polar moment of weld group

8/18/2019 Polar Moment of Weld Group

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Chapter 12

Designing for Combined Loads:

Welded Connections

his is the third of three chapters that discusses the design for combinedloads. The previous two chapters dealt with the design of members sub-

jected to the combined loads of axial and bending or torsion and bending.This chapter discusses the design of fillet welds. The distance along one ormore sides of a member or the perimeter around the member influences thesize of the weld, which has an upper limit. There are design situations wherea cross-sectional shape is selected in order to obtain adequate weld length.The base length of most brackets is defined by the length of the weld re-quired to support the loading.

The actual design of the members becomes an iterative process: performthe structural analysis, select a cross-sectional shape, and design the connec-tion. If the welded connection requires a change in the cross-sectional shape,select a new member. Plane frames and plane grids, however, are staticallyindeterminate and any change in the cross-sectional properties requires anew structural analysis. The procedure goes on until all of the stress and de-flection ratios satisfy the design criteria. Since the length requirements of aweld influences the selection of a cross section, the material related to theanalysis and design of welded joints is an integral part of the design forcombined loadings.

The design of welded connections is covered in several machine designand structural design books, but there is a lack of consistency in the presen-tations. Every book discusses the analysis of welded connections subjectedto shear and torsion loads while fewer books discuss the analysis of connec-tions subjected to an axial force and a bending moment similar to what oc-curs in a beam. Even fewer books discuss the analysis of connections

T

Segerlind, Larry J. 2010. Designing for Combined Loads: Welded Connections.

Chapter 12 in Designing Structural Components for Machines, 361-386. St. Joseph,

Michigan: ASABE. Copyright © 2010 American Society of Agricultural and

Biological Engineers. ASABE Order Number 801M0310, Textbook Number 21.

ISBN 1-892769-76-X.


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362 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSsubjected to the combined loadings of axial, shear, and bending, which arevery common in plane frames, and the combined loadings of torsion, shear,and bending, which occur in plane grids. Several books give the basic ingre-dients for analyzing these loading situations, but they do not work completeexamples nor do they combine the design of a weld with the design of amember.

A systematic procedure for analyzing and designing welded joints subjected to combined loadings is presented in this chapter. The approach isunique in several ways and the information presented is used in the designs presented in Chapter 13 and the design projects outlined in Chapter 14.

12.1 Welds for Structural Tubes

The design of connections for structural tubes is covered in the HollowStructural Sections Connections Manual (AISC, 1997), referred to as theAISC CM. It covers the design of welded and bolted connections involvinghollow structural sections (tubes) subjected to different types of loadings.Even though the LRFD design procedure (discussed in Chapter 2) is used inthis manual, it does contain general information that is useful with the al-lowable stress design procedure. None of the information in the AISC CMmanual is in SI units even though the publication date is many years afterthe adoption of SI units in the United States.

The AISC CM discusses the use of fillet welds, complete-joint- penetration (CJP) groove welds, and partial-joint-penetration (PJP) groovewelds. Members in a plane frame or plane grid that are connected together atangles less than 90° can be connected using complete-joint-penetration

groove welds. These welds have the same shape and thickness as the members and are assumed to develop the strength of the base metal. The AISCCM states that no allowance for the presence of such welds needs to bemade in proportioning the connections of structural members for any type ofstatic loading.

The analysis and design procedures in this chapter are for fillet welds.These are used to attach brackets to structural members, attach the membersto supports, and to attach the members to each other when they are not ma-chined to use a groove weld. The basic information for fillet welds was in-troduced in Section 6.2 (pp. 158-164) and all of the rules presented there areapplicable here. The AISC CM does state that structural tube sections madefrom steel grades with F y up to 50 ksi are normally welded with an E70 elec-trode, which has a nominal tensile strength F

u of 70 ksi. The AISC CM

manual also states that fillet welds loaded transverse to the weld axis are50% stronger than welds loaded parallel to the weld axis. This increasedstrength is ignored in design procedures. As noted above, the design proce-dures for fillet welds were introduced in Section 6.2. The same design equa-tions are used in this chapter. The equations for the E60 and E70 rods, (6.17)


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SHEAR FLOW IN FILET WELDS 363

through (6.20), are repeated here as (12.1) through (12.4).The allowable shear flow equations for E60 rods are:

42.4USCS: kips/inall

w Q =

N (12.1)

0.293SI: kN/mmall

w Q =

N (12.2)

The allowable shear flow equations for E70 rods are:

USCS: kips/in5.49

N

wQall = (12.3)

0.341SI: kN/mmall

w Q =

N (12.4)

The dimension of the weld leg, w, is known in analysis problems and is theone of the variables in a design problem.

