Quikjoint - Results
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kwc=0.70
kwc=0.70
Quikjoint - Comprehensive Edition 8.90
Untitled1.qj2
email: [email protected]
Designed : ALEX
Checked : ALEX
EX: EAVES APEX CONNECTION
: EAVE APEX CONNECTION TRIAL
Title
:1 : Untitled1
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: 09/03/18
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2077
160 x 20 FL x 755 LG No end plate stiffs M20 Grade 8.8 Bolts All fittings grade S275
Shear rows=2
360
80
80
75
Printed
81.63 degrees
Haunch C/F 356x171x51UB S275 Length=2300 Depth=513
406x140x39UB S275 Rafter pitch=0.000 degrees No web stiffener kwc=0.70
[+20]
80
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8
6
6
8
Combination: Reversal Shear -75.00 kN. Moment -160.00 kN.m. Horizontal 30.00 kN.
457x152x52UB S275 No cap plate Flange stiffs 70 x 10 FL x 140 LG Web stiffs C/F 70 x 10 FL top Web stiffs C/F 70 x 12 FL bot Morris stiff C/F 70 x 10 FL
[+20]
Combination: Dead+IMPOSED Shear 380.00 kN. Moment 330.00 kN.m. Horizontal -75.27 kN.
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Designed : ALEX
Checked : ALEX
National Annex: UK National Annex EN 1993-1-1: 2005 General rules and rules for buildings 3.2.1(1)
fy and fu taken from product standard
6.1(1)
Partial factors = 1.000 = 1.000 = 1.100
6.3.2.3(1)
resistance of cross section resistance of members to instability resistance of cross-sections in tension fracture
For rolled sections, hot-finished and cold-formed hollow sections: _ LT,0 = 0.400 = 0.750 For welded sections _ LT,0 = 0.200 = 1.000 Table 6.5 taken from UK National Annex
6.3.3(5)
Determination of interaction factors kyy, kyz, kzy, kzz For doubly symmetrical sections use: Method 1 given in Annex A For mono-symmetrical sections use: Method 2 given in Annex B
EN 1993-1-8: 2005 Design of joints 2.2(2)
Partial factors = = = = ,ser = u =
1.250 1.250 1.250 1.250 1.100 1.100
resistance of bolts resistance of welds resistance of plates in bearing slip resistance at ultimate limit state (Category C) slip resistance at serviceability limit state (Category B) partial factor for tying resistance
EN 1992-1-1: 2004 Design of concrete structures - General rules and rules for buildings C = 1.500 partial factor for concrete S = 1.150 partial factor for reinforcing steel cc = 0.850 is the coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied
ct
= 1.000
is the coefficient taking account of long term effects on the tensile strength and of unfavourable effects, resulting from the way the load is applied
CRd,c
= 0.180/C
as defined in 6.2.2(1)
Untitled1.qj2
EN 1994-1-1: 2004 Design of composite steel and concrete structures - General rules and rules for buildings V = 1.250 partial factor for design shear resistance of a headed stud
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EX: EAVES APEX CONNECTION Designed : ALEX
Checked : ALEX
Untitled1.qj2
EN 1995-1-1: 2004 Design of timber structures - Common rules and rules for buildings = 1.300 Partial factor for material properties - Solid timber. Table 2.3 = 1.250 Partial factor for material properties - Glued laminated timber. Table 2.3 kcr = 0.670 Crack factor for shear resistance - Solid timber. 6.1.7(2) kcr = 0.670 Crack factor for shear resistance - Glued laminated timber. 6.1.7(2)
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Designed : ALEX
Checked : ALEX
WARNING End plate is narrower than the flange width. These calculations assume that the end plate is at least as wide. Summary of results Moment capacity Bolts in shear Tension flange weld Compression flange weld Web weld tension zone Web weld shear zone Rafter web transverse compression Tension stiffeners - column side Stiffener weld Flange stiffeners - shear Flange stiffeners - web shear Flange stiffeners - web weld Web tension - additional check Compression stiffeners Column web fitted Reversal Moment capacity Bolts in shear Tension flange weld Compression flange weld Web weld tension zone Web weld shear zone Tension stiffeners - column side Stiffener weld Compression stiffeners Column web fitted
Unity
Verdict
0.92 0.71 0.77 1.00 0.77 0.89 0.60
Passed Passed Passed Passed Passed Passed Passed
0.59 0.48 0.57 0.50
Passed Passed Passed Passed N/A
Minimum 6 fillet weld
N/A
Minimum 6 fillet weld
0.72 0.11 0.84 0.77 0.89 0.77
Passed Passed Passed Passed Passed Passed
Full strength Full strength Full strength
0.21
Passed
Minimum 6 fillet weld
N/A
Minimum 6 fillet weld
Overall
Comment
Full strength Bearing fit assumed Full strength Full strength
Single sided
Passed
Note: Connection is single sided. = 1.00
5.3(8) and Table 5.4
Detailed calculations in accordance with EN 1993-1-8:2005
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Connection geometry Column side: m = w/2 - twc/2 - 0.8rc e = ( 152.4 - 80)/2 = 40 - 3.8 - 8.2 = 36.2 mm. = 28.0 mm. n = lesser of 36.2, 40.0 and 1.25 x 28.0 = 35.1 mm. Beam side: m = w/2 - twb/2 - 0.8s e = ( 160.0 - 80)/2 = 40 - 3.2 - 4.8 = 40.0 mm. = 32.0 mm. n = lesser of 36.2, 40.0 and 1.25 x 32.0
Quikjoint - Comprehensive Edition
Table 6.2
ew=
8.2 mm.
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Designed : ALEX
Checked : ALEX
= 36.2 mm. Bolt capacity Using Table 3.4 tension resistance of M20 Grade 8.8 bolt Ft,Rd = k2 fub As / M2 where k2 = 0.90 and fub= 800 N/mm2 = 0.90 x 800 x 245.0 / 1.25 = 141.1 kN
See table 3.1
Bolt row 1 Row 1 alone Column flange bending Calculate equivalent effective length Leff of T-stub. leff,cp = 2.Pi.m leff,nc = au.m + al.m - (4m + 1.25e) leff,nc = a.m leff,nc = a.m leff,nc = 4m + 1.25e leff,cp leff,nc leff,1 leff,2
Table 6.4, 6.5, 6.6
= = = = =
= [i] = Max[iv,iii(m2u),iii(m2l),ii]] = Min[leff,cp,leff,nc] = leff,nc
176 mm 174 mm 162 mm 169 mm 157 mm
Pattern(i)
= = = =
176 mm. 174 mm. 174 mm. 174 mm.
Pattern(iv) Pattern(iii) Pattern(iii) Pattern(ii)
Equivalent T-stub resistances Mpl,1,Rd = 0.25 leff,1 t2 fy / = 0.25 x 174 x 10.92 x 275 / 1.00
=
1421 kN.mm.
= 0.25 leff,2 t2 fy / = 0.25 x 174 x 10.92 x 275 / 1.00
=
1421 kN.mm.
Mpl,2,Rd
Table 6.2
Find critical bolt failure mode FT,1,Rd = (8n - 2ew) Mpl,1,Rd / (2mn - ew(m + n))
=
259
kN.
Mode 1
FT,2,Rd
= (2.Mpl,2,Rd+ n Ft,Rd) / (m + n)
=
202
kN.
Mode 2
FT,3,Rd
= Ft,Rd = 2 x 141
=
282
kN.
