Theoretical development for the state of stress along weld lines connecting plane rigid surfaces

0045.7949/89 53.00 + 0.00

Con~pur~ws& Strucrur~s Vol. 32. No. 6. pp. 1447-1453. 1989

Pergamon Press plc

Printed in Great Britain.

THEORETICAL DEVELOPMENT FOR THE STATE OF STRESS ALONG WELD LINES CONNECTING PLANE RIGID SURFACES GARY BURGESS School of Packaging, Michigan State University, East Lansing, MI 48823-1223, U.S.A. (Received 9 Se~te~~~ 1987) Abstrnct-A method for determining the complete state of stress in welds connecting two plane rigid surfaces subjected to full static loading is first developed. The method accounts for the significant contribution of the unwelded portion of the connected surfaces forced into contact by the applied loads to the overall strength of the assembly. The nonlinearity introduced into the system of descriptive equations by the contact nature of the problem is handled numerically using a rapidly convergent iteration technique suitable for programming on a small computer. A compact program written in BASIC for the method is used to illustrate the technique on two example probiems. With only minor modifications. the method and program may be used to obtain the familiar section m~ulus for fuily loaded beams with unsymmet~cal thin walls or solid cross-sections.

ET-_-

l-r”IIl

The use of fillet welds to secure beams to connecting members, walls or foundations as in Fig. 1 is a fairly common practice. There are several 1eveIs of analysis aimed at detaining the efastic stress distribution along the weld line which range from the analytically simple but inaccurate to the very complicated [i-4]. The purpose of this paper is to outline a method which combines the important ideas from each of these to obtain the desired result while maintaining a reasonable degree of simplicity. The starting point for the simpler methods is the strength of materials result [l] for the normal stress dist~bution on the beam cross-section,

_-..:*:I-.:..cqulrlo‘lurr,

,...^~f~~_,.r:~~” L”‘l31wx~LI”‘I~

^_ “1‘

.

a

“..-*-+

WL;‘ILCIIb

,.&-rL_ “1 1115;

weld, this would then also be the average shear stress on the adjacent leg in the case of a weld with equal legs. For a 45” fillet weld which has a right triangular cross-section, the suggested design stress is bfi which takes into account the reduced area at the throat of the weld [2]. The problem with eqn (2), however, is that it considers every point on the beam cross-section to be materially connected to the wall with all points above the neutral axis in tension and

where M is the applied bending moment, I is the cross-sectional moment of inertia about the neutral axis, and y is the distance from the neutral axis to a point on the cross-section. Considering the weld to be of the same material as the beam and using the midline of the weld as the border of the solid cross-section in Fig. 1, the moment of inertia in eqn (1) is

b

a+2t*

I = (a f t)(b + t)3/12. The maximum distance y = (b + t)/2 from the neutral axis to the weld midline defines the critical weld location. When used in eqn (I), these values predict an average normal stress on the leg of the weld of 6M C7=

(0 -t t)(b + z)~’

(2) 1447

Fig. 1. Solid rectangular beam welded to wall.

GARYBURGESS

1448

I:_

l fEuTR&

l

_

h

-

-

-

_

_

_

-

-

AXIS

_ _ - Eontact_ _ _ _

_ - - _ _ _ _ x_

k--a+

loading remain plane after loading, is assumed to hold for the sections defined by the beam end and the wall. The second, stating that the linear relation between stress and strain holds for both tension and compression, must be modified for those points on the beam cross-section which are not in compression. The first assumption implies that the normal displacements are linearly distributed, so that the strain is also. Together with the equilibrium conditions, the second implies that there exists a neutral axis above which only the weld is in tension and below which both the weld and enclosed portion of the crosssection are in compression as indicated in Fig. 2, with the stress distribution for responses in the elastic range given by

Fig. 2. Beam cross-section for the contact problem.

all points below in compression. However, points above the neutral axis cannot accommodate this condition as no physical connection between the beam and wall exists except of course for the weld itself. This approach therefore accounts for more material than is actually available to carry the load and consequently underestimates the critical stress. The conventional approach [2, 31 recognizes this and uses only the weld area as the true cross-section. Under these conditions,

u =k<

(4)

where k is constant and { is the distance from the new neutral axis located somewhat below the halfway point. Using the notation in Fig. 2 with r = y - h, d.r positive, t small compared to the beam dimensions and t . ds as the element of area for a segment of weld, equilibrium requires that the location h of the neutral axis and the constant k in eqn (4) satisfy

c&=4weldk(_y -h)(t

ds)

