short course worked example
TRANSCRIPT
Helsinki University of Technology Laboratory of Steel Structures Publications 33
Teknillisen korkeakoulun teräsrakennetekniikan laboratorion julkaisuja 31
Espoo 2007 TKK-TER-33
DESIGN OF STRUCTURAL CONNECTIONS TO EUROCODE Worked Examples
F. Wald
AB TEKNILLINEN KORKEAKOULUTEKNISKA HÖGSKOLANHELSINKI UNIVERSITY OF TECHNOLOGYTECHNISCHE UNIVERSITÄT HELSINKIUNIVERSITE DE TECHNOLOGIE D’HELSINKI
1
List of Contents
page
1...Welded connection of tension member ........................................................................... 3
2...Welds of a fin plate connection ........................................................................................ 4
3...Welded truss connection ................................................................................................. 6
4...Welded connection of T cantilever ................................................................................... 8
5...Bolted connection of tension member ............................................................................. 10
6...Bolted truss connection .................................................................................................... 11
7...Bolted slip resistant connection ....................................................................................... 13
8...Bolted long connection ..................................................................................................... 14
9...Bolted connection with packing ....................................................................................... 15
10..Single lap connection with one bolt ................................................................................. 17
11..Bolted beam splices ......................................................................................................... 18
12..Beam-to-column welded joint .......................................................................................... 18
13..Pin connection ................................................................................................................. 21
14..Simple column base ......................................................................................................... 25
15..Fixed column base ........................................................................................................... 27
16..Fin plate connection ......................................................................................................... 30
17..Header plate connection .................................................................................................. 32
18..Bolted extended end plate connestion ............................................................................. 34
19..Extended end plate semirigid joint ................................................................................... 34
2
1 Welded Connection of Tension Member Check the resistance of the connection of the flat section, shown in Fig. 1.1, loaded in tension by the factored force FSd = 330 kN. The steel is Grade S460N. The material partial safety factors are
10,10M =γ and 2512 ,M =γ .
80
170
330 kN
10
6
3
33
Fig. 1.1 Drawing of the connection _____________________________________________________________________________________________________ The structural welds should be (i) longer than 40 mm, and (ii) longer than 6 aw = 6 ∗ 3 = 18 mm. Both of these are satisfied. The full length of the weld can be taken into account in the strength calculation, because mmmmaw 170450350150 >=∗= .
IIσaw
1,4 aw
τI_
I_ τ
Fig. 1.2
τII
FSd
τIILwIIawII
Fig. 1.3
Longitudinal welds In the longitudinal welds is . Based on the fillet welds resist
σ τ⊥ ⊥= = 0ance
( )2
222 3Mw
u//
fγβ
ττσ ≤++ ⊥⊥ and 2M
ufγ
σ ≤⊥
is the shear strain
23 Mw
uRd.II
fγβ
τ = .
The design resistance is
N,,,
LaF wIIwIIRd.IIRd.w.II310125917023
2510135502 ∗=∗∗∗
∗∗== τ
. Front weld The equation for the resistance may be at the front weld
( 0=Rd.IIτ and 2
) rewritten:
wστσ == ⊥⊥
2
22
23
2 Mw
uww fγβ
σσ≤⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ .
he front weld design strain is
T
2u
Rd.wf
γβσ = .
2Mw
FSd
σ
τ
wσLw
aw
Fig. 1.4
The design resistance of the front weld is
3
N,,,
Lw310774803
225101550
∗=∗∗∗∗
=⊥
onnehe connection resistance is
FkN,,,FFF SdRd.w.Rd.w.IIRd.w 83337741259
aF wRd.wRd.w. = ⊥⊥ σ
CT
ction resistance
kN330== >+=+= ⊥
.
The connection resistance is satisfactory.
170
3
33
P 6 - 300 x 800
80 x 10 - 1 500
Scale: 1 : 10Steel S460N J2
Fig. 1.5 Design drawing of the connection _____________________________________________________________________________________________________ ote: N
1) The llows:
weld resistance may conservatively be checked independent of the loading direction as fo
kNFN, Sd 3301000 3 =<∗ . ,,
)(LafF
Mw
wwuRd.w 32
3251018017023550
32
=∗∗
+∗∗∗==
γβ
The welds are not satisfactory uder this model.
)
2 The tension resistance of a member is
kNN,,
fAN
M
yRd.u 330105334
1014601080 3
0
>∗=∗∗
==γ
.
The tension resistance is satisfactory.
e = 60 mm. The steel is Grade S235, and the material partial safety factors
2 Welds of a Fin Plate Connection Check the resistance of the fillet-welded connection of the fin plate, shown in Fig. 5.2.1. The connection is subject to the vertical factored force VSd = 250 kN acting at an eccentricity ,
V = 250 kNSd
e = 60
4L = 300 t =15p
are 1510 ,M = and 5012 ,M . γ γ =
Fig. 2.1 Drawing of the connection
The struc 4tural welds should be (i) longer than 0 mm, and (ii) longer than 6 aof these are satisfied. The full length of the weld can be take
u = 6 ∗ 4 = 24 mm. Both n into account in the strength calculation,
ecause mmmmaw 300600450150 >b =∗= . The shear stress perpendicular to the weld cross-section is
4
MPa,La
VSdII 2104
3002410250
2
3
=∗∗∗
==τ . w
al stress parallel to the weld cross-section, based on an elastic distribution of ending stresses is
The maximum normb
MPa,La
eVW
M
w
Sd
w.elw 0125
300426010250
2 2
3
2 =∗∗
∗∗===σ ,
66
he shear across the critical plane (the weld throat) nd the normal stress perpendicular to this plane:
which may be decomposed (see Fig. 5.2.2) into ta
MPa,w 488125==== ⊥⊥
σστ . 22
V
τ
Sd
II
e
L
σ
σ
τ
ττ II σI_ I_
104,2 88,4 88,4
w
⊥
⊥
Fig. 2.2
Check of the weld design resistance:
( ) ( ) MPa,,,
fMPa,,,, u// 0300
51803606252210448834883 222222 =
∗=<=+∗+=++ ⊥
Mww γβττσ ,
⊥
and
MPa240,
fMPa, u ==<=⊥ 501360488
γσ .
M 2
The weld strength is satisfactory.
4
P 15 - 90 x 300
Scale: 1 :10
Steel S235 J2
Fig. 2.3. Design drawing of the connection
Note: 1) The weld resistance may conservatively be checked independent of the loading direction as
follows:
MPa,,,
fMPa,,, uIIw 2173
35180360
3716221040125 2222 =
∗∗=<=+=+
γβτσ .
Mww
2) The plate’s resistance in shear is
kNVN,
fAV Sd
yvRd.pl 25010531
315123530015
33 =>∗=
∗
∗∗=
∗=
γ.
M0
5
and in bending:
15,1/2356
/fW 0Myel ∗== γ 30015M2
Rd.c∗
shear need not be checked, because the shear resistance is more than ouble
531 10 2 266 10 2503 3∗ = ∗ >/ N kN .
of the welds may be performed, based on the expression
Nmm1015=6010250=M>Nmm 100, 63Sd
6 ∗∗∗∗= 46 . The interaction of bending and d the shear force acting: 3) The elastic distribution of stresses in the welds is used because the above is an elastic checkthe fin-plate connection. A plastic check of
4
Welded Truss Connection
welds, shown in Fig. 3.1. The steel is Grade S275, and the material partial safety factors are
La2
eVW
M2
w
Sd
w.plw ==σ .
3 Connect the tension member of the double equal-leg angle section 2 L 50x5 to the plate P 10 by
1,10=M0γ , 251,= .
Sd
2 L 50 x 5
3
3
F
P10 - 100100
2Mγ
Fig. 3.1 Drawing of the connection
____________________________________________________________ ________________________________________
he tension member resistance (area of a angle A = 480 mm2) is
_ _ T
N,,
fAN y
Rd.pl3100240
1012754802
∗=∗∗
==γ
. M0
r than 3 mm, and thinner an
Weld at connected leg At the connected flange of the angle may be provided as a fillet weld thicke
3 2 4 2 5, mm . The the force in the weld by the connected leg is ∗ = <th
b2
.
he weld shear resistance is limited by
eNF Rd.pl
b.w =
T
32Mwbb.w
ub.wb.w
fLa
Fγβ
τ ≤= ,
b e
Fig. 3.2
6
which can be rewritten as
mm,,,,afb
eNL
b.wu
MwRd.plb 947
34303251850
50014
2102403
2
32 =
∗∗∗
∗∗∗
==γβ
Provide Lb = 50 mm. ab
1,41 aw.b
Fig. 3.3 Weld at free leg The force in the weld by the free leg
kN,)(b
)eb(NF Rd.pl
a.w 48650
14502
2402
=−
∗=−
= .
the length may be estimated as mm150a/aL5,2L abby ≈≅ . The shear stress in the weld
MPa,,La
F
aa.w
a.wa.II 0192
150310486 3
=∗
∗==τ ,
The bending moment due to the eccentricity of weld is
σ
τ
τ
⊥.α
⊥.α
//.a
τ //.b
a
b
Fig. 3.4
kNm,,b/FM Sda.w 51
8050240
222
=∗
=∗= .
The bending stress is
MPa,,La
M
aa.w
a.wa.a. 394
1503261051
6
12 2
6
2 =∗∗
∗∗=== ⊥⊥ στ .
The check of stresses:
( ) ( ) =++=++ ⊥⊥222222 019239433943 ,,,a.IIa.a. ττσ
MPa,,
fMPa,Mw
u 404251850
43033822
=∗
=<=γβ
MPa3,
fMPa,M
u 36251
4207672
==<=⊥ γσ .
