Analysis of planar welds

Transcription

1 Dr Andrei Lozzi Design II, MECH Analysis of planar elds School of Aerospace, Mechanical and Mechatronic Engineering University of Sydney, NSW 2006 Australia lecture eld ne b References: Blodget, Design of Weldments, J Lincoln oundation Hall et al, Machine Design, McGra Hill (Schaum) Shigley et al, Mechanical Engineering Design, eds 4 to 7, McGraa Hill Australian Standard Association AS1554 elding & AS3990 steel structures A quick overvie: There are many aspects of elding hich ought to be knon by engineers, e ill begin ith a simple method of calculating the idth of a fillet eld. That is, ho ide should e make a eld once the outline of the eld has been decided on and all the loads transmitted to the eld are knon. The creative part here is reduced to just inventing the shape and lengths of the eld outline. This is not as dumb as it may sound, for example, for the eld shon in ig 1 belo, one may select the C outline that goes around the outside, shon on ig 1, or just 2 horizontal runs, top and bottom, or one of many others. Having chosen the outline you then calculate the eld idth. If there appears to be problems ith that idth, then nominate another outline and redo the calcs. Continue this for a hile and you ill be able to tell from the loads and the beam hat for example is likely to be the eld outline that ill requires least eld deposit. The eld deposit is the thing. The eld belo joins 2 adjacent parts, the beam and the base plate, but the eld is the only component being analysed here. It is assumed that the base plate is flat and that the beam meets it ith a flat cross section. The beam does not have to be perpendicular to the plate, and the shape and size of the beam from the eld to the load has no effect on the eld calculation. The beam can be curved, bent and joined to other components. The only thing that matters is the relative location in space of the load to the Centre of Area (CofA) of the eld outline. The exploded vie belo can be replaced ith ig 2 here the location of the force and the eld outline are shon schematically. igs 2 & 3 sho ho e arrive at the shear force, moment and torque that the eld has to be designed to carry. illet eld deposit Beam CofA of Weld outline Applied orce Base plate ig 1 End plate 1

2 Y Origin at CofA of eld z=a X x=b Z ig 2. A schematic diagram reproducing ig1, shoing force and its location in space ith respect to the CofA of the eld outline. Because the eld is on the outside of the beam, the eld outline is taken to be the outer boundary of the beam. Note that it is the CofA of the eld run not the CofA of the beam, that is relevant here. Note again that it does not matter hat the shape of the components are that connect force to the eld. They have to be considered separately, together ith many other things such as the compatibility of eld deposit to the plates thicknesses. Moment a orce relocated to CofA of eld Torque b ig 3 The stresses in the eld are determined by firstly relocating the force from its point of application to the CofA of the eld outline. Moving the force to the Z axis through distance b generates the torque b, moving it don the Z axis to the CofA of the eld generates the moment a. Here is assumed parallel to the base plate and the Y axis, but this need not be the case. We can no arrive at the eld idth - from the loads that it has to carry: that is the torque b, moment a, and the shear force. Practical short-cuts taken to simplify eld analysis. The analysis presented here is based on that of the American Welding Association (AWS) as ritten up in Hall and Blodget. The objective is to provide a simple means of calculating fillet idths for any combination of shear force, moment and torque. It has been developed to reduce engineer s time and ith it engineering costs. 1 We ill consider shear forces and shear stresses only this ill simplify calculations, it is nearly alays the most conservative approach, but here in part some of the theoretical modelling is less than conservative, a correction ill be made. 2 We ill only consider equal legged fillet elds (45 fillet) these are possibly the most common and their analysis can be readily extended (by you) to unequal legged elds. 2

