Bolt and Weld Connections-1

7/27/2019 Bolt and Weld Connections-1

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Structural Design FTUI 2012


2/82

Connections of structural steel members are of critical

importance. An inadequate connection, which can be the

,

failures.

a ure o structura mem ers s rare; most structura

failures are the result of poorly designed or detailed

connections.

Modern steel structures are connected by welding or

bolting (either high-strength or common bolts) or by

.

Welding has several advantages over bolting. A welded

,

any, holes. Connection that are extremely complex with

fasteners can become very simple when welds are used.


3/82

Riveted Welded


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5/82

Welding can be done in the shop and bolting in the field. Example : in the single-plate beam-to-column connection below,

the plate is shop welded to the column flange and field boltedto the beam web.

h


6/82

Tension member connectionand splice. It subjects theo s o orces a en o

shear the shank.

Beam end simple connection.

It subjects the bolts to forcesr .

Han er connection. Thehanger connection puts thebolts in tension


7/82

The bolts are subjected to shear or tension loading.

In most bolted connection the bolts are sub ected to

shear. Bolts can fail in shear or in tension.

Simple connection: If the line of action of the force

acting on the connection passes through the center ofgravity o t e connection, t en eac o t can eassumed to resist an equal share of the load.

he stren th of the sim le connection will be e ual tothe sum of the strengths of the individual bolts in theconnection.


8/82

Design Concept : Rn > Puwhere : R

n

= Factored design strength

Pu = factored load.

We need to examine the various possible failuremodes and calculate the corresponding designstrengths (Rn )

Possible failure modes are:1. Shear failure of the bolts2. Failure of connected member (failure mode of tension member)3. Edge tearing or fracture of the connected plate (gusset plate)4. Excessive bearing deformation at the bolt hole (bearing


9/82

Shearing stress in the bolt : fv = P/A = P/( db2/4)

reng o e o : = v x b

Where :

A = area of the bolt and db is its diameterfv

= shear yield stress = 0.6Fy

Bolts can be in singleshear or doubleshear as shownbelow.

When the bolt is in double shear, two cross-sections are effective inresisting the load. The bolt in double shearwill have the twice theshear strength of a bolt in single shear.


10/82

When the bolt is in double shear, two cross-sections areeffective in resisting the load. The bolt in double shearwill

.


11/82

Common bolts, also known as unfinished bolts, are designated as ASTM A307,which differ from high-strength bolts not only in material properties, but also

.

The nominal shear strength:bvn

AFR .

Where

Fv = ultimate shearing stressb = cross-sec ona area o e un rea e par o e o

The design strength: ).(75.0bvn

AFR

For ASTM A307, Fv=24 ksi


12/82

Two type of grades:1. A325

- - . .

2. A490Fy= 115 - 130 ksi Fy = 803- 908 MPa measured at 0.2% offset.

In certain cases, A325 and A490 bolts are installed to such degree of tightnessthat the are subjected to extremely large tensile force. The purpose of such a

large tensile force is to achieve the clamping force.

The total compressive force acting on the connected part is numerically equalto the tension in the bolt. A list of minimum tension values, for those connectionin which a minimum tension is required, is given in AISC Table J3.1.

If an external load P is applied, a friction force will develop between theconnected parts.The maximum ossible value of this force is NF .


13/82

The design shear strength of HTB is: ).(bvn

AFR

where = 0.75

The nominal shear strength of HTB is given by ultimate shearing stress timesthe nominal bolt area. shear rather than use a reduced cross-sectional area. Approximately, the threaded area = 0.75 x unthreaded area.

Fastener Nominal Shear Strength

Rn = FvAb

,

A325X, threads not in plane of shear

A490N, threads in plane of shear

A490X, threads not in plane of shear

60Ab60Ab75Ab


14/82

Hole is slightly larger than

fastener is loosely placedin hole

The stress will be highesta e ra a con ac po n(A).

calculated as : the appliedforce divided by theprojected area of contact

fp = P/(db t)

where P = force applied to the.


15/82

The bearing stress state can becomplicated at the nearby/edge bolt.

he bolt spacing and edge distance willhave an effect on the bearing stress.

of the bolt type because the bearing

stress acts on the connected plate not.

