Comparison of the Design of a Portal Frame

Eurobuild in Steel

Comparison of the Design of a Portal Frame

EUROPEAN COMPARISON OF THE DESIGN OF A PORTAL FRAME EUROCODE 3 – NATIONAL STANDARD

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CONTENTS:

1 INTRODUCTION: ................................................................................................................................................................................................................4

2 COMPARISON OF THE DESIGNS...................................................................................................................................................................................6

2.1 FRANCE............................................................................................................................................................................................................................................................................. 6 2.2 GERMANY ........................................................................................................................................................................................................................................................................ 7 2.3 SWEDEN............................................................................................................................................................................................................................................................................ 8 2.4 UNITED KINGDOM ......................................................................................................................................................................................................................................................... 9

3 FRANCE: CALCULATION PROCEDURE COMPARED TO CM66 .........................................................................................................................13

3.1 CROSS-SECTIONAL PROPERTIES ........................................................................................................................................................................................................................................ 13 3.2 CHECK, IF EFFECTS OF DEFORMED SHAPE HAVE TO BE CONSIDERED (2ND ORDER EFFECTS) ............................................................................................................................................. 14 3.3 INTERNAL FORCES ACC. TO 1ST ORDER THEORY ................................................................................................................................................................................................................ 15 3.4 CLASSIFICATION OF CROSS-SECTIONS (LOCAL BUCKLING CHECK) ................................................................................................................................................................................... 16 3.5 ULTIMATE LIMIT STATES................................................................................................................................................................................................................................................... 17 3.6 SERVICEABILITY LIMIT STATES ......................................................................................................................................................................................................................................... 26

4 GERMANY: CALCULATION PROCEDURE COMPARED TO DIN 18800.............................................................................................................27

4.1 CROSS-SECTIONAL PROPERTIES ........................................................................................................................................................................................................................................ 27 4.2 CHECK, IF EFFECTS OF DEFORMED SHAPE HAVE TO BE CONSIDERED (2ND ORDER EFFECTS) ............................................................................................................................................. 28 4.3 INTERNAL FORCES ACC. TO 1ST ORDER THEORY ................................................................................................................................................................................................................ 29 4.4 CLASSIFICATION OF CROSS-SECTIONS (LOCAL BUCKLING CHECK) ................................................................................................................................................................................... 30 4.5 ULTIMATE LIMIT STATES................................................................................................................................................................................................................................................... 31 4.6 SERVICEABILITY LIMIT STATES ......................................................................................................................................................................................................................................... 36

5 SWEDEN: CALCULATION PROCEDURE COMPARED TO BSK99 .......................................................................................................................38

5.1 CROSS-SECTIONAL PROPERTIES ........................................................................................................................................................................................................................................ 38 5.2 CHECK, IF EFFECTS OF DEFORMED SHAPE HAVE TO BE CONSIDERED (2ND ORDER EFFECTS) ............................................................................................................................................. 39 5.3 INTERNAL FORCES ACC. TO 1ST ORDER THEORY ................................................................................................................................................................................................................ 40 5.4 CLASSIFICATION OF CROSS-SECTIONS (LOCAL BUCKLING CHECK) ................................................................................................................................................................................... 41


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5.5 ULTIMATE LIMIT STATES................................................................................................................................................................................................................................................... 42 5.6 SERVICEABILITY LIMIT STATES ......................................................................................................................................................................................................................................... 49

6 U.K.: CALCULATION PROCEDURE COMPARED TO BS 5950-1 ...........................................................................................................................50

6.1 FRAME GEOMETRY ..................................................................................................................................................................................................................................................... 50 6.2 LOADING ........................................................................................................................................................................................................................................................................ 51 6.3 ULTIMATE LIMIT STATE ANALYSIS ....................................................................................................................................................................................................................... 52 6.4 CALCULATE, αCR, FOR STABILITY............................................................................................................................................................................................................................ 54 6.5 COLUMN DESIGN: IPE 600 .......................................................................................................................................................................................................................................... 57 6.6 RAFTER DESIGN: IPE 400 ............................................................................................................................................................................................................................................ 67 6.7 HAUNCH DESIGN.......................................................................................................................................................................................................................................................... 81

ANNEX A: CHARACTERISTIC VALUES OF INTERNAL FORCES ...............................................................................................................................95

ANNEX B: PLASTIC FAILURE OF PORTAL FRAME.......................................................................................................................................................96


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1 Introduction: The typical structure of a single-storey hall in most countries in Europe is the portal frame. This document provides a comparison of the calculation methods acc. to Eurocode 3 (EN 1993-1-1) and the national standard of the respective country. The calculation is performed for a typical portal frame structure in Europe as shown in Figure 1. The frame uses hot rolled I-sections of steel grade S235 for the rafter and columns. Frame span is 30m and eaves height 5m, which are typical dimensions for small and medium sized industrial halls in Europe. Haunches are used for the eaves by providing additional hot-rolled sections welded onto the bottom flange of the rafter. The eaves connection as well as the apex is typically bolted and assumed rigid for this example, so that effects of rotations in the connections do not have to be taken into account. Column bases are pinned as no increased horizontal stiffness, e.g. for cranes, has to be provided.

Figure 1: Sketch of the treated portal frame structure


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Frame spacing is 5.0m, which allows the profile sheeting to span between the frames directly. For this case in this example it is assumed that the sheeting provides sufficient stiffness to avoid lateral torsional buckling of the rafter section. In addition to that the bottom flange of the haunch is assumed lateral restrained. Alternatively in the U.K. calculation rafters are provided to show the effect of single lateral and torsional restraints on lateral torsional buckling of the frame. In the following Figure 2 the load scheme as well as the characteristic loads is shown. The resulting internal forces are given in the respective chapters for the individual countries.

Figure 2: Load scheme and characteristic loads

In all European countries covered by the project, except the U.K., for single-storey portal frames usually elastic design for the actions is performed, and then compared to the plastic resistances of the sections, where allowed by the section classes. Therefore ANNEX A provides the internal forces, determined elastically for the unfactored loads, separated by the individual load cases. Due to the current practice in the U.K. a plastic method of global analysis of EN 1993-1-1 is compared to the national code. For this calculation method ANNEX B describes the design for plastic failure of portal frames in general. The left column in the tables below contains the calculation according to Eurocode 3, whereas in the right column the respective national standard is covered. Where possible, the structure of the calculation is kept similar for both designs, i.e. the calculation is divided into the same subchapters for both standards. For the used equations and values references to the respective chapter in EN 1993-1-1 and the respective national code are given.


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This ensures a quick comparison of the calculation procedure for the portal frame, so that similarities and differences can be found out at first sight.

2 Comparison of the designs This chapter summarises the main similarities and/or differences in the design of portal frames according to EN1993-1-1 and the respective national standard. For detailed comparison see relevant chapter in the design tables below.

2.1 FRANCE

General issues: CM66 "Constructions Métalliques – 1966" is quite an old code drafted in the mid-60's. It follows the admissible stress design approach

and does not inform for plastic design. In 1980 "Additif 80" was published with limited guidance for plastic design, both on the plastic analysis and members check aspect. In that time it was considered as a very first step to plastic design. Additif 80 introduces plastic inches, the ULS and SLS approach in design.

The use of both codes conducts to hybrid design situations, not fully admissible stress and not fully plastic approach. The code takes count of this situation and gives guidance to check in both situation from point to point. Usually the designer makes his decision before starting to design "CM66, pure elastic design" or "CM66 + Additif 80", plastic design with corrective factors on formula originally set for elastic design.

Safety factors: CM66 set safety factors on the loads. They are quite the same as Eurocode. There are NO safety factors on material. When needed as for example for buckling phenomenon's, safety factor are directly introduced in the formula but not really showed as.

Load combinations: Load combinations are not the same than Eurocode. Loads are weighted depending of the combinations.

Design of members: Structural analysis: In additive 80 the second effect order is taken in count with a critical factor. The control is made by comparison

between the loaded situation and the situation when the collapse occurs. A "distance factor" is required. Formula of additive 80 and Eurocode show differences but the final results appear to be in a relative close range.


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Buckling: When checking buckling with interaction between forces, Additif 80 uses k factors which are always greater than 1, increasing the influence of the related force. It appears in calculation with Eurocode that k factors lower than 1, decreasing the effect of the related force are accepted (see § 5.1.2.4.). Additif 80 always require k>1.0. χ factors are not concerned by this remark.

Buckling length: There is an important gap of interpretation in clause 3.5.1.2.1 of Eurocode between European countries. In France we still use the approach as expressed in the ENV and NAD document version of Eurocode 3. The following approach is used: When αcr is greater than 10, the buckling length may be limited to the real geometrical length of the element.

Buckling check: Additif 80 allows to consider a pinned column base with a rigidity factor of 0.05 (0.00 if purely pinned) taking into account a relative limited restrain at the considered base (a joint is never absolutely pinned) and at the opposite consideration a purely fixed base with a rigidity factor of 0.95 (purely fixed would be 1.00).

Section class: There is no section class in CM66 but from clause to clause the code gives provisions to limit the slenderness ratio of the elements or make the step to design accounting for local buckling check. Specific limits are given for the slenderness ratio: as an

example, section webs for plastic check are limited as: eftb

σ23542<

. If this condition is not met, local buckling check shall be made. CM66 + additive 80 provide a direct calculation method for Ncr for longitudinal and transversal buckling and Mcr for lateral torsional

buckling. In the current draft of Eurocode3 there is no direct provision on this calculation.

Design of joints and constructive issues: All joint designs are quite different and there is too much to say in a short description. SLS: CM66 requires complying with vertical deflexion limits. It does not require complying with horizontal displacement limits. The

current drafts of Eurocodes do not require any limits. For Eurocode, provisions should be introduced in the national annex or in the tender document.

2.2 GERMANY

General issues: Generally almost the same design of portal frames, differences only in details. Nearly comparable amount of effort in design of portal frames. Almost the same usage of terms and symbols, only slight differences in indices.


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Same semi-probabilistic safety-concept as DIN 18800, only load combinations in EN 1993-1-1 are more detailed. Safety factor on material reduced from γM = 1.1 in DIN 18800 to γMi = 1.0 in EN 1993-1-1 (general part) currently, but National Annex not

available yet.

Design of members: Basically the same design procedure, small differences in details. Both give the opportunity to use detailed calculations for the structure, e.g. 2nd order effects, as well as simplified methods. The more

detailed the methods, the more effort in design. EN 1993-1-1 provides another method for the interaction between shear and axial stress of members, which leads to ultimate loads

being significantly higher.

Design of joints and constructive issues: Basically the same design procedure for bolted and welded joints, small differences in equations. EN 1993-1-8 provides a design method for structural bolted joints using custom dimensions and gives the normative possibility of

considering the effect of semi-rigid joints on the global structure. Both, DIN18800 and EN1993-1-1, provide similar constructive specifications for the consideration of sheeting and or purlins for lateral

torsional buckling.

2.3 SWEDEN

EN 1993-1-1 contains much more detailed information and more complete rules for design of steel structures than BSK. The technical content in EN 1993-1-1 is quite similar to that of BSK. They are based on the same semi-probabilistic safety-concept and the differences are mainly in details.

The safety factor in EN 1993 is moved from the resistance to the loads compared to BSK. The Swedish design method includes safety classes and this will be retained when we apply the Eurocodes. In the lowest safety class, applicable for members that are unlikely to harm people if failing, the design loads are reduced with a factor 0,83.

The classification of cross-sections is extended to four instead of three classes in BSK. The Swedish Class 2 is split into two classes, Class 2 and 3 in EN 1993-1-1.


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There is one additional buckling curve in EN 1993-1-1 that is valid for high-strength steel. The other curves are the same except for very slender bars where EN 1993-1-1 gives 10 % higher values.

For axial force and bending moment EN 1993-1-1 gives two different possibilities where Method 1 is chosen in the Swedish National Annex. The calculations needed are extensive compared to BSK, and sometimes it is easier to include second order effects to avoid this interaction.

For lateral torsional buckling the differences in buckling curves are insignificant with the choice done in the Swedish NA.. Small differences in the use of terms and symbols.

2.4 UNITED KINGDOM

Plastic analysis assumes that plastic hinges occur at points in the frame where the value of the applied moment is equal to the plastic moment capacity of the member provided. Failure is deemed to have taken place when sufficient hinges have formed to create a mechanism.

‘Plastic’ design produces the lightest and hence the most economical form for a portal frame when using hot rolled sections. In the UK, single storey portals frames are almost always designed using ‘plastic’ design techniques. For ‘plastic’ designed portal frames in order to prevent local buckling, it is essential that Class 1 plastic sections are selected at hinge

positions that rotate, Class 2 compact sections can be used elsewhere. ‘Plastic’ design methods result in relatively slender frames and checking frame stability is a basic requirement of the method. Both in-

plane and out-of-plane stability of both the frame as a whole and the individual members must be considered. In addition, it is essential that local buckling and lateral distortion are also checked, because of the large strains at the hinge positions. Onset of plasticity normally occurs at loads well above those at the serviceability limit state and the plastic rotations are small. The effect of axial load on the classification of members should be considered. However, in many members, the axial force is so small

compared with the bending moment that the classification is not affected. Haunches provide for an economical bolted connection between the rafter and the column. However, haunches should be designed

(i.e. proportioned) to prevent plastic hinges forming within their length. Purlins and side rails are used to provide intermediate lateral restraint to the rafters and columns and support the roofing and wall

sheeting.


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Structural cladding or decking (cladding that spans between frames and where no purlins or side rails are present) is not typically used in the UK. This is because for ‘plastic’ design, structural cladding on its own cannot provide torsional restraint to the frame members. Extra restraint steel has to be provided to the bottom/inner flange of the rafter/column, thereby increasing costs.

Elastic design is a common design method in all countries. However, ‘plastic’ design may be accepted with reluctance and with onerous requirements for stability or for analysis methods, e.g. second-order analysis only.

‘Elastic’ design is used in the UK where deflection is the governing criteria. ‘Elastic’ design is recommended for the design of tied portal frames. Design of a portal frame can be undertaken using BS 5950 by elastic or ‘plastic’ design techniques.

For plastic design of portal frames using EC3-1-1 the following procedure is suggested. More detailed information can be obtained from SCI Publication P164 Design of steel portal frames for Europe Chapter 17 Design procedures.

1. Define frame geometry, determine loads, load combinations, γ factors and ψ factors.

2. Choose trial sections and trial haunch lengths by selecting beam sections that have resistances at least equal to the following:

Rafter Mpl = wL2/24 Haunch Mpl = wL2/10

Column Mpl = wL2/12 x (height to bottom of haunch / height to centre of rafter)

where w is the maximum ULS gravity load/unit length along the span and L is the span of the portal.

The haunch length should be chosen to optimise the overall portal structure. A length of L / 10 from the column face is a reasonable initial choice, but the proportions of the haunch generally depend of the characteristics of each individual building, especially the size of the rafter. A haunch length of L / 10 will normally place the first hinge in the top of the column. A rather longer haunch will place the first hinge at the sharp end of the haunch.

3. Calculate frame imperfection equivalent forces. This can be by a preliminary frame analysis (which is necessary for all but the simplest buildings) or by a suitable approximation.

4. Perform plastic analysis of the frame assuming:


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a.) No reduction in plastic moment of resistance from coexistent axial and shear forces. This approach may need modifying, where axial loads are high, e.g. in tied portals, in portals with steep slopes and in portals with heavy roofing loads.

b.) A trial value of VSd / Vcr = 0,12 unless a better estimate is possible. Note that in uplift cases, the members might be subject to axial tension. In this case, there will be no destabilisation of the frame and VSd / Vcr can be taken as zero.

5. Calculate an accurate value of VSd / Vcr.

6. If (accurate VSd / Vcr) > (trial VSd / Vcr) or if a more refined design is required, return to Step 4.

Note: For relatively slender frames, it is often wise to check the deflections at the serviceability limit state (SLS), before checking the buckling resistance.

7. For the columns check that:

a.) The classification is Class 1 or Class 2 as appropriate. b.) The cross sectional resistance is adequate. c.) The resistance of the member to minor axis buckling between lateral restraints is adequate. d.) Resistance of member to minor axis buckling between torsional restraints is adequate.

8. For the rafters check that:

a.) The classification is Class 1 or Class 2 as appropriate. b.) The cross sectional resistance is adequate. c.) The resistance of the member to minor axis buckling between lateral restraints is adequate. d.) The resistance of the member to minor axis buckling between torsional restraints is adequate.

9. For the haunches check that:

a.) The classification is Class 1 or Class 2 as appropriate, b.) The cross sectional resistance is adequate, as step 7b above, but checking at several cross sections within the length of

the haunch (both ends, quarter, mid-span and three quarter points) is recommended.