The weld strength is satisfactory. The welds are acceptable from the structural point of view: longer than 40 mm and than 6 a = 6∗3 = 18 mm, shorter than mma 4503150150 =∗= . (No reduction due to the long weld is needed).
7
Scale: 1 :10Steel S 275 J22 L 50 x 5 - 1 500
3 50
3 150
P10 - 100 x 200
150
25
Fig. 3.5 Design drawing of the connection
_________________________________________________________________________________ Note: The gusset plate in tension is
kN6,229NN1025010,1
27510100f
AN Rd.pl3
0M
yRd.t =>∗=∗∗==
γ,
which is satisfactory.
4 Welded connection of T cantilever Check the resistance of the welded connection of the cantilever of the T section, shown in Fig. 4.1. The plate is P 10. The column is IPE 300. The connection is subject to the factored force FSd = 25 kN. The steel is Grade S275. The weld is designed 3 mm. The partial safety factor is 50,1Mw =γ .
F = 25 kNSd
300
5P 10 - 350 x 120
P 10 - 350 x 110
IPE 300
3
Fig. 4.1 Drawing of the connection ______________________________________________________________________________________________________ The effective area of the column flange is calculated based on the effective width:
mm112117112
275275
107,1071521,7
7,1071521,7
ff
tt
7r2t
t7r2t
minb 2
yp
yf
p
2f
w
fw
eff =⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
∗+∗+
∗+∗+=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
++
++
= .
The effective width is shorter compare to the width of the column flange, see Fig. 5.16.2 ceff b150112b =<= , but longer than 0,7 times the width of the column flange mm1051507,0112beff =∗>= . Hence the column need not to be stiffen in tension and compression.
b
r =15
t =7,1wc
t =10,7fc
eff150
Fig. 4.2
8
The cross-section of the filled weld:
( ) 2we.eff mm10701011251125A =−+∗= ,
2
we.v mm105021055A =∗∗= ,
112
10584,2
5
5
e =we
10
23,3
31,7
38,3A
A v.we
eff.we
ττ IIσI_
23,8 144,9
I_
Fig. 4.3
( ) mm2,8410501070
2/105251051105101121205112ewe =+
∗∗∗+∗∗−+∗∗= ,
( ) 432223
we mm10299851123,385101123,2351057,31212/51102 ∗=∗∗+∗−∗+∗∗∗+∗∗=I , . 333
we mm106,352,84/102998W ∗=∗= The shear and normal stress: MPa8,231050/25000A/F w.vSdII ===τ ,
MPa0,1492106,35
105,72W
M3
6
w
Sd =∗
∗=== ⊥⊥ τσ .
The resistance of the filled weld is
( ) ( ) MPa3,3375,185,0
430fMPa8,3008,230,14930,1493
Mww
u2222II
22 =∗
=<=++=++ ⊥⊥ γβττσ .
The welded connection is satisfactory.
300
5P 10 - 350 x 120P 10 - 350 x 110
IPE 300 - 3300
2 P 8 - 110 x 503
250
Scale: 1 : 10Steel S 275 J2
3
Fig. 4.4 Design drawing of the connection _________________________________________________________________________________ Note: The resistance of the web-to-flange filled welds may be checked as:
The section moment of the area of the connected part (flange) is
33fbeff mm103,337,2910112ztbS ∗=∗∗== .
9
112
11085,3
10
10
3
z = 29,7
The shear stress is
MPa7,4432103107
103,331025a2ISV
3
33
w
SdSd.II =
∗∗∗
∗∗∗==τ .
The shear resistance of the web-to-flange welds ( ): 0== ⊥⊥ τσ
( )Mww
u2II
22 f3
γβττσ ≤++ ⊥⊥ ,
MPa7,44MPa7,194350,185,0
4303
fSd.II
Mww
uRd.II =>=
∗∗== τ
γβτ .
The weld strength is satisfactory.
10
5 Bolted connection of tension member Check the resistance of the bolted connection of a tie, shown in Fig. 5.1, loaded in tension by the factored force FSd =130 kN. The steel is Grade S355. The bolts M16, Grade 5.6, are the not preloaded. The shear plane passes through the unthreaded portion of the bolts. The material partial safety factors are γ M 2 =1,30 and γ Mb =1 45, .
27,5
55
27,5
40
130 000 N
8
5
e1 e1p1
e2
e2
p2
27,5 27,5
Fig. 5.1 Drawing of the connection
_____________________________________________________________________________________________________ The bolt spacing is satisfying the design rules 20 e5,25175,1d5,1 ≤=∗= . 20 p0,51170,3d0,3 ≤=∗=
The force per bolt is N500324000130
F Sd.v == .
The shear resistance per a bolt M 16 (at one shear plane) is
N50032N106,4145,1
4165006,0Af6,0F 3
2
Mb
ubRd,v >∗=
∗∗∗
==
π
γ.
Fig. 5.2 The factor of the bearing resistance is
539,01735,27
d3e
0
1 =∗
==α (limit),
73,041
17350
41
d3p
0
1 =−∗
=−=α ,
98,0510500
ff
u
ub ===α ,
α =1,0 .
p 1e1
d
d
0
Fig. 5.3
The bearing resistance per bolt in the plate 5 mm is
kN5,32FN109,3745,1
516510539,05,2tdf5,2F Sd.v
3
Mb
uRd,b =>∗=
∗∗∗∗==
γα
.
The resistance of the member net section is
kN130N102,13430,1
510)172110(59,0
fA9,0N 3
2M
unetRd.u >∗=
∗∗−∗==
γ.
The connection resistance is satisfactory.
11
27,5
55
40 150 x 8 - 3 200
P 5 - 320 x 6006 17∅
4 M 16
Scale: 1 : 10
Steel S 355 J2Bolts 5.6
6 17∅
27,5
27,527,5
Fig. 5.4 The design drawing of the connection 6 Bolted Truss Connection Check the bolted connection of the tension member of two equal-leg angles, shown in Fig. 6.1. The connection is subject to the factored force FSd = 135 kN. Plate is P 15. The steel Grade S275. The fully threaded bolts M 20, Grade 8.8, are in holes 22 mm. The material partial safety factors are γ M0 =1,15 , γ M 2 =1,30 and γ Mb =1 45, .
SdF = 135 kN
10
50 4070
2 L 90 x 6
110
15
Fig. 6.1 Drawing of the connection
_____________________________________________________________________________________________________ Shear resistance The design shear resistance per two bolts with two shear plane, if the shear is not passing through threaded part of the bolt, is
kN135N104,32445,1
2458006,022Af6,0
n2F 3
Mb
subRd,v >∗=
∗∗∗∗==
γ.
Fig. 6.2
Bearing resistance The bolt bearing resistance of the plate:
61,0223
40d3
e
0
1 =∗
==α (limit)
81,041
22370
41
d3p
0
1 =−∗
=−=α
86,1430800
ff
u
ub ===α
α =1,0
d d0
p 1 e1
4070
Fig. 6.3
12
N107,13545,1
152043061,05,2tdf5,2F 3
Mb
uRd.b ∗=
∗∗∗∗==
γα kN135F Sd.v => .
The bolts shear resistance is satisfactory.
The bolt bearing resistance of the angles:
76,0223
50d3
e
0
1 =∗
==α (limit)
81,041
22370
41
d3p
0
1 =−∗
=−=α
86,1430800
ff
u
ub ===α
α = 1,0
p 1e1
d d0
50 70
Fig. 6.4
N102,13545,1
622043076,05,2tdf5,2F 3
Mb
uRd.b ∗=
∗∗∗∗∗==
γα
kN135F Sd.v => .
The bolts resistance is satisfactory.
Section resistance The reduction factors for the angles connected by one leg (linear transition) is
( ) ( ) 482,0225,270225,2225
3,04,0d5,2pd5,2d5
4,07,04,0 0100
2 =∗−∗−∗
+=−−−
+=β .
The resistance at the net section with two bolts in the force direction is
( ) kN2/135N104,14630,1
4306221050482,0fAN 3
2M
unet2Rd.u >∗=
∗∗−∗==
γβ
.
The member resistance in tension is
N102,50215,1
27510502fAN 3
0M
yRd.pl ∗=
∗∗==
γkN135F Sd.v => .
The net area of the connection plate is
( ) kN135N100,28030,1
27522120159,0fA9,0N 3
2M
ynetRd.pl >∗=
∗−∗∗==
γ.
The section resistance in connection is satisfactory.
Scale: 1 : 10
Steel S 275 J2Bolts 8.8
2 L 90 x 6 - 1200
P15 -120 - 220
10
50 4070
2 22
2 M 20
∅
Fig. 6.5 The design drawing of the connection
13
________________________________________________________________________________________________________
Note: 1) The eccentricity of the fasteners at the end connections and the effects of the spacing and edge
distances of the bolts may be taken into account by analyss. The design resistance of the net section is in this case checked directly.
2) The bolt end distances are in the connections often the same (e.g. 50 + 70 + 50). In this case the bearing resistance is checked for both sides of the connection together, e.g. just for the sum of the thinner plates acting in one direction.
7 Bolted Slip Resistant Connection Evaluate resistance of the bolted connection of a tie, shown in Fig. 7.1, with the slip on the ultimate limit state. The steel is Grade S235. The bolts M 20, Grade 10.9, are preloaded (the bolt net area As = 245 mm2) in holes ∅ 22. The material partial safety factors are γ γMs ult. =1,30 , M 2 =1,30 and γ Mb =1 45, .
