.06 207.021 4.00 61.788 102.981

(*) preponderant ULS (**) preponderant SLS

55

9.1.2 Comparison of the values obtained by the design by testing method and the values of the polish Technical Agreement and EN 14509 9.1.2.1 With the characteristic values and calculated rigidities

Elastic approach without creep – Bending and shear rigidities calculated –

Span (m) iL

Limit states

Value of (daN/m²)

(Optimized value)

iq

Value of (daN/m²)

Polish Technical

Agreement

iq Value of (daN/m²) EN

14509

iq

(to be completed by

is Mainz) ULS -284.908 - 217

3.00 SLS - 124.06 - 125 ULS -172.711 - 157

4.00 SLS -61.788 - 74 Elastic approach without creep– Bending and shear rigidities issued of design by testing -

Span (m) iL

Limit states

Value of (daN/m²)

(Optimized value)

iq

Value of (daN/m²)

Polish Technical

Agreement

iq Value of (daN/m²) EN

14509

iq

(to be completed by

is Mainz) ULS - 284.908 - 217

3.00 SLS -136.823 - 125 ULS -172.711 - 157

4.00 SLS -70.823 - 74 Conclusion The results of the design by testing are very close with the polish agreement without creep.

56

10) Interpolations and extrapolations of the load tables below the following combinations:

At ULS : 1.50 Q + 1.5 x 0.6 ΔT 1.50 ΔT + 1.5 x 0.6 Q

At SLS :

1.00 Q + 0.6 ΔT 1.00 ΔT + 0.6Q


57

10 Determination of the interpolated values of iq 10.1 Expression of and sB cc AG We retain the value by range of spans: a) By calculation

Span (m) iLBending and shear rigidities -

characteristic values 2 < < 4 (Without ΔT) iL 2 < < 4(With ΔT) iL

)( fixedskB (daN.m²) 22439.3583 22439.3583

)(1 fixedFB (daN.m²) 1478.4 1478.4

)( fixedckc AG (daN) 19690 19690

b) by testing

Span (m) iLBending and shear rigidities -

characteristic values 2 < < 4 iL

)( fixedskB (daN.m²) 34123.60 34123.60

)(1 fixedFB (daN.m²) 1478.4 1478.4

)( fixedckc AG (daN) 19690 19690

10.2 Expression of bending and shear capacities We retain the value by range of spans:

Span (m) Strength caracteristic value 2<span< 3.5 3.5<span<4.5

TRdTUM Δ,2 (daN.m) 471.630 494.933

TRdTUV Δ,2 (daN) 623.854 600.376

URdF (daN) 741.412 741.412

TRdTEM Δ,2 (daN.m) 239.397 323.909

TRdTEV Δ,2 (daN) 522.629 703.372

ERdF linear (daN) 682.878 682.878

ERdF washer (daN) 452.295 452.295

G = 11.46 daN/m² for the panel of 3m G = 11.6 daN/m² for the panel of 4m

L/200 for the deflection

58

11 Conclusions: Final complete load tables following an elastic behaviour and comparison with the Polisch agreement below the following combinations:

At ULS : 1.50 Q + 1.5 x 0.6 ΔT 1.50 ΔT + 1.5 x 0.6 Q

At SLS :

1.00 Q + 0.6 ΔT 1.00 ΔT + 0.6Q

Roof sandwich panel on 2 supports in depression

59

Final load table for a profiled PU roof panel on 2 supports: Nominal Thickness 60 mm 2 facings

0.50/0.50 in succion – calculated rigidities

Values of q in daN/m² obtained with characteristic values without creeping – 4 fasteners / line of support with k = 0.85

Span (m) Limit states

2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50

ULS Polska Agrement 0.5/0.4 with ΔT 50 SLS

- - 272 183

241 150

217125

197104

181 88

157 74

157 74

135 64

11755

ULS Polska Agrement 0.50/0.50?) with ΔT40 SLS

213 181

160 152

124130

99 111

81 96

67 83

57 72

49 63

42 55

ULS SLS

Design By testing 0.50/0.50 with ΔT

Min

424 299

377 235

340 187

310 151

285124

246103

223 86

195 72

173 62

154 53

13846

ULS SLS

+14+23

+17+23

+19+23

+11+23

+5 +23

+14 +23

-1 +4

+2 +4

+4 +0

DBT/AgP%

Min ULS/SLS

+23 +23 +23 +23 +23 +23 +4 +4 0

Conclusions There is a good correlation of the load capacities of the panel between the polish Technical Agreement and the new design by testing method.