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c.) The resistance of the member to minor axis buckling between lateral restraints is adequate, but giving special attention to the effect of the taper.

d.) The resistance of the member to minor axis buckling between torsional restraints is adequate, but giving special attention to the effect of the taper.

10. Check web buckling resistance to shear and transverse forces.

11. Check the connections.

12. Check the restraints.

A summary of the ‘plastic’ design procedure using British Standard BS 5950 Part 1 : 2000 is given annexed to the British calculation in § 6 . More detailed information can be obtained from SCI Publication P325 Introduction to Steelwork Design to BS 5950-1:2000 Chapter 12 Plastic design of portal frames.


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3 France: Calculation procedure compared to CM66

EN 1993-1-1 Règles CM66 + Additif 80 3.1 Cross-sectional properties

3.1.1 Column: IPE 600 (S235)

fy = 235N/mm²; γM0 = γM1 = 1,0 (EC3, 6.1) σe = 235N/mm²

b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm A = 156cm² ; Iy = 92080cm4 ; Iz = 3390cm4 ; Wpl,y = 3520cm³ A = 156cm² ; Ix = 92080cm4 ; Iy = 3390cm4 ; Zx = 3520cm³ Iω = 2846000cm6; IT = 165cm4 J = 165cm4

kN3666f

AN0M

yRd,pl =

γ⋅=

(EC3, 6.2.3 (2)) kNAN eeffp 3666=⋅= σ (Add.80 § 4.2)

kN11373

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3)) 674412562 =×=wA

kN9191000/235674458,0A58,0V ewpy =××=σ××= (Add.80 § 4.4)

kNm2,827f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNmZM ep 20,8271000/2353520 =×== σ (Add.80 § 4.3)

3.1.2 Rafter: IPE 400 (S235)

fy = 235N/mm²; γM0 = γM1 = 1,0 (EC3, 6.1) σe = 235 N/mm2

b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm A = 84,5cm² ; Iy = 23130cm4 ; Iz = 1320cm4 ; Wpl,y = 1308cm³ A = 84,5cm² ; Ix = 23130cm4 ; Iy = 1320cm4 ; Zx = 1308cm³ Iω = 490000cm6; IT = 51,1cm4 J = 51,1cm4

kN8,1985f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) kNAN eeffp 80,19851000/2358450 =×=⋅= σ (Add.80 § 4.2)

kN7,5793

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3))

232086,8373 mmAw =×=

kNAV ewpy 4371000/235320858,058,0 =××== σ (Add.80 § 4.4)


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kNm4,307f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNmZM expx 40,3071000/2351308 =×=⋅= σ (Add.80 § 4.3)

3.1.3 Eaves haunch: IPE 400 & haunched IPE 500 (S235)

fy = 235N/mm²; γM0 = γM1 = 1,0 (EC3, 6.1) σe = 235N/mm²

b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm A = 159,6cm² ; Iy = 134868cm4 ; Iz = 2390cm4 ; Wpl,y = 3920,7cm³ A = 159,6cm² ; Ix = 134868cm4 ; Iy = 2390cm4 ; Zx = 3920,7cm³ Iω = 2694000cm6; IT = 101,7cm4 J = 101,7cm4

kN6,3750f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) kNAN eeffp 6,37501000/23515960 =×=⋅= σ (Add.80 § 4.2)

kN8,12773

fA7,579V

0M

y500IPE,wRd,z,pl =

γ⋅⋅+= (EC3, 6.2.6 (3))

2500,400, 798247743208 mmAAA IPEwIPEww =+=+=

kNAV ewpy 10881000/235798258,058,0 =××=××= σ (Add.80 § 4.4)

kNm4,921f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNmZM expx 40,9211000/23570,3920 =×=⋅= σ (Add.80 § 4.3)

3.2 Check, if effects of deformed shape have to be considered (2nd order effects)

Sway mode failure may be checked with first order analysis for portal frames with shallow roof slopes if the following criteria is satisfied:

)analysisplasticfor(15hVH

Ed,HEd

Edcr ≤

δ⋅

=α (EC3, 5.2.1 (4))

With HEd = design value of horizontal reaction to all horizontal loads VEd = total design vertical loads δH,Ed = horizontal displacement at the eave due to all horizontal loads h = storey height

159,254,27

60004,270

0,32cr ≥=

⋅

=α → First order analysis sufficient

Second order effects do not have to be considered if: αcr > 5 ( Add.80. § 7)

αp ≥

crα11

1

−

with αcr = force amplification factor for elastic critical buckling αp = force amplification factor for collapse mechanism αcr = 10,79 > 5 (determined by computer analysis)

αp min = 1,39 > 102,1

79,1011

111

1=

−=

−crα

(determined at the top of column)

→ First order analysis sufficient


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3.3 Internal forces acc. to 1st order theory

Loading combinations acc. to EN 1990, 6.4.3.2 (3) and Annex A: LC 1: pImsnowdeadload EE5,1E35,1 +⋅+⋅

LC 2: pImwinddeadload EE5,1E35,1 +⋅+⋅

LC 3: pImwindsnowdeadload EE6,05,1E5,1E35,1 +⋅⋅+⋅+⋅

LC 4: pImsnowwinddeadload EE5,05,1E5,1E35,1 +⋅⋅+⋅+⋅

LC 1 is decisive because of the relieving effect of the wind undertow on the roof.

Loading combinations: (Add.80 § 3.41)

LC 1: snowdeadload EE ⋅+⋅ 5,133,1 DECISIVE !

LC 2: winddeadload EE ⋅+⋅ 5,133,1

LC 3: windsnowdeadload EEE ⋅+⋅×+⋅ 42,15,042,133,1

LC 4: snowdeadload EE ⋅+⋅ 67,11

LC 5: winddeadload EE ⋅+⋅ 75,11

LC 6: windsnowdeadload EEE ⋅+⋅×+⋅ 75,15,067,11


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3.4 Classification of cross-sections (local buckling check)

3.4.1 Column: IPE 600 (S235)

Flange: ψ = α = 1 (constant pressure) c/t = (220/2-12/2–24) / 19 = 4,21 < 9 → class 1 (EC3, table 5.2)

Non plate buckling conditions? (Add.80 § 5.1)

Compressed flange: b/tf = 220/19 = 11,58 < 2023520 =eσ

OK

Web: plastic stress distribution α = 0,547 c/t = (600-2⋅19-2⋅24) / 12 = 42,83 < 396 / (13 ⋅ 0,547-1) = 64,8 → class 1 (EC3, table 5.2)

Partially or totally compressed Web? If: (with N > 0 for traction)

32,015600674475,075,0 −=×−=−

AAw ≤ 037,0

366619,134

−=−

=pN

N ≤ 43,0156006744

==AAw

⇒ 18,642353

100678,46 =

+≤=

epww

w

NN

AA

th

σ OK

3.4.2 Rafter: IPE 400 (S235)

Flange: ψ = α = 1 (constant pressure) (180/2-8,6/2–21) / 13,5 = 4,79 < 9 → class 1 (EC3, table 5.2)

Non plate buckling conditions? (Add.80 § 5.1)

Compressed flange b/tf = 180/13,5 = 13,33 < 2023520 =eσ

OK

Web: plastic stress distribution α = 0,585 c/t = (400-2⋅13,5-2⋅21) / 8,6 = 38,5 < 396 / (13 ⋅ 0,585-1) = 60,0 → class 1 (EC3, table 5.2)

Partially or totally compressed Web: If: (with N > 0 for traction)

28,08450320875,075,0 −=×−=−

AAw ≤ 054,0

8,1985108

−=−

=pN

N ≤ 38,084503208

==AAw

⇒ 2,622353

100674,43 =

+≤=

epww

w

NN

AA

th

σ OK


Flange: ψ = α = 1 (constant pressure) (IPE 500 only, IPE 400 see above) Flange: element in compression (IPE 500 only, IPE 400 see above)


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(200/2-10,2/2–21) / 16 = 4,62 < 9 → class 1 (EC3, table 5.2) b/tf = 200/16 = 12,5 < 2023520 =

eσ OK (Add.80 § 5.1)

Web: IPE 400-part: in tension Haunched IPE 500 part : α = 1 (constant pressure) c/t = (420-16-21) / 10,2 = 37,5 < 38 → class 2 (EC3, table 5.2)

Section for eaves haunch is class 2-section, therefore the internal forces have to be determined elastically, but plastic resistances can be activated.

Web:

38,015960798275,075,0 −=×−=−

AAw ≤ 030,0

60,37505,111

−=−

=pN

N ≤ 50,0159607982

==AAw

⇒ 652353

100679,452,10

468=

+≤==

epww

w

NN

AA

th

σ OK

3.5 Ultimate limit states

3.5.1 Column IPE600 (S235)

3.5.1.1 Resistance of cross-section

3.5.1.1.1 Check for shear force

Internal shear force: |Vz,Ed| = 101,1kN Check for shear buckling:

721172728,42

12514

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))

→ No shear buckling → shear resistance Vpl,z,Rd

0,1089,00,1137

1,101VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))

Check < 0,5 → no interaction between Vz and My (EC3, 6.2.8 (2))

Check if the shear force shall also be included in the checking for axial force and bending moment?

3.5.1.1.2 Check for axial force and bending moment

Internal forces: |NEd| = 135,5kN, |My,Ed| = 606,8 kNm Axial force is compression → holes need not be considered (EC3, 6.2.4 (3))

Internal forces: |Vy| = 99,1 kN ; |N| = 125,5 kN ; |Mx| = 594,25 kNm


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0,1034,00,3666

5,125NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))

Check < 0,25 and kN7,850fth5,0

|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))

→ no interaction between N and My

0,1734,02,8278,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

If : 6,0115,03666

35,1242,0919

1,992,00 <=×+=×+≤ppy

y

NN

VV (Add.80 § 4.61)

And 18,0034,03666

35,1240 <==≤pN

N

⇒ Mx = 594,25 kNm ≤ Mpx = 827,20 kNm

3.5.1.2 Buckling resistance of member

3.5.1.2.1 In plane flexural buckling (about y-axis)

Flexural buckling length: Lcr,y = 37,08m (determinable by literature)

Elastic critical flexural buckling force: kN0,13883708

9208021000N2

2

y,cr =⋅⋅π

=

Class 1 section: Slenderness: 625,10,13880,3666

y ==λ (EC3, 6.3.1.2 (1))

Rolled section; h/b >1,2; tf < 40mm → buckling curve a (EC3, table 6.2) Reduction value: χy = 0,324 (EC3, 6.3.1.2 (1))

Flexural buckling length: From computer analysis (with αcr = 10,79): ⇒ LKx = 36,41 m (close from Eurocode results) Hand checking using the proposed method of CM66 “Rigidity factor KAKB method"

LK

KLA

AKx

4,26,1 += (CM 66 § 5.134)

(Considering that the columns are pinned at their bases: ⇒ KB = 0)

048,015347771

771

692080

3023130

3023130

=+

=+

=

+

=

∑ ∑

∑

Ac

c

Ab

b

Ab

b

A

LI

LI

LI

K

mLLLKx 87,3598,5048,0

048,04,26,1==

×+=

Additive 80 of CM66 allows to considered that the base of column is not perfectly pinned: ⇒ KB = 0,05


PAGE 19 OF 98

LK

KLA

AKx +

+=

04,092,135,1

(Add.80 § 5.33)

mLLLKx 29,24048,4048,004,0

048,092,135,1==

+×+

=

To keep comparison with National Standard, we use the buckling length determined by the computer analysis

84,149243

36410===

x

Kxx i

Lλ (Add.80 § 5.31)

⇒ 595,1210000

23584,149=×==

πσ

πλλ

Eex

x

3.5.1.2.2 Out of plane flexural buckling (about z-axis)

Flexural buckling length: Lcr,z = 6,00m (lateral restraints at endings)


339021000N2

2

z,cr =⋅⋅π

=


z ==λ (EC3, 6.3.1.2 (1))

Rolled section; h/b >1,2; tf < 40mm → buckling curve b (EC3, table 6.2) Reduction value: χz = 0,395 (EC3, 6.3.1.2 (1))


76,1286,46

6000===

y

Kyy i

Lλ (Add.80 § 5.31)

⇒ 371,1210000

23576,128=×==

πσ

πλ

λE

eyy

2222

2

04))2,0(1()2,0(1

2

λλλαλλα

λ

−+−+−+−+=k

Table B gives values for I sections

595,1max == xλλ ⇒ k0 = 3,25

3.5.1.2.3 Lateral torsional buckling

Elastic critical moment from literature or ENV1993-1-1, Annex F: Elastic critical moment


PAGE 20 OF 98

kNm8,1431IEIGL

II

LIE

CM21

z2

T2

z2

z2

1cr =

⋅⋅π

⋅⋅+⋅

⋅⋅π⋅= ω (ENV 1993-1-1, Annex F, 1.3 (3))


LT ==λ (EC3, 6.3.2.2 (1))

Rolled section → chapter 6.3.2.3 applicable (EC3, 6.3.2.3 (1)) Rolled section and h/b > 2 → buckling curve c (EC3, table 6.5)

Reduction value: χLT = 0,789 (EC3, 6.3.2.3 (1))

Modified reduction value 901,0876,0789,0

fLT

mod,LT ==χ

=χ (EC3, 6.3.2.3 (2))

2

*2

*2

12

2

+⋅

⋅⋅=

hL

EIJG

LhEI

CM D

yD

yD π

ςπ

(Add.80 § 5.22)

LD = 6,00 m 1=ς (I section)

C1 = 1,88 (triangular moment diagram) h* = H - tf = 600 – 19 = 581 mm

62

4

4

2

42

10/1.581

6000210.3390210000

8100010.165160002

58110.339021000088,1

××

××

+⋅×

××⋅⋅=

ππ

DM

kNmM D 1434=

3.5.1.2.4 Interaction for compression force and bending

Internal forces: |NEd| = 135,5kN, |My,Ed| = 606,8 kNm No second order analysis, uniform member with double-symmetric section → stability check with interaction factors (EC3, 6.3.3 (1))

Portal frame is “frequent structure” → “standard case” → Annex B Equivalent uniform moment values: Cmy = 0,9 (sway buckling mode) Cm,LT = 0,6 (EC3, Annex B, table B.3)

Member susceptible to torsional deformations → EC3, Annex B, table B.2

( ) 982,0N

N8,01C046,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=

973,0N

N)25,0C(

1,01963,0N

N)25,0C(

1,01kRd,plz

Ed

LT,mRd,plz

Ed

LT,m

zzy =

⋅χ−−<=

⋅χ−λ⋅

−=

→ kyy = 0,982; kzy = 0,973 (EC3, Annex B, table B.2)

Check for compression force and bending: (EC3, 6.3.3 (4))

0,1914,0800,0114,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

Internal forces: |N| = 134,2 kN, |Mx| = 594,25 kNm

10 ≤+px

x

D

fx

p MM

kk

NNk (Add.80 § 5.32)

p

x

mxfx

NN

Ck2

1 λ−= (Add.80 § 5.32)

Cmx = 1 (sway portal frame)

037,03666

2,134==

pNN

595,1=xλ

104,1037,0595,11

12 =×−

=fxk


PAGE 21 OF 98

0,1886,0792,0094,0M

Mk

NN

Rd,y,plmod,LT

Ed,yzy

Rd,plz

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.62))

n

n

D

p

D

MM

k

+

=

1

1 (Add.80 § 5.22)

n = 2 (hot rolled section)

866,0

143420,8271

1

2

2=

+

=Dk

036,120,82725,594

866,0104,1037,025,30 =×+×=+

px

x

D

fx

p MM

kk

NNk ≅ 1 (Add.80 § 5.32)

3.5.2 Rafter IPE 400 (S235)

3.5.2.1 Resistance of cross section



721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))


0,1138,07,5791,80

VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


Check if the shear force shall be included in the checking for axial force and bending moment.