12.2 Shear Flow in Fillet Welds

The primary quantity in the analysis and design of welds is the shear flow qwhich is a force per unit length and is obtained by multiplying a normal orshear stress by a thickness dimension. Shear flow was introduced in Chapter6 and the allowable shear flow for fillet welds made from E60 and E70 rods

is given by (12.1) through (12.4). The shear flow equations for five types ofloadings are developed in this section. Three of the loading situations aresimple and require little explanation. The other two are more complicatedand require integral evaluations. The entire discussion, however, is moreeasily presented after we have defined a local coordinate system for awelded joint, a definition that is lacking in most discussions of welded

joints.

12.2.1 A Coordinate System

The local coordinate system for a weld joint is a right-handed system withthe x axis perpendicular to the plane of the weld. The origin of the coordi-nate system is at the centroid of the weld. The y axis is usually parallel to thelong dimension of the weld but a 90° rotation in the y-z plane does notchange the design. The orientation of the local coordinate system for twodifferent weld patterns is shown in Figure 12.1.

The loads acting on a weld are defined by the orientation of the coordi-

nate system. The axial force P x is perpendicular to the weld while the shear


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364 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONS

y

z

x

y

z

x

Figure 12.1 Local coordinate systems for a pair of welds.

forces P y and P z act in the plane of the weld. A weld can also be subjected tomoments about the x and z axes, M x and M z . The way in which a bracket isattached to a member defines the type of forces and moments acting on theweld and influences the size and shape of the weld.

Consider the brackets in Figure 12.2. These are subjected to a tensionload applied by a hydraulic cylinder. The bracket in Figure 12.2(a) is at-tached to the top of the structural section using a pair of welds that are sub-

jected to an axial force P x, a shear force P y, and a moment about the z axis, M z . This moment is similar to the bending moment that occurs in beams andthe weld pattern is said to be subjected to bending. The shear flow values forthis weld act in the x and y directions. Attaching the bracket to the side ofthe structural section changes the loads significantly. There are two shearforces, P y and P z , and the moment acts about the x axis. This moment, M x, istwisting the weld and the weld is said to be in torsion. All of the shear flow

y

x

y

z

y

z

x z

x

(a) (b)

Figure 12.2 Weld joints for two brackets.


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values for this weld are in the y-z plane. The analysis of the connection inFigure 12.2b is covered in nearly every book that discusses welded joints.The analysis of the weld joint in Figure 12.2a is discussed less often. Bothtypes of welds are important and are discussed in detail in this chapter.

12.2.2 Axial and Shear Forces

The shear flow equations for welds subjected to an axial force or either ofthe two shear forces are easy to obtain. Starting with the axial force shownin Figure 12.3a, the average normal stress is

ave

w w

P P = =

A t Lσ (12.5)

where t is the throat dimension (see Figure 6.5) and Lw is the effectivelength. The shear area Aw = tLw is used in the calculation even though it isnot perpendicular to the normal stress. The average shear stress for the shearforces P y and P z are

, ,and y z

y ave z ave

w w

P P = =

t L t Lτ τ (12.6)

x

y

z

aveσ

P x

(a)

Lw . z aveτ

y

P z

(b)

Lw

P y x

z

Lw

(c), y aveτ

Figure 12.3 Average normal and shear stress in a fillet weld.


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366 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSMultiplying through (12.5) and (12.6) by the thickness t gives the respectiveshear flow equations.

and y z x

x y z

w w w

P P P q = , q = , q =

L L L (12.7)

12.2.3 Moments about the z Axis: Bending

A pair of welds associated with a rectangular bar that supports a pure bend-ing moment is shown in Figure 12.4. Since the bar behaves like a beam andthe weld is an integral part of the bar, it is assumed that the stress distribu-tion in the weld is similar to the normal stress distribution in the beam. The

normal stress in the weld is calculated using the normal stress equation for a beam

z M y = - I

σ (12.8)

where I is calculated assuming the weld is a rectangular area with the throatdimension t and a length h. Assuming two rectangular welds,

3 32

12 6

th t h I = = (12.9)

Since the strength of the weld is assumed to be independent of the type anddirection of the stress and since equal maximum values occur at y = ± h/2,

3 32

6 3

z z M M h = =t h t h

σ ⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞⎝ ⎠⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(12.10)

M z z

x

y xσ

xσ

Figure 12.4 The normal stress distribution resulting from M z .


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The normal stress is multiplied by the throat thickness to obtain a shear flowand the rest of the denominator is defined as a weld section property Z w as

3

3

z z x

w

M M q = t = =

Z hσ

⎛ ⎞⎜ ⎟⎝ ⎠

(12.11)

The bending section property Z w for a pair of welds parallel to the y axis is

3

3w

h Z = (12.12)

The weld section property equations for common shapes that are sym-metrical relative to the z axis are presented in Table 12.1. These equationswere obtained using

21w

A

I Z = = y dA

c c

⎛ ⎞⎜ ⎟⎝ ⎠

∫ (12.13)

where each weld is a line segment of unit width and c is one-half the welddimension parallel to the y axis.

