Steel Design

Worked exampleI 61 Job No. The S t e ~ l Construction I Sheet 1 of 4 I Rev Steel Designers' Manual Wind actions to BS E N 1991-1-4 Calculation of peak velocity pressure Made by Institute Job Title Subject Silwood Park, Ascot, Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET Client DGB I Date I Dec2009 I I I Date I Calculation o f peak velocity pressure This example demonstrates the calculation o,f the peak velocity pressure for the hypothetical site shown in Figure 1, on the eastern edge o,f Norwich. This example demonstrates Approaches I , 2 and 3, none of which require knowledge of the building orientation. References to expressions, tables and figures from the U K National Annex are preceded by 'NA' - all other references are to BS EN 1991-1-4 Town Terrain Average height 8 m 24 krn closest distance to sea N Country Terrain Site altitude 45rn The site 3gure 1 Details of the site Details o,f the site are: 4 5 m above sea level Site altitude 27m Building height Terrain category: From 0" to 180" is country terrain From 180" to 360" is town terrain the average height o,f the upwind obstruction (h,,,J is 8 m and the spacing to the upwind Obstructions (x) is 30m. Within this sector, the site is at least IOkm inside the town The closest distance from the sea is 24km, at a bearing of 30" from the site. ~ 62 Worked example Sheet 2 o f 4 Example Concept Design I Rev A.p.proach I This approach is the simplest, but the most conservative, v , , , , , , ~= ~ 22.5m/s c,lt c,lt Vh,O Figure NA.1 = I +0.001A (lO/z)"z = I +0.001 ~ 4 ~5 ( 1 0 / 2 7 ) =1.037 "~ = Vh,.lap x = 1.0 C,lt NA.2b NA.1 Table NA.1 Table NA.2 4. I v ~ , , ~= 22.5 x 1.0.?7 = 2.?..?m/s cd, Vh (the maximum from any direction) L'h,O = cscasm~c,br v~, = 1.0 x 1.0 x2.7.3 =2.?..?m/s = 0.61.7~~~ qh 4.10 and NA.2.18 NA.2. I I x 10-j = 0..7.3kN/m2 qk, = 0.613 ~2.3.3~ Terrain: country, when wind blowing ,from the east Distance from the sea: 24 k m minimum height z = 27m c,(z) =3.1 qp(z) = c d z ) qh qp(z)=.?.I xO..?.~ =I.02kN/mZ This peak velocity pressure may he used to determine ,forces on the rtructure in each orthogonal direction. Figure NA.7 A.p.proach 2 This approach demands knowledge of the upwind terrain all around the site. Examples of two directions, 330" and 60" are shown below, and then the full details of each direction in tabular format. 3.70" 60" v[,,o ~d,, = 23.3m/s (as above) v h , ~ =23.3m/s (as above) cdlr = 0.82 = 0. 7.7 Table NA.1 Table NA.2 ~scaro,, = 1.0 Vh vL, = cscasm~c,br = 19.1 m/s L'h,O Vh vL, = cs,,,,,, c,br = 17.0m/s Vh,O 4.1 = 0.82 x 1.0 x 23.3 = 0. 7.7 x 1.0 ~ 2 3 . 3 T,, ?k, = 0.613~~: = 0.613 x 19.12 x lo-.' = 0.22 kN/mZ q,, q,, = O.613vk,' 4.10 and NA.2.18 = 0.613 x 17.p x 10" =0.18kN/m2 Terrain: country Annex AS Terrain: town b , =8m x = 30m (the spacing to the ~ i wind p obstructions) 2 x h,,,, < x < 6 x ha,, ? x 8 c30 <6 x 8 16 <30 <48 Worked example Example Concept Design Sheet 3 of 4 Rev 63 Therefore, hd,,=min(l.2h,,,, - 0 . 2 ~ ;0.6h) hdL, =min(l.2 x 8 - 0.2 x30; 0.6 x 27) h,,,, = min(3.6; 16.2) =3.6m Distance from the sea = 42 km c,(z) = 2.94 (at z - h,,,, =23.4m) Distance inside town = IOkm = 0.88 (at z - hdri =23.4m) Distance from the sea = 25 km c,(z) = 3.10 (at z = 2 7 m ) Interpolated ,from Figure NA.7 Figure NA.8 NA.2.17 csl = 0.57kN/m2 Bearing Vh.() Cdlr (m/s) C,'.,,,, vh (m/s) q,, (kN/m') 0 23.33 0.78 1.0 18.20 0.20 32 30 23.33 0.73 1.0 17.03 0.18 24 60 23.33 0.73 1.0 17.03 0.18 25 3.1 90 150 180 120 2.?.3.? 23.33 2.?.3.? 23.33 0.74 0.73 0.80 0.85 1.0 1.0 1.0 1.0 17.26 I 7.03 18.66 19.83 0.18 0.18 0.21 0.24 27 3.08 34 3.06 0.55 47 3.04 0.64 7.? 3.00 0.72 210 23.33 0.93 1.0 21.70 0.29 >I00 2.88 0.88 0.73 240 23.33 1.00 1.0 23.33 0.33 >I00 2.88 0.88 0.84 270 23.33 0.99 1.0 23.10 0.33 >I00 2.88 0.88 0.84 300 2.?.3.? 0.91 1.0 21.23 0.28 66 2.91 0.88 0.72 330 2.?.3.? 0.82 1.0 19.13 0.22 42 2.94 0.88 0.57 Distance from seu (km) CdZ) 3.07 3.1 Ce.r q,>(z) (kN/m2) 0.61 0.56 0.56 0.57 Following Approach 2 results in a maximum peak velocity pressure of 0.84kN/mZ,compared to 1.02 k N / m 2from Approach 1. The peak velocity pressure o,f 0.84kN/m2 may he used to determine forces on the structure in each orthogonal direction. A.p.proach 3 In Approach 3, the most onerous values of any factor are taken from any direction within the chosen quadrant. Quadrants may be chosen judiciously to produce the lowest peak velocity pressure. The lowest peak velocity pressure ,found by Approach 3 will never he smaller than that from Approach 2, but is generally less than the peak velocity pressure from Approach I . Two quadrants are demonstrated in detail, and then the results presented in summary form. Assume that the quadrants are 0" to 90" inclusive, 90" to 180" inclusive, etc, and taking the quadrant from 90" to 180" as an example. With reference to the Table in Approach 2: vh.0 =23.3m/s maximum cdlr from within the quadrant 90" to 180" = 0.85 (at 180") CS,,,,,, vh = 1.0 =c~eawn cdir vh,O Table NA.1 Table NA.2 4.1 4.10 and NA.2.18 NA.2.11 vh ~0.85 x1.0 x23.3 =19.8m/s 8l~ 0 - j =0.24kN/mZ qb =O.6I.Zvk,' qh =0.613 ~ 1 9 . x Terrain: country Closest distance from the sea within the sector 90" to 180" = 2 7 k m (at 90") height z = 2 7 m 64 Worked example Sheet 4 of 4 Rev Example Concept Design c,(z) =.?.08 qp(z) = c d z ) qb q p ( z )= 3.08 x 0.24 = 0.74kN/mZ Taking the quadrant from 21O"to 300"as a second example v~,,~ = 2.7.3 m / s maximum cd,,from within the quadrant 210"to 300" = 1.0 (at 240") c,,:,,,,, vh Interpolated f r o m Figure N A . 7 Na.2.17 = 1.0 = c,,,,,,,, C,br h,,o Table NA.1 Table NA.2 4. I 4.10 and NA.2.18 NA.2.11 Annex A S v,, =1.0 x1.0 x23.3 =23.3m/s qb =O.6I.7vk,' qb =0.613 x2.7.3' xIO:' =0.33kN/mZ Terrain: Town h,,,, = 3.6m Closest distance ,from the sea within the sector 210" to 300" = 66km (at 3000) c,(z) =2.91 (at z - h,,,, =23.4m) cc,T = 0.88 (at z - h,,,, = 23.4m) qp(z) = c e ( z ) cc,l q h qp(z) = 2.91 x 0.88 x 0.33 = 0.85 k N / m z Interpolated from Figure NA. 7 Interpolated from Figure NA.8 NA.2.17 The summary results are shown below: Sector peak velocity pressure (kN/m') 0.57 0.88 0.85 0.86 30 to 120 inclusive 120" to 210" inclusive 210" to 300" inclusive 300" to 30" inclusive When the quadrants are chosen as above, the maximum peak velocity pressure o,f I.OOkN/m' may be used to determine forces on the structure in each orthogonal direction. Sector peak velocity pressure ( k N / m 2 ) 0.63 1.00 0" to 90" inclusive 90" to 180" inclusive 180" to 270" inclusive 270 to 0 inclusive When the quadrants are chosen as above, the maximum peak velocity pressure of 0.88kN/m2 may be used to determine forces on the structure in each orthogonal direction. In both these examples of the application of Approach 3, the resulting maximum peak velocity pressure is less than that from Approach 1, but more than that from Approach 2. The example demonstrates the beneficial effects of judicious choice of quadrants. 128 Worked example I The SteEl Conmruction Job No. Job Title Subject Client Sheet 1 of 6 Worked Example Portal Frame Design Chapter 4 Made by Checked by GWO Date Rev Institute Silwood Park, Ascot, Berks SLS 7QN Telephone: (01344) 623345 Fax: (0 1344) 622944 CALCULATION SHEET Portal Frame Design Example: Design of portal frame using plastic analysis This example introduces the design o,f a portal ,frame f o r a s i n g l e - s t o p building, using a plastic method of global analysis. The frame uses hot-rolled I-sections f o r rafters and columns. This example presents the overall ,frame geometry (including restraint position), definition of loads and selection of load combination. 