Mode 3
Column web tension Not applicable stiffener/Flange in tension zone Beam end plate bending Calculate equivalent effective length Leff of T-stub. leff,cp = 2.Pi.m leff,nc = 4m + 1.25e leff,nc = a.m
Untitled1.qj2
leff,cp leff,nc leff,1 leff,2
= [i] = Max[ii,iii] = Min[leff,cp,leff,nc] = leff,nc
Quikjoint - Comprehensive Edition
Table 6.4, 6.5, 6.6
= = =
201 mm 178 mm 187 mm
Pattern(i)
= = = =
201 mm. 187 mm. 187 mm. 187 mm.
Pattern(ii) Pattern(iii)
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Equivalent T-stub resistances Mpl,1,Rd = 0.25 leff,1 t2 fy / = 0.25 x 187 x 20.02 x 265 / 1.00
=
4956 kN.mm.
= 0.25 leff,2 t2 fy / = 0.25 x 187 x 20.02 x 265 / 1.00
=
4956 kN.mm.
Mpl,2,Rd
:6
Find critical bolt failure mode = (8n - 2ew) Mpl,1,Rd / (2mn - ew(m + n)) FT,1,Rd
Table 6.2
=
771
kN.
Mode 1
FT,2,Rd
= (2.Mpl,2,Rd+ n Ft,Rd) / (m + n)
=
295
kN.
Mode 2
FT,3,Rd
= Ft,Rd = 2 x 141
=
282
kN.
Mode 3
Beam web tension Not applicable stiffener/Flange in tension zone Ft,Rd(row1) = 202 kN. Ft,Rd(row1) = 202 kN.
Column flange mode 2 governs
And so on... Calculations for Moment Connections tend to be very lengthy. Full calculations for all bolt lines are available by selecting Options|Results contents and deselecting Reduce detail in the Results contents form. Ft,Rd(row2) = 201 kN. Ft,Rd(row1 + 2) = 507 kN. Ft,Rd(row2) = 201 kN. Ft,Rd(row3) = 197 kN. Ft,Rd(row2 + 3) = 371 kN. Ft,Rd(row1 + 2 + 3) = 719 kN. Ft,Rd(row3) = 171 kN. Resistance of column web Effective width of web in compression beff,c,wc = 52.7 + 62.6 + 52.7 = 168.1 mm.
Untitled1.qj2
Column flange mode 2 governs
Column flange mode 2 governs Column flange mode 1 governs Column flange mode 1 governs
6.2.6.2(1) (6.10)
where
Reduction factor for shear interaction =1.00 = 1 / (1 + 1.3 ( beff,c,wc . twc / Avc )2 )
Quikjoint - Comprehensive Edition
Column flange mode 2 governs
52.7 = 2.5 x ( 10.9 + 10.2 )
Table 6.3
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Designed : ALEX = 1 / (1 + 1.3 ( 168.1 x
7.6 /
3647 )2 )
=
Checked : ALEX
0.93
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative
6.2.6.2(2)
Resistance of column web in bearing: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 0.93 x 0.70 x 168.1 x 7.6 x 275 / 1.00
(6.9)
(6.14)
=
228.4 kN.
Resistance of: 2 - 70 x 12 web stiffeners fy,web = 275 fy,stiff = 275 Use fy = 275 N/mm2 and = 0.924 Effective width of stiffener beff,stiff = 70 mm.
but less than: 14..tstiff
=
70.0 mm.
Effective area of stiffeners in contact with flange = 2 x 59.8 x 12 = 1435 mm2. Anet Length of web acting with stiffener lw,1 = 15..tw Therefore: lw = 222.8 mm. Aeff,s = 3373 mm2. Ieff,s = 322 x 104 mm4. ieff,s = 30.9 mm.
= 15 x 0.924 x 7.6
= (Lcr / iz) x (1 / x (1 / 93.91) = 407.6 / 30.9 x 1 / 0.924 x 1 / 93.91
From Figure 6.4 using: = = 1.000
Length of web
=
0.152 and buckling curve 'c'
Resistance to transverse compression is lesser of: Fc,Rd,1 = Anet fy / + Fc,wc,Rd,1 = 1435 x 275 x 10-3 / 1.00 + 228.4 Fc,Rd,2
= 105.4 mm.
= Aeff,s fy / = 1.000 x 3373 x 275 x 10-3 / 1.00
0.152 (see Table 6.2)
Governing bearing
=
623.1 kN.
=
927.6 kN.
Governing buckling
Haunch flange and web in compression 6.2.6.7(1) and P398 Step 2 Rafter resistance with coexisting: axial = 75.3 kN. shear = 380.0 kN. Mc,Rd = 456.8 kN.m. QuikEC3 Fc,fb,Rd = Mc,Rd x 103 / (h - tf) = 456.8 x 103 / ( 712.0 - 11.5 ) = 652.1 kN. Height of haunched rafter exceeds 600 mm. 6.2.6.7(1) and SN041a-EN-EU Fc,fb,Rd < tf . bf . fy x 10-3 / 0.8 < 11.5 x 171.5 x 275 x 10-3 / 0.8 = 678.0 kN. Therefore Fc,fb,Rd = 652.1 kN.
Untitled1.qj2
Resistance of column web panel in shear
Quikjoint - Comprehensive Edition
6.2.6.1(2)
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Designed : ALEX Vwp,Rd
Sheet
= 0.9 x fy,wc . Avc / ( 3 x ) = 0.9 x 275 x 3647.3 / ( 3 x 1.00 )
Resistance of column web and Morris stiffener Asn = 2 x bsn . ts = 2 x 70.0 x 10 = 1400 mm2. Vwps,Rd = Asn . fy . Cos() + Vwp,Rd where = 59.0 = 1400 x 275 x 0.515 / 1000 + 521.2
=
521.2 kN.
=
719.5 kN.
=
268.1 kN.
Checked : ALEX
Plastic distribution of bolt forces Row Force hr kN. mm. Ft,Rd(row1) 201.9 651 Ft,Rd(row2) 200.6 571 170.7 491 Ft,Rd(row3) Ft,Rd(row) = 573.1 kN. Modification of bolt row forces 1.9 x Ft,Rd = 1.9 x 141.1 Plastic distribution valid Axial force NEd > 0.05 x Npl,Rd
6.2.3(2)
NEd will be accounted for using the methods outlined in P398 Step 4.
Equilibrium Ft,Rd(row) = 573.1 kN. = 623.1 kN. Lesser of: 623.1, 652.1, Fc,Rd NEd = -75.3 kN. Negative being compression 547.8 kN. Fc,Rd + NEd = Ft,Rd(row)
> Fc,Rd + NEd
6.2.7.2(9)
719.5 /
Surplus capacity in bolts. Reduce starting at bottom row and working up until equilibrium holds
Moment capacity Row
Ftr,Rd hr Ftr,Rd x hr kN. mm. kN.m. Ft,Rd(row1) 201.9 651 131.4 Ft,Rd(row2) 200.6 571 114.6 Ft,Rd(row3) 145.4 491 71.4 Ft,Rd(row) = 547.8 Mj,Rd = 317.4
Untitled1.qj2
Modify applied moment for effect of axial load MEd,mod = MEd + NEd x hN Where MEd = Applied moment NEd = Axial tension hN = Distance from centre line of beam to centre of compression
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Designed : ALEX = Mj,Rd
> MEd,mod
Applied shear VEd =
330.0 -
:9
75.3 x 0.507
=
Checked : ALEX
291.8 kN.m.
Pass 380.0 kN.
