Ad validation guide 2013 en parte2

ADVANCE DESIGN VALIDATION GUIDE

501

5.52 EC2 Test 40: Verifying a square concrete column subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1)

Test ID: 5153

Test status: Passed

5.52.1 Description Verifies a square cross section column made of concrete C30/37 subjected to a small compression force and significant rotation moment to the top - Bilinear stress-strain diagram (Class XC1)

Nominal rigidity method - The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a symmetrically reinforced section.

The column is considered connected to the ground by a fixed connection and free to the top part.

5.52.2 Background Nominal rigidity method.

Verifies the adequacy of a rectangular cross section made from concrete C30/37.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity, and then calculate the frames by considering a section symmetrically reinforced.

5.52.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Loadings from the structure: ► 15kN axial force ► 150kMm rotation moment applied to the column top ► The self-weight is neglected

■ Exploitation loadings: ► 7kN axial force ► 100kNm rotation moment applied to the column top

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Concrete cover 5cm ■ Transversal reinforcement spacing a = 40cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ The column is considered isolated and braced


502

Units

Metric System

Geometry

Below are described the beam cross section characteristics:

■ Height: h = 0.50 m, ■ Width: b = 0.50 m, ■ Length: L = 5.80 m, ■ Concrete cover: c = 5cm

Boundary conditions

The boundary conditions are described below:


Loading

The beam is subjected to the following load combinations:

■ Load combinations: The ultimate limit state (ULS) combination is:

NEd =1.35*15+150*7=30.75kN=0.03075MN

MEd=1.35*150+1.50*100=352.50kNm=0.352MNm

■ m...

NM

eEd

Ed 4511030750

35200 ===

5.52.2.2 Reference results in calculating the concrete column

Geometric characteristics of the column:

The column has a fixed connection on the bottom end and is free on the top end, therefore, the buckling length is considered to be:

mll 60.11*20 ==

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.2(1); Figure 5.7 b)

Calculating the slenderness of the column:

37.8050.0

60.11*32*32 0 ===alλ


503

Effective creep coefficient calculation:

The creep coefficient is calculated using the next formula:

( )Ed

EQPef M

Mt ., 0∞= ϕϕ

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.4(2)

Where:

( )0,t∞ϕ creep coefficient

EQPM serviceability firs order moment under quasi-permanent load combination

EdM ULS first order moment (including the geometric imperfections)

First order eccentricity evaluation:

ieee += 01

mei 03.0=

The first order moment provided by the quasi-permanent loads:

meNM

eee iEqp

Eqpi 56.1030.0

7*30.015100*30.0150

0

001 =+

++

=+=+=

kNNEqp 10.177*30.0151 =+=

MNmkNmeNM EqpEqp 181.058.18056.10*10.17* 111 ====

The first order ULS moment is defined latter in this example:

The creep coefficient ( )0,t∞ϕ is defined as follows:

)(*)(*),( 00 tft cmRH ββϕϕ =∞

72.2830

8.168.16)( =+

==cm

cm ffβ

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t

tβ (for t0= 28 days concrete age).

213

0

***1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH

MPafcm 35> therefore: 944.0383535 7.07.0

1 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα and

984.0383535 2.02.0

2 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα


504

( ) 72.1984.0*944.0*250*1.0

100501

1250500500*2500*500*2*2

30 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=⇒=

+== RHmm

uAch ϕ

28.2488.0*72.2*72.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ

The effective creep coefficient calculation:

( ) 17.1352.0181.0*28.2*, 0 ==∞=

Ed

EQPef M

Mtϕϕ


The second order effects; The buckling calculation:

For an isolated column, the slenderness limit check is done using the next formula:

nCBA ***20

lim =λ

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 5.8.3.1(1)

Where:

0062.020*²50.0

031.0*

===cdc

Ed

fANn

( ) 81.017.1*2.01

1*2,01

1=

+=

+=

ef

Aϕ

1.1*21 =+= ωB because the reinforcement ratio in not yet known

70.07,1 =−= mrC because the ratio of the first order moment is not known

42.1580062.0

7.0*1.1*81.0*20lim ==λ

42.15837.80 lim =<= λλ Therefore, the second order effects can be neglected.

Calculation of the eccentricities and solicitations corrected for ULS:

The stresses for the ULS load combination are:

NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN

MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525MNm

Therefore, we must calculate:

■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections


505

Initial eccentricity:

mNMeEd

Ed 46.1103075.03525.0

0 ===

Additional eccentricity:

mlei 03.0400

6.11400

0 ===

The first order eccentricity: stresses correction:

The forces correction, used for the combined flexural calculations:

MNNEd 03075.0=

meee i 49.1101 =+=

MNmNeM EdEd 353.003075.0*49.11*1 ===

0*eNM Ed=

⎩⎨⎧

=⎪⎩

⎪⎨⎧

=⎪⎩

⎪⎨⎧

== mmmmmmmm

mmhmm

e 207.16

20max

3050020

max30

20max0

According to Eurocode 2 – EN 1992-1-1 -2004; Chapter 6.1.(4)

Reinforcement calculation in the first order situation:

The theoretical reinforcement will be determined by the following diagram

The input parameters of the diagram are:

141.020*50.0*50.0

353.0** 22 ===

cd

Ed

fhbMμ

00615.020*5.0*5.0

03075.0**

===cd

Ed

fhbNν


506

Therefore:

35.0=ω

The reinforcement area will be:

∑ == 22

25.4078.434

20*50.0* cmAsω

which means 20.13cm2 per face.

The total area will be 40.25cm2.

Finite elements modeling

■ Linear element: S beam, ■ 7 nodes, ■ 1 linear element.

Theoretical reinforcement area(cm2)

(reference value: 19.32cm2)


507

Theoretical value (cm2)

(reference value: 38.64 cm2)

5.52.2.3 Reference results

Result name Result description Reference value Az Reinforcement area [cm2] 19.32 cm2

R Theoretical reinforcement area [cm2] 38.64 cm2

5.52.3 Calculated results

Result name Result description Value Error Az Az -19.32 cm² 0.0000 %


508

5.53 EC2 Test 41: Verifying a square concrete column subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1)

Test ID: 5195

Test status: Passed

5.53.1 Description Verifies a square cross section concrete column made of concrete C30/37 subjected to a significant compression force and small rotation moment to the top - Bilinear stress-strain diagram (Class XC1)

Nominal rigidity method.

The purpose of this test is to determine the second order effects by applying the method of nominal rigidity and then calculate the frames by considering a symmetrically reinforced section.


5.53.2 Background Nominal rigidity method.

Verifies the adequacy of a rectangular cross section made from concrete C30/37.





■ 3,02 =ψ

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Concrete cover 3cm and 5cm ■ Transversal reinforcement spacing a=40cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ The column is considered isolated and braced


509

Units

Metric System

Geometry

Beam cross section characteristics:

■ Height: h = 0.50 m, ■ Width: b = 0.50 m, ■ Length: L = 5.80 m, ■ Concrete cover: c = 5cm

Boundary conditions



Loading



NEd = 1.35*150+1.50*100 = 352.50kN = 0.035MN

MEd = 1.35*15+1.50*7 = 30.75kNm = 0.03075MNm

■ m..

.NM

eEd

Ed 08703530

0370500 ===




mll 60.11*20 ==



37.8050.0

60.11*32*32 0 ===alλ


510



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:





ieee += 01

mei 03.0=


meNM

eee iEqp

Eqpi 125.030.0

100*30.01507*30.015

0

001 =+

++

=+=+=

kNNEqp 180100*30.01501 =+=

MNmkNmeNM EqpEqp 0225.050.22125.0*180* 111 ====


MNmMEd 041.01 =


)(*)(*),( 00 tft cmRH ββϕϕ =∞

72.2830

8.168.16)( =+

==cm

cm ffβ

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t


213

0

***1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


1 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα and 984.0

383535 2.02.0

2 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα

( ) 72.1984.0*944.0*250*1.0

100501

1250500500*2500*500*2*2

30 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=⇒=

+== RHmm

uAch ϕ


511

28.2488.0*72.2*72.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ


( ) 25.1041.0

0225.0*28.2*, 0 ==∞=Ed

EQPef M

Mtϕϕ



For an isolated column, the slenderness limit verification is done using the next formula:

nCBA ***20

lim =λ


Where:

071.020*²50.0

353.0*

===cdc

Ed

fANn

( ) 80.025.1*2.01

1*2,01

1=

+=

+=

ef

Aϕ



24.46071.0

7.0*1.1*80.0*20lim ==λ

24.4637.80 lim =>= λλ Therefore, the second order effects must be taken into account.


The stresses for the ULS load combination are: ■ NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN ■ MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm Therefore, we must calculate:

■ The eccentricity of the first order ULS moment, due to the stresses applied ■ The additional eccentricity considered for the geometrical imperfections Initial eccentricity:

mNMeEd

Ed 087.0353.0

03075.00 ===


mlei 03.0400

6.11400

0 ===


512



MNNEd 353.0=

meee i 117.001 =+=

MNmNeM EdEd 041.0353.0*117.0*1 ===

0*eNM Ed=

⎩⎨⎧

=⎪⎩

⎪⎨⎧

=⎪⎩

⎪⎨⎧

== mmmmmmmm

mmhmm

e 207.16

20max

3050020

max30

20max0



To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concrete section will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum percentage area.

The reinforcement will be determined using a compound bending with compressive stress. The determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:

MNmhdNMM Gua 112.0250.045.0*353.0041.0)

2(*0 =⎟

⎠⎞

⎜⎝⎛ −+=−+=

Verification about the partially compressed section:

494,0)45,050,0*4,01(*

45,050,0*8,0)*4,01(**8,0 =−=−=

dh

dh

BCμ

055.020*²45.0*50.0

112.0*²*

===cdw

uacu fdb

Mμ

494.0055.0 =<= BCcu μμ therefore the section is partially compressed


513

The calculation for the tensioned steel in pure bending:

055.0=cuμ

[ ] 071,0)055,0*21(1*25,1 =−−=uα

mdz uc 437,0)071,0*4,01(*45,0)*4,01(* =−=−= α

²89,578,434*437,0

112,0*

cmfz

MAydc

ua ===

The calculation for the compressed steel in bending:

For the compound bending:

24 23.278.434

353.010*89.5 cmFNAAyd

−=−=−= −

The minimum column percentage reinforcement must be considered:

²81.078.434

353.0*10.0*10,0min, cm

FNA

yd

Eds ===

Therefore, a 5cm2 reinforcement area will be considered.

5.53.2.3 Calculation of the second order effects:

Estimation of the nominal rigidity:

It is estimated the nominal rigidity of a post or frame member from the following formula:

sssccdc IEKIEKEI **** +=

Where:

2.1cm

cdEE =

MpaMpaff ckcm 388 =+=

MpafE cmcm 57.32836

1038*22000

10*22000

3.03.0

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

MpaEE cmcd 27364

2.132837

2.1===

4343

10.208,51250.0

12* mhbIc

−===inertia of the concrete section only

MpaEs 200000=

sI : Inertia

002.050.0*50.0

10.5 4

===−

c

s

AAρ


514

01.0002.0 <=≤c

s

AAρ

22.12030

201 === ckfk

071.020*²50.0

353.0*

===cdc

Ed

fANn

20.00336.0170

37.80*071.0170

*2 ≤===λnk

45242

10.205.0250.0*

210*5*2

2*

2*2 mchAI s

s−

−

=⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=

1=sK and 018.025.110336.0*22.1

1* 21 =

+=

+=

efc

kkKϕ

Therefore:

²56.610*2*200000*110*208,5*27364*018.0 53 MNmEI =+= −−

Stress correction:

The total moment, including second order effects, is defined as a value plus the moment of the first order:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+=

11*0

Ed

BEdEd

NNMM β

MNmM Ed 041.00 = (moment of first order (ULS) taking into account geometric imperfections)

MNNEd 353.0= (normal force acting at ULS).

In addition:

0

²cπβ =

and 80 =c because the moment is constant (no horizontal force at the top of post).

234.18²

==πβ

MNlEINB 48.0

²60.1156.6*²*² 2

0

=== ππ


515

It was therefore a moment of 2nd order which is:

MNmMEd 182.01

353.048.0234.11*041.0 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+=

There is thus a second order moment of 0.182MNm

Calculation of the flexural combined reinforcement

The theoretical reinforcement will be determined by the following diagram:

MNmMEd 182.0=

MNNEd 353.0=


073.020*50.0*50.0

182.0** 22 ===

cd

Ed

fhbMμ

071.020*5.0*5.0

353.0**

===cd

Ed

fhbNν

Therefore:

11.0=ω



516

∑ === 22

65.1278.434

20*50.0*11.0*** cmf

fhbAyd

cds

ω

This means a total of 12.65cm2

The initial calculations must be repeated by increasing the section; a 6cm2 reinforcement section will be considered.

Additional iteration:

One more iteration by considering an initial section of 6.5cm ²

Estimation of the nominal rigidity:

sssccdc IEKIEKEI **** +=

Where:

2.1cm

cdEE =

MpaMpaff ckcm 388 =+=

MpafE cmcm 57.32836

1038*22000

10*22000

3.03.0

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

MpaEE cmcd 27364

2.132837

2.1===

4343

10.208,51250.0

12* mhbIc

−=== considering only the concrete section only

MpaEs 200000=

sI : Inertia

0026.050.0*50.0

10*5.6 4

===−

c

s

AAρ

01.0002.0 <=≤c

s

AAρ

22.12030

201 === ckfk

071.020*²50.0

353.0*

===cdc

Ed

fANn

20.00336.0170

37.80*071.0170

*2 ≤===λnk

45242

106.205.0250.0

210*5.6*2

2*

2*2 mchAI s

s−

−

×=⎟⎠⎞

⎜⎝⎛ −=⎟

⎠⎞

⎜⎝⎛ −=

1=sK and 018.0

25.110336.0*22.1

1* 21 =

+=

+=

efc

kkKϕ


517

Therefore:

²78.710*6.2*200000*110*208,5*27364*018.0 53 MNmEI =+= −−

Stress correction:

The total moment, including second order effects, is defined as a value plus the moment of the first order:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−+=

11*0

Ed

BEdEd

NNMM β

MNmM Ed 041.00 = (moment of first order (ULS) taking into account geometric imperfections)

MNNEd 353.0= (normal force acting at ULS).

In addition:

0

²cπβ =

and 80 =c because the moment is constant (no horizontal force at the top of post).

234.18²

==πβ

MNlEINB 57.0

²60.1178.7*²*² 2

0

=== ππ

It was therefore a moment of 2nd order which is:

MNmMEd 123.01

353.057.0234.11*041.0 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+=

There is thus a second order moment of 0.123MNm


518

Calculation of the flexural compound reinforcement

The theoretical reinforcement will be determined, from the following diagram:

MNmMEd 123.0=

MNNEd 353.0=


049.020*50.0*50.0

123.0** 22 ===

cd

Ed

fhbMμ

071.020*5.0*5.0

353.0**

===cd

Ed

fhbNν

Therefore:

05.0=ω


∑ === 22

75.578.434

20*50.0*05.0*** cmf

fhbAyd

cds

ω

This means a total of 5.75cm2


519




(reference value: 5.75cm2=2*2.88cm2)







Result name Result description Value Error Az Az -3.01 cm² -0.0000 %


520

5.54 EC2 Test 44: Verifying a rectangular concrete beam subjected to eccentric loading - Bilinear stress-strain diagram (Class X0)

Test ID: 5213

Test status: Passed

5.54.1 Description Verifies a rectangular cross section beam made of concrete C30/37 subjected to eccentric loading - Bilinear stress-strain diagram (Class X0).

The verification of the bending stresses at ultimate limit state is performed.

Simple Bending Design for Ultimate Limit State

During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.

- Support at start point (x = 0) fixed connection

- Support at end point (x = 5.00) fixed connection


521

5.55 EC2 Test 45: Verifying a rectangular concrete beam supporting a balcony - Bilinear stress-strain diagram (Class XC1) Test ID: 5225 Test status: Passed

5.55.1 Description Verifies the adequacy of a rectangular cross section beam made of concrete C25/30 supporting a balcony - Bilinear stress-strain diagram (Class XC1). Simple Bending Design for Ultimate Limit State Verifies the column resistance to rotation moment along its length. During this test, the determination of stresses is made along with the determination of the longitudinal and transversal reinforcement.

5.55.2 Background Simple Bending Design for Ultimate Limit State

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist rotation moment along its length. During this test, the calculation of stresses is performed, along with the calculation of the longitudinal and transversal reinforcement.

5.55.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. ■ The geometric dimension of the beam are:

The following load cases and load combination are used:

■ Concrete type: C25/30 ■ Reinforcement type: S500B ■ Exposure class: XC1 ■ Balcony load: 1kN/m2 ■ The weight of the beam will be considered in calculation ■ Concrete density: 25kN/m3 ■ The beam is considered fixed at both ends ■ Concrete cover: 40mm ■ Beam length: 4.00m


522

Units

Metric System

Geometry


■ Height: h = 0.75 m, ■ Width: b = 0.25 m, ■ Length: L = 4.00 m, ■ Section area: A = 0.1875 m2 , ■ Concrete cover: c=4cm

Boundary conditions


■ Outer: ► Support at start point (x=0) fixed connection, ► Support at end point (x = 4.00) fixed connection

■ Inner: None.

5.55.2.2 Reference results in calculating the concrete beam

The ULS load calculation:

The first step of the calculation is to determine charges transmitted to the beam:

■ Vertical loads applied to the beam (kN/ml) from the load distribution over the balcony ■ Rotation moment applied to the beam (kN/ml) from the load distribution over the balcony Each action (self-weight and distributed load) is determined by summing the resulting vertical loads along the eaves and the torsional moment by multiplying the resultant by the corresponding lever arm.

CAUTION, different lever arm must be considered from the center of the beam (by adding therefore the half-width).