12567b.wmf Frame Geometry G 30000 4 Spacing ofportal frames = 7.2 m * ~ torsional restrainl Portal Frame Geometry The cladding to the roof and walls is supported by purlins and side-rails as indicated. The positions and spacing of the purlins and side rails were chosen as follows: Torsional restraints are to he provided at both ends of the haunch (i.e. one at the column end and one at the rafter end). A torsional restraint is to be provided at a n intermediate position along the rafter. The spacing of the purlins and side-rails should generally he ahout 1 8 0 0 m m for lateral stability of the rafters and columns. The spacing will he less in regions where the frame members are close to their plastic moment of resistance. Uniform spacings are adopted within the above constraints. - torsional restraint .2 nz.franzes = * 7.Worked example Portal Frame Design Sheet 2 of 6 Rev 129 p 1 5725 Spacing ofportal. (ii) Upward wind pressure with minimum gravity loads. so both must be checked.7 generally. causing sagging moments at midspan of the rafter and hogging moment in the haunches.for some geometries o.for single-storey portals.f axial force on the plastic moment of resistance.f building.000 12.1 is 0.0 and a. Note that in portal frames with small roof slopes. Column plastic moment: I P E 500 has tt < 4 0 m m . The value to be used must he .355. = 355 N/mmz Load combination No.for simplicity. These assumptions are checked a.a. The value in BS EN 1990:2002 Table Al.fter the analysis has been completed. The following assumptions are made: (i) The sections are assumed to he Class I for the global analysis.3-1-1 .1: dead +snow The second-order design bending moment diagram at Ultimate Limit State is shown below. ' a. BS EN19931-1 c 1 Frame imperfections = 1/200 Taking a. Therefore the critical design combinations are usually: (i) Gravity loads without wind.. = 1. (iii) The design must also be checked for gravity plus wind as this may be the critical case .898 1.= 1. = 1/2000 x 1.. but 1. The worst wind case might be f r o m either transverse wind or longitudinal wind.2 (3) In this example the bases have been assumed to be truly pinned for simplicity. .0 . The load factors at the formation of each hinge are as follows: Load factor Fraction of ULS 0.for ignoring the effect o. Steel grade is S.0 = 1/200 It is simplest to consider the frame imperfections as equivalent horizontal forces.0 for structures supporting storage loads.070 Hinge number I 2 Member Right hand column Lefi hand rafier Position (m) 5.07. but a simple calculation based on plan areas is suitable .0 x 1. Ultimate Limit State Analvsis 5.130 Worked example Portal Frame Design I Sheet 3 of 6 Rev Loads Load combinations Values for the combination are given in BS EN 1990.3. causing maximum reversal of moment compared with case (i). J. The column loads could be calculated by a frame analysis. Therefore the section sizes are suitable for this load combination. (I9 = (b(.found f r o m the National Annex to BS E N 1990. . (ii) The axial compression is assumed to be within the limit in BS EN 199. the wind load may reduce the effects of roof load..041 A mechanism is not formed until the second hinge has formed at a load factor of 1. Portal Frame Design $ Portal I I I M = 366.8 kN V = 154.l kN / h ! V = 155.6 kN M = 494.7 kN i l J k M = O k N m g 30 m k Sheet 4 of 6 F Worked example & Rev 131 .lkN V = 117.3 kN V= OkN M = 487 2 kNm N = 154.1 kNm N = 155.'3 kN V = 13.7 kNm N = 168. ] Source: 12495.132 Worked example Portal Frame Design Load combination N0.1. hut member stability must be checked because the moments are in the opposite sense to load case no. which is greater than in load case no. Therefore this load case is not critical for cross-sectional resistance. Therefore the frame imperfection .3.50 The collapse load factor = 6.00 and yo = 1.2: dead I Sheet 5 of 6 + transverse wind Rev The load combination results in an uplift load case causing tension in the members which does not destabilise the structure.22. shear and axial load .factor EC 3 C15.factors for loads are 7% = 1. wmf Load combination 2: Bending moment.2 may be omitted from this load combination. The partial safety . shear and axial load . Source: 12496. 2.3: dead 133 I Sheet 6 of 6 + longitudinal wind Rev In this case the wind loads applied to the structure result in a net upward force (except L H Column) on the roof as in load case no. which is greater than in load case no. but member stability f o r this case must be checked because the moments are in the opposite sense t o load case no.Worked example Portal Frame Design Load combination No.].1.factor =2.69. wmf E Load combination 3: Bending moment. The collapse load . Therefore this load case is not critical f o r cross-sectional resistance. within limit Actual c.+ = 20.f a 20.4 iteel Suilding Vesign: %sign Data SS E N '0025-2 Material properties: Yield strength f .& I42 = 0.4/0.2./t. within limit : .92 = 7.Z0mm2. ti = 12. Section is not Class 4 . The problem is as shown in the sketch below: Partial factors: 7/uo = 1.Jt. Subject Sheet 1 of 2 Rev Institute Silwood Park.3 x 5 2 UC in grade S275 steel to withstand a design axial compressive load of 1150kN over an unsupported height of 3.18 c m = 51.3. 7/ui = 1.0.3-1-1.2 275 Actual ct/tF = 7.92 SS E N 1 99.04.7 x20.04/0.I . ct/tj = 7.I ruble 5. Berks SL5 7QN Telephone: (01344) 623345 Fax: (01344) 622944 Compression Members Client I Made bv CALCULATION SHEET I LG DAN I Date Checked by I 2010 Date 2010 Compression members Rolled Universal Column design Problem Check the ability o.0 LIK N A to SS E N 19931-1 Geometric properties: A = 66.8mm.498 Worked example The S t e ~ l Construction Job No.92 = 20./t.6m assuming that both ends of the member are pinned. c.3 cm2 = 66. i2 = 5. Design to BS E N 199.62.E = 20.5mm. Ascot. =275N/mm2 since ti I 1 6 m m Check cross-section class@cation under pure compression: Need only check that section is not Class 4 (slender) For outstand j a n g e cr/tie I 1 4 For web c. 0.80' 9 s EN '993-1-1 71 6. Use 203 x 203 x 52 UC in grade S275 steel It should be noted that the same answer could have been obtained directly by the use of Reference I .97' .0 L = 1.0 9 s EN 1993-1-1 Z1 6.2.97 I I 0.0.3.0.80'1 = 0.3. PS EN '99.2) + 0. .0 x3600 = 3600mm O n the assumption that minor axis jexural buckling will govern.97 + 40.2) + I f ] = 0.1.2 x i = QZ + .49(0.I . use buckling curve 'c'./-= = 0.4 Member buckling resistance: Take effective length L.80 .5[ I +a(& .66 < 1..3-1-1 Table L 2 Oz = 0. = 1.Worked example Rolled IJniversal Column Design Sheet 2 of 499 2 3ev 3s EN 199.5[ I + 0. I : .3-1 C1 6. 0. Subject PUB 809 Sheet 1 of 3 Rev Institute Silwood Park.7-1-1.0m column with p i n ends and intermediate lateral braces provided restraint against minor axis buckling at third points along the column length.3-1-1 .500 Worked example The S t e ~ l Construction Job No.0 UK NA to BS EN 199. 