A possible failure mode resulting fromexcessive bearin close to the ed e of

the connected element is shear tear-out. It can also occur between two

load.


16/82

Upper limit : to prevent excessive deformation of the hole Rn = C x Fu x bearing area

= C. F .d .tWhere :

C = 3 (If deformation is not a concern > 0.25in)C = 2.4 (If deformation is a concern < 0.25 in.

= .

THEN :

If deformation is a concern < 0.25 in.

R = 1.2 L t F 2.4 d t F

If deformation is not a concern > 0.25inRn = 1.5 Lc t Fu 3.0 db t Fu

Lc = the clear distance in the load direction, from the edge of the bolt holeto the edge of the adjacent hole or to the edge of the material

F = specified tensile strength of the connected materialt = thickness of connected material


17/82

The upper limit will become effectivewhen 1.2 L t F = 2.4 d t F

then Lc = 2 db

If Lc < 2 db, Rn = 1.2 Lc t FuIf L

c> 2 d

b, R

n= 1.4 d

bt F

u


18/82

In a simple connection,all bolts share the loadequally.

T/n T/nT/n T/n

The shear strength of allbolts = shear strength ofone bolt x number of

T

TT/n T/n

T/n T/n

T

TT/n T/n

T/n T/n

The bolts are subjected toshear and the connecting

TT

TT

connec e p a es aresubjected to bearingstresses.Bearing stresses in plate

T

Bearing stresses in plateBearing stresses in plate

T

The bearing resistanceshall be taken as the sumof the bearing resistancesof the individual bolts.

Bolt in shear

Bearing stresses in plate

T

Bolt in shear


Bolt in shear


T


19/82

AISC AISC Specification SNIThe minimum s acin s 2 d or3d is referred 3 x diameter baut

The maximum spacing 24 times thickness of connected

part but not more than 12

15 tp atau 200mm

The minimum edge

distance (Le)

See AISC Table J3.4 1.5 x dimeter baut

The maximum edge

distance (Le)

12 times the thickness of the

connected part (but not more

4 tp + 100 atau

200mm

than 6 in.).

=


20/82

AISC SpecificationS asi minimum s 3 * d

Spasi maksimum 24 times the thickness of the thinner

part (but not more than 12 in.).

The minimum edge distance (Le) See AISC Table J3.4

The maximum edge distance (Le) 12 times the thickness of the

connected part (but not more than 6

in.).


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Le s s

Le

(a) (b)


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Hole Dimensions

Diameter Standard

(Dia.)

Oversize

(Dia.)

Short-slot

(Width x Length)

Long-slot

(Width x Length)

9/16 5/8 9/16 x 11/16 9/16 x 1

5/8

7/8

11/16

13/16

15/16

13/16

15/16

1 1/16

11/16 x 7/8

13/16 x 1

15/16 x 1 1/8

11/16 x 1 9/16

13/16 x 1 7/8

15/16 x 2 3/16

1 1/8

d+1/16

d+5/16

x

(d+1/16)x(d+3/8)

x

(d+1/16)x(2.5xd)


23/82

Nomina Rivet or Bo tDiameter (in) At Sheared Edges

At Ro e o P ates, S apes orBars, or Gas Cut Edges [c]

7 8

5/8

7/8

1 1/8

1 1/4

1 1/2 [d]

7/8

1

1 1/8

1

1 1/8

1 1/4

1 3/4 [d]

2

2 1/4

1 1/4

1 1/2

1 5/8

[a] Lesser edge distance are permitted to be used provided Equation from J3.10, asa ro riate are satisfied.

[b] For oversized or slotted holes, see Table J3.8.

[c] All edge distance in this column are permitted to be reduced 1/8-in. when theholes at a point where stress does not exceed 25 percent of the maximum designs reng n e e emen .

[d] These are permitted to be 11/4-in. at the ends of beam connection angles and

shear end plates.


24/82

A36 5 x _

3/8 in.

A36 5 x _

3/8 in.