The results are displayed in the table below:

Load calculation:

From previously calculated results, the following stresses can be determined:

■ Shear: EdV

■ Bending moment: EdM

■ Torque: EdT


523

Shear and bending moment:

One can determine the load at ULS taken over by the beam:

mKNPu /14.212*5,144.13*35,1 =+= For a beam fixed on both ends, the following values will be obtained:

Maximum shear (ULS):

MNKNlPV uEd 042,03,42

24*14,21

2*

====

Bending moment at the supports:

MNmKNmlPM uEd 028,02,28

12²4*14,21

12²*

====

Maximum Moment at middle of span:

MNmKNmlPM uEd 014,01,14

24²4*14,21

24²*

====

Torsion moment:

For a beam subjected to a torque constant:

mMNmmKNmmtu /015,0/97,1425,2*5,159,8*35,1 ==+=

MNmlmT tuEd 03,024*015,0

2* ===

5.55.2.3 Bending rebar

Span reinforcement (at bottom fiber)

007,067.16*²70.0*25,0

014,0==cuμ

( ) 0088,0007,0*211*25,1 =−−=uα

( ) mzc 697,00088,0*4,01*70.0 =−=

²46.0²10*62.478,434*697,0

014,0 4 cmmAu === −


524

Minimum reinforcement percentage verification: ⎪⎩

⎪⎨⎧

=db

dbff

MaxA

w

wyk

effct

s

**0013.0

***26.0 ,

min,

Cracking matrix required (calculation hypothesis): Mpaff ctmeffct 56.2, ==

²27.2²27.270.0*25.0*0013.0**0013.0

²27.270.0*25.0*500

56.2*26.0***26.0 ,

min, cmcmdb

cmdbff

MaxA

w

wyk

effct

s =⎪⎩

⎪⎨⎧

==

===

Therefore, it retains 2.27 cm ².

Reinforcement on supports (at top fiber)

014,067.16*²70.0*25,0

028,0==cuμ

( ) 018,0014,0*211*25,1 =−−=uα

( ) mzc 695,0018,0*4,01*70.0 =−=

²93.0²10*27.978,434*695,0

028,0 4 cmmAu === −

It also retains 2.27 cm ² (minimum percentage).

Shear reinforcement

MNVEd 042,0=

The transmission to the support is not direct; it is considered a connecting rod inclined by 45˚, therefore 1cot =θ

Concrete rod verification:

θαθ

²cot1cotcot****max, +

+= cdwRd fvzbV

According to: EC2 Part 1,1 EN 1992-1-1-2004 Chapter 6.2.3(4)

mzz c 695.0== (from the design in simple bending of support)

Vertical frames:

54.0250251*6,0 =⎥⎦

⎤⎢⎣⎡ −=v

MNtg

fvzbfvzbV cdcwcdcwRd 78.0

267.16*54.0*695.0*25.0

cot***

²cot1cotcot****max, ==

+=

++

=θθθ

αθ

θθ cot***1

max, +=

tgbzfvV wucd

Rd


MNVMNV RdEd 78.0042.0 max, =<=


525

Calculation of transverse reinforcement:

mlcmfztgV

sA

ydu

Edsw /²39.178.434*695.0

042.0**. ===

θ

(over shear)

From the minimum reinforcement percentage:

αρ sin..min, wwsw bsA

≥


With:

0008.0500

25*08,0*08,0min, ===

yk

ckw f

fρ

mlcmsAsw /²225.0*0008.0 =≥

Therefore:

mlcmsAsw /²2≥

Torsion calculation:

Torsion moment was calculated before: MNmTEd 03,0=

Torsional shear stress:

kief

Edit At

T**2 ,

, =τ


cmcmuA

cmct ief 375.9375.9

)7525(2)75*25(8*2

max, =⎪⎩

⎪⎨⎧

=+

=

==

²1025.0²1025)375.975(*)375.925( mcmAk ==−−=

Mpa..*.*

.A*t*

T

ki,ef

Edi,t 561

102500937502030

2τ ===

Concrete verification:

Calculate the maximum allowable stress in the rods:

θθα cos*sin******2 ,max, iefkcdcwRd tAfvT =


54.0250

1*6,0 =⎟⎠⎞

⎜⎝⎛ −= ckfv

MNTRd 085.070.0*70.0*09375.0*1025.0*67.16*54.0*2max, ==

Because of the combined share/moment effect, we must calculate:


526

0,1max,max,

≤+Rd

Ed

Rd

Ed

VV

TT


0,1399.078.0

042.0085.003.0

≤=+

Torsion longitudinal reinforcement

θcot***2

*

ydk

kEdl fA

uTA =Σ


mcmthtbu efefk 625.15.162)]375.975()375.925[(*2)]()[(*2 ==−+−=−+−=

²47.578.434*1025.0*2

625.1*03.0 cmAl ==Σ

Torsion transversal reinforcement

mlcmfAT

sA

ydk

Ed

T

swT /²36.378.434*1025.0*2

03.0cot***2

===θ

Therefore mlcmsA

T

swT /²36,3≥ for each face.




527

ULS load combinations(kNm)

Torsional moment (Ted=29.96kNm)

Longitudinal reinforcement (5.46cm2)

Transversal reinforcement (3.36cm2/ml)


Result name Result description Reference value Mx Torsional moment [kNm] 29.96 cm2

Al Longitudinal reinforcement [cm2] 5.46 cm2

Ator,y Transversal reinforcement [cm2/ml] 3.36 cm2/ml


Result name Result description Value Error Mx Mx 29.9565 kN*m 0.0000 % Al Al 5.4595 cm² 0.0000 % Ator,y / face Ator,y/face 3.35969 cm² 0.0001 %


528

5.56 EC2 Test 47: Verifying a rectangular concrete beam subjected to a tension distributed load - Bilinear stress-strain diagram (Class XD2)

Test ID: 5333

Test status: Passed

5.56.1 Description Verifies a rectangular cross section beam made of concrete C25/30 subjected to a tension distributed load - Bilinear stress-strain diagram (Class XD2).

The verification of the bending stresses at ultimate stress limit and serviceability limit state is performed.

Simple Bending Design for Ultimate and Service State Limit

During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.57 EC2: Verifying the longitudinal reinforcement area of a beam under a linear load - Inclined stress strain behavior law

Test ID: 4522

Test status: Passed

5.57.1 Description Verifies the longitudinal reinforcement area of a beam under a linear load - inclined stress strain behavior law.

Verification is done according to Eurocodes 2 norm with French Annex.


529

5.58 EC2 Test 3: Verifying a rectangular concrete beam subjected to uniformly distributed load, with compressed reinforcement- Bilinear stress-strain diagram

Test ID: 4976

Test status: Passed

5.58.1 Description Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the determination of stresses is made along with the determination of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The objective is to verify:

- The stresses results

- The longitudinal reinforcement

- The verification of the minimum reinforcement percentage

5.58.2 Background Simple Bending Design for Ultimate Limit State

Verifies the adequacy of a rectangular cross section made from concrete C25/30 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

5.58.2.1 Model description ■ Reference: Guide de validation Eurocode 2 EN 1992-1-1-2002; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combinations are used:

■ Loadings from the structure: G = 70 kN/m, ■ Exploitation loadings (category A): Q = 80kN/m,

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q The objective is to verify:

■ The stresses results ■ The longitudinal reinforcement ■ The verification of the minimum reinforcement percentage

Simply supported beam


530

Units

Metric System

Geometry


■ Height: h = 1.25 m, ■ Width: b = 0.65 m, ■ Length: L = 14 m, ■ Section area: A = 0.8125 m2 , ■ Concrete cover: c = 4.50 cm ■ Effective height: d = h-(0.6*h+ebz) = 1.130 m; d’ = ebz = 0.045m

Materials properties

Rectangular solid concrete C25/30 and S500B reinforcement steel is used. The following characteristics are used in relation to this material:

■ Exposure class XD1 ■ Concrete density: 25kN/m3 ■ Stress-strain law for reinforcement: Bilinear stress-strain diagram ■ Cracking calculation required

■ Concrete C25/30: MPa,,

ff

c

ckcd 6716

5125

γ===

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.6(1); 4.4.2.4(1); Table 2.1.N

■ MPa.*.f*.f //ckctm 56225300300 3232 ===

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.1.3(2); Table 3.1

■ Steel S500 : MPa,,

ff

s

ykyd 78434

151500

γ===

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 3.2

■ MPa*f

*E..

ckcm 31476

1082522000

108

220003030

=⎟⎠

⎞⎜⎝

⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=


Boundary conditions


■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 14) restrained in translation along Z.

■ Inner: None.

Loading



Cmax = 1.35 x G + 1.5 x Q=1.35*70+1.5*80=214.5*103 N/ml

Characteristic combination of actions:

CCQ = 1.0 x G + 1.0 x Q=70+80=150*103 N/ml

Quasi-permanent combination of actions:

CQP = 1.0 x G + 0.3 x Q=70+0.3*80=94*103 N/ml


531

■ Load calculations:

Nm*.²*.MEd6102555

8145214

==

Nm*.²*MEcq6106753

814150150 ==

Nm*.²*MEqp6103032

81494

==

5.58.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:

■ )t(*)f(*)t,( cmRH 00 ββ=∞

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.2

■ 9252825

816816β ..f.)f(

cmcm =

+==


■ 488028101

101β

2002000

0 ..t.

)t(..

=+

=+

= at t0=28 days


The value of the φRH coefficient depends of the concrete quality:

■ 2130

αα10

1001

1 **h*.

RH

RH⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

■ 121 == αα if MPafcm 35≤

■ If not: 70

135α

.

cmf ⎟⎟⎠

⎞⎜⎜⎝

⎛= and

20

235α

.

cmf ⎟⎟⎠

⎞⎜⎜⎝

⎛=

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c

■ In our case we have: 1338 21 ==⇒=+= ααMpaMpaff ckcm

■ 6616342710

100501

16342712506502125065022

30 ..*.

mm.)(*

**uAc*h RH =

−+=⇒=

+==


■ 375248809252661ββ 00 ..*.*.)t(*)f(*)t,( cmRH ===∞

The coefficient of equivalence is determined by the following formula:

■

Ecar

Eqp

cm

se

MM

*)t,(

EE

01

α

∞+

=


532

Defined earlier:

NmMEcar310*3675=

NmMEqp310*2303=

This gives:

8215

1039751023037021

31476200000α

3

3

.

***.

e =

+

=

5.58.2.3 Reference results in calculating the concrete beam reduced moment limit

For the calculation of steel ULS, we consider the moment reduced limit of 372.0=luμ for a steel grade 500Mpa.

Therefore, make sure to enable the option "limit )500/372.0( Sμ " in Advance Design.

Reference reinforcement calculation at SLU:

The calculation of the reinforcement is detailed below:

■ Effective height: d=0.9*h=1.125 m ■ Calculation of reduced moment:

383,067,16*²125.1*65,0

10*25.5255*²* 2

3

===MPamm

Nmfdb

M

cdw

Edcuμ

372.0383.0 =<= lucu μμ therefore the compressed reinforced must be resized and then the concrete section must be adjusted.

Calculation of the tension steel section (Section A1):

The calculation of tensioned steel section must be conducted with the corresponding moment of :

NmfdbM cdwluEd6

1 10*10.567.16*²125.1*65.0*372.0*²** === μ

■ The α value: [ ] [ ] 61803720211251μ211251α .).*(*.)*(*. lulu =−−=−−=

■ Calculation of the lever arm zc: m.).*.(*.)*.(*dz lulu 851061804011301α401 =−=−=

■ Calculation of the reinforcement area:

²35.13778.434*851.0

10*10.5.

61

1 cmMPam

Nmfz

MAydlu

Ed ===−


533

Compressed steel reinforcement reduction (Section As2):

Reduction coefficient:

( ) 00327.0045.0130.1*618.0130.1*618.0*1000

5.3)'*(**1000

5,3=−=−= dd

d lulu

sc αα

ε

MPaf ydscydsc 78.43400217.000327.0 ==⇒=>= σεε

Compressed reinforcement calculation:

²32.278.434)045.0130.1(

15.5255.5)'(

12 cm

ddMMA

sc

EdEds =

×−−

=−

−=

σ The steel reinforcement condition:

²32.2.22 cmf

AAyd

scs ==

σ

Total area to be implemented:

In the lower part: As1=A1+A2=142.42 cm2

In the top part: As2=2.32 cm2

Reference reinforcement calculation at SLS:

The concrete beam design at SLS will be made considering a limitation of the characteristic compressive cylinder strength of concrete at 28 days, at 0.6*fck

The assumptions are:

■ The SLS moment: Nm*.²*MEcq6106753

814150150 ==

■ The equivalence coefficient: 8215α .e =

■ The stress on the concrete will be limited at 0.6*fck=15Mpa and the stress on steel at 0.8*fyk, or 400MPa Calculation of the resistance moment MRd for detecting the presence of the compressed reinforcement:

m..*.

*.d.*

*x

sce

ce 41901251400158215

158215σσα

σα1 =×

+=

+=

N*..*.**x*b*F cwc6

1 1004215419065021σ

21

=×==

m...x

dzc 9850341901251

31 =−=−=

Nm*..**.z*FM ccrb66 10012985010042 ===

Therefore the compressed reinforced established earlier was correct.


534

Theoretical section 1 (tensioned reinforcement only)

NmMM rb6

1 10*01.2==

372.0125.1419.01

1 ===dxα

mxdzc 985.03419.0125.1

31 =−=−=

²01.51400985.0

01.211 cm

zMA

sc

=×

=×

=σ

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

NmMMM rbcqser66

,2 10*665.110*)01.2675.3( =−=−=

Compressed reinforcement stresses:

ddd

cesc *'***

1

1

αασασ −

=

MPasc 78.211125.1*372.0

045.0125.1*372.0*15*82.15 =−

=σ

Compressed reinforcement area:

( ) ²92.7178.211*045.0125.1

645.1*)'(

' 2 cmddMA

sc

=−

=−

=σ

Complementary tensioned reinforcement area:

²08.38400

78.211*92.71*' cmAAs

scs ===

σσ

Section area:

Tensioned reinforcement: 51.01+38.08=89.09 cm2

Compressed reinforcement: 71.92 cm2

Considering an envelope calculation of ULS and SLS, it will be obtained:

Tensioned reinforcement ULS: A=141.42cm2

Compressed reinforcement SLS: A=71.92cm2

To optimize the reinforcement area, it is preferable a third iteration by recalculating with SLS as a baseline amount of tensioned reinforcement (after ULS: Au=141.44cm2)

Reference reinforcement additional iteration for calculation at SLS:

For this iteration the calculation will be started from the section of the tensioned reinforcement found when calculating the SLS: Au=141.44cm2.

For this particular value, it will be calculated the resistance obtained for tensioned reinforcement:

MPaAA

sELU

ELSs 252400

42.14118.89

=×=×= σσ

This is a SLS calculation, considering this limitation:

Calculating the moment resistance MRb for detecting the presence of compressed steel reinforcement:


535

mdxsce

ce 55.0130.1*25215*82.15

15*82.15**

*1 =

+=

+=

σσασα

NxbF cwc6

1 10*67.215*55.0*65.0*21***

21

=== σ

mxdzc 94.0355.0125.1

31 =−=−=

NmxdxbzFM cwccrb61

1 10*52.2355.0125.1*15*55.0*65.0*

21

3****

21* =⎟

⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−== σ

According to the calculation above Mser,cq is grater then Mrb the compressed steel reinforcement is set.

Theoretical section 1 (tensioned reinforcement only)

NmMM rb6

1 10*52.2==

488.0125.155.01

1 ===dxα

mxdzc 94.03548.0125.1

31 =−=−=

²38.10625294.0

52.211 cmzMA

sc

=×

=×

=σ

Theoretical section 2 (compressed reinforcement and complementary tensioned reinforcement)

NmMMM rbcqser66

,2 10*155.110*)52.2675.3( =−=−=

Compressed reinforcement stresses:

ddd

cesc *'***

1

1

αασασ −

=

MPasc 73.217125.1*485.0

045.0125.1*485.0*15*82.15 =−

=σ

Compressed reinforcement area:

( ) ²12.4973.217*045.0125.1

155.1*)'(

' 2 cmddMA

sc

=−

=−

=σ

Complementary tensioned reinforcement area:

²44.42252

73.217*12.49*' cmAAs

scs ===

σσ

Section area:

Tensioned reinforcement: 106.38+42.44=148.82 cm2

Compressed reinforcement: 49.12 cm2


536

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

⎪⎩

⎪⎨⎧

=db

dbff

MaxA

w

wyk

effct

s

**0013.0

***26.0 ,

min,

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2

Where:

MPaff ctmeffct 56.2, == from cracking conditions

Therefore:

⎪⎩

⎪⎨⎧

==

==−

−

²73.9²10*51.9125.1*65.0*0013.0

²10*73.9125.1*65.0*500

56.2*26.0max4

4

min, cmm

mAs



ULS and SLS load combinations(kNm)

Simply supported beam subjected to bending

ULS (reference value: 5255kNm)

SLS –Characteristic (reference value: 3675kNm)

SLS –Quasi-permanent (reference value: 162.94kNm)


537


(reference value: A’=148.82cm2 A=49.12cm2)

Minimum reinforcement area(cm2)



Result name Result description Reference value My,ULS My corresponding to the 101 combination (ULS) [kNm] 5255 kNm My,SLS,c My corresponding to the 102 combination (SLS) [kNm] 3675 kNm My,SLS,q My corresponding to the 103 combination (SLS) [kNm] 2303 kNm Az (A’) Theoretical reinforcement area [cm2] 148.82 cm2

Az (A) Theoretical reinforcement area [cm2] 49.12 cm2

Amin Minimum reinforcement area [cm2] 9.73 cm2


Result name

Result description Value Error

My My USL -5255.25 kN*m 0.0000 % My My SLS cq -3675 kN*m 0.0000 % My My SLS pq -2303 kN*m 0.0000 % Az Az -147.617 cm² -0.0003 % Amin Amin 9.79662 cm² 0.0000 %


538

5.59 EC2 Test 2: Verifying a rectangular concrete beam subjected to a uniformly distributed load, without compressed reinforcement - Bilinear stress-strain diagram

Test ID: 4970

Test status: Passed



539

5.60 EC2 Test 6: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram

Test ID: 4979

Test status: Passed

5.60.1 Description The purpose of this test is to verify the software results for the My resulted stresses for the USL load combination and for the results of the theoretical reinforcement area Az.

5.60.2 Background This test performs the verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.


■ Loadings from the structure: G = 45 kN/m ■ Exploitation loadings (category A): Q = 37.4kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q

■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to verify:

■ The theoretical reinforcement area results



540

Units

Metric System

Geometry


■ Beam length: 8m ■ Concrete cover: c=3.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.435 m; d’=ebz=0.035m



■ Exposure class XC1 ■ Concrete density: 25kN/m3 ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram

■ Concrete C25/30: MPaffc

ckcd 67.10

5,125

===γ


■ MPaff ckctm 56.225*30.0*30.0 3/23/2 ===


■ Steel S500B : MPaf

fs

ykyd 78.434

15,1500

===γ


Boundary conditions


■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X.

■ Inner: None.