3 3 In the calculations e ill use the leg length, not the minimum cross section, that is the throat t, because e cannot easily measure t but is readily available. A correction ill be made for this. 4 We assume that the point here the load is being applied is sufficiently far enough aay from the eld so as not to cause any stress concentration at the eld. y ig 4. shos a short straight fillet eld of equal leg and length l. The 2 plates shon transmit force through the eld. To of the components of the force lie transverse to the eld, ie x & y, the third component z lines up parallel ith the eld run. z l x The rule is: if any component of a force at a eld is parallel to the eld then the hole force ill be assumed to be parallel to the eld. Most of the time this ill forgo a small advantage of transverse elds (about 15%) for the sake of reducing the cost of anlysing elds. If the deposit is particularly large and expensive the advantages of transverse elds may be considered. Also for the sake of simplification e ill also use l as the nominal stress area. The fact that the minimum stress area is less than l ill be alloed for ith a simple coefficient later. Shear orce. We ill no examine the effect of the shear force transmitted by the eld outline. We ill characterise this by a force per mm of the eld length due to shear force - f s. shear _ force Eq 1 Shear stress is: S assuming is completely in shear and Area Eq 2 at nominal stress area: S S is the actor of Safety, hich may be 1.2 to 2. l Eq 3 rearrange to give : orce S l S Eq 4 e ill use the form: all l S Eq 5 or: f s all all all is the alloable shear stress of the eld deposit This gives eld idth for a shear force and length l Since the terms in the first bracket are predetermined by the choice of materials and S, the second bracket (/l) or force per unit length of eld, determines the eld idth. The expression /l is substituted by the term f s, hich is called the force per unit length (mm) of eld run due to the shear force. 3

4 Note that f s is a vector, because it comes from dividing, a vector, by a scalar l. Therefore f s points parallel to, in this example in the Y direction and in the plane of the base plate, ie f s has only a y component. Also f s exists undiminished (does not change) all around the eld outline. Left like that this analysis ould not be conservative because shear stress, due to a shear force, is not equally distributed across a section, it is a maximum at the neutral axis and is 0 at the most distant fibres. We ill come back to this later. Eq 6 Eq 5 can be reordered to give all f s (temporarily omitting the S). l This form tells us that if e can evaluate all for the moment and the torque that is transmitted through the eld, M and T on fig 3, it ill give us the force per mm of eld length resulting from each of those loads. inally, the vector sum of all the 3 forces per unit length, hen divided by ill give us the idth of the eld. all Bending moment We ill no deal ith the shear stress at the eld due to the bending moment. We ill characterise this by a force per mm of the eld length due to bending - f b. Eq 7 Normal stress due to bending: Eq 8 using modulus of section Z: Eq 9 assuming, e can let: M y I M I here Z is tabled in catalogues of metal sections Z y M here Z Z, that is Z is the moduls of Z section for that eld outline, for a idth = 1 mm. The Z for the actual section ill be larger than Z unless the idth < 1 mm. This analysis ill be conservative for all but very very thin elds. M Eq 10 Eq 9 gives us f b f b force per mm of eld length due to bending. Z f b Y y X ig 5. Just as the normal stress created by the bending moment is a vector, so is the force per unit length due to bending f b is also a vector, hich as derived from it. Note that f b ranges beteen a maximum positive to max negative value, at the largest distance y from the neutral axis X. f b varies linearly to 0 at the neutral axis. Moment M = a O The eld outline is a reversed C. rom ig 1 e kno that it attaches a C channel to a plate, but the analysis of this outline ould be no different if it ere used on 3 sides of a rectangular hollo channel (RHS) or any other beam that has 3 suitable sides. f b () Z Note also that f b is perpendicular to the plate, and that in this coordinate system it has only a z component. 4