Bolts in shear Using Table 3.4 shear resistance of M20 Grade 8.8 bolt Fv,Rd = v fub A / M2 where v = 0.60 and fub= 800 N/mm2 = 0.60 x 800 x 245.0 / 1.25 = 94.1 kN. Residual shear resistance of bolt fully loaded in tension Fvs,Rd = (1 - 1/1.4) x Fv,Rd = 26.9 kN. Bolt bearing end plate Fb,Rd,p = 1.0 x k1.b.fu.d.t /
See table 3.1
Table 3.4
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 40
= 315.6 kN. Bolt bearing column Fb,Rd,c = 1.0 x k1.b.fu.d.t /
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 36
= 172.0 kN. Shear capacity of bolt group Fv,Rd = Lesser of: Fv,Rd, Fb,Rd,p and Fb,Rd,c Fvs,Rd = Lesser of: Fvs,Rd, Fb,Rd,p and Fb,Rd,c
= =
94.1 kN. 26.9 kN.
VRd,b
=
537.6 kN.
VRd,b
= 4 x Fv,Rd + 6 x Fvs,Rd > VEd
Pass
Calculate thickness factor for full strength double fillet welds SN014a, TC10, P358 Annex C fy . w . weld 275 x 0.85 x 1.25 from Table 4.1 a / t,ply = = 0.504 . fu . 2 1.00 x 410 x 2 w
Tension flange weld Size of full strength fillet weld s = 8.6 x 0.504 / 0.7 Pass Tension flange weld is full strength
=
Design shear strength of end plate weld fu / 3 410 / 3 fvw,d = = 222.8 N/mm2. w 0.85 x 1.25
6.2 mm.
w
from Table 4.1
(4.4)
Untitled1.qj2
Compression flange weld
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Designed : ALEX
Checked : ALEX
Fc,Rd Capacity of
= 623.1 kN. 8 fillet weld to compression flange = 2 x 171.5 x 0.7 x 8 x 222.8 = 427.9 kN. Pass Compression flange weld can not transmit full compressive force Provide a bearing fit with nominal 6 fillet weld Web welds in tension zone Size of full strength fillet weld s = 6.4 x 0.504 / 0.7 Pass Web weld in tension zone is full strength. Weld must continue past the last tension bolt a minimum of 1.73W/2 = 69 mm.
=
4.6 mm.
Web welds shear zone Pass Web weld to shear zone is full strength Rafter web in transverse compression Effective width of web in compression beff,c,wc = 47.0 + 76.1 + 47.0 where = 170.1 mm.
6.2.6.2(1) (6.10)
47.0 = 2.5 x ( 8.6 + 10.2 )
Reduction factor for shear interaction =0.00 = 1.00
Table 6.3
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative Reduction factor for plate buckling p = 0.932 (( beff,c,wc . dwc . fy,wc ) / ( E . tw 2)) p> 0.72 therefore = ( p - 0.2) / p2 Resistance to transverse compression is lesser of: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 1.00 x 0.70 x 170.1 x 6.4 x 275 / 1.00 Fc,wc,Rd,2
6.2.6.2(2) (6.14) (6.13c)
=
1.30
=
0.65
(6.13b)
(6.9) Governing web bearing
=
209.6 kN.
= . kwc . . beff,c,wc . twc . fy,wc / M1 = 1.00 x 0.70 x 0.65 x 170.1 x 6.4 x 275 /1.00 =
136.0 kN.
Force in haunch flange Fc,fh,Ed = Fc,Rd / Cos( 8.69) = 630.3 kN But less than capacity 171.5 x 11.5 x 275 / = 542.4 kN. Force normal to rafter Fc,wb,Ed = 542.4 x Sin( 8.69 - 0.00) Fc,wb,Rd,2 > Fc,wb,Ed
=
81.9 kN
Pass
Tension stiffeners - Column side Force in stiffener Fs,Ed is greater of: Fs,Ed,1 and Fs,Ed,2 Web tension: Flange bending
Untitled1.qj2
Governing web buckling
Quikjoint - Comprehensive Edition
P398 Check 6A Note: Fs,Ed is for a pair of stiffeners
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Designed : ALEX
Fs,Ed,1 > (Fri,Rd + Frj,Rd) - ( Lwt x twc x fy,c) / Asn
: 11
Checked : ALEX
Fri,Rd Frj,Rd ( + m1 + m2U m1 + m2L
Fs,Ed,2 > m1 x
= Fs,Ed / ( fy,s / )
Weld size
= Fs,Ed / (4 x fvw,d x bsg x 0.7)
Column side fy,c = 275 N/mm2 m1 = 28.0 mm. Stiffener
Asn,act
Fitted
Fri,Rd Frj,Rd
m2L m2U
0.0
0.0
201.9 201.9
40.9 30.2
200.6
30.2
1196
Rib
1196
Lwt
Fs,Ed,1
> Fs,Ed
> Fs,Ed
> Fw,Ed
Verdict
0
82.1
298.5
6FW
Passed
189
7
193.7
704.5
6FW
Passed
P398 Step 6A
=
400.1 kN.
Pass P398 Step 6A
337.9 kN.
Pass
Partial depth stiffener - Web weld Lw = 140 - 6 L1 = ( 80 - 7.6) /2 L2 = 2/3 x Lw Ft = 0.5 x Fs,Ed x L1/L2 t = 2 Ft / Lw s = Fs,Ed / (2 x Lw) Fw,Ed = ( s2 + t2 ) / 2 Fw,Rd = 0.7 x 6 x 222.8 Fw,Rd
Weld
115
Partial depth stiffener - Check shear capacity of column web Fv,Rd = 2 x Ls x tw x fy / ( 3 x ) = 2 x 140 x 7.6 x 275 / ( 3 x 1.00 ) = Fv,Rd
Asn (required)
Partial depth stiffener - Check shear capacity of stiffener Fv,Rd = 2 x 0.9 x Ls x ts x fy,s / ( 3 x ) = 2 x 0.9 x 140 x 10 x 275 / ( 3 x 1.00 ) Fv,Rd
Fs,Ed,2
P398 Figure 2.18
= = = = = =
134.0 mm. 36.2 mm. 89.3 mm. 39255.1 N 585.9 N/mm. 722.9 N/mm.
Weld length Lever arm Lever arm Total tension Weld tension Weld shear
= =
465.3 N/mm. 935.7 N/mm.
Pass
Compression stiffener - assume bearing fit to column flange Fs,Ed = 1.0 x 623.1 Minimum weld size to column web = 623.1 / (4 x ( 449.8 - 2 x 10.9) x 0.7 x 223) Use 6 fillet weld (Minimum)
= =
P398 Step 6B
623.1 kN. 2.3 mm.
Untitled1.qj2
Reversal
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Designed : ALEX
Bolt capacity Using Table 3.4 tension resistance of M20 Grade 8.8 bolt Ft,Rd = k2 fub As / M2 where k2 = 0.90 and fub= 800 N/mm2 = 0.90 x 800 x 245.0 / 1.25 = 141.1 kN Resistance of column web Effective width of web in compression beff,c,wc = 0.0 + 56.6 + 52.7 = 109.3 mm.
Checked : ALEX
See table 3.1
6.2.6.2(1) (6.10)
where
52.7 = 2.5 x ( 10.9 + 10.2 )
Reduction factor for shear interaction =1.00 = 1 / (1 + 1.3 ( beff,c,wc . twc / Avc )2 ) = 1 / (1 + 1.3 ( 109.3 x 7.6 / 3647 )2 )
Table 6.3
=
0.97
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative
6.2.6.2(2)
Resistance of column web in bearing: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 0.97 x 0.70 x 109.3 x 7.6 x 275 / 1.00
(6.9)
(6.14)
=
154.8 kN.
Resistance of: 2 - 70 x 10 web stiffeners fy,web = 275 fy,stiff = 275 Use fy = 275 N/mm2 and = 0.924 Effective width of stiffener beff,stiff = 70 mm.
but less than: 14..tstiff
=
70.0 mm.
