Topic 18 Welds and weld group PDF

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Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.1

TOPIC 18: WELDS AND WELD GROUP



Trahair,N.S. & Bradford,M.A.,”The behaviour and design of steel structures to AS4100”, 1998, Chapter 9.

REQUIRED 18.1

General There are two types of welds: butt welds and fillet welds. Butt welds:

All the examples shown are “complete penetration” butt welds. Fillet welds:

Butt welds v. Fillet weld Full penetration butt weld has at least the full strength of the joined components (e.g. splices), and therefore does not require design check. Butt weld usually requires edge preparation while fillet weld does not.

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.2

The following example shows that fillet welds may be more economical for small size welds (about < 16 mm) than butt welds.

Cost

16mm

Plate thickness (or weld size)

Welds may be made in two categories of quality “general purpose” (GP) and “structural purpose” (SP) (highest quality) Complete penetration butt welds are assumed to have the full strength of the cross-sections joined by the welds. For butt welds

 = 0.6 for GP butt  = 0.9 for SP butt

For fillet welds

 = 0.6 for GP fillet  = 0.8 for SP fillet

The dimensioning of fillet welds requires some calculation. 18.2

Strength of fillet welds per unit length Throat thickness tt and potential failure plane tw

tw  tw

 tw

tw“ = leg length” (size of weld)

The welding code prohibits values of  less than 60 and greater than 120. The leg length may not be greater than the thickness of the thinnest part joined. Preferred sizes of a fillet weld less than 15mm are 3, 4, 5, 6, 8, 10 and 12mm, of which 6 and 8mm are most commonly specified. Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.3

In practice  is normally 90. Welds for structural purposes should be at least 6 or 8mm. Some fabricators can make welds of 10 or 12mm in a single pass, but if more passes are necessary cost is greatly increased. Welds generally fail across the “throat thickness” by shear failure. If the tensile yield stress of the weld material is fuw then the yield stress in shear is fuw/3 = 0.577 fuw. The code takes 0.6fuw. Hence the shear strength per unit length of weld is 0.6fuw  tt  1. In long welds, the shear stress distribution is not uniform, being much higher near the ends. The code introduces a factor kr < 1.0 to allow for the reduction in strength. Thus the nominal shear flow capacity (i.e. force per unit length) [9.7.3.10]

vw = 0.6 fuw tt kr

The limit state requirement for weld is vw*  vw where  = 0.8 for SP category or 0.60 for GP. The reduction factor kr is applied to the strength of welds in lap joints of more than 1700mm length (lw). For lw > 8000mm, kr = 0.62. For 1700mm < lw  8000mm, kr = 1.10 – 0.06lw /1000. Otherwise kr = 1.0. Grades of rod are: Manual

Submerged arc Flux cored arc W40X W50X

E41 XX E48XX

fuw (MPa) 410 480

The force vw* is the resultant of all the forces acting on the failure plane with thickness tt. That is, F1 L F2

tt F3

v  (F1 )  (F2 )  (F3 ) * w

2

2

2

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.3

18.4

Examples

18.3.1 Example 1 - Tee Joint SP category weld with P = 1260 N/mm, tw = 10mm. P

P 12mm

fuw

=

410 MPa

tt

=

0.707 x 10 = 7.07mm

Load on joint = 1260 N/mm For each weld v*w = 630 N/mm vw = 0.8  0.6  410  7.07  1 = 1391 N/mm Weld okay although much bigger than necessary

18.3.2 Example 2 - Lapped plates with longitudinal welds

N* = 500kN N* = 500kN

1.25kN/mm 200mm

Category GP welds 10mm fillet v*w 

Total length = 400mm

500 1.25kN / mm 400

vw = 0.6  0.6  7.07  1  410  10-3 = 1.04 kN/mm Weld is unsafe Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.5

18.3.3 Example 3 - Lapped plates with transverse welds Weld 10mm. Total length 400mm. Calculation the same as for example 2.

N* = 500kN

200mm N* = 500kN

Weld underneath

18.3.4 Example 4 - Mixed longitudinal and transverse welds

N* = 500kN

200mm N* = 500kN

Weld underneath

Total length = 2 (L1 + L2) = 800mm v *w 

18.4

500  0.625 kN / mm < 1.04 kN/mm 800

Weld is safe

Weld Groups P

18.4.1 In-plane loading Weld groups are sometimes subjected to bending moment as well as direct force. In the example shown here, the BM (torsion) is in the plane of the weld group. The bracket is assumed to act as a rigid body and rotate about the centroid of the weld group.

