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Definicion:Definicion: La soldadura es el proceso de conectar piezas de metal entre si por medio de la aplicación de calor, ya sea con presión o sin presión.El tipo de soldadura mas común en trabajos de acero estructural es la de fusión (método para conectar piezas por medio de metal fundido). Se sujeta un alambre o varilla especial a un calor intenso en su extremo (1480 C) generando que esta se funda y deposite sobre el metal base, el cual se funde localmente y se une al metal depositado formando una conexión soldadaProcesos precalificadosProcesos precalificados:

• Soldadura de arco metálico protegido (SMAW)• Soldadura de arco metálico sumergido (SAW)• Soldadura de arco metálico con gas (GMAW)• Soldadura de arco con núcleo fundente (FCAW)

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Soldadura de arco eléctricoSoldadura de arco eléctricoEl calor se genera por medio de un arco eléctrico formado entre un electrodo de acero y las partes que se van a soldar. El calor del arco funde simultáneamente el metal base y el electrodo, y el campo electromagnético conduce el metal fundido de la varilla de soldadura (electrodo) hacia el metal base, mientras que el operador mueve el electrodo, manual o automáticamente, a lo largo de la soldadura con una velocidad adecuada y depositando la cantidad necesaria de metal de aportación.

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Soldadura de arco eléctricoSoldadura de arco eléctrico

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1 – Soldadura de arco metalico protegido( Shielded Metal Arc Weld) (SMAW):La mas común de las técnicas de soldaduraLos electrodos están recubiertos con ciertos compuestos minerales y parte de estos se funden formando:

•Atmosfera gaseosa protectora alrededor del arco metálico (estabiliza el arco y protege de los gases atmosféricos)•Una capa fluida de escoria de menor densidad que el metal fundido, sube a la superficie, retardando la rapidez de enfriamiento del metal de soldadura y protegiéndolo de una exposición indeseable a los gases atmosféricos.

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1 – Soldadura de arco metalico protegido( Shielded Metal Arc Weld) (SMAW):

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2 – Soldadura de arco sumergido (SAW):Es un proceso automático en que el arco esta cubierto por un

montículo de material granular fundible. Un electrodo metálico

desnudo es alimentado desde un carrete, es fundido y depositado

como material de relleno. El electrodo, la funde potencia y una

tolva de fundente están unidos aun bastidor que se coloca sobre

rodillos y se mueve a cierta velocidad conforme se forma el cordón

de soldadura

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2 – Soldadura de arco sumergido (SAW):

Submerged arc welding (SAW)

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A – A tope:Requires full penetration “groove welding”, but it

eliminates eccentricity and more pleasant to view.Suitable for shop welding.

B – De traslape:Is the most common, due to ease of fitting, requires

“fillet weld”, and can be double-sided.

C – Junta en T:Useful for fabricating

“Built-up” sections

D – DE esquina:-

E – De borde:-

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A) Soldaduar de preparacion (tope): Se usan siempre cuando las partes a conectar están alineadas en el mismo plano, también se pueden usar en juntas en T

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A) Soldaduar de preparacion (tope):

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B) Soldadura de filete:Se usan en juntas de traslape, en T y de Esquina

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C) Soldadura de ranura:

They can be used in addition to fillet welds when more areas of weld are required.

D) Soldadura de muesca: Same as above

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W-10

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Aw = te Lw {AISC J 2.1.1a. Page 16.1-93}

where:

Aw = Área efectiva de la soldadura. te = Espesor efectivo de la garganta.Lw = Longitud efectiva.

• El espesor efectivo de garganta de una soldadura a tope con penetración parcial será el indicado en la Tabla,(table J 2.1 Page 16.1 -44)• El espesor efectivo de garganta de una soldadura a tope de forma acampanada, (table J 2.3 page 16.1-95).

t1t2 t t

te = t1

te = t

El área efectiva de la soldadura a tope será determinado como el producto de la longitud efectiva de la soldadura por el espesor efectivo de garganta.

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Fillet welds are assumed to have a cross section of 45 right triangle, as

shown. The size of the weld is the leg of the triangle, Denoted as (w). The failure plane (the weakest section) is along the “throat” of the weld, denoted as

(t); Where (t = 0.707 w). The length of the weld (L)is the length of the shear plane along the weld.

Effective Area of Fillet Welds:

Aw = te Lwwhere,Aw = Effective area of weld.

te = Effective throat thickness = 0.707 wLw = Effective Length of weld.