J

I = (a + 2t)(b + 2r)3/12 - ab3/12 Hid

which replaces the result in eqn (2) with 6M(b + t)

=

Mx =fvdd N_v - h)(t ds) l-h

d=(a+2t)(b+2t)3-ab3’

The problem with eqn (3) now is that even though it correctly accounts for the fact that points above the neutral axis are incapable of supporting tension, it ignores the substantial contribution of points below the neutral axis which do in fact support compression. This approach therefore utilizes less material than is actually available to carry the load and consequently overestimates the critical stress. Given a choice, eqn (3) is a better design criterion than eqn (2) since it incorporates a built-in safety factor. The size of the safety factor is not known, however, since the true stress is not available for comparison. The correct way to view the situation is as a contact problem between two bodies, the beam and the wall, which requires the use of the more rigorous theory of elasticity [4] for an accurate solution. But this approach is too complicated to be practical even for the simplest of problems [5]. What is needed is a method which allows for contact and yet retains the straightforward strength of materials approach. Examination of the reasoning behind eqn (1) reveals that there are only two assumptions which need to be reconsidered. The first, stating that sections originally plane before

ykO, - h)(a dy) = M.

+

(6)

J.0 The circuit integral around the beam perimeter and the definite integral in each of these equations represent the effects of the entire weld line and beam cross-section below the neutral axis respectively. After integration, eqn (5) gives a quadratic equation in h whose solution is h=&5-1 b

e

(7)

where 1 ab e =Tta+b

As expected, h/b is bounded between 0 and 0.5 for any value of e. Equation (6) then gives k=

6M 2bt[b(3a + 26) - 3h(a + b)] - ah3’

(9)

For purposes of illustration, take a = 1, b = 2 and t = 0.1. The approach which underestimates the

Theoretical development for stress along weld lines

1449

Since both plane surfaces are rigid by assumption, any relative motion between the two is equivalent to the rotation of one about some point C as in Fig. 3. Considering only the twisting moment for now, rotation about C sets up a shear stress distribution given by =rs = 9r

\

where q is constant and r is the distance from C to a point on the weld line. Friction caused by relative motion under pressure between the two surfaces may alter this situation somewhat but will be ignored here. With ds as the positive element of arc length along the weld line, equilibrium of forces in the x-direction requires that tEul7AL AXIS

(qr)sin fi(r ds) = 0. 6 J weld

Fig. 3. Generalization of the contact problem.

stress by treating the entire beam cross-section and weld as the stress carrier gives h = 1 and e = 1.237M from eqn (2). The conventional approach which overestimates the stress by treating only the weld line as the carrier gives h = 1 and o = 2.637M from eqn (3). Equations (7), (8) and (9) give h = 0.6490 and a = 2.0065M.

Equilibrium in the y-direction gives a similar result with sin p replaced by cos /?. Since r ’ sin /I = x -x, and r . cos /I = y, - y in Fig. 3, these two conditions together require that point C be the centroid of the weld line. Now since the twisting moment is the resultant of the distribution, a sum of moments about C requires that

6weldr(qr)(t h) = M, J

GENERAL ANALYSIS

The generalization to arbitrary beam cross-sections and weld line configurations subjected to any applied loading requires only that the two members connprtprl art .e.. SQ turn . ..a”.“_ hv “_I the .*._ wdd I,_._ .._. I.,_ nlan~ y..e..- rioid .-m-v g~~rfpg 511 the connection so that the assumption that plane sections remain plane is satisfied. This assumption is reasonably satisfied in a number of application areas other than beams. A good example is the caset of a crane with its turntable and rigid base plate welded to the bed of a split-beam trailer for transportation. Figure 3 shows the general situation of a weld line and the contact area between the two surfaces __--__-- con____ netted by the weld. The forces F,, FY and F,, the bending moments M, and MY, and the twisting moment M, at the origin of a conveniently placed xy coordinate system in the plane of the two surfaces represent the static resultant of all externally applied loads and are equivalent to the shear and normal stress distributions r2,., rrY, Q and cr caused by the action of one surface on the other under the influence of the applied loads. Since there are really three separate problems here, individual treatment will be given first to the twisting moment acting alone, then to the forces F, and FY acting alone, and finally to the remainder F,, M, and MY acting alone. The individual effects can then be superposed. 7 Failure analysis investigation performed by the author for an insurance company.

