Welding 2

Weld M4.xmcd [Pg 1 / 8]

(pg 784)WeldingDue to lower initial cost, many structural parts of a machinery formerly made by casting are nowfabricated by welding. In the Figure 1 below, smaller components are fabricated separately andlater welded together to form the complete assembly.

Fig. 1 Welded Parts

As a fasterner, welding usually provide greater strength at a reduction in weight compare to bolts,rivets and others - an important advantage for moving parts of machines and transportequipment. A common type of welding used is arc welding and gas welding. This topicconcentrate on the selection of the correct welding size for arc welding type.

Fig. 2 Arc Welding with coated electrode

Objective:to do stress analysis on the group of welding.•to decide the correct size of the welding under static loading.•

Scope:Only arc-welding is considered in this analysis.•Fillet welding only. Butt welding is not included.•


Sample of Welding symbols applied on work-piece

Fig. 3 FILLET WELD(a) number indicate leg(h) size and (b) symbol indicate that the welds are intermittentand staggered 60 mm along a 200 mm centers

Fig. 4 FILLET WELD The circle on the weld symbolindicates that the fillet weldingis to go all around.

Fig. 5 BUTT or GROOVE WELD(a) square butt-welded on both sides;(b) single V with 600 bevel and root opening of 2 mm;

FORCE AND STRESS ANALYSIS (Fillet Weld)- DIRECT AND SHEAR LOADINGS

If load, whether direct or shear, is applied to weld, failure will occurs at the section ofsmallest cross-sectional area ABCD which is called the throat area i.e along line ABwhich is called the throat line.Loads applied in welding can be categorized into Primary and Secondary Loads.


PRIMARY LOADS

i) Axial loads (Tensile or Compressive)

If l is the length of the weld, t is throat size, F is load applied, then in diagram below for Weldresisting tensile load ,

σF

AreaABCD=

FA

=Ft l⋅

=

σF

0.707 w⋅ l⋅=

1.414Fw l⋅

=

ii) Shear loads

For diagram below, shear load is applied to the weld and the failure occurs along the line AD.


τV

AreaABCD=

Vt l⋅

=

τV

0.707 w⋅ l⋅=

1.414Vw l⋅

=

SECONDARY LOADS

iii) Bending Load

Secondary SHEAR stress due to bending , σsM yi⋅

I=

iv) Torsional Load

Secondary SHEAR stress due to twisting , τsT ri⋅

J=


Using STANDARDS (to find weld size)

The objectives of the analysis is to employ several tables and figures in specifying the length of thelegs of the fillet, the pattern of the weld and the length of the weld. The tables and figures areprovided in the standards (but only a few is shown in this book).

Allowable shear stresses and forces on welds

The allowable shear stress and force per in of leg is obtained from Table 20-3 Pg 785.

Formula for Type of Loading

Direct tension & compression fP

Aw= Eqn 20-4

Direct transverse shear fV

Aw= Eqn 20-5

Bending fMSw

= Eqn 20-6

Twisting fT c⋅Jw

= Eqn 20-7

Geometry Factors for weld analysis

Various formulaes for different welding patterns are shown in Fig 20-8 Pg 786.

Minimum weld sizes for thick plate

Minimum leg size for fillet weld is influenced by plate thickness. See Table 20-4 Pg 787.


Procedure to analyze welding1. Draw freebody diagram

2. Identify the geometry of the weld according to Figure 20-8 pg 786.

3. Identify the types of Loading/stresses to which the joint is subjected(bending, twisting, vertical shear, direct tension or compression).

4. Analyze the joint to determine the magnitude and the direction of the forceon the weld due to each type of load.

5. Combine the forces vectorially at the point(s) of the weld where the forcesappear to be maximum.

Sample 1 Find Aw, centroid,Sw, M, Jw, T, c (ifappropriate) for welding patterns 1a & 1b.

1b) bending load (weld region E-F-G-H)(given EF=140mm, FG=200mm)

1a) twisting load

Aw 200 2 140( )⋅+=

Aw 2 100( )⋅=

y1402

200 2 140( )+=

Jw1003 3 100( )⋅ 802

⋅+

6=

Swt2 200( )⋅ 140( )⋅ 1402

+

3=

Swb1402 2 200( )⋅ 140+⎡⎣ ⎤⎦⋅

3 200 140+( )=


Sample 2

The bracket in (as in Sample 1b) with F = 7.5kN and horizontal distance,L = 12cm) is shown in figure below. Determine the safety factor if leg sizew = 6mm of AWS E70 and the bracket is located in a building.

ASTM A36, A441 for building structures &Electrode AWS E60, E70, etc

Given: force F 7.5 kN⋅= w 6 mm⋅=

Distance L 120 mm⋅= b 60 mm⋅= d 120 mm⋅=

Bending moment, M F L×= M 9 105× N mm⋅=

PRIMARY LOADS: Direct Vertical Shear

Geometric Factors for shear,Welding length :

Aw b 2d+= Aw 300 mm=

Max force applied per mm thk fsF

Aw= fs 25

Nmm

=

SECONDARY LOADS: Bending Force (secondary)

GeometricFactors forBending :

At top pt. B: Swt2b d⋅ d2+

3= Swt 9.6 103

× mm2=

At bot pt. A: Swbd2 2b d+( )⋅

3 b d+( )= Swb 6.4 103

× mm2=

fbtM

Swt= fbt 93.75

Nmm

=Max forceapplied permm thk

At top pt. B:

fbbM

Swb= fbb 140.625

Nmm

= ( critical )At bot pt. A:

Combine both secondary and primary loads

The equivalent force: fe fbb2 fs

2+= fe 142.83N

mm=


Calculating allowable load (strength) of welding rod

For E70 : Allowable force per inch of leg 11200lbfin

1961N

mm=

Fall 1961N

mm⎛⎜⎝

⎞⎟⎠

w( ) in of leg1( ) in of leg

⋅= 1961N

mm⎛⎜⎝

⎞⎟⎠

w( ) mm⋅

25.4 mm of leg⋅= 77.2

Nmm

⎛⎜⎝

⎞⎟⎠

w( ) mm⋅ of legmm of leg

⋅=

. 77.2N

mm⎛⎜⎝

⎞⎟⎠

w⋅=

Welding rod size, w 6mm=

Conclusion : Allowable Load, Fall196125.4

⎛⎜⎝

⎞⎟⎠

6⋅= Fall 463.228N

mm=

ηFallfe

=463.228142.83

= η 3.243=

Welding 2

Documents

Transcript of Welding 2