Plastic Design Portal Frame to Ec3

Department of of Civil Civil and Structural Engineering

Plastic Design of of Portal Portal frame to Eurocode 3 Worked Example

University of Sheffield

University University of Sheffield Sheffield Department Department of Civil Civil Structural Engineering Engineering

Contents GEOMETRY .......................................................................................................................................................................3 1 GEOMETRY ....................................................................................................................................................................... 2 DESIGN BRIEF ..................................................................................................................................................................4 3 DETERMINING LOADING ON FRAME .......................................................................................................................5 3.1Combination factors ψ.................................................................................................................................................. 5 3.2Snow loading................................................................................................................................................................ loading ................................................................................................................................................................ 6 3.3Self weight 3.3Self weight of steel of steel members........................................................................................................................................ members ........................................................................................................................................ 7

4 INITIAL SIZING OF MEMBERS..................................................................................................................................... MEMBERS .....................................................................................................................................8 5 LOAD COMBINATION (MAX VERTICAL LOAD) (DEAD + SNOW).................................................................. SNOW)..................................................................10 10 5.1Frame imperfections equivalent horizontal forces....................................................................................................... forces....................................................................................................... 10 5.2 Partial safety factors and second order effects ........................................................................................................... 11 5.2.1 Sway buckling mode Stability ( αcr,s,est). ............................................................................................................14 5.2.2 Snap‐through buckling stability (αcr,r,est ) ..........................................................................................................16 5.3.2Accounting Second Order effects.......................................................................................................................... effects ..........................................................................................................................17

6 MEMBER CHECKS ........................................................................................................................................................ ........................................................................................................................................................20 20 6.1 Purlins ....................................................................................................................................................................... 20 6.2 Column (UB 610 x 229 x 101)...................................................................................................................................... 101) ...................................................................................................................................... 21 6.2.1Classification.......................................................................................................................................................... 6.2.1Classification.......................................................................................................................................................... 21 6.2.2Cross section resistance ........................................................................................................................................ 21 6.2.3Stability against lateral and torsional buckling (EN 1993‐1‐1: 2005 (E) Sec BB3.2.1):........................................... BB3.2.1):...........................................22 22 6.3Rafter (UB457 x 191 x 89)............................................................................................................................................ 89)............................................................................................................................................ 29 6.3.1Section Classification ............................................................................................................................................. 29 6.3.2 Cross‐section Resistance. .....................................................................................................................................29 6.3.3 Check rafter buckling in apex region ....................................................................................................................31 6.3.4 Stability check for lower bending moments .........................................................................................................32 32 6.4Haunch (UB 457 x 191 x 89)......................................................................................................................................... 89)......................................................................................................................................... 35 6.4.1Classification.......................................................................................................................................................... 6.4.1Classification.......................................................................................................................................................... 35 6.4.2Haunch Stability..................................................................................................................................................... Stability..................................................................................................................................................... 36 6.4.3 Cross‐section resistance. ......................................................................................................................................41

7. COMPARISON BETWEEN DIFFERENT CODES.................................................................................................... CODES....................................................................................................43 43

8APPENDIX ...................................................................................................................................................................... ......................................................................................................................................................................44 44


Contents GEOMETRY .......................................................................................................................................................................3 1 GEOMETRY ....................................................................................................................................................................... 2 DESIGN BRIEF ..................................................................................................................................................................4 3 DETERMINING LOADING ON FRAME .......................................................................................................................5 3.1Combination factors ψ.................................................................................................................................................. 5 3.2Snow loading................................................................................................................................................................ loading ................................................................................................................................................................ 6 3.3Self weight 3.3Self weight of steel of steel members........................................................................................................................................ members ........................................................................................................................................ 7

4 INITIAL SIZING OF MEMBERS..................................................................................................................................... MEMBERS .....................................................................................................................................8 5 LOAD COMBINATION (MAX VERTICAL LOAD) (DEAD + SNOW).................................................................. SNOW)..................................................................10 10 5.1Frame imperfections equivalent horizontal forces....................................................................................................... forces....................................................................................................... 10 5.2 Partial safety factors and second order effects ........................................................................................................... 11 5.2.1 Sway buckling mode Stability ( αcr,s,est). ............................................................................................................14 5.2.2 Snap‐through buckling stability (αcr,r,est ) ..........................................................................................................16 5.3.2Accounting Second Order effects.......................................................................................................................... effects ..........................................................................................................................17

6 MEMBER CHECKS ........................................................................................................................................................ ........................................................................................................................................................20 20 6.1 Purlins ....................................................................................................................................................................... 20 6.2 Column (UB 610 x 229 x 101)...................................................................................................................................... 101) ...................................................................................................................................... 21 6.2.1Classification.......................................................................................................................................................... 6.2.1Classification.......................................................................................................................................................... 21 6.2.2Cross section resistance ........................................................................................................................................ 21 6.2.3Stability against lateral and torsional buckling (EN 1993‐1‐1: 2005 (E) Sec BB3.2.1):........................................... BB3.2.1):...........................................22 22 6.3Rafter (UB457 x 191 x 89)............................................................................................................................................ 89)............................................................................................................................................ 29 6.3.1Section Classification ............................................................................................................................................. 29 6.3.2 Cross‐section Resistance. .....................................................................................................................................29 6.3.3 Check rafter buckling in apex region ....................................................................................................................31 6.3.4 Stability check for lower bending moments .........................................................................................................32 32 6.4Haunch (UB 457 x 191 x 89)......................................................................................................................................... 89)......................................................................................................................................... 35 6.4.1Classification.......................................................................................................................................................... 6.4.1Classification.......................................................................................................................................................... 35 6.4.2Haunch Stability..................................................................................................................................................... Stability..................................................................................................................................................... 36 6.4.3 Cross‐section resistance. ......................................................................................................................................41

7. COMPARISON BETWEEN DIFFERENT CODES.................................................................................................... CODES....................................................................................................43 43

8APPENDIX ...................................................................................................................................................................... ......................................................................................................................................................................44 44

University University of Sheffield Sheffield Department Department of Civil Civil Structural Engineering Engineering Date 16/02/2009

Geometry Geometry of t he Frame Frame

Reference

Calculation

1 Geometry

Sheet No 2


Revised Revised by:

EC3 EC3 Plastic Portal Frame Design Design

Prepared Prepared b y: Cia06mh

Date 16/02/2009

Client Client brief

Sheet No

Reference

Calculation

3

2 Design Brief A client requires a single-storey building, building, having a clear floor area 30 m x 80 m, with a clear height to the underside of the roof steelwork of 5 m. The slope of the roof member is to be at least 6o.