Internal forces: |Vy| = 79,2 kN ; |N| = 107,87 kN ; |Mx| = 208 kNm


PAGE 22 OF 98

0,1055,08,1985

7,108NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1705,04,3076,216

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

If : 6,0192,080,198587,1072,0

4372,792,00 <=×+=×+≤

ppy

y

NN

VV

(Add.80 § 4.61)

And 18,0054,080,198587,1070 <==≤

pNN

⇒ Mx = 208 kNm ≤ Mpx = 307,40 kNm



Elastic critical flexural buckling force: Ncr,y = 1117,7kN (from computer analysis)


y ==λ (EC3, 6.3.1.2 (1))


Flexural buckling length: For the determination of the buckling length along xx, a buckling analysis is performed to calculate the buckling amplification factor αcr for the load combination giving the highest vertical load, with a lateral restrain at top of column:

⇒ Ncr,x = 3250 kN (from computer analysis)

⇒ mLKx 14,12=

40,735,165

12140===

x

Kxx i

Lλ (Add.80 § 5.31)

781,0210000

23540,73=×==

πσ

πλλ

Eex

x

3.5.2.2.2 Out of plane flexural buckling (about z-axis) and lateral torsional buckling

For this example it is assumed, that the roofing made of profile sheeting provides sufficient continuous lateral and torsional restraints. PrEN 1993-1-1 provides formulas for this check in Annexes BB.2.1 and BB.2.2. → No out of plane flexural buckling and lateral torsional buckling (χz = χLT = 1,0)

For this example it is assumed, that the roof is made of profile sheeting providing sufficient continuous lateral and torsional restraints. (cf. check with EN 1993-1-1)



PAGE 23 OF 98

Internal forces: |NSd| = 108,7kN, |My,Sd| = 216,6 kNm No second order analysis, uniform member with double-symmetric section → stability check with interaction factors (EC3, 6.3.3 (1))

Cmy = 0,9 (sway buckling mode) (EC3, Annex B, table B.3)

Member not susceptible to torsional deformations → EC3, Annex B, table B.1

( ) 987,0N

N8,01C024,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=

→ kyy = 0,987; kzy = 0 (no out of plane failure) Check for compression force and bending: (EC3, 6.3.3 (4))

0,1818,0695,0123,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

Internal forces: |Vy| = 79,2 kN ; |N| = 107,87 kN ; |Mx| = 208 kNm

2222

2

04))2,0(1()2,0(1

2

λλλαλλα

λ

−+−+−+−+=k (Add.80 § 5.31)

Table B gives values for I sections

781,0max == xλλ ⇒ k0 = 1,35

10 ≤+px

x

D

fx

p MM

kk

NNk

px

mxfx

NN

Ck2

1 λ−= (Add.80 § 5.32)

Cmx = 1 (sway portal frame)

054,080,198587,107

==pN

N

781,0=xλ

034,1054,0781,01

12 =

×−=fxk

kD = 1 (Add.80 § 5.22)

771,04,307

2081034,1054,035,10 =×+×=+

px

x

D

fx

p MM

kk

NNk < 1 OK (Add.80 § 5.32)

3.5.3 Eaves haunch IPE 400 & haunched IPE 500 (S235)



Internal shear force: |Vz,Ed| = 114,5kN Check if the shear force shall be included in checking for axial force and bending


PAGE 24 OF 98

Check for shear buckling:

721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (IPE 400) (EC3, 6.2.6 (6))

721172725,37

2,10383

th

w

w =⋅=ηε

⋅<== (haunched IPE 500) (EC3, 6.2.6 (6))


0,1090,08,1277

5,114VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


moment.



0,1030,06,3750

3,112NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1659,04,9218,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

Internal forces: |Vy| = 113,24 kN ; |N| = 111,46 kN ; |Mx| = 594,25 kNm

If : 6,011,060,3750

46,1112,01088

24,1132,00 <=×+=×+≤ppy

y

NN

VV

(Add.80 § 4.61)

And 18,003,060,3750

46,1110 <==≤pN

N

⇒ Mx = 594,25 kNm ≤ Mpx = 921,40 kNm


Stability failures of member do not have to be considered: In plane flexural buckling is covered safely by check of IPE 400. Out of plane flexural buckling and lateral torsional buckling is excluded (see Rafter). For lateral torsional buckling of the haunch, where the lower flange is in compression, for this example it is assumed, that lateral restraints are provided.

For the verification of the haunch, the compressed part of the cross-section is considered isolated composed of the compressed flange and 1/6 of the web with a buckling length along the yy-axis equal to 3,70m (length between the top of column and the end of haunch with a torsional restraint).


PAGE 25 OF 98

Properties of the compressed part: Section at the mid-length of the haunch including 1/6th of the web depth Section area A = 45 cm2 Second moment of area /xx Ix = 693 cm4 Second moment of area /yy Iy =1068 cm4

⇒ cmiz 87,445

1068==

98,757,48

3700===

y

fzy i

Lλ (Add.80 § 5.31)

809,0210000

23598,75=×==

πσ

πλλ

Eex

x

54,10 =k Table C (T section) (Add.80 § 5.31)

Compression in the bottom flange:

kNN f 50,7134500100010.7,3920

100059425015960450046,111 3 =×

××

+×=

Verification of buckling resistance of the bottom flange:

039,12354500

71350054,1,0 =

××=

Rk

fEd

NN

k ≅ 1 (Add.80 § 5.31)

200 mm

130 mm x

y


PAGE 26 OF 98

3.6 Serviceability limit states

3.6.1 Vertical deflection

Maximum vertical displacement at apex: uz = 132mm PrEN 1993-1-1 (11.2003) does not specify allowable displacements. ENV 1993-1-1 (1993) gives max. uz < L/200 for roofs in general. (ENV, table 4.1)

132mm = L/227 < L/200

Maximum vertical displacement at apex: uy = 132 mm CM66 requires max. uy < L/200 for roofs in general. ( Additif 80 § 6.1 and CM66 § 5.25) uy = 132 mm = L/227 < L/200

3.6.2 Horizontal deflection

Maximum horizontal displacement at eaves: ux = 19,9mm PrEN 1993-1-1 (11.2003) does not specify allowable displacements. ENV 1993-1-1 (1993) gives max. uz < H/300 for single-storey buildings (ENV, 4.2.2 (4))

19,9mm = H/302 < H/300

Maximum horizontal displacement at eaves: ux = 19,9 mm CM66 do not requires any limits for horizontal displacements ENV 1993-1-1 DAN requires max. ux < H/150 for portal frame (ENV, 4.2.2 (4))

ux = 19,9 mm = H/302 < H/150


PAGE 27 OF 98

4 Germany: Calculation procedure compared to DIN 18800

EN 1993-1-1 DIN 18800 4.1 Cross-sectional properties

4.1.1 Column: IPE 600 (S235)

fy = 235N/mm²; γM0 = γM1 = 1,0 (EC3, 6.1) fy,k = 240N/mm²; γM = 1,1

b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm A = 156cm² ; Iy = 92080cm4 ; Iz = 3390cm4 ; Wpl,y = 3520cm³ A = 156cm² ; Iy = 92080cm4 ; Iz = 3390cm4 ; Wpl,y = 3520cm³ Iω = 2846000cm6; IT = 165cm4 CM = 2846000cm6; IT = 165cm4

kN3666f

AN0M

yRd,pl =

γ⋅=

(EC3, 6.2.3 (2)) kN6,3403

fAN

M

k,yd,pl =

γ⋅=

kN11373

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3)) kN0,912

3

f)tb2A(V

M

k,yfd,z,pl =

γ⋅⋅⋅⋅−=

kNm2,827f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNm0,768

fWM

M

k,yy,pld,y,pl =

γ⋅=

4.1.2 Rafter: IPE 400 (S235)


b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm A = 84,5cm² ; Iy = 23130cm4 ; Iz = 1320cm4 ; Wpl,y = 1308cm³ A = 84,5cm² ; Iy = 23130cm4 ; Iz = 1320cm4 ; Wpl,y = 1308cm³ Iω = 490000cm6; IT = 51,1cm4 Iω = 490000cm6; IT = 51,1cm4

kN8,1985f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) kN6,1843

fAN

M

k,yd,pl =

γ⋅=

kN7,5793

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3)) kN2,452

3

f)tb2A(V

M

k,yfd,z,pl =

γ⋅⋅⋅⋅−=


PAGE 28 OF 98

kNm4,307f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNm4,285

fWM

M

k,yy,pld,y,pl =

γ⋅=



b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm A = 159,6cm² ; Iy = 134868cm4 ; Iz = 2390cm4 ; Wpl,y = 3920,7cm³ A = 159,6cm² ; Iy = 134868cm4 ; Iz = 2390cm4 ; Wpl,y = 3920,7cm³ Iω = 2694000cm6; IT = 101,7cm4 Iω = 2694000cm6; IT = 101,7cm4

kN6,3750f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) kN2,3482

fAN

M

k,yd,pl =

γ⋅=

kN8,12773

fA7,579V

0M

y500IPE,wRd,z,pl =

γ⋅⋅+= (EC3, 6.2.6 (3)) kN1,995

3

f)tbtb2A(V

M

k,y500IPE,f500IPE400IPE,f400IPEd,z,pl =

γ⋅⋅⋅−⋅⋅−=

kNm4,921f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2)) kNm4,855

fWM

M

k,yy,pld,y,pl =

γ⋅=




Ed,HEd

Edcr ≤

δ⋅

=α (EC3, 5.2.1 (4))


159,254,27

60004,270

0,32cr ≥=

⋅


Second order effects do not have to be considered if either

1,01NN

d,Kid,Ki

Sd ≤η

= with ηKi,d = force amplifier for elastic critical buckling

(from literature or computer analysis)

or Sd

d,ply,K

NN

3,0 ⋅≤λ with y,Kλ = slenderness for elastic critical buckling

or dy

Sdy,Ky,K )EI(

Ns ⋅=ε⋅β with sK,y = elastic critical buckling length

1,0096,044,10

11NN

d,Kid,Ki

Sd <==η

= → First order analysis sufficient


PAGE 29 OF 98







Loading combinations: LC 1: pImsnowdeadload EE5,1E35,1 +⋅+⋅


LC 3: pImwindsnowdeadload EE9,05,1E9,05,1E35,1 +⋅⋅+⋅⋅+⋅


PAGE 30 OF 98


4.4.1 Column: IPE 600 (S235)


Flange : ψ = α = 1 (constant pressure) vorh (b/t) = (220/2-12/2–24) / 19 = 4,21 < 9 → plastic-plastic analysis allowed


Web: plastic stress distribution α = 0,547 vorh (b/t) = (600-2⋅19-2⋅24) / 12 = 42,83 < 32 / α = 58,5 → plastic-plastic analysis allowed

4.4.2 Rafter: IPE 400 (S235)


Flange : ψ = α = 1 (constant pressure) vorh (b/t) = (180/2-8,6/2–21) / 13,5 = 4,79 < 9 → plastic-plastic analysis allowed


Web: plastic stress distribution α = 0,585 vorh (b/t) = (400-2⋅13,5-2⋅21) / 8,6 = 38,49 < 32 / α = 54,7 → plastic-plastic analysis allowed


Flange: ψ = α = 1 (constant pressure) (IPE 500 only, IPE 400 see above) (200/2-10,2/2–21) / 16 = 4,62 < 9 → class 1 (EC3, table 5.2)

Flange: ψ = α = 1 (constant pressure) (IPE 500 only, IPE 400 see above) vorh (b/t) = (200/2-10,2/2–21) / 16 = 4,62 < 9 → plastic-plastic analysis allowed

Web: IPE 400-part: in tension Haunched IPE 500 part : α = 1 (constant pressure) c/t = (420-16-21) / 10,2 = 37,5 < 38 → class 2 (EC3, table 5.2)


Web: IPE 400-part: in tension Haunched IPE 500 part : α = 1 (constant pressure)

vorh (b/t) = (420-16-21) / 10,2 = 37,5 > 37 / α = 37 Web of haunched IPE 500 is class 3-section, so only elastic cross-section resistances can be used.


PAGE 31 OF 98


4.5.1 Column IPE600 (S235)




721172728,42

12514

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))


0,1089,00,1137

1,101VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


Internal shear force: |Vz,Sd| = 101,1kN Check for shear buckling:

708,4212514

tb

<==

→ No shear buckling → shear resistance Vpl,z,d

0,1111,00,9121,101

VV

d,z,pl

Sd,z <==

Check < 0,33 → no interaction between Vz and My



0,1034,00,3666

5,125NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1734,02,8278,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

Internal forces: |NSd| = 125,5kN, |My,Sd| = 606,8 kNm Axial force is compression & hole clearance < 1mm→ holes need not be considered

0,1037,06,3403

5,125NN

d,pl

Sd <==

Check < 0,10 → no interaction between N and My

0,1790,00,7688,606

MM

d,y,pl

Sd,y <==


PAGE 32 OF 98





9208021000N2

2

y,cr =⋅⋅π

=


y ==λ (EC3, 6.3.1.2 (1))




9208021000N2

2

y,Ki =⋅⋅π

=

Slenderness: 642,10,13880,3744

y,K ==λ

Rolled section; h/b >1,2; t < 40mm → buckling curve a Reduction value: κy = 0,318




339021000N2

2

z,cr =⋅⋅π

=


z ==λ (EC3, 6.3.1.2 (1))




339021000N2

2

z,Ki =⋅⋅π

=

Slenderness: 385,17,19510,3744

z,K ==λ

Rolled section; h/b >1,2; t < 40mm → buckling curve b Reduction value: κz = 0,388


Elastic critical moment from literature or ENV1993-1-1, Annex F:

kNm8,1431IEIGL

II

LIE

CM21

z2

T2

z2

z2

1cr =

⋅⋅π

⋅⋅+⋅

⋅⋅π⋅= ω (ENV 1993-1-1, Annex F, 1.3 (3))


LT ==λ (EC3, 6.3.2.2 (1))


Elastic critical moment from DIN 18800, Part 2:

kNm1,1348I

IL039,0CL

EIM

z

T2

M2

z2

y,Ki =⋅⋅+

⋅⋅π

⋅ζ=

Slenderness: 792,01,1348

8,844M ==λ

Rolled section and moment gradient ψ = 0 < 0,5 → kn = 1,0 → n = 2,5 Reduction value: κM = 0,897


PAGE 33 OF 98



fLT

mod,LT ==χ

=χ (EC3, 6.3.2.3 (2))





( ) 982,0N

N8,01C046,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=

973,0N

N)25,0C(

1,01963,0N

N)25,0C(

1,01kRd,plz

Ed

LT,mRd,plz

Ed

LT,m

zzy =

⋅χ−−<=

⋅χ−λ⋅

−=



0,1914,0800,0114,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

0,1886,0792,0094,0M

Mk

NN

Rd,y,plmod,LT

Ed,yzy

Rd,plz

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.62))

Internal forces: |NSd| = 135,5kN, |My,Sd| = 606,8 kNm Uniform member with double-symmetric section, approx. constant axial force and no torsional load → stability check with interaction factors

Values for moment distribution: 66,0893,011d,Ki

m >=η

−=β ; 8,1y,M =β

Coefficient: 1,0030,0N

N1

NN

n 2y,K

2y

d,ply

Sd

d,ply

Sd <=λ⋅κ⋅

⋅κ−⋅

⋅κ=∆

Coefficient: 9,0224,015,015,0a y,Mz,Ky <=−β⋅λ⋅=

Interaction factor: 0,1977,0aN

N1k y

d,plz

Sdy <=⋅

⋅κ−=

Check for compression and bending:

In plane: 0,1861,003,0706,0125,0nM

MN

N

d,y,pl

Sd,ym

d,ply

Sd <=++=∆+⋅β

⋅+⋅κ

Out of plane: 0,1964,0861,0103,0M

Mk

NN

d,y,plM

Sd,yy

d,plz

Sd <=+=⋅κ

⋅+⋅κ

4.5.2 Rafter IPE 400 (S235)






PAGE 34 OF 98

721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))


0,1138,07,5791,80

VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


705,386,8

331tb

<==


0,1177,02,4521,80

VV

d,z,pl

Sd,z <==




0,1055,08,1985

7,108NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1705,04,3076,216

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

Internal forces: |NSd| = 108,7kN, |My,Sd| = 216,6 kNm Axial force is compression & hole clearance < 1mm→ holes need not be considered

0,1059,06,1843

7,108NN

d,pl

Sd <==

Check < 0,10 → no interaction between N and My

0,1759,04,2856,216

MM

d,y,pl

Sd,y <==





y ==λ (EC3, 6.3.1.2 (1))


Elastic critical flexural buckling force: NKi,y = 1117,7kN (from computer analysis)


y,K ==λ

Rolled section; h/b >1,2; t < 40mm → buckling curve a Reduction value: κy = 0,445



PAGE 35 OF 98


For this example it is assumed, that the roofing made of profile sheeting provides sufficient continuous lateral and torsional restraints. DIN 18800 provides the same formulas for this check as PrEN 1993-1-1 in Annexes BB.2.1 and BB.2.2, → No out of plane flexural buckling and lateral torsional buckling





( ) 987,0N

N8,01C024,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=


0,1818,0695,0123,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

Internal forces: |NSd| = 108,7kN, |My,Sd| = 216,6 kNm Uniform member with double-symmetric section, approx. constant axial force and no torsional load → stability check with interaction factors Value for moment distribution: ψ = -0,283 → 0,1m =β

Coefficient: 1,0041,0N

N1

NN

n 2y,K

2y

d,ply

Sd

d,ply

Sd <=λ⋅κ⋅

⋅κ−⋅

⋅κ=∆

Check for compression and bending:

In plane: 0,1932,0041,0759,0132,0nM

MN

N

d,y,pl

Sd,ym

d,ply

Sd <=++=∆+⋅β

⋅+⋅κ





721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (IPE 400) (EC3, 6.2.6 (6))

721172725,37

2,10383

th

w

w =⋅=ηε

⋅<== (haunched IPE 500) (EC3, 6.2.6 (6))



705,386,8

331tb

<== (IPE 400)

705,372,10

383tb

<== (haunched IPE 500)



PAGE 36 OF 98

0,1090,08,1277

5,114VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


0,1115,01,9955,114

VV

d,z,pl

Sd,z <==




0,1030,06,3750

3,112NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1659,04,9218,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

Internal forces: |NSd| = 112,3kN, |My,Sd| = 606,8 kNm Axial force is compression & hole clearance < 1mm→ holes need not be considered Check for elastic cross-section resistance due to local buckling:

²cmkN08,18)63,38(

13486860680

6,1593,112z

IM

AN

.min oy

Sd,ySdSd −=−⋅+

−=⋅+=σ

²cmkN81,1837,43

13486860680

6,1593,112z

IM

AN

.max uy

Sd,ySdSd =⋅+

−=⋅+=σ

0,1862,082,2181,18

f||.max

d,y

Sd <==σ



Stability failures of member do not have to be considered: In plane flexural buckling is covered safely by check of IPE 400. Out of plane flexural buckling and lateral torsional buckling is excluded (see Rafter).