541

Loading



Cmax = 1.35 x G + 1.5 x Q=1.35*45+1.5*37.4=116.85kN/ml


CCQ = 1.0 x G + 1.0 x Q=45+34.7=79.7kN/ml


kNmMEd 8.9348

²8*85.116==

5.60.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:

kNmfh

dhbM cdf

feffbtu 967.16*215.0435.0*15.0*00.1*

2** =⎟

⎠⎞

⎜⎝⎛ −==⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Comparing Mbtu with MEd:

Therefore, the concrete section is not entirely compressed;

Therefore, the calculations considering the T section are required.

5.60.2.3 Reference reinforcement calculation:

Theoretical section 2:

The moment corresponding to this section is:

kNm

hdfhbbM f

cdfweffEd

720215.0435.0*67.16*15.0*)20.01(

2***)(2

=

=⎟⎠⎞

⎜⎝⎛ −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

According to this value, the steel section is:

²00.468.347*

215.0435.0

720.0

*2

22 cm

fh

d

MA

ydf

Ed =⎟⎠⎞

⎜⎝⎛ −

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

kNmMkNmM Edbtu 8.934900 =<=


542


The theoretical section 1 corresponds to a calculation for a rectangular shape beam section.

kNmMMM EdEdEd 21572093521 =−=−=

341.067.16*²435.0*20.0

215.0*²*1 ===

cdw

Edcu Fdb

Mμ

For a S500B reinforcement and for a XC1 exposure class, there will be a: 3720μμ .lucu => , therefore there will be no compressed reinforcement (the compression concrete limit is not exceeded because of the exposure class)

There will be a calculation without considering compressed reinforcement:

[ ] ( )[ ] 544.0341.0*211*25.1)*21(1*25.1 =−−=−−= cuu μα

mdz uc 340.0)544.0*40.01(*435.0)*4.01(*1 =−=−= α

²52.1478.347*340.0

215.0*1

11 cm

fzMA

ydc

Ed ===


In conclusion, the entire reinforcement steel area is A=A1+A2=46+14.52=60.52cm2






For Class B reinforcement steel ductility (reference value: A=60.52cm2)


Result name Result description Reference value Az (Class B) Theoretical reinforcement area [cm2] 60.52 cm2




543

5.61 EC2 Test 7: Verifying a T concrete section, without compressed reinforcement- Bilinear stress-strain diagram

Test ID: 4980

Test status: Passed

5.61.1 Description The purpose of this test is to verify the My resulted stresses for the USL load combination and the results of the theoretical reinforcement area Az.

This test performs verification of the theoretical reinforcement area for the T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.

5.61.2 Background Verifies the theoretical reinforcement area for a T concrete beam subjected to the defined loads. The test confirms the absence of the compressed reinforcement for this model.


■ Loadings from the structure: G = 0 kN/m (the dead load is not taken into account) ■ Exploitation loadings (category A): Q = 18.2kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q

■ ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram The objective is to test:

■ The theoretical reinforcement area



544

Units

Metric System

Geometry


■ Beam length: 8m ■ Concrete cover: c=3.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.482 m; d’=ebz=0.035m



■ Exposure class XC1 ■ Concrete density: 25kN/m3 ■ Reinforcement steel ductility: Class B ■ The calculation is made considering bilinear stress-strain diagram


ckcd 67.10

5,125

===γ


■ MPaff ckctm 56.225*30.0*30.0 3/23/2 ===


■ Steel S400B : MPaf

fs

ykyd 83.347

15,1400

===γ


Boundary conditions


■ Outer: ► Support at start point (x = 0) restrained in translation along X, Y and Z, ► Support at end point (x = 8) restrained in translation along Y and Z, and restrained rotation along X.

■ Inner: None.


545

Loading



Cmax = 1.35 x G + 1.5 x Q=1.35*0+1.5*18.2=27.3kN/ml


CCQ = 1.0 x G + 1.0 x Q=0+18.2=18.2kN/ml


kNmMEd 40.2188

²8*3.27==

kNmMEcq 60.1458

²8*20.18==

5.61.2.2 Reference results in calculating the concrete beam moment At first it will be determined the moment resistance of the concrete section only:

kNmfh

dhbM cdf

feffbtu 92867.16*212.0482.0*12.0*10.1*

2** =⎟

⎠⎞

⎜⎝⎛ −==⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Comparing Mbtu with MEd:

Therefore, the concrete section is not entirely compressed; This requires a calculation considering a rectangular section of b=110cm and d=48.2cm.

Reference longitudinal reinforcement calculation:

051.067.16*²482.0*10.1

218.0*²*1 ===

cdw

Edcu Fdb

Mμ

[ ] ( )[ ] 066.0051.0*211*25.1)*21(1*25.1 =−−=−−= cuu μα

mdz uc 469.0)066.0*40.01(*482.0)*4.01(* =−=−= α

²38.1383.347*469.0

218.0*

cmfz

MAydc

Ed ===



kNmMkNmM btuEd 92840.218 =<=


546




For Class B reinforcement steel ductility (reference value: A=13.38cm2)


Result name Result description Reference value Az (Class B) Theoretical reinforcement area [cm2] 13.38 cm2




547

5.62 EC2 Test 8: Verifying a rectangular concrete beam without compressed reinforcement – Inclined stress-strain diagram

Test ID: 4981

Test status: Passed


The purpose of the test is to verify the results using a constitutive law for reinforcement steel, on the inclined stress-strain diagram.



- The longitudinal reinforcement corresponding to Class A reinforcement steel ductility

- The minimum reinforcement percentage

5.62.2 Background Verifies the adequacy of a rectangular cross section made from concrete C30/37 to resist simple bending. During this test, the calculation of stresses is made along with the calculation of the longitudinal reinforcement and the verification of the minimum reinforcement percentage.

The purpose of this test is to verify the software results for using a constitutive law for reinforcement steel, on the inclined stress-strain diagram.


■ Loadings from the structure: G = 25 kN/m (including the dead load) ■ Exploitation loadings (category A): Q = 30kN/m,

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.6 x Q ■ Reinforcement steel: Class A ■ The calculation is performed considering inclined stress-strain diagram ■ Cracking calculation required The objective is to verify:

■ The stresses results ■ The longitudinal reinforcement corresponding to doth Class A reinforcement steel ductility ■ The minimum reinforcement percentage


548


Units

Metric System

Geometry


■ Height: h = 0.65 m, ■ Width: b = 0.28 m, ■ Length: L = 6.40 m, ■ Section area: A = 0.182 m2 , ■ Concrete cover: c=4.50 cm ■ Effective height: d=h-(0.6*h+ebz)=0.806 m; d’=ebz=0.045m


Rectangular solid concrete C30/37 and S500A reinforcement steel is used. The following characteristics are used in relation to this material:

■ Exposure class XD3 ■ Concrete density: 25kN/m3 ■ Reinforcement steel ductility: Class A ■ The calculation is performed considering inclined stress-strain diagram ■ Cracking calculation required


ckcd 20

5,130

===γ


■ MPaff ckctm 90.230*30.0*30.0 3/23/2 ===


■ MPa*f

*E..

ckcm 32837

1083022000

108

220003030

=⎟⎠

⎞⎜⎝

⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

■ Steel S500 : MPa,,

ff

s

ykyd 78434

151500

γ===



549

Boundary conditions


■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z, ► Support at end point (x = 6.40) restrained in translation along Y and Z, and restrained rotation along X.

■ Inner: None.

Loading



Cmax = 1.35 x G + 1.5 x Q=1.35*(25)+1.5*30=78.75kN/ml


CCQ = 1.0 x G + 1.0 x Q=25+30=55kN/ml

Quasi-permanent combination of actions:

CQP = 1.0 x G + 0.6 x Q=25+0.6*30=43kNm


kNmMEd 20.4038

²40.6*75.78==

kNmMEcq 60.2818

²40.6*55==

kNmMEqp 16.2208

²40.6*43==

5.62.2.2 Reference results in calculating the equivalent coefficient To determine the equivalence coefficient, we must first estimate the creep coefficient, related to elastic deformation at 28 days and 50% humidity:

)(*)(*),( 00 tft cmRH ββϕϕ =∞



at t0=28 days



550

The value of the φRH coefficient depends of the concrete quality:

2130

αα10

1001

1 **h*.

RH

RH⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.3a

1αα 21 == if MPafCM 35=

If not: 70

135α

.

cmf ⎟⎟⎠

⎞⎜⎜⎝

⎛= and

20

235α

.

cmf ⎟⎟⎠

⎞⎜⎜⎝

⎛=

According to: EC2 Part 1,1 EN 1992-1-1-2002; Annex B; Chapter B.1(1); B.8c

In this case, MpaMpaff ckcm 388 =+=

9440383535α

7070

1 .f

..

cm=⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

9840383535α

2020

2 .f

..

cm=⎟

⎠

⎞⎜⎝

⎛=⎟⎟⎠

⎞⎜⎜⎝

⎛=

7819840719510

100501

17195650280265028022

30 ..*.*.

mm.)(*

**uAc*h RH =

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=⇒=

+==


3724880732781ββ 00 ..*.*.)t(*)f(*)t,( cmRH ===∞

The coefficient of equivalence is determined using the following formula:

4017

60281162203721

32837200000

1

α

0

.

.

.*.MM

*)t,(

EE

Ecar

Eqp

cm

se =

+

=

∞+

=

5.62.2.3 Reference results in calculating the concrete beam reduced moment limit Due to exposure class XD3, we will verify the section having non-compressed steel reinforcement, determining the reduced moment limit, using the formula defined by Jeans Roux in his book “Practice of EC2”:

This value can be determined by the next formula if MPafck 50< valid for a constitutive law to horizontal plateau:

)*62.7969.165(*)*66.162.4(*)(

γγαμ

−+−=

ck

ckeluc f

fK

( ) ( )24 ***10 eee cbaK ααα ++= −

The values of the coefficients "a", "b" and "c" are defined in the following table:

Diagram for inclined tier Diagram for horizontal plateau

a 8,189*3,75 −ckf 108*2,71 +ckf

b 5,874*6,5 +− ckf 4,847*2,5 +− ckf

c 13*04,0 −ckf 5,12*03,0 −ckf


551

This gives us:

22069818925375 ..*.a =+=

570658743025 ..*,b =+−=

8111330040 .*,c −=−=

( ) ( ) 087.1²60.17*80.1160.17*5.7062.206910 4 =−+= −eK α

Then:

425,12.30776.437

==γ

272.0)*62.7969.165(*)*66.162.4(

).( =−+−

=γγ

αμck

ckeluc f

fK

Calculation of reduced moment:

225,020*²566.0*50,0

10*403.0*²* 2

3

===MPamm

Nmfdb

M

cdw

Edcuμ

272.0225.0 =<= luccu μμ therefore, there is no compressed reinforcement

Reference reinforcement calculation at SLS:

The calculation of the reinforcement is detailed below:

■ Effective height: d = 0.566m ■ Calculation of reduced moment:

225.0=cuμ

■ Calculation of the lever arm zc:

mdz uc 493,0)323,0*4,01(*566,0)*4,01(* =−=−= α

■ Calculation of the reinforcement area:

²68,11²10*68,1178,434*458,0

233,0*

4 cmmfz

MAydc

Edu ==== −

■ Calculation of the stresses from the tensioned reinforcement:

Reinforcement steel elongation:

35.75.3*323,0

323,01*12 =

−=

−= cu

u

usu ε

ααε ‰

Tensioned reinforcement efforts:

MPaMPasusu 45471.43900735.0*38,95271,432*38,95271,432 ≤=+=+= εσ

■ Reinforcement section calculation:

²60.18²10*60.1871.439*493.0

403.0*

4 cmmfz

MAydc

Edu ==== −


552

Reference solution for minimal reinforcement area

The minimum percentage for a rectangular beam in pure bending is:

⎪⎩

⎪⎨

⎧=

d*b*.

d*b*f

f*.

MaxA

w

wyk

eff,ct

min,s00130

260

According to: EC2 Part 1,1 EN 1992-1-1-2002 Chapter 9.2.1.1(1); Note 2

Where:

MPa.ff ctmeff,ct 902== from cracking conditions

Therefore:

⎪⎩

⎪⎨⎧

==

==−

−

²39.2²10*06.2566.0*28.0*0013.0

²10*39.2566.0*28.0*500

90.2*26.0max4

4

min, cmm

mAs



ULS and SLS load combinations (kNm)


ULS (reference value: 403.20kNm)

SLS (reference value: 281.60kNm)


For Class A reinforcement steel ductility (reference value: A=18.60cm2)


553

Minimum reinforcement area(cm2)



Result name Result description Reference value My,ULS My corresponding to the 101 combination (ULS) [kNm] 403.20 kNm My,SLS My corresponding to the 102 combination (SLS) [kNm] 281.60 kNm Az (Class A) Theoretical reinforcement area [cm2] 18.60 cm2

Amin Minimum reinforcement area [cm2] 2.39 cm2


Result name Result description Value Error My My USL -394.8 kN*m 2.0833 % My My SLS -275.733 kN*m 2.0833 % Az Az -18.4844 cm² 0.6248 % Amin Amin -2.38697 cm² 0.0001 %


554

5.63 EC2 Test 4: Verifying a rectangular concrete beam subjected to Pivot A efforts – Inclined stress-strain diagram

Test ID: 4977

Test status: Passed


The purpose of this test is to verify the software results for Pivot A efforts. For these tests, the constitutive law for reinforcement steel, on the inclined stress-strain diagram is applied.



- The longitudinal reinforcement corresponding to both Class A and Class B reinforcement steel ductility

- The minimum reinforcement percentage


555

5.64 EC2 Test 39: Verifying a circular concrete column using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1)

Test ID: 5146

Test status: Passed

5.64.1 Description Verifies the adequacy of a concrete (C25/30) column with circular cross section using the simplified method – Professional rules - Bilinear stress-strain diagram (Class XC1).

Simplified Method

The column is considered connected to the ground by an articulated connection (all the translations are blocked). To the top part the translations along X and Y axis are blocked and the rotation along the Z axis is also blocked.

5.64.2 Background Simplified Method

Verifies the adequacy of a circular cross section made from concrete C25/30.


■ Loadings from the structure: ► 5000kN axial force ► The self-weight is neglected

■ Concrete cover 5cm ■ Concrete C25/30 ■ Steel reinforcement S500B ■ Relative humidity RH=50% ■ Buckling length L0=5.00m


556

Units

Metric System

Geometry


■ Radius: r = 0.40 m, ■ Length: L = 5.00 m, ■ Concrete cover: c=5cm

Boundary conditions


The column is considered connected to the ground by a articulated connection (all the translations are blocked) and to the top part the translations along X and Y axis are blocked and the rotation along Z axis is also blocked.

Loading


■ Load combinations:

kNNED 67505000*35.1 ==


Scope of the method:

According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(1)

L0 :buckling length L0=L=5m

According to Eurocode 2 EN 1992-1-1 (2004): Chapter 5.8.3.2(2)

i :the radius of giration of uncraked concrete section

I : second moment of inertia for circular cross sections

A : cross section area

, therefore:


557

The method of professional rules can be applied as:

■

■

■

Reinforcement calculation:

6025 ≤=λ , therefore:

722.0

62251

84.0

621

84.022 =

⎟⎠⎞

⎜⎝⎛+

=

⎟⎠⎞

⎜⎝⎛+

=λ

α

)**81(*)*5.070.0( δρ−+= Dkh

If the ρ and δ values are unknown, the term will be considered:

95.0)**81( =− δρ , therefore:

045.195.0*)8.0*5.070.0()**61(*)*5.070.0( =+=−+= δρDkh

95.0500500*65.06.1

500*65.06.1 =−=−= yk

s

fk

MPaff ckcd 67.16

5.125

5.1===

MPaf

f ykyd 33.333

5.1500

5.1===

²14.3167.16*4.0*722.0*95.0*045.1

750.6*33.333

1****

*1 22 cmfrkkN

fA cd

sh

ed

yds =⎟

⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−= ππ

α






558


Result name Result description Reference value Ay Reinforcement area [cm2] 17.43cm2





559

5.65 EC2 Test 43: Verifying a square concrete column subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)

Test ID: 5211

Test status: Passed

5.65.1 Description Verifies the adequacy of a square cross section column made of concrete C30/37 subjected to a small rotation moment and significant compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.

Nominal curvature method.



5.65.2 Background Nominal curvature method.

Verify the adequacy of a rectangular cross section made from concrete C30/37.



■ Loadings from the structure: ► 150 kN axial force ► 15 kMm rotation moment applied to the column top ► the self-weight is neglected

■ Exploitation loadings: ► 100 kN axial force ► 7 kNm rotation moment applied to the column top

■ ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Concrete cover 3cm and 5cm ■ Transversal reinforcement spacing a=40cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ The column is considered isolated and braced


560

Units

Metric System

Geometry


■ Height: h = 0.50 m, ■ Width: b = 0.50 m, ■ Length: L = 5.80 m, ■ Concrete cover: c = 5 cm

Boundary conditions



Loading



NEd =1.35*150+1.50*100=352.50kN=0.353MN

MEd=1.35*15+1.50*7=30.75kNm=0.03075MNm

■


561




mll 60.11*20 ==



37.8050.0

60.11*32*32 0 ===alλ



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:





ieee += 01

mei 03.0=


meNM

eee iEqp

Eqpi 125.030.0

100*30.01507*30.015

0

001 =+

++

=+=+=

kNNEqp 180100*30.01501 =+=

MNmkNmeNM EqpEqp 0225.050.22125.0*180* 111 ====


MNmMEd 041.01 =


)(*)(*),( 00 tft cmRH ββϕϕ =∞

72.2830

8.168.16)( =+

==cm

cm ffβ

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t



562

213

0

***1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


1 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα and 984.0

383535 2.02.0

2 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα

( ) 72.1984.0*944.0*250*1.0

100501

1250500500*2500*500*2*2

30 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=⇒=

+== RHmm

uAch ϕ

28.2488.0*72.2*72.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ


( ) 25.1041.0

0225.0*28.2*, 0 ==∞=Ed

EQPef M

Mtϕϕ



nCBA ***20

lim =λ


Where:

071.020*²50.0

353.0*

===cdc

Ed

fANn

( ) 80.025.1*2.01

1*2,01

1=

+=

+=

ef

Aϕ



24.46071.0

7.0*1.1*80.0*20lim ==λ

24.4637.80 lim =>= λλ

Therefore, the second order effects cannot be neglected.