Figure 1‐ Plan view of of the the frame

Figure 2‐ 3 Dimensional view of of the the building (Plum, 1996)


Revised Revised by:

EC3 EC3 Plastic Portal Frame Design Design

Prepared Prepared b y: Cia06mh

Date 16/02/2009

Determining load on the frame

Sheet No

Reference

Calculation

4

3 Determining loading on frame 3.1Combination factors ψ

EN 1991-11:2002 (E) Annex A 1

The combination ψ must be found from Eurocode 1 (EN1991-1) or relevant NAD. Note that because most portal frame designs are governed by gravity (dead + snow) loading, so in this worked example only maximum vertical load combination is considered. Therefore, the combination factor ψ is never applied in this example, but for full analysis the following load combination should be considered

See Supporting Notes Sec 6.4

1) Maximum gravity loads loads without wind, causing causing maximum sagging sagging moment in the rafter and maximum hogging moments in the haunches. 2) Maximum wind loading loading with minimum gravity loads, causing maximum maximum reversal of moment compared with case 1. The worst wind case might be from either transverse wind or longitudinal wind so both must be checked.

Figure 3‐ Frame spacing (SX016, Matthias Oppe)

Basic data : • • • • •

Total length: Spacing: Bay width: Height (max): Roof slope:

b = 72 m s = 7.2 m d = 30 m h = 7.577m o α = 6

University of Sheffield Department of Civil Structural Engineering

Revised by:

EC3 Plastic Portal Frame Design

Prepared b y: Cia06mh

Date 16/02/2009

Determining loading on the frame

Sheet No

Reference

Calculation

5

3.2Snow loading General EN 1991-1-3 Sec 5.2.2 Eq.5.1

Snow loading in the roof should be determined as follow

     µ

Where: µ is the roof shape coefficient is the exposure coefficient usually taken as 1 is the thermal coefficient set to 1 for nominal situations Is the characteristic value of ground snow load for relevant altitude.

 

EN 1991-1-3 Sec 5.3 Table 5.1 See Appendix A Table A1

Roof shape coefficient Shape coefficients are needed for an adjustment of the ground snow load to a snow load on the roof taking into account effects caused by non-drifted and drift ed snow loading. The roof shape coefficient depends on the roof angle so

Snow load on the ground

EN 1991-1-3 Annex C See Appendix A Table A2

030



µ =0.8

For the snow load on the ground; the characteristic value depends on the climatic region; for site in the UK the following expression is relevant Sk=0.140z-0.1+(A/501) Where: Z is the( zone number /9 ) depending on the snow load on sea level here in Sheffield z=3 A is the altitude above sea level A=175m


Revised by:



Date 16/02/2009

Determining loading on the frame

Sheet No

Reference

Calculation

6

Snow load on the roof Sk = 0.8 x 1 x 1 0.67 = 0.54 KN/m2 Spacing = 7.2 m For internal frame UDL by snow = 0.54 x 7.2 = 3.89 m

Figure 4‐ Distributed load due to snow per meter span (SX016, Matthias Oppe)

3.3Self weight of steel members

Self-weight estimated needed to be checked at the end

Assume the following weight by members, • • •

Roofing = 0.2 KN/m2 Services = 0.2 KN/m2 Rafter and column self weight = 0.25 KN/m2

Total self weight

_____________ 0.65 KN/m2


Revised by:



Date 16/02/2009

Initial sizing if members

Sheet No

Reference

Calculation

7

4 Initial Sizing of members TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3 See Appendix B for the method Figure 5‐ Dimensions of portal (The institutionof Structural Engineers, TP/08/43 EC3/08/16)

a) L/h = 30/6 = 5 r/L = 1.577/30=0.0526 b) Loading 1) Gravity loading Snow loading = 0.54 x 7.2 = 3.80KN/m Self weight = 0.65 x 7.2 = 4.68 KN/m 2) Factored load w= (4.68 x 1.35 ) + (3.80 x 1.5 ) = 12.0 KN/m c) Finding Mp for the sections 1) Total load on the frame (wL)= 12.0 x 30 = 360.5KN 2) Parameter wl2 = 12.0 x 302 = 10816 KNm

3) From Graphs (Figure B2) obtain horizontal force ratio (0.36) H= 0.36 x 360.5 = 129.8 KN 4) From Graphs (Figure B3) obtain rafter Mp ratio (0.034) Mrafter,,Rd = 0.034 x 10816 = 367.7 KNm


Revised by:



Date 16/02/2009

Initial sizing if members

Sheet No

Reference

Calculation

8

5) From Graphs (Figure B4) obtained column Mp ratio (0.063) Mcolumn, Rd = 0.063 x 10816 = 681.4 KNm. 6) Selecting members a) Wpl (rafter),required = (367.7 x 106) / 275 = 1337 x 10 3 cm3 Try UB 457x152x74 b) Wpl(column),required = (681.4 x 106)/275= 2478 x 103 cm3 Try UB 533 x 210 x 109

Section Tables of Universal Beams

•

Properties Rafter Section UB 457x152x74 G=74.2 Kg/m tw=9.6mm d=428mm Iy= 32670 x 104 mm4 iy=186 mm Iz = 1047 x 104 mm4 It = 66.18 x 104 mm4

