Job No. Sheet 1 of 6 Rev B. Made by IR Date Oct Checked by FH/NB Date Oct Revised by MEB Date April 2006

Transcription

1 Job No. Sheet 1 of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April 006 DESIGN EXAMPLE 6 BOLTED JOINT A 0 0 angle loaded in tension is to be connected to a gusset plate of mm thickness. Stainless steel grade is used for the angle and the gusset plate. Eight austenitic 16 mm diameter bolts of property class 50 are used in a staggered line to connect one leg of the angle to the gusset plate. The proposed joint is shown in the following figure. It is required to determine the design resistance of the joint. 0x0x angle mm thick gusset plate 0 8 M16 property class 50 bolts 18 mm diameter holes x0 The connection is a Category A: Bearing Type connection. The design ultimate shear load should not exceed the design shear resistance nor the design bearing resistance. EN ,.4.1 Material properties The angle and plate is of material grade : The 0,% proof stress = 0 N/mm and the tensile strength is = 50 N/mm Table.1 Take f y as the 0,% proof stress = 0 N/mm and f u = 50 N/mm Section..4 The bolt material is of property class 50. Take f yb as the 0,% proof stress = N/mm and f ub = 500 N/mm Table. Partial safety factors Partial safety factor on gross section resistance: = M1 = 1,1 Partial safety factor on net section resistance: = 1,5 Partial safety factor on bolt resistance in shear and in bearing: = 1,5 Table.1 Position and size of holes Section 6.. For M16 bolts a hole diameter d 0 = 18mm is required. End distances e 1 = 0 mm and edge distances e = 5 mm e 1 and e < 4t 40 = 80 mm and > 1,d 0 = 1, 18 = 1,6 mm For the staggered bolt rows : - spacing p 1 = 60 mm >,d 0 = 9,6 mm 14

2 Job No. Sheet of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April distance between two bolts in staggered row 0 5 = 46,1 mm >,4d 0 = 4, mm - therefore, spacing for staggered rows p = 5 mm > 1,d 0 = 1,6 mm Note: For compression loading, e and p 1 should be checked that they satisfy local buckling requirements for an outstand element and an internal element respectively. Checks on both the angle and gusset plate are required. Design resistance of the angle gross cross-section in tension Section 6.. Gross cross-sectional area of the angle A g = 1915 mm Design plastic resistance Ag fy N pl, Rd = = = 8 kn Eq ,1 Design resistance of the angle net cross-section in tension Section For staggered holes the net cross-sectional area should be taken as the lesser of: - gross area minus the deduction for non-staggered holes s - Ag t nd 0 4 p Deductions for non-staggered holes Ag td0 = = 175 mm Net cross- sectional area through two staggered holes, n =, s = 0 mm and p = 5 mm: Section p= 5 18mm holes s = 0 s = 0 s 0 A net = A g t nd 0 = 1915 ( 18) 4 p 4 5 = 1915 (6-6.4) = 1619 mm Therefore A net = 1619 mm² Conservatively the reduction factor for an angle connected by one leg with a single row of bolts may be used. By interpolation for more than bolts in one row: β = 0,57. Table

3 Job No. Sheet of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April 006 Design ultimate resistance of the net cross-section of the angle: β N u,rd = A net f u 0, = = 91, kn 1, 5 Section 6.. Eq. 6.8 Design resistance of the angle in block tearing The expressions for block tearing are taken from EN rather than EN since EN explicitly covers angles. 60 Design resistance in block tearing considering rows as staggered: V eff,,rd = 0,5 fu A nt f y A nv 0,5 50 (60 18) 0 ( ) = = = 8 kn 1,1 Design resistance in block tearing considering rows as if non staggered: 0,5 fu A fy A nt nv V eff,,rd = 0,5 50 ( ) = = = 74 kn 40 0 ( ) 1,1 EN ,..() Eq.. EN ,..() Eq..9 Design resistance of the gross cross-section of the gusset plate Section 4.7. Gross cross-sectional area towards the end of the angle: A g = ( ) = 400 mm Design plastic resistance Eq. 4. Ag fy N pl,rd = = = 480 kn 1,1 Design resistance of the net cross-section of the gusset plate Section 4.7. Net cross-sectional area towards the end of the angle (where the applied load is greatest) through one hole non symmetrically placed on an element of width : b = = 40 mm: A net = Ag d0t = = 0 mm 145

