Tutorial 7 - Connections - Part A

8/12/2019 Tutorial 7 - Connections - Part A

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TUTORIAL 7

CONNECTIONSPart 1

Bolted

27 MARCH 2012


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Problem 1 :Calculate the design resistance of the slip connections as shownon figure bellow. The cover plates arte made S430 Steel and connected witheither (a) non-preloaded bolts of diameter 20mmm and grade 4.6 or (b) pre-stressed bolts also of diameter 20mm and grade 4.6. assume that in bothcases, the shear plane passes through the untreated portions of the bolts.


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SOLUTION:

a). Non- Preloaded Bolts

STEP 1: Design Shear Resistance( F v,Rd , per shear plane):

f ub = 400N/mm 2(Table 3.1 EC 3-1-8 ), Mb=1.25 and A= 20 2/4=314.16mm 2

All 4 bolts in double shear resistance:

F sd = 4 x ( 2 x 60.32) = 482.56 kN

kN A f

F M

ub Rd v 32.60

100025.1

16.3144006.06.0

2

,


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STEP 2: Bearing Resistance( F b,Rd , )

Bearing failure tends to take place in cover plates since they are thinner

is the smallest of:

1: e 1/3d 0 = 35/(3x22)=0.530 ; 2 : p 1/3d 0 = [70 /(3x22)]- = 0.810

3: f ub /f u = 400/430 = 0.930

Thus = 0.530

Bearing Resistance( 1 bolt)

All of bolt group:Fbd = 4 x ( 2 x 54.7) = 437.6 kN

kN dt f

F M

u Rd b 7.54100025.1

620430530.05.25.2

2

,


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STEP 3: Tensile Resistance of cover plates( F b,Rd , )

Net Area of cover plate, A net , is:

Anet = 6 x 140 - 2 x 6x 2 2 =576mm 2

Design resistance(1 plate):

Both Plates:F t,Rd = 2 x 178.33= 356.66 kN

Hence design resistance of the connection is (the lowest) = 356 kN

kN f A

N M

unet Rd u 33.178100025.1

4305769.09.0

2,


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b). Prestressed Bolts

STEP 1: Preloading force( F p,Cd ):

F p,Cd = 0.7xf ub x As=(0.7 x400 x 245)/1000 = 68.6 kN

STEP 2: Slip Resistance( F s,Rd )

Assuming the surfaces have been shot blasted, i.e.. Class A, = 0.5 , for 2

surfaces n = 2, standard clearance k s = 1.0

Slip resistance, F s,Rd (each bolt):

For all bolts

F s,Rd = 4 x 54.88 = 219.52 kN

Hence design resistance of the connection with prestressed bolts is (the

lowest, compared to the Plates Tension Resistance) = 219.52 kN

kN F k

F Cd s Ms

s Rd v 88.546.6825.1

5.020.1,,


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Problem 2 : If a beam is to be connected to a column using 8no. Grade 4.6,M16 bolts as shown[ next slide ], calculate the maximum shear resistance ofthe connection.

SOLUTION:STEP 1: Check Positioning of Holes for Bolts

Diameter of bolt, d = 16mm ; diameter if bolt hole, d 0 = 18mm; End distances e 1 = 30mm; Edge distance, e 2 = 35mm; Spacing between center of bolts in the direction of load transfer p

1 = 60 mm;

Spacing between rows of bolts, p 2 = 115mm; Thickness End Plates, t p = 10 mm; Condition to be met(Table 3.3 EC3-1-8):

OK mmd mme 6.21182.12.130 01

OK mmd mme 27185.15.135 02OK mmd mm p 6.39182.22.260 01

OK mmd mm p 2.43184.24.2115 02


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STEP 2: Shear resistance of Bolt Group

Cross-sectional Area:

Shear of Bolt Group, Vv,Rd

STEP 3: Bearing resistance of Bolt Group

is the smallest of:1: e 1/3d 0 = 30/(3 x 18)=0.555; 2 : p 1/3d 0 = [60 /(3x 18)]- = 0.8613: f ub /f u = 400/430 = 0.930

Thus = 0.555

Bearing of Bolt Group is:

22

06.201416

mm A

kN V

A f F V

Rd v

Mb

ub Rd v Rd v

83.308

100025.106.2014006.0

86.0

88

,

,,

kN xdt f

x xF Mb

pu Rd b 94.610

100025.1

1016430555.05.28

5.288 ,


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STEP 4: Resistance of welded connection between Beam and End Plate:

STEP 5: Shear Resistance of End Plate

Bolts holes in the plate do not need to be taken into account provided that

where:

and

Substituting the above gives:

Hence Shear Resistance of End Plates(2 planes) is:

kN

f A xV

M

yv Rd pl

81.725

100005.1)3/275(2400

2)3/(

220

,

u yvnet v f f A A //,

kN 62.4481000 / )24063.934( 2a ) L F ( 2V Rd ,w Rd ,w

2240024010 mm x Av

64.0430/275/7.02400/1680 u y f f

2, 168010842400 mm x x A net v


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STEP 6: Local Shear Resistance of Beam Web:

Shear Ares of the Web is:

Shear Resistance of Web

The maximum Shear Resistance of the connection is controlled by the

resistance of the bolt group( the lowest of all ) and is equal to 308.83kN .

kN

f AV

M

ywebvWeb Rd pl

24.330

100005.1)3/275(2184)3/(

0

,,,

2, 21841.9240 mm x Lt A wwebv


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Problem 3 : Show that the double angle web cleat beam-to-columnconnection details as shown [ fig.],is suitable to resist the design shear force,Vsd = 225 kN . Assume the steel grade S430 and the bolts are of grade 5.8and diameter 16mm .