1/2

1.25

2.50

1.2565 k

A361.25

2.50

1.2565 k

A36

60 k

1.25 2.50 1.25

_ in. bolts

1.25 2.50 1.25

_ in. bolts3/4 A325N

I. The design shear strength of bolt in shear :

Rn= FnAb = 0.75 x 48 x x 0.752/4 = 15.9 kips per bolt

Shear stren th of connection = 4 x 15.9 = 63.6 ki s

II. Check bolts spacing :

minimum edge distance = 1 in. for rolled edges of plates (see Table J.3.4)

edge distances (1.25 in.) > 1 in OKmin.spacing = 2.67 db = 2.67 x 0.75 = 2.0 in.

preferred spacing = 3.0 db = 3.0 x 0.75 = 2.25 in.

given spacing (2.5 in.) > 2.25 in. OK


25/82

A36 5 x _

3/8 in.

A36 5 x _

3/8 in.

1/2hole diameterdh = 3/4 + 1/16 = 13/16 in

1.25

2.50

1.2565 k

A361.25

2.50

1.2565 k

A36

60 k

A36 : Fu = 58 ksi

1.25 2.50 1.25

_ in. bolts

1.25 2.50 1.25

_ in. bolts3/4

III. Bearing strength at bolt holes (plate 1/2in thickness)

edges : Lc = 1.25 dh/2 = 1.25 (13/16)/2 = 0.844 in.

Rn = 0.75 (1.2 Lc t Fu) = 0.75(1.2 x 0.844 x 0.5 x 58) = 22.02 kipsupper limit : 0.75 (2.4 db t Fu) = 0.75(2.4 x 0.75 x 0.5 x 58)= 39.15 kips

other holes, s = 2.5 in, Lc = 2.5 dh = 1.688 in.

Rn = 0.75 x (1.2 Lc t Fu) = 0.75 x (1.2 x 1.688 x 0.5 x 58) = 44.05 kipsUpper limit : 0.75 (2.4 db t Fu) = 39.15 kips.

Bearing strength at holes = 2 x 22.02 + 2 x 39.15

= 122.34 kips


26/82

A36 5 x _

3/8 in.

A36 5 x _

3/8 in.

1/2hole diameterdh = 3/4 + 1/16 = 13/16 in

1.25

2.50

1.2565 k

A361.25

2.50

1.2565 k

A36

60 k

A36 : Fu = 58 ksi

1.25 2.50 1.25

_ in. bolts

1.25 2.50 1.25

_ in. bolts3/4

IV. Bearing strength at Gusset Plates (3/8 in)

edges : Lc = 1.25 dh/2 = 1.25 (13/16)/2 = 0.844 in.

Rn = 0.75 (1.2 Lc t Fu) = 0.75(1.2 x 0.844 x 0.375 x 58) = 16.52 kips. . b u = . . . . = .

other holes, s = 2.5 in, Lc = 2.5 dh = 1.688 in.

R = 0.75 x 1.2 L t F = 0.75 x 1.2 x 1.688 x 0.375 x 58 = 33.04 ki s

Upper limit : 0.75 (2.4 db t Fu) = 29.36 kips.

Bearing strength at gusset = 2 x 16.52 + 2 x 29.36 = 91.76 kips

Bearing strength of the connection is the smaller = 91.76 kips


27/82

onnec on reng Shear strength = 63.3 kips

Bearing strength (plate) = 122.34 kips

Bearing strength (gusset) = 91.76 kips

Rn = 63.3 ki s > 60 ki s factored load

Connection strength (Rn) > applied factored loads (Pu). OK.


28/82

Step I. Select 2L3 x 2 x 3/8 with Pn = 112 kips (yielding) and 113 kips(fracture)

Step II. Select size and number of boltsThe bolts are in double shear for this design (may not be so for other designs)

Use four 3/4 in. A325 bolts in double shear ; Rn = 31.8 x 4 =127 kips

Step III. Design edge distance and bolt spacing

Le min = 1 in. for 3/4 in. diameter bolts in rolled edges.Select Le = 1.25 in.

Minimum spacing = 2.67 db = 2.0 in.

Preferred spacing = 3.0 db = 2.25 in.

e ec spac ng = . n., grea er an pre erre or m n mum spac ng

Step IV. Check the bearing strength at bolt holes in angles

Angle thickness = 3/8 in.

e = . .