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1 – Minimum Size of Fillets Welds :

2 – Maximum Size of Fillets Welds :

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3 – Minimum Effective Length of Fillet Welds :

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Select a suitable fillet weld size for the connection shown below:

Solution :-

Maximum weld sizeFor ½” thick plate:

in.167

161

21

161t

in. 83 between-in size welda Select

in. 165 plate thick 1 the

for size weldMinimum

Note : Maximum weld size controlled by thin plate, Minimum weld size controlled by thick plate.

½”

1”

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1 – Strength of Groove welds : The strength of full penetration groove weld is superior

to that of the base metal, if matching filler is selected (Page W-15).

The Nominal strength of weld per one inch length (Rnw)

is based on yielding of base metal rather than on weld material:

Rnw = te Fy (Tension or Compression)

Rnw = te (0.6 Fy) (Shear)

where : te = effective thickness of weld.

Fy = yield stress of base metal.

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2 – Strength of Fillet Welds :The strength of fillet welds are always controlled by shear on the

weld plane or shear on the base metal. Thus:

Rnw = te (0.6 FEXX) weld metal.

or

Rnw = t (0.6 Fu) base metal.

where:

Rnw = Resistance (strength) of weld per one inch length.

te = Effective throat thickness of weld.

t = Thickness of the base metal.

FEXX = Ultimate tensile strength of electrode.

Fu = Ultimate tensile strength of metal.

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weld.of length unit per load factoredR :whereR R Φ

: WeldsFor R Φ

:general In

u

unw

n

iiQ

1 – Groove Welds (Full penetration):

A – Tension or compression (Table J2.5)

material weldF t0.90ΦR

material base F t0.90ΦR

ywenw

yenw

B – Shear on Effective Area (Table J 2.5)

material weld)F (0.6 t 0.90ΦR

material base )F 0.6( t 0.90

t 0.90ΦR

EXXenw

ye

yenw

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2 – Fillet WeldsThe design strength per unit length of fillet weld is based on the

shear resistance through the effective throat of the weld (Table J2.5):-

but should not be greater than the shear capacity of the base metal:EXXe

EXXenw

F t0.45 material eld w)F (0.6 t0.75ΦR

(J4.2) F t 0.45 material base )F (0.6 t 0.75ΦR

u

unw

where:

te = Effective throat thickness.

FEXX = Tensile strength of electrode.

t = Thickness of the base metal.

Fu = Tensile strength of metal

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Calculate the capacity of the connection shown below, consideringE70XX weld and (A-36) bare metal?

Solution:-

1) Check weld size :- Minimum weld size = 1/8 in (Table J.2.4) Maximum weld size = ¼ - 1/16 = 3/16 in size is ok.

in. 0.133163 x 0.7070.707wte

(controls) kips/in. 4.2 70 x 0.6 x 0.133 x 0.75

)F (0.6 t75.0R EXXenw

2) Capacity of weld :-

Example W-2:

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kips/in. 6.5

58 x 0.6 x 41 x 0.75

)F (0.6 t75.0R unw

3) Capacity of base metal :-

4) Capacity of Connection:-

kips/in. .43 8 x 4.2

L x ΦR R nwn

3

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Select a plate thickness and design a full penetration weld(A 572 Grade 50 steel) SAW process?

Solution:Solution:6"

DL = 12k

LL = 60kPu

1) Pu = 1.2 12 + 1.6 60 = 110 kips.

2) Determine required thickness of plate:

Tn = 0.9 Fy Ag = 0.9 50 6 t = 270 t (controls)

or Tn = 0.75 Fu Ae = 0.75 65 6 t =292.5 t. . thickinchsayintrequired

167410

270110

3) Select electrode from “matching table” F7XX-EXXX (FEXX = 70 ksi)

No check required for groove welds.

Example W-3:

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Design the size and length of Fillet weld for the lap joint shown below,Use SMAW E 70XX process, plates are A-36 steel ?

Solution:Solution:

Factored load = 1.2 x 25 + 1.6 x 70 = 142 kips.Minimum weld size = ¼ inch (AISC table J 2.4)Maximum weld size = 5/8 – 1/16 = 9/16 (AISC 2 – 2b Page 95).Select size (5/16 inch) since you can get it in a single pass.

te = 0.707 x 5/16 = 0.22 inch

Example W-4:

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continued:continued:1) Weld capacity per inch = Rnw = 0.75 te (0.6 FEXX)

= 0.75 x 0.22 x 0.6 x 70` = 6.96 kips per inch (controls)

2) Plate shear-rupture = Rnw = t (0.6Fu)

= 0.75 x 0.625 x 0.6 x 58 = 16.3 kips per inch

3) Total length of weld required:

Use weld around plate as shown.(Use three sides of 7 inch each = 21 inch).

inchesRPL

nw

uw 420

906142 ..