The evaluation of (12.13) is illustrated by evaluating the section property Z w when the weld pattern consists of two welds parallel to the z axis with thedimensions b and h as shown in Table 12.1 (Case 3). The section property Z w is

21w

A Z = y dAc ∫

Since there is a pair of welds and c = h/2 while dA = (1)dz , the integral for Z w is

2(1)

2 2

2b/2

w

-b/2

h Z = dz

h/

⎛ ⎞⎜ ⎟⎝ ⎠

∫

which equals w Z = b h (12.14)

Weld patterns that lack symmetry relative to the z axis can be analyzedusing (12.8) where

2

A I = y dA∫ (12.15)has to be evaluated on an individual basis. The location of the maximumshear flow also has to be determined by inspection.


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368 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSTable 12.1

Section properties of fillet welds.

Shape Bending Torsion

y

z h

6

2

wh

= Z 12

3

wh

= J

h z

b

3

2

wh

= Z 2 23

6w

h( b + h ) = J

b

h

y

z

w= b h Z 23

6

3

w

+ bhb = J

2

(2 )

b z

b h=

+

h

b

y

z z

6

2

wh

= b h + Z

32

12 2

22

w

(b + h) (b + h )b= - J

( b + h)

y

z h

b

3

2

wh

= b h + Z 6

3

w

(b + h ) = J

r

y

z

2w= r Z π

32w= r J π


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12.2.4 Moments about the x

Axis: TorsionA pair of horizontal welds subjected to a twisting moment, M x, are shown inFigure 12.5. The shear stress in a weld subjected to a twisting moment iscalculated using

x M r = J

τ (12.16)

where 2 2 2( ) A A

J = r dA = x y dA+∫ ∫ (12.17)

is the polar moment of inertia. Equation (12.16) parallels the analysis of acircular bar in torsion and does not describe what happens in a welded jointto a high degree of accuracy. This is one reason for the relatively high safety

factor for welded joints. Since welds are represented by rectangles of thick-ness t and a length, the elemental area becomes dA = (t )dy or dA = (t )dz de- pending on the orientation of the weld. The thickness dimension can betaken outside the integral and (12.16) can be written as

2

x

A

M r =

t r dAτ

∫ (12.18)

where the dA is now dA = (1)dy or dA = (1)dz . Multiplying through by thethickness produces the shear flow equation

x

w

M r t = q =

J τ (12.19)

where 2w A

J = r dA∫ (12.20)

+ z

b

h

y

dz

r

M x

Figure 12.5 A pair of welds subjected to a twisting moment.


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370 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSand dA = dy or dA = dz . The quantity J w is the section property that is usedin the analysis and design of welded joints subjected to a twisting moment.The shear flow q is assumed to act perpendicular to the radial arm from thecentroid to a point on the weld. A radial distance to a location must be sub-stituted into (12.20) before a value for q can be determined.

The section property J w is relatively easy to evaluate because most weldsconsist of horizontal and vertical line segments. Consider the pair of weldsin Figure 12.5. The elemental area is dA = (1)dz and the integral in (12.20) becomes

2 2 2( ) 2 2

2b/2 b/2

w A

-b/2 -b/2

h J = y z dA = z dz + dy

2

⎛ ⎞+ ⎜ ⎟

⎝ ⎠∫ ∫ ∫

which equals3

36

whb + b J = (12.21)

Equations for J w for several common weld patterns are given in Table 12.1.A more extensive list is given in Hall et al. (1961).

The section properties Z w and J w as given in Table 12.1 have differentunits because the location of the maximum shear flow is built into Z w while(12.19) is left in terms of a radial distance r . The shear flow q is assumedto act perpendicular to r and must be resolved into y and z components be-fore it can be combined with other shear flow values. Consider the locationshown in Figure 12.6. The radial distance r has an angle θ relative to the yaxis. The y and z components of the shear flow are

sincos

y

z

q = qq = q

θ

θ − (12.22)

Substituting (12.19) for q produces

sin x x y

w w

M r M z q = =

J J

θ (12.23)

andcos x x

z

w w

M r M yq = - = -

J J

θ (12.24)

because r sin θ is a distance in the y direction and r cos θ is a distance in the z direction. The values of y and z are the coordinates of a point on the weldrelative to the centroid of the pattern. The sign on the shear flow is usuallydetermined by visual inspection and the negative signs in (12.22) and(12.24) are deleted.


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q

q y q

q z

θ

a+

M x

y

z

θ r

a

Figure 12.6 Components of the shear flow for torsion.