(10)

or

q’+ J,

where J, is the polar moment of inertia of the thin weld line about C. The results in eqns (10) and (11) are seen to be identical to those in the elementary theory of torsion [ 11. Next, consider the effects of the forces F, and F, acting together. Transferring these from the chosen nricrin tn ---0’---

the cpp_!mid ----

C

gep_ergt~s 8:

miditinnal ..U”...“...A.

twisting moment Fly, - F,x, at C which must be combined with the applied twisting moment and used in place of M, in eqn (11). The forces themselves now at C are expected to generate constant shear stress distributions along the weld line,

9)

tds weld

P

tds weld

That this is the case is easily confirmed by verifying that the resultants of these shear stress distributions at point C are simply F, and F,. Finally, consider the effects of the remaining loads, F,, M, and M, acting together. Following the reasoning behind eqns (5) and (6), out of all the areas of possible contact between the two rigid surfaces, only

GARY BURGESS

1450

Equations (14), (16) and (17) form a system of three independent relations in the unknowns h, 0 and k. The entire analysis up to this point is equivalent to the standard treatment given to beams with unsymmetrical cross-sections in texts on the advanced strength of materials [6] with the exception that the integrals in eqns (15) and (18) are not known a priori since by definition of A, they are functions of h and 0. This makes the system of equations highly nonlinear which suggests a rearrangement into a new set of relations which will admit an iterative solution. Elimination of k between eqns (14), (16) and (17) gives

a portion of these on one side of the neutral axis in Fig. 3 will actually be in contact under compression. This portion will be referred to as the contact region. The location of the neutral axis is specified by its perpendicular distance h from the origin and by its inclination 0 relative to the x-axis. Equation (4) now applies with 5 as the perpendicular distance from the neutral axis to a point on the weld line or contact region taken positive in the direction of increasing 8. Using geometric arguments, it may be shown that in terms of the xy coordinates of the point, { = h + y cos 0 - x sin 0.

(13) k=

There are three unknowns, h, 0 and k, between eqns (4) and (13) which must be determined from equilibrium conditions. A force balanced in the x-direction requires that adA=F,.

-M, = QJh + IXYcos 0 - Iy) sin 0’

B Using eqns (4) and (13) in this gives Ah +Q,YcosO

-Q,sinO

=2

(14)

where A is the total contact area composed of the weld line and contact region and Q, and Q, are the corresponding first moments about the x- and yaxes, A=$,,

QT=I

ydA

F: Ah +Q,cos8-Q,sintI

Q,,=$I~ xdA.

(19)

Solving for h from the first and second equalities and again from the first and third and equating the results gives h

=

fFJry - MxQ,lsin0 - fFJx, - MxQxlcos0 FzQx- MxA PA,, + M,Q,lsin 0 - fFJ,, + M,Q&os -9

(20)

=

FzQy+ M,.A

(15) Finally, solving for tan 0 from eqn (20),

tan

e

=

fF.L - M,Q.rlfFzQy+M,.Al- [FJ,,. + M,Q,lfFzQ, - MxAl fF,!r, - M,Q,.IfF,Q,.+ M, Al - fF,I,:, + M,.Q.,lfF.Q, - MJI

The remaining nontrivial equilibrium equations moment balances about the x- and y-axes, yadA

iA

=M,

iA

xadA

are

=-MI.

Again, using eqns (4) and (13) in these gives Q, h + I,, cos 0 - I.Y,sin 0 = $

(16)

and (17) where the moments of inertia are I.T!,,=

y=dA

IXY=

xy dA

Iv,, =

x2dA.