557055
505050
FRd
8 x M 20 -10.9
P 16 - 180 x 1200
816
F
2 P 8 - 180 x 340507070
Rd
e1p1e1
p2
e2
e2
Fig. 7.1 Drawing of the connection
_____________________________________________________________________________________________________ Slip resistance The design preloading force (net section of the bolt As = 245 mm2) is . N105,17124500017,0Af7,0F 3
subCd.p ∗=∗∗== The slip factor is for surface blasted with shot μ = 0,5 (Class A surface). There are two slip surfaces, as shown in Fig. 5.6.2. The design slip resistance of one bolt under the ultimate limit state (Connection Category C):
kN9,1315,17130,1
50,020,1Fnk
F Cd,pult.Ms
sRd.s =
∗∗==
γ
μ.
F
Fs.Rd
p.Cd
Fig. 7.2
Bearing resistance The bearing resistance is derived for the plate 16 mm:
758,0223
50d3
e
0
1 =∗
==α (limit
811,041
22370
41
d3p
0
1 =−∗
=−=α ,
778,23600001
ff
u
ub ===α ,
508,0=α ,
Fb.Sd
bearing
Fig. 7.3
N 10150,6 3∗=∗∗∗∗
==45,1
1620360758,05,2tdf5,2F
Mb
uRd.b γ
α.
14
Net section The design resistance of net cross section is
N103,54230,1
360)222180(169,0
fA9,0N 3
2M
unetRd.net ∗=
∗∗−∗∗==
γ.
The connection resistance is 4 ∗ 131,9 = 527,6 kN .
557055
505050
8 x M 20
P 16 - 180 x 1200
2 P 8 - 180 x 340
507070
8 22
4 22P 16 - 180 x 10004 22
Scale: 1 : 10Steel S 235 J2
Bolts 10.9 Contact surfaces blasted with shot, no paintingø
ø
ø
Fig. 7.3 Design drawing of the connection
8 Bolted Long Connection Evaluate the design resistance of the bolted connection, shown in Fig. 8.1. The steel is Grade S235. The non-preloaded bolts M16 are Grade 5.6 in holes ∅ 17 mm. The shear passes through the unthreaded portion of the bolts. Material partial safety factors are 1,10=0Mγ , 1,25=2Mγ ,
25,1Mb =γ .
55
70
55
50 4 x 70 50
12
20
180
e1
FRd FRd
Fig. 8.1 ____________________________________________________________________________________________________
The reduction factor of a long connection:
9875,016200
16152801d200d15L
1 jLf =
∗∗−
−=−
−=β
; p 75,0≥β . 280
e1 e1L j
50 Fig. 8.2
,
p 0,1≤β
The shear resistance of a long connection with ten bolts is
15
N105,47625,1
416
5006,09875,010
Af6,010F 3
2
Mb
bubLfRd.v ∗=
∗∗∗∗=∗=
π
γβ .
Design bearing resistance is calculated for connection factor:
98,0173
50d3e
0
1 =∗
==α ; 12,141
17370
41
d3p
0
1 =−∗
=−=α ; 39,1360500
ff
u
bu ===α ; 1,0=α
for ten bolts is
N108,135425,1
1216360980,05,210tdf5,2
10F 3
Mb
uRd.b ∗=
∗∗∗∗∗=∗=
γα
.
The net section resistance is
N101,45425,1
36012)172180(9,0
fA9,0N 3
2M
unetRd.u ∗=
∗∗∗−∗==
γ.
The connection resistance is 454,1 kN.
55
70
55
50 4 x 70 50 180 x 12 -1 200
P 20 - 300 x 600
10 17∅
10 17∅
10 M 16
Scale: 1 : 10Steel S 235 J2Bolts 5.6
Fig. 8.2 Design drawing of the connection
_____________________________________________________________________________________________________ Notes: 1) The reduction of the resistance starts at the connection length 15 d, see Fig. 8.3 .
0
0,2
0,4
0,6
0,8
1
0 15d 65d
β Lt
L j
L j
0,75
Fig. 8.3 2) The design plastic resistance of section is
N105,46110,1
23512180fAN 3
0M
yRd.pl ∗=
∗∗==
γ.
9 Bolted Connection with Packing Evaluate the resistance of the bolted connection with a packing, shown in Fig. 9.1. The steel is Grade S355. The non-preloaded bolts M 16 are Grade 8.8 in holes ∅ 17 mm. The shear passes through untreated portion of the bolt. Material partial safety factors are 1,25=2Mγ and 25,1Mb =γ .
16
50
70
50
50 70 50 5
10
5
10
P10 -170 x 2000 P5 -170 x 2000P10 -170x170
4 M 16 - 8.8
t tpe p1 1
e
p
2
2
e2
e1
Fig. 9.1 Drawing of the connection ___________________________________________________________________________________________________
The reduction factor of the packing is
911,0103168
169t3d8
d9
pp =
∗+∗∗
=+
=β , 0,1p ≤β
The shear resistance of four bolt M 16 (each acting in two shear planes) is
N107,56225,1
416
8006,0911,024
Af6,024F 3
2
Mb
ubpRd.v ∗=
∗∗∗
∗∗∗=∗=
π
γβ .
The bearing resistance per four bolt in plate 10 mm:
98,0173
50d3
e
0
1 =∗
==α ; 12,141
17370
41
d3p
0
1 =−∗
=−=α ; 980,0510500
ff
u
ub ===α ; 1,0=α ,
N107,63925,1
1016510980,05,24tdf5,2
4F 3
Mb
uRd,b ∗=
∗∗∗∗∗=∗=
γα
.
The net section resistance is
N104,49925,1
51010)172170(9,0
fA9,0N 3
2M
unetRd.u ∗=
∗∗∗−==
γ.
The connection resistance is 499,4 kN.
____________________________________________________________________________________________________ Notes: 1) The reduction of the bolt shear resistance is for packing tp ≥ d / 3 shown in Fig. 9.2.
β
t p
0,5
1,0
0 0,3 d 1,5 d
p
1,0 d
Fig. 9.2
17
2) The additional bolt row may be optionally placed in an extension of the packing. see Fig. 9.3.
Fig. 9.3 10 Single Lap Connection with One Bolt Check the resistance of the single lap connection. The connection is subject to the factored force FSd = 40 kN, shown in Fig. 10.1. The steel is Grade S355. The no-preloaded bolt M16 is Grade 5.6 in hole ∅ 17 mm. The shear passes through the untreated portion of the bolt. The material partial safety factors are γ M 2 =1,30 , γ Mb =1 45, .
30 30
M 16 - 5.6P5 - 60 x 840
58FSd
Fig. 10.1 Drawing of the connection _____________________________________________________________________________________________________ The design shear resistance of one bolt M 16 of one shear plane is
kN40N106,4145,1
4165006,0Af6,0
F 3
2
Mb
ubRd,v >∗=
∗∗∗
==
π
γ .
The design bearing resistance of one bolt for plate 5 mm is calculated for the smallest α
588,0173
30d3
e
0
1 =∗
==α (limit); 980,0510500
ff
u
bu ===α ; 1,0=α
as
kN40N104,4145,1
516510588,05,2tdf5,2F 3
Mb
uRd,b >∗=
∗∗∗∗==
γα
.
The bearing resistance of single lap joint with one bolt is
kN40N102,4245,1
5165105,1tdf5,1F 3
Mb
uRd.b >∗=
∗∗∗=≤
γ .
The net section resistance is
kN40N109,7530,1
510)175605(9,0
fA9,0N 3
2M
unetRd.u >∗=
∗∗−∗∗==
γ.
The connection is satisfactory.
18
_____________________________________________________________________________________________________ Note: The connections with single bolt are used for structural purposes in special cases only, e.g. for the transmission electric towers and less important structures. In this type of connection is the washer positioned under the bolt nut as well as bolt head due to the shear and bending of the bolt.
11 Bolted Beam Splices Check the resistance of the beam splices joint, shown in Fig. 11.1. The joint is subject to the factored moment MSd = 180 kNm and the factored shear force VSd = 280 kN. Beams are IPE 450. The steel is Grade S235. The bolts M 20, Grade 8.8, are designed in the holes ∅ 22 mm. Material partial safety factors are γ M0 =1,15 ,γ M 2 =1,30 , and γ Mb =1 45, .
IPE 450 M 20 - 8.8
40
40
3 x 90
40 4080
40 4070 7080
P 15 - 190 x 300
2P8 - 160 x 350
14,6
15
9,4
21
450378,8
88
190
M V V MSd Sd SdSd
P 15 - 190 x 300
Fig. 11.1 Drawing of the connection
The internal forces in the connection has to follow the material distribution in the connecting elements. Hence the splices on the beam flanges are designed of the area of the beam flanges and the beam splices on the beam web of the area of the beam web. The splices on the beam flanges from both sides bring the most smooth transfer, see Notes below. The splices for the hot rolled beams till 500 mm are designed with the splices on one side only to economise the solution in the building design. The distribution of internal forces in joint may follow elastic, elastic-plastic or plastic distribution based on the global analysis of the whole structure, of the connected elements and on economy of design. The splices at the flanges carry the forces based on the bending moment (the plastic internal force distribution in the connection)
kN4,4130146,045,0
1800146,045,0
MN Sd
Sd.f =−
=−
= .
flange web
Nf.Sd
Nf.Sd
VSdFw.Sd
FFF
w.Sd
w.Sd
w.Sd
Fig. 11.2 Splices The tension splice cross-section area is affected by holes ∅ 22 mm: , 2
net mm19022215219015A =∗∗−∗= . kN4,413NN108,54530,1/36019029,0/fA9,0 Sd.f
32Munet =>∗=∗∗=γ
The resistance of the compressed splice, may be taken without holes,
19
. kN4,413NN104,58215,1/23519015/fA Sd.f
30My =>∗=∗∗=γ
The shear resistance of the beam web splices is calculated for the whole splice area: , 2
net mm600535082A =∗∗= , 2
net.v mm1924)224350(82A =∗−∗∗=
2v
u
y2net.v mm36565600
360235A
ff
mm4192A =∗=>= ,
175 135
45
Fig. 11.3
N107,660315,1
23556003
fAV 3
0M
yvRd.pl.w ∗=
∗
∗==
γ. kN280V> Sd =
The splices are satisfactory.