7/ui = 1. Check the adequacy of the column. Ascot.f 1250 k N The problem is as shown in the sketch below: NEd NEd I 31 \ \ 1 I E 9 I I 2 I I I I I I I I Partial factors: 7/uo = 1. to carry a design axial compressive load o. according to BS EN 199. Berks SL5 7QN Telephone: (01344) 623345 Fax: (01344) 622944 Compression Members Client I Made bv CALCULATION SHEET I LG I Date Checked by DAN I 2010 Date 2010 Compression members Pinned column with intermediate lateral restraints Problem A 254 x254 x 89 UC in grade S275 steel is t o be used as a 12. Jt.38.Worked example Pinned Column with Intermediate Lateral Restraints SO1 I Sheet 2 of 3 3ev iteel Building yesign: yesign Data Geometric properties: A = 113 cmz = 11300mmz.i.55 cm = 65.z= 1.4 Material properties: Yield strength J. Non-dimensional slendernesses: Buckling curves.+= 19. = 6.0 L = 1.5mm. c. Yuo 1.77./t.94 Actual cr/tie = 7. use buckling curve '6' For minor axis buckling.94 = 20.e = 20.0 OK 9 s EN '993-1-1 71 6.94 = 6. ct/t+=6.Section is not Class 4 Cross-section compression resistance: N .2 cm = 112mm.2 x: . For major axis buckling.. Rii = -= 11300x265x10-~7 = 2 9 9 5 k N > 1 2 5 0 k N = N . use buckling curve 'c'.0 x 12000 = 12000mm for buckling about the y-y axis L.! = 1.2. within limit Actual c. = 11./t. tt = 17.0 L = 1. = 265 N / m m z since 16 > tt 2 4 0 m m Check cross-section classification under pure compression: Need only check that section is not Class 4 (slender) For outstandjange cr/tiE 5 14 9 s EN '0025-2 9 s EN '993-1-1 rable 5.4 Effective lengths: L.7. within limit :. i.4/0.r. Buckling reduction .3mm.2 For web c.04/0.0 x 4000 = 4000mm for buckling about the z-z axis.factors 3s EN '993-1-1 rable 6.& 5 42 E=& 235 - = 0. 692 X? = O? +.34(1.0.40 = 0.1.2) +I : ] = 0.2)+ Rev BS EN 199..3.I C16. .I .69.O BS E N 199.0.SO2 Worked example Sheet 3 of 3 Pinned Column with Intermediate Lateral Restraints @> = 0.49(0.0.5[1+ .2 I:] = 0.0.21'1 = 1.2 = 0. . .402 -1.3.69'1 = 0.3. Use 254 x 254 x 89 UC in grade S275 steel.5[1+0.86 + 40.0 BS E N 199.1 : . / = 0. +.47 _C XY @? 1 = @.86' = 0.2)+ 0.31-1 CI 6.212 1.2)+ 1.5[1+0.31-1 C1 6. .1.5[1+ .(I.21./=1 1 1.0.86 1 .3.40+41.73 _C 1.1. 92 From Table 5. page C-66. Cross-section is Class I BS EN 199.58/0. Ascot. from Reference 2...73 x 210 x 82 U K B when used as a beam in ( I ) S275 steel and (2) S355 steel.1 of BS EN 1993-1-1: For a Class I outstand flange in compression: ct/t+& I9 For a Class 1 web in bending: Actual cI/tl" = 6.1 c .12. BS EN 1993-1-1 Table 5. / t .522 Worked example The S t e ~ l Job No.58.2mm 1 1 6 m m E= BS EN 10025-2 6 235 - = 0.92 = 7..RII = 566 k N m Alternatively.. within = 53. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET I Beam example 1 Made by Beam example 1 Rolled Universal Beam using Design Tables Problem Determine the cross-section classification and moment resistance for a 5. c J t . cross-section is Class I and Mc.2.Kd = 731 k N m .5(2) (2) Using S3SS steel From Reference 2. page B-4: cl/tr = 6.6/0. (1) Using S27S steel From Reference 2. = 49. Job Title Subject Client Sheet 1 of 2 Rev Construction Institute I Silwood Park..7-1-1 C16. W.92 within limit : .. = 2060 cm' Yield strength f ..6. = 275N/mmz since ti = 13. page 0-66 cross-section is Class I and Mc.. = ~ 49. ~I72 limit Actual c . / t .7. within limit Actual c . Cross-section is Class I BS EN 1993-1-1 C16. within limit BS EN 1993-1-1 Table 5.6.0.1 : .09. from Reference 2.5(2) .e 5 9 For a Class 1 web in bending: c./t. & = 49.58/0.+ / t .Worked example Beam example 1 Alternatively.2mm I 16mm Sheet 2 of 2 523 Rev BS EN 10025-2 From Table 5.2. /t.& 5 72 Actual ct/tt&= 6.58.6/0. c. = 49. page B-4: ct/tt= 6.81 = 8.1 of BS EN 1993-1-1: For a Class 1 outstandjange in compression: c. W pl.81 = 61.= 2060 cm' Yield strength J = 355 N / m m 2 since ti = 13. / t . 22/8 kNm Design ultimate shear force Vbd= 55.flange. there is no possibility of lateral torsional buckling.5 ~ 2 5 .2/2 = 198.2 x 7.2m Due to the lateral restraint from the concrete slab.function as a simply supported beam carrying a 140mm thick solid concrete slab together with an imposed load of 7. 0 +11. Assume a concrete density o. 2 =55.0 k N / m Z . Job Title Sheet 1 of 3 Rev Institute Silwood Park.6 kN Initial sizing Adopting S275 steel and assuming no material is greater than 16mm thick (to be confirmed later).0 x 3 6 = (1. so design the beam for: i) ii) Cross-section bending resistance Shear resistance iii) Dejections Loading Assume self-weight o.2 ) kN/m = 357.5 x 7.f 2400 kg/m-' and a deflection limit of span/360.524 Worked example I The S t e ~ l Construction Job No.81 x 0. The slab may be assumed capable of providing continuous lateral restraint to the beam's top .2 k N / m + (1. the nominal yield strength f.6 m intervals. 6 = 11.2m and beams are spaced at 3. Ascot. BS EN 10025-2 .9 k N / m Variable imposed loading Design combination at U L S Design ultimate moment Mbd = 7. I ) 7.14 x 10-3 ~ 3 .2 = 25.f beam = 1.0 k N / m Permanent load (concrete slab) =2400 x9.9]) = 55.is 275 N/mmz. Berks SLS 7QN Telephone: (01344) 623345 Fax: (0 1344) 622944 CALCULATION SHEET Client Beam example 2 Laterally restrained Universal Beam Problem Select a suitable U K B in S275 steel to .35 x l l .The beam span is 7. .8kNm>..for class I or 2 cross-sections BS EN 1993-1-1 C16. With 11 = 1. BS EN 199. a 457 x 1 5 2 x 6 7 U K B has a value of W.& 5 72 Actual c. 0 m m : .30 x l b m m ' =1300 cm' From section tables. within : .. =275 N / m m 2 is OK. = A -2btl + (t..0mm I 1 6 .H. for all U K B and UKC.49..5 k N m 1.7-1-5.t.0. f.= wpiyfy .5 YWI M. .7-1-1: For a Class I outstand flange in compression: ct /ti&I9 For a Class 1 web in bending: c. loaded parallel to the web.& = 4. and a self-weight less than that assumed.6 Shear resistance For a rolled 1-section./t. > q h ..I ./t.3/0.3.3-1-5 . is given by: A..Worked example Beam example 2 525 I Sheet2of3 I I Rev Assuming that the cross-section is Class 1 or 2 in bending (to be confirmed when a section is chosen): Required W. . BS EN 10025-2 Check cross-section classification From Table 5. Bending resistance M .92 = 4. 11 = 1.flange thickness ti = 15. t .1 of BS EN 199.+ 2r)tl (but not less than qh.357.=357..2.) From U K N A to BS EN 199./tle Actual c.I (26. of 1450 cm'.92 = 49. A .0. ~ .15/0.oo : .2 within limit limit = 45. Cross-section is Class I . BS EN 1993-1-1 Table 5.5 x106/275 =1. U K N A to BS EN 199. the shear area A.2. Cross-section resistance in bending is OK. = 1 4 5 0 x 1 0 ' x 2 7 5 x 1 0 4 =398..0. Maximum component thickness is. .(2 x153. Use 457 x 152 x 67 U K B in Grade S275 steel.I Assumed deflection limit is span/360 = 7200/360 = 20.2]) x15.526 Worked example Sheet 3 of 3 Beam example 2 Rev : . Dejections OK : . =8560 . A.5mm c 20.3.2 x 7200' = 14.0 +[2 x10.0 mm 384 x 210000 x 28900 x l o 4 : .0) + (9. Shear resistance is OK Dejections Check deflection under unfactored variable loads.I . U K N A to BS EN I 99.0 =4387mm2 I : .8 x15.0mm Actual deflection: ij--= 5 WL' 384 E I 5 x 25. = 3680 x 1O'mm' t. Steel Building Design: Design Data B S EN 10025-2 Maximum component thickness is j a n g e thickness ti = 19.0mm : .89/0.0mm.94 From Table 5.19. tr=19. Check cross-section classification: E=& 235 - = 0. an iterative design approach is therefore required. Ascot. within limit B S EN I 99.94 = 48. iL= 4. It is not now possible to arrange the calculations in such a way that a direct choice of section can be made (since susceptibility to lateral torsional buckling is not yet known).f Reference 7)..I .7mm. lateral torsional buckling slenderness is defined as: .>.97 cm = 49.0/0.6mm. W.E = 46. within limit Actual c . =11.6mm > 16.! = 3680 cm' b =229. Job Title Sheet 1 of 2 Institute Silwood Park.7-1-1: For a Class 1 outstand j a n g e in compression: cr/trE I9 For a Class 1 web in bending: c. c+/tt= 4.