563

Calculation of the eccentricities and solicitations corrected for ULS :


NEd= 1.35*150 + 1.50*100= 352.5kN = 0,3525 MN

MEd= 1.35*15 + 1.50*7= 352.5kN= 0.03075MNm

Therefore it must be determined:


mNMeEd

Ed 087.0353.0

03075.00 ===


mlei 03.0400

6.11400

0 ===



MNNEd 353.0=

meee i 117.001 =+=

MNmNeM EdEd 041.0353.0*117.0*1 ===

0*eNM Ed=

⎩⎨⎧

=⎪⎩

⎪⎨⎧

=⎪⎩

⎪⎨⎧

== mmmmmmmm

mmhmm

e 207.16

20max

3050020

max30

20max0



To apply the nominal rigidity method, we need an initial reinforcement area to start from. For this, the concrete section will be sized considering only the first order effect.

The Advance Design calculation is different: it is iterating as many time as necessary starting from the minimum percentage area.

The reinforcement is determined using a compound bending with compressive stress. The determined solicitations were calculated from the center of gravity of the concrete section alone. Those stresses must be reduced to the centroid of tensioned steel:


564

MNmhdNMM Gua 112.0250.045.0*353.0041.0)

2(*0 =⎟

⎠⎞

⎜⎝⎛ −+=−+=

Verification about the partially compressed section:

494,0)45,050,0*4,01(*

45,050,0*8,0)*4,01(**8,0 =−=−=

dh

dh

BCμ

055.020*²45.0*50.0

112.0*²*

===cdw

uacu fdb

Mμ

494.0055.0 =<= BCcu μμ therefore the section is partially compressed

The calculation for the tensioned steel in pure bending:

055.0=cuμ

[ ] 071,0)055,0*21(1*25,1 =−−=uα

mdz uc 437,0)071,0*4,01(*45,0)*4,01(* =−=−= α

²89,578,434*437,0

112,0*

cmfz

MAydc

ua ===

The calculation for the compressed steel in bending:

For the compound bending:

The minimum column percentage reinforcement must be considered:

Therefore a 5cm2 reinforcement area will be considered

5.65.2.3 The nominal curvature method (second order effect):

The curvature calculation:

Considering a reinforcement of 5cm ² (considered symmetric), one can determine the curvature from the following formula:

0

1**1r

KKr r ϕ=

According to Eurocode 2 – EN 1992-1-1(2004): Chapter 5.8.8.3(1)

1

0

0107.045.0*45,0

20000078.434

)*45,0()*45,0(1 −==== m

dEf

drs

yd

ydε


565

Kr is the correction coefficient depending of the normal force:

1≤−−

=balu

ur nn

nnK


0706.020*²50.0

353.0*

===cdc

Ed

fANn

0435.020*²50.0

78.434*10*5** 4

===−

cdc

yds

fAfA

ω

0435.10435.011 =+=+= ωun

4,0=baln

51.140.00435.1

0706.00435.1=

−−

=rK

Condition: 1≤rK , therefore we consider: 1=rK

ϕK creep coefficient: 1*1 ≥+= efK ϕβϕ


,

Therefore:

1

0

0107.00107.0*1*11**1 −=== mr

KKr r ϕ

Calculation moment:

The moment of calculation is defined by the formula:

20 MMM EdEd +=

Where:

EdM 0 is moment of the first order including geometric imperfections.

2M is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

22 *eNM Ed=

mcl

re 144.0

10²60.11*0107.0*1 2

02 ===

MNmeNM Ed 051.0144.0*353.0* 22 ===

MNmMMM EdEd 09.0051.0041.020 =+=+=


566

The reinforcement must be sized considering the demands of the second degree effects, as follows:

Reinforcement calculation according the second order effects:

The interaction diagram will be used.

The input parameters in the diagram are:

According to the diagram, it will be obtained a minimal percentage reinforcement:

2min, 5*002.0²81.0

78.434353.0*10.0*10,0 cmAcm

fNA c

yd

Eds =≥===




567


(reference value: Au=5cm2)


(reference value: 5 cm2)


Result name Result description Reference value Amin Reinforcement area [cm2] 5 cm2

R Theoretical reinforcement area [cm2] 5 cm2


Result name


Amin Amin 5 cm² 0.0000 %


568

5.66 EC2 Test 46 I: Verifying a square concrete beam subjected to a normal force of traction - Inclined stress-strain diagram (Class X0)

Test ID: 5232

Test status: Passed

5.66.1 Description Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Inclined stress-strain diagram (Class X0).

Tie sizing

Inclined stress-strain diagram

Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm


- Support at start point (x=0) fixed connection

- Support at end point (x = 5.00) translation along the Z axis is blocked

5.66.2 Background Tie sizing

Bilinear stress-strain diagram / Inclined stress-strain diagram

Verifies the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.


■ Loadings from the structure: Fx,G = 233.33 kN The dead load is neglected

■ Exploitation loadings (category A): Fx,Q = 56.67kN ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q

Units

Metric System


569

Geometry


■ Height: h = 0.15 m, ■ Width: b = 0.15 m, ■ Length: L = 5.00 m, ■ Section area: A = 0.00225 m2 , ■ Concrete cover: c=3cm ■ Reinforcement S400, Class: B ■ Fck=20MPa ■ The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm

Boundary conditions


■ Outer: ► Support at start point (x=0) fixed connection, ► Support at end point (x = 5.00) translation along the Z axis is blocked

■ Inner: None.

5.66.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then a inclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

²50,11²10*50,11

15.1400400.0 4

,, cmmNA

Us

EdUs ===≥ −

σ

It will be used a 4HA20=A=12.57cm2

Calculations according a inclined stress-strain diagram

MPaClassBS Us 373400 , =⇒− σ

²70,10²10*72,10373400.0 4

,, cmmNA

Us

EdUs ===≥ −

σ

It can be seen that the gain is not negligible (about 7%).

Checking the condition of non-fragility:

yk

ctmcs ffAA *≥

MPaff ckctm 21,220*30.0*30.0 3/23/2 ===

²0225.015.0*15.0 mAc ==

²24.1400

21.2*0225.0* cmffAAyk

ctmcs ==≥


570



ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75 cm2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)


Result name Result description Reference value Az,1 Longitudinal reinforcement obtained using the bilinear stress-strain

diagram [cm2] 5.75 cm2

Az,2 longitudinal reinforcement obtained using the inclined stress-strain diagram [cm2]

5.37 cm2


Result name


Az Az-ii -5.36526 cm² 0.0001 %


571

5.67 EC2 Test 42: Verifying a square concrete column subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1)

Test ID: 5205

Test status: Passed

5.67.1 Description Verifies a square cross section column made of concrete C30/37 subjected to a significant rotation moment and small compression force to the top with Nominal Curvature Method - Bilinear stress-strain diagram (Class XC1).

The verification of the axial stresses and rotation moment, applied on top, at ultimate limit state is performed.

Nominal curvature method.



5.67.2 Background Nominal curvature method.

Verifies the adequacy of a square cross section made from concrete C30/37.





■ 3,02 =ψ

■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q ■ Concrete cover 3cm and 5cm ■ Transversal reinforcement spacing a=40cm ■ Concrete C30/37 ■ Steel reinforcement S500B ■ The column is considered isolated and braced


572

Units

Metric System

Geometry


■ Height: h = 0.50 m, ■ Width: b = 0.50 m, ■ Length: L = 5.80 m, ■ Concrete cover: c=5cm

Boundary conditions



Loading



► NEd =1.35*15+1.50*7=30.75kN=0.03075MN ► MEd=1.35*150+1.50*100=352.50kNm=0.352MNm

■ m...

NM

eEd

Ed 4511030750

35200 ===




mll 60.11*20 ==



37.8050.0

60.11*32*32 0 ===alλ


573



( )Ed

EQPef M

Mt ., 0∞= ϕϕ


Where:





ieee += 01

mei 03.0=


meNM

eee iEqp

Eqpi 56.1030.0

7*30.015100*30.0150

0

001 =+

++

=+=+=

kNNEqp 10.177*30.0151 =+=

MNmkNmeNM EqpEqp 181.058.18056.10*10.17* 111 ====


MNmMEd 3523.01 =


)(*)(*),( 00 tft cmRH ββϕϕ =∞

72.2830

8.168.16)( =+

==cm

cm ffβ

488.0281.0

11.0

1)( 20.020.00

0 =+

=+

=t


213

0

***1.0100

11 ααϕ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=

h

RH

RH


1 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα and 984.0

383535 2.02.0

2 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

cmfα


574

( ) 72.1984.0*944.0*250*1.0

100501

1250500500*2500*500*2*2

30 =⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛ −+=⇒=

+== RHmm

uAch ϕ

28.2488.0*72.2*72.1)(*)(*),( 00 ===∞ tft cmRH ββϕϕ


( ) 17.1352.0181.0*28.2*, 0 ==∞=

Ed

EQPef M

Mtϕϕ




nCBA ***20

lim =λ


Where:

0062.020*²50.0

031.0*

===cdc

Ed

fANn

( ) 81.017.1*2.01

1*2,01

1=

+=

+=

ef

Aϕ



42.1580062.0

7.0*1.1*81.0*20lim ==λ

42.15837.80 lim =<= λλ

Therefore, the second order effects can be neglected.


575



■ NEd= 1.35*15 + 1.50*7= 30.75kN = 0,03075 MN ■ MEd= 1.35*150 + 1.50*100= 352.5kN= 0.3525 MNm Therefore it must be determined:


mNMeEd

Ed 46.1103075.03525.0

0 ===


mlei 03.0400

6.11400

0 ===



MNNEd 03075.0=

meee i 49.1101 =+=

MNmNeM EdEd 353.003075.0*49.11*1 ===

0*eNM Ed=

mm)mm.;mmmax(mm;mmmaxh;mmmaxe 207162030

5002030

200 ==⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛=



The theoretical reinforcement will be determined by the following diagram:


576


141.020*50.0*50.0

353.0** 22 ===

cd

Ed

fhbMμ

00615.020*5.0*5.0

03075.0**

===cd

Ed

fhbNν

Therefore:

35.0=ω


∑ == 22

25.4078.434

20*50.0* cmAsω

which means 20.13cm2 per face.

The total area will be 40.25cm2.

The nominal curvature methods (second order effect):

The curvature calculation:

Considering a reinforcement of 40.25cm² (considered symmetric), one can determine the curvature from the following formula:

0

1**1r

KKr r ϕ=


1

0

0107.045.0*45,0

100078.434

)*45,0()*45,0(1 −==== m

dEf

drs

yd

ydε

Kr is the correction coefficient depending of the normal force:

1≤−−

=balu

ur nn

nnK


00615.020*²50.0

03075.0*

===cdc

Ed

fANn

350.020*²50.0

78.434*10*25.40** 4

===−

cdc

yds

fAfA

ω

350.1350.011 =+=+= ωun

4,0=baln

41.140.0350.1

00615.0350.1=

−−

=rK

Condition: 1≤rK , therefore it will be considered: 1=rK

ϕK creep coefficient: 1*1 ≥+= efK ϕβϕ



577

therefore

Calculation moment:

The moment of calculation is estimated from the formula:

20 MMM EdEd +=

Where:

EdM 0 is moment of the first order including geometric imperfections.

2M is nominal moment of second order.

The 2nd order moment is calculated from the curvature:

22 *eNM Ed=

mcl

re 144.0

10²60.11*0107.0*1 2

02 ===

MNmeNM Ed 00443.0144.0*03075.0* 22 ===

MNmMMM EdEd 357.000443.0353.020 =+=+=

The reinforcement must be sized considering the demands of the second degree effects, as follows:

MNNEd 03075.0=

MNmMEd 357.0=


578

Reinforcement calculation according the second order effects:

The interaction diagram will be used.

The input parameters in the diagram are:

35.0=ω will be obtained, which gives: ∑ ==== ²25.40078.434

20*²50.0*35.0*** cmf

fhbAyd

cds

ω

Meaning a 20.13 cm2 per side (which confirms the initial section)




579


(reference value: 40.25cm2=2*20.13cm2)









580

5.68 EC2 Test 46 II: Verifying a square concrete beam subjected to a normal force of traction - Bilinear stress-strain diagram (Class X0)

Test ID: 5231

Test status: Passed

5.68.1 Description Verifies a square cross section beam made of concrete C20/25 subjected to a normal force of traction - Bilinear stress-strain diagram (Class X0).

Tie sizing

Bilinear stress-strain diagram

Determines the armature of a pulling reinforced concrete, subjected to a normal force of traction.

The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm


- Support at start point (x=0) fixed connection

- Support at end point (x = 5.00) translation along the Z axis is blocked

5.68.2 Background Tie sizing

Bilinear stress-strain diagram / Inclined stress-strain diagram

Determine the armature of a pulling reinforced concrete, C20/25, subjected to a normal force of traction.


■ Loadings from the structure: Fx,G = 233.33 kN The dead load is neglected

■ Exploitation loadings (category A): Fx,Q = 56.67kN ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q


581

Units

Metric System

Geometry


■ Height: h = 0.15 m, ■ Width: b = 0.15 m, ■ Length: L = 5.00 m, ■ Section area: A = 0.00225 m2 , ■ Concrete cover: c=3cm ■ Reinforcement S400, Class: B ■ Fck=20MPa ■ The load combinations will produce the following rotation efforts:

NEd=1.35*233.3+1.5+56.67=400kNm

Boundary conditions


■ Outer: ► Support at start point (x = 0) fixed connection, ► Support at end point (x = 5.00) translation along the Z axis is blocked

■ Inner: None.

5.68.2.2 Reference results in calculating the concrete beam There will be two successive calculations, considering a bilinear stress-strain diagram constitutive law and then a inclined stress-strain diagram constitutive law.

Calculations according a bilinear stress-strain diagram

²50,11²10*50,11

15.1400400.0 4

,, cmmNA

Us

EdUs ===≥ −

σ

It will be used a 4HA20=A=12.57cm2

Calculations according a inclined stress-strain diagram

MPaClassBS Us 373400 , =⇒− σ

²70,10²10*72,10373400.0 4

,, cmmNA

Us

EdUs ===≥ −

σ

It can be seen that the gain is not negligible (about 7%).


582

Checking the condition of non-fragility:

yk

ctmcs ffAA *≥

MPaff ckctm 21,220*30.0*30.0 3/23/2 ===

²0225.015.0*15.0 mAc ==

²24.1400

21.2*0225.0* cmffAAyk

ctmcs ==≥



ULS load combinations (kNm)

In case of using the bilinear stress-strain diagram, the reinforcement will result: (Az=11.50cm2=2*5.75 cm2)

In case of using the inclined stress-strain diagram, the reinforcement will result: (Az=10.70cm2=2*5.35 cm2)


Result name Result description Reference value Az,1 Longitudinal reinforcement obtained using the bilinear stress-strain

diagram [cm2] 5.75 cm2

Az,2 Longitudinal reinforcement obtained using the inclined stress-strain diagram [cm2]

5.37 cm2


Result name Result description Value Error Az Az-i -5.75001 cm² 0.0000 %

6 General Application


584

6.1 Verifying 2 joined vertical elements with the clipping option enabled (TTAD #12238)

Test ID: 4480

Test status: Passed

6.1.1 Description Performs the finite elements calculation on a model with 2 joined vertical elements with the clipping option enabled.

6.2 Verifying the precision of linear and planar concrete covers (TTAD #12525)

Test ID: 4547

Test status: Passed

6.2.1 Description Verifies the linear and planar concrete covers precision.

6.3 Defining the reinforced concrete design assumptions (TTAD #12354)

Test ID: 4528

Test status: Passed

6.3.1 Description Verifies the definition of the reinforced concrete design assumptions.

6.4 Verifying the synthetic table by type of connection (TTAD #11422)

Test ID: 3646

Test status: Passed

6.4.1 Description Performs the finite elements calculation and the steel calculation on a structure with four types of connections. Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. The model connections: columns base plates, beam - column fixed connections, beam - beam fixed connections and gusset plate. Live loads, snow loads and wind loads are applied.

6.5 Importing a cross section from the Advance Steel profiles library (TTAD #11487)

Test ID: 3719

Test status: Passed

6.5.1 Description Imports the Canam Z 203x10.0 section from the Advance Steel Profiles library in the Advance Design list of available cross sections.


585

6.6 Creating and updating model views and post-processing views (TTAD #11552)

Test ID: 3820

Test status: Passed

6.6.1 Description Creates and updates the model view and the post-processing views for a simple steel frame.

6.7 Creating system trees using the copy/paste commands (DEV2012 #1.5)

Test ID: 4164

Test status: Passed

6.7.1 Description Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is in the source system.

6.8 Creating system trees using the copy/paste commands (DEV2012 #1.5)

Test ID: 4162

Test status: Passed

6.8.1 Description Creates system trees using the copy/paste commands in the Pilot. The source system consists of several subsystems. The target system is on the same level as the source system.

6.9 Creating a new Advance Design file using the "New" command from the "Standard" toolbar (TTAD #12102)

Test ID: 4095

Test status: Passed

6.9.1 Description Creates a new .fto file using the "New" command from the "Standard" toolbar. Creates a rigid punctual support and tests the CAD coordinates.

6.10 Verifying mesh, CAD and climatic forces - LPM meeting

Test ID: 4079

Test status: Passed

6.10.1 Description Generates the "Description of climatic loads" report for a model consisting of a concrete structure with dead loads, wind loads, snow loads and seism loads. Performs the model verification, meshing and finite elements calculation. Generates the "Displacements of linear elements by element" report.


586

6.11 Generating liquid pressure on horizontal and vertical surfaces (TTAD #10724)

Test ID: 4361

Test status: Passed

6.11.1 Description Generates liquid pressure on a concrete structure with horizontal and vertical surfaces. Generates the loadings description report.

6.12 Changing the default material (TTAD #11870)

Test ID: 4436

Test status: Passed

6.12.1 Description Selects a different default material for concrete elements.

6.13 Verifying the objects rename function (TTAD #12162)

Test ID: 4229

Test status: Passed

6.13.1 Description Verifies the renaming function from the properties list of a linear element. The name contains the "_" character.

6.14 Launching the verification of a model containing steel connections (TTAD #12100)

Test ID: 4096

Test status: Passed

6.14.1 Description Launches the verification function for a model containing a beam-column steel connection.

6.15 Verifying the appearance of the local x orientation legend (TTAD #11737)

Test ID: 4109

Test status: Passed

6.15.1 Description Verifies the color legend display. On a model with a planar element, the color legend is enabled; it displays the elements by the local x orientation color.


587

6.16 Verifying geometry properties of elements with compound cross sections (TTAD #11601)

Test ID: 3546

Test status: Passed

6.16.1 Description Verifies the geometry properties of a steel structure with elements which have compound cross sections (CS4 IPE330 UPN240). Generates the geometrical data report.