(18)

(21)

The replacement eqns (19), (20) and (21) now allow for an-iterative solution as follows. Step I. Arbitrarily locate the neutral axis by choosing h and 8. Step 2. Evaluate the integrals in eqns (15) and (18). Step 3. Find the new h as the average of the two values from eqn (20). Step 4. Find the new 0 from eqn (21). Step 5. Repeat Steps 2, 3 and 4 until convergence is obtained. Step 6. Determine the three values for k from eqn (19) and compare to verify convergence. The question of convergence is difficult to discuss on theoretical grounds because of the complexity of the equations involved and consequently will not be dealt with on this level except to say that the likelihood of convergence is good since 6 must remain bounded as the iteration proceeds by virtue of eqn

Theoretical development for stress along weld lines (21) although oscillation remains a possibility. The reason for using the average of the two values for h in Step 3 as the next generation guess is to equally weigh the influences of M, and My since one value depends on M, alone and the other on My alone. When both F, and one of the bending moments are zero together, one of these values for /I is indeterminate, which presents a problem numerically. This is most easily handled without using branch statements in the algorithm by assigning small dummy values, say 0.00001, to the zero loads and relying on the averaging process for the next generation h to stabilize the iteration against the unreliable value obtained from the originally indeterminate equation. Ultimately, convergence of the iteration as a whole may be checked by comparing the three values for k obtained from eqn (19). NUMERICAL

PROCEDURE

The primary purpose of this section is to work out the details of Step 2 in the iteration as this involves the greatest amount of computation. In the process, all information relevant to the shear stress distributions will be made available automatically. The region of integration, A, for the integrals in eqns (15) and (18) is composed of two parts, the thin band centered on the weld line and the contact region below the neutral axis where r < 0. The integrals over the weld area are first expressed as the weld thickness t multiplied by the line integral along the weld. The line integral is then approximated by the sum of elemental contributions from straight line segments joining a number NW of specified nnintc L’y’v”‘...... rl=nrl=rPntino ucncrin ______, fmnl y”‘...” D the weld line. In_ o_____-_ I = t

“c(ai)“uf)”

lines may be improved using the finite element approach to integration [7] if the regions are simple shapes, but overall accuracy will eventually be lost at the neutral axis in spite of this, since regions intersected there by the neutral axis are discarded entirely by the cutoff criterion involving <. The computer program in the Appendix performs the iteration according to the method outlined above and prints all information relevant to the complete state of stress along the weld line. Written in BASIC for use on small computers, it contains intentional syntax errors in several statements in order to limit program size. The data required are the NW weld line coordinates x and y stored in the two columns of the matrix W (NW + 1, 2) with the last point identical to the first, the NR region coordinates stored in R (NR, 2), the region areas DR (NR), the fillet weld leg dimension T, part of the load resultant FZ, MX, MY, and the initial guess H, THETA, which begins the iteration. The contributions of the weld line to the six integrals in eqns (15) and (18) are denoted by WO to W5 and the integrals themselves by A0 to A5. The parameters El to E6 make up the various terms found in eqn (21). The information required in eqns (lo), (11) and (12) for the evaluation of the shear stresses produced by the remainder of the load resultant is provided using already determined quantities, x,= WI/W0 y, = w2/wo J,=

W3+

W5-(Wl*+

W2*)/WO

ds, RESULTS

1-l

where m and n are the integers 0, 1 or 2 appropriate to the particular integral, fi and yi are the coordinates of the midpoint of the ith line segment, and ds, is the lenoth ____o___ of __ the ____ semnent __5)_______.Since these Inteerals 51----- are --- in&-

pendent of h and ~9,this part should be done before the iteration begins. In a similar manner, the integrals over the contact region can be approximated by I = f

1451

(n,)m(jji)” dRi(l - sign 5,)/2

(23)

i-l

once the entire region of potential contact above and below the neutral axis has been subdivided by grid lines forming NR elemental regions of area dR, centered on (Zi, jj,). Note that the term involving ti here counts an elemental contribution only if it lies below the neutral axis. Since the integrals in eqn (23) do depend on h and 8 through 5 via eqn (13), they must be recalculated for every iteration and added to the results in eqn (22) each time. Accuracy of integration over the elemental regions formed by the grid

Two example problems are presented here to illustrate the performance of the algorithm for the iteration method. The first is the previous example in Fig. ir _3 .and ...” the I.._ r~rnnrl “II”.._ .I thP . ..” rnlirl “VS... rirm,lar W.IW..IU. hpsm “.,U‘ll in 1stFin a ‘6’ A 7, Example