12.2.5 Section Properties for Intermittent Welds

The intersection of members in complicated three-dimensional machinesmay make it difficult or nearly impossible to use the weld patterns given inTable 12.1. It may be necessary to use several short welds. The welds, how-ever, still need to be analyzed to make sure they are adequate, which in-volves the evaluation of the section properties Z w and J w.

The properties of lines and curves are discussed in most statics books andthe first and second moments of curves have the same interpretations as thefirst and second moments of areas. The first moment defines the centroid

L L y dL y dL=∫ ∫ (12.25)

which becomes ( )i i i y L y L=∑ ∑ (12.26)for a shape composed of several line segments. The second moment is

2 z

L I z dL= ∫ (12.27)

This equation also has a parallel-axis theorem which is

2 z o I I Ld = + (12.28)

where I o is (12.27) about the centroid of the line, L is the length of the line,and d is the distance between the centroid of the line and the z axis.

The evaluation of the polar moment of inertia for a line segment parallelsthe evaluation for an area

2 2 2( ) L L

J r dL y z dL= = +∫ ∫ (12.29)with the parallel-axis theorem of

2 2 2o o J J Lr J Ly Lz = + = + + (12.30)


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372 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSwhere r is the radial distance from the point to the centroid of the line seg-ment, and J o is the polar moment of inertia for the line segment about itscentroidal axis. The value for J o is the first selection in Table 12.1 and(12.30) becomes

32 2

12w

L J Ly Lz

⎛ ⎞= + +⎜ ⎟

⎝ ⎠∑ (12.31)

for a group of line segments.

Example 12.1 ■■■■ Evaluate J w for the weld pattern consisting of three straight line segments asshown. The weld size is 6 mm.

1 5 0 m m

100 80

Solution:The first step is to determine the dimensions of the effective length of eachweld. This is done by keeping the centroid of each line segment at its origi-nal position. The y′ and z ′ axes, shown below, are used as the reference pointfor determination of the centroid of the weld pattern which has a horizontalaxis of symmetry. The location of horizontal centroid z is given by

140(138) (0)(68) (0)(68)70.5 mm

(138 2(68))

i i

i

z L z

L

Σ + += = =

Σ +

The reference point for determining the centroid and the final set of dimen-sions for the effective weld pattern are shown below.

z

y

1 3 8 m m

7 5

7 5

69.5 70.5

68

y'

z'

1 3 8 m m

140

68

1 5 0

The section property J w is given by (12.31). Substitution of the appropriatevalues gives


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3

2 2

3 32 2 2

3

( ( )12

138 68138(0 69.5) 2 68(75 70.5 )

12 12

2 379 000 mm

w

w

w

L J L y z

J

J

= + +

⎛ ⎞= + + + + +⎜ ⎟

⎝ ⎠

=

∑

■■■■

Example 12.2 ■■■■ Evaluate Z w for the symmetric weld pattern shown. The bracket has an openarea to accommodate a structural component. The weld size is 1/4 inch.

Solution:The dimensions of the original (left) and effective weld (right) are shown below.

z

y

z

3 in3 in

103.5

2.5 2.5

3.5

The centroid of each weld segment retains its original value. Note that theweld segments are symmetrical about the z axis, therefore, the parallel-axistheorem must be used with each segment. Since each segment has the samedimensions

( )' 24 z z I I Ld = +

where'

z I is the area moment about its centroidal axis. The evaluation of Z w follows the calculation of I z . Substituting the appropriate values gives

323

23

in127.7)in3.5(in)2.5(12

)in2.5(4

124 =⎥

⎦

⎤⎢⎣

⎡+=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ += hd

h I z


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374 DESIGNING FOR COMBINED LOADS: WELDED CONNECTIONSThe section property Z w = I z /c where c is the distance from the z axis to theouter most location or

( )[ ]2

3

in9.26in2/5.25.3

in7.127=

+

==c

I Z z w

■■■■

12.2.6 Resultant Shear Flow

Shear flow has the units of kip/in or kN/mm. If we specify a very small dis-tance around any point, the shear flow values become forces that can beadded as vectors. The final step in the analysis or design procedure is toevaluate the resultant shear flow using

2 2 2 R x y z Q = q + q + q (12.32)

which is similar to the equation used to determine the magnitude of a forcewhen the values of its components are known.

The shear flows resulting from positive values of P x and M z are shown inFigure 12.7. There are one or more locations on the weld where q x resultingfrom the two loadings add together. The same is true when one of the load-ings is negative or when both loadings are negative. The equation for thelargest value of q x is

x z x

w w

P M q = +

L Z

⎛ ⎞± ⎜ ⎟

⎝ ⎠ (12.33)

Since q x is squared in (12.34), the sign that denotes tension or compressionis usually deleted.