DIN 18800 as well as the standards dealing with building construction in steel as DIN 18801 give no information concerning the limitation of vertical deflections.


PAGE 37 OF 98

132mm = L/227 < L/200



19,9mm = H/302 < H/300

As for vertical deflections no limitation of horizontal deflections exists in German standards.

EUROPEAN COMPARISON OF THE DESIGN OF A PORTAL FRAMEEUROCODE 3 – NATIONAL STANDARD

PAGE 38 OF 98

5 Sweden: Calculation procedure compared to BSK99

EN 1993-1-1 BSK99 5.1 Cross-sectional properties

5.1.1 Column: IPE 600 (S235)

fy = 235N/mm²; γM0 = γM1 = 1,0 (EC3, 6.1) fyk = 235N/mm²; fuk = 340N/mm2; γM = 1,0; γn = 1,1 (BSK99, 2:21 and 3:42)

b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm b = 220mm; h = 600mm; tf = 19mm; tw = 12mm; r = 24mm A = 156cm² ; Iy = 92080cm4 ; Iz = 3390cm4 ; Wpl,y = 3520cm³ A = 156cm²; Iy = 92080cm4; Iz = 3390cm4; Wpl,y = 3520cm³, Wpl,z = 486cm3,

Wel,y = 3069cm³, Wel,z = 308cm³

Iω = 2846000cm6; IT = 165cm4 Iω = 2846000cm6; IT = 165cm4

kN3666f

AN0M

yRd,pl =

γ⋅=

(EC3, 6.2.3 (2)) yk

pl,dM n

3333kNf

N Aγ γ

= ⋅ =⋅

kN11373

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3)) yk

pl,y,d pl,yM n

752,0 kNmf

M Wγ γ

= ⋅ =⋅

(BSK99, 6:243)

kNm2,827f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2))

5.1.2 Rafter: IPE 400 (S235)


b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm b = 180mm; h = 400mm; tf = 13,5mm; tw = 8,6mm; r = 21mm A = 84,5cm² ; Iy = 23130cm4 ; Iz = 1320cm4 ; Wpl,y = 1308cm³ A = 84,5cm²; Iy = 23130cm4; Iz = 1320cm4; Wpl,y = 1308cm³; Wel,y = 1160cm³ Iω = 490000cm6; IT = 51,1cm4 Iω = 490000cm6; IT = 51,1cm4

kN8,1985f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) yk

pl,dM n

1805kNf

N Aγ γ

= ⋅ =⋅

(BSK99, 6:22)


PAGE 39 OF 98

kN7,5793

f)t)r2t(tb2A(V

0M

yfwfRd,z,pl =

γ⋅⋅⋅++⋅⋅−= (EC3, 6.2.6 (3)) yk

pl,y,d pl,yM n

279,4 kNmf

M Wγ γ

= ⋅ =⋅

(BSK99, 6:243)

kNm4,307f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2))



b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm b = 180(200)mm; h = 820mm; tf = 13,5(16)mm; tw = 8,6(10,2)mm; r = 21mm A = 159,6cm² ; Iy = 134868cm4 ; Iz = 2390cm4 ; Wpl,y = 3920,7cm³ A = 159,6cm²; Iy = 134868cm4; Iz = 2390cm4; Wpl,y = 3920,7cm³; Wel,y = 3451,1cm³ Iω = 2694000cm6; IT = 101,7cm4 Iω = 2694000cm6; IT = 101,7cm4

kN6,3750f

AN0M

yRd,pl =

γ⋅= (EC3, 6.2.3 (2)) yk

pl,dM n

3410kNf

N Aγ γ

= ⋅ =⋅

(BSK99, 6:22)

kN8,12773

fA7,579V

0M

y500IPE,wRd,z,pl =

γ⋅⋅+= (EC3, 6.2.6 (3)) yk

pl,y,d pl,yM n

837,6 kNmf

M Wγ γ

= ⋅ =⋅

(BSK99, 6:243)

kNm4,921f

WM0M

yy,plRd,y,pl =

γ⋅= (EC3, 6.2.5 (2))

Comments: The steel grade normally used in Sweden is S355. The walls would have been used to brace the columns. The factor, γn depends on the risk for bodily injury in case of a collapse.




Ed,HEd

Edcr ≤

δ⋅

=α (EC3, 5.2.1 (4))


Second order effects are indirectly considered in the interaction equations of K18:51 – 53 when the bending moment is calculated according to the theory of elasticity and the frame is one storey high. (K18:55)


PAGE 40 OF 98

159,254,27

60004,270

0,32cr ≥=

⋅







The load combination used for the ultimate limit states design is 1,0 Dead load 1,3 Snow load 0, 25 Wind load⋅ + ⋅ + ⋅

The load from sway imperfection is not used together with the sway buckling length.


PAGE 41 OF 98



5.4.1 Column: IPE 600 (S235)


Flange:

k

yk

/ =(220/2-12/2-24) / 19 =4,21 0,3 9Ec tf

≤ ⋅ = → class 1 (BSK99,table 6:211a)


Web:

k

yk

/ =(600-2 19-2 24) / 12 =42,83 2, 4 71,7Ec tf

⋅ ⋅ ≤ ⋅ = → class 1

(BSK99,table 6:211a)

5.4.2 Rafter: IPE 400 (S235)


Flange:

k

yk

/ =(180/2-8,6/2-21) / 19 =4,62 0,3 9Ec tf

≤ ⋅ = → class 1 (BSK99,table 6:211a)


Web:

k

yk

/ =(400/2-2 13,5-2 21) / 8,6 =38,5 2,4 71,7Ec tf

⋅ ⋅ ≤ ⋅ = → class 1

(BSK99,table 6:211a)


Flange: ψ = α = 1 (constant pressure) (IPE 500 only, IPE 400 see above) (200/2-10,2/2–21) / 16 = 4,62 < 9 → class 1 (EC3, table 5.2)

Flange: ψ = α = 1 (constant pressure) (IPE 500 only, IPE 400 see above)

k

yk

/ =(200/2-10,2/2-21) / 16 =4,62 0,3 9Ec tf

≤ ⋅ = → class 1 (BSK99,table 6:211a)

Web: IPE 400-part: in tension Web: IPE 400-part: in tension


PAGE 42 OF 98

Haunched IPE 500 part : α = 1 (constant pressure) c/t = (420-16-21) / 10,2 = 37,5 < 38 → class 2 (EC3, table 5.2)


Haunched IPE 500 part: α = 1 (constant pressure)

k

yk

/ =(420-16-21) / 10,2=37,5 1, 46 43,6Ec tf

≤ ⋅ = → class 1 (BSK99, table 6:211a)


5.5.1 Column IPE600 (S235)




721172728,42

12514

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))


0,1089,00,1137

1,101VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


Internal shear force: |Vz,Ed| = 92kN Slenderness and reduction value

ykww v

w k

0,35 0,50 0,67fb

t Eλ ω= ⋅ ⋅ = → = (BSK99, 6:261d)

The shear resistance is the smallest of:

ykz,Rd v w

M n

1036kNf

V Aωγ γ

= ⋅ ⋅ =⋅

(BSK99, 6:261a)

z,Ed

pl,z,Rd

92,3 0,089 1,01036

VV

= = <

Class 1 → No interaction between Vz and My



0,1034,00,3666

5,125NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))

Internal forces: |NEd| = 110,5kN, |My,Ed| = 493,6 kNm

pl,yy

el,y

1,15WW

η = = (BSK 99, 6:242)

20 y 1,32γ η= = (K18:51 e)


PAGE 43 OF 98


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1734,02,8278,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

0 1,32SydSd

Rd Ryd

110,5 493,6 0,67 1,03333 752,0

MNN M

γ + = + = <

(K18:51 a)





9208021000N2

2

y,cr =⋅⋅π

=


y ==λ (EC3, 6.3.1.2 (1))



Elastic critical flexural buckling force: 2

cr,y 2

21000 92080 1388 kN3708

N π ⋅ ⋅= =

Class 1 section: Slenderness: ykyc

cr

3666 1,631388

A fN

λ⋅

= = = (BSK 99, 6:233a)

Rolled section; h/b >1,2; tf < 40mm → buckling curve a (BSK 99, table 6:233)

1 0,21 4,20β α→ = → = (BSK 99, 6:233)

Reduction value: 2 2

cyc 2

c

4, 40,30

2, 2α α λ

ωλ

− −= = (BSK 99, 6:233b)

Ed

yc pl,d

1NNω

≤⋅

110,5 0,11 1

0,30 3333= <

⋅

Comments: In Sweden the handbook K18 Dimensionering av stålkonstruktioner is a complement to the Swedish code BSK 99. The flexural buckling length for columns in frames calculated as shown in K18:38 gives, Lcr,y= 23,1m. The differences in flexural buckling length can be explained with that the columns connection to the foundation is assumed to be partially restrained for rotation in K18:38. With Lcr,y according to K18:38 the reduction value, ωc=0,63.


PAGE 44 OF 98




339021000N2

2

z,cr =⋅⋅π

=


z ==λ (EC3, 6.3.1.2 (1))



Elastic critical flexural buckling force: 2

cr,z 2

21000 3390 1951,7 kN600

N π ⋅ ⋅= =

Class 1 section: Slenderness: ykzc

cr

3666 1,371952

A fN

λ⋅

= = = (BSK 99, 6:233a)

Rolled section; h/b >1,2; tf < 40mm → buckling curve b (BSK 99, table 6:233)

1 0,34 3,46β α→ = → = (BSK 99, 6:233)


czc 2

c

4,40,37

2,2α α λ

ωλ

− −= = (BSK 99, 6:233b)

Ed

zc pl,d

1NNω

≤⋅

110,5 0,09 1

0,37 3333= <

⋅


Elastic critical moment from literature or ENV1993-1-1, Annex F:

kNm8,1431IEIGL

II

LIE

CM21

z2

T2

z2

z2

1cr =

⋅⋅π

⋅⋅+⋅

⋅⋅π⋅= ω (ENV 1993-1-1, Annex F, 1.3 (3))


LT ==λ (EC3, 6.3.2.2 (1))




fLT

mod,LT ==χ

=χ (EC3, 6.3.2.3 (2))

Elastic critical moment from literature or ENV1993-1-1, Annex F 1

2 2 2ωz T

cr 1 2 2z z

1432kNmIE I L G IM CL I E I

ππ

⋅ ⋅ ⋅ ⋅= ⋅ ⋅ + = ⋅ ⋅

(ENV 1993-1-1, Annex F, 1.3 (3))

2m

10,8 0,2 0,8M

Mχ = + ≥

(BSK 99, 6:2442)

2 m0 0,8M χ= → =

Class 1 section: Slenderness:

c el,y yk 3b

m cr

1,15 3069 235 10 0,850,8 1432

W fM

ηλ

χ−⋅ ⋅

= = ⋅ =⋅

(BSK 99, 6:2442a)


PAGE 45 OF 98

Rolled section → b 4b

1,02 1,01

ωλ

= ≤+

(BSK 99, 6:2442b)

b 4

1,02 0,801 0,95

ω = =+

,

, ,

1y Ed

b pl y d

MMω

≤⋅

0,8 493,6 0,66 1

0,80 752,0⋅

= <⋅





( ) 982,0N

N8,01C046,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=

973,0N

N)25,0C(

1,01963,0N

N)25,0C(

1,01kRd,plz

Ed

LT,mRd,plz

Ed

LT,m

zzy =

⋅χ−−<=

⋅χ−λ⋅

−=



0,1914,0800,0114,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

0,1886,0792,0094,0M

Mk

NN

Rd,y,plmod,LT

Ed,yzy

Rd,plz

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.62))

Double-symmetric section Interaction for flexural buckling

20 y 1γ η= ≥

20 1,15 1,32γ = = (K18:51 e)

yc 0 yc 0,8γ γ ω= ≥ yc 0,8γ = (K18:52 d)

yc

SydSd

Rycd Ryd

1,00MN

N M

γ

+ ≤

(K18:52 a)

0,8110,5 493,6 0,83 1

0,30 3333 752,0

+ = ≤ ⋅

Comments: With the different flexural buckling length and reduction value in section

3.5.1.2.1 the interaction yc

SydSd

Rycd Ryd

1,00MN

N M

γ

+ ≤

and yc 0,83γ = gives the result

0,83110,5 493,6 0,74 1

0,63 3333 752,0

+ = ≥ ⋅


PAGE 46 OF 98

5.5.2 Rafter IPE 400 (S235)




721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (EC3, 6.2.6 (6))


0,1138,07,5791,80

VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


Internal shear force: |Vz,Ed| = 65,8kN Slenderness and reduction value

ykww v

w k

0,35 0,45 0,67fb

t Eλ ω= ⋅ ⋅ = → = (BSK99, 6:261d)


ykz,Rd v w

M n

513,9 kNf

V Aωγ γ

= ⋅ ⋅ =⋅

(BSK99, 6:261a)

z,Ed

z,Rd

65,8 0,13 1,0513,9

VV

= = <

Class 1 → No interaction between Vz and My



0,1055,08,1985

7,108NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1705,04,3076,216

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

Internal forces: |NEd| = 89,6kN, |My,Ed| = 173,3kNm

pl,yy

el,y

1,13WW

η = = (BSK 99, 6:242)

20 y 1,27γ η= = (K18:51 e)

0 1,27SydSd

Rd Ryd

89,6 173,3 0,64 11805 279,4

MNN M

γ + = + = <

(K18:51 a)


PAGE 47 OF 98





y ==λ (EC3, 6.3.1.2 (1))



Class 1 section: Slenderness: ykyc

cr

1986 1,331118

A fN

λ⋅

= = = (BSK 99, 6:233a)

Rolled section; h/b >1,2; tf < 40mm → buckling curve a (BSK 99, table 6:233)

1 0, 21 3,19β α→ = → = (BSK 99, 6:233)


cc 2

c

4, 40, 42

2, 2α α λ

ωλ

− −= = (BSK 99, 6:233b)



For this example it is assumed, that the roofing made of profiled sheeting provides sufficient continuous lateral and torsional restraints. → No out of plane flexural buckling and lateral torsional buckling





( ) 987,0N

N8,01C024,1N

N2,01CkRdy

Edmy

Rdy

Edymyyy =

⋅χ+>=

⋅χ−λ+=


0,1818,0695,0123,0M

Mk

NN

Rd,y,plmod,LT

Ed,yyy

Rd,ply

Ed <=+=⋅χ

⋅+⋅χ

(EC3, eq. (6.61))

Double-symmetric section, member not susceptible to torsional deformations Interaction for flexural buckling

yc 0 yc 0,8γ γ ω= ≥ yc yc1,27 0,42 0,53 0,8γ γ= ⋅ = → = (K18:52 e)

yc

SydSd

Rycd Ryd

1,00MN

N M

γ

+ ≤

(K18:52 a)

0,889,6 173,3 0,80 1

0, 42 1805 279,4 + = ≤ ⋅


PAGE 48 OF 98





721172725,38

6,8331

th

w

w =⋅=ηε

⋅<== (IPE 400) (EC3, 6.2.6 (6))

721172725,37

2,10383

th

w

w =⋅=ηε

⋅<== (haunched IPE 500) (EC3, 6.2.6 (6))


0,1090,08,1277

5,114VV

Rd,z,pl

Ed,z <== (EC3, 6.2.6 (1))


Internal shear force: |Vz,Ed| = 94,1kN Slenderness and reduction value IPE 400

ykww v

w k

0,35 0,45 0,67fb

t Eλ ω= ⋅ ⋅ = → = (BSK99, 6:261d)


ykz,Rd v w

M n

513,9 kNf

V Aωγ γ

= ⋅ ⋅ =⋅

(BSK99, 6:261a)

Haunched IPE 500

ykww v

w k

0,35 0,44 0,67fb

t Eλ ω= ⋅ ⋅ = → = (BSK99, 6:261d)


ykz,Rd v w

M n

616,9 kNf

V Aωγ γ

= ⋅ ⋅ =⋅

(BSK99, 6:261a)

z,Ed

z,Rd

94,1 0,08 1,0513,9 616,9

VV

= = <+



Internal forces: |NEd| = 92,5kN, |My,Ed| = 493,6 kNm

pl,yy

el,y

1,14WW

η = = (BSK 99, 6:242)


PAGE 49 OF 98

0,1030,06,3750

3,112NN

Rd,pl

Ed <== (EC3, 6.2.4 (1))


|N|0M

ywwEd =

γ

⋅⋅⋅< (EC3, 6.2.9.1 (4))


0,1659,04,9218,606

MM

Rd,y,pl

Ed,y <== (EC3, 6.2.5 (1))

20 y 1,29γ η= = (K18:51 e)

0 1,29SydSd

Rd Ryd

92,5 493,6 0,60 13410 837,6

MNN M

γ + = + = <

(K18:51 a)



Stability failures of member do not have to be considered: In plane flexural buckling is covered safely by check of IPE 400. Out of plane flexural buckling and lateral torsional buckling is excluded (see Rafter).