Bolts at flange The design shear resistance (four bolts M 20, each one shear plane) is
N100,41645,1
420
8006,014
Af6,0n4F 3
2
Mb
ubRd.v ∗=
∗∗∗
∗∗==
π
γkN4,413N Sd.f => .
The design bearing resistance per four bolts (the beam flange of thickness tf = 14,6 mm) is
61,0223
40d3e
0
1 =∗
==α (limit); 81,041
22370
41
d3p
0
1 =−∗
=−=α ; 22,2360800
ff
u
ub ===α ; 0,1=α ,
N 102,42445,1
6,142036061,05,24tdf5,2
4F 3
Mb
uRd.b ∗=
∗∗∗∗∗=∗=
γα
kN4,413N Sd.f => .
The resistance of the bolts at the flanges is satisfactory.
Bolts at web The bearing resistance of four bolt at the web splices:
61,0223
40d3e
0
1 =∗
==α (limit);
11,141
22390
41
d3p
0
1 =−∗
=−=α ;
22,2360800
ff
u
ub ===α ; 0,1=α ,
e = 401
p = 901
e = 901.w
Fig. 11.4
N 106,48445,1
822036061,05,24tdf5,2
4F 3
Mb
uRd.b.p ∗=
∗∗∗∗∗∗=∗=
γα
kN280V> Sd = .
The bearing resistance of the beam web is calculated based on the pitch distance p1 only (α = 1,0):
N108,46645,1
4,9203600,15,24tdf5,2
4F 3
Mb
uRd.b.w ∗=
∗∗∗∗∗=∗=
γα
kN280V> Sd = .
The design shear resistance (four bolts M 20, each two shear planes) is
20
N100,83245,1
420
8006,024
Af6,0n4F 3
2
Mb
ubRd.v ∗=
∗∗∗∗=∗=
π
γkN280V> Sd = .
The resistance of the bolts at the web is satisfactory
Beam net section The holes at the beam web are not taken into account , 2
fwfv mm4,08256,14)2124,9(6,1419028809t)r2t(tb2AA =∗∗++∗∗−=++−=
, 2wvnet.v mm8,66844,92224,0825t224AA =∗∗−=∗∗−=
because
2
u
yv
2net.v mm7,3173
3602354,0825
ff
Amm8,6684A =∗=>= .
The shear resistance of the I section is
N106,599315,12354,5082
3
fAV 3
0M
yvRd.pl.f ∗=
∗
∗==
γkN280V> Sd = .
The shear resistance is satisfactory.
The bending resistance is not reduced by the shear force, because kN3,2992/6,5992/V Rd.pl == kN280V> Sd = . The holes in beam flanges and web shall be taken into account:
0M
2M
u
y
f
net.f
ff
AA
9,0γγ
< ,
74,015,130,1
36023569,0
1902221909,0 =∗<=
∗−∗ .
The reduced second moment of area (IPE 450, Iy = 337,4∗106 mm4), the holes are taken into account at the beam web as well as at the both flanges:
−∗∗−∗∗−∗= 336net 226,9
1246,1422
124104,337I
46222 mm101,268224,9452224,913526,1422)2
6,14450(4 ∗=∗∗∗−∗∗∗−∗∗−
∗− .
The bending resistance of the net cross-section is
kNm180MNmm105,24315,1
2450
235101,268
2h
fIM Sd
66
0M
ynetRd.el =>∗=
∗
∗∗==
γ.
The bending resistance is satisfactory. The joint is satisfactory.
Steel S 235 J2Bolts 8.8Scale: 1 : 10
8 22∅
IPE 450 - 2 200
24 M 20
40
40
3 x 90
40 4080
40 4070 7080 P 15 - 190 x 300
2P8 - 160 x 350
IPE 450 - 4 650P 15 - 190 x 300
8 22∅
12 22∅ 12 22∅
8 22∅
1004545
M 20
M 20
M 20
Fig. 11.5 Design drawing of the joint
________________________________________________________________________________ Notes:
21
1/ The distribution of the internal forces in joint may be based on elastic, elastic-plastic or plastic distribution. The plastic distribution is presented. For the elastic distribution is the bending moment transferred by the flange cover plate as well as by the web cover plates. The bolts forces in the flanges are taken the part of the bending moment. This bending moment may be estimated by reducing the bending moments by the second moment of inertia of the flages
4623f mm100,2632)7,2176,141906,14190
121(I ∗=∗∗∗+∗∗=
divided by the second moment of inertia of the I cross-section
y
fSdSd.f I
IMM = .
The force in flanges is
14,6
420,8
14,6
217,7457,4
217,7
Fig. 11.6
/Sd.f
Sd.fh
MN = .
The cover plates and the bolts on the beam web are transferring the rest of the bending moment Sd.fSdSd.w MMM −=and the shear force VSd. 2/ The splices on the beam flanges from both sides bring the most smooth transfer but the solution is expensive compare to a single side splices.
IPE 450 M 20 - 5.6
50
50
3 x 90
50
100
50 5070 70100
P 8 - 190 x 340
2P8 - 160 x 3509,4
450378,8
66
190
M V V MSd Sd SdSd
P 8 - 190 x 34014,68
8
50
Fig. 11.7 Double side splices on the beam flange
12 Beam-to-Column Welded Joint Design the beam-to-column welded joint, shown in Fig. 12.1. The joint is subject to the factored internal forces M k evaluated by the plastic global analysis. The steel is Grade S235. Nm VSd Sd= =30 150; kNThe material partial safety factors are γ γ =1 50, . =1,15 and MwM0
b = 110fb
h =220 w
a f
HE 200 B
IPE 220
t =15fcr =18c
t = 9,2fb
t = 9,0wc
wb
b
b
fa
z
M SdVSd
10
10
t = 5,9
r =12a
200
Fig. 12.1 Drawing of the connection _________________________________________________________________________________
_
22
Beam web The beam shear resistance is calculated based on the shear effective area:
( ) ( ) 23fbbwbfbbbvb mm57015,81229,52,911021034,3tr2ttb2AA =∗∗++∗∗−∗=++−= ,
kN150VN107,18715,132351570
3
fAVF Sd
3
0M
wb.yb.vRd.plRd.wc.v =>∗=
∗
∗===
γ.
The beam web weld can be calculated based on the beam shear resistance is
( ) mm7,23602,922202
350,180,0107,187f)t2h(23F
a3
ufbb
MwwSdw =
∗∗−∗∗∗∗
=−
=γβ
.
The overloading of the welds due to the strain hardening of the steel of the section is taken into account by strengthening of the welds in case of the plastic global analysis:
mm5=mm4,59=2,71,7 wa≈∗ .
Beam flange
The beam flange compression resistance is
( ) N3102772,922015,1
2353104,285)]th(/[fWFfbb0MyplRd.fb.c
∗=−∗∗∗
=−= γ .
The weld throat width at the beam flange is based on the full beam flange resistance:
mm9,53601102
25,18,0277fb2
2Fa
ufb
MwwRd.fb.cf =
∗∗∗∗∗
==γβ
.
The possible overloading of the welds is taken into account for the plastic global analysis:
mm10,0=1,7 9,5∗ . Column web
The joint design resistance is limited (In case of the column with web stiffened in tension and in compression) by the shear resistance of the column web (which is not important for the hot rolled H shapes) and by the resistance of the beam flange: . ( ) kNm30VNmm104,582,922010277zM Sd
63Rd =>∗=−∗∗== c.fb.RdF
The welded joint resistance is satisfactory.
23
13* Pin Connection Design a pin, shown in Fig. 13.1. subject to the factored tensile force Fv.Sd = 110 kN. The steel of the plates is Grade S275, pin Grade S355. The material partial safety factors 15,10M =γ , 30,12M =γ .
FSd
d = 30
t1 = 10
c = 1t2 = 18
d3 = 20
t = 101
c = 1
Fig. 13.1 Drawing of the connection _________________________________________________________________________________
_ The pin diameter is chosen . mm30d = Shear resistance The shear resistance for two shear planes
=∗
∗∗∗∗==
30,14510306,02
fA6,02F
2
2M
upRd.v
πγ SdFkN110kN8,332 =>
The pin shear resistance is satisfactory. Bearing resistance The bearing resistance for the minimal yeld strength
and for the minimal thickness ( ) MPa275355;275minfy ==
( ) ( ) mm1818;102mint;t2mint 21 =×== may be checked
Sd0M
yRd.b FkN110kN7,193
15,127518305,1ftd5,1
F =>=∗∗∗
==γ
The pin bearing resistance is satisfactory. Bending resistance The bending moment resistance
Nmm61265415,132
355308,032
fd8,0fW8,0M
3
0M
yp3
0M
ypelRd =
∗∗∗∗
===π
γπ
γ
is higher compare to acting moment
( ) ( )14181028000110c4tt2
8FM 21
SdSd ∗++∗=++=
RdMNmm612654Nmm500577 =<=
The pin bending resistance is satisfactory.
t1cc
t1 t2
MSd
Fig. 13.2
The bending moment and shear interaction
189,0800332000110
612654500577
FF
MM
222
Rd.v
Sd2
Rd
Sd <=⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡.
is satisfactory. The pin resistance is satisfactory.