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.6

The force per unit length is assumed proportional to the radius from the centroid and to act at right-angles to it. [AS4100-9.8.1.1] Unit length Unit throat thickness of weld (for derivation of formula) F = kri

ri C

C = centre of rotation = centroid of weld group

M  F.ri  kri 2 ; Mk ri 2 k

M ; 2  ri

F

Mri

r

2

i

For design, F is the force/unit length acting on the failure plane and is to be resisted by weld of suitable size. For welds under forces acting in different directions, it is always preferable to group the forces perpendicular to each other so that resultant force on the weld can be calculated. The theory is thus similar to that for bolt group. That is,

x Mx Fv  Fcos  F.  2 ri  r i y My FH  Fsin  F.  2 ri ri



The term  ri2 is in fact the polar moment of inertia of the weld based on a unit throat thickness. That is,  ri2 = Ixx + Iyy. The effect of the direct force P/L (where L is total length) must be added to that due to torsional moment to find the most highly loaded location in the weld group.

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.4.2 Out-of-plane loading

18.7

[AS4100-9.8.2]

A typical case of this is when the end of a beam is attached to the face of a column. There are two theoretical approaches. One is to assume that the end of the beam remains rigid, so that the weld group is subjected to a distribution of shear flow similar to elastic stress distribution in a member. The other, described as an “alternative” method is AS4100 is to work out the total force in each flange M/d and design the flange welds for this force. The web welds are designed to take the shear force. The first method is described below. Consider the following example: P Beam

Welded connection A shear force of P is applied at the connection. We may consider two types of beam cross-section (1)

Rectangle

(2) y

I-beam y

parabolic



End view of beam

Shear stress distribution w.r.t. depth



End view of beam

On the basis of the above shear stress distributions the following distributions of force/length on the welds (due to shear force of P) are suggested:

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

Rectangular beam welds

18.8

I-beam welds

y

y 

all shear force assumed to be resisted by web fillet welds with uniform shear stress (reasonable  approximation)

The welds must resist the bending moment as well as the shear force. For beams a section modulus of Zxx can be calculated such that the maximum bending stress, M where M is the bending moment. Similarly we can max, is given by  max  Z xx define a “section modulus” Zw associated with the welds connecting the beam such that: M f Zw where f is the maximum force/length acting on the weld, due to the bending moment M. The method of calculating Zw is outlined as follows: Replace this bf

d

with this

bf

d

M/Zw

N.A.

so that each weld is replaced by a strip of the appropriate length and a width or throat thickness of 1 (unity).

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.9

For flange fillet welds: (i.e. one of them) I NA 

bf 13 d  b f 1.  12 2

2

Since there are four flange welds INA for all flange welds



bf  bf d2 3

1d 3 d 3  2  12 6 bf d3 I NA   bf d2 3 6

INA for web welds Hence total

Z w 

INA 2bf d 2    2b f d d / 2 3d 3

NB: These terms appear to have inconsistent dimensions because of the hidden 13 and 1 representing unit throat thickness which have “disappeared”. For a weld throat thickness of tt, modulus = ttZw and stress  Hence force/unit length = stress  tt  1 = tt

M t t Zw

M M  t t Zw Zw

Thus, although Zw is calculated for a weld of unit throat thickness, when the BM is divided by Zw, the result is force per unit length on the weld material (at the top and bottom).

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.10

18.4.3 Example: Cantilever beam supported by column Check the strength of the bracket-to-column connection shown below if the fillet welds are of 6mm leg length.

W* = 70 kN 190

Column 454 460 UB 67

1m

To obtain the force per unit length we treat the weld group as follows: Using the expression for Zw derived earlier we have 190

Zw 

2b f d2   2bf d 3d 3

454 2.4110 5 mm3 ! (Note the unit)

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.11

Hence the force per unit length on the flange welds due to bending moment is: Moment 100070103 290 N / mm   Zw 2.41105

This value of force per unit length will be similar to that which the weld at the top of the web carries. Conservatively this is assumed to be true. These forces are shown below with the weld isolated as a freebody. The additional forces in the vertical direction result from the shear force on the connection. As noted previously this is assumed to be carried uniformly by the web welds. Hence, the force per unit length due to shear, on one weld,

=

35103 154N / mm 454 B

C 290N

154N 290N

154N

A

Critical plane

Enlarged view of the weld at the junction between the flange and the web. Similar force diagram can be drawn for the welds along the top flange. vw*  290 2 154 2  328N / mm

Assuming category GP welds,

6 1410 626N / mm O.K. 2 The reason that the weld at the top of the web was examined was that this is likely to be the most highly stressed part of the weld.

vw  0.60.6

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017

Unit CIV2225: Steel and Timber Structures Topic 18: Welds and Weld Group

18.5

18.12

Welds for plate-girder flanges and Crane Beams Shear “stress” on top of web



VA f y I xx tw

(N/mm2)

where Af = area of flange.

Af

Force per unit length of (2) welds =  t 1

y X

t X

=

VA f y I xx

(N/mm)

An alternative ultimate strength approach is to say that the force developed in the flange at the location of max BM must be transmitted from the web into the flange M between this location and the nearest point of zero BM. Force in flange  . d Therefore, the length of weld between these locations must be able to carry that force. This approach results in much smaller welds.

Department of Civil Engineering, Monash University CIV2225 Date: 2/2017...