132mm = L/227 < L/200

No vertical deflection limits



19,9mm = H/302 < H/300

No vertical deflection limits


PAGE 50 OF 98

6 U.K.: Calculation procedure compared to BS 5950-1

EN 1993-1-1 BS 5950-1 6.1 FRAME GEOMETRY

LCIPE 400 (S235)

6°

IPE 500 haunch

IPE 600 (S235)

30 m3700

6 m

Spacing of portal frames = 5.0 m

The configuration of the single storey portal frame is given below and shows steel section sizes, the eaves haunch and the arrangement of purlins and side rails. Note the steel grade is S235 for the rafter and columns.

LCIPE 400 (S235)

6°

IPE 500 haunch

IPE 600 (S235)

30 m3700

6 m

Spacing of portal frames = 5.0 m w kN/m

h

6°

V

HPinned base

H

V

rRise h = 1.58 m

L = 30 m

h= 6 m

b = 3.7 m

The proposed frame for analysis purposes is defined by the line diagram as in the figure below, where the lines represent the centrelines of the members.

w kN/m

h

6°

V

HPinned base

H

V

rRise h = 1.58 m

L = 30 m

h= 6 m

b = 3.7 m

Portal frame dimensions Portal frame dimensions

The cladding to the roof and walls is supported by purlins and side rails, or Frame span = L = 30 m


PAGE 51 OF 98

alternatively by deep profiled sheeting spanning between the frames, provided that torsional restraints are introduced at plastic hinge positions as follows: • At both ends of the haunch (i.e. one to the column, one to the rafter) • At an intermediate position along the rafter.

Eaves height = h = 6 m Frame centres = Ls = 5 m

The roof sheeting or purlins provide lateral restraint to the outer flange of the rafter and columns.

The length of the haunch is taken as 3.7 m which is 12% of the span of the frame.

6.2 LOADING

6.2.1 Vertical Loads Unfactored loads

The value of snow load that has been used in this case is 0.70 kN/m2 but the relevant national code must be used.

Dead loads: Sheeting = 0.20 kN/m2 Purlins = 0.07 kN/m2 Frame = 0.11 kN/m2 Services = 0.28 kN/m2

Roofing = 0.20 kN/m2 × 5.0 m = 1.0 kN/m on plan Services = 0.20 kN/m2 × 5.0 m = 1.0 kN/m on plan Snow = 0.70 kN/m2 × 5.0 m = 3.5 kN/m on plan

Total dead load = 0.66 kN/m2 Imposed load = 0.60 kN/m2

6.2.2 Combination factor ψ Load combiantions

Note that where the NAD specifies a value for ψ0 , this value must be used instead of the value from Eurocode 1. The value in Eurocode 0, EN 1990:2002 Table A1.1 is 0.7 generally, but 1.0 for structures supporting storage loads.

The vertical load (Dead and Imposed) at the ultimate limit state is usually used to determine the size of the members for preliminary design purposes. At the detailed design stage, other load combinations should also be checked at the ultimate and the serviceability limit states.

In this example, wind load is considered in combination with vertical loading. Total factored load w = Ls (γ fd × 0.66 + γ fi × 0.60) = 5 (1.4 × 0.66 + 1.6 × 0.60) = 9.42 kN/m

6.2.3 Global analysis

Plastic analysis of single-storey steel portal frames leads to an economical form of structure. However, in this case, member sizes are given for consistency by elastic analysis, which will lead to a higher failure load than required for this building.


PAGE 52 OF 98

6.3 ULTIMATE LIMIT STATE ANALYSIS MOMENT CAPACITY

6.3.1 Frame imperfection – equivalent horizontal forces Column (IPE 600 Grade S235)

mh0 ααφφ ⋅⋅= EN1993-1-1; §5.3.2

Where 0φ = 2001

Conservatively, 1,1 mh == αα

The moment capacity of both the column and the rafter will be reduced slightly because of the axial loads. The reduction in moment capacity is usually ignored at the preliminary design stage.

giving φ = 1 / 200 × 1.0 × 1.0 = 1 / 200 Axial Force Fc* = V = wL/2 = 9.42 × 30/2 = 141.3 kN

The column loads could be calculated by a frame analysis, but a simple calculation based on plan areas is suitable for single storey portal frames.

n = Fc/Apy = 141.3 × 103 / (156 × 102 × 235) = 0.04

Thus, the unfactored equivalent horizontal forces are given by: Span/height to eaves L/h = 30/6 = 5.0

Permanent/frame = 88.8/ 200 = 0.444 kN Variable/frame = 105.0 / 200 = 0.525 kN

Rise/span hr /L = 1.58/30 = 0.053

Vertical load wL = 9.42 × 30 = 282.6 kN wL2 = 11.3 × 302 = 8478 kNm

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

=××=×= −3yxcx 102353512fSM 825.3 kNm

As n = 0.04 it can be conservatively assumed that the reduced moment capacity of the column will be very similar to the moment capacity with no axial load acting. (For a similar UK UB sections the reduction in bending moment capacity for n = 0.04 less than 1% (0.75%)). Hence assume M.rx = 825.3 x 99.25/100 = 819.1 kNm

which is > the 525.6 kNm required.

6.3.2 Partial safety factors and second order effects Rafter (IPE 400 Grade S235)

For simplicity, when carrying out an elastic-plastic software analysis the reduced collapse load factor can be accounted by multiplying the partial load factor by:

−

cr

11

1

α

Axial force Fc* = Hcosθ + Vsinθ = 101.7 cos 6o + 141.3 sin 6o

= 115.9 kN n = Fc/Apy = 115.9 × 103 / (84.5 × 102 × 235) = 0.06


PAGE 53 OF 98

Assume for preliminary calculations, based on experience, that αcr = 9 From section tables moment capacity of column

So 125.1

911

111

1

cr

=−

=−

α

Therefore the modified partial safety factors are: γG = 1,35 × 1.125 = 1.52 γQ = 1,5 × 1.125 = 1.69

6.3.3 Analysis

In this example, the bases have been assumed to be pinned for simplicity

Steel grade is S235. Assume sections are Class 1, then check later.

Column : IPE 600 has tf ≤ 40 mm, and so fy = 235 N/ mm2

*The axial load should be that which is relevant to the load case being checked. In practice, however, the axial load is so low that the conservatism in using the largest axial load is negligible for low pitch roofs.

=××=×= −3102351307yxcx fSM 307.1 kNm

As n = 0.06 it can be conservatively assumed that the reduced moment capacity of the column will be very similar to the moment capacity with no axial load acting. (For a similar UK UB sections the reduction in bending moment capacity for n = 0.06 less than 1% (0.85%)).

Hence assume M.rx = 307.1 × 99.15/100 = 304.5 kNm

which is > the 296.7 kNm required.

kNm7.740101

23510152.36

6

MOyypl,p

=×

×⋅== γfWM

Rafters : IPE 400 has tf≤ 40 mm, and so fy = 235 N/ mm2

kNm1.307101

23510307.16

6

MOyypl,p

=×

×⋅== γfWM

Load factor Hinge number

Span no.

Member Position (m)

0.898 1 1 RH Column 5.0

1.070 2 1 LH Rafter 12.04

Although hinge 1 occurs at a load factor ≤ 1,0, a mechanism is not formed until the second hinge has formed. Therefore this combination of section sizes is suitable for


PAGE 54 OF 98

preliminary sections.

A diagram of bending moments, shear and axial forces is given opposite for a load factor 1,0, which is the condition at ultimate limit state. The bending moments in the columns are shown to reduce from the level of the bottom of the haunch to the top of the column. This is the true bending moment in the column when the haunch to column connection is a bolted connection on the inner vertical face of the column.

Load combination 1: bending moment, shear and axial load

6.4 CALCULATE, αcr, FOR STABILITY CLASSIFICATION OF SECTIONS WITH AXIAL LOAD

These checks use internal forces derived from a separate analysis using partial safety factors 1.35 and 1.5, not the increased values used to allow for second order effects in the ultimate limit state analysis.

At the detailed design stage, it is necessary to ensure that the sections can be classified as ‘plastic’ or Class 1 cross-sections. The axial load is usually so low that, providing the section can be classified as plastic when subject to bending only.

6.4.1 Load combination no.1 For column (IPE 600 Grade S235)

6.4.1.1 Sway stability check

For frames where L ≤ 8 h may be found applying a simple modification to αcr,H. In this case L = 30 m and h =6.0 m, so 30 ≤ 8 × 6 = 48, so the following formulae is used:

ε = (275/py)½ = (275/235)½ = 1.08

Flange b/T = 5.8

Hcr,

maxR,cr

E,ULSscr,e, 18.0 αα

−=

N

N

Limiting b/T value for Class 1 plastic flange = 9ε = 9.72 5.8 < 9.72 ∴ flange is classified as plastic


PAGE 55 OF 98

where :

maxR,cr

E,ULS

N

N is the maximum ratio in any rafter

R,crN is the Euler load of the rafter on the full span (assumed pinned)

= 2

2

LEIπ

= 2

42

300001023130210000 ×××π

=533kN

R,ULSN is the maximum axial compression in the rafter in the load case

= 107kN (value obtained with the analysis of the software without considering second order effects)

Web d/t = 42.8

Limiting d/t value for Class 1 plastic “Web Generally” = 11

80r+

ε but ≥ 40ε

80ε = 86.4

r1 = yw

c

dtp

F but –1 < r1 ≤ 1

= 2350.120.514103.141 3

××

×

= 0.097

1 + r1 = 1 + 0.097 = 1.097

αcr,H is the elastic critical buckling factor as by Horne(1975)

αcr,H = 3.218.281

100.6 3

ULS

EHF =×

=

=

δδhh

VH

EN1993-1-1; §5.2.1(4)B

Limiting d/t value = 097.1

4.86180

1

=+ r

ε

= 78.8

where

δh

is the minimum ratio of [column height]/[horizontal deflection of the

column top derived from first order analysis]

42.8 < 78.8 ∴ web is classified as plastic

6.133.21105031010718.018.0

max3

3

Hcr,

maxR,cr

E,ULSscr,e, =⋅

×

×−=

−= αα

N

N

Both the flange and the web are classified as plastic, so the section can be classified as plastic.

Therefore,

08.1

6.1311

111

1

cr

=−

=−

α


PAGE 56 OF 98

6.4.1.2 Rafter snap-through buckling load factor For rafter (IPE 400 Grade S235)

( ) ( )rr

rc

IIIhL

LD θ

Ωα 2tan

1

47.55cr,r

+

−

+

=

ε = (275/py) ½ = (275/235) ½ = 1.08 Flange b/T = 6.7

where D cross-section of the rafter L span of the bay h mean height of the column from base to eaves or valley) Ic in-plane second moment of area of the column(taken as zero if the

column is not rigidly connected to the rafter, or if the rafter is supported on a valley beam)

Ir in-plane second moment of area of the rafter Fyr nominal yield strength of the rafters

θr roof slope if the roof is symmetrical

Limiting b/T value for Class 1 plastic flange = 9ε = 9.72 6.7 < 9.72 ∴ flange is classified as plastic Web d/t = 34.5 The axial load in the rafter is generally so small that it can be assumed that the neutral axis is at mid-depth and the d/t limit can be taken as 80ε, but for completeness the axial load will be taken into account.

Limiting d/t value for Class 1 plastic “Web Generally” = 11

80r+

ε but ≥ 40ε

‘Snap through’ will only occur if the ratio given by Ω = 0WWr is >1.0.

Wo is the plastic failure load of the rafters as fixed ended beam of span L (Mp= )162lw ×

Wo = 16 × =LM p 16 × kN4.217)4.730(1.307 =−

Wr is the total factored vertical load on the rafters of the bay Roofing = 1.35 kN/m2 × 1.44 = 1.35 kN/m Services = 1.35 × 1.44 = 1.35 kN/m Rafter = 1.35 × 0.76 = 0.88 kN/m Snow = 1.5 × 5.04 = 5.25 kN/m Total = 8.83kN/m 8.83 × 30-7.4) = 199.6 kN

r1 = 2356.8331

109.101 3

yw

c

××

×=

dtpF

= 0.15

Limiting d/t value = 15.1

4.86180

1

=+ r

ε = 75.1

34.5 < 75.1 ∴ web is classified as plastic Both the flange and the web are classified as plastic, so the section can be classified as plastic.

Ω = 0r WW = 192.04.2176.199 ≤=

Therefore the ‘snap through’ will not occur and need not be considered further for this example.


PAGE 57 OF 98

6.5 COLUMN DESIGN: IPE 600 IN-PLANE FRAME STABILITY

Load combination no.1 is clearly the worst case for the column for axial force, bending moment and shear force as well as restraint to the compression flange. Therefore, the columns checks need to be made for only load combination no.1

This check should be carried out at the preliminary design stage, after the size of the sections is determined. It should be carried out again at the detailed design stage if the section sizes are reduced.

MEd = 540.8 kNm VEd = 108.1 kN NEd = 148.8 kN

6.5.1 Classification Check the geometry of the Frame

The section is Class 1 to permit plastic hinge formation. The web is checked from EN 1993-1-1 Sheet 1 of Table 5.2.

f /

f /

h

==

58,0

y γMO

MOy γ

Plastic stress distribution in web

αc

c

Web is under combined axial and bending forces, so find α :

Depth of stress block at yield stress resisting axial load

= NEd / ( fy × tw / γMO) = 148.8×103 / (235 × 12.0 / 1) = 52.8 mm

∴ αc = c / 2 + 52.8 / 2 ∴ α = 0.5 + [( 52.8 / 2) / 514] = 0.55

(a) L ≤ 5h

L = 30 m, 5h = 5 × 6 = 30 m

30 m ≤ 30 m ∴ OK

(b) hr ≤ 0.25L

hr = 1.58 m 0.25L = 0.25 × 30 = 7.5 m

∴ 1.58 m < 7.5 m ∴ OK

∴ geometry of the frame is within the limits

Formula Method- Vertical Loads

Check effective span to depth ratio of the rafter satisfies the condition:

DLb ≤

+ yrr

275/4

.44pLLh

Lρ

ρ

Ω

where Lb = L – hhs

h2 LDD

D

+

assuming Dh ≈ Ds


PAGE 58 OF 98

∴for Class 1, limiting w tc = 396 ε / (13 α -1 )

= 396 × 1.0 / ( 13 × 0.55 − 1 ) = 64.3 actual w tc = 514 / 12.0 = 42.8 → web is Class 1

Flange check from EN 1993-1-1 Sheet 2 of Table 5.2

∴for Class 1, limiting ftc = 9 ε = 9 × 1.0 = 9

actual f tc = 85 / 19 = 4.5 → flange is Class 1

Lh is the length of a single haunch (= 3.7 m)

∴ Lb = 30 – 3.7 = 26.3 m

ρ =

hL

II

r

c2 =

630

23130920802

××

= 39.8

Lr = L/cos θ = 30/cos 6o = 30.2 m

Ω = Wr/Wo

Wr = wL = 9.42 × 30 = 282.6 kN Wo is the maximum value of Wr that causes failure of the rafter treated as a fixed ended beam of span L

oW

30 m

Plastic hinges

Calculation of Wo – failure load of fixed-ended beam

6.5.2 Cross-sectional resistance Mp = Mcx = WoL/16

The frame analysis assumed that there is no deduction in the plastic moment resistance from interaction with shear force or axial force. This assumption must be checked because it is more onerous than checking that the cross-sectional resistance is sufficient.