Values of q in daN/m² obtained with characteristic values without creep (EN 14509) 4 fasteners / line of support with k = 0.85 Design by Testing rigidities

Span (m) Limit states

2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50

ULS Polska Agrement 0.5/0.4? with ΔT without creep

SLS - - 272

183241 150

217125

197104

181 88

157 74

157 74

135 64

11755

ULS Polska Agrement 0.50/0.50?) with ΔT40 SLS

213 181

160 152

124130

99 111

81 96

67 83

57 72

49 63

42 55

ULS SLS

Design By testing 0.50/0.50 with ΔT with creeping Min

424 311

377 247

340 200

310 165

285137

246115

223 97

195 83

172 71

154 61

13853

ULS DBT/AgP% SLS

- - +25+9

+29+10

+31+10

+25+10

+23 +10

+24 +12

+9 -5

+14-5

+18-4

Conclusions Creep creates a big decrease of the value. But it is a global value (severe) that it is used here. It will be useful to do a creep test to optimize this creep value.

60

12 Annex test results ; curves load deflexion

61

12.1 Bending test results (without thermal gradient)

62

12.2 Shear test results (without thermal gradient)

64

Statistic study of the shear test results

Test results Qc(daN) 964.8 1036.6 921.8

Mean value 974.3333

∑=

=n

ixinL

ny

1)(

1

6.88065017

( )2

1)(∑

=

−=n

ixin yLσ

964.8 7.620144667E-5 1036.6 3.975458361E-3 921.8 2.950867733E-3

( ) ⎟⎠

⎞⎜⎝

⎛−

−= ∑

=

n

ixiy yLn

n 1

2

)(11σ

0.059171477 )( yky

p ex σ−=

k=3.15 xp 807.7476093

k statistic factor design by testing

0.8290259446

69

Statistic study of the shear test results

Test results Qel,v (daN) 870 840 900

Mean value 870

∑=

=n

ixinL

ny

1)(

1

6.768096623

( )2

1)(∑

=

−=n

ixin yLσ

870 1.572828349E-7 840 1.203724359E-3 900 1.176362407E-3

( ) ⎟⎠

⎞⎜⎝

⎛−

−= ∑

=

n

ixiy yLn

n 1

2

)(11σ

0.034498145 )( yky

p ex σ−=

k=3.15 xp 780.104

k statistic factor design by testing

0.8966714624

70

12.3 Fastener test

71

12.4 Bending test results (with thermal gradient)

73

12.5 Shear test results (with thermal gradient)

74

13 Annex: Official performances of the PU roof panel following polish agreement

75

13.1 Extract of the polish agreement for the roof product Extract of the polish agreement Tab.3 Values of permissible loads – roofing panels – on basis of the tests

76

Tab.27 Maximum loads of single span roofing panels, ext. skin thickness 0,5 mm, int. skin thickness 0,4 mm, deltaT = 50C Reduction coefficents for loads detaching panel from a support Bearing capacity: coef. 0,8 => 1.25 Ridigity : coef. 0,85 => 1.176

Fastener assembly

During the test no visible deformations of external steel skin were observed Load causing the crush of the core was accepted as breaking load. The crush appeared with ca. 749 daN. Taking safety coefficient as 2 , it was determined the force on one fixing should not exceed 375 daNFixing (roofing panels) (security factor of 2) Maximum force on one fixing should not exceed 150 daN (fastener with calottte in a crown of the roofing panel)

78

14 Annex: Analyse of the tkk Helsinki test results

79

14.1 Extract of the tkk report draft 20 10 2009 Test serie B panel on 2 supports without thermal gradient

80

14.2 Collapse load

81

14.3 bending tests with thermal gradient

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