24
Plate geometry
The plate minimal area
2
y
0MSdmin mm460
27515,1000110
fFA =
∗==
γ
Influences the distance d3, see Fig.
mm6,2518460
tAd
2
min3 ==≥ is taken as mm30d3 = .
The end distances of plates
mm6,2518460
tAd
2
min1 ==≥ as mm30d3 =
mm6,2518460
tAd
2
min2 ==≥ as mm30d3 = ,
should be kept in angle ±45º from the axis.
30
19
70
30
Fig. 13.3
mm1,171846067,0
tA67,0d
2
min4 ==≥ is choosen mm19d4 = ,
to be the total with mm7032192dd2b 04 =+×=+= .
__________________________________________________________________________________ Note: If the pin is intended to be replaceable, the contact bearing stress should satisfy the check at the Serviceability Limit State
( )
ser,6M
yEd,h2
0ser,SdEd,h
f5,2f
td
ddFE591,0
γσ =≤
−= ,
where d0 is the pin hole diameter, FSd,ser the design force at serviceability and γM6,ser the partial safety factor for serviceability,
and the bearing of the plate and the pin ser,Sd,bser,6Myser,Rd,b F/ftd6,0F ≥= γ , as well as the check in bending ser,Sd0Mupelser,Rd M/fW8,0M ≥= γ .
25
14 Simple Column Base Calculate the design resistance of the column base shown in Fig. 14.1. The column cross-section is HE200B, the base plate thickness is 30 mm, and the concrete foundation block dimensions are 850 x 850 x 900 mm. The steel is Grade S235 and the concrete is Grade C20/25. The material partial safety factors are γ M0 =1,15 and γ =1 50, . c
RdF
t = 30
HE 200 B a = 850a = a = 255
b = 255340
1
b = 850b =
340 1
r
r
3xP10-40x40
30
h = 900
Fig. 14.1 Drawing of the connection For the effective cross section of the foundation block:
, mm850
42508505b51240900340ha
17003405a58502552340a2a
mina
1
r
1 =
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=∗==+=+
=∗==∗+=+
=
and, from symmetry b1 = a1. The stress concentration factor is
2,5= 340340850850 =
b ab a
= k 11j ∗
∗ .
The bearing strength of the concrete under the base-plate is
MPa22,3 =1,50
20,02,50,67=f k0,67
= fc
ckjj
∗∗γ
.
A rigid plate of effective width c, surrounding the column H-section, replaces the flexible base-plate:
mm4,5215,13,223
23530f3
f t =c
M0j
y =∗∗
∗=γ
.
The effective area (see Fig. 5.11.2) is
c c
c
c
c
c
t = 15 t = 9
h =200
wf
c
b = 200c
Fig. 14.2
)4,522152200()4,52294,522200()4,522200()4,522200(Aeff ∗−∗−×∗−−∗+−∗+×∗+=
. 2mm44980=The design resistance of the column base is . N1079413,2244980fAN 3
jeffRd ∗=∗==
26
170 170 170 170
170
170
340
550
30
240
200
40
70
50
60
70
70
70
240
120 120
50
340 200
HE 200 B
P30 - 340 x 340
20 - 950 M20
Steel S235 J2
Scale: 1:10
2 22
P10 - 40 x 40 25
3 x P10 - 40 x 240
340 340
5
Concrete C20/25
Fig. 14.3 Design drawing of the column base. The anchor bolts are designed for structural integrity.
Notes: 1) The design resistance of the column is lower than the resistance of the column base:
, where A is the column cross-sectional area.
kN7941N<N10596115,1/2358087/fAN dR3
0MyRd.pl =∗=∗== γ
2) The joint coefficient is taken as 2/3 provided that the characteristic strength of the grout is not
less than 0,2 times the characteristic strength of the concrete foundation, and the grout thickness is less than 0,2 ∗min (a; b) = 0,2 ∗ 340 = 68 mm.
3) Packing plates (see Fig. 14.3) are used to level the base plate during erection.
27
15* Fixed Column Base Calculate the design moment resistance MRd of the fixed column base shown in Fig. 15.1. The column base is subject to the factored force FSd = 500 kN. The column cross-section is HE200B. The base plate thickness is 30 mm. The concrete foundation blosk dimensions are 1600 x 1600 x 1000 mm. The steel is Grade S235. The concrete is Grade C12/15. The material partial safety factors are γ M0 =1,15 and γ =1 50, . c
1600 630
630
t = 30
HE 200 B340
1600340
MF SdRd
m = 60d = 160b
30
h = 1000
M 24
b =
Fig. 15.1 Drawing of the connection
_________________________________________________________________________________
Tension part The tension resistance of two anchor bolts M 24 (net section As = 353 mm) is
. N1011445,1/3532352/Af2F2F 3MbsybRd.sRd.s2 ∗=∗∗=== γ
The resistance of the base plate in bending loaded by the force is
Rd.s23
2
0M
y2
0M
yelRd.p FN107,137
15,160623530340
m6ftb
mfW
F >∗=∗∗
∗∗===
γγ .
Compressed part
1 600a = a = 630
b = 630340
1
r
r
1 600b =340
1
a = 1 340
b = 1 340
Fig. 15.2 For the effective cross section of the foundation block, see Fig. 15.2, is
28
, mm3401
000816005b534010001340ha
70013405a56001630*2340a2a
mina
1
r
1 =
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=∗==+=+
=∗==+=+
=
and from symmetry a1 = b1. The the stress concentration factor is
3,94= 340340
4013 =
b ab a
= k 11j ∗
∗ 3401 .
The bearing strength of the concrete under the base plate is
MPa21,1=1,50
12,00,67=f k0,67
= fc
ckjj
∗∗ 94,3γ
.
The effective area of the concrete in compression is calculated based on the force equilibrium: , Rd.s2jeffSd FfAF −=
233
j
Rd.2Sdeff mm10029
1,211011410500
fFF
A =∗+∗
=+
= .
The rigid plate of effective width c, see Fig. 5.19.3 surrounding the column section, replaces the flexible plate
mm4,5605,11,21
23530 =∗∗
∗=3f 3
f t =c
M0j
y
γ .
c = 56, 4
Fig. 15.3
The width of effective area is
mm0,934,562200
10029beff =
∗+=
56,4
56,4
200
beff
Fig. 15.4 The concrete lever arm is
mm9,1092
0,934,561002
bc2/hd eff
cc =−+=−+= .
FSd
dc
AeffF2s.Rd f j
db
Fig. 15.5 The base plate bending resistance is
29
=+= cjeffbRd.s2Rd dfAdFM
kNm4,83Nmm104,839,1091,211002914010114 63 =∗=∗∗+∗∗= .
The column base transfer under axial force 500 kN the bending moment 83,4 kNm.
_________________________________________________________________________________
Notes: 1) The column normal force resistance is . N10047315,1/23591014/fAN 3
0MyRd.pl ∗=∗== γ The column bending resistance is . Nmm1038215,1/235108691/fWM 63
0MyplRd.pl ∗=∗∗== γ The interaction of the normal force and bending moment: 00,197,05961/3,131/7,85N/NM/M Rd.plSdRd.plSd <==+ 500+
The column cross-section resistance is higher compare to the column base resistance.
2) The column base resistance is compared to the column resistance for the different base plate thickness in Fig. 15.6. For plate P 30 are shown the major points of the diagram, e.g. the pure compression, the highest bending resistance (the point of the coincidence of the neutral axis and the axis of symmetry of the cross section), the pure bending, and the pure tension.
0
1 000
100 Moment, kNm
Normal force, kN
30
40
25
1520
HE 200 Bt
MNSd
Sd
30
h = 1 000
M 24
1 600340 630
630
340
1 600
pl.Rd
pl.RdN
M
t =
Column resistance
1 596
131,3
Fig. 15.6 The column base resistance for the different base plate thickness
30
16 Fin Plate Connection The AccessSteel check is summarised below. Shear resistance of fin plate connection Mode of failure Bolts in shear* VRd,1
Fin plate in bearing* VRd,2
Fin plate in shear (gross section) VRd,3
Fin plate in shear (net section) VRd,4
Fin plate in shear (block shear) VRd,5
Fin plate in bending VRd,6
Fin plate in buckling (LTB) VRd,7
Beam web in bearing* VRd,8
Beam web in shear (gross section) VRd,9
Beam web in shear (net section) VRd,10
Beam web in shear (block shear) VRd,11
Supporting element (punching shear) VRd,12The shear resistance of the joint is the minimum of the above values. Tying resistance of fin plate connection Mode of failure Bolts in shear* NRd,u,1
Fin plate in bearing* NRd,u,2
Fin plate in tension (block tearing) NRd,u,3
Fin plate in tension (net section) NRd,u,4
Beam web in bearing* NRd,u,5
Beam web in tension (block tearing) NRd,u,6
Beam web in tension (net section) NRd,u,7
Supporting member in bending NRd,u,8The tying resistance of the joint is the minimum of the above values.
1 1 13 3
4
2 2 2
For weld For bolt group For weld For bolt group For weld
For bolt group
For bolt group
Assumed lines of shear transfer
Assumed lines of shear transfer
Assumed lines of shear transfer
Key. 1. Fin plate; 2. Supported beam; 3. Supporting column; 4. Supporting beam
31
Check the beam-to-beam fin plate connection, shown in Fig. 16.1. The connection is subject to the vertical factored force VSd = 110 kN. The steel is Grade S275. The bolts M 20 are Grade 8.8 in holes ∅ 21 mm with the thread in the shear plane. The material partial safety factors are 1,10=0Mγ , 30,12M =γ , 25,1Mb =γ , and 25,1Mw =γ .
e30
L
t
p 1
w pt
1
e1
10
IPE 200
14060
5,6
40
40
445010
V = 110 kNSd
P 10 - 140 x 100M 20 - 5.6
10
R 10 20
Fig. 16.1 Drawing of the connection ________________________________________________________________________________________________________
Bolts The design shear resistance of two threaded bolts (each with one shear plane) is
N102,18825,1
2458006,012Af6,0
n2F 3
Mb
subRd.vs ∗=
∗∗∗∗==
γ.