+= 46. select a suitable U K B in S275 steel assuming that the member must now be designed as laterally unrestrained.+ /t./t.2 : .3. the simplified method set out in NCCI SN002 and referred to as Method 3 in Reference 7 will be employed.Worked example 527 Rev ' ThsStesl Construction Job No.2mm. Cross-section is Class I Using NCCI SN002 (Method 3 o.1 of BS EN 199.I Table 5. =265 N/mmz.9mm.89.& I 7 2 Actual cr/tiE = 4. Try 610 x 2 2 9 x 125 U K B Geometric properties: h =612. c. /t.94 =5. In this example.0. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET Made by Date I 2010 Date I 2010 Beam example 3 Laterally unrestrained Universal Beam Problem For the same loading and support conditions of example 2.8. J. For rolled sections./= = 0.'.0 = 2.0 for a Class 1 or 2 section.39~)] =1. and 6 will also he satisfactory. & = 0.3.7 k N m < 357.39 Buckling curve selection: h / h = 612. M hRii =xrTW. Use 610 x229 x 125 U K B in Grade S275 steel.5[1+0.49)..39-0.0 .528 Worked example Beam example 3 ?LIT I Sheet2of2 I Rev 1 =puvxz& fi pbv = 1.46+41.44 1. Take UV = 0.2/229.4. V.7-1 -1 C16. 7 5 ~ 1 . . . For the present shape of bending moment diagram.3 =0.462 . -= 0.. U K NA to BS EN 1993-1-1 Buckling reduction factor xLI : 011 = 0 4 1+ ~ L (111 I .5 k N m = Mrcl 10' x 265 BS EN 199.1.94 x 0.67 For the case of rolled and equibalent welded sections.2.o = 424.75~1.3.9. L I T = -UVxZ& - 1 fi = 0.7 .49(1.94 Table 17.2.7-1 -1 C16. 3 9 ~ Lateral torsional buckling resistance.111 o) +/%I I BS EN 199.64 = 1.9 x 1. : . use buckling curve 'c' (a = 0.OK Since section is larger than before.75 and 0 = 0.46 1 Ol1 XIT= 1 +. for I-sections with 2-< h / b 53. 0 = 0.1 :.44 x 3680 x YMl f t I .4)+(0. Thus the actual level o. From statics.Worked example The S t e ~ l Construction 529 Rev Job No.f transfer o.f load at B and C (relative to the main beam's shear centre) will have no effect. select a U K B with Mc. it is clear f r o m the B M D that segment B C is critical (since it contains the most severe magnitude and distribution of bending moments). Therefore.RII > 406 k N m - 126 kN I 10 kN I I A 457 x 152 x 74 U K B is Class 1 and provides Mc. the bending moment diagram ( B M D ) and shear force diagram (SFD) are as follows: BMD SFD 135 kN For initial trial section. Given that each o. In this example. only segment B C need be considered for lateral torsional buckling. Steel Building Design: Design Data . assume further that ends A and D are similarly restrained.= 3. BC and C D separately. Ascot. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET Made by Date Date I 2010 I 2010 Beam example 4 Universal beam supporting point loads Problem Select a suitable U K B in S275 steel to carry a pair o.Kd = 431 k N m . the simplified method set out in N C C l SN002 and referred to as Method 3 in Reference 7 will be employed. Job Title Sheet 1 of 4 Institute Silwood Park.0 m A C 3. BC and C D ) have the same length ( L ( . 1145k~ A & & B 3.0 m X X " The cross-beams may reasonably be assumed to provide full lateral and torsional restraint at B and C.f the unrestrained segments (AB.f point loads at the third points transferred by cross-beams as shown in the accompanying sketch. the general lateral torsional buckling aspects of the design are therefore to consider the segments AB.0 m).0 m A Y Ill6 kN D + A 7 3. 1. = 42.49(0.4 = 2.3 A. For rolled sections. Table 17. t .9.6mm. / t .0mm.90. = 16.4) + (0.x 91 .7 A L.5[I + 0.93.92 XLI = QL1 +.. 0 2 ~ 1 . for I-sections with 2 < h / h 53..75 x I 0. For the ratio of end moments t y =377/406 = 0. f .75 and Buckling reduction factor xLI : =0 U K N A to BS EN 1993-1-1 1 . iz = 3.90')] = 0. 1I& = 0. cl /ti = 3. = 9. ti = 17. use buckling curve 'c' ( a = 0. Conservatively . %IT = -UVnz& I & = 0.0.4 : .. 0 = 0.3 mm.=>== 1.0mm : .66.T O = 0. tiz .49).3000133.33 cm = 33.0.TO)[email protected] ~ A 1 Ll ~ 88.4 of Reference 7).98 ~ 0 .. Wz.70 = 0.lateral torsional buckling slenderness is defined as: BS EN 10025-2 For class I or 2 sections.5.75x 0. h = 154.99 For the case of rolled and equivalent welded sections.0/154.70 x IO'mm-' I Sheet2of4 I Rev Steel Building Design: Design Data Maximum component thickness is./a= + 0.70 cm' = 16.for I-sections.0mm.0.2.530 Worked example Beam example 4 Try 457 x 152 x 74 U K B Geometric properties: h = 462.'T] I = 0. =265 N/mmz Using NCCI SN002 (Method 3 of Reference 7).4.90' .7 (Table 6. BS EN 1993-1-1 C 1 6.flange thickness tt = 19.3. p = 0. c.4mm.98 (by interpolation) from Table 16.92 40. take UV = 0..3 4 1+ a r T ( z r -n.90 Buckling curve selection: h / h = 462.6mm > 16.92' . = 1. Conservatively for I-sections.81 .0. For rolled sections. take UV = 0.3mm.9.0mm.5[1+ 0.8 .3 = o.I = 303 k N m c 406 k N m =ME() : .30 x IO’mm’.75 x 0.. 0 m m :.4) + (0.70 x lo7 x ~ YA41 I .77 Cl 6. 8 2 ~ l . : . lateral torsional buckling slenderness is defined as: Steel Building Design: Design Data BS EN 10025-2 For Class 1 or 2 sections.= 1.I Buckling reduction factor XL1: OL1= 0. = xLI W. for I-sections with 2 < h / b 53..= 18.Worked example Beam example 4 Lateral torsional buckling resistance: M.I .B& - ] BS EN 1993-1-1 = 0. 531 I S h e e t 3 o f 4 I Rev 3’ f ~ 265 = 0.t. 9 ~ 0 . use buckling curve ‘c’ (a = 0. I i.722 = 0.75 and U K N A to BS EN I 99.3.75 x 0.82 86.Zmm..f end moments =377/406 = 0.1 Not OK. p = 0.0. 0 = 0 . Maximum component thickness is.Ll 0 = 0. =37100 cm4=371 x1O6mm4.Air = .0/191.3. ratio o.7 /-1 = 2.o BS EN I 99.0mm.1.77 + 40.70 x 16.1. ll&=O.9. 2.f.1 - ~ A1 3000 I 42...23 cm = 42.49(0.‘.3 X I T = 1 OL1+ 1 d m = 0.3.4.3.98.0mm I 1 6 .72. =275 N / m m 2 A s above.0.3.3 = 2.72’)] = 0.tt=16. Wpi.b =191. Section is Class I A = l o 4 cmz =10400mmz. using NCCI SN002 (Method 3 of Reference 7).40 For the case of rolled and equivalent welded sections.) + .BN.772. .2. ’ 2.I .70 cm’ = 18. ’ = 2 L. As above.l J V ~ z & = 0 . =9. iL= 4.2. try larger section Try 457 x I91 x 82 UKB Geometric properties: h =460.5[1+ aLl (xLl xLl. C16.49). 9 8 ~ 0 .9mm.flange thickness ti = 16.. 7 2 - J c 1 Buckling curve selection: h/b = 460. Table 17. 9 + [2 x 10.3..0. is given by: A. -= 0.3mmz ~ : .+t. also segments A B and CD. A.21) x 16.7-1 -1 145+116 = 19.) U K N A to BS E N 1993-1-5 for all U K B From U K N A to B S E N 1993-1-5. . OK BS EN 199.0 f t = 409 k N m > 406 k N m = Mro : .0 mm U K N A to BS EN 199. Use 457 x 191 x 82 U K B in Grade S275 steel. q =1. = A .3 x 9000' = 21.532 Worked example Beam example 4 Lateral torsional buckling resistance: I Sheet4of4 I Rev Mh = 275 xrT W .5~9 5 wL4 384 EI 5 x 19.1 A 457 x191 x 8 2 U K B provides sufficient resistance to lateral torsional buckling for segment BC and. OK Beam is clearly satisfactory for dejection since these (approximate) calculations have used the full load and not just the imposed (variable) load. : . Assume for initial design that the point loads can be represented as a U D L and that the design loads can be factored down by 1. A. With q =l.3 x 16. the shear area A. Shear resistance: BS E N 1993-1-1 C 1 6. and UKC.t.3 k N / m 1.7-1 -1 C1 6. Shear resistance is O K Dejections: Check deflection under unfactored variable loads.5 to estimate the serviceability loads.6 For a rolled I section.2.0 mm 384 x 210000 x 371 x lo6 : . + 2r)tt (hut not less than qh.O.0) + (9. : .2 mm < 25.2htt + (t. Assumed dejection limit is span/360 Assumed serviceability U D L w = Actual deflection.0 = 474. loaded parallel to the web.. ij---= = 3000/360 = 25. The beam is therefore satisfactory in bending.2. > qhlh. by inspection. = 10400 (2 x 191.81 x 1830 x 10' -x Y'VI 1. 