The shear flow quantities q y and q z at the corners of a rectangular weldsubjected to the twisting moment M x are shown in Figure 12.8. The corners

P x

M z

q x y

q x

Due to P x

Figure 12.7 The shear flow component q x.


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q y

q y

Due to P y

q z

q y

q y

q y

q z

q z q z

+

P y

P z

M x

q z

Due to P z

Figure 12.8 Shear flow components q y and q z .

are selected because q y and q z obtain the largest values when the distances z and y in (12.23) and (12.24) have their maximums. There is at least one lo-cation where the q y value resulting from M x adds to the q y value resultingfrom P y. The combined equation is

y x

yw w

z P M = +q

J L (12.34)

This equation holds when either one of P y or M x are negative or when bothare negative. The same situation is true for q z and

x z z

w w

M y P q = +

L J (12.35)

Equations (12.33) through (12.35) are the shear flow equations for aweld subjected to a set of combined loadings. When one or more forces ormoments are zero, the corresponding terms are deleted. The joints in planeframes and the brackets welded to the top or bottom of a member usuallysupport P x, P y, and M z loadings and the terms containing P z and M x disappear. Brackets welded to the side of a member support P y , P z , and M x load-

ings and the terms containing P x and M z disappear. The joints in a plane gridsupport P y, M x, and M z and the terms containing P x and P z are absent.The location of the maximum shear flow Q R is usually determined by in-

spection when doing hand calculations. Computer analyses of welded jointsusually incorporate a sign with the magnitude of the load and provide thelocation of Q R as part of the output.


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12.3 Analysis of Welded ConnectionsThe shear flow equations for five types of loadings that can act on a filletweld were discussed in the previous section. Equations (12.33) through(12.35) and the section property equations for Z w and J w given in Table 12.1summarize the results. Our immediate objective is to calculate the resultantshear flow Q R for three different joints with combined loadings and deter-mine whether the weld is adequate using

1 R

all

Q

Q≤ (12.36)

The first example is a welded joint from a plane frame. The second ex-

ample is a joint from a plane grid. The forces acting on these joints are ob-tained from a structural analysis as discussed in Chapter 10 or 11. The thirdexample is a bracket supporting a hydraulic cylinder.

The analysis problem is relatively straightforward. You evaluate all of the parameters and check the shear flow ratio (12.36). The only item to be care-ful about is the effective length of the welds. The AISC ASD manual doesnot allow the start and stop distances to be included in the calculations of single welds, but it does allow the use of the actual length when the weld goes around a corner. We will hold true to this requirement in both theanalysis and design calculations, although many books that discuss theanalysis and design of fillet welds do not subtract the start and stop lengths.We start our calculations by determining the effective weld dimensionswhen necessary. When the weld consists of intermittent welds, we will keep

the centroid of each weld in its original location.

12.3.1 A Welded Joint in a Plane Frame

Section 72, a 3×5×0.250 RST, is connected to a support by placing a 0.1875(3/16) inch fillet weld on each side. The structural analysis of the planeframe containing this member indicates the member is subjected to theforces shown in Figure 12.9: P x = 4.91 kips, P y = 3.54 kips, and M z = 28.7kip·in at the support. The objective is to determine whether this weld is ade-quate for an E70 rod and a safety factor of N = 2.80.

The analysis problem is straightforward. We evaluate all of the parame-ters and check the shear flow ratio in (12.36). The analysis begins by calcu-lating the effective dimensions, which are the outside dimensions of the

member since the weld is continuous and goes around the corners. The di-mensions are b = 3 inches and h = 5 inches. The weld length is the distancearound the rectangular shape or

2( ) 2(3 5) 16.0 inw L b h= + = + =


17/26

ANALYSIS OF WELDED CONNECTIONS 377

P x

= 4.91 kips

P y = 3.54 kips

M z = 28.7 kip·in

#72

Figure 12.9 The support forces for a fixed joint in a plane frame.

The allowable shear flow is given by (12.1)

49.5 49.5(0.1875)3.31 kips/in2.80

all

wQ N

= = = (12.37)

The section property Z w for bending is obtained from Table 12.1 Usingthe outside dimension, since the welds goes around the corners, gives

2 225( )( ) (3)(5) 23.3 in

3 3w

h Z b h= + = + =

This completes the evaluation of the basic properties.This joint has shear flow values in the x and y coordinate directions. The

shear flow q x is a combination of values while q y results from the shear force P y. Starting with (12.33),

24.91 k ips 28.7 kip in 1.54 kips/in16.0 in 23.3 in x z

x

w w

P M q L Z

⋅

= + = + =

The shear flow in the y direction is given by (12.34) where M x = 0. Thisvalue is

3.54 kips0.221kips/in

16.0 in

y

y

w

P q

L= = =

The resultant shear flow is

2 2 2 21.54 0.221 1.55kips/in R x yQ q q= + = + = (12.38)

and the shear flow ratio is

1.55 kips/in0.47 1

3.31 kips/in

R

all

Q

Q= = < (12.39)

which is more than adequate. The engineer would investigate whether a 1/8


18/26


inch weld is adequate if this was a design problem. If the weld had not beenadequate, the engineer is faced with the decision of whether to increase theweld size or change the cross-sectional shape of the tube to increase theweld length.