Effective area of stiffeners in contact with flange Anet = 2 x 59.8 x 10 = 1196 mm2. Length of web acting with stiffener lw,1 = 15..tw Therefore: lw = 138.7 mm. Aeff,s = 2454 mm2. Ieff,s = 268 x 104 mm4. ieff,s = 33.1 mm.
= 15 x 0.924 x 7.6
= (Lcr / iz) x (1 / x (1 / 93.91) = 407.6 / 33.1 x 1 / 0.924 x 1 / 93.91
From Figure 6.4 using: = = 1.000
Untitled1.qj2
Length of web
=
0.142 and buckling curve 'c'
Resistance to transverse compression is lesser of: Fc,Rd,1 = Anet fy / + Fc,wc,Rd,1 = 1196 x 275 x 10-3 / 1.00 + 154.8 Fc,Rd,2
= 105.4 mm.
= Aeff,s fy / = 1.000 x 2454 x 275 x 10-3 / 1.00
Quikjoint - Comprehensive Edition
0.142 (see Table 6.2)
Governing bearing
=
483.7 kN.
=
674.8 kN.
Governing buckling
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Designed : ALEX
Rafter flange and web in compression Rafter resistance with coexisting: axial = 30.0 kN. shear = 75.0 kN. Mc,Rd = 456.8 kN.m. Fc,fb,Rd = Mc,Rd x 103 / (h - tf) = 456.8 x 103 / ( 712.0 - 8.6 ) = Height of haunched rafter exceeds 600 mm. Fc,fb,Rd < tf . bf . fy x 10-3 / 0.8 < 8.6 x 141.8 x 275 x 10-3 / 0.8 = Therefore Fc,fb,Rd =
Checked : ALEX
6.2.6.7(1) and P398 Step 2 QuikEC3
649.4 kN. 6.2.6.7(1) and SN041a-EN-EU
419.2 kN. 419.2 kN.
Resistance of column web panel in shear Vwp,Rd = 0.9 x fy,wc . Avc / ( 3 x ) = 0.9 x 275 x 3647.3 / ( 3 x 1.00 )
=
521.2 kN.
Resistance of column web and Morris stiffener Asn = 2 x bsn . ts = 2 x 70.0 x 10 = 1400 mm2. Vwps,Rd = Asn . fy . Cos() + Vwp,Rd where = 42.9 = 1400 x 275 x 0.732 / 1000 + 521.2
=
803.1 kN.
=
268.1 kN.
6.2.6.1(2)
Plastic distribution of bolt forces Row Force hr kN. mm. 198.7 651 Ft,Rd(row1) Ft,Rd(row2) 162.1 571 = 360.8 kN. Ft,Rd(row) Modification of bolt row forces 1.9 x Ft,Rd = 1.9 x 141.1 Plastic distribution valid Axial force NEd < 0.05 x Npl,Rd
6.2.3(2)
NEd may be ignored
Equilibrium Ft,Rd(row) = 360.8 kN. Fc,Rd = 419.2 kN. Lesser of: 483.7, 419.2, Ft,Rd(row)
< Fc,Rd
6.2.7.2(9)
803.1 /
Bolt capacity governs
Moment capacity Row
Untitled1.qj2
Ftr,Rd hr Ftr,Rd x hr kN. mm. kN.m. Ft,Rd(row1) 198.7 651 129.3 Ft,Rd(row2) 162.1 571 92.5 Ft,Rd(row) = 360.8 Mj,Rd = 221.8
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Designed : ALEX
Checked : ALEX
Applied moment MEd = 160.0 kN. Mj,Rd
> MEd
Pass
Applied shear = VEd
75.0 kN.
Bolts in shear Using Table 3.4 shear resistance of M20 Grade 8.8 bolt Fv,Rd = v fub A / M2 where v = 0.60 and fub= 800 N/mm2 = 0.60 x 800 x 245.0 / 1.25 = 94.1 kN. Residual shear resistance of bolt fully loaded in tension Fvs,Rd = (1 - 1/1.4) x Fv,Rd = 26.9 kN. Bolt bearing end plate Fb,Rd,p = 1.0 x k1.b.fu.d.t /
See table 3.1
Table 3.4
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 40
= 315.6 kN. Bolt bearing column Fb,Rd,c = 1.0 x k1.b.fu.d.t /
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 36
= 172.0 kN. Shear capacity of bolt group Fv,Rd = Lesser of: Fv,Rd, Fb,Rd,p and Fb,Rd,c = Lesser of: Fvs,Rd, Fb,Rd,p and Fb,Rd,c Fvs,Rd
= =
94.1 kN. 26.9 kN.
VRd,b
=
672.0 kN.
VRd,b
= 6 x Fv,Rd + 4 x Fvs,Rd > VEd
Pass
Calculate thickness factor for full strength double fillet welds SN014a, TC10, P358 Annex C fy . w . weld 275 x 0.85 x 1.25 from Table 4.1 a / t,ply = = 0.504 . fu . 2 1.00 x 410 x 2 w
Design shear strength of end plate weld fu / 3 410 / 3 fvw,d = = 222.8 N/mm2. w 0.85 x 1.25
Untitled1.qj2
Tension flange weld Tension capacity of the beam flange Npl,Rd = b . tf . fy / = 171.5 x 11.5 x 275 / 1.00
Quikjoint - Comprehensive Edition
w
from Table 4.1
(4.4)
SN014a Annex A
=
542.4 kN
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Designed : ALEX Total force in the top 2 bolt rows Ftr,Rd = 198.7 + 162.1 Capacity of Pass
: 15
8 fillet weld to tension flange = 2 x 171.5 x 0.7 x 8 x 222.8 Tension flange weld adequate
Compression flange weld Size of full strength fillet weld s = 8.6 x 0.504 / 0.7 Pass Compression flange weld is full strength Web welds in tension zone Size of full strength fillet weld s = 7.4 x 0.504 / 0.7 Pass Web weld in tension zone is full strength. Weld must continue past the last tension bolt a minimum of 1.73W/2 = 69 mm.
Checked : ALEX
=
360.8 kN.
=
427.9 kN.
=
6.2 mm.
=
5.3 mm.
Web welds shear zone Pass Web weld to shear zone is full strength Tension stiffeners - Column side Force in stiffener Fs,Ed is greater of: Fs,Ed,1 and Fs,Ed,2 Web tension: Flange bending Fs,Ed,1 > (Fri,Rd + Frj,Rd) - ( Lwt x twc x fy,c) / Asn
P398 Check 6A Note: Fs,Ed is for a pair of stiffeners
Frj,Rd Fri,Rd ( + m1 + m2U m1 + m2L
Fs,Ed,2 > m1 x
= Fs,Ed / ( fy,s / )
Weld size
= Fs,Ed / (4 x fvw,d x bsg x 0.7)
Column side fy,c = 275 N/mm2 m1 = 28.0 mm. Stiffener
Fitted
Asn,act
Fri,Rd Frj,Rd
m2L m2U
0.0
0.0
198.7
40.2
1435
Lwt
Fs,Ed,1
0
Compression stiffener - assume bearing fit to column flange
Untitled1.qj2
Quikjoint - Comprehensive Edition
Asn
Weld
Verdict
(required)
160
Fs,Ed = 1.0 x 360.8 Minimum weld size to column web = 360.8 / (4 x ( 449.8 - 2 x 10.9) x 0.7 x 223) Use 6 fillet weld (Minimum)
Fs,Ed,2
= =
81.7
297.0
6FW
Passed
P398 Step 6B
360.8 kN. 1.4 mm.