Wo = 16Mcx/L = 16 × 305/30 = 162.7 kN

Ω = Wr/Wo = 282.6/162.7 = 1.74

Load combination no.1 is clearly the worst load combination.

Max. shear force, VEd = 108.1 kN Max. axial force, NEd = 148.8 kN

+ yr

275/4

.44pLLh

Lρ

ρ

Ω

Check that the plastic moment of resistance, Mpl.Rd , is not reduced by the coincident shear force. EN1993-1-1; §6.2.8(2) =

235275

30/2.308.3948.39

630.

74.144

×+ = 133.6

Check EdV ≯ 0.5 Rdpl,V Lb/D = 26.3 × 103/400 = 66.8

Av = h tw = 600 × 12.0 = 7200 mm2 §6.2.6(3a) As Lb/D < 133.6, the frame is stable under gravity loads


PAGE 59 OF 98

Vpl,Rd = Av (fy / √3) / γMO EN1993-1-1; §6.2.6(2)

= 7200 ( 235 / √3) / 1 = 976 kN

∴ The required load factor for frame stability λr = 1.0 for gravity load case

∴ 0.5 Rdpl,V = 0.5 × 1045 = 523 kN

Max VEd = 108.1 kN

As VED ≯ 0.5 Rdpl,V , plastic moment of resistance is not reduced by the coexistent shear force.

For horizontal loads Required load factor λr for frame stability

λr = 1sc

sc

−λ

λ

Now check that the plastic moment of resistance, Mpl.Rd , is not reduced by the coincident axial force’ EN 1993-1-1; §6.2.9(4)

Check: λsc =

+ yrrb

275/4

220pLLhL

DLρ

ρ

Ω

(i) EdN ≯ 0.25 Npl.Rd Eqn. 6.33

∴ 0.25 Npl.Rd = 0.25 × 15600 × 235 / 1 = 916.5 kN =

×+××

××

235275

30/2.308.3948.39

3.26674.1304.0220

(ii) EdN ≯MO

yww5.0

γ

fth Eqn. 6.34

= 10.2

kNfth

2.7950.1

2352.105645.05.0

MO

yww =×××

=γ

∴ λr = 12.10

2.10−

= 1.11

Max NEd = 148.8 kN ≤ 795.2 kN

Checks (i) and (ii) show that the plastic moment of resistance is not reduced by the coexistent axial force. Therefore the frame analysis assumption is validated.

Therefore, for this load case, λp must not be less than 1.11. The actual value of λp would depend on the magnitude of the applied horizontal loads, but generally λp would be greater than λr. In this example it is assumed that vertical load case is critical.

6.5.3 Buckling between intermediate restraints

6.5.3.1 Upper section analysis

For members with plastic hinges, EC3 gives guidance for member buckling check. The critical column bending moment diagram is from load combination no.1 in this structure, causing a plastic hinge to occur at the underside of the haunch. Therefore,


PAGE 60 OF 98

find the stable length with a plastic hinge.

0 kNm

Bending momentsbelow haunch

Geometry

Moments, forces and restraints

3650

1350

6000

N = 148,8 kNV = 108,1 kN

N = 148,8 kNV = 108,1 kN

540,8 kNm

394.8 kNm

Stable length of column: EN1993-1-1; BB3.1.1

2y

t

2ypl,

21

m

2357561

4.571

38

+

=

f

AI

W

CAN

iL

Ed

z

where NEd = 148.8 kN A = 15600mm2

Wpl,y = 3512 × 103mm3 It = 165 × 104mm4

fy = 235 N/mm2 iz = 46.6 mm

C1 is dependant on ψ based on the shape of the bending moment diagram.

NCCI SN003a-EN-EU; Table 3.1

ψ = 1532.173.050003650

1 =→= C


PAGE 61 OF 98

( ) 2

4

23

2

3m

235235

1016515600103512

1532.17561

15600108.148

4.571

6.4638

××

×

×+

×

×=L

=2196 mm Therefore, 1350 mm is acceptable.

6.5.3.2 Lower section (3650 mm) LAYOUT OF PURLINS AND SIDE RAILS

(a) Calculate slenderness λ and λLT

Assume side rail depth = 200 mm

At this stage, a more detailed assessment of the frame geometry can be made. It is also useful to determine a layout of purlins and side rails that can provide restraint to plastic hinges and adjacent lengths.

==

200

400

600

==

Cross-section through column

s

s

s4

s

s

1P P2P3

P4P5

P6P7

P8P9

P10

6

5

3

2

1

6

l6

110015083 183

6 @ 1630

= =

6000

620

4 @

130

018

0

h

Distance from columns shear centre to centre of side rail, a Purlin and side rail spacing

a = 600 / 2 + 200/ 2 = 400 mm is2 = iy2 + iz2 + a2

is2 = 2432 + 46.62 + 4002 = 221221 mm2

The value of the bending moment can be found at any point in the rafter from the formula:

Mx = VΡx – H hx – w Ρ x2/2

Distance between shear centres of flanges hs = 600 −19 = 581 mm

where Ρx is the horizontal distance to the point considered


PAGE 62 OF 98

α =

+

2s

z

w2

iIIa

hx is the height of the point considered

= h + Ρx tanθ

V is the vertical reaction at the base due to w

H is the horizontal reaction at the base due to w

w is the load per unit length of the frame (factored loading)

Using the simplification for doubly symmetric I sections;

Iw = Iz ( hs / 2)2

α = ( )

+2

s

2s

2 2h

i

a

= ( )

+221221

2581400 22

= 1.1

The slenderness of the column is given by :

λ = ( )[ ] 5.02

sz22

tt

zt

6.2 iILI

iL

πα +

= ( )[ ] 5.07225 2212211039.36.23650105.161.1

6.463650

××××××+ π

= 71.2

Note:λ has been calculated using the slenderness method, λLT has been calculated using the Mcr method to show the different approached for obtaining these parameters.

λ = 1λλ

= 9.932.71

= 0.76


PAGE 63 OF 98

LTλ = cr

yy

M

fW for Class 1 pl.yy WW = EN1993-1-1; §6.3.2.2

Where BB3.3.1

( ) ( )

=+

−+

=

+=

+

+=

++=

×=××==

++==

→=

crT

crE2

10

2t2t1o

m

12242zw

t2t

w2

2t

2z

2

2s

crcr

2s

cr,o

cr,0m2cr

2015.0

15.0

10

5

201101

11025.124841021422

12

1

NNB

BB

BBBC

hII

GILEI

LaEI

iNN

aiM

MCc

M

s

ηηηπ

ηπ

η

η

η

ββ

ππ

NcrT = Ncr = 6

4

2

122

2

242

10428.6

10165810003650

1086.22100003650

400103387210000

1659741

×=

××+

+×××

+

+××××

π

π

NcrE = 6

2

42

2t

z2

10269.53650

103387210000×=

×××=

ππLEI

82.010428.610269.5

6

6

crT

crE =×

×==

NNη


PAGE 64 OF 98

101.02015.0

15.0371.0

10

553.0

201101

210 =+

−+

==+

==+

+=

ηηπηπ

η

ηη BBB

and β = 0 because it is the ratio of the smaller end moment to the larger end moment in the column

Therefore,

Cm = 89.10101.00371.0528.0

1122

t2t1

=×+×+

=++ ββ BBBo

c 1=

kNm 3360177889.1111

kNm 177810428.64002

2212212

2cr,0m2cr

6cr

2

cr,o

=××==

=×××

==

MCc

M

Na

iM s

50.0103360

2351031526

3

cr

yyLT =

×

××==

M

fWλ

(b) Calculate buckling resistance for axial load

Nb, Rd = M1y γχ fA ⋅⋅ EN1993-1-1; §6.3

22min

1

λφφχ

−+=

φ = ( )[ ]22.015.0 λλα +−+

bh = 600/200 = 3.0 EN1993-1-1; Table 6.2

tf = 19 mm buckling between axis z-z Use Curve b for hot rolled I sections α = 0.34


PAGE 65 OF 98

φ = ( )[ ]276.02.076.034.015.0 +−+

= 0.884

zχ = 22 76.0884.0884.0

1

−+

= 0.45

Nb, Rd = M1y γχ fA ⋅⋅

Nb, Rd = 12351560045.0 ⋅⋅

= 1650 kN

(c) Calculate buckling resistance for bending

Mb, Rd = M1yypl,LT γχ fW

LTχ = 2

LT2

LTLT

1

λβφφ ⋅−+

LTφ = ( )[ ]2LTLT,0LTLT15.0 λβλλα ⋅+−+

LT,0λ = 0.4 ( maximum value)

β = 0.75

bh = 3.0 EN1993-1-1; Table 6.5

For hot rolled sections use Curve c for buckling. α = 0.49

φLT = ( )[ ] 62.05.075.04.05.049.015.0 2 =⋅+−+

94.05.075.062.062.0

122

=⋅−+

=LTχ

Mb, Rd = kNm 776123510351294.0 3 =×××

(d) Calculate buckling resistance to combined axial and bending


PAGE 66 OF 98

1

M1

Rkz,

Edz,Edz,yz

M1

Rky,LT

Edy,Edy,yy

M1

Rky

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

EN1993-1-1; §6.3.3(4)

1

M1

Rkz,

Edz,Edz,zz

M1

Rky,LT

Edy,Edy,zy

M1

Rkz

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

The following simplifications may be made: • Edy,M∆ and Edz,M∆ =0 for Class 1

• No bending minor axis

Buckling about y-y axis is covered by Merchant-Rankine

Therefore,

1

M1

Rky,

Edy,zy

M1

Rkz

Ed ≤+

γγχ M

Mk

NN

0.1b.Rd.yy.Edyzb.Rd.zEd ≤⋅+ MMkNN

As zλ ≥ 1 EN1993-1-1; Table B.2

kzy = ( ) ( )

−−=

−−

zRd,b,

Ed

mLTM1Rkz

Ed

mLT 25,01,01

25,01,01

NN

CNN

C γχ

ψ =0 4,04,06,0 ymLT ≥+= ψC EN1993-1-1; Table B.3

6.004,06,0mLT =⋅+=C

kzy = ( )

×

×

−

×−

3

3

101650108.148

25,06.076.01,01

= 0.98


PAGE 67 OF 98

0.159.08.7758.39498.016508.148 ≤=⋅+ ∴Column OK

6.6 RAFTER DESIGN: IPE 400 DETERMINATION OF THE PLASTIC FAILURE LOAD

For the rafter it is not clear which load case gives the worst load combination, specifically for buckling. Therefore all load combinations should be checked.

The rafter checks for cross-sectional resistance need to be made for the worst case load effects. This is produced by load combination 1 near the apex of the roof.

At the preliminary design stage, it is not necessary to calculate the plastic failure load. At the detailed design stage, however, the plastic failure load (w′) will be used instead of the applied factored loading (w). Assume that the plastic hinges are located in the column at the bottom of the eaves haunch and in the rafters at the second purlin from the ridge (i.e. P9 in the figure below.

MEd = 252.6 kNm VEd = 107.4 kN NEd = 117.2 kN

6.6.1 Classification

The moment in the rafter at P9 is then given by:

M@P9 = V ′Ρ9 - H′ h9 - 2w′

(Ρ9)2

and the moment in the column @ the bottom of the eaves haunch is given by:

Ensure the section is Class 1 to accommodate plastic hinge formation. Web check from EN 1993-1-1 Sheet 1 of Table 5.2 :

f /

f /

h

==

58,0

y γMO

MOy γ

Plastic stress distribution in web

αc

c

The web is under combined axial and bending forces, so find α :

M@S6 ≈ H1(h – Dh – Ds/2) w′ is the collapse load V ′and H′ are the reactions at the base due to w′ At the point of failure, the moment M@P9 and M@S6 must be equal to the reduced moment capacities of the rafter and column sections provided (see Section 5 of these calculations).

Thus M@P9 = Mr.rx = 304.5 kNm M@S6 = Ml.rx = 819.1 kNm


PAGE 68 OF 98

Depth of stress block at yield stress resisting axial load

= NEd / ( fy × tw / MOγ ) = 117.2 ×103 / (235 × 8.6 / 1) = 58.0 mm

∴ α c = c / 2 + 50.6 / 2 ∴ α = 0.5 + [( 58.0/ 2) / 331.0] = 0.59 ∴ for Class 1, limiting w tc = 396 ε / (13 α -1 )

= 396 × 1.0 / ( 13 × 0.59 − 1 ) = 59.4 actual c/t w = 331 / 8.6 = 38.5 → web is Class 1

Flange check from EN 1993-1-1 Sheet 2 of Table 5.2

∴for Class 1, limiting f tc = 9 ε = 9 × 1.0 = 9

actual c/t f = 64.7 / 13.5 = 4.8 → flange is Class 1

w'

M9

9

M

h

l.rx

r.rx

6m

0.62

Determination of failure load w′

6.6.2 Cross-sectional 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 checking that the cross-sectional resistance is sufficient.

Ρ9 = 13.72 m h9 = 7.44 m H′ = (M@S6)/(h@S6) = 819.1/5.38 = 152.2 kN V′ = w′ L/2 = 15w′

Load combination no.1 is clearly the worst load combination

Max. shear force, VEd = 107.4 kN Max. axial force, NEd = 117.2 kN Therefore the plastic moment resistance is not reduced by coincident shear.

Check that the plastic moment of resistance, Mpl.Rd , is not reduced by the coincident shear force. EN1993-1-1; §6.2.8(2)

Check EdV ≯ 0.5 Rdpl,V EN1993-1-1; §6.3.4

Av = h tw = 400 × 8.6 = 3440 mm2

Vpl,Rd = Av (fy / √3) / γMO = 3440 ( 235 / √3) / 1 EN1993-1-1; §6.2.6(2)

Substituting: M@P9 = 15w′ × 13.72 – 152.2 × 7.44 – w′ × 13.722/2 Equating M@P9 to Mr.rx

304.5 = 111.7w’ – 1132.4 w′ =12.86 kN/m Collapse load w′ = 12.86 kN/m (compare with the applied factored load w of = 9.42 kN/m calculated originally)


PAGE 69 OF 98

= 389 kN

∴ 0.5 Rdpl,V = 0.5 × 389 = 194 kN

Max VEd = 107.4 kN

This is less than 194 kN, and therefore the plastic moment is not reduced by co-existent shear.

The corresponding base reactions are: H′ = 152.2 kN V′ = 192.9 kN

8p = w'w

= 42.986.12

= 1.37

Check that the plastic moment of resistance, Mpl.Rd , is not reduced by the coincident axial force: EN1993-1-1; §6.2.9(4)

Check:

(i) EdN ≯ 0.25 Npl.Rd Eqn. 6.33

∴ 0.25 Npl.Rd = 0.25 × 8450 × 235 / 1 = 496 kN

(ii) EdN ≯MO

yww5.0

γ

fth Eqn. 6.34

kN3770.1

2356.83735.05.0

MO

yww =×××

=γ

fth

Max NEd = 117.2 kN ≤ 377 kN

Checks (i) and (ii) show that the plastic moment of resistance is not required by the coexistent axial force.

Therefore, the effects of shear and axial force on the plastic resistance moment can be neglected according to EC3 and the frame analysis assumption is validated.

6.6.3 Buckling between intermediate restraints COLUMN STABILITY

By inspection, the worst case is near the apex in the left hand rafter, because this has the highest bending moment in the rafter.

6.6.3.1 Stable length check for high bending moment

The critical rafter bending moment diagram is obtained from load combination no.1 in


PAGE 70 OF 98

this structure, causing a plastic hinge very close to the apex, and so the stable length with a plastic hinge is calculated according to the following criteria:.

MEd = 343.1 kNm VEd = 0 kN NEd = 107.8 kN

Rafter under highest bending moment

s

s4

s

s

5

3

2

s6

1

Stay

Plastic hingeposition 62

04

@ 1

300

180

6000

1300

4080

819.1 kNm

621.2 kNm

Bending moment diagram and restraints to column

C1 = 1 because the bending moment is approximately uniform between restraints. The stable length of the rafter is given by:

The hinge position will be torsionally restrained by the provision of a rafter stay at the base of the haunch to side rail S6.