6.17 Verifying material properties for C25/30 (TTAD #11617)

Test ID: 3571

Test status: Passed

6.17.1 Description Verifies the material properties on a model with a concrete (C25/30) bar. Generates the material description report.

6.18 Verifying element creation using commas for coordinates (TTAD #11141)

Test ID: 4554

Test status: Passed

6.18.1 Description Verifies element creation using commas for coordinates in the command line.

7 Import / Export


590

7.1 Verifying the export of a linear element to GTC (TTAD #10932, TTAD #11952)

Test ID: 3628

Test status: Passed

7.1.1 Description Exports a linear element to GTC.

7.2 Exporting an Advance Design model to DO4 format (DEV2012 #1.10)

Test ID: 4101

Test status: Passed

7.2.1 Description Launches the "Export > Text file" command and saves the current project as a .do4 archive file.

The model contains all types of structural elements, loads and geometric objects.

7.3 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)

Test ID: 4193

Test status: Passed

7.3.1 Description Exports the analysis model to ADA (through GTC) with:

- Export results: enabled

- Export meshed model: disabled

7.4 Importing GTC files containing elements with haunches from SuperSTRESS (TTAD #12172)

Test ID: 4297

Test status: Passed

7.4.1 Description Imports a GTC file from SuperSTRESS. The file contains steel linear elements with haunch sections.

7.5 Exporting an analysis model to ADA (through GTC) (DEV2012 #1.3)

Test ID: 4195

Test status: Passed

7.5.1 Description Exports the analysis model to ADA (through GTC) with:

- Export results: disabled

- Export meshed model: enabled


591

7.6 Exporting linear elements to IFC format (TTAD #10561)

Test ID: 4530

Test status: Passed

7.6.1 Description Exports to IFC format a model containing linear elements having sections of type "I symmetric" and "I asymmetric".

7.7 Importing IFC files containing continuous foundations (TTAD #12410)

Test ID: 4531

Test status: Passed

7.7.1 Description Imports an IFC file containing a continuous foundation (linear support) and verifies the element display.

7.8 Importing GTC files containing "PH.RDC" system (TTAD #12055)

Test ID: 4548

Test status: Passed

7.8.1 Description Imports a GTC file exported from Advance Design. The file contains the automatically created system "PH.RDC".

7.9 Exporting a meshed model to GTC (TTAD #12550)

Test ID: 4552

Test status: Passed

7.9.1 Description Exports a meshed model to GTC. The meshed planar element from the model contains a triangular mesh.

7.10 Verifying the load case properties from models imported as GTC files (TTAD #12306)

Test ID: 4515

Test status: Passed

7.10.1 Description Performs the finite elements calculation on a model with dead load cases and exports the model to GTC. Imports the GTC file to verify the load case properties.


592

7.11 Verifying the releases option of the planar elements edges after the model was exported and imported via GTC format (TTAD #12137)

Test ID: 4506

Test status: Passed

7.11.1 Description Exports to GTC a model with planar elements on which the edges releases were defined. Imports the GTC file to verify the planar elements releases option.

7.12 System stability when importing AE files with invalid geometry (TTAD #12232)

Test ID: 4479

Test status: Passed

7.12.1 Description Imports a complex model containing elements with invalid geometry.

7.13 Verifying the GTC files exchange between Advance Design and SuperSTRESS (DEV2012 #1.9)

Test ID: 4445

Test status: Passed

7.13.1 Description Verifies the GTC files exchange (import/export) between Advance Design and SuperSTRESS.

7.14 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)

Test ID: 4388

Test status: Passed

7.14.1 Description Imports a GTC file from SuperSTRESS. The file contains elements with circular hollow section. Verifies if the cross sections are imported from the attached "UK Steel Sections" database.

7.15 Importing GTC files containing elements with circular hollow sections, from SuperSTRESS (TTAD #12197)

Test ID: 4389

Test status: Passed

7.15.1 Description Imports a GTC file from SuperSTRESS when the "UK Steel Sections" database is not attached. The file contains elements with circular hollow section. Verifies the cross sections definition.

8 Connection Design


594

8.1 Deleting a welded tube connection - 1 gusset bar (TTAD #12630)

Test ID: 4561

Test status: Passed

8.1.1 Description Deletes a welded tube connection - 1 gusset bar after the joint was exported to ADSC.

8.2 Creating connections groups (TTAD #11797)

Test ID: 4250

Test status: Passed

8.2.1 Description Verifies the connections groups function.

9 Mesh


596

9.1 Verifying the mesh for a model with generalized buckling (TTAD #11519)

Test ID: 3649

Test status: Passed

9.1.1 Description Performs the finite elements calculation and verifies the mesh for a model with generalized buckling.

9.2 Verifying mesh points (TTAD #11748)

Test ID: 3458

Test status: Passed

9.2.1 Description Performs the finite elements calculation and verifies the mesh nodes of a concrete structure.

The structure consists of concrete linear elements (R20*20 cross section) and rigid supports; the loads applied on the structure: dead loads, live loads, wind loads and snow loads, according to Eurocodes.

9.3 Creating triangular mesh for planar elements (TTAD #11727)

Test ID: 3423

Test status: Passed

9.3.1 Description Creates a triangular mesh on a planar element with rigid supports and self weight.

10 Reports Generator


598

10.1 Verifying the modal analysis report (TTAD #12718)

Test ID: 4576

Test status: Passed

10.1.1 Description Generates and verifies the modal analysis report.

10.2 Verifying the shape sheet strings display (TTAD #12622)

Test ID: 4559

Test status: Passed

10.2.1 Description Verifies the shape sheet strings display for a steel beam with circular hollow cross-section.

10.3 Verifying the shape sheet for a steel beam (TTAD #12455)

Test ID: 4535

Test status: Passed

10.3.1 Description Verifies the shape sheet for a steel beam.

10.4 Verifying the global envelope of linear elements stresses (on the start point of super element) (TTAD #12230)

Test ID: 4502

Test status: Passed

10.4.1 Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the start point of super element by generating the "Global envelope of linear elements stresses report".

The model consists of a concrete portal frame with rigid fixed supports.

10.5 Verifying the Max row on the user table report (TTAD #12512)

Test ID: 4558

Test status: Passed

10.5.1 Description Verifies the Max row on the user table report.


599

10.6 Verifying the steel shape sheet display (TTAD #12657)

Test ID: 4562

Test status: Passed

10.6.1 Description Verifies the steel shape sheet display when the fire calculation is disabled.

10.7 Verifying the EC2 calculation assumptions report (TTAD #11838)

Test ID: 4544

Test status: Passed

10.7.1 Description Verifies the EC2 calculation assumptions report.

10.8 Verifying the shape sheet report (TTAD #12353)

Test ID: 4545

Test status: Passed

10.8.1 Description Generates and verifies the shape sheet report.

10.9 Verifying the model geometry report (TTAD #12201)

Test ID: 4467

Test status: Passed

10.9.1 Description Generates the "Model geometry" report to verify the model properties: total weight, largest structure dimensions, center of gravity.

10.10 Verifying the global envelope of linear elements forces result (on end points and middle of super element) (TTAD #12230)

Test ID: 4488

Test status: Passed

10.10.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on end points and middle of super element by generating the "Global envelope of linear elements forces result report".



600

10.11 Verifying the global envelope of linear elements forces result (on start and end of super element) (TTAD #12230)

Test ID: 4489

Test status: Passed

10.11.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start and end of super element by generating the "Global envelope of linear elements forces result report".


10.12 Verifying the global envelope of linear elements stresses (on the end point of super element) (TTAD #12230, TTAD #12261)

Test ID: 4503

Test status: Passed

10.12.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on the end point of super element by generating the "Global envelope of linear elements stresses report".


10.13 Verifying the global envelope of linear elements displacements (on each 1/4 of mesh element) (TTAD #12230)

Test ID: 4492

Test status: Passed

10.13.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on each 1/4 of mesh element by generating the "Global envelope of linear elements displacements report".


10.14 Verifying the global envelope of linear elements displacements (on all quarters of super element) (TTAD #12230)

Test ID: 4493

Test status: Passed

10.14.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on all quarters of super element by generating the "Global envelope of linear elements displacements report".



601

10.15 Verifying the global envelope of linear elements forces result (on all quarters of super element) (TTAD #12230)

Test ID: 4487

Test status: Passed

10.15.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on all quarters of super element by generating the "Global envelope of linear elements forces result report".


10.16 Verifying the global envelope of linear elements displacements (on the start point of super element) (TTAD #12230)

Test ID: 4496

Test status: Passed

10.16.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the start point of super element by generating the "Global envelope of linear elements displacements report".


10.17 Verifying the global envelope of linear elements displacements (on the end point of super element) (TTAD #12230, TTAD #12261)

Test ID: 4497

Test status: Passed

10.17.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on the end point of super element by generating the "Global envelope of linear elements displacements report".


10.18 Verifying the global envelope of linear elements forces result (on the end point of super element) (TTAD #12230, #12261)

Test ID: 4491

Test status: Passed

10.18.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the end point of super element by generating the "Global envelope of linear elements forces result report".



602

10.19 Verifying the global envelope of linear elements displacements (on start and end of super element) (TTAD #12230)

Test ID: 4495

Test status: Passed

10.19.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on start and end of super element by generating the "Global envelope of linear elements displacements report".


10.20 Verifying the global envelope of linear elements forces result (on each 1/4 of mesh element) (TTAD #12230)

Test ID: 4486

Test status: Passed

10.20.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on each 1/4 of mesh element by generating the "Global envelope of linear elements forces result report".


10.21 Verifying the global envelope of linear elements forces result (on the start point of super element) (TTAD #12230)

Test ID: 4490

Test status: Passed

10.21.1Description Performs the finite elements calculation and verifies the global envelope of linear elements forces results on the start point of super element by generating the "Global envelope of linear elements forces result report".


10.22 Verifying the global envelope of linear elements displacements (on end points and middle of super element) (TTAD #12230)

Test ID: 4494

Test status: Passed

10.22.1Description Performs the finite elements calculation and verifies the global envelope of linear elements displacements on end points and middle of super element by generating the "Global envelope of linear elements displacements report".



603

10.23 Verifying the global envelope of linear elements stresses (on each 1/4 of mesh element) (TTAD #12230)

Test ID: 4498

Test status: Passed

10.23.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on each 1/4 of mesh element by generating the "Global envelope of linear elements stresses report".


10.24 Verifying the global envelope of linear elements stresses (on start and end of super element) (TTAD #12230)

Test ID: 4501

Test status: Passed

10.24.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on start and end of super element by generating the "Global envelope of linear elements stresses report".


10.25 Verifying the global envelope of linear elements stresses (on end points and middle of super element) (TTAD #12230)

Test ID: 4500

Test status: Passed

10.25.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on end points and middle of super element by generating the "Global envelope of linear elements stresses report".


10.26 Verifying the Min/Max values from the user reports (TTAD# 12231)

Test ID: 4505

Test status: Passed

10.26.1Description Performs the finite elements calculation and generates a user report containing the results of Min/Max values.


604

10.27 Verifying the global envelope of linear elements stresses (on all quarters of super element) (TTAD #12230)

Test ID: 4499

Test status: Passed

10.27.1Description Performs the finite elements calculation and verifies the global envelope of linear elements stresses on all quarters of super element by generating the "Global envelope of linear elements stresses report".


10.28 Creating the rules table (TTAD #11802)

Test ID: 4099

Test status: Passed

10.28.1Description Generates the "Rules description" report as .rtf and .txt file.

The model consists of a steel structure with supports and a base plate connection. Two rules were defined for the steel calculation.

10.29 Creating the steel materials description report (TTAD #11954)

Test ID: 4100

Test status: Passed

10.29.1Description Generates the "Steel materials" report as a .txt file.

The model consists of a steel structure with supports and a base plate connection.

10.30 System stability when the column releases interfere with support restraints (TTAD #10557)

Test ID: 3717

Test status: Passed

10.30.1Description Performs the finite elements calculation and generates the systems description report for a structure which has column releases that interfere with the supports restraints.

The structure consists of steel beams and steel columns (S235 material, HEA550 cross section) with rigid fixed supports.


605

10.31 Generating the critical magnification factors report (TTAD #11379)

Test ID: 3647

Test status: Passed

10.31.1Description Performs the finite elements calculation and the steel calculation on a structure with four types of connections. Generates the "Synthetic table by type of connection" report.

The structure consists of linear steel elements (S275) with CE505, IPE450, IPE140 and IPE500 cross section. The model connections: columns base plates, beam - column fixed connections, beam - beam fixed connections and gusset plate. Live loads, snow loads and wind loads are applied.

10.32 Modal analysis: eigen modes results for a structure with one level

Test ID: 3668

Test status: Passed

10.32.1Description Performs the finite elements calculation and generates the "Characteristic values of eigen modes" report.

The one-level structure consists of linear and planar concrete elements with rigid supports. A modal analysis is defined.

10.33 Generating a report with modal analysis results (TTAD #10849)

Test ID: 3734

Test status: Passed

10.33.1Description Generates a report with modal results for a model with seismic actions.

11 Seismic Analysis


608

11.1 Verifying the spectrum results for EC8 seism (TTAD #11478)

Test ID: 3703

Test status: Passed

11.1.1 Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 French standards.

11.2 Verifying the spectrum results for EC8 seism (TTAD #12472)

Test ID: 4537

Test status: Passed

11.2.1 Description Performs the finite elements calculation and generates the "Displacements of linear elements by load case" report for a concrete beam with rectangular cross section R20*30 with fixed rigid punctual support. Model loads: self weight and seismic loads according to Eurocodes 8 Romanian standards (SR EN 1998-1/NA).

11.3 Verifying the combinations description report (TTAD #11632)

Test ID: 3544

Test status: Passed

11.3.1 Description Generates a dead load case, two live load cases and a snow load case, defines the concomitance between the generated load cases and generates the combinations description report.

11.4 EC8 : Verifying the displacements results of a linear element according to Czech seismic standards (CSN EN 1998-1) (DEV2012 #3.18)

Test ID: 3626

Test status: Passed

11.4.1 Description Verifies the displacements results of an inclined linear element according to Eurocodes 8 Czech standard. Performs the finite elements calculation and generates the displacements of linear elements by load case and by element reports.

The concrete element (C20/25) has R20*30 cross section and a rigid point support. The loads applied on the element: self weight and seismic loads (CSN EN 1998-1).

11.5 Verifying signed concomitant linear elements envelopes on Fx report (TTAD #11517)

Test ID: 3576

Test status: Passed

11.5.1 Description Performs the finite elements calculation on a concrete structure. Generates the "Signed concomitant linear elements envelopes on Fx report".

The structure has concrete beams and columns, two concrete walls and a windwall. Loads applied on the structure: self weight and a planar live load of -40.00 kN.


609

11.6 EC8 French Annex: verifying torsors on walls

Test ID: 4803

Test status: Passed

11.6.1 Description Verifies the torsors on walls. Eurocode 8 with French Annex is used.

11.7 EC8 French Annex: verifying torsors on grouped walls from a multi-storey concrete structure

Test ID: 4810

Test status: Passed

11.7.1 Description Verifies torsors on grouped walls from a multi-storey concrete structure. EC8 with French Annex is used.

11.8 Verifying the damping correction influence over the efforts in supports (TTAD #13011).

Test ID: 4853

Test status: Passed

11.8.1 Description Verifies the damping correction influence over the efforts in supports. The model has 2 seismic cases. Only one case uses the damping correction. The seismic spectrum is generated according to the Eurocodes 8 - French standard (NF EN 1993-1-8/NA).

11.9 EC8 French Annex: verifying seismic results when a design spectrum is used (TTAD #13778)

Test ID: 5425

Test status: Passed

11.9.1 Description Verifies the seismic results according to EC8 French Annex for a single bay single story structure made of concrete.

11.10 EC8: verifying the sum of actions on supports and nodes restraints (TTAD #12706)

Test ID: 4859

Test status: Passed

11.10.1Description Verifies the sum of actions on supports and nodes restraints for a simple structure subjected to seismic action according to EC8 French annex.

11.11 Seismic norm PS92: verifying efforts and torsors on planar elements (TTAD #12974)

Test ID: 4858

Test status: Passed

11.11.1Description Verifies efforts and torsors on several planar elements of a concrete structure subjected to horizontal seismic action (according to PS92 norm).


610

11.12 EC8 Fr Annex: Generating forces results per modes on linear and planar elements (TTAD #13797)

Test ID: 5455

Test status: Passed

11.12.1Description Generates reports with forces results per modes on a selection of elements (linear and planar elements) from a concrete structure subjected to seismic action (EC8 French Annex).

12 Steel Design


612

12.1 EC3 Test 2: Class section classification and share verification of an IPE300 beam subjected to linear uniform loading

Test ID: 5410

Test status: Passed

12.1.1 Description Classification and verification of an IPE 300 beam made of S235 steel.

The beam is subjected to a 50 kN/m linear uniform load applied gravitationally.

The force is considered to be a live load and the dead load is neglected.

12.1.2 Background Classification and verification of sections for an IPE 300 beam made from S235 steel. The beam is subjected to a 50 kN/m linear uniform load applied gravitationally. The force is considered to be a live load and the dead load is neglected.

12.1.2.1 Model description ■ Reference: Guide de validation Eurocode 3 Part1,1 EN 1993-1-1-2001; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Exploitation loadings (category A): Q = -50kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System



613

Boundary conditions


■ Outer: ► Support at start point (x=0) restrained in translation along X, Y and Z axis, ► Support at end point (x = 5.00) restrained in translation along Y and Z axis and rotation restrained on X

axis ■ Inner: None.

12.1.2.2 Reference results for calculating the cross section class According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2

In this case the stresses distribution along the section is like in the picture below:

■ compression for the top flange ■ compression and tension for the web ■ tension for the bottom flange


614

To determine the web class it will be used the Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.

The section geometrical properties are described in the picture below:

According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:


615

Therefore:

This means that the column web is Class 1.

To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2


According to the Table 5.2 and the beam section geometrical properties, the next conclusions can be found:


616

276.57.1045.56

==mmmm

tc

92.0=ε

Therefore:

28.892.0*9*9276.57.1045.56

==≤== εmmmm

tc

this means that the column flanges are Class 1.

A cross-section is classified by quoting the heist (least favorable) class of its compression elements.

According to the calculation above, the beam section have a Class 1 web and Class 1 flanges; therefore the class section for the entire beam section will be considered Class 1.


