Connections-F2007-PPP (1).pdf

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CIVE 374 / ENVE 471CIVE 374 / ENVE 471

Design ofDesign ofsimplesimple

connectionsconnectionsSections 9.1 to 9.9 of Reference TextSections 9.1 to 9.9 of Reference Text

Riveted connections

Bolted connections Weldedconnections

It all started with rivetsIt all started with rivets RivetsRivets


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Bolts replaced rivets in the 60Bolts replaced rivets in the 60 s after about 30s after about 30years of researchyears of research

Bolted Connections

Documents Manufacture of Bolts

Deals with: Chemical composition Strength properties (tensile strength, toughness,..) Physical dimensions, tolerances, etc

Documents Use of Bolts in Structures

www.boltcouncil.org

Documents Assistance with the Specifications

www.boltcouncil.org

Documents Assistance with the Assistance


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Structural BoltsStructural Bolts

ASTM A307 bolts common bolts made of low carbon steel. Tensile strength, F u, about 415 MPa. Tightening of these bolts is done only through

the ordinary effort of the worker, resulting in alow level of tension developed in the bolt.

ASTM A325 bolts Structural high strength bolts made of carbon

steel with a tensile strength of 825 MPa. Made both in metric and imperial sizes. The metric size bolts are designated as A325M

and have a tensile strength of 830 MPa

Structural BoltsStructural Bolts

ASTM A490 bolts High strength low alloy steel bolts with a

tensile strength of 1035 MPa. Metric size bolts of this grade, designated as

A490M, have a tensile strength of 1040 MPa. Galvanizing of these bolts for corrosion

protection is not allowed.

Structural BoltsStructural BoltsHow can we determine the grade of a bolt?

Comparison of Bolts: Direct TensionComparison of Bolts: Direct Tension

0.05 0.10 0.15 0.20

20

40

60

80

7/8 in. dia. A490 bolt7/8 in. dia. A490 bolt



elongation (inches)

bolttension

kips

Imperial size bolts

BuildingsBridges

Page of 6-158 of Steel Design Handbook


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Metric size bolts


Metric size bolts


Bolt lengths

Grip length


Types of bolted connections

Bearing type connections Bolt threads in the shear plane Bolt threads not in the shear plane

Friction type connections Direct tension connections

Bolt pretensionBolt pretension

To = 0.70 (0.75 A b) Fub

Bolts need to be pretensioned when (Clause 22.2.2 of S16):1. Slip-critical connections where slippage cannot be

tolerated (connections subjected to load reversals,connections in structures with rigorous deflection limitstates);

2. Shear connections in seismic areas;3. All elements resisting crane loads;4. Connections subject to impact or cyclic loading;5. Connections where bolts are subjected to tension loading;6. Connections with oversized or slotted holes unless

specifically designed to accommodate movement.

Shear planes

Bearing ConnectionsBearing ConnectionsP

P/2P/2


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Bolts in shearBolts in shear

r b b ubV = 0.6 n m A F

For bolt with threads not in the shear plane,

where: 0.6 is a shear correction factor b = 0.80n number of bolts in the connectionm number of shear planesA b nominal bolt areaFub tensile strength of the bolt

Bolts in shearBolts in shear

If threads are intercepted by shear plane,

r b b ubV = 0.6 n 0m A F 7.

30% reduction in strength

br = 0.67t = thickness of material in bearingd = bolt diameter n = number of boltsFu = tensile strength of plate

ur br B 3 t dn F=

Load BearingLoad Bearing Sample ProblemSample ProblemDetermine the number of bolts required for the connectionillustrated below.

Vr = 0.7 x 0.6 b n m A b Fub

= 0.7 (0.8) (0.6) (1 bolt) (2 shear planes) (314 mm 2) (825 MPa) = 174 kN N = 465kN/174kN/bolt = 2.7 bolts required Use 4 bolts

Bolt ShearGrip Length = 17.3 + 2x9.5 = 36.3 mmFrom Table in page 6-161 in Handbook, it can be observed that boltlength from 60 to 70 mm would do.Choosing 70 mm bolts, thread length = 36+7.5 = 44 mm

70 44 = 26 mm unthreaded length

The shear planes are at 9.5 mm and 26.8 mm. Therefore the threads arein a shear plane.

Bolt lengths


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Metric size bolts What if threads not in the shear planes?A b = (20)

2 / 4 = 314.2 mm 2 ; Fub = 825 MPa

V r = 0.6 b n m A b Fub = 0.6 (0.8) (1 bolt) (2 shear planes) (314 mm2) (825 MPa)

V r = 249 kN/bolt (Here only 2 bolts would be required)

BearingWhen the tension member is loaded the bolts will bear against the gusset

plate in one direction and against two angles in the other direction. Since thecombined thickness of two angles is larger than the thickness of the gusset

plate, the gusset plate will govern bearing resistance.

B r = 3 br t d n F u = 3 (0.67) (17.3 mm) (20 mm) (1 bolt) (450 MPa)

B r = 313 kN/bolt

No. of bolts required = 465 kN / 313 kN/bolt = 1.49 Need at least 2 boltsfor bearing.