Rn = 44.0 x 3/8 = 16.5 k


29/82

BS at non-ed e holes s = 3 in. = R = 78.3 x 3 8 = 29.4 kBearing strength at bolt holes in each angle = 16.5 + 3 x 29.4 = 104.7 kips

Bearing strength of double angles = 2 x 104.7 kips = 209.4 kips

Step V. Check the fracture and block shear strength of the tension member

Step VI. Design the gusset plate

The plates must be designed for the limit states of yielding and rupture

Limit state of yielding

Rn = 0.9 Ag Fy > 100 kips - Ag = L x t > 3.09 in2

Assume t = in - L > 6.18 in. Design gusset plate = 6.5 x in.

Limit state for fracture

An = Ag (db+1/8) x t

An = 6.5 x 0.5 (3/4 + 1/8) x 0.5 = 2.81 in2

But, An 0.85 Ag = 0.85 x 3.25 = 2.76 in2

Rn = 0.75 x An x Fu = 0.75 x 2.76 x 58 = 120 kips

Design gusset p ate = 6.5 x 0.5 in.


30/82

Step VII. Bearing strength at bolt holes in gusset plates

e . . Plate thickness = 1/2 in.

BS at the edge holes = Rn = 44.0 x 1/2 = 22.0 k

- = = = n . .

BS at bolt holes in gusset plate = 22.0 + 3 x 39.15 = 139.5 kips

Connection Member Gusset Plate

Shear strength = 127 kips Yielding = 113 kips Yielding = 105.3 kips

BS = 209.4 kips (angles) Fracture = ? Fracture = 120 kips

BS = 139.5 (gusset) Block Shear = ?

Overall Strength is the smallest = 105.3 kipsGusset plate yielding controls

Nominal Strength > Factored Load (100 kips).Design is acceptable


31/82

The classification of a connection with high-strength bolts:1. Slip-critical connections: no slippage is permitted2. Bearing type connections: slip is acceptable

Theoretically, SCC are not subject the shear and bearing, but they musthave sufficient shear and bearing strength in the event of an overload that

.

To prevent slip, the service loadshear on the fastener, must not exceed:

).(bvn

AFR

where = 1.0 for standard, oversized, short-slotted, and long-slotted holes

the long slot is perpendicular to the line of force = 0.85 for long-slotted holes the long slot is parallel to the line of

force,


32/82

High strength (A325 and A490) bolts can be installed with such adegree of tightness that they are subject to large tensile forces.

These large tensile forces in the bolt clamp the connected platestogether. The shear force applied to such a tightened connection

P

P

P

P

P

P

TightenedTightenedTightened


33/82

There are currently four authorized procedure for installation ofhigh-strength bolts:

1. The turn-of-the-nut method

.

3. Alternate design bolts

4. Direct tension indicator


34/82

N =Tb

= b

PN =TbN =Tb

= b

P

= b

P

Tb

N =Tb

F=N

=

TbTb

N =Tb

F=N

N =Tb

F=N

==

Tb

N = Tb

N =Tb

P

Tb

N = Tb

Tb

N = Tb

N =Tb

P

N =TbN =Tb

P

Slip-critical bolted connections can be designed to resist the applied

N = TbN = TbN = TbN = Tb

shear forces using friction.

If the applied shear force is less than the friction that developsbetween the two surfaces, then no slip will occur between them.


35/82

Slip will occur when the friction force is less than the applied shearforce. After slip occurs, the connection will behave similar to the

- .

Table J3.1 summarizes the minimum bolt tension that must be- .

High strength bolts in slip-critical connections can be designed topreven s p e er as a serv cea y m s a e or a e requ restrength limit state.

However, the connection must also be checked for shear strengthand bearing strength.


36/82

A325

Fy=81- 92 ksi or Fy= 566- 643 MPa measured at 0.2% offset..

Fy= 115 - 130 ksi Fy = 803- 908 MPa measured at 0.2% offset.

In certain cases, A325 and A490 bolts are installed to such degree of tightnessthat the are subjected to extremely large tensile force. The purpose of such alarge tensile force is to achieve the clamping force.

The total compressive force acting on the connected part is numerically equalto the tension in the bolt. A list of minimum tension values, for those connectionin which a minimum tension is required, is given in AISC Table J3.1.

If an external load P is applied, a friction force will develop between theconnected parts.The maximum ossible value of this force is NF .