5 Torque We ill no deal ith the force per mm of the eld length, due to torque - fj. Eq 11 Shear stress due to torque: Eq 12 let polar moment of area J: Eq 13 T r J J J T r J outline, for a idth = 1 mm. The J for the actual section ill be larger than normally this equation applies to circular sections only here J is the polar moment of area of the eld J unless as indicated above the idth < 1 mm. This analysis ill also be conservative for all but very very thin elds. T r Eq 14 The above gives us f j: f j force per mm eld length due to torsion. J f j r2 r 1 Y f j 2 X Torque b ig 6. This figure shos graphically ho fj is evaluated around a eld outline. fj is a vector quantity and ill typically have a different value for each point around the outline. The applied torque T ill be a constant as ill J and Z. ig 7 on page 6 gives expressions for J and Z for common outlines. The only variable in the calculation of f j ill then be r i, the radial distance from the CofA of the eld to the point of interest on the eld. j being in the same direction as the shear stress, ill be in the plane of the plate and perpendicular to r i. Here fj i ill only have x and y components. Eq 11 applies only to circular sections (tubes etc) cannot be applied to the beam of C section but may be applied to this eld because the inner portion of the square plate ill behave approx like part of a circular plate Vector sum of f s, f b and f j. These 3 values are often mistakenly added like scalars ie (fs + fb + fj). This ill overstate their sum except for the situation hen they are collinear to each other. In contrast adding them as if they are perpendicular to each other (fs 2 + fb 2 + fj 2 ) 1/2 ould understate their sum. The vector sum of f s, f b and f j is of course: Eq 15. ft = ((fsx+fbx+fjx) 2 +(fsy+fby+fjy) 2 +(fsz+fbz+fjz) 2 ) 1/2. In most situations many of these components are 0. or the eld outline shon on ig 1 only 4 of the 9 components are non 0. S Eq 16. inally the eld idth is arrived at by a version of Eq 5: f t all given a suitable alloable shear stress for the eld deposit and an appropriate S to the elded joint. 5

6 ig 7. Table of equations to evaluate Z and J for some common outlines, of unit idth. 6

7 Y Weld runs: top & bottom of RHS An example (sort of) X f b ln r f jy f j f jx b d f s ig 8. 1 The selected eld outline is to be the 2 horizontal elds at top & bottom of the RHS (rectangular hollo section). Z 2 Select a point on the outline that may be the most heavily loaded location on the outline. Here e ill try the bottom LH corner. 3 given b & d, e get that l b d and e calculate Z & J from ig 7, 3 rd ro don. 4 Given perpendicular distance ln from force to CofA of the eld, the moment about the eld is ln 5 Torque about CofA of eld is d/2. Note: CofA of eld and of the section of the beam coincide. 6 orce per unit length due to shear is: f s, a vector made up of the components: (0, -f s, 0). l 7 orce per unit length due to bending is: M f b, it is composed of the components (0, 0, -f b) Z 8 orce per unit length due to torsion is: Tr f j, its components are: (-f j cos(), f j sin(), 0). J 9 The vector sum of f s, f b & f j is: f t = (( -f j cos) 2 + (f j sin -f s) 2 + ( -f b) 2 ) 1/2. S 10 inally the eld idth is to be: f t, try S ~1.5 and all ~100 N/mm 2 all 11 All this ork has to be repeated for a number of other likely locations around the outline. The rule is: the eld idth required at the most heavily loaded location is to be used everyhere on the outline. Only if it is particularly expensive may the idth be varied around the outline. 12 The value for needs to be compared to the all thickness of the RHS and the thickness of the plate being elded to. The eld idth should be less than or equal to the smaller all thickness. 13 The above calcs should be repeated for different eld outlines, to arrive at possibly the outline that requires the least eld deposit. Depending on the loads some outlines ill be more effective than others. 7