The bearing resistance of the fin plate in the vertical direction is calculated for tminimum α factor:
e = 40
p = 60
e = 60
e = 40
1
1
1
Fig. 15.2
635,0213
40d3e
0
1 =∗
==α ; 702,041
21360
41
d3p
0
1 =−∗
=−=α ; 86,1430800
ff
u
ub ===α , 1,0=α ,
for two bolts
N101,26225,1
1220430635,05,22tdf5,2
2F 3
Mb
uRd.b ∗=
∗∗∗∗∗==
γα
.
The bearing resistance of the beam web is evaluated for minimum α:
794,02131040
d3e
0
1 =∗+
==α ; 702,041
21360
41
d3p
0
1 =−∗
=−=α ; 86,1430800
ff
u
ub ===α , 1,0=α ,
for two bolts
N109,15225,1
6,520430794,05,22tdf5,2
2F 3
Mb
uRd.b ∗=
∗∗∗∗∗=∗=
γα
.
The bolts are satisfactory. Fillet weld The fillet weld resistance is calculated based on normal stress:
MPa9,146
614010
601080
6Lt
eVWM
2
3
2p
Sd
we
Sd =∗
∗∗===σ .
32
The shear and normal stresses in the weld thread:
MPa8,1292424,12210
2a2
t
we
p =∗∗
∗=== ⊥⊥
σστ
MPa4,7114042
1080La2
V 3
we
SdII =
∗∗∗
==τ .
The stresses in the throat of the weld:
V
τ
Sd
II
σ
σ
τ
ττ II τ σ e
L
σ
wea weσ
pt
τ a
σ
Fig. 16.3
( ) ( ) MPa7,40485,025,1
430fMPa6,2874,718,12938,1293
Mww
u2222II
22 =∗
=<=++=++ ⊥⊥ γβττσ ,
MPa44325,1
430fMPa8,129
Mw
u ==<=⊥ γσ .
The weld strength is satisfactory. Block tearing of beam web Tensile failure occurs along the horizontal limit of the block, and shear plastic yielding occurs along the left vertical limit of the block. For beam end with a shear force acting eccentric relative to the bolt group the design value for block tearing Veff,2,Rd should be determined from =+= 0Mnvy2MntuRd.2.eff /Af)3/1(/Af5,0V γγ .
70
5060
50
Fig. 16.4
=∗−+∗∗∗+∗−∗∗∗= 15,1/)215,26050(6,5275)3/1(30,1/)215,050(6,54305,0
kN80VN100,81 Sd3 =>∗
The block tearing of the beam web is satisfactory. The shear resistance of the fin plate is
kN80VN101,202310,127514010
3
fAV Sd
3
0M
yvRd.pl =>∗=
∗
∗∗==
γ.
The net section, see Fig. 5.15.5, of the fin plate is not necessary to be taken into account, because , uyvnet.v f/fA/A >
64,043027584,0
14010)212140(10
=>=∗
∗−∗ .
Connection is satisfactory.
140
Fig. 16.5
33
Steel S 275 J2Bolt 8.8
Scale: 1 : 10
2 21∅M 20
IPE 200 - 2 100
30
10
60
40
4450
10
R 10 20
40
2 21∅P 12 - 140 x 100
Figure16.5 Design drawing of the connection 16 Header Plate Connection The AccessSteel check is summarised below. Shear resistance of end plate connection Mode of failure Bolts in shear VRd,1
End plate in bearing VRd,2
Supporting member in bearing VRd,3
End plate in shear (gross section) VRd,4
End plate in shear (net section) VRd,5
End plate in shear (block shear) VRd,6
End plate in bending VRd,7
Beam web in shear VRd,8The shear resistance of joint is the minimum of the above values. Tying resistance of end plate connection Mode of failure Bolts in tension NRd,u,1
End plate in bending* NRd,u,2
Supporting member in bending NRd,u,3
Beam web in tension NRd,u,4The tying resistance of joint is the minimum of the above values.
1 11
2 2
3 3
4
2
Assumed line of shear transfer
(face of column) Assumed line of shear transfer
(face of web) Assumed line of shear transfer
(face of web)
Key: 1. End plate; 2. Supported beam; 3. Column; 4. Supporting beam
34
Derive the resistance of the simple beam-to-beam connection, shown in Fig. 17.1. The steel is Grade S355. The bolt M 20, Grade 5.6, is full threaded bolts in holes ∅ 21 mm. The material partial safety factors are 1,10=0Mγ , 25,1Mb =γ , and 1,25=Mwγ .
2 x 6040 40
200
35
35
45
70
10
4
4 x M 20 - 4.6
10
140
6,2
IPE 240
SdV
P 10 - 140 x 200
Fig. 17.1 Drawing of the connection
Shear resistance of the beam web The shear resistance of a part of a beam web (in the height of the end plate and the welds) is
N108,17410,13
355)242140(2,63
fAV 3
0M
yvRd.pl.b ∗=
∗
∗∗∗+∗==
γ.
140 + 2 x 4 x 2
Fig. 17.2
Bolts resistance The shear resistance in the threaded part of the bolts, four bolts, one shear plane per bolt,
N102,23525,1
2455006,014Af6,0
n4F 3
Mb
subRd.v ∗=
∗∗∗∗==
γ.
The bearing resistance of four bolts at the end plate, which has lower resistance compare to the beam web of the same thickness, may be calculated for
556,0213
35d3e
0
1 =∗
==α (limit); 861,041
21370
41
d3p
0
1 =−∗
=−=α ; 980,0510500
ff
u
ub ===α ; α =1,0 ,
as N108,31525,1
1020355556,05,24tdf5,2
4F 3
Mb
uRd,b ∗=
∗∗∗∗∗==
γα
.
Weld resistance The resistance of filled welds is
N101,293325,19,0
140245103
LafF 3
Mww
weweuRd,w ∗=
∗∗
∗∗∗==
γβ.
140
Fig. 17.3
Shear resistance of the end plate The net area is taken into account for calculation the shear resistance of the end plate because
35
uyvnet,v f/fA/A >
696,0510355700,0
14010)212140(10
=>=∗
∗−∗ .
The shear resistance
N105,749310,1
5101401023
fAV 3
0M
yvRd.pl.p ∗=
∗∗∗==
γ.
140
Fig. 17.4
The connection resistance
VSd = min (174,8 kN; 235,2 kN; 315,8 kN; 293,1 kN; 749,5 kN) = Vb.pl.Rd = 174,8 kN. is driven by the shear resistance of the beam web.
2 x 6040 40
35
35
45
704
4 x M 20
IPE 240 - 2 600
Steel S 355 J2Bolts 5.6Scale: 1 : 10
P 10 - 500 x 2200
4 21 ∅
30
10
R 10
P 10 - 140 x 200
Fig. 17.5 Design drawing of the connection
36
18 Bolted Extended End Plate Connestion Joint resistance of eaves moment connection in symbols according to AccessSteel materials. Resistance
Potential resistance of bolt rows in the tension zone Rdt,F
Compression resistance Rdc,F
Shear resistance of the column web panel Rdwp,V
Moment resistance Rdj,M
Shear resistance for vertical forces RdV
Check the resistance of the extended end plate beam-to-column joint, shown in Fig. 18.1. The joint is subject to the factored bending moment Msd = 110 kNm and to the factored shear force VSd = 100 kN. The steel is Grade S235. The bolts M 20 Grade 5.6, are not preloaded. The elastic global analysis was applied. The material partial safety factors are γ M 0 = 1,15 , γ Mb =1 45, , and γ Mw =1 50, .
wbp
wawa 1
e
e
e
e
1
2
3
4
hp
u
40120
40
40
105
340
75
560
20
35
14,5
9,4
twc
wbt
d p
HE 500B
M 20 - 5.6
30
P 30 - 200 x 520
P 12 - 116 x 65
28
14,6
IPE 450
MSdVSd
t fc
t fb
12
12
P 12 - 60 x 60
tpc
d s425
t pc
35
bcf = 300= 200 Point of contact
in connection
Fig. 18.1 Drawing of the connection
__________________________________________________________________________ Bending resistance of bolted part It is expected that the plate rotates as a rigid body round the centre of the compressed flange of the beam, see Fig. 5.10.2. The bending moment equilibrium: . )F0,4927+F0,3877+F(0,04772=Fr2=M 1.Sd2.Sd3.Sd
ii.SdiSd ∑
The forces depends on the distance to the centre of rotation (the elastic internal force distribution):
0,4927
F0,0477=
rFr
=F 1.Sd
1
1.Sd33.Sd ,
0,4927F0,3877
=rFr
=F 1.Sd
1
1.Sd22.Sd .
The force in the top bolt row:
37
.kN 137,1 =0,4927+0,3877+0,0477
0,4927110 = F2
0,4927+0,4927
0,3877+0,4927
0,0477(F2 =110
F0,4927+F0,4927
0,3877+F0,4927
0,0477(2 = M
2221.Sd
22
1.Sd
1.Sd1.Sd
2
1.Sd
2
Sd
∗
)
)
2 F
3
2 F2
2 F
1
47,7
387,7
492,7
Fig. 18.2
The bolt tension resistance is
kN1,137kN0,152N100,76245,1
2455009,02Af9,02F2B2 3
Mb
subRd.tt >=∗∗=
∗∗∗===
γ.