37/82

=mean slip coefficientWhere : = coefficient of static friction between connected parts,N = the normal compressive force acting on the inner surfaces.

: depend on the surface condition of the steel: Painted or Rusted

a For Class A surface un ainted clean mill scale steel surface orsurface with Class A coating on blast-cleaned steel), = 0.33

(b) For Class B (unpainted blast-cleaned steel surface or surface

with Class B coating on blast-cleaned steel), = 0.50- , = 0.40

Thus, each bolt in the connection is capable of resisting a load of P=F, even if

the bolt shank does not bear on the connected part. As long as this frictionalforce does not exceeded, there is no bearing or shear. If P is greater than F andslippage occurs, shear and bearing will then exist and will affect the capacity of

.


38/82

Slip resistance = Rn = m * Du * hsc * Tb * Nsw ere,

= 1.0 for connections at serviceability limit 0.85 for connections at the required strength level.

m = mean slip coefficient for Class A or B surfaces

=0.35 for Class A surfaces (upainted clean mill scale)

= 0.50 for Class B surfaces (unpainted blast cleaned surfaces

Du = 1.13 reflects the ratio of the mean installed bolt pretension

to the specified minimum bolt pretension.

hsc= hole factor= 1.00 for STD, 0.85 for OVS and SSLT, 0.70 for LSLT)

Tb= minimum bolt tension given in Table J3.1

Ns= number of slip planes


39/82

Splice plateSplice plate

W8 x 28 W8 x 28W8 x 28 W8 x 28

Splice plateSplice plate

Step I. Service loads = D + L = 200 kips.Step II. Slip-critical splice connection, assume class A surface, Standard holes (STD)Rn of one fully-tensioned slip-critical bolt = 0.35x1.13x 1 x TbNs

I db = 7/8 in. From Table J.3.1 -- Tb=39kips

Rn of one bolt = 1.0 x 0.35 x 1.13 x 1.00 x 39 x 1 = 15.4 kips

Rn of n bolts = 15.4 x n > 200 kips (splice must be slip-critical at service)

,

Choose 16 fully tensioned 7/8 in. A325 bolts on each side of the splice

3 3 3 1.25


40/82

Step III. Layout of splice connection C.L.

Minimum edge distance (Le) = 1-1/8in. from Table J3.4

Design edge distance Le = 1.25 in. Minimum spacing = s = 2-2/3 db = 2.67 x 7/8 = 2.336 in. (Spec. J3.3)

= = . b = . = . . . .

Design spacing = 3 inch

Ste IV. Connection stren th at factored loads =300ki s

The splice connection should be designed as a normal shear/bearing connection

Beam flange thickness : tf= 0.465 in and flange width bf= 6.535 in.

The shear strength of bolts = (0.75x48xAb) /bolt x 16 = 345.6 kips

Bearing strength at edge holes (Le = 1.25 in.) = 40.8 kips/in. thickness

- = . = . .

Bearing strength = 4 x 40.8 x 0.465 +12 x 91.3 x 0.465 = 673 kips


41/82

Step V. Design the splice plate

Tension yielding: 0.9 Ag Fy > 300 kips; Ag > 6.66 in2

Tension fracture: 0.75 A F > 300 ki sAn =Ag - 4 x (7/8 +1/8) x t > 6.15 in.

Beam flange width = 6.535 in.ssume . n. w e sp ce p ates w t t c ness = . n.

The strength of the splice plate=

Fracture = 329 kips

Check Block shear (Homework)

Step VI. Check member strength (yield, fracture and block shear)


42/82

e ua an eser :Rn = 0.75 * m * rt * Fu * Ab

dimana :

m = jumlah bidang geser

Fu = Tegangan tarik ultimate (putus) dari baut(Lihat tabel tipe-tipe baut)

rt = 0.5 untuk baut tanpa ulir pada bidang geser

= 0.4 untuk baut dengan ulir pada bidang geser

Ab = luas area baut

Kekuatan TumpuUntuk Le > 1.5 db dan s > 3db

Rn = 0.75 * 2.4 * db * t * FuFu = Tegangan tarik ultimate (putus) terkecil antara baut dan pelat