•

Iy= 66820 x 104 mm4 iy=218.7 mm Iz = 2692 x 104 mm4 It = 101.6 x 104 mm4

 

b=154.4mm A=94.48 x 102 mm2

Wpl,y=1627 x 103 mm3 iz = 33,3 mm Wpl,z = 213.1 x 103 mm3 Iw = 516.3 x 106 mm6

Properties Column Section UB 533x210x109 G=109 Kg/m tw=11.6mm d=510.9mm

EN 1993-1-1: 2005 (E) Table 3.1

h= 462mm tf =17mm

h= 539.5mm tf =18.8mm

b=210.8mm A=138.9 x102 mm2

Wpl,y=2828 x 103 mm3 iz = 45.7 mm Wpl,z = 399.4 x 103 mm3 Iw = 1811 x 106 mm6

Steel grade is S275 Assume Sections Class1, then check


Revised by:



Date 16/02/2009

Ultimate Limit State Analysis

Sheet No

Reference

Calculation

9

5 Load Combination (Max vertical Load) (Dead + Snow) 5.1Frame imperfections equivalent horizontal forces

ØØ   Ø √  0.5 1 . 5 Ø

EN 1993-11:2005 (E) Sec 5.3.2 See Supporting Notes Section 9

=

= 3.54 x 10-3

The column loads could be calculated by a frame analysis, but a simple calculation based on plan areas is suitable for single storey portals (i)

Permanent loads ( un-factored ): Rafter = (74.5 x 15 x 9.8) / 103 = 11 KN Roofing = (15 x 0.2 x 7.2) = 21.6 KN Services = (15 x 0.2 x 7.2) = 21.6 KN _________ Total = 54.2 KN

(ii)

Variable loads ( un-factored ) Snow load = 15 x 0.54 x 7.2 = 58.3 KN

Thus the un-factored equivalent horizontal forces are given by: (i)

Permanent/column = 3.54 x 10-3 x 54.2 = 0.19 KN

(ii)

Variable/column = 3.54 x 10-3 x 58.3 = 0.21 KN

Note EC3 requires that all loads that could occur at the same time are considered together, so the frame imperfection forces and wind loads should be considered as additive to permanent loads and variable loads with the appropriate load factors.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

10

Figure 6‐Frame imperfections equivalent horizontal forces

5.2 Partial safety factors and second order effects For Second Order effects See Supporting Notes Section 7.1 & Section 7.2

Second order effects increases not only the deflections but also the moments and forces beyond those calculated by the first order. Second-order analysis is the term used to describe analysis method in which the effects of increasing deflections under increasing load are considered explicitly in the solution method. The effects of the deformed geometry are assessed in EN 1993-1-1 by calculating alpha crit (αcrit) factor. The limitations to the use of the first-order analysis are defined in EN 1993-1-1 Section 5.2.1 (3) as αcrit 15 for plastic analysis. When a second order analysis is required there are two main methods to proceed:



1) Rigorous 2nd order analysis (i.e. using appropriate second order software). 2) Approximate 2nd order analysis (i.e. hand calculation using first order analysis with magnification factors). Although the modifications involve approximations, they are sufficiently accurate within the limits given by EN 1993-1-1.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

11

•

See Supporting Notes Section 7.3

•

Carrying first order analysis to obtain first order moments and member forces using partial safety factors (γG=1.35) and (γQ=1.5) with loading calculated above. Then Checks if second order effects are relevant by calculating the following αcr,est =min ( αcr,s,est , αcr,r,est ) where αcr,s,est = estimated of αcr for sway buckling mode αcr,r,est =estimated of αcr for rafter snap-through buckling mode.

Figure 7 ‐ Bending moment diagram for first order analysis (Burgess, 20/01/1990)


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

12

Load Factor 1.02 1.14

Hinge number 1 2

Member RHC LHR

Hinge status Formed Formed

Table 1‐ Position of Hinges and Load factors

Figure 8‐ – Member forces (Burgess, 20/01/1990)


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

13

5.2.1 Sway buckling mode Stability ( αcr,s,est).

See Supporting Notes Section 7.3.2.1

αcr,s,est =  

,,

0.8 1,, , ,, is the axial force in rafter {see figure 8 (150.8KN)} is the Euler load of rafter full span

  ,    



Where

is the in-plan second moment of area of rafter L is the full span length.







•

See Figure 9

,     , ,, , ,

= 752 KN

is the minimum value for column 1 to n

is the horizontal deflection for top of column as indicated in Appendix is the axial force in columns {see figure 8 (207.5KN , 208.1KN)}

,

As can be seen that is the lateral deflection at the top of each column subjected to an arbitrary lateral load HEHF then here an arbitrary load HEHF can be chosen and using analysis software the deflection at top of each column can be obtained.

,,

1) Arbitrary load HEHF=50KN 2) = 98mm = 98mm


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

14

Figure 9‐ Sway mode check (Burgess, 20/01/1990)

So either

.   .   , ,, 0.8 1.  14.75  9.5 OR

Min Thus αcr,s,est =

= min(14.75 , 14.75 ) = 14.75


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

15

5.2.2 Snap-through buckling stability (αcr,r,est )

See Supporting Notes Section 7.3.2.2 & Section 7.2

αcr,r,est =        

Thus

.  tan2

D cross-section depth of rafter (462mm). L span of the bay (30m). h mean height of the column (6m). in-plane second moment of area of column (66820 x 104 mm4) in-plane second moment of area of rafter (32670 x 104 mm4 ) nominal yield strength of the rafter (275 N/mm2) roof slope if roof is symmetrical (6o) Fr /Fo the ratio of the arching effect of the frame where Fr = factored vertical load on the rafter ( 432 KN see section 3) F0 = maximum uniformly distributed load for plastic failure of the rafter treated as a fixed end beam of span L

   

   ,, 16162710 275 238.6  3010

= 1.81

αcr,r,est =

  .              .   tan26

αcr,r,est = 6.2

Hence αcr,est =min ( αcr,s,est , αcr,r,est ) = 6.2

•


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

16

•

Although the snap-through failure mode is critical mode as shown in calculation above, but because this example is for designing single bay portal frames, the snapthrough mode of failure is irrelevant but included to show complete design steps for simple portal frame design. Snap-through failure mode can be critical mode in three or more spans, as internal bay snap-through may occur because of the spread of the columns inversion of the rafter (The institutionof Structural Engineers, TP/08/43 EC3/08/16) see figure 10.

