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Title4. Steel Joint Design (Double Angle)
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Page 1

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 

Clause Remarks
Bolted Beam to Column Connection using Web Cleats


Show that the double angle web cleat beam-to-column
connection detail shown below to resist the design shear force,
VEd, of 200 kN. Take steel grade as S275 and the bolts are class 8.8
and having diameter of 16 mm.





Configuration Beam to Column Flange
Column 254 x 254 x 89 UC, S275
Beam 406 x 140 x 46 UB, S275
Connection Double Angle connection using non-preload

bolts, class 8.8, M16
Double Angle 90 x 90 x 10mm thk. – 260mm, S275

Page 2

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 

1. JOINT DETAILS AND SECTION OF PROPERTIES


COLUMN 254 x 254 x 89kg/m UC, S275

Web thickness, twc 10.3 mm
Flange thickness, tfc 17.3mm
Yield Strength, fyc 275 N/mm2
Ultimate Strength, fuc 430 N/mm2


BEAM 406 x 140 x 46 kg/m UB, S275

Web thickness, twbl 6.8 mm
Flange thickness, tfbl 11.2 mm
Yield Strength, fybl 275 N/mm2
Ultimate Strength, fubl 430 N/mm2

DOUBLE ANGLE, 90 x 90 x 10mm thk, S275

Depth, hp 260 mm
Thickness, tp 10 mm
Yield Strength, fyp 275 N/mm2
Ultimate Strength, fup 430 N/mm2

Direction of Load Transfer



Number of bolts row, n1 5
Plate Edge to first bolt row, e1 30 mm
Pitch between bolts row, p1 50 mm

Direction to perpendicular to load transfer



Numbers of vertical lines of bolts, n2 1
Plate Edge to first bolt line, e2 45 mm
Gauge, p2 96.8 mm


BOLTS, NON PRELOAD, M16 CLASS 8.8 BOLTS

Gross Section of Bolt, A (un-threaded portion) 201 mm2
Tensile Stress Area, As (threaded portion) 157 mm2
Diameter of shank, d 16 mm
Diameter of Holes, do 18 mm
Yield Strength, fyb 640 N/mm2
Ultimate Strength, fub 800 N/mm2

Page 3

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 

2. POSITIONING OF HOLES FOR BOLTS


3.5(1)


Table 3.3




Minimum and maximum spacing and end and edge distance for
bolts and rivets are given in Table 3.3

Minimum Maximum

End distance, e1 = 1.2 do = 4tp + 40mm
= 1.2 (18) = 4(10) + 40mm
= 21.6 mm = 80 mm

Edge distance, e2 = 1.2 do = 4tp + 40mm
= 1.2 (18) = 4(10)+ 40mm
= 21.6 mm = 80 mm

Spacing, p1 = 2.2 do = smaller of 14tp or 200mm
= 2.2 (18) = 14 (10)
= 39.6 mm = 140 mm


Since – 21.6mm < 30mm < 80mm .: End distance, e1 satisfied
- 21.6mm < 45mm < 80mm .: Edge distance, e2 satisfied
- 39.6mm < 50mm < 140mm .: Spacing, p1 satisfied




3. BOLTED CONNECTION








Table 3.4


















Shear Resistance Of Bolts Group (at Supporting Column)

For this example, we assume that the shear plane passes through
threaded portion of the bolt. Hence, tensile stress area of bolt As =
157mm2 and for bolt class 8.8, αv = 0.6

Fv,Rd = (αv fub As ) / γM2
= (0.6 x 800 x 157) / 1.25
= 60.28kN

Hence, shear resistance of bolt group, Vv,Rd

Vv,Rd = Fv,Rd x n
=60.28kN x 10
=602.8kN

Page 4

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 




Table 3.4




Bearing Resistance Of Bolt Group (at Supporting Column)