1

For comparison purposes, the problem in Figs 1 and 2 where the beam with the solid rectangular cross-section was given the dimensions a = 1, b = 2 and t = 0.1 will be solved using the program in the Appendix. The cross-section was divided into NR = 100 rectangular elements each with constant area DR = 0.2 and the region coordinates relative to the xy axes in Fig. 2 were placed at the element centers. The weld line was represented by NW = 40 points, 10 per side, with regular spacing and the resultant of the load on the cross-section was taken as MX = 1 and FZ = MY = 0.00001. Table 1 shows the results of the iteration beginning with the intentionally poor initial guess, h = 1000 and 6 = 90”, which led to convergence after only five iterations.

1452

GARY BURG= Table

1. Results

for examole

I

Table 2. Results

h

Iteration

Iteration

0 1 2 3 4 5 6

1000 0.95293 - 0.99997 -0.68748 -0.64998 -0.64997 -0.64997

k, = 2.0201

k, = 2.0202

2.45 2.45 2.18 2.56 2.56 2.56

90 x 10-j x lO-3 x lO-3 x 10-j x lO-3 x lO-3

k, = 2.0200

The exact values from eqns (7) and (9) are h = 0.6490, 0 = 0, and k = 2.0065. The negative value for h in Table 1 indicates that the neutral axis is on the opposite side of the origin in Fig. 3 as it should be. The results of the program for a number of cases in which NW, NR and the initial guess were varied indicated that five iterations appears to be the upper limit on the number required for this problem, regardless of initial guess or coarseness of subdivision of the weld line and contact region, and that accuracy improves only slightly with increasing NW and NR beyond a certain limit, most likely due to round-off error accumulation. Example

2

The cross-section of the solid circular beam in Fig. 4 was divided into NR = 300 elemental areas formed by the intersection of 60 radial lines placed 6” apart and five concentric circles spaced 0.4 apart, with the region coordinates taken at mid-radius and mid-angle for each element. The weld line was represented by NW = 60 points on the radial lines at the

for example

h

0

0

1 2 3 4 5 6

1.0523 I .6344 2.2892 3.1818 2.9733 2.9733

2

80

0 - 16.976 - 37.289 - 57.683 -63.435 -63.435 -63.435

k, = k. = k. = 2.6111

boundary and the load resultant was taken as FZ = 10, A4X = 3 and MY = -6. Table 2 shows the results of the iteration beginning with the initial guess h = 0 = 0. Again, only five iterations were required for convergence. The exact value for 0 coincides with the direction of the resultant moment, 0 = arctan( - 6/3) = -63.435, and the result h = 2.4733 indicates that the neutral axis lies outside the cross-section where it should be since the dominant load FZ = 10 leads to a general state of tension which eliminates the contact region and forces the weld to carry the entire load. Superposing the stress due to the axial load F, and the bending stress from eqn (I) with M as the resultant moment gives the exact solution

Using the vaiues for F,, MX, My, R and t in the above and forcing it to fit the form of eqn (4) requires that h = 2.9814 and k = 2.6691 for the exact solution. Under the new loading, FZ = 10 and MX = MY = 0, the iteration converges rapidly to h = 3.92 x 109, 0 = 33.15” and k = 2.03/109. These results for h and k are understandable since the only way in which the stress distribution of eqn (4) can represent the uniform state of tension implied by the loading is if h + 00 and k -+O with the product of h and k being FJ2nRt = 7.9577. The actual product is 7.9576. CONCLUSIONS

Fig. 4. Solid circular

beam welded to wall.

The new contribution to the analysis of weld stresses presented here lies in the determination of the effect of the contact region between the connected surfaces on the location of the neutral axis, the normal stress distribution, and ultimately on the strength of the connection. The iteration technique developed for the solution of the nonlinear equilibrium equations has been shown to converge rapidly by example. The additional benefit gained by this approach is that the computer program in the Appendix may be used with only minor modifications to obtain the familiar section modulus [l, 61 for beams with unsymmetrical cross-sections. For thin walled beams, the weld thickness T becomes the beam wall thickness and the contact area influence is

Theoretical development for stress along weld lines either by setting DR = 0 or by bypassing the area loop in the program entirely. For solid beams, T = 0 and the IF statement in the program is eliminated. Of course, in each of these two cases, the desired results may be obtained by noniterative methods [6] which require the solution of a cubic equation whose roots are the eigenvalues of the moment of inertia tensor composed of the terms in eqn (18). In comparison to the iteration method, this approach is expected to involve a similar amount of programming effort and computation associated mainly with the solution of the cubic equation. removed

REFERENCES 1.