See Supporting Notes 7.3.3 Figure 10‐ Snap through failure mode critical for 3 bay or more

5.3.2Accounting Second Order effects

To account for second order effects the partial safety factors can be modified by the following criteria 1) 2)

γG  γQ 

= 1.50 = 1.68


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

17

•

See Supporting Notes 7.3.3

Re-analyze the first order problem with the modified safety factors using same initial sized sections gives the following results,

Load Factor 0.92 1.02

Hinge number 1 2

Member RHC LHR

Hinge status Formed Formed

Table 1‐ Hinges obtained from analysis

•

•

It could be seen that using Sections UB 533 x 210 x 109 and UB 457 x 152 x 74 is suitable, although hinge 1 occurs at a load factors ≤ 1 , a mechanism is not formed until the second hinge is formed. Therefore this combination of section sizes is suitable

Hence size of member initially estimated is suitable and can withstand second-order effects. Note that if the load factors in positions 1 and 2 were less than 1, then the members size needs to be increased to sustain second order effects as the initially sized members cannot sustain second order effects.

Figure 11‐Bending moment diagram for first order analysis (Burgess, 20/01/1990)


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

18

Figure 12‐Member forces for first‐order analysis (Burgess, 20/01/1990)


Revised by:



Date 16/02/2009

Member checks – Purlins

Sheet No

Reference

Calculation

19

6 Member checks See Supporting notes section 10.4

Note. Here the safety factors are used as indicated in King span load table

6.1 Purlins Today the design of the secondary members is dominated by cold formed sections. The ‘design’ of cold formed members consists of looking up the relevant table for the chosen range of sections. The choice of a particular manufacture’s products is dependent on clients or designer’s experiences and preferences. Table (Appendix C) illustrates a typical purlin load table based on information from manufacture’s catalogue (King span) for double span conditions. As the overall distance between columns is 30 meters, which is assumed to be divided to 18 equal portions would gives purlin centers 1.67 meters (on the slope). The gravity loading (dead (cladding Load plus snow load) is w= (0.1x 1.4) + (0.54 x 1.6) = 1.004 KN/m2. From the Table (Appendix C), knowing the purlin length of 7.2 m, purlin spacing of 1.25m and the gravity load to be supported by purlin 1.004KN/m2, the M175065120 section seems adequate.

See Appendix C

Purlin

Rafter

Figure 13‐ Connection between rafter section and purlins

Figure 14‐ Purlin cross‐section (Kingspan)


Revised by:



Date 16/02/2009

Member checks – Colu mn

Sheet No

Reference

Calculation

20

6.2 Column (UB 610 x 229 x 101) See Figure11

-

MEd = 904.7 KNm VEd = 150.1 KN NEd = 208.2 KN

6.2.1Classification 

EN 1993-1-1: 2005 (E) Section 5.5

Web ( Bending + Axial ) ε =

 275/235 .. 44.04 72ε =1.08

actual (d/tw ) = 

Flanges ( Axial Compressive )

actual (c/tf )= •

Class 1

. .. . 4. 6 19ε

Class1

So the column sections are overall class 1

6.2.2Cross section resistance

The frame analysis assumed that there is no reduction in the plastic moment resistance from interaction with shear force or axial force. This assumption must be checked; 6.2.2.1Shear force effects of Plastic moment resistance (EN 1993-1-1: 2005 (E) Sec 6.2.6)

See supporting notes section 12.3

VEd < 0.5 Vpl,Rd

6508./5√ 2375/ /γ√ 3 /1.1 10 

Av = 1.04 h tw = 1.04 x 539.5 x 11.6 = 6508.5 mm2 Vpl,Rd = Vpl,Rd = 0.5 Vpl,Rd = 469.7 KN •

939.4 KN

VEd < 0.5 Vpl,Rd so the plastic moment of resistance is not reduced by the coexistence of axial force


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

21

6.2.2.2Axial force effects of Plastic moment resistance (EN 1993-1-1: 2005 (E) Sec 6.2.9)

Check i.


If

 ., 

   . ... 

NEd <

208.2 <

208.2 < 727.8 ii.

If NEd < 0.25 Npl,Rd NEd < 0.25 plastic tensile resisitance of the section

.  . .. 

NEd <

208.2 <

208.2 < 868.1 •


Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN1993-1-1: 2005.

6.2.3Stability against lateral and torsional buckling (EN 1993-1-1: 2005 (E) Sec BB3.2.1):

. The design of the frame assumes hinge forms at the top of the column member, immediately below the haunch level. The plastic hinge position must be torsionally restraint in position by diagonal stays. With the hinge position restraint, the hinge stability is ensured by EC3 by limiting distance between hinge and the next lateral restraint to Lm.

        38 / 57.4  7561   235


Revised by:



Date 16/02/2009

Member checks - Colum n

Sheet No

Reference

Calculation

22

 3845.7/ ...     . .  

= 1.53 m

•

Thus there must be a lateral restraint at a distance from the hinge not exceeding (1.53m).

Figure 15‐ Column member stability (Plum, 1996)

•

Therefore if 1.5 meters spacing assumed, this would ensure the stability between the intermediate restraints at the top of the column where maximum bending moment occurs, then the spacing of 1.8 meters is OK for sheeting rails below 2.4 meters from the top of the column, where the moment is l ower.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

23

It must be checked that the column buckling resistance is sufficient, so that the column is stable between the tensional restraint at S2 and the base. This part of the column would be checked using slenderness calculated.