35 x 0.925 ~ 2 0 0 + ) (1. For the design loading spec@ed below. I A 9350 1 u 11300 i 9350 D 1 30000 Plate girder span and loading 1 UDL Point loads Actions (loading) Permanent actions: gk =25 k N / m G k = 200 k N Variable actions: UDL Point loads qk =40 k N / m Qk = 450 k N Partialvfactorsvforactions Partial factor for permanent actions Partial . Ascot.factor for variable actions Reduction factor for permanent actions BS EN 1990 x.50 5 =0.925 ~ 2 5 + ) (1.5 x 40) = 91.5 x 450) = 925 k N F d = (1.Worked example The S t e ~ l Construction 553 Rev Job No.2 k N / m 0. design a transversely stiffened plate girder in S275 steel.= 1.fully laterally restrained along its length. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET Client Made by Date Date I I Plate girders Plate girder design example Design brief The plate girder shown below is .925 ULS combination qf actions Design U D L Design point loads F d = (1.35 x . Job Title Sheet 1 of 10 Institute Silwood Park.35 x)= 1. 7 mm). bending moments and stiffener spacing Initial sizing ofplate pirder The recommended span/depth ratio for simply supported non-composite plate girders ranges between 12 for short span girders and 20 for long span girders.554 Worked example Plate girder design example I Sheet 2 of 10 I Rev Design shear-force and bending moment diagrams The shear force and bending moment diagrams corresponding to the U L S design loads are shown below.e. J 1 2 4 ~ to ensure a non-slender section.5 of the depth. 2 h . The minimum web thickness for plate girders in buildings is usually t . BS E N 10025-2 For non-composite plate girders. = 1 5 m m .3 and 0. . Assume afEange 750 x50 =37500mm2. Assume the web thickness t . Estimate flange area assuming Lt = 255 N/mmz (i. the flange width is usually within the range of 0. assuming 4 0 m m <ti < 6 . slightly less than that required for a Class 3 section due to proposed transverse stiffening. Herein the depth is assumed to be span/l5. 1550 3900 3900 L L 5650 L 5650 L 3900 3900 L 1550 8 30000 A B C D Shear force kN w 1 _ 1 J / 11142 17453 Bending moment kNm 18909 Design shear forces. bending moments and stiffener spacing Initial sizing ofplate pirder The recommended span/depth ratio for simply supported non-composite plate girders ranges between 12 for short span girders and 20 for long span girders. .554 Worked example Plate girder design example I Sheet 2 of 10 I Rev Design shear-force and bending moment diagrams The shear force and bending moment diagrams corresponding to the U L S design loads are shown below.e. Estimate flange area assuming Lt = 255 N/mmz (i.5 of the depth. slightly less than that required for a Class 3 section due to proposed transverse stiffening. J 1 2 4 ~ to ensure a non-slender section. Assume afEange 750 x50 =37500mm2. Herein the depth is assumed to be span/l5. 7 mm). 2 h . = 1 5 m m . BS E N 10025-2 For non-composite plate girders. The minimum web thickness for plate girders in buildings is usually t . Assume the web thickness t . 1550 3900 3900 L L 5650 L 5650 L 3900 3900 L 1550 8 30000 A B C D Shear force kN w 1 _ 1 J / 11142 17453 Bending moment kNm 18909 Design shear forces. assuming 4 0 m m <ti < 6 . the flange width is usually within the range of 0.3 and 0. = 2000mm = 114. Flange: For tt = 50mm.Worked example Plate girder design example 555 I Sheet 3 of 10 I I Rev Cross-section classtfication lnitial sizing proposed a plate girder with 750 x 5 0 m m janges and 2000 x l 5 m m web..35 tf 50 BS EN I 99.56.> 7 ~ 2 ~ =7 2= ~66. J.2 Limiting slenderness for Class 1 jange is 9~ = 8. the web must be checked for shear buckling.2 : . = 15mm.+=275 N / m m 2 BS EN 10025-2 E = 6 235 - = 0.35 : .2. Since h.5 = 7.f plate width. Flange is Class I Web: For t .+ = 255 N / m m 2 BS EN 10025-2 Ignoring weld size in determination o.f plate width.92 Ignoring weld size in determination o. Check cross-section classification.63 < 133 Limiting slenderness . Cf = - 750-15 =367. L.I .I Table 5.64 > 7.for Class 3 web in bending is 1 2 4 ~ BS EN 1993-1-1 Table 5.3. c .6(6) .5 mm 2 9= 367. Web is Class 4. w t rl BS EN 1993-1-1 C16. .3(1) BS EN 1993-1-5 c 1 5. it may be assumed that the bending moment is resisted by the janges alone while the web is designed to carry the shear only.. =152-=-=8. = 1550/2000 = 0. h.0mm 250 250 OK To avoid the..+Iaj2 = 4... LN.?4(h..83 =-=-=0.) = (750 x50 x 2 5 5 ) x (2000 + 50)/106 = 1960.556 Worked example Plate girder design example I Sheet 4 of 10 I Rev Dimensions qf web and-flanges Assume stiffener spacing a > h.: BS EN 1993-1-5 C15.+ 1. The bending resistance of the janges alone MI.7) Web buckling reduction factor (for non-rigid end post): I>.Hd = Ahl x (h.3 ( I ) Shear resistance of web V.77 A. The contribution o.89 Web slenderness: 2000 = 1. = 2000mm.x 0.? k N m OK MrKd = 19603 > M..flanges will be ignored. = 2293 kN Web aspect ratio a / h . = 275 N / m m 2 .7-1 -5 C1 A..83 0.08 :.00 +5..00 + 5.f the. Minimum web thickness to avoid serviceability problems: h 2000 t .07 BS EN 1993-1-5 C15.2(1) . = 4. + t. a = 1550mm.78 Buckling coefficient (for a / h .07 BS EN I 99. = 18909 k N m Shear bucklinp resistance cf end (anchor) panel A B The shear resistance of panel A B will be calculated assuming a non-rigid end post.. < 1): k .?(. with the reduced capacity compensated for by closer stiffener spacing. as described above..V.flanges buckling into the web: BS E N I 99. <1.7-1 -5 C18(1) Bending moment resistance For sections with class 1-3flanges but a class 4 web.34(2000/1550)2 = 12. .95 Buckling coefficient (for a/h.00(2000/5650)2 = 5.37 1.7+1.+2 I): k .Z7. BS E N 1993-1-5 C15. 2 1): k .53 .(0.92J584 C15.4 x 15 x 0 .83 Buckling coefficient (for a/h.3-1-5 h 37. = 5.: BS EN 1993-1-5 C15.?) I"= 0'76g= Vb.N = 275 N/mmz..Worked example Plate girder design example 557 I Sheet 5 of 10 I I Rev Shear buckling resistance qf panel BC The shear panel resistance of the end panel B C is calculated assuming a rigid end post but ignoring the contribution of the flange.3-1 -5 C15. =3900/2000 = 1.RII: Web buckling reduction factor (for rigid end post). h. = 2152 kN Web aspect ratio a / h .. = 2000mnz..34 + 4.7+1.Z(I) BS EN 199.2(1) Shear buckling resistance of panel DE The shear resistance of panel D E is calculated assuming a rigid end post bul ignoring the contribution of the jange.3-1-5 Cl A.+).39 Web slenderness: BS EN 199.60 37.3(1) 1 ) " 2 1.+ = 275 N/mm2. = 5650/2000 = 2. la)' = 5.. = 5.3(1) Shear resistance of web BS EN OK 199.4x15~0.34 + 4 . h .84 Web slenderness: "f'W BS EN 1993-1-5 Cl A.08 : . xbV = ~ 1. l ~=5.2(1) .34 )~ + 4.+. f. V.60 (0.60) Shear resistance of web V..?(..~&- - "v=o'76&= 2000 = 1..3(1) BS EN h. = 515 k N Web aspect ratio a / h . 9 2 m 1993-1-5 Cl 5. 