Untuk baut dengan ukuran lobang besar :Rn = 0.75 * 2 * db * t * Fu


43/82

TIPETIPEBAUT

TipeBaut DiameterBaut(mm)

ClampingForces/ProofStress(Mpa) TeganganLeleh(Mpa)(Fy) KuatTarikMin(Fu)(MPa) Material

A307 6.35 10.4 NA 60 Kadarkarbonrendah

A325 12.7 25.4 585 560 630 825 Bajakartbon

A490 12.7 38.1 825 790 900 1035 BajaAlloy

Gayayangdibutuhkan

untuk mengencangkan

u


44/82

3@70mm 2L 120x120x85040

Gussett=12mmD22 A308

DuaprofilsikuL120.120.8disambungdengan4bautberdiameter22mm(7/8in)jenisA308(Fv=168Mpa).

a.Cekapakahlayoutsambungantersebutmemenuhipersyaratanjarak

yangdiaturAISC

.

c.JikabautditukardengantypeA490N(Fv=420Mpa);hitungpersentase

peningkatankekuatanGESERsambungan.


45/82


46/82

125


47/82

Pelatpenyambung6mm 125

WF250x125250

TampakSamping Potongan

125Disainlah sambungan pada sayap (flange)

profil WF250x125(tf=9mmdan tw=6mm)

TampakAtas

beban mati adalah 100kN,beban hidup

70kNdan beban angin 50kN,dengan

menggunakan baut berdiameter 16mm

n mana s pt a n an

terjadi.Sket hasil disain dalam gambar kerja

yanglengkap (tampak samping,tampak

atas dan oton an !


48/82


49/82

Structural welding is a process whereinthe parts to be connected are heated

molten metal added to the joint

(a) (b)


50/82

(SMAW) is usually donemanually and is the processuniversally used for fieldwelds. For shop welding, anautomatic or semi automatic

process is usually used.Foremost among these is thesubmerged arc welding

, gas s e e meta arc, flux cored arc, andelectro-slag welding


51/82

(1) Fillet weld, which are defined as those placed in a corner formed by twoparts in contact. Example a lap joint & a tee joint.

roove we , are ose epos e n a gap, or groove, e ween wo par sto be connected. They are most frequently used for butt, tee, and cornerjoints. In most cases, one or both of the connected parts will have beveleded es called re ared ed es althou h relativel thin materials can begroove welded with no edge preparation.

Type: Complete penetration groove welds.

(3) Plug or slot weld


52/82


53/82

Fillet welds are most common and used in all structures. Weld sizes are specified in 1/16 in. increments

A fillet weld can be loaded in an direction in shear, com ression, ortension. However, it always fails in shear.

Fillet weldFillet weldFillet weld

i et wei et wei et we


54/82

Assumption the cross section (A) =Effective throat thickness 0.707 a x length L

= o= o

a

a

= 0.707 a

a

a

= 0.707 a

L

The shear failure ofthe fillet weld occursalong a planethrough the throat ofthe weld

Failure Plane


55/82

wLav

707.0

The nominal Design Strength:

wwn a .

ww Laf 707.075.0R

n

where Fw= the weld ultimate shear strength =


56/82

Fw depends on the weld metal used that is, it is a function of the typeelectrode.

Ultimate tensile strength of Electrode: 60, 70, 80, 90, and 110ksi.

E60XX / E60 = an electrode with an ultimate tensile strength of 60 ksi.This is the standard terminology for weld electrodes

Electrodes should be selected to match the base metal:E70xx electrodes : for steels with Fy < 60 - 65 ksi

- . E70XX is the most popular electrode used for fillet welds made by the SMAW

method.

Fw in a fillet weld = 0.60 times the tensile strength of the weld metal,

denoted by FEXX.