Different countries have different procedure to calculate the slenderness of the column and check the susceptibility of this part to lateral tensional buckling. Thus the designer must refer to the national Annex. In this example the procedure used in for assessing the significance of the mode of failure is taken from (King, Technical Report P164).

Figure 16‐ Column between tensional restraints (King, Technical Report P164)

(a) Calculate slenderness λ and λ LT Assume side rail depth = 200 mm

Figure 17‐ Column/ Sheeting rails cross‐section (King, Technical Report P164)


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

24

(King, Technical Report P164)

Distance from column shear center to center of the side rail, a a = 607.3/2 + 200/2 = 369.75 mm is2 = iy2 + iz2 + a2 is2 = 218.72 + 45.72 + 369.75 2 = 186633 mm 2

Distance between shear center of flanges hs = h – t f = 539.5 – 18.8 = 520.7mm

α =

   

using the simplification for doubly symmetrical I sections Iw = Iz ( hs / 2 )2

α =

α

=

  .   .       

= 1.122

The slenderness of the column is given by:

  

  .   .   1  .... .

= 64.35


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

25

.     ,    .     Appendix D Figure D1

Where: mt is moment factor obtained from appendix D . Because loads combination considered there is no lateral loads applied to the walls, so there are no intermediate loads ψ = 0 / 603.1 = 0 y = 82.632 / ( Lt / iz ) = 82.632 / (4000 / 45.7 ) = 0.944 mt = 0.53  c =1 for uniform depth members

.      .     0.53. 1 .    64.35

(b) Calculate buckling resistance for axial force

EN1993-11:2005 Sec 6.3.2.2

    /  ,     .       Ф0.  1/ФФ 5  1  0.2   h/b = 539.5/210.8 = 2.56

curve b for hot rolled I sections α= 0.34

EN1993-11:2005 Table 6.3

   64. /35/82.8 0.78 82.8

= 42.1


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

26

Xz =

Ф0.5  1 0.340.0.780.90 2  0.78  1/0.90 0.90  0.78. ,    /  = 0.741

2

3

= (0.741 x 138.9 x 10 x 275) / (1.1 x 10 )= 2574.13KN

(c) Calculate buckling resistance for bending Mb,Rd,y = EN1993-11:2005 Sec 6.3.2.2

EN1993-11:2005 Table 6.3

 , 

  /86.8

= 42.1/ 86.8 = 0.485

   Ф0. 5  1  0. 2       Ф0.5  1 0.2100..4850.65 2  0.485     .  .  1/ 1/ФФ 0.65 0.65  0.485   = 0.92

Mb,Rd,y =

. .     650.4 


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

27

(d) Calculate buckling resistance to combined axial and bending

,,  ,,, 1


Ψ = 0

βM,LT = 1.8 – 0.7 Ψ = 1.8 – 0.7 (0) = 1.8

LTLT  0.0.1155 0. M.78LT1.0.810.5 but15 LTbut0.LT 0. LT 0.0606  1 LT 0.0606208.   1.2100   1    1. 0  0. 7 41138. 9 10 275 0.996 μ μ

β

9

μ

μ μ

9

μ

.  .  .   . . •

= 0.91

The column is OK and stable over the section considered (between restraint S o and S2).


Revised by:



Date 16/02/2009

Member c hecks - Rafter

Sheet No

Reference

Calculation

28

6.3Rafter (UB457 x 191 x 89) 6.3.1Section Classification

Ensure the section is class 1 to accommodate plastic hinge formation.

EN 1993-1-1: 2005 (E) Section 5.5

ε = 

 275/235

=1.08

Web ( combined axial and bending )

actual (d/tw ) =

. 44.6

44.6 ≤72 ε Class 1



Flanges ( Axial Compressive )

actual (c/tf )= •

. . . 3. 6 69ε

Class1

The rafter section is Class 1

6.3.2 Cross-section Resistance.

The frame analysis assumed that there is no reduction in the plastic moment resistance from interaction with shear force or axial force. This assumption must be checked because it is more onerous than that the cross-sectional resistance is sufficient. See Figure11 (Burgess, 20/01/1990)

-

Max. shear force VEd = 160.5 KN at haunch tip

-

Max. axial force NEd = 166.9 KN

at haunch tip


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

29

6.3.2.1Shear force effects of Plastic moment resistance (EN 1993-1-1: 2005 (E) Sec 6.2.6)

VEd < 0.5 Vpl,Rd See supporting notes section 12.3

Av = 1.04 h tw = 1.04 x 462 x 9.6 = 4613 mm2 Vpl,Rd = Vpl,Rd =

/√ 3 /γ  4613 275/√ 3 /1.1 10  

666 KN

0.5 Vpl,Rd = 333 KN •

VEd < 0.5 Vpl,Rd so the plastic moment of resistance is not reduced by the coexistence of axial force.

6.3.2.2Axial force effects of Plastic moment resistance (EN 1993-1-1: 2005 (E) Sec 6.2.9)

Check i) If


.  ,

   . ... 

NEd < 166.9 <

166.9 < 531.4 ii)

If NEd < 0.25 Npl,Rd

. 

NEd < 0.25 plastic tensile resistance of the section NEd < 231.1 <

•

. .. 

166.9 < 590.5 Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN1993-1-1: 2005.


Revised by:



Date 16/02/2009

Member c hecks – Rafter

Sheet No

Reference

Calculation

30

6.3.3 Check rafter buckling in apex region

Another highly stressed region is the length of rafter in which the ‘apex’ hinge occur see fig below. Under (dead + snow) loading, the outstand flange is in tension, while compression flange is restrained by purlin/rafter connection.