12.3.2 A Welded Joint in a Plane Grid

One of a pair of wheels mounted behind a machine such as a forage moweris shown in Figure 12.10. This configuration produces a shear force P y andtwo moments at the connection. The primary component is #9, a 76 ×76×6.4RST, connected to the structure using an E70 rod and a 4 mm weld that goescompletely around the tube. Our objective is to determine whether this weldis adequate when the safety factor is N = 3.0.

The applied loadings can be calculated from the information in Figure12.10, which gives

P y = 6.70 kN, M x = 1.68 kN·m, and M z = 5.03 kN·m

These values are shown on a segment of the member in Figure 12.10. Theallowable shear flow is given by (12.2)

0.341 0.341(4)0.455 kN/mm

3.0all

wQ

N = = = (12.40)

M z = 5.03 kN·m M x = 1.68 kN·m

P y = 6.70 kN

0.250 m

6.70 kN

0.750 m

6.70 kN

#9 x

Figure 12.10 A welded joint in a wheel assembly.


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The effective dimensions of the weld are b = h = 76 mm. The length ofthe weld is

2( ) 2(76 76) 304 mmw L b h= + = + =

while the section property for bending is

2 227676(76) 7700 mm

3 3w

h Z bh= + = + =

The section property for torsion is

3 33( ) (76 76) 585 300 mm

6 6w

b h J

+ +

= = =

Three different shear flows occur in the weld, q x, q y, and q z . The shearflow in the x direction results from the bending moment and is

2

5 030 kN·mm0.653kN/mm

7 700 mm

z x

w

M q

Z = = = (12.41)

The shear flow in the y direction is composed of the two components givenin (12.34). The largest value occurs at the corner. We use values of 38 mm(76/2) for the y and z directions in (12.34) and (12.35).

3

6.70 kN (1680 kN·mm)(38 mm)0.131kN/mm

304 mm 585 300 mm

y x y

w w

y

P M z q

L J

q

= +

= + =

(12.42)

The third shear flow, q z , results from the torsion moment and is

2

3

(1680)(38)kN·mm0.109 kN/mm

585300mm

x z

w

M yq

J = = = (12.43)

Combining the three shear flow values using (12.32) produces

2 2 2

2 2 20.653 0.131 0.109 0.675 kN/mm

R x y z

R

Q q q q

Q

= + +

= + + =

(12.44)



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0.675kN/mm1.48 1

0.455kN/mm R

all

Q

Q= = > (12.45)

The weld is not adequate. This result was obvious from (12.41). Anytimeone of the shear flow components exceeds the allowable value, the weld isinadequate. You should do all of the calculations, however, in order to de-termine which loading governs the design. In this example, the bendingmoment governs the design and a rectangular tube is probably more appropriate for the primary structural component.

12.3.3 A Weld for a Bracket

The bracket shown in Figure 12.11 supports a 32 kN force and is welded to

the side of a structural member. A similar bracket and loading exists on theother side of the member. The weld is made using an E60 rod and has a legdimension of 4 mm. Determine whether the weld is adequate when the safe-ty factor is N = 2.75. Assume the weld is continuous around the corners.

50

20°

8 0

180 mm

7 5

64 kN

40

z

y

30 kN

10.9 kN

180 mm

7 6

17.4 1 3 7 . 6A

•

y

Figure 12.11 The weld for a side bracket.

Since the weld is on the side of the member, the x axis is perpendicular tothe page and the applied force can be resolved into y and z components of10.9 kN and 30 kN, respectively. The effective values for b and h are b =180 mm and h = 80 – 4 = 76 mm. The weld size is only subtracted from theends of the vertical welds. The distance in the y direction from the weld tothe centroid of the weld pattern can be obtained from Table 12.1 by inter-changing the variable names or from Hall et al. (1961). The location of thecentroid from the base, b, is

( )

( )

22 76 mm17.4 mm

2 180 2 76 mm

h y

b h= = =

+ ⎡ ⎤+⎣ ⎦

The force components and the dimensions for the effective weld are also


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shown in Figure 12.11. The shear force values are P y = 10.9 kN and P z = -30kN. The bottom or horizontal portion of the weld remains 155 mm from theapplied load. The moment about the x axis is

+ (40 mm)(10.9 kN) (137.6 mm)(30 kN) 3 692 kN·mm x M = − = −∑ A visual inspection indicates the location A is the critical location where the y and z shear flow components resulting from M x add to the components produced by the two shear forces.