4 Job No. Sheet 4 of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April 006 Net cross-sectional area towards the end of the angle through two staggered holes with s = 0 mm and p = 5 mm : s t 0 A net = A g d0t = p 4 5 = = 4 mm Therefore A net = 4 mm² Design ultimate resistance of the net cross-section of the gusset plate near the end of the angle: kr Anet f Eq. 4. u N u,rd = Reduction factor k r : k r = [ 1 r( d0 / u 0,) ] but < 1.0 u = e p therefore, u = 5 mm Eq kr = 1 0, = 1,16> 1,0 8 5 Take k r = 1,0 1, N u,rd = = 89, 1 kn It is advisable to check the resistance of net cross-sections at intermediate cross-sections along the gusset plate. Cross-section at the 1 st bolt hole near the gusset plate edge (Where b = 0 0/ = 117,5 mm) A net = Ag d0t = 117,5 18 = 995 = 995 Section mm This cross-section must be capable of transmitting the load from one bolt Design ultimate resistance at the section: kr Anet f u 1, N u,rd = = = 41, 88 kn It is obvious that there is no need to check any other cross-sections of the gusset plate as the load applied cannot exceed the design resistance of the angle itself which has been shown to be smaller than the above value. Eq

5 Job No. Sheet 5 of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April 006 Design resistance of the gusset plate in block tearing 5 40 Design resistance to block tearing considering rows as staggered: f y nv V eff,1,rd = u A f A nt 50 (5 9) 0 ( ) = 1,1 = 1, 98,4 = 508,6 kn Design resistance to block tearing considering rows as if non staggered: fu A f nt y Anv V eff,1,rd = EN ,..() Eq..9 EN ,..() Eq..9 = 50 (5 9) 0 ( ) 1,1 = 7,1 408,8 = 480,9 kn Design resistance of the bolts in shear The bolts are in single shear. Consider the shear to be in the plane of the threads. Section 6..4 Therefore, α = 0,5 Sectional area of M16 bolt: A s = 157 mm² Design resistance of class 50 M16 bolt of sectional area A s = 157 mm² : α fub As 0, F v,rd = = = 1,4 kn Eq. 6.9 Design resistance of the bolt group in shear: n b F v,rd = 8 1,4 = 51, kn 147

6 Job No. Sheet 6 of 6 Rev B, Route de Limours Tel : (0) Fax : (0) Revised by MEB Date April 006 Design resistance of the bolts/ply in bearing Section 6.. k1α b fu,reddt F b,rd = Eq. 6. Design resistance in bearing on the mm thick ply of the end M16 bolt where end distances e 1 = 0 mm, edge distances e = 5 mm ( > 1,d 0 = 1,6 mm) and bolt spacings p 1 = 60 mm and p = 5 mm The deformations at the serviceability load level: reduced design bearing stress f u,red is used here in order to avoid excessive bolt hole f u,red = 0,5 fy 0,6 fu but fu 0,5 0 0,6 50 = 48 N/mm < f u α d = e1 0 = d 0 18 = The reduction factor for the end bolt nearest the ends where e 1 = 0 mm, p 1 = 60 mm: α b = min ( α d ; fub / fu,red ; 1,0 ) = min (0,556; = 1,17; 1,0) = 0, 556 k 1 = e min,8 d 0 1,7;,5 5 = min,8 1,7 =,;,5 =, 18 The bolt itself is not critical in bearing, since f ub /f ur = 500/48 = 1,17 > 1,0 Design bearing resistance for the end bolt : k F b,rd = 1 αb f u,red dt, 0, = = 67,0 kn Design resistance of the joint in bearing: n b F b,rd = 8 67,0 = 56 kn Note :The critical mode of failure for all of the bolts in the joint is shear. Eq. 6.1 Eq. 6. Design resistance of the joint at the Ultimate Limit State The smallest design resistance found was that for block tearing of the connected angle leg : Design resistance to block tearing of the connected angle leg: N Rd = V eff,,rd = 74 kn Sheet 148