Fb,Rd = (k1 αb fu d t ) / γM2

Where αb is the smallest of αd , fub/fup , 1.0

For end bolts αd = e1/3do
= 30 / 3(18)
= 0.56

For inner bolts αd = (p1/3do) – (1/4)
= (50/3(18)) – (1/4)
= 0.68

fub/fup = 800/430
= 1.86

.: smallest αd = αb = 0.56

And k1 for edge bolts is smallest of:

(2.8e2/d0) – 1.7) , (1.4(p2/do) – 1.7 and 2.5

(2.8e2/d0) – 1.7 = (2.8(45)/18) – 1.7
= 5.3

(1.4(p2/do) – 1.7 = (1.4(96.8/18) – 1.7
= 5.83

And k1 for inner bolts is smallest of:
(1.4(p2/do) – 1.7 or 2.5

(1.4(p2/do) – 1.7 = (1.4(96.8/22) – 1.7
= 4.46

.: smallest k1 = 2.5

.: Fb,Rd = (k1 αb fup d tp ) / γM2
= (2.5 x 0.56 x 430 x 16 x 10) / 1.25
= 77.1kN

Bearing Resistance of bolt Groups, Vb,Rd

Vb,Rd = Fb,Rd x n
=77.1 kN x 10
= 771 kN

Page 5

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 

Shear Resistance Of Bolts Group (at beam web)

Shear force per bolt in vertical direction, Fv,Ed

Fv,Ed = VEd /n
= 200 kN/5
= 40 kN

Maximum shear force on bolt assembly in horizontal direction,Fh,Ed




Fh,Ed = MEd /np1
= (VEd x s)/ np1
= (200 x 45)/ 5(50)
= 36 kN

Maximum design resultant shear force, Fv,Ed = √(Fv,Ed)2 + (Fh,Ed)2
= √(40)2 + (36)2
= 53.81 kN

Since the bolts are in double shear the total shear resistance is

Fv,Rd x 2 = 60.28 kN x2
= 120.56 kN

Since 120.56 kN > 53.81 kN .: shear resistance on bolt group
connecting cleat to web of beam is adequate.


4. SHEAR RESISTANCE OF CLEATS CONNECTED TO SUPPORTING
COLUMN




6.2.6(2)
(EC3-1)




Vpl, Rd = Av (fyp/√3) /γMo

Gross Section

Av = hp tp
= 260(10)
= 2600 mm2

Vpl, Rd = Av (fyp/√3) /γMo
= (2600)(275/√3) / 1.0
= 412.81 kN

Consider for 2 plates .: Vpl, Rd = 412.8 kN x 2
= 825.6 kN

Page 6

STRUCTURAL STEELWORK & TIMBER DESIGN - ECS328
STEEL JOINT DESIGN
NOV13 – MAC14 MA.T


 

Net Section

Avnet = tp (hp – n1do)
= 10 (260 – (5x18))
= 1700 mm2

Vnet, Rd = Av (fup/√3) /γM2
= (1700)(430/√3) / 1.25
= 337.63 kN

Consider for 2 plates .: Vnet, Rd = 337.636 kN x 2
= 675.26 kN


5. BEARING RESISTANCE OF BEAM WEB


Table 3.4



Fb,Rd = (k1 αb fu d t ) / γM2

Where αb is the smallest of αd , fub/fup , 1.0

For end bolts αd = e1bl/3do
= 35/3(18)
= 0.65

fub/fup = 800/430
= 1.86

.: smallest αd = αb = 0.65

And k1 for edge bolts is smallest of:

(2.8e2bl/d0) – 1.7) , (1.4(p2/do) – 1.7 and 2.5

(2.8e2bl/d0) – 1.7 = (2.8(60)/18) – 1.7
= 7.63

.: smallest k1 = 2.5

.: Fb,Rd = (k1 αb fubl d twbl ) / γM2
= (2.5 x 0.65 x 430 x 16 x 6.8) / 1.25
= 60.82 kN


Since Fb,Rd = 60.82 kN > Fv,Ed = 53.81kN .: Bearing resistance of web
of supported beam is adequate.

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