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SOLUTION:

STEP 1: Check Positioning of Holes for Bolts

d = 16mm ; d 0 = 18mm; e 1 = 30mm; e 2 = 45mm; p1 = 50 mm; t p = 10 mm; Condition to be met(Table 3.3 EC3-1-8):

STEP 2: Shear resistance of Bolt Group

Assuming that the shear plane passes through a threaded portion of the bolt,Tensile stress Cross-sectional Area of the bolt is A s = 157mm 2

f ub = 500 N/mm 2 (table 3.1 EC3-1-8 for grade 5.8 )

OK mmd mme 6.21182.12.130 01

OK mmd mme 27185.15.145 02

OK mmd mm p 6.39182.22.250 01


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Shear of Bolt Group, Vv,Rd

STEP 3: Bearing resistance of Bolt Group

is the smallest of:

1: e 1/3d 0 = 30/(3 x 18)=0.555; 2 : p 1/3d 0 = [50 /(3x 18)]- = 0.6763: f ub /f u = 500/430 = 1.163

Thus = 0.555

Bearing of Bolt Group is:

OK KN V KN V

A f F V

sd Rd v

Mb

ub Rd v Rd v

225314

100025.11575005.0

105.0

1010

,

,,

OK KN V KN

xdt f x xF

sd

Mb

pu

Rd b

22568.763

100025.11016430555.05.2105.21010 ,


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STEP 4: Shear Resistance of Legs Cleats

Bolts holes need not to be allowed provided that

where:

and

Substituting the above gives:

Hence Shear Resistance of both Legs Cleats is:

OK KN V KN

f A xV

sd

M

yv

Rd pl

2253.786

100005.1)3/275(2600

2)3/(

220

,

u yvnet v f f A A //, 2260026010 mm x Av

64.0430/275/654.02600/1700 u y f f

2, 1700185102400 mm x x A net v


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STEP 5: Shear Action on Bolts due to Shear Force and Resultant Bending Moment

Shear Force per bolt, F v,Sd , is

Maximum Shear Force on bolt assembly in horizontal direction, F h,Sd is:

Resultant Force:

Shear Resistance per bolt:

Total Shear Resistance(by bolts in double shear):

kN nV F Sd Sd v 455/225/,

kN x x p M F Sd Sd h 5.40)505/(452255/ 1,

kN F F F Sd hSd vSd V 54.605.4045 222

,

2

,,

kN F Rd v 4.31,

OK kN F kN x F Sd v Rd v 5.608.624.312 ,,


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STEP 6: Bearing Resistance of Legs Cleats connected to Web of Beam

is the smallest of:1: e 1/3d 0 = 30/(3 x 18)=0.555; 2 : p 1/3d 0 = [50 /(3x 18)]- = 0.6763: f ub /f u = 500/430 = 1.163

Thus = 0.555

STEP 7: Bearing Resistance Web of Beams

is the smallest of:1: e 1/3d 0 = 35/(3 x 18)=0.648; 2 : p 1/3d 0 = [50 /(3x 18)]- = 0.676

3: f ub /f u = 500/430 = 1.163 Thus = 0.648

OK kN F kN

xdt f

x xF

Sd v

Mb

pu

Rd b

5.6074.152

100025.11016430555.05.2

25.2

22

,

,

OK kN F kN

dt f F

Sd v

Mb

webu Rd b

5.6052.61

100025.19.616430648.05.25.2

,

,


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STEP 8: Block Shear Resistance ( V eff,Rd )

Thickness of web of beam =6.9

and

mm xd mma L 8016556011

mmaand mm L v 6.95 _ _ 200 3 mm x f f kd a L yut 6.40)275/4300)(185.035()/)(( .022

mm xmm

f f nd aa Laa L L yuvvv

3.415)275/4300)(185.06.9560200(6.3556.9560200

)/)(( .031313

3

321,

6.3006.4060200 Lmm

L L L L L veff v2

,,

2074

6.3009.6

mm

xtL A eff veff v

OK KN V kN

f AV

Sd

M

yeff v

Rd eff

22561.313

100005.1)3/275(2074)3/(

0

,

,

bl l f b l


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Problem 4 : In-Plane Design Resistance of a bolt group .Problem: Determine the design resistance of the bolt group in the web fin plate

joint shown[ next slide ]

SOLUTION: For 20mm grade 8.8 bolts, A s = A t = 245mm 2

STEP 1:

v= 0.6 (Table 3.4 EC3-1-8), ; f ub = 400N/mm 2 ( Table 3.1 EC3-1-8), Mb=1.25

so that

kN A f

F Mb

sub Rd v 08.90100025.1

2458006.06.0,

08.94370.0 vQ kN Q v 23.254


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P bl 5 Pl B i R i


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Problem 5 : Plate Bearing Resistance Problem: Determine the bearing resistance of the bolt group in the web fin

plate joint.