2000 = 1. a = 3900mm.34 + 4.37 = 0.4t.3(3) Web buckling reduction factor (for rigid end post): BS EN 1993-1-5 C15. J. 0 0 ( h .Vr.00(2000/3900)2 = 6.00(h. a = 3900mm. = 2000mnz.4tbV&&= 37. J. l l 3 .f the stiffeners. = (24 x[15 + 2 x280]'/12) + (208 x15'/12) =380. at the end of the plate girder. =24mm) Check outstands: For t.force.558 Worked example Plate girder design example I Sheet 6 of 10 I Rev Design qf bearing stijfener at A The bearing stiffener at A should be designed for the compressive force due to the support reaction equal to 229.7-1 -5 C1 9. =15 xO. The effective stiffener section is shown below: BS EN I 99. such that there is no bending moment induced. I t .94 x 24 = 294mm Since h . web material is available on one side of the stiffeners only. It is assumed that both ends of the stiffeners are fixed laterally such that their effective length may he taken as 0.1 (8) 24 f ..7 k N The single (double-sided) stiffener constitutes a non-rigid end post and does not therefore need to be designed to resist any tensile anchorage . = 24mm. The effective stiffener section comprises the area of the stiffeners themselves.f the effective stiffener section. l l 4 .94 h. 0 e which is the class 3 limit for a compressed outstand. = 13 x 0. = 280mm. where such material is available.e.75h. = (2 x 280 x 24) + (I208 + 241 x 15) = 16920mm2 I.92 x15 =208mm (where E relates to the web material) either side o.2. A t location A.. Effective stiffener properties: A. The effective stiffener section will therefore he designed to resist an axial compression equal to the support reaction Nbd =2293 kN. Try double-sided stiffening consisting of two j u t s 280 x24mm (i. t. h. = 265 N/mmz BS EN 10025-2 235 E=& - = 0. 0 ~ torsional buckling is avoided. plus an effective web width equal to 15a.W. as is local buckling since h 3 I t .28 x10'mm4 . = 28Oll. i i z Effective stiffener section at A It is assumed that the support reaction acts at the centroid o. .3.75h.3-1 -5 Cl 9. =275N/mm2 BS EN 10025-2 E= 6 235 - = 0. t. = 1 5 m m .92 h.+=0.0 265 x = 4484 kN > 2293 kN = Nro OK Stiffener buckling resistance: a.f two flats 80 x 15mm (i. 80mm. as is local buckling since h .for locations B and C is shown below: BS EN 199.1 (8) .2(4) Design qf intermediate transverse stijfeners at B and C The stiffeners at B and C should be designed to have a minimum stiffness and sufficient buckling resistance to withstand a compressive axial force. = 80113&t.92 x 15 = 180mm Since h . = 1 5 m m ) Check outstands: Fort.i = -= 16920 Yuo 1. a = 3900mm. where such material is available. 0 ~which is the Class 3 limit for a compressed outstand. The effective stiffener section . buckling effects may be ignored and only cross-section checks apply. Try double-sided stiffening consisting o.92 x 15 = 208mm (where E relates to the web material) either side of the stiffeners.1. l t .2. i t .e. arising from the tension field. The effective stiffener section comprises the area of the stiffeners themselves plus an effective web width equal to 1 5 ~ t = .75 x2000 = 1500mm BS EN 1993-1-5 C19. f . 0 s torsional buckling is avoided. Pro.4(2) Since 1c 0. 15 x 0.R. =0. l l 4 . h. l l 3 .4 L.= 13 x 0. BS EN 1993-1-1 C16. = rfi J F = = 88. For panels BC and CD.Worked example Plate girder design example 559 I Sheet 7 of 10 Rev Stiffener cross-section resistance: N.2. for a 2 h.82x106mma ) Actual I. 0.7) Actual I .7-1 -5 Annex A.Jajz = 5.34 + 4. .2.v’) b [‘r = 190000 [ir = 190000 2000 = 10. Vkd= 2152 kN L = kzor 0 k = - 12(1. k. 2 1. = ( 1 5 ~ [ 1 5 + 2 ~ 8 0 ] ..3(. =0.00(200/.where.00(h.h.560 Worked example Plate girder design example I Sheet 8 of 10 I Rev Y Y Z Effective stiffener section at B and C Check minimum stiffness: = 0 .t5: BS EN 1993-1-5 c 1 9. = 6.06x106 5’ mm4 Minimum I.0.3. = 5.39 BS EN I 99..3 PA* = Vb.34 + 4. 7 5 ~ 2 0 0 0 ~ 1=5.82 x 1O6 > Required I..75h.8x68. .8z.7 qkd = 0 for a symmetric section.t.80.7 Nimm’ For a / h .: BS EN 1993-1-5 C 1 NA.2x2000x15x10-‘) =514 kN J I - 0. = 5. rii ~ .‘ /+ 12 (2 ) x 2 0 8 ~ 1 5 ‘ / 1 2=6. I =2152-(0.7900)2= 6.06 x 106mm4 Douhle-sided intermediate transverse stiffeners are designed to resist a compressive axial force Pbd. 1.75h.622]= 0. L. = 80mm. = (15 x 11.0.field.1. .3.+= 0. t. = 0.802. h.2 X= 1 0 + -= =-= 1 0. which has a n imperfection factor a = 0.3. The minimum stiffness requirements are satisfied as before. BS EN 199. = ( 2 x 8 0 ~ 1 5 + ) (12 x 2 0 8 +15] ~ 1 5 =8865mmz ) I.O NhHd Af.\/0. R.80 BS EN 1993-1-1 Cl 6.2) + 1 ' 1 = 0.77II. Try double-sided stiffening consisting of two j u t s 80 x 1 5 m m (i.4(2) 0 = 0.82 x 106mm4 Stiffener cross-section resistance: N .77~8865~275 x10~i=1881kN>514kN=Pbd 1.80 + . use buckling curve 'c'.1 Design of intermediate load-bearing stiffener at D The stiffener at D should be designed to have a minimum stiffness and sufficient buckling resistance to withstand the externally applied load at D of 92 kN plus a compressive axial .4(2) For buckling of stiffeners..62 .2) + 0.o = 2438 k N > 514 k N = Pro OK Stiffener buckling resistance.0 OK BS EN 1993-1-1 Cl 6.3-1 -5 Cl 9.0.e.49..Worked example Plate girder design example 561 I Sheet 9 of 10 I I Rev Effective stiffener properties: A. as employed at B and C .62' =0.5[1+a(n. = 15mm).0.75 x 2000 = 1500mm BS EN 1993-1-5 C19.49(0.5 + 2 x 80/-'/12) + (2 x 208 x I f ' / l 2 ) = 6.see above Figure.force Pro arising f r o m the tension .5[1+0.i = -= Yuo X275 x lo-' I . YWl 0. 1550 / 30000 Final plate girder details .Ja)2 = 5.3 q r = k Z o b=5.2.84x10.00(h.force to he resisted is therefore 925 + 0 = 925 k N Buckling resistance of the stiffener (as above) NI. The total compressive . the compressibe axial force Pro is given by: BS EN 1993-1-5 C 1 NA.4Nlmmz =1440-(0. the final plate girder dimensions and details are as below: All intermediate stiffeners 1550 J ' 3 c 13900 13900 c c 1 5650 c 5650 c 13900 13900 c c ( 1 . Vbd= 1440 kN L = kzor For a/h.7kN :.84 BS EN 1993-1-5 Annex A.+2 I .34 + 4. a = 5650mm..00(200/5650)2 = 5.7 orbd = 0 for a symmetric section.562 Worked example Plate girder design example I Sheet 10 of 10 I Rev For panels DE.4x2000x15x10") =-96. k . = 5.8x62.7=62. For a 2 h.Hd = 1881 k N > 925 kN OK Final girder dimensions and details Based on the above calculations.Take Pro = 0.34 + 4... 8mm. i.586 Worked example The S t e ~ l Construction Job No. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET Client Made by I Date I 2010 Beam-column example 1 Rolled Universal Column Problem Select a suitable UKC in S275 steel to carry safely a combination of 840kN in direct compression and a uniform bending moment about the minor axis of 12 kNm.i. c . / t . = 8. ti =14. iz = 5.0mm. cl /ti = 6. = 209.Hd of approximately 1400 kN will provide correct sort o.4mm. WIJl. over an unsupported height of 3.? = 305 cm' = 305 x l O'mm ' . t . Job Title Sheet 1 of 3 Rev Institute Silwood Park.2 . = 275 N/mmz since ti < 16mm B S EN 10025-2 Check cross-section classijkation: For a Class I outstand flange in compression ct /tie 5 9 BS EN 1993-1-1 Table 5.f margin to carry the moment.20 cm WIJl.0 U K N A to BS EN 1993-1-1 Geometric properties: h b = 205. = 9.4 cm2 = 7640mm2. = 656 cm' = 656 x 10' mm ' . Ascot. = 52.3 x 60 UKC in S275 steel member capacity tables suggest a minor axis buckling resistance Nh. 7 x20. =17. Steel Building Design: Design Data Material properties..1.6mm. 7/Ul = 1. ~ Steel Building Design: Design Data Partial factors: 7/uo = 1.0. : _I W t Try 2 0 . Yield strength f .96 cm = 89.2mm.6mm. A = 76.6 m.20. within limit Actual c.96’ .2) = 0.65 + 40.0 Oz = 0.0 L = 1.2) +%.2 .6/205.5[1+0. L c .66 5 1 .96 + 40. = 1.92 = 6.& =17.6 =o..70...8 A=_=’ A 1 - A.34) For minor axis buckling.0.O .!I 1 - 1 BS EN 1993-1-1 C 1 6.5.80’ +.46 86.z=1. A. use buckling curve ‘6’ (a = 0.&= 6.5[1+a(/l.8 = 1.Z.802]= 0.0.7-1 -1 C16...bE 5. = 0. For major axis buckling. - 3600189.r. + JG0.:I = 0. /-= = 0.0.1.96 BS EN 199.5’ E=& 587 I Sheet2of3 Rev 235 - = 0./t.46’ = 0.5[1+a(%. 