2y

t

2ypl,

21

Ed

zm

2357561

4.571

38

+

=

f

AI

W

CAN

iL EN1993-1-1; BB3.1.1The distance to the next lateral restraint to the compression flange from the plastic hinge position can be determined by one of four approaches:

1. Calculate the limiting distance Lm. 2. For IPE section, calculate the modified distance Lm to account for

moment gradient. 3. For IPE section, calculate limiting distance Ls. 4. Use Appendix G of BS5950-1: 2000

( ) 2

4

23

2

3m

235235

103.518450101307

17561

8450108.107

4.571

5.3938

××

×

×+

×

×=L

=1741 mm

The first approach assumes no restraint to the tension flange and is conservative. The second approach also assumes no restraint to tension flange, but involves more work because it takes account of the shape of bending moment and hence permits use of a greater length between restraints. The third approach is a simplification of Appendix G for IPE sections. The fourth approach is generally too complex for manual design and gives little advantage for the column as usually only one restraint is required between the


PAGE 71 OF 98

plastic hinge position and the base. Therefore, 1200 mm purlin spacing provides stability.

In many cases, it will be adequate to provide an adjacent lateral restraint to the compression flange at a distance Lm from the plastic hinge position. The section between this restraint and the base should then prove adequate when checked to BS 5950-1: Clause 4.8.3.3.

6.6.3.2 Combined axial and moment check for lower bending moments Check length between side rails S6 and S5

Where bending moment is lower, the purlin spacing can be increased. BS 5950-1: 2000 Clause 5.3.3(a) is used to check the length between restraints at side rails S6 and S5. Assume restraint is provided at S6 and S5 by means of column stays.

Rafter under lower bending moments The critical case is in right hand rafter. Try purlin spacing at 2200 mm centres.

Check for lateral torsional buckling between purlins. Load combination no.1 gives the biggest moments:

Limiting length Lm is given by:

Lm = 2

12

y2

c

y

27536130

38

+

pxf

r

fc = V′/A = 192.9 × 103/156 × 102 = 12.4 N/mm2

x = ( )JATD −566.0

D=600 mm T=19.0mm A=156.0 cm2 J=165 cm4 x = 32.0

Max. moment, MEd = 252.6 kN Max. axial force, NEd = 117.2 kN

(a) Calculate buckling resistance to axial force

Firstly the slenderness should be calculated

59.09.93

15.39

22001

1z

cr =×==λ

λiL

LTλ = cr

yy

M

fW ⋅ EN1993-1-1; §6.3.2.2

Lm = 2

122

275235

360.32

1304.12

6.4638

+

× = 2160 mm

Thus, the length of 1550 mm from the plastic hinge position at side rail S6 to the column stay at side rail S5 is stable.


PAGE 72 OF 98

z2

t2

z

w2

z2

1rEIGIL

II

LEICMc

π

π+= SN003a-EN-EU

C1 comes from the table indicated for the correspondent value of ψ 1veConservati 1 =→ C Check the length between side rail S5 and the base (1)

In order to obtain Iw , the distance between shear centres of flanges must be calculated; hs = 400 −13.5 = 386.5 mm

There is no plastic hinge in the length between S5 and the base and a restraint to the compression flange has been provided at side rail S5 by means of a column stay.

( )( ) 1224

2zw

1049.025.386101318

2

×=××=

== shII

For external columns, the relevant check is for out-of-plane buckling only, because in-plane member stability is assured by the in-plane frame stability checks given in Section 7 of this example. It is therefore required that:

kNm11931032.12100001013.5807692200

1032.11049.0

22001032.12100001

72

52

7

12

2

72

cr

=×××

×××+

×

×××××=

ππM

b

LTLT

cy

c

MMm

PF

+ ≤ 1^

MLT = M@S5 = 621.2 kNm

LTλ = =×

××6

6

1013022351031.1

0.51 Fc = V′ = 192.9 kN mLT = 0.6 for β = 0.0 Pz = AgPy = 156 × 235= 3666 kN

(a) Calculate buckling resistance to bending

Nb, Rd = M1y γχ fA ⋅⋅ EN1993-1-1; §6.2.9.1(5)

22min

1

λφφχ

−+= EN1993-1-1; §6.3.1.2

φ = ( )

+−+

22.015.0 λλα

bh = 180400 = 2.22

tf = 13.5 mm Buckling between axis z-z; EN1993-1-1; Table 6.2

Use Curve b for hot rolled I sections α = 0.34 EN1993-1-1; Table 6.1

z

c

PF

= 3666

9.192 = 0.05

which is significantly less than the limit for Class 2 or compact sections (typically between 0.25 and 0.35).

For LEY = 4.08 m ry = 4.66 cm 6.876.464080 ==λ

∴ pc = 117 N/mm2

=×××== −32 1010156117gccy ApP 1825 kN

6.876.464080cm66.4m08.4EY ==== λyrL

wLT βλλ vu=


PAGE 73 OF 98

φ = ( )[ ]251.02.051.034.015.0 +−+

= 0.68

zχ = 22 51.068.068.0

1

−+

= 0.84


Nb, Rd = 1235845084.0 ⋅⋅

= 1767kN

(b) Calculate buckling resistance for bending

Mb, Rd = M1yypl,LT γχ fW ⋅⋅

LTχ = 2

LT2

LTLT

1

λβφφ ⋅−+

LTφ = ( )[ ]2LTLT,0LT15.0 λβλλα ⋅+−+

6.3119600 === T

Dx with 9.0=u

0.1=wβ for UB’s and UC’s

( )

92.0

6.316.8705.01

125.02

=

+

=v

730.16.8792.09.0LT =×××=λ

164b =p N/mm2 =xS 3512 cm3

=×××== −63xbb 10103512164SpM 576 kNm

∴ b

LTLT

cy

c

MMm

PF

+ = 576

2.6216.01825

9.192 ×+ = 0.08 + 0.65 = 0.73 < 1

∴ No further column restraints are required between side rail S5 and the base.


β = 0.75

bh = 2.22

Use Curve c for hot rolled I sections α = 0.49

φLT = ( )[ ] 62.051.075.04.051.049.015.0 2 =⋅+−+ EN1993-1-1; Table 6.5

LTχ = 95.051.075.062.062.0

122

=⋅−+

EN1993-1-1; Table 6.3

Mb, Rd = kNm 28912351031.195.0 6 =×××


PAGE 74 OF 98

(c) Calculate buckling resistance to combined axial and bending

1

M1

Rkz,

Edz,Edz,yz

M1

Rky,LT

Edy,Edy,yy

M1

Rky

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

EN1993-1-1; §6.3.3(4)

1

M1

Rkz,

Edz,Edz,zz

M1

Rky,LT

Edy,Edy,zy

M1

Rkz

Ed ≤+

++

+

γ

∆

γχ

∆

γχ M

MMk

MMM

kN

N

There are the following simplifications: 1) Edy,M∆ and Edz,M∆ = 0 for Class 1

2) No bending minor axis 3) Buckling about y-y axis is covered by Merchant-Rankine

Therefore,

1

M1

Rky,

Edy,zy

M1

Rkz

Ed ≤+

γγχ M

Mk

NN

0.1b.Rd.yy.Edzyb.Rd.zEd ≤⋅+ MMkNN

As zλ < 1

kzy = ( ) ( )

−−=

−−

zRd,b,

Ed

mLT

2z

M1RkzmLT

2z

25,01,01

25,01,01

NN

CNN

CEd λ

γχλ

ψ=1 Conservative 4,04,06,0 ymLT ≥+= ψC

114,06,0mLT =⋅+=C

kzy = ( )

×

×

−

×−

3

3

101668107.140

25,0157.01,01 = 0.996 EN1993-1-1; §6.3.2.2, BB3.1.1


PAGE 75 OF 98

0.194.02896.252996.016682.117 ≤=⋅+ ∴Rafter OK

The rafter is stable between intermediate restraints to compression flange.

6.6.4 Buckling between torsional restraints RAFTER STABILITY BELOW THE APEX

Where the bottom flange is in compression, the stability must be checked between torsional restraints (e.g. restraint to bottom and top flanges). For first trial, assume rotational restraints are positioned at approximately quarter span intervals(see diagram in section 1.1 of this worked example)

The plastic hinge (at purlin P9) will be restrained by the purlin P9 and a rafter stay to give torsional restraint.

The rafter near the apex is subject to a sagging moment in this load case. The compression flange is stabilised by the purlin.

6.6.4.1 Load combination no.1 Check the length between purlins P9 and P8

Worst case between RH haunch tip and quarter span rotational restraint between 2992 mm and 9217 mm from the intersection of the rafter and the column.

Taking hogging moment as positive for this calculation:: MEd.haunch = 252.6 kN MEd.2 = 99.5 kN MEd.3 = -31.7 kN MEd.4 = -139.0 kN MEd.restrarint = -222.5 kN

1P P2P3

P4P5

P6P7

P8P9

P10

110015083 183

=

6 @ 1630

=

Purlin spacing

Tors

iona

l res

train

t

Tors

iona

l res

train

t

Worst buckling from gravity loads (a) Calculate slenderness λ and λLT

Lm = 2

12

y2

c

y

27536130

38

+

pxf

r

where fc = F/A F@P8 = 115.9 kN

fc = 115.9 × 103/(84.5 × 102) = 13.2 N/mm2

( ) ( ) 3.515.845.13400566.0566.0 ×−×=−= J

ATDx


PAGE 76 OF 98

The purlins provide intermediate restraints x = 28.1

Assume side rail depth = 200 mm

==

200

300

=

400

=

Cross section through rafter

Lm = 2

122

275235

361.28

1302.13

5.3938

+

× = 2030 mm

The distance between purlins P9 and P8= 1630 mm. This is less than Lm, therefore no additional purlin is required between P9 and P8.

Distance between the centroid of the rafter and the centroid of the purlins, a

a = 400 / 2 + 200/ 2 = 300 mm is2 = iy2 + iz2 + a2

is2 = 1652 + 39.52 + 3002 = 118785 mm2

Distance between shear centres of flanges hs = 400 −13.5 = 386.5 mm

α =

+

2s

z

w2

iIIa

Using the simplification for doubly symmetric I sections: Iw = Iz ( hs / 2)2

α = ( )

+2

s

2s

2 2

iha


PAGE 77 OF 98

= ( )

+

118785

25.386300 22

= 1.072

(a) calculate slenderness for λ and LTλ

λ = ( )[ ] 5.0222 6.2 sztt

zt

iILI

iL

πα +

=( )[ ] 5.07225 1187851032.16.262251013.5072.1

5.396225

××××××+ π

= 126.1

λ = 1λλ

= 9.931.126

= 1.34

LTλ = λ5.0

2s

ypl,5.0

n

2 1

ia

AW

CC

EN1993-1-1; BB3.3.2

where

Cn = ( )[ ]Es RRRRRRR −+++++ 2343

12

54321

R1 to R5 are the values of R according to ypl,y

EdEdy,

Wf

NaMR

+= at ends, quarter

points and mid-length. Only positive values are included.

Conservatively, taking NED positive at all positions:

NED = 168.7 kN


PAGE 78 OF 98

3

36

ypl,y

EdEdy,1

101310 23530010117.2 108.222

××

××+×−=

+=

Wf

NaMR <0 so Omitted

3

36

ypl,y

EdEdy,2

101310 23530010117.2 100.139

××

××+×−=

+=

Wf

NaMR <0 so Omitted

02.0 101310235

30010117.2 107.31

3

36

ypl,y

EdEdy,3 =

××

××+×−=

+=

Wf

NaMR

44.0 101310235

30010117.2 105.99

3

36

ypl,y

EdEdy,4 =

××

××+×=

+=

Wf

NaMR

94.0 101310235

30010117.2 107.254

3

36

ypl,y

EdEdy,5 =

××

××+×=

+=

Wf

NaMR

5

51

studiedlength thein R of value max94.0,max

RRRRR

s

E

====

[ ]21.5

94.044.0302.0412

n =+×+×

=C

Conservatively C = 1

LTλ = 51.034.1118785

30028450

1031.1 121.51

5.065.0

=×

××

×××

(b) Calculate buckling resistance to axial load

Nb, Rd = M1y γχ fA ⋅⋅ EN1993-1-1; §6.3

22min

1

λφφχ

−+=


PAGE 79 OF 98

φ = ( )

+−+

22.015.0 λλα

bh = 180400 = 2.22 EN1993-1-1; Table 6.2

tf = 13.5 mm buckling about z-z axis → Curve b for hot rolled I sections → α = 0.34

φ = ( )[ ]234.12.034.134.015.0 +−+

= 1.59

zχ = 22 34.159.159.1

1

−+

= 0.41


Nb, Rd = 1235845041.0 ⋅⋅

= 812 kN


Mb, Rd = M1yypl,LT γχ fW ⋅⋅

LTχ = 2

LT2

LTLT

1

λβφφ ⋅−+

φLT = ( )[ ]2LTLT,0LT15.0 λβλλα ⋅+−+


β = 0.75

h/b = 2.5 EN1993-1-1; Table 6.5

For hot rolled sections use Curve c α = 0.49


PAGE 80 OF 98

φLT = ( )[ ] 59.046.075.04.046.049.015.0 2 =⋅+−+

LTχ = 975.046.075.059.059.0

122

=⋅−+

Mb, Rd = kNm 3.3001235101310975.0 3 =×××


1

M1

Rkz,

Edz,Edz,yz

M1

Rky,LT

Edy,Edy,yy

M1

Rky

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

EN1993-1-1; §6.3.3(4)

1

M1

Rkz,

Edz,Edz,zz

M1

Rky,LT

Edy,Edy,zy

M1

Rkz

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

There are the following simplifications: ∴ Edy,M∆ and Edz,M∆ =0 for Class 1

∴No bending minor axis

∴Buckling about y-y axis is covered by Merchant-Rankine

So,

1

M1

Rky,

Edy,zy

M1

Rkz

Ed ≤+

γγ

χ MM

kN

N

1b.Rd.yy.Edzyb.Rd.zEd ≤⋅+ MMkNN

As zλ ≥ 1 EN1993-1-1; Table B.2

kzy = ( ) ( )

−−=

−−

zRd,b,

Ed

mLT

2z

M1Rkz

Ed

mLT

2

25,01,01

25,0

1,01

NN

CNN

Cz λ

γχ

λ


PAGE 81 OF 98

Conservatively taking ψ=1 EN1993-1-1; Table B.3

4,04,06,0 ymLT ≥+= ψC

114,06,0mLT =⋅+=C

kzy = ( )

×

×

−−

3

3

10812102.117

25,011,01

= 0.981

0.197.03.3006.252981.08122.117 ≤=⋅+ ∴Rafter is acceptable.

6.7 HAUNCH DESIGN EAVES HAUNCH STABILITY

The bending moments and plastic modulus (and effective elastic modulus where appropriate) of the section are required at end and quarter points for the stability checks, so these points are also used to check the stresses along the haunch.

Determine section properties of the haunched beam-at four positions along haunch:

This section includes the length up to purlin P3. Four design options exist for the situation where the tension flange is restrained between restraints to the compression flange.

IPE 600

A

A

Haunch geometry

1 2 3 4 5Section A - A

5°

620

3720930930930930

From IPE 500

IPE 400

1. Ignore the tension flange restraints and design to Clause 4.8.3.3.1 providing restraints to the compression flange as necessary.

2. Limit the length between compression flange restraint to Lm given by Clause 5.3.3.

3. Limit the length between compression flange restraints to Ls as given by Clause 5.3.4.

4. Check the length according to Appendix G of BS 5950-1:2000. Method 2 will be conservative as it ignores the restraint to the tension flange between torsional restraints. Method 3 is relatively straightforward but using the limiting length Ls requires the distance between tension flange restraints to be adequate when checked to Clause 4.8.3.3 (or Clause I.1) Method 4 would not normally be carried out manually although it can be shown that rafter stays to the compression flange at purlins P3 and P5 would be adequate.


PAGE 82 OF 98

Position Distance from

column face (mm)

Areagross

(mm2) Areaeff (mm2)

Areaweb,eff(mm2)

Wpl,y (mm3)

Weff,ply β=

ypl,

eff.pl.y

W

W

1 0 11478 11142 6282 3176×103 2986×103 0.94

2 925 10656 10656 5796 2647×103 2647×103 1.0

3 1850 9831 9831 4971 2156×103 2156×103 1.0

4 2775 9010 9010 4150 1706×103 1706×103 1.0

Method 3 will be demonstrated here. Method 3, Clause 5.3.4 approach (Simple Method) Provided the geometrical limitations are complied with, the spacing Ly between restraints to the compression flange should not exceed the limiting spacing Ls. For S275 (assume acceptable for S235 steel grade)

Ls = 5.02

1

y

10072

620

−

xK

r

Moments acting at the four positions are as follows:

MEd1 = 577.8 kNm at 0.302 m from intersection of rafter and column





Assumedelastic neutralaxis

Elastic stress distribution in web

420

224,610,4

235

400

820

410

410

ry and x for the un-haunched section (i.e. rafter)

s

h

DD

≈ 1, ∴ K1 = 1.25

∴Ls = 5.02

1.281007225.1

5.39620

−

× mm

∴Ls = 2543 mm The length of the haunch (between the column face and purlin P3) is 3700 mm. This is greater than Ls, therefore an additional stay at purlin P2 would be required. Alternatively the length between the face of the column and purlin P3 could be checked according to Appendix G.2.2.