SOLUTION: For 20mm grade 8.8 bolts, A s = A t = 245mm 2

STEP 1:

Fp,C = 0.7 x 800 x 245 = 137.2kN

Ks = 1.0 (Table 3.6, EC3-1-8); = 0.4 v= 0.6 (Table 3.7 EC3-1-8) ; M3,serv =1.10

so that89.49370.0 sLQ kN Q sL 83.134

kN F k

F C p Ms

s Rd s 89.492.137

1.1

4.00.10.1,,

P bl 6 O Pl R i f b l G


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Problem 6 : Out-Plane Resistance of bole Group Problem : The flexible end plate joint is to transmit factored design actions

equivalent to the downwards force of Q kN acting at the centroid of the boltgroup and an out-of-plane moment M x = -0.05Q kNm. Determine themaximum bolt forces, and design resistance of the bolt group.

SOLUTION

STEP 1: Bolt Group Analysis :

The maximum bolt tension given by:

Average bolt shear is:

QkN N xQx

F Ed t 107.049000)7035()1005.0( 6

.

2222 490003541054 mm x x z i

QkN N Qx F Ed v 125.08/)1005.0( 6

.


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STEP 2: Plastic Resistance of Plate :

If the web thickness is 8.8mm,

And so: hence

STEP 3: Bolt Resistance :

At plastic collapse of the plate, the prying force causes a plastic hinge at the bolt line,so that the prying force = 0.712 x 10 6 / 30N = 23.7kN .

Bolt Tension = 0.107 x 443.0 + 23.7 = 70.6kN

Using the solution of E.g.5, F v,Rd = 94.08kN .

8.6813 __ 8.3568.02/8.82/90 F EC mm xm

4.6813 __ 4.1253530625.08.352 T EC mm x xl eff

2.6813 __ 712.00.1/35584.12525.0 2,1, T EC kNmm Nmm x x x Rd pl

2.6813 __ 9.9430/10712.04 6,1, T EC kN N x x F Rd T

Q x F Rd T 712.02,1, kN Q 0.443


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For 20mm Grade 8.8 bolts, A t = 245mm 2 and f ub = 800N/mm 2,

Hence:

And so the bolt resistance does not govern.

Thus:

4.3813 __ 1.14125.1/2458009.0, T EC kN N x x F Rd t

0.1946.01.1414.1

6.701.94

0.443125.0 x

x

kN Q 0.443

P bl 7 Pl t B i g R i t


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Problem 7 : Plate Bearing Resistance Problem: Determine the bearing resistance of the end plate.

SOLUTION:

is the smallest of:1: e 1/3d 0 = 35/(3 x 22)=0.530; 2: f ub /f u = 800/430 = 1.569 Thus = 0.530

k1 = 2.8x30/22 = 3.82 > 2.5 so that k 1 = 2.5

Using the analysis solution,

so that

kN Q 3.98125.0

kN dt f

F Mb

pu Rd b 3.9810001.1

820510530.05.25.2,

)(0.4438.786 pb QkN kN Q

Problem 8: Design of pin joint for roof truss[ next slide ] The forces size of


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Problem 8: Design of pin joint for roof truss[ next slide ]. The forces, size ofangles are and tees have been obtained from analysis at ultimate loadassuming pin joints and axial forces in the members. The centroidal axes ofthe members intersect so there is no eccentricity to be taken into account.The thickness of the gusset plate is at least 6mm, and at least equal to theminimum thickness of the angle or tee(6.1mm). Use a 10mm thick gradeS275steel.

SOLUTION:

STEP 1: Tensile Resistance of the tee S275:

OK kN kN x

x Af N

M

y

Rd pl 2155.709

100000.12752580

0

,

OK kN kN x x x x f A N M unet

Rd u 21515.785100025.1 430)2222580(9.09.0 2,


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Member 31 structural tee cut from 165x152UB20 welded to gusset plate


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Member 31, structural tee cut from 165x152UB20, welded to gusset plate,design force Ned = 125kN

STEP 2: Resistance of 4-M20 class 4.6 bolts in single shear

STEP 3: Resistance of 2-M20 class 4.6 bolts in bearing on the web tee section (t=

6.1mm)

Where: For an end bolt b = e 1/(3d 0)= 60/(3x22)=0.909 For an edge bolt k 1 = 2.8e 2/d 0 -1.7 = 2.8x55/22 -1.7 =5.3 > 2.5 The Strap increases out of the plane stiffness of the truss.

OK kN N kN F

kN x A f

n F

Ed Rd v

M

ubb Rd v

12516.188

16.188100025.12454006.0

46.0

,

2

,

OK kN N kN F

kN x x

xdt f k

n F

Ed Rd b

M

ubb sb Rd b

12544.177

44.177100025.1

1.620400909.05.22

,

2

,

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Tutorial 7 - Connections - Part A

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