= ---- =E70XX: Fw = 0.75 [0.60 (70)] = 31.5 ksi

E80XX: Fw = 0.75 [0.60 (80)] = 36 ksi


57/82

axisofweld

Fillet weld strength that account for load direction :

Fw = 0.6FExx (1.0 + 0.50 sin

1.5

)

Fw = 0.6 FExx is valid only if = 0

= ,

strength is 50% higher


58/82

AISCJ2.4cspecifiesthatthelargernominalstrengthfrom

thefollowingtwooptions:

33333

1.Usethebasicstrengthforboth:Rn =Rwl +RwtR

wl

=Rwt

=0.6FExx

2.Usethe50%higheroftransverseweldsbutreduce

t e as cstrengt o ong tu na we s

Rn =Rwl +Rwt

n = . . Exx + . . Exx


59/82

The design shear strength of base metal:

sheartosubjectmetalbaseofareaxFRn BM

gFRn BM where= 0.90FBM = Fv=0.60 Fy

Ag = the area subject to shear

AISC specification J4.2

Base metal area can fail by shear yielding or rupture. The smaller of the

.

Shear yielding;

n . . y

Shear rupture;Rn = 0.75 x 0.6 Fu x net area of base metal subjected to shear

, y u .


60/82

Example :

TT

ElevationPlan

ElevationPlan

Stren th of weld in shear Stren th of base metal

= 0.75 x 0.707 x a x Lw x fw = min {1.0 x 0.6 x Fy x t x Lw0.75 x 0.6 x Fu x t x Lw}


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1. Always check weld metal and base metal strength. Smaller valuegoverns. In most cases, the weld metal strength will govern.

2. In weld design problems, it is advantageous to work with strength

per unit length of the weld or base metal.

Example:

t

The design strength of weld per inch of length:

WnFxsizexR 707.0

The capacity of the bracket plate in shear per inch of length:

tFRnBM


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a. Berdasarkan Kekuatan Las

Wn FaxxR 70775.0

b. er asar an e ua an a an asar ase e a

(Pelat yangdisambung)

FuaxxRn

6.070775.0


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Minimum Size of Fillet Welds amin:Material Thickness of Thicker Minimum Size of Fillet

Part Joined (in.) Welds[a] (in.)

To inclusive

Over to

1/8

3/16

Over to

Over 3/4

5/16

[a] Leg dimension of fillet welds. Single pass welds must be used.

. .

Teballasminimum(t,mm) Ukuranlassudutminimumpalingtebal (amin)

t


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- function of the thickness of the thickest connected late:

- plates with thickness 0.25 in., amax- = t or 0.25 in. (the smallest).- plates with thickness 0.25 in., amax = t - 1/16 in.

L Lw- Lw 4 a otherwise aeff= Lw / 4

- Intermittent fillet welds: Lw-min = 4 a and 1.5 in.

When a weld extends to the corner of a member, it must be continued aroundthe corner to avoid stress concentration.

Length of end returns > two times the weld size. = 2a

The maximum size than can be made with a single pass ofe ectro e s approx mate y nc , an mu t p e passes

will add to the cost.


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Pe at engan te a 6.4mm;amks =t 1.6mm


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1/4 66 Near side (arrow side)

1/4 6 Ot her side

1/4 6Bot h side

Weld all arround

1/4 6E70

e erence e we


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t = 0.5 in.t = 0.5 in.

Step I. Check the limitations

tmin = 3/8 in. (member)5 in.

4 in x 3/8 in.a = 0.25 in.

5 in.

4 in x 3/8 in.a = 0.25 in.

tmax = 0.5 in. (gusset)

Then, amin = 3/16 in.

amax = 3/8 - 1/16 = 5/16in.5 in.

0.5 in.

. .

5 in.

0.5 in.

. .

weld size = a = 1/4 in. OK!

Lw-min = 1.0 in.

. . min.

OK!

End returns :

ep . es gn s reng o e we

Weld strength = fx 0.707 x a x 0.60 x FEXX x Lw= 0.75 x 0.707 x 0.25 x 0.60 x 70 x 10 = 55.67 kips

Min size = 2 a = 0.5 in. OK!

Step II. Design strength of the weld


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. y w . u w

= min {1.0 x 0.6 x 50 x 10 x 3/8 ; 0.75 x 0.6 x 65 x 10 x 3/8}

= min {112.5 ; 109.7 kips}

.

Step III. Tension strength of the member

-n . .

Rn = 0.75 x Ae x 65 - tension fracture

Ae = U A

2 g .U = 0.75 , since connection length (Lconn) < 1.5 w

Therefore, Rn = 54.8 kips

The design strength of the member-connection system = 54.8 kips.Tension fracture of the member governs.

.