The design resistance of the top bolt row is calculated based on effective T stub resistance of the column flange, which is thinner compare to the end plate. The T stub geometry is shown in Fig. 5.10.3:
mm8,522
5,141202
twm wc1 =
−=
−= ,
. mm35m2 =
For 51,0508,52
8,52em
m
1
11 =
+=
+=λ
and 41,0508,52
35em
m
1
22 =
+=
+=λ ,
The effective length of the stiffened T stub is calculated based on the factor α = 5,7, see the graph [8]. The effective length of the T stub at the top bolt row
m2
e m
w1
twc
≅
≅40
35
point
120
14,5
of contact
Fig. 18.3
. mm0,3010,301
6,331min
8,527,5
8,522min
m
m2minLeff =
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
∗=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
π
α
π
The column flange resistance in bending is . == 0My
2feffRd.pl /ftL25,0M γ Nmm1005,1215,1/235280,30125,0 62 ∗=∗∗∗
The resistance of the top bolt row in case of i) the failure of the column flange is
==M4
F Rd.plRd.cf m
N102,9131005,124 36
∗=∗∗
8,52,
ii) the failure of the column flange and bolts is (The size of end plate limits the width of the T stub by
n = min (e ; 1,25 m) = (40; 1,25 ∗ 52,8) = 40 mm)
38
=+
+=
Af9,0n2M2
F Mb
subRd.pl
Rd.cfγ
nm
N103,325408,52
0,764021005,122 36
∗=+
∗∗+∗∗= ,
iii) the failure of the bolts is
N100,152100,762B2F 33tRd.cf ∗=∗∗== .
The bolt rupture shows the minimum resistance.
The bolt resistance is satisfactory in bending. Shear resistance of bolted part It is expected, that each bolt row carries the same shear force under the elastic distribution of the internal force. The shear resistance per a bolt in treated part for one shear plane
N107,5045,1
2455006,0Af6,0F 3
Mb
subRd.v ∗=
∗∗==
γ.
6 ∗ 50,7 = 304,1 kN > 100 kN. The bolts subject to the combined shear and tension:
1F4,1
FFF
Rd.t
Sd.t
Rd.v
Sd.v ≤+ ,
197,02/0,1524,1
2/1,1377,50
6/100≤=
∗+ .
The bolt bearing resistance:
635,0213
40d3e
0
1 =∗
==α ; 417,141
213105
41
d3p
0
1 =−∗
=−=α ; 39,1360500
ff
u
ub ===α ; 1,0=α .
kN7,166/1006/VN105,23645,1
3020360635,05,2tdf5,2F Sd
3
Mb
uRd.b ==>∗=
∗∗∗∗==
γα
.
The bolt resistance is satisfactory in shear.
Welds Provide the weld 4 mm based on the structural point of view. The shear force at the beam web fillet welds is
MPa7,29)6,142450(42
10100)t2h(a
V 3
fbbw
Sd2.II =
∗−∗∗∗
=−
=τ .
zz12
4
14,6450
14,6
4
1
2
9,4
190 Fig. 18.4
39
The stresses perpendicular to weld cross-section is calculated based on the second moment of area of filled weld cross-section:
−∗−∗−∗−−∗+−∗+∗+= 33we )426,142450)(424,942190(
121)42450)(42190(
121I
46233 mm101,205)2/)6,14450((6,1419026,14190122)6,142450(
124,9
∗=−∗∗∗−∗−∗−− ,
based on the elastic stress distribution:
MPa8,792101,205
2/)6,142450(101102IzM
2/6
6
we
2Sd2.w2.2. =
∗
∗−∗∗==== ⊥⊥ σστ .
The stress check in the throat section:
( ) ( ) MPa0,3005,18,0
360fMPa7,1677,298,7938,793
Mww
u22222.II
22.
22. =
∗=<=++=++ ⊥⊥ γβ
ττσ ,
MPa50,1
360fMPa8,79
Mw
u2. 240==<=⊥ γ
σ .
The welds at the web flange are loaded by shear only:
MPa8,862101,205
2/)42450(101102IzM
2/6
6
we
1Sd1.w1.1. =
∗∗
∗+∗∗==== ⊥⊥ σστ .
The stress check in the throat section is
MPa0,3005,18,0
360fMPa6,1738,8638,863
Mww
u2221.
21. =
∗=<=∗+=+ ⊥⊥ γβ
τσ ,
MPa50,1
360fMPa8,86
Mw
u1. 240==<=⊥ γ
σ .
The weld strength is satisfactory.
The joint resistance is satisfactory.
40
41
19 Extended end plate semirigid joint Calculate the design bending Rd.jM and shear RdV resistance of the extended end plate semirigid joint shown Fig. 19.1. Steel S 235, nonpreloaded bolts M 16 class 8.8 The shear plane passes through the unthreaded portion of the bolts. Calculate the length, for which the point may be claasified as rigid in a brace frame. The material partial safety factors are γ M 0 = 1,15 , γ Mb =1 45, , and γ Mw =1 50, .
a = 351
w = 80
b = 140p
wa1
e = 70
e = 140
e = 55
1
2
3
4
p
u = 10
t = 12p
IPE 220
HE 140B
M 16 - 8.8
a = 4w
P 12 - 140 x 305
P 12 - 116 x 65
M VSd Sd
IPE 220
MSdVSd
t = 12fc
fb
t = 7wc
wb
a = 5f
aw = 30
a = 5f
a = 5w
a = 5w
t = 9,2
t = 5,9h = 305
e =40
Fig. 19.1 Drawing of the connection
a) The design resistance of the column web in compression Rd.wc.cF and in tension Rd.wc.tF are not influenced the resistance, if the stiffeners have at least same the resistance as the beam flanges. The beam flange is in symmetrical joints symmetrically loaded not loaded activated and need not be checked Rd.wpV .
F F
F
F
F
F
F
F
b1.Rd
b2.Rd
b3.Rd
p1.Rd
p2.Rd
Fbf.Rd
cf1.Rdt.wc.Rd
wp.Rd
c.wc.Rd
V
Fcf2.Rd
Fig. 19.2 Internal forces in the joint b) The resistance of a bolt row in the column flange may be calculated for the effective T stub for the distances m1 a m2 according to Fig. 19.3
mm,,m 9262
128027801 =
∗∗−−=
mm,**,m 3232
5280212702 =
∗−−=
m2
0,8* 2*aw
e m 0,8 r 1
7 80
1270
Fig. 19.3
42
Factor α may be evaluated for 1λ and 2λ
47030926
9261
11 ,
,,
emm
=+
=+
=λ
41030926
3231
22 ,
,,
emm
=+
=+
=λ
16,=α For the calculation of the resistance of the effective T stub is necessary to evaluate the length of the effective T stub, which is for the top bolt row
mm,,**mL cp.eff 016992622 === ππ mm,,*,mL op.eff 816692616 === α
and the resulting effective length are
( ) ( ) mm,,;,minL;LminL op.effcp.eff.eff 8166816601691 === mm,LL op.eff.eff 81662 ==
he resistance of the flange Rd.pl.cm of the unite length is
mm/Nmm,,**,ft
mM
yfcRd.pl.c 57356
1514235012
4
2
0
2
===γ
Leff
Leffp
e m 0,8 r
tfc
1
Fig. 19.3 Column stiffened flange
The resistance of one bolt in tension is
kN,N*,,
**,fA,B
Mb
ubsRd.t 07810078
4511578009090 3 ====
γ
( ) ( ) mm,*,;minm,;eminn 3092625130251 1 ===
Failure mode 1
kN,N*,
,,*,*
mmL
F Rd.pl.c.effRd.cf
4182104182926
57356816644
3
11
==
==
Failure mode 2
kN,N*,,
*,**,*,*nm
BnmLF Rd.tRd.pl.c.eff
Rd.cf
412510412530926
100783025735681662
22
3
3
21
==
++
=
+
+=
Failure mode 3
kN,N*,,*BF Rd.tRd.cf 015610015607822 31 ====
The effective length for the second bolt row, under the stiffener, may be calculated as mm,,**mL cp.eff 016992622 === ππ
mm,,*pmL cp.eff 5224140926 =+=+= ππ mm,,*,mL op.eff 816692626 ===α
43
( )( )
mm,*,,**,,*,
e,mp,mL op.eff
216430625092621405092626
6250250
=+−+=
+−+= α
The resulting effective lengths and resistances are
( ) ( ) mm,,;,;,;,minL;LminL op.effcp.eff.eff 216421648166522401691 === ( ) ( ) mm,,;,minLminL op.eff.eff 2164216481662 ===
Failure mode 1
kN,N*,
,,*,*
mmL
F Rd.pl.c.effRd.cf
6179106179926
57356216444
3
12
==
==
Failure mode 2
kN,N*,,
*,**,*,*nm
BnmLF Rd.tRd.pl.c.eff
Rd.cf
712410712430926
100783025735621642
22
3
3
22
==
++
=
+
+=
Failure mode 3
kN,N*,,*BF Rd.tRd.cf 015610015607822 32 ====
c) The resistance of the end plate in tension may be evaluated as for flange based on the T stub model. The model for extended part of the end plate is taking into account the bending round the beam upper flange
mmex 40=
mm,**,a,am fx 3295280352801 =−=−= ( ) ( ) mm,,*,;minm,;eminn xx 63632925140251 ===
The length of effective T stub in the unstiffened end plate for the top bolt row is
b
wL eff
t p
em
xx
0,8 2 a f
e e
p
Fig. 19.4
mm,,,,
min,,
,*min
emwm
mminL
x
x
x
cp.eff 0152015201721184
302329803293292
2
2=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
∗+∗+∗