Figure 18‐ Member stability apex region (Plum, 1996)

Therefore, the buckling resistance of the rafter member between purlins in the apex region needs to be checked. Because the ‘apex’ hinge is the last to form in order to produce a mechanism (which is true for low pitched portal frame under dead + snow loading), then adequate rotation capacity is not a design requirement, i.e. hinge is required only to develop Mp not to rotate. It is set by EC3 EN1993-1-1: 2005 that if the value of Lm (as defined in BB.3.1.1) is not exceeded by restraint lateral torsional buckling can be ignored. So assuming that the purlins act as restraint because of their direct attachment to the compression flanges in the apex hinge region, then the purlin spacing should not exceed EN 1993-1-1: 2005 (E) Sec BB3.2.2

         38 / 57.4  7561   235   3833.3 166.910     1 627  10  275 / 57.494.4810  7561  94.4810  66.1810  235

See section 6.1

Lm = 1221 mm = 1.221m •

As purlin spacing is 1.67m (on slope), thus because Lm has been smaller than 1.67m then the purlin spacing would have to be reduced in the apex region to 1.2m.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

31

6.3.4 Stability check for lower bending moments

Where bending moment is lower, the purlin spacing can be increased:

Figure 19‐ Rafter under lower bending moments

Next critical case is in right hand rafter. Try purlin spacing at 1670 mm centres Check for lateral torsional buckling between purlins: See Figure11 (Burgess, 20/01/1990)

o o

MEdmax.y = 345.6 kNm NEd.max = 166.9 kN

at haunch tip at haunch tip

(a) Calculate buckling resistance to axial force EN1993-11:2005 Sec 6.3.2.2 EN1993-11:2005 Table 6.3

L =1670 mm λ z = L/iz = 1670/33.3 = 50.15 λ z = λ z / 86.8 = 50.15 / 86.8 = 0.578 2 Ф = 0.5 [ 1 + α ( λ – 0.2 ) + λ ] =0.5 [ 1 + 0.34 ( 0.578 – 0.2) + (0.578)2 ] = 0.7313 Xz =

  Ф√ Ф  .√ . .

Nb,Rd,z =

=

   ..  =

= 2003.0 KN

= 0.8480


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

32

(b) Calculate buckling resistance to bending moment EN1993-11:2005 Sec 6.3.2.2 •

               ,      

Take C1 = 1 (conservative)

      1  2 10000  1047  10 5 16. 3  10 1 670  81000  66. 1 8  10   1670  10  1047  10    210000  1047  10   648.1  •

λ LT = λ LT =

EN1993-11:2005 Table 6.3

•

•

= 0.831

ФLT = 0.5 [ 1 + α ( λ – 0.2 ) + λ2 ] 2 ФLT = 0.5 [ 1 + 0.21( 0.831 – 0.2 ) + 0.831 ] = 0.912

XLT =

XLT =

•

 ,   .

 Ф√ Ф  .√ . .

Mb,Rd,y =

Mb,Rd,y =

 ,  . . 

= 0.80

= 325.4KNm


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

33

(c) Calculate buckling resistance to combined axial and bending

Check that

,, .

+ kLT

Take kLT = 1 ( conservative )

+1x 1

,,  1 ..  1

1

The value is slightly greater than one but due to the conservative assumption of KLT=1 the rafter can be assumed to be stable between intermediate restraint (purlin/sheeting rails) and purlin spacing could be increased to 1.67m between apex and hunch region shown in figure 19. Otherwise if the value was significantly greater than 1 the purlin spacing (1.67m) should be reduced.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

34

6.4Haunch (UB 457 x 191 x 89) The depth of a haunch is usually made approximately twice depth of the basic rafter sections, as it is the normal practice to use a UB cutting of the same serial size as that of the rafter section for the haunch, which is welded to the underside of the basic rafter (UB 457x191x 89). Therefore it is assumed that the haunch has an overall depth at connection is 0.90 m.

EN 1993-1-1: 2005 (E) Section 5.5

6.4.1Classification

ε =

 275/235



=1.08

Web The web can be divided into two, and classified according to stress and geometry of each. actual (d/tw ) =

.  44.6

web 1 ( bending ) -------web 2 ( Compressive) ---

44.44.66

≤72 ε Class 1 ≤38 ε Class 2

Figure 20‐ Haunch region cross‐section classification (King, Technical Report P164) 

Flanges ( Axial Compressive

actual (c/tf )= •

. .  .  3.66  9ε

Thus the haunch section is a class 2.

Class1


Revised by:



Date 16/02/2009

Member ch ecks - Haunch

Sheet No

Reference

Calculation

35

6.4.2Haunch Stability

See Supporting notes section 13.1 & 13.3

First, check the stability of the haunched portion of the rafter ( from eaves connection to the haunch/ rafter intersection) as this represents one of the most highly stressed lengths, and with its outstand flange (inner) in compression, this part of the rafter is the region most likely to fail due to instability. As it has already decided to stay the inside corner of the column/haunch intersection (column hinge position), assume that the haunch/rafter intersection is also effectively torsionally restrained be diagonal braces, giving an effective length of 3m as indicated in Fig below.

Figure 21‐ Member stability haunch‐rafter region (Plum, 1996)


Revised by:



Date 16/02/2009

Member ch ecks – Haunch

Sheet No

Reference

Calculation

36

EN 1993-1-1: 2005 (E) clause BB.3.1.2 (3)B

It would appear that clause BB.3.1.2 (3)B is the most appropriate creation to check the stability of the haunched portion, as there is three flanged haunch, so the distance between rotational restraint should be limited t o

   

Where: L k is length limit specified where lateral torsional buckling effects can be ignored where the length L of the segment of a member between restraint section at a plastic hinge location and adjacent torsional restraint.

Lk

Lk

  .     .

    .      .       . 