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Step I.Assume material properties :

Fy= 36ksi for member & gusset plates and E70XX electrode for the fillet welds.

Step II. Design the tension member

Select 2L5 x 3 x 1/2 made from 36 ksi steel.

Yield strength = 260 kips Fracture strength = 261 kips.

tep . es gn t e we e connect on

amin = 3/16 in. amax = 1/2 - 1/16 in. = 7/16 in.

Design, a = 3/8 in. = 0.375 in.

Shear strength of weld metal = Rn = 0.75 x 0.60 x FEXX x 0.707 x a x Lw= 8.35 Lw kips

Strength of the base metal in shear =

min {1.0 x 0.6 x Fy x t x Lw ; 0.75 x 0.6 x Fu x t x Lw}

min { 10.8 Lw ; 14.62 Lw} kips

Shear strength of weld metal governs, Rn = 8.35 Lw kips

Rn > 250 ips 8.35 Lw > 250 ips

Lw > 29.94 in.

use Lw = 30.0 in.


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Welding on both

sides of gusset. L1 a

(b)

Welding on both

sides of gusset.

Welding on both

sides of gusset. L1 aL1 aL1 aL1 a

(b)

2a2a2a2a

L2

(c)

L2L2L2L2

(c)

Lw= 30 in. for two angles

Assume Lw for each = 15.0 in.E70XX fillet weld can be placed in

L3L3L3L3L3

AISC S ec re uires that the fillet weld(a)(a)


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AISC S ec. re uires that the fillet weld( )( )

terminate at a distance greater than thesize (1/2 in.) of the weld. For this option,L1 will be equal to 7.5 in.

The fillet weld can be returnedcontinuously around the corner for a

L1 a

(b)

L1 aL1 aL1 aL1 a

(b)

. .

L2 can be either 6.5 in. or 7.5 in.However, the value of 7.5 in. is

preferred.

2a2a2a2a

L3 will be equal to 5.75 in.

L2

(c)

L2L2L2L2

(c)

33333

Step V. Fracture strength of the member

A = U A


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A = U A

U = 1- x/L Asssume case a ------ U = 1-0.901/7.5 = 0.88 0.9

Rn = 0.75 x 0.88 x 8.00 x 58 = 306.24 kips > 250 kips OK

Step VI. Design the gusset plate Rn > Tu

Tension yielding limit state0.9 x A x 36 > 250 ki s A > 7.71 in2

Tension fracture limit state

0.75 x An x Fu > 250 kips

AISC s ecification A 0.85 A 1/2 7.5 in.(a)

1/2 7.5 in.(a)

An > 5.747 in2 Ag > 6.76 in

2

Design gusset plate :

1/2 7.5 in.

Gusset plate 8 x in.

1/2 7.5 in.

Gusset plate 8 x in.

thickness = 1.0 in.

width = 8.0 in.

1/2 7.5 in.1/2 7.5 in.Two 5 x 3.5 x 1/2 in1/2 7.5 in.1/2 7.5 in.Two 5 x 3.5 x 1/2 in

E707 225


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7 120E70

120.120.8

7 75

Hitung beban maksimum yang

.


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D=80kN

L=140kN Akibat pembebanan yang

ada disainlah sambun an


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antara profil siku 100x100x8dengan gussetplatesetebal

3/8inc(9.5mm)

menggunakan tebal las

minimumyang

disyaratkan.

Sambungan hanya pada 2

32.4

sisi saja (arah longitudinal)

dan memperhitungkan efek

eksentrisitas akibat titik

10032.4

.disain dalam gambar kerja

yanglengkap!

100


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UnbalancedCondition

. uran as yangsama

pada sisi atas dan

bawah,men akibatkan

terjadinya

ketidaksetimbangan

pada gayagaya

. sambungan bersifat

eccentric

Disain Sambungan Laspada Profil yangunsymetri(profil siku)


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Dengan menerapkan konsep kesetimbangan,

1.maka :P1 P2

2.Akan diperoleh panjang las yangberbeda antara sisi

atas dan sisi bawah L1 L2


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Sambungan Baut : 7.4.4 ; 7.4.6 ;

Sambungan Las : 7.11.2; 7.11.4

Bolt and Weld Connections-1

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Transcript of Bolt and Weld Connections-1