∗=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
+
+=ππ
π
πππ
mm,
,,,,
min
,,,,,
,,,,
min
e,mw,b,
e,mee,m
minL
xx
p
xx
xx
op.eff
070
612307061132167
40625032928050014050
406250329230402513294
625025050
625022514
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
∗+∗+∗∗
∗+∗+∗+∗
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
++
++
+
=
The resulting effective lengths are
( ) ( ) mm,,;,minL;LminL op.effcp.eff.eff 07007001521 === mm,LL op.eff.eff 0702 ==
44
The design resistance of the plate in bending for unite length Rd.pl.pm may be calculated as
mm/Nmm,,**,ft
mM
ypRd.pl.p 57356
1514235012
4
2
0
2
===γ
Failure mode 1
kN,N*,
,,**
mmL
F Rd.pl.p.effRd.p
37010370329
573567044
3
11
==
==
Failure mode 2
kN,N*,,,
*,*,*,**nm
BnmLF Rd.tRd.pl.p.eff
Rd.p
3102103102636329
10078636257356702
22
3
3
21
==
++
=
+
+=
Failure mode 3
kN,N*,,*BF Rd.tRd.p 015610015607822 31 ====
d) The design resistance of the second bolt row of the end plate in tension may be evaluated form the
values
mm,**,,m 5322
4280295801 =
∗−−=
mm,**,,m 12052802935702 =−−−=
52030532
532
1
11 ,
,,
emm
=+
=+
=λ
32030532
120
1
22 ,
,,
emm
=+
=+
=λ
06,=α ( ) ( ) mm,*,;minm,;eminn 3053225130251 === The length of the effective T stub is for the second bolt row
mm,,mL cp.eff 220453222 =∗== ππ , mm,,pmL cp.eff 1242140532 =+∗=+= ππ
mm,,,mL op.eff 019553206 =∗== α ( )
( ) mm,*,,**,,,
e,mp,mL op.eff
3181306250532214050532066250250
=+−+∗=
+−+= α
Leff
p
e m
t p
0,8 2 a
m2
m1
e
w1
Fig. 19.5 End plate
For the bolt row are governing the lengths
( ) ( ) mm,,;,;,;,minL;LminL op.effcp.effeff 318131810195124222041 === , ( ) ( ) mm,,;,minLminL op.effeff. 31813181019522 === ,
The minimal resistance is for basic failure modes
Failure mode 1
kN,N*,
,,*,*
mmL
F Rd.pl.p.effRd.p
2164102164532
57356318144
3
12
==
==
45
Failure mode 2
kN6,117N10*6,117
305,3210*0,78*30*25,7356*3,181*2
nmBn2mL2
F
3
3
Rd.tRd.pl.p2.effRd.2p
==
++
=
+
+=
Failure mode 3
kN,N*,,*BF Rd.tRd.p 015610015607822 32 ====
e) The resistance of the beam flange in compression may be calculated as
( ) ( ) kN,N,,,
,th
fWth
MF
fbbM
ypl
fbb
Rd.plRd.fb 7276107276
29220151235104285 3
3
0
=∗=−∗∗∗
=−
=−
=γ
.
f) The design resistance of each bolt row may be calculated as a minimal resistance of bolt row. For top row ( ) ( ) kN,,;,minF;FminF Rd.pRd.cfRd.t 3703704125111 ===
For middle row ( ) ( ) kN,,;,minF;FminF Rd.pRd.cfRd.t 611761177124222 ===
The tensile resistance of the third bolt row is not taken into account third row.
Fcf2.Rd
Fcf1.Rd
F
F
p1.Rd
p2.Rd
Fbf.Rd
Summary Column End plate
F
Ft1.Rd
t2.Rd 1
Result
124,7 kN
70,3 kN
276,7 kN
70,3 kN
z =250,4mm117,6 kN z =180,4 mm
2117,6 kN
125,4 kN
of resistances:
Fig. 19.6 Tension forces
The lever arms of the bolt rows to the centum of compressions are
mm,,z 4250229352201 =−+= , mm,,z 4180
229352202 =−−= .
The design bending resistance may be calculated as
kNm,,,,,zFzFM Rd.tRd.tRd.j 838180406117250403702211 =∗+∗=+= .
The plastic force distribution is expected. For the welds on flanges is taken into account the bending moment only
( ) ( ) kN,N,
,,,tbafF
Mww
wbfbfuRd.wfb 4237104237
35018009514025360
32 3 =∗=
∗∗−∗∗∗
=−
=γβ
The design bending resistance of the weld connecting the end plate is
( ) ( )rd.j
fbbRd.wfbRd.wfb
MkNm,kNm,Nmm*,
,*,thFM
=>==
−=−=
83805010050
292201042376
3
46
The design bending resistance of the welds is higher compare to the joint resistance. g) The connection shear resistance may be calculated from the third bolr row shear resistance from the M 16 bolt in shear
Fwfb.Rd
bfb
fa
twb
tfb
Fig. 19.7
kN,N,,
,Af,F
Mb
bubRd.v 56610566
4512018006060 3 =∗=
∗∗==
γ
The bolt bearing resistance for plate 12 mm will be lower for pitch
001001222502
001360800
41
173140
001
41
3 0
1
,,,,
min
,
min
,ffdp
minu
ub =⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧ −∗
=
⎪⎪⎪
⎭
⎪⎪⎪
⎬
⎫
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧−
=α F
Fp1.Rd
p2.Rd
Fb3.Rd
Fig. 19.8
kN,N,,
,,tdf,F
Mb
uRd.b 2119102119
4511216360015252 3 =∗=
∗∗∗∗==
γα
For the plastic distribution of internal forces may be expected the welds on web carrying the shear force only. The weld resistance is
( )
kN,N,
,,,fLaF
Mww
uweweRd.wbw
3279103279
350180036029222024
33 =∗=
∗∗
∗∗−∗∗==
γβ
Fwbw.Rd
Lwe
Fig. 19.9 The connection design shear resistance may be calculated as the minimum of bolts and shear resistance
kN,,,
,,
minFFF
minF
Rd.wbw
Rd.b
Rd.v
Rd.b 213366623279
211926662
22
3 =∗=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧∗∗
=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧= .
h) The initial stiffness of the connection may be calculated based on the deformation stiffness of the deformable components, the column web, bolts and end plate, see Fig. 19.10. The stiffness of the components in the first bolt row may be calculated from the stiffness coefficient of the component beam flange in bending
mm,,
,,m
tL,k fceff 58612
926128166850850
3
3
3
314 =
∗∗== .
The deformation stiffness coefficient of the bolt row is
mm,,LA
,kb
s 9786621212
157616110 =∗++
== .
The stiffness coefficient of the end plate is
mm,,
,m
tL,k peff 0884
3291270850850
3
3
3
315 =
∗∗== .
z
k k10 5
Lb
k4
Fig. 19.10
47
The resulting stiffness is
mm,
,,,kkk
keff 1402
08841
97861
586121
1111
1
1510
14
1 =++
=++
=
For the second bolt row it is for column flange
mm,,
,,m
tL,k fceff 39012
926122164850850
3
3
3
324 =
∗∗== ,
End plate
mm,,
,,m
tL,k peff 7577
532123181850850
3
3
3
325 =
∗∗== .
The second bolt row
mm,
,,,kkk
keff 8332
75771
97861
390121
1111
1
2510
24
2 =++
=++
= .
The lever arm may be estimated in the centre of the bolt rows mm,,z 821029220 =−= or more accurate from equation
mm,,,,,,,,,
zk
zkz
i ii.eff
i ii.eff2216
41808332425014024180833242501402 22
2
=∗+∗∗+∗
==∑∑ .
The stiffness of the part of the connection in tension is
mm,,
,,,,z
zkk i ii.eff
eq 84242216
4180833242501402=
∗+∗==
∑.
The connection initial stiffness may be predicted as
rad/kNmrad/Nmm,
,
,
k
zES ini.j 529471052947
84241
22160002101
922
=∗=∗
==
∑
i) The joint stiffness at ultimate level state may be reduced as
98828388385151
72
,,,*,
MM
,,
Rd
Sd =⎟⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞⎜⎜⎝
⎛=
ψ
μ
rad/kNmrad/Nmm,,
,SS ini.j
j 9071510907159882
1052947 99
=∗=∗
==μ
.
The moment rotation diagram of the joint is shown at Fig. 19.11
48
M =38,8 kNm
M
φ
j.Rd
S =47 529 kNm/radj.ini
S =15 907 kNm/radjoint
M =58,3 kNmpl.Rd
Fig. 19.11
The connected beam plastic resistance is
kNm,Nmm,,
,fWM
M
yplRd.pl 35810358
151235104285 6
3
0
=∗=∗∗
==γ
.
j) For given beam span may be the connection classified. The length, at which is the joint still rigid may
be calculated as
mm,
,S
IEL
ini.j
bb 979
10529471072700021088
9
6
=∗
∗∗∗== .
0 φ
pl.Rd
1,00
pl,Rd / (L M . )E Ib b
0,66
Rigid for beam longer than 0,979 m
M/M
Lb
bI
Lb
Fig. 19.12 Connection classification
Notes: 1) The software may calculate with the bolt rows interaction, which is for hand calculation too long. 2) The welds are designed for elastic frame global analyses. 3) This joint is rigid for braced frames for beam longer than 0,98 m. The stiffness may be
neglected.