slide1 l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
W-1 PowerPoint Presentation
Download Presentation
W-1

Loading in 2 Seconds...

play fullscreen
1 / 35

W-1 - PowerPoint PPT Presentation


  • 196 Views
  • Uploaded on

Welding. Chap. (5). Definition: A process of joining two metal pieces by heating to a fluid slate (melting), with or without filler material or pressure. History of Welding: 5500 B.C. Egyptians recognize “forge welding” for copper and gold.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'W-1' - locke


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Welding

Chap. (5)

Definition: A process of joining two metal pieces by heating to a fluid

slate (melting), with or without filler material or pressure.

History of Welding:

5500 B.C. Egyptiansrecognize “forge welding” for copper and gold.

800 A.D. “Damascus Sword”-Layers of metal forge welded to high strength.

1877 A.D. Elihu Thompson, Electric resistance welding possibility

recognized.

1889 A.D. P. Strohmeyer introduced the concept of coated electrodes

(Shielded Arc Weld), (SMAW).

1903 A.D. Fouche & Picard introduced oxyacetylene cutting and welding

techniques.

1932 A.D. Gas Metal Arc Welding (GMAW).

1990 A.D. Welding Automation (Robotics).

W-1

slide2

Basic Welding Processes:-

1 – Shielded Metal Arc Weld (SMAW):

Most popular welding technique (stick welding).

The electrode coating performs the following:

A – Produce gaseous shield to exclude oxygen.

B – Introduce deoxider material to improve grain.

C – Produce a blanket of slag to retard cooling and prevent

oxidation.

The SMAW process is

designated by AWS as

“E6OXX” or “E7OXX”

e.g. (E 6013)

W-2

slide3

2 – Submerged Arc Weld (SAW):

In this process the automatically fed arc (spool) is protected by a blanket of granular material called “flux”.

This flux material acts to improve weld quality and to protect it from the air. Most (SAW) processes are used in welding

shops and not at site. (SAW) is designated by AWS as:

(“FXXX-EXXX”)

Submerged arc welding (SAW)

W-3

slide4

3 – Gas Metal Arc Weld (GMAW):

The electrode is fed constantly through a spool, and the protection is provided constantly by some gas mixture.

Initially, inert gases were used such as Argon, hence called “MIG” process. Carbon Dioxide (CO2) is a popular gas with this process.

Recently, mixture of Argon, Helium and CO2 are used for better weld characteristics.

W-4

slide5

Weld base metal

(steel)

1mm

Inner flux

  • 4 – Flux Arc Weld (FCAW):
  • Developed in 1958, very similar to (GMAW), but the filler material is a tabular rod on a spool that contains the flux material inside it. The internal flux assumed similar role for the shield on the SMAW process.
  • Electrodes Welding (EGW).
  • Electro slag Welding (ESW).
  • Stud Welding.

W-5

slide6

Type of Weld Joints:

A – Butt Joints:

Requires full penetration “groove welding”, but it eliminates eccentricity and more pleasant to view.

Suitable for shop welding.

B – Lap Joints:

Is the most common, due to ease of fitting, requires “fillet weld”, and can be double-sided.

C – Tee Joints:

Useful for fabricating

“Built-up” sections

D – Corner Joints:-

E – Edge joints:-

W-6

slide7

(a) Groove weld.

(b) Fillet weld.

Type of Welds:

A) Groove Welds :

Is used to weld members aligned in the same plane. The weld should have the same strength of the base material

(or even greater).

B) Fillet Welds :

More popular as they require no special alignment of pieces, nor edge preparation.

W-7

slide8

C) Slot Weld :

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

D) Plug Weld :

Same as above

W-8

slide11

Effective Area of Groove Welds

Aw = te Lw

{AISC J 2.1.1a. Page 16.1-93}

where:

Aw = Effective area of weld.

te = Effective throat thickness.

Lw = Length of weld.

t2

t1

t

t

te = t

te = t1

  • For partial joint penetration groove welds,
  • see (table J 2.1 Page 16.1 -44)
  • For minimum size of partial penetration groove welds,
  • see (table J 2.3 page 16.1-95).

W-11

slide12

FILLET WELDS

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  Lw

where,

Aw = Effective area of weld.

te = Effective throat thickness = 0.707 w

Lw = Effective Length of weld.

W-12

slide13

Size and length Limitations for Fillet Welds

1 – Minimum Size of Fillets Welds :

2 – Maximum Size of Fillets Welds :

W-13

slide14

Size and length Limitations for Fillet Welds

3 – Minimum Effective Length of Fillet Welds :

W-14

slide15

Example W-1

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

Solution :-

Maximum weld size

For ½” thick plate:

½”

1”

Note : Maximum weld size controlled by thin plate,

Minimum weld size controlled by thick plate.

W-15

slide17

Nominal Strength of Welds

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.

W-17

slide18

Nominal Strength of Welds

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.

W-18

slide19

Load & Resistance Factor Design - Welds

1 – Groove Welds (Full penetration):

A – Tension or compression (Table J2.5)

B – Shear on Effective Area (Table J 2.5)

W-19

slide20

Load & Resistance Factor Design - Welds

2 – Fillet Welds

The 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:

where:

te = Effective throat thickness.

FEXX = Tensile strength of electrode.

t = Thickness of the base metal.

Fu = Tensile strength of metal

W-20

slide21

Capacity of Fillet Welds

Example W-2:

  • Calculate the capacity of the connection shown below, considering
  • E70XX weld and (A-36) bare metal?
  • Solution:-
  • 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.

2) Capacity of weld :-

W-21

slide22

3) Capacity of base metal :-

4) Capacity of Connection:-

W-22

slide23

DL = 12k

6"

Pu

LL = 60k

Design of Groove Welds

Example W-3:

Select a plate thickness and design a full penetration weld

(A 572 Grade 50 steel) SAW process?

Solution:

  • Pu = 1.2  12 + 1.6  60 = 110 kips.
  • 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

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

No check required for groove welds.

W-23

slide24

Design of Fillet Welds

Example W-4:

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:

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

W-24

slide25

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).

W-25

slide26

Balanced Welded Connection

Angles are eccentric in carrying axial loads as their center of gravity

x, or y are denoted by their properties. But we can balance the weld

To 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

W-26

slide27

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).
  • 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.

Design of Balanced Weld

Example W-5

Design a balanced weld for angle

shown below, (A 572-Gr 50 steel)

and SMAW process specified

Solution:

W-27

slide28

Design of Balanced Weld

contd.

  • 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.
  • Determine weld length required:
  • F2 =  Rnw Lw
  • = 5.57 x 6 = 33.4 kips

W-28

slide29

Eccentric Shear Welded Connections

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:

W-29

slide30

Py

1 inch

P

M=P·e

Px

Elastic Vector Analysis

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):-

W-30

slide31

Elastic Vector Analysis

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.

where:

This can be vector analyzed to:

W-31

slide32

Elastic Vector Analysis

Once these four stress components are evaluated for an

extreme point of the weld. They can be added vectorially:

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

Fv =  0.707 w x 1 x (0.6 FEXX)

W-32

slide33

Eccentric Shear Connection

Example W-6

Determine the size of the weld required

for the bracket shown. A-36 steel is

used, SMAW process?

Solution:

W-33

slide34

Eccentric Shear Connection

Example (W-6) - Continued

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

Take point (A):

W-34

slide35

Eccentric Shear Connection

Example (W-6) - Continued

Select E70XX as given by matching table.

Use ½ inch weld size

W-35