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Steel Design Tutorial - Connections - SolutionQuestion 1: Determine a suitable bolt size in grade 8/S for the lap joint shown below. The tensile force on the joint is 300kN and the tensile strength of the plates is 410MPa (grade of plate is 250MPa). Plate thickness = 8mm.300 kN40654040656540P* = 300...

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Steel Design Tutorial - Connections - Solution Question 1: Determine a suitable bolt size in grade 8.8/S for the lap joint shown below. The tensile force on the joint is 300kN and the tensile strength of the plates is 410MPa (grade of plate is 250MPa). Plate thickness = 8mm. 40 65

300 kN

65 40 40

65

40

P* = 300 kN (a)

Check Min design action §9.1.4 (b) (iv) Member design capacity in tension ØNt = lesser of Agfy or 0.85 ktAnfu Agfy

= 210 x 8 x 280 MPa = 470 kN ↖Table 2.1, flats and sections, t < 11mm

0.85ktAnfu

= 0.85 x 1 (210 – 3 x 26) x 8 x 410 = 368 kN ↖guess M24 bolts so 26 mm holes

 ØNt = 0.9 x 368 kN = 331 kN > 300 kN OK 0.3 ØNt = 99.4kN = min. design action (not critical)  P* = 300 kN governs

(b)

Check bolts in shear Shear / bolt = 300 kN / 6 = 50 kN From AISC design capacity tables Adopt or

(c)

M24-4.6/S (ØVfn = 64.3 kN threads included) M16-8.8S (ØVfn = 59.3 kN threads included)

Check plate bearing and tear-out ØVb = Ø3.2dftpfup = 0.9 x 3.2 x 16 x 8 x 410 = 151 kN > 50 kN OK Or ØVb = Øaetpfup = 0.9 x 40 x 8 x 410

= 118 kN > 50 kN OK

 Adopt 6-M16-8.8/S bolts (or 6-M24-4.6/S bolts) Question 2: If the connection in Question 1 is made using two fillet using E43xx electrodes determine the size of welds required. N* = 300 kN Lw = Weld length = 2 * 210 = 420 mm Shear stress =

N * 300 *10 3 = = 714 N/mm Lw 420

For GP weld ØVw = Ø0.6 fuw tt kr

=

↙Table 9.7.3.10(1) 0.6 * 0.6 * 430 * tt * 1.0 = 154.8 tt N/mm 154.8tt = 714  tt  4.6mm Leg length required = 4.6/0.7 = 6.6 mm

 Adopt 8mm fillet welds, GP category – Too large for 8mm plate If use E49xx Electrode, fuw = 490MPa, leg length = 6.6 x 430 / 490 = 5.8mm ie 6mm CFW E49XX (SP) CFW = Continuous Fillet Weld

Question 3 300

Determine if the joint shown can safely support a total vertical load of P* = 440 kN. The bolts are 16mm 8.8/S grade and the tensile strength of the plate is 410MPa.

40 3 @ 100 = 300 100 40 12 mm thick plate 40

P* = 220 kN each face of the UC Direct shear/bolt = 220/8 = 27.5 kN Indirect shear due to eccentricity of the load. M* = 220 * 0.3 = 66 kNm

The bolts are assumed to carry shear loads perpendicular to the line from the centroid of the bolt group AND in proportion form the distance away from the centroid of the bolt group ie the further away from the centroid the more load they carry.

r1 =

150 2 + 50 2 = 158.1mm

r2 =

50 2 + 50 2 = 70.7mm

 P2 =

70 .7 P1 = 0.447 P1 158 .1

 (P1 * 0.158 + 0.447P1 * 0.0707) x 4 = 66  P1 = 87.0 kN Vertical component of P1 = 87 *

50 = 27.5 kN 158

Horizontal component of P1 = 87 *

150 = 82.6 kN 158

Resultant max shear on bolt =

( 27.5 + 27.5) 2

+ 82.62 = 99.2 kN

Capacity of M16-8.8/S = 59.3kN  too low. Would require 8-M24 – 8.8/S (ØVfm = 133 kN/bolt) Plate tearout capacity = 190kN (12mm plate, ae = 40 mm) Plate bearing capacity = >304kN OK

Question 4 If the bolts in Question 3 are to be replaced by 2 horizontal fillet welds, determine the size of the welds required. Use E49xx electrode. Weld length = 2 * 180mm horizontal welds Area welds = 2 * 180 x tt Ix = [180tt *1902] x 2 Iy = 2 [tt * 1803/12] J = Ix + Iy

= = = =

360tt mm2 13 * 106 tt mm4 972 * 103tt mm4 13.97 * 106tt

Direct shear stress Fs =

220 *103 = 611/tt N/mm2 360t t

Torsion shear stress max

66 *10 * 210 Tr = = 992/tt N/mm2 J 13.97 *10 6 tt

6

Vertical component

= 992/tt x 90/210

= 425/tt

Horizontal component = 992/tt x 190/210 = 898/tt Max stress at corner

2 =  (611 + 425) + 898 2  / t t = 1371/ tt  

GP weld Vv/tt

= Ø0.6 fuwkr =0.6*0.6*490*1= 176.4N/mm2 from previous

 tt = 1371/176.4

= 7.77 mm

tw =

7.77 0.7

= 11.1 mm

Use  Need 12mm weld – This is very large weld – normally limit to 6 or 8mm. Or if SP welds need Vv/tt = 235.2 N/mm2 (Ø = 0.8)  tt = 5.83 tw = 8.3  9mm weld Better solution would be to use GP weld but on 3 sides

Question 5 Determine a suitable bolt size in grade 8.8/S for the bolted bracket connection shown. The bracket carries an ultimate vertical load of 550kN. Plate thickness is 8mm.

250 40

5@60 = 300

P* = 550 kN M* = 550 x 0.25 = 137.5 kNm

40 180

Considering rotation about the bottom bolt 2 *[F * 0.3 + F *

240 * 0.24 + F * 0.6 * 0.18 + F * 0.4 * 0.12 300

+ F * 0.2 * 0.06] = 137.5  F = 104 kN Shear/bolt =

= tensile force/bolt N*tf

550 kN = 45.8 kN 12

Using interaction graphs in AISC  M20-8.8/S OK Check bearing capacity and tearout for 8mm plate = 127 kN OK

Question 6 If this connection is to be made with two horizontal lines of fillet welds at the top and bottom of the existing bracket, determine the size of the welds required. Use E49xx electrode. Area welds

=

2 * 180 * tt

= 360 tt

Direct shear stress

=

550 *10 3 360t t

Ix weld group

=

2 * 180tt * 1902 = 13 * 106 * tt mm4

=

1528 N/mm2 tt

6

Shear stress due to bending =

Resultant shear stress

=

137.5 * 10 *190 2010 My = = N/mm2 I tt 13 * 10 6 t t

1528

2



+ 20102 / tt = 2525 / tt N / mm2

For GP category, capacity  tt required =

2525 176 .4

= Ø 0.6 fuw kr =0.6*0.6*490*1= 176.4 N/mm2 = 14.3

tw = 20.4 mm  21mm fillet weld required (too high – should weld all round)...