Release 8.90
kwc=0.70
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Untitled1.qj2
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Designed : ALEX
Checked : ALEX
EX: EAVES APEX CONNECTION
: EAVE APEX CONNECTION TRIAL
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80
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2077
160 x 20 FL x 755 LG No end plate stiffs M20 Grade 8.8 Bolts All fittings grade S275
Shear rows=2
360
80
80
75
Printed
81.63 degrees
Haunch C/F 356x171x51UB S275 Length=2300 Depth=513
406x140x39UB S275 Rafter pitch=0.000 degrees No web stiffener kwc=0.70
[+20]
80
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8
6
6
8
Combination: Reversal Shear -75.00 kN. Moment -160.00 kN.m. Horizontal 30.00 kN.
457x152x52UB S275 No cap plate Flange stiffs 70 x 10 FL x 140 LG Web stiffs C/F 70 x 10 FL top Web stiffs C/F 70 x 12 FL bot Morris stiff C/F 70 x 10 FL
[+20]
Combination: Dead+IMPOSED Shear 380.00 kN. Moment 330.00 kN.m. Horizontal -75.27 kN.
314
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Designed : ALEX
Checked : ALEX
National Annex: UK National Annex EN 1993-1-1: 2005 General rules and rules for buildings 3.2.1(1)
fy and fu taken from product standard
6.1(1)
Partial factors = 1.000 = 1.000 = 1.100
6.3.2.3(1)
resistance of cross section resistance of members to instability resistance of cross-sections in tension fracture
For rolled sections, hot-finished and cold-formed hollow sections: _ LT,0 = 0.400 = 0.750 For welded sections _ LT,0 = 0.200 = 1.000 Table 6.5 taken from UK National Annex
6.3.3(5)
Determination of interaction factors kyy, kyz, kzy, kzz For doubly symmetrical sections use: Method 1 given in Annex A For mono-symmetrical sections use: Method 2 given in Annex B
EN 1993-1-8: 2005 Design of joints 2.2(2)
Partial factors = = = = ,ser = u =
1.250 1.250 1.250 1.250 1.100 1.100
resistance of bolts resistance of welds resistance of plates in bearing slip resistance at ultimate limit state (Category C) slip resistance at serviceability limit state (Category B) partial factor for tying resistance
EN 1992-1-1: 2004 Design of concrete structures - General rules and rules for buildings C = 1.500 partial factor for concrete S = 1.150 partial factor for reinforcing steel cc = 0.850 is the coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied
ct
= 1.000
is the coefficient taking account of long term effects on the tensile strength and of unfavourable effects, resulting from the way the load is applied
CRd,c
= 0.180/C
as defined in 6.2.2(1)
Untitled1.qj2
EN 1994-1-1: 2004 Design of composite steel and concrete structures - General rules and rules for buildings V = 1.250 partial factor for design shear resistance of a headed stud
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EX: EAVES APEX CONNECTION Designed : ALEX
Checked : ALEX
Untitled1.qj2
EN 1995-1-1: 2004 Design of timber structures - Common rules and rules for buildings = 1.300 Partial factor for material properties - Solid timber. Table 2.3 = 1.250 Partial factor for material properties - Glued laminated timber. Table 2.3 kcr = 0.670 Crack factor for shear resistance - Solid timber. 6.1.7(2) kcr = 0.670 Crack factor for shear resistance - Glued laminated timber. 6.1.7(2)
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Designed : ALEX
Checked : ALEX
WARNING End plate is narrower than the flange width. These calculations assume that the end plate is at least as wide. Summary of results Moment capacity Bolts in shear Tension flange weld Compression flange weld Web weld tension zone Web weld shear zone Rafter web transverse compression Tension stiffeners - column side Stiffener weld Flange stiffeners - shear Flange stiffeners - web shear Flange stiffeners - web weld Web tension - additional check Compression stiffeners Column web fitted Reversal Moment capacity Bolts in shear Tension flange weld Compression flange weld Web weld tension zone Web weld shear zone Tension stiffeners - column side Stiffener weld Compression stiffeners Column web fitted
Unity
Verdict
0.92 0.71 0.77 1.00 0.77 0.89 0.60
Passed Passed Passed Passed Passed Passed Passed
0.59 0.48 0.57 0.50
Passed Passed Passed Passed N/A
Minimum 6 fillet weld
N/A
Minimum 6 fillet weld
0.72 0.11 0.84 0.77 0.89 0.77
Passed Passed Passed Passed Passed Passed
Full strength Full strength Full strength
0.21
Passed
Minimum 6 fillet weld
N/A
Minimum 6 fillet weld
Overall
Comment
Full strength Bearing fit assumed Full strength Full strength
Single sided
Passed
Note: Connection is single sided. = 1.00
5.3(8) and Table 5.4
Detailed calculations in accordance with EN 1993-1-8:2005
Untitled1.qj2
Connection geometry Column side: m = w/2 - twc/2 - 0.8rc e = ( 152.4 - 80)/2 = 40 - 3.8 - 8.2 = 36.2 mm. = 28.0 mm. n = lesser of 36.2, 40.0 and 1.25 x 28.0 = 35.1 mm. Beam side: m = w/2 - twb/2 - 0.8s e = ( 160.0 - 80)/2 = 40 - 3.2 - 4.8 = 40.0 mm. = 32.0 mm. n = lesser of 36.2, 40.0 and 1.25 x 32.0
Quikjoint - Comprehensive Edition
Table 6.2
ew=
8.2 mm.
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Designed : ALEX
Checked : ALEX
= 36.2 mm. Bolt capacity Using Table 3.4 tension resistance of M20 Grade 8.8 bolt Ft,Rd = k2 fub As / M2 where k2 = 0.90 and fub= 800 N/mm2 = 0.90 x 800 x 245.0 / 1.25 = 141.1 kN
See table 3.1
Bolt row 1 Row 1 alone Column flange bending Calculate equivalent effective length Leff of T-stub. leff,cp = 2.Pi.m leff,nc = au.m + al.m - (4m + 1.25e) leff,nc = a.m leff,nc = a.m leff,nc = 4m + 1.25e leff,cp leff,nc leff,1 leff,2
Table 6.4, 6.5, 6.6
= = = = =
= [i] = Max[iv,iii(m2u),iii(m2l),ii]] = Min[leff,cp,leff,nc] = leff,nc
176 mm 174 mm 162 mm 169 mm 157 mm
Pattern(i)
= = = =
176 mm. 174 mm. 174 mm. 174 mm.
Pattern(iv) Pattern(iii) Pattern(iii) Pattern(ii)
Equivalent T-stub resistances Mpl,1,Rd = 0.25 leff,1 t2 fy / = 0.25 x 174 x 10.92 x 275 / 1.00
=
1421 kN.mm.
= 0.25 leff,2 t2 fy / = 0.25 x 174 x 10.92 x 275 / 1.00
=
1421 kN.mm.
Mpl,2,Rd
Table 6.2
Find critical bolt failure mode FT,1,Rd = (8n - 2ew) Mpl,1,Rd / (2mn - ew(m + n))
=
259
kN.
Mode 1
FT,2,Rd
= (2.Mpl,2,Rd+ n Ft,Rd) / (m + n)
=
202
kN.
Mode 2
FT,3,Rd
= Ft,Rd = 2 x 141
=
282
kN.
Mode 3
Column web tension Not applicable stiffener/Flange in tension zone Beam end plate bending Calculate equivalent effective length Leff of T-stub. leff,cp = 2.Pi.m leff,nc = 4m + 1.25e leff,nc = a.m
Untitled1.qj2
leff,cp leff,nc leff,1 leff,2
= [i] = Max[ii,iii] = Min[leff,cp,leff,nc] = leff,nc
Quikjoint - Comprehensive Edition
Table 6.4, 6.5, 6.6
= = =
201 mm 178 mm 187 mm
Pattern(i)
= = = =
201 mm. 187 mm. 187 mm. 187 mm.