8 ~ -L=- A A.0 L =1.. L.Worked example Beam-column example 1 For a Class I web in compression c. use buckling curve ’c’ ( a = 0. -0.80 .65 BS EN 1993-1-1 Table 6. l i . within limit Cross-section is Class I under pure compression. so will also be Class I under the more favourable stress distribution arising from compression plus bending.0 x3600 =3600mm for buckling about the y-y axis L.2 X? = I OZ I 0.3.0 = 0.2 XY = 0.80 A 1 86.49) Buckling reduction factors X: O.1/0.46’] = 0.02 < 1.0 x3600 =3600mm for buckling about the z-z axis Non-dimensional column slendernesses: fi - -=a \ 1 F = 8 6 .92 =18.92 Actual ct/t. .0.. Major and minor axis column buckling resistances Effective lengths: L.8 Buckling curves: h / b =209.2) +0.65’ + %./t.90 51.2) + 0.49(0.20/0. z l i z.3600152.1.34(0.46 -0.2.5[1+0. 34 = 0.f kiLmay be taken from Table 18. RO M.3. =2.62 of BS EN 199. =2...7 x20.'. as k.bd= 0.M 7bd = 840 + 2.21=0. = 0. =0. I OK Minor axis bending resistance Combined axial load plus bending: To verify resistance under combined axial loading plus bending.44-=0.44+0.0 from Table B.1 BS EN 1993-1-1 C1 6.4 ~ 1 . C.44 4 BS EN 1993-1-1 Table B..zRij +kz2.6k.3 of BS EN 1993-1-1.3.4.4..Nld +k.4C.I C16.z = 1.9 ~ : . Maximum (conservative) values o.1.3.6 ~ 2 .OK BS EN 1993-1-1 Equation 6. .O M c z ~ i j 1395 83. OK . .& Ni.95 < 1.. Adopt 20. For t y = 1. The major axis bending term is absent in this example since M..12 = 0.9 :.61 and 6.7-1-1 must be satisfied. kzz= 2. =1.65<1.I .60 + 0.0 1892 83. both Equations 6.3 .Nil...588 Worked example Beam-column example 1 I Sheet3of3 I Rev OK BS EN 199.61 BS EN 1993-1-1 Equation 6. bd I = ~ RO 12 840 +1.1...4C.. 0 k j Z=0.6kzi.7 x60 UKC .I .kzz=1..62 . . Ascot. out of plane) and rotation.0. ct/tt = 6.. Material properties: Yield strength J = 355 N/mmz since ti < 16mm Steel Building Design: Design Data BS E N 10025-2 . W..i. Job Title Sheet 1 of 8 Institute Silwood Park.factors.+= 49. = . =4. Bending is about the major axis. A = 105 cm2 = 10500mm2.55.7-1 -1 Geometric properties.3 cm =213mm. Check initially over the full column height. =21.8mm.Worked example The S t e ~ l Construction 589 Rev Job No.. t .e. tt = 13. i.f the columns are adequately restrained against lateral displacement (i.Z. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET I Subject Client I Beam-column example 2 I Date I 2010 Beam-column example 2 Rolled Universal Beam Problem Check the suitability of a 533 x210 x 8 2 U K B in S355 steel for use as a column in a portal frame of clear height 5. = 9. The ends o.= 1.3 mm.6 mm. Use Annex B of BS EN 1993-1-1 to determine the beam-column interaction .8mm. yv.Jt.6.0 U K N A to BS EN 199. W.6 m if the axial compression is 160kN.2mm. = 1. c.. h = 528.38cm =4. h = 208. Partial factors: y v ( . =2060 cm' =2060 x IO'mm'..ZOO x I 0' mm ' .ZOO cm' = .. the moment at the top of the column is 530kNm and the base is pinned. 3-1-1 Table 5.34 Buckling curves: h / b = 528.55/0.2 @> = o ..2)+1..57 . use buckling curve 'b' ( a = 0..572 1 1 .34' = 0. use buckling curve 'a' (a = 0.6/0.81 Actual cl/tlc = 6..15 BS EN 1993-1-1 Cl 6. = 10500 x355/10' =3727kN) that section is Class I under combined loading.672]=2.21(0.2)+/zf]=0.590 Worked example Beam-column example 2 Check cross-section classification: For a Class 1 outstand jange in compression cl/t. = 1. within limit Actual c.0 L = 1.53 > 1.4 = 0.2)+0. li.2)+~~]=0.Jt.342]=0.E I 72 E= BS EN 199.f.34) Buckling reduction factors BS EN 1993-1-1 Table 6. Major and minor axis column buckling resistances Effective lengths: L.2 x: BS EN 1993-1-1 Cl 6.81 = 61. X Y = 0.0 x 5600 = 5600mm for buckling about the z-z axis Non-dimensional column slendernesses: A. - 5600/213 76.1.97 I I .67-0. L. = 1.2 . + d m = 0. S [ l + ~ ( x .34-0.05.+/t. -0.E= 49.3/208.2 6 235 - = 0.0 x5600 =5600mm for buckling about the y-y axis L. within limit for pure bending Assume since compression of 160kN is very low compared with axial resistance of cross-section (A.5[l+0.81 = 8. For major axis buckling.57 + 40.O O2=OS[I+a(/ZL -0..0.Z.Z..21) For minor axis buckling.5[1+0.& I 9 I Sheet 2 of 8 I Rev For a Class I web in bending c.8 = 2..34(1.1.0 L = 1.0.2.. 75 and A.15 + 42.29 5 1.5[1+ 0.5[1+ a1 T(xI - 1. 16 + BS E N 1993-1-1 C 1 6.0. use buckling curve 'c' ( a = 0.0 : . 1I& = 0. I BS EN 199.1.49(1. Buckling reduction factor xL1: - U K N A to BS E N 1993-1-1 Or = 0.7-1 -1 C16.= 1.49). -= 0.9. = 1 1. 13')] = I . Conservatively for I-sections.= 0.75 . For rolled sections.3.0.75 x1.0.2.3. ")+ @ f T ] = 0.56 Lateral torsional buckling resistance: Mb = X r T W. for I-sections with 2 < h / b 53. For the ratio of end moments v =0.0 .4) (0. . .3.1 Clearly the lateral torsional buckling resistance is insufficient.BN.16 +41.56 x YWI f t 2060 x 10 x = 412 kNm c 530 kNm = Mbd 355 10 1.3 XIT = 1 OL1+ / . .from Table 6. - For the case of rolled and equivalent welded sections.. take UV = 0.1.13 ..7-1 -1 C16.1 Lateral torsional buckling resistance Non-dimensional beam slendernesses is determined using the simplified approach given in NCCI SN0026and Reference 2 as follows: NCCI SN002 For Class 1 or 2 sections.2. This may be Lmproved by adding bracing to reduce the minor axis buckling length and hence LIT.Worked example Beam-column example 2 1 O? 591 I 1 = 0.162 .67' Sheet 3 of 8 Rev X? = +d n = 2.13' = 0.15' Column buckling resistances: BS E N 1993-1-1 C 1 6. p = 0.1.I .4.75 x I .4 of Reference 2 see also Chapter 17.3. Member fails BS EN 199.. 34) Buckling reduction factor BS EN 1993-1-1 Table 6.53 > 1.0.6 m .34(0..8 = 2. the bending moment varies linearly . L. the bending moment varies linearly from 530kNm to 379kNm (= 530 x 4/5.2) +0.0 m for the lower part.2 Oz= 0. while for the lower part of the member. = 0.5[1+a(I? .0 x1600 =1600mm for buckling about the z-z axis Non-dimensional column slenderness.3. For the upper part o f the member: *530 kNm E 2 I I I I \ Minor axis column buckling resistance Effective length: L. = 1..2) +I:] = 0.. =1..592 Worked example Beam-column example 2 Add bracing to reduce minor axis buckling length Estimate suitable bracing location as 1.482]= 0.from 379 k N m to zero.6rn.48 .0 L =1. h / b = 528.for the upper part of the member and 4. = 5.L ~ ~ .5[1+0.2 x: BS EN 1993-1-1 Cl 6.66 .6 m from top of column I Sheet 4 of 8 I Rev For minor z-z axis buckling and lateral torsional buckling (LTB).4 Buckling curve.3/208.21ii -1600143.6). For the upper part o. Major axis buckling resistance is therefore unaltered.8 L.2. For minor axis buckling. while for major y-y axis buckling.1. L. while minor axis buckling resistance and L T B resistance are increased. use buckling curve 'b' (a = 0. 76.48 ..0. A L.f the member. Since xLl c 0. there is n o reduction for lateral torsion buckling and M h K d M ..12 (a) For j Nro ~ - 160= 0.71..7-1-1 must be satisfied.and kz....?0 = 0.are determined graphically from Figure 19.89 = 0. =0.= 0. Conservatively for I-sections.71. Values of k.Kd k . 7 1 BS EN 199. .9./-= 0.89 Figure 18.= 0'89x10500x355 x10-' =3332kN . Ril f..12 (6) .0.o = 731kNm >530kNm= Mr(1 :. l l f i = 0. 3604 C.Worked example Beam-column example 2 I Oz + 593 I I . The minor axis bending term is absent in this example since M.~=0.61 and 6. take W = 0.see also Chapter 17. . = 1.. M . For t. k = 1.0..01. = = WZ.Rii =X.01x 0.. I Lateral torsional buckling resistance Non-dimensional beam slendernesses is determined using the simplified approach given in NCCI SN0026 and Reference 2 as follows: NCCI SN002 For Class I or 2 sections..4 of Reference 2 .91 (by interpolation) from Table 6.3 Figure 18. Nh.4.OK Combined axial load plus bending: To verify resistance under combined axial loading plus bending.62 of BS E N 199.Af. = 1. both Equations 6.I .12..o =NEo OK BS E N 1993-1-1 C 1 6. = 0 . 