The section properties for the weld are

2 180 2(76) 332 mmw L b h= + = + =

and from Hall et al. (1961),

3 2 2

3 2 23

( 2 ) ( )

12 ( 2 )

332 76 (76 180)1909 362 mm

12 332

w

w

b h h b h J

b h

J

+ +

= −

+

+

= − =

The shear flow component q y is

10.9 3692(90)0.207 kN/mm

332 1 909 362

y x y

w w

P M z q

L J = + = + =

while q z is

30 3 692(76 17.4) 0.204 kN/mm332 1909 362

x z z

w w

M y P q L J

−= + = + =

The resultant shear flow is

2 2 2 20.207 0.204 0.290 kN/mm R y z Q q q= + = + =

The allowable shear flow is

0.293 0.293(4)0.426 kN/mm

2.75all

wQ

N = = =


0.290 0.68 10.426

R

all

QQ

= = <

The weld is adequate.


22/26


12.4 Design of Welded ConnectionsThe design of welded connections is an interactive process that involves allof the steps in the analysis procedure. You make an educated guess at theoverall dimensions and the leg size and then check to see if the weld is ade-quate. If the weld is oversized, you reduce the dimensions and check the de-sign again. If the weld is undersized, you increase the dimensions and do theanalysis calculations again.

There are no design assistance tables for weld design because of thecomplicated interactions between the leg sizes, the effective values of b andh, and because Q R is related to the square root of the sum of squared quanti-ties, (12.32). The best design assistance for weld design is to place the calcu-lations in a spreadsheet and do a group of what-if calculations involving b,

h, and the leg dimension, w.Since the design calculations are basically the same as analysis calcula-

tions, we will not go through a design example. There are several comments,however, that can be made relative to the design process. First, select themembers in a plane frame or plane grid problem before doing weld designcalculations. The outside dimensions of the members define b and h for theweld. If b and h are not adequate, select a new member. The usual procedureis to increase h and decrease b because the strength of the member and theweld are a function of h squared and linear in b. Remember that planeframes and plane grids are statically indeterminate and a change in themember size changes the support forces, which change the final dimensionsof the weld.

Second, you can not reverse the last step in the analysis calculations and

come up with a leg size. Return to the example in Figure 12.10. The resul-tant shear flow in (12.44) was Q R = 0.675 kN/mm, and the allowable shearflow is defined by (12.4). If we equate these two relationships, we can solvefor the leg size by

0.3410.675 kN/mmall R

wQ Q

N = = = (12.46)

Since N = 2.75, the solution yields w = 5.44 mm. It appears that a 6 mmweld is adequate. The mistake in this approach is that both section proper-ties, Z w and J w, are a function of the start and stop distances. An increase inthe leg size decreases the effective values of b and h, which decreases thevalues of Z w and J w. The final result is that Q R increases and the 0.675

kN/mm value used in (12.46) is not correct. The calculation defined in(12.46) can be used to make an educated decision about the adjustment ofthe values for b, h, and w.

The third comment relates to unsymmetrical brackets similar to the onein Figure 12.11. The design of unsymmetrical brackets is complicated by thefact that the torsion moment M x is a function of the weld dimensions. Each


23/26

PROBLEMS 383

time the value of b and/or h is changed, the location of the weld centroidchanges and M x must be re-evaluated. The evaluation of M x can be incorpo-rated into a spreadsheet with proper planning. The spreadsheet is usuallydeveloped specifically for this type of bracket rather than for the shape ofthe weld.

Problems

12.1 Verify the equation for J w given in Table 12.2 for a pair of vertical

welds.

12.2 Verify the equations for J w given in Table 12.2 for the circle.

12.3 Evaluate the section property equation Z w for the group of four welds

shown. Each weld has a length of h and the spacing between each pair

is h. Do the evaluation by (a) evaluating the integral in (12.13) and

(b) using the parallel-axis theorem in (12.28).

z

h

h

b

h x

z

bb

h

b

x

P12.3 P12.4

12.4 Determine the section property equation J w for the group of four

welds in P12.4 above. Each weld has a length b and the spacing be-

tween each pair is b. Do the evaluation by (a) using the integral in

(12.29) and (b) using the parallel-axis theorem in (12.31).

12.5 Rectangular tubes welded to the supports are subjected to the support

forces given in the table below. The tube is defined by the section

code SC while the weld leg dimension is given by w. Determine

whether the welds are adequate when N = 2.75 and an E60 rod is

used. The member is welded on each side but the welds are discon-

tinuous at each corner. The effective length of each weld is the mem-

ber dimension minus the start-and-stop distances. The positive

direction for each force or moment is in the positive coordinate direc-


24/26


tions as shown in Figure 12.1. The 1-1 axis of the section is aligned

with the z axis. Information is provided in each system of units. The

reader is encouraged to develop a spreadsheet to solve the problems.