Pattern(ii) Pattern(iii)
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Designed : ALEX
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Equivalent T-stub resistances Mpl,1,Rd = 0.25 leff,1 t2 fy / = 0.25 x 187 x 20.02 x 265 / 1.00
=
4956 kN.mm.
= 0.25 leff,2 t2 fy / = 0.25 x 187 x 20.02 x 265 / 1.00
=
4956 kN.mm.
Mpl,2,Rd
:6
Find critical bolt failure mode = (8n - 2ew) Mpl,1,Rd / (2mn - ew(m + n)) FT,1,Rd
Table 6.2
=
771
kN.
Mode 1
FT,2,Rd
= (2.Mpl,2,Rd+ n Ft,Rd) / (m + n)
=
295
kN.
Mode 2
FT,3,Rd
= Ft,Rd = 2 x 141
=
282
kN.
Mode 3
Beam web tension Not applicable stiffener/Flange in tension zone Ft,Rd(row1) = 202 kN. Ft,Rd(row1) = 202 kN.
Column flange mode 2 governs
And so on... Calculations for Moment Connections tend to be very lengthy. Full calculations for all bolt lines are available by selecting Options|Results contents and deselecting Reduce detail in the Results contents form. Ft,Rd(row2) = 201 kN. Ft,Rd(row1 + 2) = 507 kN. Ft,Rd(row2) = 201 kN. Ft,Rd(row3) = 197 kN. Ft,Rd(row2 + 3) = 371 kN. Ft,Rd(row1 + 2 + 3) = 719 kN. Ft,Rd(row3) = 171 kN. Resistance of column web Effective width of web in compression beff,c,wc = 52.7 + 62.6 + 52.7 = 168.1 mm.
Untitled1.qj2
Column flange mode 2 governs
Column flange mode 2 governs Column flange mode 1 governs Column flange mode 1 governs
6.2.6.2(1) (6.10)
where
Reduction factor for shear interaction =1.00 = 1 / (1 + 1.3 ( beff,c,wc . twc / Avc )2 )
Quikjoint - Comprehensive Edition
Column flange mode 2 governs
52.7 = 2.5 x ( 10.9 + 10.2 )
Table 6.3
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Designed : ALEX = 1 / (1 + 1.3 ( 168.1 x
7.6 /
3647 )2 )
=
Checked : ALEX
0.93
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative
6.2.6.2(2)
Resistance of column web in bearing: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 0.93 x 0.70 x 168.1 x 7.6 x 275 / 1.00
(6.9)
(6.14)
=
228.4 kN.
Resistance of: 2 - 70 x 12 web stiffeners fy,web = 275 fy,stiff = 275 Use fy = 275 N/mm2 and = 0.924 Effective width of stiffener beff,stiff = 70 mm.
but less than: 14..tstiff
=
70.0 mm.
Effective area of stiffeners in contact with flange = 2 x 59.8 x 12 = 1435 mm2. Anet Length of web acting with stiffener lw,1 = 15..tw Therefore: lw = 222.8 mm. Aeff,s = 3373 mm2. Ieff,s = 322 x 104 mm4. ieff,s = 30.9 mm.
= 15 x 0.924 x 7.6
= (Lcr / iz) x (1 / x (1 / 93.91) = 407.6 / 30.9 x 1 / 0.924 x 1 / 93.91
From Figure 6.4 using: = = 1.000
Length of web
=
0.152 and buckling curve 'c'
Resistance to transverse compression is lesser of: Fc,Rd,1 = Anet fy / + Fc,wc,Rd,1 = 1435 x 275 x 10-3 / 1.00 + 228.4 Fc,Rd,2
= 105.4 mm.
= Aeff,s fy / = 1.000 x 3373 x 275 x 10-3 / 1.00
0.152 (see Table 6.2)
Governing bearing
=
623.1 kN.
=
927.6 kN.
Governing buckling
Haunch flange and web in compression 6.2.6.7(1) and P398 Step 2 Rafter resistance with coexisting: axial = 75.3 kN. shear = 380.0 kN. Mc,Rd = 456.8 kN.m. QuikEC3 Fc,fb,Rd = Mc,Rd x 103 / (h - tf) = 456.8 x 103 / ( 712.0 - 11.5 ) = 652.1 kN. Height of haunched rafter exceeds 600 mm. 6.2.6.7(1) and SN041a-EN-EU Fc,fb,Rd < tf . bf . fy x 10-3 / 0.8 < 11.5 x 171.5 x 275 x 10-3 / 0.8 = 678.0 kN. Therefore Fc,fb,Rd = 652.1 kN.
Untitled1.qj2
Resistance of column web panel in shear
Quikjoint - Comprehensive Edition
6.2.6.1(2)
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Designed : ALEX Vwp,Rd
Sheet
= 0.9 x fy,wc . Avc / ( 3 x ) = 0.9 x 275 x 3647.3 / ( 3 x 1.00 )
Resistance of column web and Morris stiffener Asn = 2 x bsn . ts = 2 x 70.0 x 10 = 1400 mm2. Vwps,Rd = Asn . fy . Cos() + Vwp,Rd where = 59.0 = 1400 x 275 x 0.515 / 1000 + 521.2
=
521.2 kN.
=
719.5 kN.
=
268.1 kN.
Checked : ALEX
Plastic distribution of bolt forces Row Force hr kN. mm. Ft,Rd(row1) 201.9 651 Ft,Rd(row2) 200.6 571 170.7 491 Ft,Rd(row3) Ft,Rd(row) = 573.1 kN. Modification of bolt row forces 1.9 x Ft,Rd = 1.9 x 141.1 Plastic distribution valid Axial force NEd > 0.05 x Npl,Rd
6.2.3(2)
NEd will be accounted for using the methods outlined in P398 Step 4.
Equilibrium Ft,Rd(row) = 573.1 kN. = 623.1 kN. Lesser of: 623.1, 652.1, Fc,Rd NEd = -75.3 kN. Negative being compression 547.8 kN. Fc,Rd + NEd = Ft,Rd(row)
> Fc,Rd + NEd
6.2.7.2(9)
719.5 /
Surplus capacity in bolts. Reduce starting at bottom row and working up until equilibrium holds
Moment capacity Row
Ftr,Rd hr Ftr,Rd x hr kN. mm. kN.m. Ft,Rd(row1) 201.9 651 131.4 Ft,Rd(row2) 200.6 571 114.6 Ft,Rd(row3) 145.4 491 71.4 Ft,Rd(row) = 547.8 Mj,Rd = 317.4
Untitled1.qj2
Modify applied moment for effect of axial load MEd,mod = MEd + NEd x hN Where MEd = Applied moment NEd = Axial tension hN = Distance from centre line of beam to centre of compression
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Designed : ALEX = Mj,Rd
> MEd,mod
Applied shear VEd =
330.0 -
:9
75.3 x 0.507
=
Checked : ALEX
291.8 kN.m.
Pass 380.0 kN.
Bolts in shear Using Table 3.4 shear resistance of M20 Grade 8.8 bolt Fv,Rd = v fub A / M2 where v = 0.60 and fub= 800 N/mm2 = 0.60 x 800 x 245.0 / 1.25 = 94.1 kN. Residual shear resistance of bolt fully loaded in tension Fvs,Rd = (1 - 1/1.4) x Fv,Rd = 26.9 kN. Bolt bearing end plate Fb,Rd,p = 1.0 x k1.b.fu.d.t /
See table 3.1
Table 3.4
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 40
= 315.6 kN. Bolt bearing column Fb,Rd,c = 1.0 x k1.b.fu.d.t /
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 36
= 172.0 kN. Shear capacity of bolt group Fv,Rd = Lesser of: Fv,Rd, Fb,Rd,p and Fb,Rd,c Fvs,Rd = Lesser of: Fvs,Rd, Fb,Rd,p and Fb,Rd,c
= =
94.1 kN. 26.9 kN.