1P P2P3

P4P5

** 3.70 m

Haunch restraints


PAGE 83 OF 98

Therefore stresses available to resist bending:

2NMOyM mmN6.2244.101235 =−=−= σγσ f

(b) Classify web assuming σ M = 224.6 N/mm2

∴Coexistent stress at top of haunch would be

σ = σ M – σ N = 224.6 – 10.4 = 214.2

Total depth = rafter + web + bottom flange = 5.796

∴Depth from neutral axis to underside of middle flange

mm8.12

5.796400 =

−=

∴bending and axial stress at top of haunch cutting:

( ) 2mmN40.114.102.2148.16.224 =+×=

Distance from assumed elastic neutral axis to top of root radius on bottom flange of haunch cutting

8.3635.132125.796 =−−= mm

∴bending and axial stress at top of root radius on bottom flange of haunch cutting

0.216= N/mm2

For class 3 check, 053.0=ψ

Depth of web excluding roof radius = 404-21=383 mm

Class 3 limit for ψ >-1:

1.61053.033.067.0

0.14233.067.0

42w =×+

×=

+≤

ψεtc


PAGE 84 OF 98

382.10

383w ==tc So, web class 3

6.7.1 Cross-sectional resistance RAFTER STABILITY ABOVE THE HAUNCH (BETWEEN PURLINS P3 AND P5)

For the stability checks given in this document for tapered haunches to remain valid, the tapered haunch must not contain a plastic hinge.

The length between purlin P3 and P5 is mostly in the hogging region and does not contain a plastic hinge. Restraints (stays) are provided to the compression flange at purlins P3 and P5

6.7.1.1 Shear Using 4.8.3.3.2(a) out-of-plane buckling because the in-plane stability of the rafter is assured by the in-plane frame stability.

The depth of the web between flanges is not greater than in the rafter, so shear buckling is not a problem in the haunch.

The shear in the rafter has been checked in 5.2 above , showing VED ≯0.5Vpl.Ed

In the haunch, the shear area Av increases more than the applied shear VED, so the shear force has no effect, by inspection.

b

LTLT

cy

c

MMm

PF

+ ≤ 1

The tables provided below give the axial and moment resistance of the haunch section at various positions from the column face. A series of checks is carried out to determine whether the cross-sectional moment resistance Mc.Ed is reduced by coexistent axial force. Positions 1 to 5 are checked to find MED/Mc.Ed

191.5 kNm 60.9 kNm


PAGE 85 OF 98

6.7.1.2 Axial and bending

Position Distance (mm)

NEd

(kN) Aeff

(mm2) Npl.Rd

(kN) Aweb,eff

(mm2) Aweb,eff fy

1 0 119.5 11142 2618 6282 1476×103

2 925 118.9 10656 2504 5796 1362×103

3 1850 118.3 9831 2310 4971 1168×103

4 2775 117.7 9010 2117 4150 975×103

5 3700 117.2 8450 1986 3208 754×103

Is NEd> Position Distance (mm)

MEd

(kNm) 0.5Aweb.efffy 0.25Npl.Rd

Mc.Rd

(=Mpl.y.Rd)

Is MEd>Mc.Rd?

1 0 577.8 No No 702 No

2 925 488.3 No No 622 No

3 1850 403.5 No No 507 No

4 2775 323.1 No No 401 No

5 3700 247.8 No No 307 No The loading on the haunch is a combination of axial load, shear and bending. By inspection, the applied shear force is small relative to the shear capacity of the section and need not therefore be considered. Check positions 1 to 5 to find

c.RdEd MM where c.RdM is the cross-sectional moment of resistance. With low

coexistent axial force and shear force, pl.Rdc.Rd MM =

i. Position 1: 82.0702

8.577= < 1 ∴OK

ii. Position 2: 79.0622

3.488= < 1 ∴OK

β ≈ 5.1915.69

− = – 0.36

∴ mLT = 0.47 MLT = M@P3 = 191.5 kNm


PAGE 86 OF 98

iii. Position 3: 80.0507

5.403= < 1 ∴OK

iv. Position 4: 81.0401

1.323= < 1 ∴OK

v. Position 5: 81.0307

8.247= < 1 ∴OK

→ No plastic hinges in the haunch

b

LTLT

cy

c

MMm

PF

+ ≤ 1

5.2235.19147.0

10489.192 ×

+ = 0.18 + 0.40 = 0.58 ∴ OK

6.7.2 Buckling between intermediate restraints

Assuming a purlin is positioned at the mid-length of the haunch, intermediate buckling should be checked between the column and purlin, and between purlin and haunch tip. Overall buckling between checks should be carried out for the haunch as a whole.

The following effective section properties are required: 1. Effective area 2. Effective plastic section modulus

Table 6.2.2.1 gives effective section properties at the start and mid span of the haunch, calculated for the haunch neglecting the “middle” flange, but remembering its stabilising effect on the web.


PAGE 87 OF 98

6.7.2.1 Check between column flange and mid-haunch purlin

1850 mm

15432

Buckling on deep end of haunch

From 6.2 above, Position 1 is the most critical cross-section, having 82.0c.RdEd =MM . Therefore, the resistance is calculated using the area and modulus at this cross section, together with the axial force and bending moment at this cross section

Note that zχ and LTχ are calculated at the deepest end because this gives the most conservative results where the flanges are of constant section and the web is of constant thickness along the haunch.

(a) Calculate buckling resistance to axial force

M1yeffzb.Rd.z γχ fAN =

( )

( ) 574.091.93971.071.54

71.5499.33185081.331147810312.1

flange middle the neglecting 10312.1

971.01147811142

mm1850

5.01

5.0A

5.07z

7z

ffA

=×==

===×=

×=

===

=

λβλλ

λ

β

i

I

AA

l

e

To obtain zχ as above the following steps should be done.


PAGE 88 OF 98

minχ = 22

1

λφφ −+

φ = ( )

+−+

22.015.0 λλα

bh = 1805.796 = 4.43

tf = 13.5 mm Buckling between axis z-z use Curve b for hot rolled I sections α = 0.34

φ = ( )[ ]2574.02.0574.034.015.0 +−+

= 0.73

zχ = 22 574.073.073.0

1

−+

= 0.85

Nb, Rd = M1yffz γχ fAe

Nb, Rd = 12351114285.0 ××

= 2225 kN

(b) Calculate buckling resistance to bending moment

M1yeff.pl.yLTb.Rd.z γχ fWM =

485.0574.09.09.0

94.031762986

zLT

pl.yeff.pl.yw

=×==

===

λλ

β WW

LTχ = 2

LT2

LTLT

1

λβφφ ⋅−+

LTφ = ( )[ ]2LTLT,0LT15.0 λβλλα ⋅+−+



PAGE 89 OF 98

β = 0.75

bh = 53.36

For hot rolled sections use Curve c α = 0.49

LTφ = 0.60

LTχ = 0.97

Mb, Rd = kNm 297123510986.297.0 6 =×××


1

M1

Rkz,

Edz,Edz,yz

M1

Rky,LT

Edy,Edy,yy

M1

Rky

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

1

M1

Rkz,

Edz,Edz,zz

M1

Rky,LT

Edy,Edy,zy

M1

Rkz

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

Which reduces to -

1b.Rd.yy.Edzyb.Rd.zEd ≤⋅+ MMkNN

As zλ ≥ 1

kzy = ( ) ( )

−−=

−−

zRdb

EdEd

NN

CNN

C ,,mLTM1RkzmLT 25,01,01

25,01,01

γχ

kzy = Conservatively

0.189.02978.247

21382.117 ≤=+ OK


PAGE 90 OF 98

6.7.3 Buckling between torsional restraints

3720 mm

1 2543

Haunch buckling length

Section 6.2 of this example shows that the critical section in the haunch is Position 1 because MEd/MRd is maximum at that point. Therefore, check the resistance based on forces, moments and resistances at that point.

(a) Calculate slenderness λ and λLT

zχ and LTχ are calculated the constants accounting for the effect of the taper are based on the shallow end of the haunch

l = 3720 mm Iz = 1.312 × 107 mm4 (ignoring the middle flange) iz = (1.312 × 10y/8450)0.5 = 39.4 α = 1.072

The axial slenderness (for restrained tension flange) is given by:

λ = ( )[ ] 5.02

sz22

tt

zt

6.2 iILI

iL

πα +

= ( )[ ] 5.07225 11878510312.16.237201013.5072.1

4.393720

××××××+ π

= 63.7


PAGE 91 OF 98

λ = 1λλ

= 9.937.63

= 0.678

LTλ = λ5.0

2ypl,

5.0

n

2 1

sia

AW

cC

EN1993-1-1; BB3.3.2

where nC = 1.2

and since

h/tf = 6.295.13400 =

h/b = 22.2180400 =

c = 32

min

max

f

1

9

31

−

−

+hh

th

EN1993-1-1; BB 3.3.3

hmax = 654 hmin = 400

c = 1.11400654

95.13

40031

32

=

−

−

+

LTλ = 654.0678.0118785

30028450

103.1 1.12.1

15.065.0

=×

××

×××

(b) Calculate buckling resistance to axial load

Nb, Rd = M1y γχ fA EN1993-1-1; §6.3


PAGE 92 OF 98

minχ = 22

1

λφφ −+

φ = ( )

+−+

22.015.0 λλα

bh = 1805.796 = 4.43 EN1993-1-1; Table 6.2

tf = 14.6 mm Buckling between axis z-z Curve b for hot rolled I sections α = 0.34

φ = ( )[ ]2678.02.0678.034.015.0 +−+

= 0.811

zχ =22 678.0811.0811.0

1

−+

= 0.796

A = 11142 mm2 (neglecting the middle flange)

Nb, Rd = M1y γχ fA

Nb, Rd = 123511142796.0 ××

= 2084 kN


Mb, Rd = M1yypl,LT γχ fW

LTχ = 2

LT2

LTLT

1

λβφφ −+

LTφ = ( )[ ]2LTLT,0LT15.0 λβλλα +−+


β = 0.75


PAGE 93 OF 98

h/b = 2.22 EN1993-1-1; Table 6.5

Use Curve c for hot rolled I sections α =0.49

LTφ = 0.723

LTχ = 0.854

Mb, Rd = kNm 5991235102986854.0 3 =×××


1

M1

Rkz,

Edz,Edz,yz

M1

Rky,LT

Edy,Edy,yy

M1

Rky

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

EN1993-1-1; §6.3.3(4)

1

M1

Rkz,

Edz,Edz,zz

M1

Rky,LT

Edy,Edy,zy

M1

Rkz

Ed ≤+

++

+

γ

∆

γχ

∆

γ

χ MMM

kM

MMk

NN

Which reduces to – 1b.Rd.yy.Edzyb.Rd.zEd ≤⋅+ MMkNN

As zλ ≤ 1 EN1993-1-1; Table B.2

kzy = ( ) ( )

−−=

−−

zRd,b,

Ed

mLT

2

M1Rkz

Ed

mLT

2

25,0

1,01

25,0

1,01

NN

CNN

Czz λ

γχ

λ

ψ = 1 C mLT = 0,6 + 0,4 11 =⋅ EN1993-1-1; Table B.3

kzy = 0.98

0.199.05998.57798.020845.119 ≤=×+ ∴ Acceptable.


PAGE 94 OF 98

General design procedure according o BS 5950-1: 2000: BS 5950-1: 2000

1. Select steel grade and trial sections

2. Check in-plane stability of frame (λp ≥ λr) using: • Sway check method, or • Amplified moments method, or • Second-order analysis

Cl. 5.5.3 Cl. 5.5.4.1 Cl. 5.5.4.4 Cl. 5.5.4.5

3. Check out-of-plane stability of frame Cl. 5.5.1

4. Check in-plane stability of members Cl. 5.2.3.1

5. Check out-of-plane stability of members Determine limiting segment length for: (a) Segment adjacent to plastic hinge (Lm) (b) Member or segment with one flange restrained (Ls) using:

– Simple method, or – Annex G approach

Cl. 5.3.1

Cl. 5.3.3

Cl. 5.3.4 Annex G.3

6. Check deflections Table 8

7. Design connections and bases to transmit forces and moments.


PAGE 95 OF 98

ANNEX A: Characteristic values of internal forces

Characteristic values of internal forces due to the particular load cases determined by first order analysis, see load scheme in Figure 2.

Figure 3: Characteristic internal forces due to dead load incl. profiles

Figure 4: Characteristic internal forces due to snow load

Figure 5: Characteristic internal forces due to wind load

Figure 6: Characteristic internal forces due to sway imperfection


PAGE 96 OF 98

ANNEX B: Plastic failure of portal frame

Plastic failure of a portal frame subject to uniformly distributed loading may be analysed by considering the development of either: pairs of plastic hinges in the rafters, or a plastic hinge in the rafter and also at the top of the column. For the first case, the positions of the plastic hinges occur at the tip of the haunch, at distance a into the span, and at a variable position x from the apex (see Figure 3(a)). For the second case, the plastic hinge occurs in the column below the haunch and at position x from the apex. Equilibrium can be established for each plastic hinge position in terms of the horizontal reaction R at the base of the column, the applied loading and the plastic resistance of the rafter (beam), as illustrated in Figure 3(b). These equilibrium equations can be solved to establish the minimum value of the load w, at failure. A plastic hinge will form in the column for low-rise long span portal frames.

Figure 3: Plastic hinge mechanisms in portal frame

Case 1: Plastic hinges in rafters

Equilibrium is established at the plastic hinge positions, according to: At point A:

R(H + a tan θ) = pl

2

22MwawLa

+−

At point B:


PAGE 97 OF 98

R(H + (0.5L – x) tan θ) = pl

2

2222MxLwxLwL

−

−−

−

Solving these two equations by eliminating the reaction R leads to the following relationship between the applied moment and bending resistance of the rafter.

( )[ ]θ

θ

tan5.02

tan2

4441

8

2

2

2

2

2

2

2

xaLH

La

LaxLH

Lx

La

La

wLM pl−++

−

−+

−+−

=

where w is the load per unit length applied to the rafter and Mpl is the plastic bending resistance of the rafter L is the span of the portal frame H is the column length a is the length of the haunch θ is the slope of the rafter This unique equation is a function of x, which can be solved by selecting different values of x and establishing the minimum value of w (or maximum value of Mpl). The term in brackets represents the deviation from the free bending moment, wL2/8. Example: L = 25 m θ = 15° a = 2 m H = 7.5 m Try x = 3 m and x/L = 0.12:

( ) ( )( )[ ]( )[ ]27.0325.125.72

27.008.008.035.125.712.0408.0408.0418

2222

pl×−++×

×−−+××−×+×−=

wLM

= 0.28 (wL2/8) Try x = 1 m and x/L = 0.04: Mpl = 0.27(wL2/8) This shows that there is a relatively small variation in Mpl for different values of x, and it may be found that x ≈ 2 m. The horizontal reaction R is established from the previous equations, and for this value of x is given by: R = 0.45 (wL/2)


PAGE 98 OF 98

Case 2: Plastic hinges at top of column and the rafter

A plastic hinge may occur at the top of the column below the haunch in which case, equilibrium is established from: R(H –b) = Mpl,col where b is the depth of the haunch Mpl,col is the plastic bending resistance of the column The equation for equilibrium at point B in the rafter is the same as the previous case. Solving the two equations leads to the following relationship between the applied moment and the bending resistances of the rafter and column:

( )

−=

−

−+

+ 2

22

,41

8

tan2

LxwL

bH

xLHMM colplpl

θ

For x/L ≈ 0.08 and b/H = 0.1 and Mpl = 0.28 ( )8/2wL , it follows that Mpl,col = 0.42 ( )8/2wL . Therefore, the bending resistance of the column should be 51% higher than that of the rafter in order for Case 1 to control. If Mpl,col < 1.51 Mpl, then Case 2 will control. Note: The analysis provided above is for the vertical loading case only. It does not consider horizontal loads or wind loading which can cause non-symmetrical loading. This approach is satisfactory for BS 5950 however care should be taken when designing using Eurocode 3-1-1 as all load combinations must be taken into account.

Comparison of the Design of a Portal Frame

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