2.

3. 4. 5.

6.

7.

B. I. Sandor, Strength of Materials. Prentice-Hall, Englewood Cliffs, NJ (1978). J. Giachino, W. Weeks and G. Johnson, Welding Technology, 2nd Edn. American Technical Publishers, IL (1973). A. Blake, Design of Mechanical Joints. Marcel Dekker, New York (1985). L. Landau and E. Lifshitz, Theory of Elasticity. Pergamon Press, London (1959). x* r I”IuskL,eIIJII”III, I,.. _,.?--,:^l-..:,: So,x .&sic. 1D”A.,^...^ ,ftl., 11,1L IY. 1. l”“‘rrr‘J “, ‘,lC1*x”‘,‘ematical Theory of Elasticity, 3rd Edn. P. Noordhoof, Holland (1953). A. Boresi, 0. Sidebottom, F. Seely and J. Smith, Advanced Mechanics of Materials, 3rd Edn. John Wiley, New York (1978). L. Segerlind, Applied Finite EIemenf Analysis. John Wiley, New York (1976).

APPENDIX

10 DIM W(NW + l,Z),R(NR,Z),DR(NR) 20 INPUT NW,ALL OF W(NW + l,Z),NR,ALL OF R(NR,Z) AND DR(NR) 30 INPUT T,FZ,MX,MY,H,THETA 40 WO=Wl=W2=W3=W4=W5=0

1453

50 FORI= TONW 60 DS=SQR((W(I+l,l)-W(I,I))A2+(W(I+l,2) - W(V)) A 2) 70 X=(W(I+ l,l)+W(I,l))/2 80 Y = (W(I + 1,2) + W(I,2))/2 90 WO = WO + TzDS 100 WI = Wl + X*T*DS 110 W2 = W2 + YtT+DS 120 W3 = W3 + X*X*T*DS 130 W4 = W4 + XsY*T+DS 140 W5 = W5 + Y+Y*T*DS 150 NEXT I 160 FOR ITERATION = 1 TO 10 170 AO=WO : Al=Wl : A2=W2 AS=WS 180 A3=W3 : A4=W4: 190 ST = SIN(THETA) : CT = COS(THETA) 200 FORI=lTONR 210 IF H + R(I,2)zCT - R(I,I)+ST > 0 THEN 280 220 A0 = A0 + DR(I) 230 Al = Al + R(I,I)*DR(I) 240 A2 = A2 + R(I,Z)*DR(I) 250 A3 = A3 + R(I,l)cR(I,l)*DR(I) 260 A4 = A4 + R(I.l)*R(I.ZkDR(I) 270 A5 = A5 + R(I,Zj*R(I;Zj*DR(Ij 280 NEXT 1 290 El = FZ*A3 - MX*AI 300 E2 = FZ*A2 + MY*AO 310 E3 = FZ*A4 + MY*AI __^ _ MX*AO 320 E4 = FZ,Al 330 E5 = FZ+A4 - MX*A2 340 E6 = FZ+AS + MYsA2 350 Hl = (ES&T - El+CT)/E4 360 H2 = (E6+ST - E3+CT)/E2 370 H=(Hl +H2)/2 380 THETA = ATN((El*E2 - E3rE4)/(E5+E2 - E6rE4)) 390 PRINT H,THETA+l80/3.14159 400 NEXT ITERATION 410 K 1= FZ/(AO*H + A 1*CT - AZ&T) 420 K2 = MX/(Al*H + A3*CT - A4tST) 430 K3 = - MY/(AZ*H + A4*CT - AS&T) 440 PRINT K I ,K2,K3 450 XC=Wl/WO : YC= w2/wo 460 JC = W3 + W5 - (Wl+Wl + W2+W2)/Wo 470 PRINT WO,XC,YC,JC 480 END