= 3738 mm

Lk = 3.738m

c EN 1993-1-1: 2005 (E) section BB.3.1.3 See supporting notes section Appendix B

is the taper factor (shape factor) which accounts for the haunching of the restraint length (BB.3.3.3)

    1          1       = 1.15


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

37

Figure 22‐ Dimensions defining taper factor (BS EN 1993‐1‐1:2005)



Is the modification factor for non-linear moment gradient (BB.3.3.2).

The plastic moduli are determined for five cross-sections indicated on the figure below, the actual cross-section considered are taken as being normal to the axis of the basic rafter (unhaunched member). The plastic moduli together with the relevant information regarding the evaluation of the ratios Ni/Mi are given in the following table. The worst stress condition at the hunch/rafter intersection (location 5) is taken.


Revised by:



Date 16/02/2009

Member ch ecks – Haunch

Sheet No

Reference

Calculation

38

Figure 23‐ Member stability haunch region (Plum, 1996)

Position on haunch (FIG ) Distance from the eaves (m) Depth of bottom web (mm) Factored moment (My,Ed) (KNm) Factored axial force (NEd) (KN) Moment capacity (KNm) Plastic modulus (cm3) Ratio (N/M) a Value for (R) calculation (mm) R Value

1

2

3

4

5

0

0.75

1.5

2.25

3

428

321

214

107

0

904.5

764.8

625.1

485.3

345.6

171.1

170.1

169.1

168.1

167.1

1157

1011

784

686

447

4209

3677

2849

2495

1627

0.19

0.22

0.27

0.35

0.48

532.5

479.0

425.5

332.0

231.0

0.86

0.84

0.89

0.79

0.86

Table 2‐ Member forces at locations indicated in figure 18


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

39

The modification factor

See supporting notes section Appendix A



is determined by the form;

  

in which R1 to R5 are the values of R according to equation below at the ends, quarter points and mid-length ( R values at positions 1 to 5 indicated in Table 2)

 

In addition, only positive values of ( ) should be included where, - RE is the greater of R1 and R5 - Rs is the maximum value of R anywhere ( R1 to R5 ) -

    ,    , 

Where (a) is the distance between the centroid of the member and the centroid of restraining members (such as purlins restraining rafter). Here for simplicity a conservative value of (a) is found by conservative method of ignoring the middle flange as shown if figure (19).

Figure 24‐ Simplification for distance between centriod of rafter and purlin sections


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

40

  3 4 312  2   0.86  30.84  40.89 1230.79 0.8620.890.86         1.17

•

•

  √ ...  3.4  3

Thus this portion of the rafter is stable over the assumed restrained length of 3 m (haunch length), as Ls is around 3m.



If the value was found to be less than the haunch length then a torsion restraint should be provided in the haunch region as shown below

See supporting notes section 12.3 6.4.3 Cross-section resistance.

6.4.3.1Shear force effects of Plastic moment resistance

The shear in the rafter has been checked above, showing that VEd < 0.5 Vpl,Rd . In the haunch, the shear area Av increases more than the applied shear VEd, so the shear force has no effect on the plastic moment capacity of the haunch.


Revised by:



Date 16/02/2009


Sheet No

Reference

Calculation

41

6.4.3.2Axial force effects of Plastic moment resistance,

The tables provided below gives the axial and moment resistances of the haunch section at points 1to 5 shown in figures 18. A series of checks is carried out to determine whether the cross-sectional moment resistance MN,Rd is reduced by coexistence of axial force.


•

Position

Distance (mm)

NEd (KN)

A (mm2)

Npl,Rd (KN)

Aweb (mm2)

(Aweb, f y )/ymo (KN)

1

0

171.1

16092

4023

8216

2054

2

0.75

170.1

15064

3766

7190

1798

3

1.5

169.1

14038

3510

6163

1541

4

2.25

168.1

13010

3253

5136

1284

5

3

167.1

11983

2996

4109

1027

Npl,Rd = A f y / ymo

and

f y=275N/mm

2

Table 3 – Axial force at positions indicated in figure 18 for haunch

Is NEd > 0.5 Aweb f y 0.25 /ymo Npl,Rd

Does Axial force affect plastic bending resistance

Position

Distance (mm)

MEd (KNm)

1

0

950

No

No

No

2

0.75

850

No

No

No

3

1.5

751

No

No

No

4

2.25

652

No

No

No

5

3

553

No

No

No

Table 4‐ Checking the significance of axial force on plastic moment of resistance •

Therefore, the effect of shear and axial on the plastic moment resistance of the column sections can be neglected according to EC3 EN1993-1-1: 2005.


Revised by:



Date 16/02/2009

Comparison between out comes of different codes

Sheet No

Reference

Calculation

42

7. Comparison between Different Codes As the dimensions of portal frame designed in this worked example were deliberately chosen to be exactly the same as worked-example in (King, Technical Report P147) a comparison was done between the carried out worked example to (BS EN 1993-1-1:2005), (BS9590-1:2000) and (ENV1993-1-1:1992). The following is a summary of different outcomes and source of design, Design no

Design Code

Column Section size

Rafter Section Size

Haunch Length (m)

Purlin Spacing (m)

Design Source

1

BS EN 1993-11:2005

533x210 UB 109

457x152UB 74

3

1.67

Worked example

2

BS9590-1:2000

610x229 UB 113

533x210 UB 82

3

1.85

3

ENV1993-11:1992

610x229 UB 113

457x191UB 74

3

1.85

Table 5‐ Comparison between different code outcome

(King, Technical Report P164) (King, Technical Report P164)


Revised by:



Date 16/02/2009

Ap pendi x A

Sheet No

Reference

Calculation

43

8Appendix EN 1991-13:2003 (E) Section 5.3.2

A.1 Ro of sh ape c oef fi ci ent The values given in table A1 apply when the snow is not prevented from sliding off the roof. Where the snow fences or other obstruction exists or where the lower edge of the roof is terminated with a parapet, then the snow load shape coefficient should not be reduced below 0.8.