VRd,b
=
537.6 kN.
VRd,b
= 4 x Fv,Rd + 6 x Fvs,Rd > VEd
Pass
Calculate thickness factor for full strength double fillet welds SN014a, TC10, P358 Annex C fy . w . weld 275 x 0.85 x 1.25 from Table 4.1 a / t,ply = = 0.504 . fu . 2 1.00 x 410 x 2 w
Tension flange weld Size of full strength fillet weld s = 8.6 x 0.504 / 0.7 Pass Tension flange weld is full strength
=
Design shear strength of end plate weld fu / 3 410 / 3 fvw,d = = 222.8 N/mm2. w 0.85 x 1.25
6.2 mm.
w
from Table 4.1
(4.4)
Untitled1.qj2
Compression flange weld
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Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX
Checked : ALEX
Fc,Rd Capacity of
= 623.1 kN. 8 fillet weld to compression flange = 2 x 171.5 x 0.7 x 8 x 222.8 = 427.9 kN. Pass Compression flange weld can not transmit full compressive force Provide a bearing fit with nominal 6 fillet weld Web welds in tension zone Size of full strength fillet weld s = 6.4 x 0.504 / 0.7 Pass Web weld in tension zone is full strength. Weld must continue past the last tension bolt a minimum of 1.73W/2 = 69 mm.
=
4.6 mm.
Web welds shear zone Pass Web weld to shear zone is full strength Rafter web in transverse compression Effective width of web in compression beff,c,wc = 47.0 + 76.1 + 47.0 where = 170.1 mm.
6.2.6.2(1) (6.10)
47.0 = 2.5 x ( 8.6 + 10.2 )
Reduction factor for shear interaction =0.00 = 1.00
Table 6.3
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative Reduction factor for plate buckling p = 0.932 (( beff,c,wc . dwc . fy,wc ) / ( E . tw 2)) p> 0.72 therefore = ( p - 0.2) / p2 Resistance to transverse compression is lesser of: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 1.00 x 0.70 x 170.1 x 6.4 x 275 / 1.00 Fc,wc,Rd,2
6.2.6.2(2) (6.14) (6.13c)
=
1.30
=
0.65
(6.13b)
(6.9) Governing web bearing
=
209.6 kN.
= . kwc . . beff,c,wc . twc . fy,wc / M1 = 1.00 x 0.70 x 0.65 x 170.1 x 6.4 x 275 /1.00 =
136.0 kN.
Force in haunch flange Fc,fh,Ed = Fc,Rd / Cos( 8.69) = 630.3 kN But less than capacity 171.5 x 11.5 x 275 / = 542.4 kN. Force normal to rafter Fc,wb,Ed = 542.4 x Sin( 8.69 - 0.00) Fc,wb,Rd,2 > Fc,wb,Ed
=
81.9 kN
Pass
Tension stiffeners - Column side Force in stiffener Fs,Ed is greater of: Fs,Ed,1 and Fs,Ed,2 Web tension: Flange bending
Untitled1.qj2
Governing web buckling
Quikjoint - Comprehensive Edition
P398 Check 6A Note: Fs,Ed is for a pair of stiffeners
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Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX
Fs,Ed,1 > (Fri,Rd + Frj,Rd) - ( Lwt x twc x fy,c) / Asn
: 11
Checked : ALEX
Fri,Rd Frj,Rd ( + m1 + m2U m1 + m2L
Fs,Ed,2 > m1 x
= Fs,Ed / ( fy,s / )
Weld size
= Fs,Ed / (4 x fvw,d x bsg x 0.7)
Column side fy,c = 275 N/mm2 m1 = 28.0 mm. Stiffener
Asn,act
Fitted
Fri,Rd Frj,Rd
m2L m2U
0.0
0.0
201.9 201.9
40.9 30.2
200.6
30.2
1196
Rib
1196
Lwt
Fs,Ed,1
> Fs,Ed
> Fs,Ed
> Fw,Ed
Verdict
0
82.1
298.5
6FW
Passed
189
7
193.7
704.5
6FW
Passed
P398 Step 6A
=
400.1 kN.
Pass P398 Step 6A
337.9 kN.
Pass
Partial depth stiffener - Web weld Lw = 140 - 6 L1 = ( 80 - 7.6) /2 L2 = 2/3 x Lw Ft = 0.5 x Fs,Ed x L1/L2 t = 2 Ft / Lw s = Fs,Ed / (2 x Lw) Fw,Ed = ( s2 + t2 ) / 2 Fw,Rd = 0.7 x 6 x 222.8 Fw,Rd
Weld
115
Partial depth stiffener - Check shear capacity of column web Fv,Rd = 2 x Ls x tw x fy / ( 3 x ) = 2 x 140 x 7.6 x 275 / ( 3 x 1.00 ) = Fv,Rd
Asn (required)
Partial depth stiffener - Check shear capacity of stiffener Fv,Rd = 2 x 0.9 x Ls x ts x fy,s / ( 3 x ) = 2 x 0.9 x 140 x 10 x 275 / ( 3 x 1.00 ) Fv,Rd
Fs,Ed,2
P398 Figure 2.18
= = = = = =
134.0 mm. 36.2 mm. 89.3 mm. 39255.1 N 585.9 N/mm. 722.9 N/mm.
Weld length Lever arm Lever arm Total tension Weld tension Weld shear
= =
465.3 N/mm. 935.7 N/mm.
Pass
Compression stiffener - assume bearing fit to column flange Fs,Ed = 1.0 x 623.1 Minimum weld size to column web = 623.1 / (4 x ( 449.8 - 2 x 10.9) x 0.7 x 223) Use 6 fillet weld (Minimum)
= =
P398 Step 6B
623.1 kN. 2.3 mm.
Untitled1.qj2
Reversal
Quikjoint - Comprehensive Edition
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Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX
Bolt capacity Using Table 3.4 tension resistance of M20 Grade 8.8 bolt Ft,Rd = k2 fub As / M2 where k2 = 0.90 and fub= 800 N/mm2 = 0.90 x 800 x 245.0 / 1.25 = 141.1 kN Resistance of column web Effective width of web in compression beff,c,wc = 0.0 + 56.6 + 52.7 = 109.3 mm.
Checked : ALEX
See table 3.1
6.2.6.2(1) (6.10)
where
52.7 = 2.5 x ( 10.9 + 10.2 )
Reduction factor for shear interaction =1.00 = 1 / (1 + 1.3 ( beff,c,wc . twc / Avc )2 ) = 1 / (1 + 1.3 ( 109.3 x 7.6 / 3647 )2 )
Table 6.3
=
0.97
Reduction factor for longitudinal compressive stress kwc = 0.7 Conservative
6.2.6.2(2)
Resistance of column web in bearing: Fc,wc,Rd,1 = . kwc . beff,c,wc . twc . fy,wc / M0 = 0.97 x 0.70 x 109.3 x 7.6 x 275 / 1.00
(6.9)
(6.14)
=
154.8 kN.
Resistance of: 2 - 70 x 10 web stiffeners fy,web = 275 fy,stiff = 275 Use fy = 275 N/mm2 and = 0.924 Effective width of stiffener beff,stiff = 70 mm.
but less than: 14..tstiff
=
70.0 mm.
Effective area of stiffeners in contact with flange Anet = 2 x 59.8 x 10 = 1196 mm2. Length of web acting with stiffener lw,1 = 15..tw Therefore: lw = 138.7 mm. Aeff,s = 2454 mm2. Ieff,s = 268 x 104 mm4. ieff,s = 33.1 mm.