12.1.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd shall be determined as follows:

0,

3*

M

yv

Rdpl

fA

Vγ

=


Where:

Av: section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=

A: cross-section area A=53.81cm2

b: overall breadth b=150mm

h: overall depth h=300mm

hw: depth of the web hw=248.6mm

r: root radius r=15mm

tf: flange thickness tf=10.7mm

tw: web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=


fy: nominal yielding strength for S275 fy=275MPa

0Mγ : partial safety coefficient 10 =Mγ

Therefore:

kNMN

fA

VM

yv

Rdpl 7.4074077.01

3275*10*68.25

3* 4

0, ====

γ


617

For more:

Verification of the shear buckling resistance for webs without stiffeners:

εη*72≤

w

w

th


20.111*20.1*20.1

0

1 ===M

M

γγη

92.0275235235

===yf

ε

91.9392.020.1*72*7201.35

1.76.248

==≤==εη

w

w

th

There is no need for shear buckling resistance verification

According to: EC3 Part 1,5 EN 1993-1-5-2004 Chapter 5.1(2)




618

Finite elements results

The Advance Design steel calculation results are found in the Shape Sheet of the element:


Result name Result description Reference value Work ratio % 31 %


Result name Result description Value Error Fz Fz -125 kN -0.0000 % Work ratio - Fz Work ratio Fz 0 % 3.2258 %


619

12.2 EC3 Test 6: Class section classification and combined biaxial bending verification of an IPE300 column

Test ID: 5424

Test status: Passed

12.2.1 Description Classification and verification on combined bending of an IPE 300 beam made of S235 steel.

The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis.

The beam is subjected to a -10 kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis.

Both forces are considered as live loads.

The dead load will be neglected.

12.2.2 Background Classification and verification on combined bending of sections for an IPE 300 beam made from S235 steel. The beam is connected to its ends by a connection with all translation blocked and on the other end by a connection with translation blocked on the Y and Z axis and rotation blocked along X axis. The beam is subjected to a -10kN/m uniform linear force applied along the beam gravitational along the Z local axis, and a 10kN/m uniform linear force applied along the beam on the Y axis. Both forces are considered live loads. The dead load will be neglected.


■ Exploitation loadings (category A): Q1 = -10kN/m, Q2 = 10kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


620

Units

Metric System


Boundary conditions


■ Outer: ► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (x = 5.00) restrained in translation and rotation along Y, Z axis and rotation

blocked along X axis. ■ Inner: None.


In this case, the beam is subjected to linear uniform equal loads, one vertical and one horizontal, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below:


621

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.


According to Table 5.2 and the column section geometrical properties, the next conclusions can be found:


622

Therefore:

This means that the beam web is Class 1.

Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the flanges class.


According to Table 5.2 and the beam section geometrical properties, the next conclusions can be found:


623

Therefore:

This means that the beam left top flanges are Class 1.

Overall the beam top flange cross-section class is Class 1.

In the same way will be determined that the beam bottom flange cross-section class is also Class 1


According to the calculation above, the column section have a Class 1 web and Class 1 flanges; therefore the class section for the entire column section will be considered Class 1.


12.2.2.3 Reference results for calculating the combined biaxial bending

1,

,

,

, ≤⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡βα

EdNz

Edz

RdNy

EdY

MM

MM

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.9.1(5)

In which α and β are constants, which may conservatively be taken as unity, otherwise as follows:

For I and H sections:

2=α

)1;max(n=β

00

,,

===RdplRdpl

Ed

NNNn therefore 1=β

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

anM

M RdyplRdNy *5.01

)1(*,,, −

−= but RdplyRdNy MM ,, ≤


00

,,

===RdplRdpl

Ed

NNNn

5.0403.010*81.53

)0107.0*15.0*210*81.53()**2(4

4

≤=−

=−

= −

−

AtbA

a f

8.0)403.0*5.01(,,,,

,,RdyplRdypl

RdyN

MMM =

−=

RdyplRdyNRdyplRdyN MMMM ,,,,,,,,*8.0 >⇒= but RdplyRdNy MM ,, ≤


624

Therefore, it will be considered:

Bending around Y:

For cross-sections without bolts holes, the following approximations may be used for standard rolled I or H sections and for welded I or H sections with equal flanges:

anM

M RdzplRdNz *5.01

)1(*,,, −

−= but RdplzRdNz MM ,, ≤


RdzplRdzNRdzplRdzN MMMM ,,,,,,,,*8.0 >⇒= but RdplzRdNz MM ,, ≤ therefore it will be considered:

MNmfw

MMM

yzplRdzplRdzN 030.0

1235*10*20.125* 6

0

,,,,, ====

−

γ In conclusion:

109.1029375.003125.0

148.003125.0 12

,

,

,

, >=⎥⎦⎤

⎢⎣⎡+⎥⎦

⎤⎢⎣⎡=⎥

⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡βα

EdNz

Edz

RdNy

EdY

MM

MM




625




Result name Result description Reference value Combined oblique bending

% 111 %


Result name Result description Value Error Work ratio - Oblique Work ratio-Oblique 0 % 0.9009 %


626

12.3 EC3 Test 3: Class section classification and share and bending moment verification of an IPE300 column

Test ID: 5411

Test status: Passed

12.3.1 Description Classification and verification of an IPE 300 column made of S235 steel.

The column is connected to the ground by a fixed connection and is free on the top part.

In the middle, the column is subjected to a 200 kN force applied on the web direction, defined as a live load.


12.3.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 200kN force applied on the web direction, defined as a live load. The dead load will be neglected.


■ Exploitation loadings (category A): Q = -200kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


627

Units

Metric System


Boundary conditions


■ Outer: ► Support at start point (x = 0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (z = 5.00) free.

■ Inner: None.


In case the column is subjected to a lateral load, the stresses distribution on the most stressed point (the column base) is like in the picture below:


628

To determine the web class, we will use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.


According to the Table 5.2 and the column section geometrical properties, the next conclusions can be found:


629

Therefore:


To determine the flanges class it will be used the Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2.



630


Therefore:

This means that the column flanges are Class 1.



According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2(6).

12.3.2.3 Reference results in calculating the shear resistance Vpl,Rd The design resistance of the cross-section Vpl,Rd , is determined as follows:

0,

3*

M

yv

Rdpl

fA

Vγ

=


Where:

Av section shear area for rolled profiles fwfv trttbAA *)*2(**2 ++−=

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf=10.7mm

tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=


fy nominal yielding strength for S235 fy=235MPa

0Mγ partial safety coefficient 10 =Mγ

Therefore:

kNMN

fA

VM

yv

Rdpl 42.3483484.01

3235*10*68.25

3* 4

0, ====

γ


631

12.3.2.4 Reference results in calculating the bending moment resistance

%50%4.5742.348

200

,

>==Rdpl

Ed

VV

The shear force is greater than half of the plastic shear resistance. Its effect on the moment resistance must be taken into account.

According to: EC3 Part 1,1 EN 1993-1-1-2001 Chapter 6.2.8(1)(2)

Where:

0223.01348.0

200.0*212 22

,

=⎟⎠⎞

⎜⎝⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

Rdpl

Ed

VVρ


fwfv trttbAA *)*2(**2 ++−=

Av section shear area for rolled profiles

A cross-section area A=53.81cm2

b overall breadth b=150mm

h overall depth h=300mm

hw depth of the web hw=248.6mm

r root radius r=15mm

tf flange thickness tf=10.7mm

tw web thickness tw=7.1mm 268.2507.1*)5.1*271.0(17*15*281.53*)*2(**2 cmtrttbAA fwfv =++−=++−=




Therefore:

MNmf

tAw

MM

yw

vpl

RdVy 146.01

235*0071.0*4

)²10*68.25(*0223.010*40.628*4

²* 46

0,, =

⎟⎟⎠

⎞⎜⎜⎝

⎛−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

=

−−

γ

ρ

MNmkNmkNmMEd 5.0500200*5.2 ===

%342146.0500.0

,,

==RdVy

Ed

MM


632






Result name Result description Reference value Shear z direction work ratio % 57 % Combined oblique bending % 341 %


Result name Result description Value Error Work ratio - Fz Work ratio Fz 0 % 1.7544 % Work ratio - Oblique Work ratio - Oblique 0 % 0.2933 %


633

12.4 EC3 Test 5: Class section classification and combined axial force with bending moment verification of an IPE300 column

Test ID: 5421

Test status: Passed



The column is subjected to a 500 kN compressive force applied on top and a 5 kN/m uniform linear load applied on all the length of the column, on the web direction, both defined as live loads.


12.4.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. The column is subjected to a 500kN compressive force applied on top and a 5kN/m uniform linear load applied for all the length of the column, on the web direction, both defined as live loads. The dead load will be neglected.


■ Exploitation loadings (category A): Q1 = 500kN, Q2 = 5kN/m, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q

Units

Metric System


634


Boundary conditions


■ Outer: ► Support at start point (x=0) restrained in translation and rotation along X, Y and Z axis, ► Support at end point (z = 5.00) free.

■ Inner: None.


In this case the column is subjected to compression and lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.


635

Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.




636

Therefore:

Therefore:




637



Therefore:





Cross sections for class 3, the maximum longitudinal stress should check:


In this case:

In absence of shear force, for Class 3 cross-sections the maximum longitudinal stress shall satisfy the criterion:


This means:


638






Result name Result description Reference value Combined oblique bending % 87 %


Result name Result description Value Error Work ratio - Oblique Work ratio- Oblique 0 % 1.1494 %


639

12.5 EC3 test 4: Class section classification and bending moment verification of an IPE300 column

Test ID: 5412

Test status: Passed



In the middle, the column is subjected to a 50 kN force applied on the web direction, defined as a live load.


12.5.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. In the middle, the column is subjected to a 50kN force applied on the web direction, defined as a live load. The dead load will be neglected.


■ Exploitation loadings (category A): Q = 50kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


640

Units

Metric System


Boundary conditions



■ Inner: None.


In this case, the column is subjected to a lateral load, therefore the stresses distribution on the most stressed point (the column base) is like in the picture below.


641

The Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2 determines the web class.




642

Therefore:





643


Therefore:





12.5.2.3 Reference results in calculating the bending moment resistance

1,

≤Rdc

Ed

MM


0,

*

M

yplRdc

fwM

γ= for Class1 cross sections


Where:

340.628 cmwpl =



Therefore:

MNmfw

MM

yplRdVy 147674.0

1235*10*40.628* 6

0,, ===

−

γ

MNmkNmkNmMEd 125.012550*5.2 ===

%84148.0125.0

,,

==RdVy

Ed

MM




644




Result name Result description Reference value Combined oblique bending % 85 %


Result name Result description Value Error Work ratio - Oblique Work ratio - Oblique 0 % 1.1765 %


645

12.6 EC3 Test 1: Class section classification and compression verification of an IPE300 column

Test ID: 5383

Test status: Passed



On top, the column is subjected to a 100 kN force applied gravitationally, defined as a live load.


12.6.2 Background Classification and verification of sections for an IPE 300 column made from S235 steel. The column is connected to the ground by a fixed connection and is free on the top part. On top, the column is subjected to a 100kN force applied gravitationally, defined as a live load. The dead load will be neglected.


■ Exploitation loadings (category A): Q = -100kN, ■ The ultimate limit state (ULS) combination is: Cmax = 1 x Q


646

Units

Metric System


Boundary conditions



■ Inner: None.

12.6.2.2 Reference results for calculating the cross section class In this case, the column is subjected only to compression, therefore the distribution of stresses along the section is like in the picture below:


647

To determine the web class, we use Table 5.2 sheet 1, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2



1=ε Therefore:



648

To determine the flanges class, we will use Table 5.2, sheet 2, from EC3 Part 1,1 EN 1993-1-1-2001 Chapter 5.5.2



276.57.1045.56

==mmmm

tc

1=ε

Therefore:

9*9276.57.1045.56

=≤== εmmmm

tc

this means that the column flanges are Class 1.





649

12.6.2.3 Reference results in calculating the compressive resistance Nc,Rd The design resistance of the cross-section force Nc,Rd shall be determined as follows:

For Class 1, 2 or 3 cross-section 0

,

*

M

yRdc

fAN

γ=

Where:

A section area A=53.81cm2

Fy nominal yielding strength for S235 fy=235MPa


Therefore:

kNMNfA

NM

yRdc 54.1264264535.1

1235*10*81.53* 4

0, ====

−

γ







650


Result name Result description Reference value Work ratio % 8 %


Result name Result description Value Error Work ratio - Fx Work ratio Fx 8 % 0.0000 %


651

12.7 Generating the shape sheet by system (TTAD #11471)

Test ID: 3575

Test status: Passed

12.7.1 Description Generates shape sheets by system, on a model with 2 systems.

12.8 Verifying the calculation results for steel cables (TTAD #11623)

Test ID: 3560

Test status: Passed

12.8.1 Description Performs the finite elements calculation and the steel calculation for a steel model with cables (D4) and a static nonlinear case. Generates the steel analysis report: data and results.

12.9 Verifying the shape sheet results for a fixed horizontal beam (TTAD #11545)

Test ID: 3641

Test status: Passed

12.9.1 Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a horizontal steel element, verifies the cross section class.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support at one end and a rigid support with translation restraints on Y and Z and rotation restraint on X at the other end. A linear live load of -10.00 kN on FX and a punctual live load of 1000 kN on FX are applied.

12.10 Verifying the shape sheet results for a column (TTAD #11550)

Test ID: 3640

Test status: Passed

12.10.1Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet report for a vertical steel element.

The steel element (S235 material, IPE300 cross section) has a rigid fixed support. A vertical live load of -200.00 kN is applied.

12.11 Verifying the shape sheet results for the elements of a simple vault (TTAD #11522)

Test ID: 3612

Test status: Passed

12.11.1Description Performs the finite elements calculation and the steel calculation. Generates the shape sheet results report.

The structure is a simple vault consisting of three steel elements (S235 material, IPEA240 cross section) with two rigid fixed supports. The loads applied on the structure: self weight and a linear live load of -10.00kN on FZ.


652

12.12 Verifying the cross section optimization according to EC3 (TTAD #11516)

Test ID: 3620

Test status: Passed

12.12.1Description Verifies the cross section optimization of a steel element, according to Eurocodes 3.

Performs the finite elements calculation and the steel calculation. Generates the "Envelopes and shapes optimization" report.

The steel bar has a IPE360 cross section, a rigid hinge support at one end and a rigid support with translation restraints on X, Y and Z and rotation restraint on X. Two loads are applied: a punctual dead load of -1.00 kN on FZ and a punctual live load of -40.00 kN on FZ.

12.13 Verifying shape sheet on S275 beam (TTAD #11731)

Test ID: 3434

Test status: Passed

12.13.1Description Performs the steel calculation for two horizontal bars and generates the shape sheets report.

The bars have cross sections from different catalogues (1016x305x487 UKB and UKB1016x305x487). They are made of the same material (S275); each is subjected to a -500.00 kN linear horizontal dead load and has two rigid supports at both ends.

12.14 Verifying results on square hollowed beam 275H according to thickness (TTAD #11770)

Test ID: 3406

Test status: Passed

12.14.1Description Performs the steel calculation for two vertical bars with different thicknesses and generates the shape sheets report.

The bars are made of the same material (S275 H - EN 10210-1), have rectangular hollow cross sections, but with different thicknesses (R80*40/4.1 and R80*40/3.9). Each is subjected to a -150.00 kN vertical live load and has a rigid support.

12.15 Verifying the shape sheet for a steel beam with circular cross-section (TTAD #12533)

Test ID: 4549

Test status: Passed

12.15.1Description Verifies the shape sheet for a steel beam with circular cross-section when the lateral torsional buckling is computed and when it is not.

12.16 Changing the steel design template for a linear element (TTAD #12491)

Test ID: 4540

Test status: Passed

12.16.1Description Selects a different design template for steel linear elements.


653

12.17 Verifying the "Shape sheet" command for elements which were excluded from the specialized calculation (TTAD #12389)

Test ID: 4529

Test status: Passed

12.17.1Description Verifies the program behavior when the "Shape sheet" command is used for elements which were excluded from the specialized calculation (chords).

12.18 EC3: Verifying the buckling length results (TTAD #11550)

Test ID: 4481

Test status: Passed

12.18.1Description Performs the steel calculation and verifies the buckling length results according to Eurocodes 3 - French standards. The shape sheet report is generated.

The model consists of a vertical linear element (IPE300 cross section, S275 material) with a rigid fixed support at the base. A punctual live load of 200.00 kN is applied.

12.19 EC3 fire verification: Verifying the work ratios after performing an optimization for steel profiles (TTAD #11975)

Test ID: 4484

Test status: Passed

12.19.1Description Runs the Steel elements verification and generates the "Envelopes and optimizing profiles" report in order to verify the work ratios. The verification is performed using the EC3 norm with Romanian annex.

12.20 CM66: Verifying the buckling length for a steel portal frame, using the roA roB method

Test ID: 3814

Test status: Passed

12.20.1Description Verifies the buckling length for a steel portal frame with one level, using the roA roB method, according to CM66.

Generates the "Buckling and lateral-torsional buckling lengths" report.

12.21 CM66: Verifying the buckling length for a steel portal frame, using the kA kB method

Test ID: 3813

Test status: Passed

12.21.1Description Verifies the buckling length for a steel portal frame with one level, using the kA kB method, according to CM66.



654

12.22 EC3: Verifying the buckling length for a steel portal frame, using the kA kB method

Test ID: 3819

Test status: Passed

12.22.1Description Verifies the buckling length for a steel portal frame, using the kA kB method, according to Eurocodes 3.


12.23 Verifying the steel shape optimization when using sections from Advance Steel Profiles database (TTAD #11873)

Test ID: 4289

Test status: Passed

12.23.1Description Verifies the steel shape optimization when using sections from Advance Steel Profiles database. Performs the finite elements calculation and the steel elements calculation and generates the steel shapes report.

The structure consists of columns with UKC152x152x23 cross section, beams with UKB254x102x22 cross section and roof beams with UKB127x76x13 cross section. Dead loads and live loads are applied on the structure.

12.24 Verifying the buckling coefficient Xy on a class 2 section

Test ID: 4443

Test status: Passed

12.24.1Description Performs the finite elements calculation and the steel elements calculation. Verifies the buckling coefficient Xy on a class 2 section and generates the shape sheets report.

The model consists of a vertical linear element (I26*0.71+15*1.07 cross section and S275 material) with a rigid hinge support at the base and a rigid support with translation restraints on X and Y and rotation restraint on Z, at the top. A punctual live load is applied.

13 Timber Design


656

13.1 Modifying the "Design experts" properties for timber linear elements (TTAD #12259)

Test ID: 4509

Test status: Passed

13.1.1 Description Defines the "Design experts" properties for a timber linear element, in a model created with a previous version of the program.