8 Shear stress in fillet elds loaded in parallel Here e ill endeavour to find a closer approximation to the shear stress in a fillet eld. To make the situation simple e ill consider mirror image elds. This ill remove an unbalanced bending moment. On ig 9 belo the loads are parallel to the fillets, ie in and out of the paper, ith each filet transmitting force. We ill determine the max shear stress at a throat t, at angle θ. t (sin cos ) from sine rule Eq 17 Area t l +2 (sin cos ) Eq 18 l Applying calculus to get the max value of τ gives the unsurprising anser that it occurs at the min throat, ie for θ = ig 9. illets eld loaded in parallel The min tensile stress of elding rod steel may taken to be S u 410 N/mm 2 By AWS convention the max alloable shear Eq 22 stress is to be t all 2 Eq 19 l 2 Eq 20 all l Thus if e use as parameter to measure be fillet elds hen loaded in parallel e can use the form: l l Eq 21 SU all. Effectively for parallel elds all = 96.6 N/mm 2 3 Shear stress in fillet elds loaded in transverse fig 10 belo shos a pair of symmetric fillet elds carrying a load of 2 perpendicular to the eld runs. At a general throat t, defined as in ig 9, force may be divided in a normal components N and a shear component S. rom the geometry: 2 2 S N ig 10. illet elds loaded in transverse θ S = sinθ Eq 23 Taking only the shear force into account, from Eq 18 e deduce: (sin cos ) sin Eq 24 l Since the above expression for τ is different to Eq 18 max occurs at = max Eq 25 l Eq l 113 l Effectively for transverse elds all = 113 N/mm 2 ie capable of about 15% greater loads than parallel elds. 8

9 As mentioned on p3 (ig 4) an AWS rule is that if any component on any part of a eld outline is loaded in parallel; then the hole eld is to be analysed as if it is all in parallel. Obviously if it ere orth the time, at the cost of an engineer s rate of pay, adjustments could be done for different part of a eld and the fillets could be tapered from point to point. It all could be quite aesthetically pleasing. Specific application of elding. There are many applications of elding such as pressure vessels, pipe ork, bridges and many others, hich are covered by particular codes. Take care to make yourself aare of them and adhere to them. The procedure examined here may be used for machine frames and similar mechanical applications. Welding typical involves human judgement to a greater degree than most other machine operations. As a consequence there are precautions taken in important jobs that are not often seen in other operations. To ensure a high degree of safety it may be orth considering the use of more than one joint to carry an significant load, instead of applying a large actor of Safety to a single joint. ig 11 At right shos some of the undesirable realities of elded joint that may go undetected. ig 12 belo shos one of the reasons that hen a joint fails it is usually not the eld deposit that fails. The penetration has provided a greater deposit than alloed for in the calcs. Weld deposit Inclusions Heat effected zone Undercut Incomplete penetration ig 13 Left and belo. These figures from Reshtov give an indication of the stress concentration that takes place in ide range of elded joints. Normal and shear stresses are shon together ith the effect of partial and full eld penetration. 9

10 Modes of eld analysis. 1 Analyse the eld as described here or in more detail. Try alternate eld outlines to arrive at a preferable eld hich has some clear advantages eg length, eight, least preparation. A programmable CAD system could moderately easily be used to automate the analysis. 2 Design a structure so that its components can carry their loads. Then eld all around the joins to the depth of the all of the components. Do not analyse the elds. All around elds are employed in many situation to seal the joining faces from corrosion or other intrusions. 3 Use a EA method of analysis. Prescribe elds at joints to the depth and idth that they loer stresses in their neighbours belo safe limits. Some Precautions 1 Every joint generates some stress concentration. A ell prepared butt joint that is machined flat after elding may cause the least, but it too may raise the stresses by at least 10%. 2 Plate or sections joined together should not vary in thickness by more than about 50%. If the elder is suitably experienced a larger difference may be used, but it is better to have section that provides for a transition in thickness. This is indicated by the 2 figures belo: ig 14. Indications of limits in disparity in thicknesses 3 When designing a elded joint that has to be relatively stiff under load or has to be relatively safe against failure, reinforce the joint ith plates and other section to spread the load into a larger portion of the parent materials. Examples of this are shon on page 11: a igs 1 to 5 sho ho an end flange on a shaft is reinforced ith larger and double elds figs 2 & 3, And by a scre thread and an interference fit in figs 4 & 5. b A tensioned component in fig 6 is trapped by a shoulder in 7, and by a eb in compression in 24 to 25. c The loads are spread at the joint in figs 19 to 23, and in 26 to 30, again in figs 33 to

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12 ig 15 A preliminary spreadsheet developed to solve for the eld idth of the example shon on ig 8. Be arned that this spreadsheet has not been checked and it almost certainly contains faults. 12