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Angles are eccentric in carrying axial loads as their center of gravityx, or y are denoted by their properties. But we can balance the weldTo comply with the angle center of gravity (CG).

Since (F2) is known (Leg length x weld capacity),

Then (F1) is found, also:

F3 = T – F1 – F2

2

00

21

221

Fd

yTF

FdFyTM dA

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Example W-5Example W-5Design a balanced weld for angleshown below, (A 572-Gr 50 steel)and SMAW process specified

Solution:Solution:

1) Section capacity: Tn = 0.90 Fy Ay = 0.9 x 50 x 3.61 = 162 kips

or Tn = 0.75 Fu Ae = 0.75 x 65 x 0.85 x 3.61 = 150 kips (controls).

2) Steel weld size:Minimum weld size = 3/16 inch (Table J2.4)Maximum weld size =

Use ¼ inch fillet weld with E 70XX electrodes.te = 0.707 x ¼ = 0.177 inch.

.inch165

161

83

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contd.contd.3) Evaluate weld capacity:

Rnw = te (0.6 FEXX) = 0.75 x 0.177 x 0.6 x 70 = 5.57 kips per inch. (controls)

or Rnw = t (0.6 Fu) = 0.75 x 0.375 x 0.6 x 65 = 11 kips per inch.

4) Determine weld length required:

F2 = Rnw Lw

= 5.57 x 6 = 33.4 kips

inch. 15.25.5784.8L

inch. 5.75.5731.8L

kips. 84.831.833.4150FFTF

kips. 31.86

333.461.94150F

w3

w1

213

1

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Eccentric shear welded connection are popular for column brackets supporting gantry crane loadings or mezzanine floors. The connection below produces (shear + torsion) on the weld. The LRFD method for (shear + torsion) is both difficult and lengthy, A more conservative “elastic vector analysis” method is allowed by AISC:

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The eccentric load is transferred to the “centroid” of the weld group. The transferred load shall consist of a load plus a moment (Mz = P·e), where:

e = eccentrity.

The actual weld thickness resisting shear and torsion forces is (0.707w), but for simplicity, we consider effective throat thickness = 1.0 inch.

From direct shear only (due to P):-

yx PPP ~~~

LP

f

LPf

LPf

y1y

x1x

1

1 inch

Py

Px

PM=P·e

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JdMf .

2

The transferred moment (M) causes additional shear stresses depending on the critical locations of the weld (far edges of the weld).

where: M = moment due to eccentricity = (P.e) d = distance to the farthest weld point . J = Polar moment of inertia.

yyxx

A A

A

A

II

dAydAx

dAyx

dArJ

22

22

2where:

This can be vector analyzed to:

JMxf

JMyf

2y

2x

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Once these four stress components are evaluated for anextreme point of the weld. They can be added vectorially:

22

21

21

)()( yxv

yyy

xxx

fff

ffffff

Then the strength of 1 inch wide weld is evaluated as:

)F (0.6 0.707 w required

EXXvf

Fv = 0.707 w x 1 x (0.6 FEXX)

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Example W-6Example W-6Determine the size of the weld requiredfor the bracket shown. A-36 steel isused, SMAW process?

Solution:

.in 915195720IIJ

in 195.0(1.714)8112

1x(8)286)(1x12)(2.2I

in 720288288144

8x(6)8x(6)12

x(12)1I

in. k 942.8 5.714)(10 x 60 .M

inch 2.286 88124 x 8 x 2x

4yyxx

423

2yy

4

223

xx

eP

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Example (W-6) - Example (W-6) - ContinuedContinued

ksi 2.1432860

12)81x(860f

forces). horizontal (no 0 f

1y

1x

By inspection points (A $ B) are more critical than points (B & C);Take point (A):

ksi 10.131 5.886)(2.143(6.181)

)f(f)(ff

ksi 5.886915

5.714942.8J

x M)(f

ksi 6.181915

6942.8J

y M)(f

22

2A2y1y

2A2xv

A2y

A2x

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EXX

vreq. F 0.6 1 0.707

w

f

Select E70XX as given by matching table.

inch 0.455700.610.7070.75

10.131 w req

Use ½ inch weld size

inch) / (t inch 83 size weldminimum

inch / 161

169 size weldmaximum Since

161.9

f

21

Example (W-6) - Example (W-6) - ContinuedContinued