Table A1- Snow load shape coeff ici ents (BS EN 1991-1-3:2003)

EN 1991-13:2003 (E) Table C1

A.2 Snow load relationships The snow load on ground; the characteristic value depends on the climatic region; the following table gives different expressions for different regions,

Table A2- Altitude-Snow load relationships (BS EN 1991-1-3:2003)


Revised by:



Date 16/02/2009

Ap pendi x A

Sheet No

Reference

Calculation

44

Where: Sk A Z

is the characteristic snow load on the ground (KN/m2) is the site altitude above the sea level (m) is the zone number given on the map ( see fig A1 )

The following maps gives the zone number Z for UK and republic of Ireland if other Z values for regions mentioned in Table A2 refer to EN 1991-1-3 Annex C pages ( 41 to 52 ).

EN 1991-13:2003 (E) Figure C.4

Figure A1 UK , Republic of Ireland : snow loads at sea level (BS EN 1991-1-3:2003)


Revised by:



Date 16/02/2009

Ap pendi x B

Sheet No

Reference

Calculation

45

B1 Initial sizing using Weller’s charts TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3

The method described relies for its simplicity on a series of three charts developed by Alan Weller. The chart has been constructed with the following assumptions and which leads to reasonably economic solution (Note. This is not a rigorous design method; it is a set of rules to arrive at initial size). 1) 2) 3) 4) 5)

The rafter depth is approximately span / 55 The hunch length is approximately span /10 The rafter slope lies between 0o and 20o. The ratio of span to eaves height is between 2 and 5. The hinges in the mechanism are formed at the level of the underside of the haunch in the column and close to the apex.

Each chart requires a knowledge of the geometry of the frame and the design loading as input data in order to determine approximate sizes for the column and rafter members Using of charts

Figure B1 Dimensi ons o f por tal (The insti tut iono f Struc tur al Engineers, TP/08/43 EC3/08/16) a) Calculate the span/height to eaves ratio = L/h b) Calculate the rise/span ratio = r/L c) Calculate the total design load FL on the frame and then calculate FL2, where F is the load per unit length on plan of span L (e.g. F =qs, where q is the total factored load per m2 and s is the bay spacing). d) From figure B2 obtain the horizontal force ratio HFR at the base from r/L and L/h e) Calculate the horizontal force at the base of span H=HFR W L.


Revised by:



Date 16/02/2009

Ap pendi x B

Sheet No

Reference

Calculation

46

TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3

f) g) h) i) j)

From figure B3 obtain the rafter Mp ratio MPR from r/L and L/h. Calculate the Mp required in the rafter from Mp (rafter) = MPR x W L2. From figure B4 obtain the column Mp ratio MPL from r/L and r/h. Calculate the Mp required in the rafter from Mp (rafter) = MPL x W L2. Determine the plastic moduli for the rafter Wpl,y,R and the column Wpl,y,C from Wpl,y,R =Mp,(rafter) /f y Wpl,y,C = Mp,(column) /f y Where f y is the yield strength.

Using the plastic moduli, the rafter and column sections may be chosen from the range of plastic sections as so defined in the section books.

Span to eaves height 2.0 2.5 3.0 3.5 4.0 4.5 5.0

0.2

0.15 n a p s / e 0.1 s i R

0.05

0

0.1

0.15

0.2

0.25

0.3

0.35

0.4

H/wL Figure B2- Horizontal force at the base (The institutionof Structural Engineers, TP/08/43 EC3/08/16)

0.45


Revised by:



Date 16/02/2009

Ap pendi x B

Sheet No

Reference

Calculation

47

TP/08/43 EC3/08/16 Manual for the design of steelwork building structures to EC3

Span to eaves height

4.5 3.5 2.5 5.0 4.0 3.0 2.0

0.2

0.15 n a p s / e 0.1 s i R

0.05

0 0.02

0.025

0.03

0.035

M

0.04 r /

0.045

wL²

Figure B3- Mp ratio required for the rafter (The

Span to eaves height 0.2

5.0 4.5 4.0 3.5 3.0

2.5

2.0

0.15 n a p s / e 0.1 s i R

0.05

0 0.045

0.05

0.055

0.06

0.065

0.07

0.075

0.08

Mpl / wL² Figure B4- Mp ratio required for the column (The instit utionof Stru ctu ral Engineers, TP/08/43 EC3/08/16)


Revised by:



Date 16/02/2009

Ap pendi x C

Sheet No

Reference

Calculation

48

C1 King Span Multi-beam Purlin (Load tables) King Span Website Loading Link Dead load http://www.ki Dead load restraining uplift or overturning ngspanstructur Dead load acting with wind and imposed loads combined al.com/multibe Imposed load Imposed load acting with wind load am/rp/load_ta Wind load bles.htm Wind load acting with wind and imposed load Forces due to temperature effects

Load Facto r 1.4 1.0 1.2 1.6 1.2 1.4 1.2 1.2


Revised by:



Date 16/02/2009

Ap pendi x C

Sheet No

Reference

Calculation

49


Revised by:



Date 16/02/2009

Ap pendi x D

Sheet No

Reference

Calculation

50


D1 Equivalent uniform moment factor mt for all other cases This formula, derived by (Sinhgh, 1969), is applicable in all cases, especially when the bending moment diagram is not a straight line between the tensional restraints defining the ends of the element.

Figure D1- Moment factor s (Kin g, Techni cal Report P164)


Revised by:



Date 16/02/2009

Ap pendi x D

Sheet No

Reference

Calculation

51


MEd1 to MEd5 are the values of the applied moments at the ends, the quarter points and mid- length of the length between effective torsional restraints, as shown in Figure D.2. Only positive values of MEd should be included. MEd is positive when it produces compression in the unrestrained flange.

Figure D2- Intermediate mo ment (Ki ng, Technical Report P164)

D2 Equivalent section factor c For uniform depth members, c = 1,0.

Plastic Design Portal Frame to Ec3

Recommend Documents