4 ~ 0 .04 and 1.O Column buckling resistance: Nh. ylvo 10 '355 x 10" I .160kN yu7 I .i.89.3. .34. 6 + 0 .7-1 -1 Table B.> -= 2060 x Hd.89 5 1.48' Sheet 5 of 8 Rev X? = . For the ratio of end moments ly=379/5.C.66 + 40.bd = 0.66' = 0. 30 +k..3-1-1 Table 6.34) Buckling reduction .5[1+0.38 X? = 1 O? 1 1.0 :.3-1-1 Equation 6..38+41.1 L.69 c 1.61 and 6./-= = 0.73 = 0.3.2.04 + 0.3/208. 160 1 = -+ 3.4 = 1.factor @? BS EN 199.48 51.0 BS E N 1993-1-1 Cl 6.2 x: BS EN 199..8 = 2.65 = 0.05 +0.20-0..32 530 1. For minor axis buckling. _C MI>.3.I .0 L = 1.3.K d + k. 2.1.78 c 1.07..202] =1.38’-1. A.594 Worked example Beam-column example 2 Applying the interaction equations (Equations 6..0 x 4000 = 4000mm for buckling about the z-z axis Non-dimensional column slenderness: 2.5[l+a(12 -0.89= 0.8 76. = 1.3-1-1 C16.z li.53 > 1. OK M I J H ~ 3604 731 ~ BS EN 199..2)+l2] =0.62) of BS EN 1993-1-1: I Sheet 6 of 8 I Rev * N i J i ~ d 5.34(1..20’ +.-M’ ‘(I = 160 + 0.OK For the lower part o f the member: Minor axis column buckling resistance Effective length: L.2)+1.0 : . I .61 BS E N 1993-1-1 Equation 6.71 = 0.=-=2.20 Buckling curve: h / b = 528.1 - 400014. use buckling curve ‘b’ ( a = 0.3.2 =0.62 Ni. 01 x Table B.75 x 0.60 Figure 18.1. .= 0. = 1 -uvXz& 6 = 0.81. and k.60 = 0. 1I 6 = 0..2.0.r(j= 0.5[l+arr(Xr.2.r.49(0. both Equations 6.f Reference 2 see also Chapter 17.7-1 -1 C16. take UV = 0.3 = 0.4) + (0. are determined graphically from Figure 19.12(a) 0.4.9 x 4000/43‘ 76.0/5.81’ Lateral torsional buckling resistance: M ~ R .5[1+ -~r.75 and - a.75 x 0.62 of BS EN 1993-1-1 must be satisfied. for I-sections with 2 < h / b 53..from Table 6.0.= !~rrW.84’ .. C. Conservatively for I-sections.4 ~ y .84 BS EN 199.84 + 40.0 = 0.12 (6) .r = 0.4 o. - a.6 BS E N 1993-1-1 3 k .12. use buckling curve ‘c’ ( a = 0. . p = 0. Values of k. ro = 530 x : .factor xrr.= 1..i& 1 NCCI SN002 6 For Class 1 or 2 sections. ~ YMI = 0.75 x 0. 1.Worked example Beam-column example 2 Lateral torsional buckling resistance Non-dimensional beam slendernesses is determined using the simplified approach given in NCCI SN0026and Reference 2 as follows: LLl 595 I Sheet 7 of 8 I I Rev = -UV.3 Figure 18. For the ratio of end moments = 0.76 ~ 2 0 6 x 0 I0 355 ~ 1. The minor axis bending term is absent in this example since Mz.. = 1.o)+.7-1 -1 C16.6) = 555 k N m < 379 k N m = M .9.76 XLI = QL1 1 1 +d m fl = 0.B~~rI 0.49). OK BS EN 199.812)] = 0.1 Combined avial load plus bending: To verlfy resistance under combined axial loading plus bending. = 0.BN. For = 0.61 and 6.3. @IT = 0.0.3..81 For the case of rolled and equivalent welded sections.0 x 10‘ (4. For rolled sections. U K N A to BS E N 1993-1-1 Buckling reduction .75 . 09+0.60-=0.OC..99Mil R ~ I 1792 ~ 179 = 0. but for low axial loads and low major axis non-dimensional column slenderness..62 *+ NI. the member would have failed.....04+0.SC. the value of k..62) of BS EN 1993-1-1: I Sheet 8 of 8 I Rev Nbd +k. as may he seen in Figure 19.0 3604 379 555 :. .RO ~ M'bd Mi... which has a maximum value of I.41=0.approaches I. R<I =160 +0. Adopt 5.1 were to he used.596 Worked example Beam-column example 2 Applying the interaction equations (Equations 6. M ' b d 5 1 = -+0.76 c 1. This is mainly due to the k.61 BS EN 1993-1-1 Equation 6. Nb i.46<1.33 x 210 x 82 U K B Note that if the maximum values of the interaction factors given in Table 19....67 555 = 0..factor.OK : ..0 :.12(a). R ~ I I 160 kz..61 and 6.OK BS EN 1993-1-1 Equation 6. Ascot. The design axial load in the column Nbdis 125.Jt. Berks SLS 7QN Telephone: (01344) 623345 Fax: (01344) 622944 CALCULATION SHEET I Subject Client I Beam-column example 3 Made by I Date I 2010 Beam-column example 3 Rolled Universal Column in Simple Construction Problem A 254 x254 x 73 UKC is to be assessed for use as an internal column in a simple .2 For a Class I web in compression c. =992cm7=992xIO'mm'. t . W.I Table 5.I . ct/tt = 7. tt = 14.frame (i. The column length is 5. A = 9.1 cm2 = 9.. Material properties: Yield strength J = 275 N/mmz since ti < 16mm Check cross-section classification: For a Class 1 outstand j a n g e in compression c.310mm2. designed on the assumptions of simple construction)..rO o.8mm.1 mm.f this section to carry the applied loads.f 9. i .3.bdof 2...77...0.6 mm.E 5.4kNm and a design minor axis bending moment M. = 8.2mm.3 k N Check the adequacy o. W. h = 254.0 m and the steel grade is S275.e.0 UK N A to BS EN 199.6mm. Job Title Sheet 1 of 3 Institute Silwood Park.1 c m = I I I mm.Worked example The S t e ~ l Construction 597 Rev Job No. i .3 kNm./tp I 9 Steel Building Design: Design Data BS EN 10025-2 BS EN I 99.Jt. xl41 = 1..factors: x l 4 n = 1. Connection eccentricity causes a design major axis bending moment M. = 6.7. =465 cm' =465 x l 0 ' m m ' .33 E=& 235 - = 0.3.3.+ = 2. c.92 .48 c m = 64. The simplified interaction expression f r o m columns in simple construction is as follows: Partial . = 11..3-1-1 Geometric properties: h = 254.. 49) Buckling reduction .9 x 0.9& = 0.0.0 51..3.9 and c.0.92 = 8. so will also be class I under the more favourable stress distribution arising from compression plus bending.5[1+0.BN. For minor axis buckling.06' = 0.89 = 0.I .61 5 1.factor @? BS EN I 99.49(0.2.06 + 41.92 = 25.3/0.1 .I C16.3.89 .3.2 = 0.3.& = 7. Minor axis column buckling resistance Effective length: L. take UV = 0.598 Worked example Beam-column example 3 Actual c+/t. = 0. Conservatively for I-sections. = 1. /-= 1 1. use buckling curve 'c' ( a = 0.892 X? = 1 O? +.= 1. : .1.892]= 1.0 x5000 =5000rnm for buckling about the z-z axis Non-dimensional column slenderness: Buckling curve: h / b =254. within I limit Sheet2of3 I Rev Cross-section is Class I under pure compression.1 Lateral torsional buckling resistance Non-dimensional beam slendernesses is determined using the simplified approach given in NCCI SN0026 and Reference 2 as follows: NCCI SN002 For Class 1 or 2 sections. within limit Actual c .77/0.6 =1. .0.2 x: BS EN I 99.2) + I:] = 0.80 .I . /t.2) + 0.0 L = 1.1/254.1.0.O Column buckling resistance: BS EN 1993-1-1 Cl 6.2.06 .& = 23.I Table 6.5[1+ .z = 1.40.0. 0.5[1+all = 0 4I U K N A to BS EN 1993-1-1 (xll-Ill..75 and All = 0. for I-sectionswith h / h 52.0 275 x 10-O = 128 k N m > 2.2.34).0& ] + 0.7 k N m > 9.3. For rolled sections.0 : .= Lateral torsional buckling resistance: Mb nil = X r T f. .80.bd: .82 X 992 X l o i -X 10 YWl 1. OK = 22.Adopt 254 x 254 x 73 UKC Note how the .3 = 0.4) + (0. 2 75 w. = 0.3 kNm = Mi. .04+0.5 (2) Combined axial loud plus bending For a column in simple construction.4.?.) + .81 +0.81+ 40.following simplified interaction check may be performed: = 0.I .OK :. Buckling reduction factor XL1: OL1= 0.0 :.1 Minor axis bending resistance M<.88 I 1.4 kNm = M .3.82 XrT = OL1+ = /.2.802)]= 0.Worked example Beam-column example 3 599 I Sheet3of3 I I Rev For the case of rolled and equivalent welded sections.0. OK YM 0 BS EN I 99.81 1 1 0.34(0.Kd = w~Jizfu ~ 465 1. illustrating why great precision is not required with the two bending terms and justifying the use of conservative interaction factors. bd BS EN 1993-1-1 Cl 6.0.75 x 0. the .2.802 BS EN 1993-1-1 Cl 6.I (26.3.812. use buckling curve 'b' ( a = 0. 0 = 0.75 x 0.0.first term (axial load) dominates for this arrangement.3 = 0.