Bearing ConnectionsBearing ConnectionsP

P/2P/2

The capacity o f the bolts in shear with threads in the shear planes governs the desig n ofthis connection. At least 4- A325 M20 bolts , 60 to 70 mm long are required.

Note: when designing the tension member the following block shear type failures should be checked.

SlipSlip --critical connectionscritical connections


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ExampleExample

Recall: We selected 4 A325-M20 bolts

s 1 s b uV = 0.53 c k m n A F

Assuming T f results from wind, T = 465/1.4 = 332 kN

s 1 s b uV = 0.53 c k m n A F

1 sc = 0.82 ; k 0 33.=For A325M Bolts and clean mill scale,

m = 2 shear planes; n = 4 bolts2

b uA 314 mm ; F 830 MPa= =

2sV = 0.53 0 82 0 33 2 4 314 mm 830 MPa. .

s f V = 299 000 N 299 kN V 1.4 = 332 kN= 1.3 m.94 used 20% reductio n when L > 1.3 m.

S16S16 01 uses a linear reduction with length01 uses a linear reduction with length

Reduction st arts at L > 15 dReduction st arts at L > 15 d


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Placement of Bolts in Connections

WeldedWelded

connectionsconnections

Welding processWelding process

Welding is a casting process where the consumable (if used)and a portion of the base metal are molten and allowed toflow within a joint where they solidify, thus joining thecomponents.

Welding thus involves the transfer of a sufficient amount ofenergy to the joint in order to raise the temperature of themetals above their melting point.

This energy could be provided by an electric current (arcwelding), burning of a fuel gas (brazing), friction (friction or inertia welding), and explosion, to name only a few methods.

Base metalWeld metal

Heat affected zone (HAZ)

Shielded metal arc welding (SMAW)Shielded metal arc welding (SMAW)

Commonly known as stick welding

Slow process because the electrode needs to be frequently replaced.


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Flux cored arc welding (FCAW)Flux cored arc welding (FCAW) Gas Metal Arc Welding (GMAW)Gas Metal Arc Welding (GMAW)MIG (metal inert gas) or MAG (metal activeMIG (metal inert gas) or MAG (metal active gas)weldinggas)welding

Submerged arc welding (SAW)Submerged arc welding (SAW)

Provides high quality welds.

High deposition rates can beachieved.

Used for fabricated steelmembers such as plategirders.

TIG (tungsten inert gas) welding or GTAWTIG (tungsten inert gas) welding or GTAW

Common weld typesCommon weld types Weld terminologyWeld terminology


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Steel Backing

Weld symbolsWeld symbolsBeveled Member

Weld symbolsWeld symbols

Combinations

Weld symbolsWeld symbolsCombinations


Weld Order

Weld symbolsWeld symbolsWeld Order



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Load vs. deformation responseLoad vs. deformation response

(of CSA-S16-01)

Base metal

Weld metal

Base metal

w = 0.67

Welded Connection Sample Calculations

The 12 mm plate shown below is loaded with a factored load of685 kN. Determine the size of the required fillet welds. UseE49018 electrode.

If we neglect the increase in strength of the fillets with the angle ,

Total weld length = = 666.4 mm22 (100) 2 2 50 325+ +a) Base metal unit resistance (resistance of a weld with 1mmleg and 1 mm long).

Vr = 0.67 w Am Fu = 0.67 x 0.67 x 1 mm 2 x 450 MPa= 0.202 kN/mm/mm

b) Weld metal unit resistanceVr = 0.67 w Aw Xu = 0.67 x 0.67 x 0.707 x 490

= 0.156 kN/mm/mm

We know the weld length to be 666 mm. We can determine therequired weld size from:

f

r

P 685D 6.59 mm

V L 0.156 666 = =


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Page 6-172 of CISC Handbook

Dmin : Since t 2 = 12 mm D 5 mm

Dmax : 12 mm 2 mm = 10 mm D

Choose D = 8 mm (Use a 8 mm fillet weld all around).

What if we want to account for the additional strength fromwelds b and c? S16 allows a greater strength from these weldsas given by the following:

Welds b ( = 45)

V r b = 0.67 w Aw Xu (1.00 + 0.50 sin 1.5 ) = 0.156 (1.2973)= 0.202 kN/mm/mm

This is the same as the base metal capacity (0.202 kN/mm/mm).

Weld c ( = 90),V r c = 0.67 w Aw Xu (1.00 + 0.50 sin 1.5 ) = 0.156 (1.5)

= 0.233 kN/mm/mm

This is greater than the strength of the base metal. The base metalcapacity governs the capacity of weld c (V r c = 0.202 kN/mm/mm).

Ductility and Compatibility RequirementDuctility and Compatibility Requirement

Note 1Although the strength of the welds at 45 o and those at 90 o issignificantly larger than the strength of longitudinal welds, testshave shown that when these welds are combined withlongitudinal welds, fracture of the weld group takes place

before the longitudinal welds reach their full capaci ty.

Note 2It is therefore not advisable to calculate the strength of acombination weld joint based on the increased strength of thewelds that are at an angle to the applied load. The increasedstrength given in the design standard should be used only whenthe welds at an angle to the applied load are used alone in a

joint.

Connections-F2007-PPP (1).pdf

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