= 15 x 0.924 x 7.6
= (Lcr / iz) x (1 / x (1 / 93.91) = 407.6 / 33.1 x 1 / 0.924 x 1 / 93.91
From Figure 6.4 using: = = 1.000
Untitled1.qj2
Length of web
=
0.142 and buckling curve 'c'
Resistance to transverse compression is lesser of: Fc,Rd,1 = Anet fy / + Fc,wc,Rd,1 = 1196 x 275 x 10-3 / 1.00 + 154.8 Fc,Rd,2
= 105.4 mm.
= Aeff,s fy / = 1.000 x 2454 x 275 x 10-3 / 1.00
Quikjoint - Comprehensive Edition
0.142 (see Table 6.2)
Governing bearing
=
483.7 kN.
=
674.8 kN.
Governing buckling
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Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX
Rafter flange and web in compression Rafter resistance with coexisting: axial = 30.0 kN. shear = 75.0 kN. Mc,Rd = 456.8 kN.m. Fc,fb,Rd = Mc,Rd x 103 / (h - tf) = 456.8 x 103 / ( 712.0 - 8.6 ) = Height of haunched rafter exceeds 600 mm. Fc,fb,Rd < tf . bf . fy x 10-3 / 0.8 < 8.6 x 141.8 x 275 x 10-3 / 0.8 = Therefore Fc,fb,Rd =
Checked : ALEX
6.2.6.7(1) and P398 Step 2 QuikEC3
649.4 kN. 6.2.6.7(1) and SN041a-EN-EU
419.2 kN. 419.2 kN.
Resistance of column web panel in shear Vwp,Rd = 0.9 x fy,wc . Avc / ( 3 x ) = 0.9 x 275 x 3647.3 / ( 3 x 1.00 )
=
521.2 kN.
Resistance of column web and Morris stiffener Asn = 2 x bsn . ts = 2 x 70.0 x 10 = 1400 mm2. Vwps,Rd = Asn . fy . Cos() + Vwp,Rd where = 42.9 = 1400 x 275 x 0.732 / 1000 + 521.2
=
803.1 kN.
=
268.1 kN.
6.2.6.1(2)
Plastic distribution of bolt forces Row Force hr kN. mm. 198.7 651 Ft,Rd(row1) Ft,Rd(row2) 162.1 571 = 360.8 kN. Ft,Rd(row) Modification of bolt row forces 1.9 x Ft,Rd = 1.9 x 141.1 Plastic distribution valid Axial force NEd < 0.05 x Npl,Rd
6.2.3(2)
NEd may be ignored
Equilibrium Ft,Rd(row) = 360.8 kN. Fc,Rd = 419.2 kN. Lesser of: 483.7, 419.2, Ft,Rd(row)
< Fc,Rd
6.2.7.2(9)
803.1 /
Bolt capacity governs
Moment capacity Row
Untitled1.qj2
Ftr,Rd hr Ftr,Rd x hr kN. mm. kN.m. Ft,Rd(row1) 198.7 651 129.3 Ft,Rd(row2) 162.1 571 92.5 Ft,Rd(row) = 360.8 Mj,Rd = 221.8
Quikjoint - Comprehensive Edition
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: 09/03/18
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: 14
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: Untitled1
Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX
Checked : ALEX
Applied moment MEd = 160.0 kN. Mj,Rd
> MEd
Pass
Applied shear = VEd
75.0 kN.
Bolts in shear Using Table 3.4 shear resistance of M20 Grade 8.8 bolt Fv,Rd = v fub A / M2 where v = 0.60 and fub= 800 N/mm2 = 0.60 x 800 x 245.0 / 1.25 = 94.1 kN. Residual shear resistance of bolt fully loaded in tension Fvs,Rd = (1 - 1/1.4) x Fv,Rd = 26.9 kN. Bolt bearing end plate Fb,Rd,p = 1.0 x k1.b.fu.d.t /
See table 3.1
Table 3.4
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 40
= 315.6 kN. Bolt bearing column Fb,Rd,c = 1.0 x k1.b.fu.d.t /
d0= 22 fub/fu = 800 / 410 = 1.95 See Table 3.4 and 3.7(1) d = p1 /(3.d0) - 0.25 = 0.96 but < 1.00 p1= 80 k1 = 2.8 e2 /d0 - 1.7 = 2.50 but < 2.50 e2= 36
= 172.0 kN. Shear capacity of bolt group Fv,Rd = Lesser of: Fv,Rd, Fb,Rd,p and Fb,Rd,c = Lesser of: Fvs,Rd, Fb,Rd,p and Fb,Rd,c Fvs,Rd
= =
94.1 kN. 26.9 kN.
VRd,b
=
672.0 kN.
VRd,b
= 6 x Fv,Rd + 4 x Fvs,Rd > VEd
Pass
Calculate thickness factor for full strength double fillet welds SN014a, TC10, P358 Annex C fy . w . weld 275 x 0.85 x 1.25 from Table 4.1 a / t,ply = = 0.504 . fu . 2 1.00 x 410 x 2 w
Design shear strength of end plate weld fu / 3 410 / 3 fvw,d = = 222.8 N/mm2. w 0.85 x 1.25
Untitled1.qj2
Tension flange weld Tension capacity of the beam flange Npl,Rd = b . tf . fy / = 171.5 x 11.5 x 275 / 1.00
Quikjoint - Comprehensive Edition
w
from Table 4.1
(4.4)
SN014a Annex A
=
542.4 kN
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Title
: EAVE APEX CONNECTION TRIAL EX: EAVES APEX CONNECTION
email: [email protected]
Designed : ALEX Total force in the top 2 bolt rows Ftr,Rd = 198.7 + 162.1 Capacity of Pass
: 15
8 fillet weld to tension flange = 2 x 171.5 x 0.7 x 8 x 222.8 Tension flange weld adequate
Compression flange weld Size of full strength fillet weld s = 8.6 x 0.504 / 0.7 Pass Compression flange weld is full strength Web welds in tension zone Size of full strength fillet weld s = 7.4 x 0.504 / 0.7 Pass Web weld in tension zone is full strength. Weld must continue past the last tension bolt a minimum of 1.73W/2 = 69 mm.
Checked : ALEX
=
360.8 kN.
=
427.9 kN.
=
6.2 mm.
=
5.3 mm.
Web welds shear zone Pass Web weld to shear zone is full strength Tension stiffeners - Column side Force in stiffener Fs,Ed is greater of: Fs,Ed,1 and Fs,Ed,2 Web tension: Flange bending Fs,Ed,1 > (Fri,Rd + Frj,Rd) - ( Lwt x twc x fy,c) / Asn
P398 Check 6A Note: Fs,Ed is for a pair of stiffeners
Frj,Rd Fri,Rd ( + m1 + m2U m1 + m2L
Fs,Ed,2 > m1 x
= Fs,Ed / ( fy,s / )
Weld size
= Fs,Ed / (4 x fvw,d x bsg x 0.7)
Column side fy,c = 275 N/mm2 m1 = 28.0 mm. Stiffener
Fitted
Asn,act
Fri,Rd Frj,Rd
m2L m2U
0.0
0.0
198.7
40.2
1435
Lwt
Fs,Ed,1
0
Compression stiffener - assume bearing fit to column flange
Untitled1.qj2
Quikjoint - Comprehensive Edition
Asn
Weld
Verdict
(required)
160
Fs,Ed = 1.0 x 360.8 Minimum weld size to column web = 360.8 / (4 x ( 449.8 - 2 x 10.9) x 0.7 x 223) Use 6 fillet weld (Minimum)
Fs,Ed,2
= =
81.7
297.0
6FW
Passed
P398 Step 6B
360.8 kN. 1.4 mm.
Release 8.90
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