USCS Units

Problem SC w, in P x , kips P y, kips M, kip·in

(a) 58 0.2500 6.40 3.20 22.5

(b) 62 0.1250 8.60 3.10 12.3

(c) 71 0.1875 17.3 9.50 22.7

(d) 86 0.2500 15.6 12.4 52.4

SI Units

Problem SC w, mm P x , kN P y, kN M z, kN·mm

(a) 58 6 28.5 14.2 2540

(b) 62 4 38.3 13.8 1390

(c) 71 5 77.0 42.3 2560

(d) 84 6 69.4 55.2 5920

Problems 12.6 – 12.7 The brackets for these problems, below, have a pair of

E60 welds subjected to an axial force, a shear force, and a bending moment

about the z axis. The location of the hole in each bracket has the same rela-

tive dimensions for each problem. Use the brackets shown below. The di-

mension L is the specified length and not the effective length. Use a safetyfactor of N = 3.0 for each problem. This group of problems is best solved by

developing a spreadsheet that incorporates the q x equation in (12.32) and the

q y equation in (12.7). Calculate the force and moment components from the

input data for the applied load and angle, which is always less than 90°.

There is a single bracket in each case.

L

4 i n

1 0 0 m m

θ

P

(a)

x

y z•

L

P

θ

6 i n

1 5 0 m m1.25 in

32 mm

(b)


25/26

PROBLEMS 385

12.6 The weld length and leg dimension as well as information on the type

of bracket and applied load are presented in the table below. Deter-

mine whether the weld is adequate.

USCS Units

Problem Bracket w, in L, in P , kips θ

(a) (a) 0.1875 7.00 6.83 30

(b) (a) 0.2500 6.00 7.45 38

(c) (b) 0.2500 6.00 6.92 45

(d) (b) 0.1875 6.50 7.80 62

SI Units

Problem Bracket w, mm L, mm P , kNθ

(a) (a) 5 180 31.2 35

(b) (a) 6 200 41.2 28

(c) (b) 5 150 29.4 51

(d) (b) 4 180 31.2 65

12.7 Calculate the leg dimension w for the weld length and loading infor-

mation presented in the table below. Use an E60 rod and a safety fac-

tor of N = 3.0. The leg dimension should increase in 1 mm or 1/16

inch increments. The ratio Q R/Qall should be greater than 0.80. The so-

lution procedure is trial and error. One procedure is to assume the full

length of the weld is the effective length and calculate the dimension

w. Select a standard dimension w and calculate the shear flow ratio

using the correct effective length.

USCS Units

Problem Bracket L, in P , kips θ

(a) (a) 5.50 8.34 55

(b) (a) 6.75 9.56 20

(c) (b) 5.00 7.83 70

(d) (b) 7.00 10.3 40

SI Units

Problem Bracket L, mm P , kN θ (a) (a) 180 38.6 42

(b) (a) 170 31.6 24

(c) (b) 140 42.3 64

(d) (b) 150 34.3 35


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12.8 The brackets in this problem are welded to the side of a member using

an E60 rod. The welds are subjected to two shear forces and a twist-

ing moment about the x axis. The location of the hole in each bracket

has the same relative dimensions for each problem. The dimension for

a is the specified length and not the effective length. The effective

length for b is the full length. Use a safety factor of N = 3.0 for each

problem. This group of problems is best solved by developing a

spreadsheet that incorporates equations (12.16), (12.32), and (12.36).

The force and moment components should be calculated from input

data for the applied load and angle which is always less than 90°. The

weld length b and the leg dimension w as well as information on the

type of bracket and applied load are presented in the table below. De-

termine whether the weld is adequate. Use the value a = 1.5 inch or a = 40 mm for all of the problems. The applied load is supported by

two brackets in each case.

P12.8(a) P12.8(b)

P

θ

a

b

3 in 75 mm

P

θ

a

b

3 in 75 mm

x

y z•

40

1.5

USCS Units

Problem Bracket w, in b, in P , kips θ

(a) (a) 0.1875 7.00 13.6 30

(b) (a) 0.2500 6.50 15.2 38

(c) (b) 0.1875 4.25 11.9 62

(d) (b) 0.2500 6.00 15.8 33

SI Units

Problem Bracket w, mm b, mm P , kN θ

(a) (a) 5 180 57.2 35

(b) (a) 6 200 82.4 28

(c) (b) 5 160 58.8 51

(d) (b) 6 200 90.0 24

polar moment of weld group

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