13.2 Verifying the timber elements shape sheet (TTAD #12337)

Test ID: 4538

Test status: Passed

13.2.1 Description Verifies the timber elements shape sheet.

13.3 Verifying the units display in the timber shape sheet (TTAD #12445)

Test ID: 4539

Test status: Passed

13.3.1 Description Verifies the Afi units display in the timber shape sheet.

13.4 EC5: Verifying the fire resistance of a timber purlin subjected to simple bending

Test ID: 4901

Test status: Passed

13.4.1 Description Verifies the fire resistance of a rectangular cross section purlin made from solid timber C24 to resist simple bending. The purlin is exposed to fire on 3 faces for 30 minutes. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.


657

13.5 EC5: Verifying a C24 timber beam subjected to shear force

Test ID: 5036

Test status: Passed

13.5.1 Description Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

13.5.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist shear. The verification of the shear stresses at ultimate limit state is performed.

13.5.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test D; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2


Units

Metric System

Geometry


■ Height: h = 0.225 m, ■ Width: b = 0.075 m, ■ Length: L = 5.00 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 16.875 x 10-3 m2 ,


Rectangular solid timber C24 is used. The following characteristics are used in relation to this material:

■ Characteristic shear strength: fv,k = 2.5 x 106 Pa, ■ Service class 1.


658

Boundary conditions


■ Outer: ► Support at start point (z=0) restrained in translation along X, Y and Z, ► Support at end point (z = 5.00) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The beam is subjected to the following loadings:

■ External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

13.5.2.2 Reference results in calculating the timber beam subjected to uniformly distributed loads

In order to verify the timber beam shear stresses at ultimate limit state, the formula (6.13) from EN 1995-1-1 norm is used. Before using it, some parameters involved in calculations, like kmod, kcr, γM, kf, beff, heff, have to be determined. After this the reference solution, which includes the design shear stress about the principal y axis, the design shear strength and the corresponding work ratios, is calculated.

Reference solution for ultimate limit state verification

Before calculating the reference solution (design shear stress, design shear strength and work ratio) it is necessary to determine some parameters involved in calculations (kmod, γM, kcr, kf, beff, heff).

■ Modification factor for duration of load (medium term) and moisture content: kmod = 0.8 (according to table 3.1 from EN 1995-1-1)

■ Partial factor for material properties: γM = 1.3

■ Cracking factor, kcr : kcr = 0.67 (for solid timber)

■ Factor depending on the shape of the cross section, kf: kf = 3/2 (for a rectangular cross section)

■ Effective width, beff: beff = kcr x b = 0.67 x 0.075m = 0.05025m

■ Effective height, heff: heff = h = 0.225m

■ Design shear stress (induced by the applied forces):

τd = Pamm

N

hbFk

effeff

dvf 6, 10485075.0225.005025.0

25.365623

×=×

×=

×

×

■ Design shear strength:

fv,d = PaPakfM

kv66mod

, 10538.13.18.0105.2 ×=××=×

γ

■ Work ratio according to formulae 6.13 from EN 1995-1-1 norm:

0.1,

≤dv

d

fτ


659



Shear force, Fz, diagram

Simply supported beam subjected to bending Shear force diagram [N]

Design shear stress diagram

Simply supported beam subjected to bending Design shear stress [Pa]

Shear strength work ratio diagram

Simply supported beam subjected to bending Work ratio S_d [%]


Result name Result description Reference value Fz Shear force [N] 3656.25 N Stress S_d Design shear stress [Pa] 485074.63 Pa Work ratio S_d Shear work ratio (6.13) [%] 32 %


660


Result name Result description Value Error Fz Shear force -3.65625 kN -0.0000 % Design shear stress 485075 Pa -0.0001 % Shear strength work ratio 0.315299 % -0.0002 %


661

13.6 EC5: Verifying a timber column subjected to compression forces

Test ID: 4823

Test status: Passed

13.6.1 Description Verifies the compressive resistance of a rectangular cross section column (hinged at base) made from solid timber C18.

13.6.2 Background Verifies the adequacy of the compressive resistance for a rectangular cross section made from solid timber C18. The verification is made according to formula (6.35) from EN 1995-1-1 norm.

13.6.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test B; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Simply supported column

Units

Metric System

Geometry

Column cross section characteristics:

■ Height: h = 0.15 m, ■ Width: b = 0.10 m, ■ Section area: A = 15.0 x 10-3 m2


662



■ Longitudinal elastic modulus: E = 0.9 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.6 x 1010 Pa, ■ Characteristic compressive strength along the grain: fc,0,k = 18 x 106 Pa, ■ Service class 1.

Boundary conditions

The boundary conditions:

■ Outer: ► Support at base (z=0) restrained in translation along X, Y and Z, ► Support at top (z = 3.2) restrained in translation along X, Y and restrained in rotation along Z.

■ Inner: None.

Loading

The column is subjected to the following loadings:

■ External: Point load at z = 3.2: Fz = N = -20000 N, ■ Internal: None.

13.6.2.2 Reference results in calculating the timber column subjected to compression force The formula (6.35) from EN 1995-1-1 is used in order to verify a timber column subjected to compression force. Before applying this formula we need to determine some parameters involved in calculations, such as: slenderness ratios, relative slenderness ratios, instability factors. After this, the reference solution is calculated. This includes: the design compressive stress, the design compressive strength and the corresponding work ratio.

Slenderness ratios

The most important slenderness is calculated relative to the z axis, as it will be the axis of rotation if the column buckles.

■ Slenderness ratio corresponding to bending about the z axis:

85.1101.0

2.31121212 =×

=×

==mm

blm

bl gc

zλ

■ Slenderness ratio corresponding to bending about the y axis (informative):

9.7315.0

2.31121212 =×

=×

==mm

hlm

hl gc

yλ

Relative slenderness ratios

The relative slenderness ratios are:

■ Relative slenderness ratio corresponding to bending about the z axis:

933.1106.01018

1.0122.31

,

12

,10

6

050

,0,

050

,0,, =

××

×××

=×

××==

PaPa

mm

Ef

blm

Ef kcgkcz

zrel πππλλ

■ Relative slenderness ratio corresponding to bending about the y axis (informative):

288.1106.01018

15.0122.31

,

12

,10

6

050

,0,

050

,0,, =

××

×××

=×

××==

PaPa

mm

Ef

hlm

Ef kcgkcy

yrel πππλ

λ

So there is a risk of buckling, because λrel,max ≥ 0.3.


663

Instability factors

In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:

βc = 0.2 (according to relation 6.29 from EN 1995-1-1)

The instability factors are: ■ kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z

2] (according to relation 6.28 from EN 1995-1-1) ■ ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y

2] (informative)

■ 2

,2,1

zrelzz

zckk

kλ−+

= (according to relation 6.26 from EN 1995-1-1)

■ 2

,2,1

yrelyy

yckk

kλ−+

= (informative)

Reference solution

Before calculating the reference solution (design compressive stress, design compressive strength and work ratio) we need to determine some parameters involved in calculations (kmod, γM).


■ Partial factor for material properties: γM = 1.3 ■ Design compressive stress (induced by the applied forces):

σc,0,d = AN

■ Design compressive strength:

fc,0,d = M

kckfγ

mod,0,

■ Work ratio:

Work ratio = 0.1,0,,

,0, ≤× dczc

dc

fkσ

(according to relation 6.35 from EN 1995-1-1)




664

Work ratio diagram

Simply supported column subjected to compression force Work ratio


Result name Result description Reference value kc,z Instability factor 0.2400246 kc,y Instability factor (informative) 0.488869 σc,0,d Design compressive stress [Pa] 1333333 Pa Work ratio Work ratio [%] 50 %


Result name Result description Value Error Kc,z Instability factor 0.240107 adim 0.0001 % Kc,y Instability factor 0.488612 adim 0.0000 % Stress SFx Design compressive stress 1.33333e+006 Pa 0.0003 % Work ratio Work ratio 50 % 0.0000 %


665

13.7 EC5: Verifying a timber beam subjected to combined bending and axial tension

Test ID: 4872

Test status: Passed

13.7.1 Description Verifies a rectangular cross section rafter made from solid timber C24 to resist combined bending and axial tension. The verification of the cross-section subjected to combined stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

13.7.2 Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending and axial tension. The verification of the deflections at serviceability limit state is also performed.

13.7.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test E; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent rafters (span): d = 0.5 m. The following load cases and load combination are used:

■ Loadings from the structure: G = 450 N/m2; ■ Snow load (structure is located at an altitude > 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 1957.5 N/m2; ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x S; ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.2 x S. All loads will be projected on the rafter direction since its slope is 50% (26.6°).

Simply supported rafter subjected to projected loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.05 m, ■ Length: L = 5.00 m, ■ Section area: A = 10 x 10-3 m2 ,

■ Elastic section modulus about the strong axis y: 322

000333.06

20.005.06

mhbWy =⋅

=×

= .


666



■ Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 2.

Boundary conditions


■ Outer: ► Support at start point (z=0) restrained in translation along Y, Z and restrained in rotation along X. ► Support at end point (z = 5.00) restrained in translation along X, Y, Z.

■ Inner: None.

Loading

The rafter is subjected to the following projected loadings (at ultimate limit state):

■ External: ► Uniformly distributed load: q = Cmax x d x cos26.6° = 1957.5 N/m2 x 0.5 m x cos26.6° = 875.15 N/m, ► Tensile load component: N = Cmax x d x sin26.6° x L = 1957.5 N/m2 x 0.5m x sin26.6° x 5.00m =

= 2191.22 N ■ Internal: None.

13.7.2.2 Reference results in calculating the timber beam subjected to combined stresses In order to verify the timber beam subjected to combined stresses at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be determined. After this the reference solution, which consists of the design tensile stress, the design tensile strength and the corresponding work ratio and also the work ratios of the combined stresses, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (the design tensile stress, the design tensile strength and the corresponding work ratio, and also the work ratios of the combined stresses) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)


■ Depth factor (“h” represents the width in millimeters because the element is tensioned):

kh = min 25.13.125.1

min3.150

150min

3.1

150 2.02.0

=⎩⎨⎧

=⎪⎩

⎪⎨

⎧⎟⎠⎞

⎜⎝⎛

=⎪⎩

⎪⎨

⎧⎟⎠⎞

⎜⎝⎛h

■ System strength factor: ksys = 1.0 (because several equally spaced similar members are subjected by an uniformly distributed load)

■ Factor considering re-distribution of bending stresses in a cross-section (for rectangular sections): km = 0.7

■ Design tensile stress (induced by the ultimate limit state force, N):

σt,0,d = PamN

AN 219122

101022.2191

23 =×

= −


667

■ Design tensile strength:

ft,0,d = Pakkf hM

kt66mod

,0, 10115.1225.13.19.01014 ×=×××=××

γ

■ Work ratio:

SFx = 0.1,0,

,0, ≤dt

dt

fσ


■ Design bending stress (induced by the applied forces):

σm,y,d = Pamm

mmN

hbLq

WM

y

y 622

22

2

2

102045.82.005.08

00.515.8756

86

×=××

××=

××××

=

■ Design bending strength:

fm,y,d = Pakkkf hsysM

km66mod

, 10615.160.10.13.19.01024 ×=××××=×××

γ


1,,

,,

,,

,,

,0,

,0, ≤++dzm

dzmm

dym

dym

dt

dt

fk

ffσσσ


1,,

,,

,,

,,

,0,

,0, ≤++dzm

dzm

dym

dymm

dt

dt

ffk

fσσσ

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

300)( LQwinst ≤

125Lwfin ≤

200,Lw finnet ≤

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are:

■ Instantaneous deflection (for a base variable action):

05.547)(00914.0)( LQwmQw instinst =⇒=

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

7.36401371.0 Lwmdw instCQinst =⇒==


668

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the deformation factor (kdef) has to be chosen:

8.0=defk (value determined for service class 2, according to table 3.2 from EN 1995-1-1)

6.97600512.00064.08.08.0 Lwmmdw creepQPcreep =⇒=×=×=

■ Final deflection:

5.26501883.000512.001371.0 Lwmmmwww fincreepinstfin =⇒=+=+=

■ Net deflection:

5.26501883.0001883.0 ,,

Lwmmmwww finnetcfinfinnet =⇒=+=+=



SFx work ratio diagram

Simply supported beam subjected to tensile forces Work ratio SFx

Strength work ratio diagram

Simply supported beam subjected to combined stresses Strength work ratio

Instantaneous deflection winst(Q)

Simply supported beam subjected to snow loads Instantaneous deflection winst(Q) [m]


669

Instantaneous deflection winst(CQ)

Simply supported beam subjected to a characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

Final deflection wfin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]

Net deflection wnet,fin

Simply supported beam subjected to characteristic load combination of actions Instantaneous deflection winst(CQ) [m]


Result name Result description Reference value SFx SFx work ratio [%] 1.808 % Strength work ratio Work ratio (6.17) [%] 51.19 % winst (Q) Deflection for a base variable action [m] 0.00914 m

dCQ Deflection for a characteristic combination of actions [m] 0.01371 m

winst Instantaneous deflection [m] 0.01371 m kdef Deformation coefficient 0.8 dQP Deflection for a quasi-permanent combination of actions [m] 0.0064 m wfin Final deflection [m] 0.01883 m wnet,fin Net deflection [m] 0.01883 m


670


Result name Result description Value Error Work ratio SFx SFx work ratio 0.0181483 % -0.0003 % Work ratio Strength work ratio 0.511941 % 0.0001 % D w_inst(Q) 0.00871459 m -4.6583 % D deflection for a characteristic combination 0.0130719 m -4.6584 % Winst instantaneous deflection 0.0137105 m 0.0003 % Kdef deformation coefficient 0.8 adim 0.0000 % D deformation for a quasi-permanent combination 0.00610023 m -4.6584 % Wfin final deflection 0.0188291 m 0.0002 % Wnet,fin net final deflection 0.0188291 m 0.0002 %


671

13.8 EC5: Verifying a timber column subjected to tensile forces

Test ID: 4693

Test status: Passed

13.8.1 Description Verifies the tensile resistance of a rectangular cross section column (fixed at base) made from solid timber C24.

13.8.2 Background Verifies the adequacy of the tension resistance for a rectangular cross section made from solid timber C24. The verification is made according to formula (6.1) from EN 1995-1-1 norm.

13.8.2.1 Model description ■ Reference: Guide de validation Eurocode 5, test A; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Column with fixed base

Units

Metric System

Geometry

Cross section characteristics:

■ Height: h = 0.122 m, ■ Width: b = 0.036m, ■ Section area: A = 43.92 x 10-4 m2 ■ I = 5.4475 x 10-6 m4.


672



■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Characteristic tensile strength along the grain: ft,0,k = 14 x 106 Pa, ■ Service class 2.

Boundary conditions

The boundary conditions:

■ Outer: Fixed at base (z = 0),

Free at top (z = 5),

■ Inner: None.

Loading


■ External: Point load at z = 5: Fz = N = 10000 N, ■ Internal: None.

13.8.2.2 Reference results in calculating the timber column subjected to tension force

Reference solution

The reference solution is determined by formula (6.1) from EN 1995-1-1. Before applying this formula we need to determine some parameters involved in calculations (kmod, γM, kh). After this, the design tensile stress, the design tensile strength and the corresponding work ratio are calculated.

■ Modification factor for duration of load and moisture content: kmod = 0.9 ■ Partial factor for material properties: γM = 1.3 ■ Depth factor (“h” represents the width, because the element is tensioned):

kh = min⎪⎩

⎪⎨

⎧⎟⎠⎞

⎜⎝⎛

3.1

150 2.0

h

■ Design tensile stress (induced by the ultimate limit state force, N):

σt,0,d = AN

■ Design tensile strength:

ft,0,d = hM

kt kkf ××γ

mod,0,

■ Work ratio:

SFx = 0.1,0,

,0, ≤dt

dt

fσ





673

Work ratio SFx diagram

Column with fixed base, subjected to tension force Work ratio SFx


Result name Result description Reference value σt,0,d Design tensile stress [Pa] 2276867.03 Pa SFx Work ratio [%] 18 %


Result name Result description Value Error Stress SFx Design tensile stress 2.27687e+006 Pa -0.0001 % Work ratio Work ratio 18 % 0.0000 %


674

13.9 EC5: Shear verification for a simply supported timber beam

Test ID: 4822

Test status: Passed

13.9.1 Description Verifies a rectangular cross section beam made from solid timber C24 to shear efforts. The verification of the shear stresses at ultimate limit state is performed.


675

13.10 EC5: Verifying a timber purlin subjected to oblique bending

Test ID: 4878

Test status: Passed

13.10.1Description Verifies a rectangular timber purlin made from solid timber C24 to resist oblique bending. The verification is made following the rules from Eurocode 5 French annex.

13.10.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to oblique bending. The verification of the bending stresses at ultimate limit state is performed.

13.10.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.3; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:

■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2; All loads will be projected on the purlin direction since the roof slope is 17°.

Simply supported purlin subjected to loadings

Units

Metric System

Geometry

Purlin cross section characteristics:

■ Height: h = 0.20 m, ■ Width: b = 0.10 m, ■ Length: L = 3.5 m, ■ Section area: A = 0.02 m2 ,

■ Elastic section modulus about the strong axis, y: 322

000666.06

20.01.06

mhbWy =⋅

=×

= ,

■ Elastic section modulus about the strong axis, z: 322

000333.06

20.01.06

mhbWz =⋅

=×

= .


676



■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 2.

Boundary conditions


■ Outer: ► Support at start point (z=0) restrained in translation along X, Y, Z; ► Support at end point (z = 3.5) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The purlin is subjected to the following projected loadings (at ultimate limit state):

■ External: ► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x

sin17° = 1101.22 N/m, ► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x

cos17° = 3601.92 N/m, ■ Internal: None.

13.10.2.2Reference results in calculating the timber purlin subjected to oblique bending In order to verify the timber purlin subjected to oblique bending at ultimate limit state, the formulae (6.17) and (6.18) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, have to be determined. After this, the reference solution (which includes the design bending stress about y axis, the design bending stress about z axis and the maximum work ratio for strength verification) is calculated.


Before calculating the reference solution (design bending stress about y axis, design bending stress about z axis and maximum work ratio for strength verification) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Modification factor for duration of load (short term) and moisture content: kmod = 0.9 (according to table 3.1 from EN 1995-1-1)


■ Depth factor (the height of the cross section in bending is bigger than 150 mm): kh = 1.0



■ Design bending stress about y axis (induced by uniformly distributed load, qz):

σm,y,d = Pam

mmN

WLq

WM

y

z

y

y 63

222

102814.8000666.08

5.392.3601

8×=

×

×=

××

=

■ Design bending stress about z axis (induced by uniformly distributed load, qy):


677

σm,z,d = Pam

mmN

WLq

WM

z

y

z

z 63

222

100638.5000333.08

5.322.1101

8×=

×

×=

×

×=


fm,y,d = fm,z,d = Pakkkf hsysM

km66mod

, 10615.160.10.13.19.01024 ×=××××=×××

γ

■ Maximum work ratio for strength verification; it represents the maximum value between the work ratios obtained with formulae 6.17 and 6.18 from EN 1995-1-1 norm:

1max

,,

,,

,,

,,

,,

,,

,,

,,

≤

⎪⎪⎭

⎪⎪⎬

⎫

+

+

⎪⎪

⎩

⎪⎪

⎨

⎧

dzm

dzm

dzm

dzmm

dym

dymm

dym

dym

f

fk

fk

fσ

σ

σ

σ




678

Stress SMy diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMy [Pa]

Stress SMz diagram

Simply supported purlin subjected to uniformly distributed load, qz Stress SMz [Pa]

Maximum work ratio for strength verification

Strength of a simply supported purlin subjected to oblique bending Work ratio [%]

13.10.2.3Reference results

Result name Result description Reference value Smy Design bending stress about y axis [Pa] 8281441 Pa SMz Design bending stress about z axis [Pa] 5063793 Pa

Work ratio Maximum work ratio for strength verification [%] 71.2 %

13.10.3Calculated results

Result name Result description Value Error Stress SMy Design bending stress about y axis 8.47672e+006 Pa 0.0000 % Stress SMz Design bending stress about z axis 5.18319e+006 Pa -0.0000 % Work ratio Maximum work ratio for strength verification 0.71153 % -0.0000 %


679

13.11 EC5: Verifying lateral torsional stability of a timber beam subjected to combined bending and axial compression

Test ID: 4877

Test status: Passed

13.11.1Description Verifies the lateral torsional stability for a rectangular timber beam subjected to combined bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.


680

13.12 EC5: Verifying a timber beam subjected to simple bending

Test ID: 4682

Test status: Passed

13.12.1Description Verifies a rectangular cross section beam made from solid timber C24 to resist simple bending. Verifies the bending stresses at ultimate limit state, as well as the deflections at serviceability limit state.

13.12.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 to resist simple bending. Verification of the bending stresses at ultimate limit state, as well as the verification of the deflections at serviceability limit state are performed.

13.12.2.1Model description ■ Reference: Guide de validation Eurocode 5, test C; ■ Analysis type: static linear (plane problem); ■ Element type: linear. The following load cases and load combination are used:

■ Loadings from the structure: G = 0.5 kN/m2, ■ Exploitation loadings (category A): Q = 1.5 kN/m2, ■ The ultimate limit state (ULS) combination is: Cmax = 1.35 x G + 1.5 x Q = 2.925 kN/m2 ■ Characteristic combination of actions: CCQ = 1.0 x G + 1.0 x Q ■ Quasi-permanent combination of actions: CQP = 1.0 x G + 0.3 x Q


Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.075 m, ■ Length: L = 4.50 m, ■ Distance between adjacent beams (span): d = 0.5 m, ■ Section area: A = 15.0 x 10-3 m2 ,

■ Elastic section modulus about the strong axis y: 322

0005.06

20.0075.06

mhbWy =⋅

=×

=


681



■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Service class 1.

Boundary conditions


■ Outer: ► Support at start point (z=0) restrained in translation along X, Y and Z, ► Support at end point (z = 4.5) restrained in translation along X, Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The beam is subjected to the following loadings:

■ External: Uniformly distributed load: q = Cmax x d = 2.925 kN/m2 x 0.5 m = 1.4625 kN/m,

■ Internal: None.

13.12.2.2Reference results in calculating the timber beam subjected to uniformly distributed loads

In order to verify the timber beam bending stresses at ultimate limit state, the formulae (6.11) and (6.12) from EN 1995-1-1 norm are used. Before using them, some parameters involved in calculations, like kmod, γM, kh, ksys, km, must be calculated. After this, the reference solution, which includes the design bending stress about the principal y axis, the design bending strength and the corresponding work ratios, is calculated.

A verification of the deflections at serviceability limit state is done. The verification is performed by comparing the effective values with the limiting values for deflections specified in EN 1995-1-1 norm.


Before calculating the reference solution (design bending stress, design bending strength and work ratios) it is necessary to determine some parameters involved in calculations (kmod, γM, kh, ksys, km).






■ Design bending stress (induced by the applied forces):

σm,d = PahbLq

WM

y

y 62

2

2

2

104039.72.0075.085.44625.16

86

×=××××

=××××

=


fm,d = Pakkkf hsysM

km66mod

, 10769.140.10.13.18.01024 ×=××××=×××

γ


682


0.1,

, ≤dm

dm

fσ


0.1,

, ≤×dm

dmm fk

σ

Reference solution for serviceability limit state verification

The following limiting values for instantaneous deflection (for a base variable action), final deflection and net deflection are considered:

300)( LQwinst ≤

125Lwfin ≤

200,Lw finnet ≤

For the analyzed beam, no pre-camber is considered (wc = 0). The effective values of deflections are the followings:

■ Instantaneous deflection (for a base variable action):

8.600)(00749.0)( LQwmQw instinst =⇒=

■ Instantaneous deflection (calculated for a characteristic combination of actions - CCQ):

45.45000999.0 Lwmdw instCQinst =⇒==

■ In order to determine the creep deflection (calculated for a quasi-permanent combination of actions - CQP), the

deformation factor (kdef) has to be chosen:

6.0=defk (calculated value for service class 1, according to table 3.2 from EN 1995-1-1)

95.157800285.000475.06.06.0 Lwmmdw creepQPcreep =⇒=×=×=

■ Final deflection:

47.35001284.000285.000999.0 Lwmmmwww fincreepinstfin =⇒=+=+=

■ Net deflection:

47.35001284.0001284.0 ,,

Lwmmmwww finnetcfinfinnet =⇒=+=+=




683

Work ratio diagram

Simply supported beam subjected to bending Strength work ratio


Result name Result description Reference value σm,d Design bending stress [Pa] 7403906.25 Pa Strength work ratio Work ratio (6.11) [%] 50 % winst (Q) Deflection for a base variable action [m] 0.00749 m

dCQ Deflection for a characteristic combination of actions [m] 0.00999 m

winst Instantaneous deflection [m] 0.00999 m kdef Deformation coefficient 0.6 dQP Deflection for a quasi-permanent combination of actions [m] 0.00475 m wfin Final deflection [m] 0.01284 m wnet,fin Net deflection [m] 0.01284 m


Result name Result description Value Error Stress Design bending stress 7.40391e+006 Pa -0.0001 % Work ratio Work ratio (6.11) 50 % 0.0000 % D Deflection for a base variable action 0.00714509 m -4.6335 % D Deflection for a characteristic combination of actions 0.00952678 m -4.6335 % Winst Instantaneous deflection 0.00998966 adim -0.0000 % Kdef Deformation coefficient 0.6 adim 0.0000 % D Deflection for a quasi-permanent combination of actions 0.00452522 m -4.6336 % Wfin Final deflection 0.0128367 adim 0.0001 % Wnet,fin Net deflection 0.0128367 adim 0.0001 %


684

13.13 EC5: Verifying a timber purlin subjected to biaxial bending and axial compression

Test ID: 4879

Test status: Passed

13.13.1Description Verifies the stability of a rectangular timber purlin made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made following the rules from Eurocode 5 French annex.

13.13.2Background Verifies the adequacy of a rectangular cross section made from solid timber C24 subjected to biaxial bending and axial compression. The verification is made according to formulae (6.23) and (6.24) from EN 1995-1-1 norm.

13.13.2.1Model description ■ Reference: Guide de validation Eurocode 5, test E.4; ■ Analysis type: static linear (plane problem); ■ Element type: linear; ■ Distance between adjacent purlins (span): d = 1.8 m. The following load cases and load combination are used:

■ The ultimate limit state (ULS) combination is: CULS = 1.35 x G + 1.5 x S + 0.9 x W; ■ Loadings from the structure: G = 550 N/m2; ■ Snow load (structure is located at an altitude < 1000m above sea level): S = 900 N/m2; ■ Axial compression force due to wind effect on the supporting elements: W = 15000 N; ■ Uniformly distributed load corresponding to the ultimate limit state combination:

Cmax = 1.35 x G + 1.5 x S = 2092.5 N/m2.

All loads will be projected on the purlin direction since its slope is 30% (17°).

Simply supported purlin subjected to loadings

Units

Metric System

Geometry


■ Height: h = 0.20 m, ■ Width: b = 0.10 m, ■ Length: L = 3.50 m, ■ Section area: A = 0.02 m2 ,


685

■ Elastic section modulus about the strong axis, y: 322

000666.06

20.01.06

mhbWy =⋅

=×

= ,

■ Elastic section modulus about the strong axis, z: 322

000333.06

20.01.06

mhbWz =⋅

=×

= .


Rectangular solid timber C24 is used. The followings characteristics are used in relation to this material:

■ Characteristic compressive strength along the grain: fc,0,k = 21 x 106 Pa, ■ Characteristic bending strength: fm,k = 24 x 106 Pa, ■ Longitudinal elastic modulus: E = 1.1 x 1010 Pa, ■ Fifth percentile value of the modulus of elasticity parallel to the grain: E0,05 = 0.74 x 1010 Pa, ■ Service class 2.

Boundary conditions


■ Outer: ► Support at start point (z = 0) restrained in translation along X, Y, Z; ► Support at end point (z = 3.50) restrained in translation along Y, Z and restrained in rotation along X.

■ Inner: None.

Loading

The purlin is subjected to the following projected loadings (corresponding to the ultimate limit state combination):

■ External: ► Axial compressive load: N =0.9 x W = 13500 N; ► Uniformly distributed load (component about y axis): qy = Cmax x d x sin17° = 2092.5 N/m2 x 1.8 m x

sin17° = 1101.22 N/m, ► Uniformly distributed load (component about z axis): qz = Cmax x d x cos17° = 2092.5 N/m2 x 1.8 m x

cos17° = 3601.92 N/m, ■ Internal: None.

13.13.2.2Reference results in calculating the timber purlin subjected to combined stresses In order to verify the stability of a timber purlin subjected to biaxial bending and axial compression at ultimate limit state, the formulae (6.23) and (6.24) from EN 1995-1-1 norm are used. Before applying these formulae we need to determine some parameters involved in calculations like: slenderness ratios, relative slenderness ratios, instability factors. After this, we’ll calculate the maximum work ratio for stability verification, which represents in fact the reference solution.

Slenderness ratios

The slenderness ratios corresponding to bending about y and z axes are determined as follows:

■ Slenderness ratio corresponding to bending about the z axis:

24.1211.0

5.31121212 =×

=×

==mm

blm

bl gc

zλ

■ Slenderness ratio corresponding to bending about the y axis:

62.602.0

5.31121212 =×

=×

==mm

hlm

hl gc

yλ


686

Relative slenderness ratios

The relative slenderness ratios are:

■ Relative slenderness ratio corresponding to bending about the z axis:

056.21074.0

102124.12110

6

05,0

,0,, =

××

==PaPa

Ef kcz

zrel ππλλ

■ Relative slenderness ratio corresponding to bending about the y axis:

028.11074.0

102162.60

,10

6

050

,0,, =

××

==PaPa

Ef kcy

yrel ππλ

λ

■ Maximum relative slenderness ratio:

( ) 056.2,max ,,max, == yrelzrelrel λλλ

So there is a risk of buckling because λrel,max ≥ 0.3.

Instability factors

In order to determine the instability factors we need to determine the βc factor. It is a factor for solid timber members within the straightness limits defined in Section 10 from EN 1995-1-1:

βc = 0.2 (according to relation 6.29 from EN 1995-1-1)

The instability factors are: kz = 0.5 [1 + βc (λrel,z – 0.3) + λrel,z

2] (according to relation 6.28 from EN 1995-1-1)

ky = 0.5 [1 + βc (λrel,y – 0.3) + λrel,y2] (according to relation 6.27 from EN 1995-1-1)

2,

2,1

zrelzz

zckk

kλ−+


2,

2,1

yrelyy

yckk

kλ−+



Before calculating the reference solution (the maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1) it is necessary to determine the design compressive stress, the design compressive strength, the design bending stress, the design bending strength and some parameters involved in calculations (kmod, γM, kh, ksys, km).

■ Design compressive stress (induced by the axial compressive load from the corresponding ULS combination, N):

σc,0,d = PamN

AN 675000

02.013500

2 ==

■ Design bending stress about the y axis (induced by uniformly distributed load, qz):

σm,y,d = Pam

mmN

WLq

WM

y

z

y

y 63

222

102814.8000666.08

50.392.3601

8×=

×

×=

××

=

■ Design bending stress about the z axis (induced by uniformly distributed load, qy):

σm,z,d = Pam

mmN

WLq

WM

z

y

z

z 63

222

100638.5000333.08

50.322.1101

8×=

×

×=

×

×=

■ Modification factor for duration of load (instantaneous action) and moisture content (service class 2):


687

kmod = 1.1 (according to table 3.1 from EN 1995-1-1)




■ Design compressive strength:

fc,0,d = PakfM

kc66mod

,0, 10769.173.11.11021 ×=××=×

γ


fm,y,d = fm,z,d = Pakkkf hsysM

km66mod

, 10308.200.10.13.11.11024 ×=××××=×××

γ

■ Maximum work ratio for stability verification based on formulae (6.23) and (6.24) from EN 1995-1-1:

1max

,,

,,

,,

,,

,0,,

,0,

,,

,,

,,

,,

,0,,

,0,

≤

⎪⎪

⎭

⎪⎪

⎬

⎫

⎪⎪

⎩

⎪⎪

⎨

⎧

+⋅+⋅

⋅++⋅

dzm

dzm

dym

dymm

dczc

dc

dzm

dzmm

dym

dym

dcyc

dc

ffk

fk

fk

ffk

σσσ

σσσ




688

Instability factor, kcy

Simply supported purlin subjected to biaxial bending and axial compression Kc,y

Instability factor, kcz

Simply supported purlin subjected to biaxial bending and axial compression Kc,z

Maximum work ratio for stability verification

Simply supported purlin subjected to biaxial bending and axial compression Work ratio [%]


Result name Result description Reference value Kc,y Instability factor, kc,y 0.67 Kc,z Instability factor, kc,z 0.21

Work ratio Maximum work ratio for stability verification [%] 71.1 %


Result name Result description Value Error Kc,y Instability factor, kc,y 0.665025 adim -0.0001 % Kc,z Instability factor, kc,z 0.212166 adim 0.0002 % Work ratio Maximum work ratio for stability verification 0.706586 % -0.0000 %


689

13.14 EC5: Verifying the residual section of a timber column exposed to fire for 60 minutes

Test ID: 4896

Test status: Passed

13.14.1Description Verifies the residual cross section of a column exposed to fire for 60 minutes. The column is made from glued laminated timber GL24 and it has only 3 faces exposed to fire. The verification is made according to chapter 4.2.2 (Reduced cross section method) from EN 1995-1-2 norm.

13.14.2Background Verifies the adequacy of the cross sectional resistance for a rectangular cross section, which is made from glued laminated timber GL24, exposed to fire for 60 minutes on 3 faces. The verification is made according to chapter 4.2.2 from EN 1995-1-2 norm.

13.14.2.1Model description ■ Reference: Guide de validation Eurocode 5, test F.1; ■ Analysis type: static linear (plane problem); ■ Element type: linear.

Timber column with fixed base

Units

Metric System

Geometry

Below are described the column cross section characteristics:

■ Depth: h = 0.60 m, ■ Width: b = 0.20 m, ■ Section area: A = 0.12 m2 ■ Height: H = 5.00 m


690


Glued laminated timber GL24 is used. The following characteristics are used in relation to this material:

■ Density: ρ = 380 kg/m3, ■ Design charring rate: βn = 0.7 x 10-3 m/min,

Boundary conditions


■ Outer: ► Fixed at base (Z = 0), ► Support at top (Z = 5.00) restrained in translation along X and Y,

■ Inner: None.

Loading


■ External: Point load at Z = 5.00: Fz = N = - 100000 N, ■ Internal: None.

13.14.2.2Reference results in calculating the cross sectional resistance of a timber column exposed to fire

Reference solution

The reference solution (residual cross section) is determined by reducing the initial cross section dimensions by the effective charring depth according to chapter 4.2.2 from EN 1995-1-2. Before calculating the effective charring depth we need to determine some parameters involved in calculations (dchar,n, k0, d0).

■ Depth of layer with assumed zero strength and stiffness: d0 = 7 x 10-3 m; ■ Coefficient depending of fire resistance time and also depending if the members are protected or not:

k0 = 1.0 (according to table 4.1 from EN 1995-1-2)

■ Notional design charring depth:

dchar,n = m.minminm.tn 0420601070β 3 =⋅⋅=⋅ − (according to relation 3.2 from EN 1995-1-2)

■ Effective charring depth: def = 00 dkd n,char ⋅+

■ Residual cross section:

Afi = )db()dh( efef ⋅−×− 2




691

Residual cross section

Column with fixed base exposed to fire for 60 minutes Afi


Result name Result description Reference value Afi Residual cross section [m2] 0.056202 m2


Result name Result description Value Error Afi Residual cross section 0.056202 adim -0.0000 %

14 XML Template Files


694

14.1 Loading a template with the properties of a planar element (DEV2012 #1.4)

Test ID: 4344

Test status: Passed

14.1.1 Description Loads an XML file as a properties template and applies it on a planar element. Saves the properties of the element on which the template was applied, in order to compare the files.

14.2 Loading a template with the properties of a linear element (DEV2012 #1.4)

Test ID: 4207

Test status: Passed

14.2.1 Description Loads an XML file as a properties template and applies it on a linear element. Saves the properties of the element on which the template was applied, in order to compare the files.

14.3 Saving the properties of a planar element as a template (DEV2012 #1.4)

Test ID: 4176

Test status: Passed

14.3.1 Description Saves the properties of a planar element as an XML file which can be later used as a template for other planar elements.

14.4 Saving the properties of a linear element as a template (DEV2012 #1.4)

Test ID: 4169

Test status: Passed

14.4.1 Description Saves the properties of a linear element as an XML file which can be later used as a template for other linear elements.

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