68416 ch 12

Chapter 12

Joining and Fastening Processes

Questions

12.1 Explain the reasons that so many differentwelding processes have been developed.

A wide variety of welding processes have beendeveloped for several reasons. There are manytypes of metals and alloys with a wide range ofmechanical, physical, and metallurgical prop-erties and characteristics. Also, there arenumerous applications involving a wide va-riety of component shapes and thicknesses.For example, small or thin parts that cannotbe arc welded can be resistance welded, andfor aerospace applications, where strength-to-weight ratio is a major consideration, laser-beam welding and diffusion bonding are attrac-tive processes. Furthermore, the workpiece maynot be suitable for in-plant welding, and thewelding process may have to be brought to thesite, such as pipelines and large structures. (Seealso Section 12.1.)

12.2 List the advantages and disadvantages of me-chanical fastening as compared with adhesivebonding.

By the student. Advantages of mechanical fas-tening over adhesive bonding:

(a) disassembly is easier (bolted connections);

(b) stronger in tension;

(c) preloading is possible; and

(d) no need for large areas of contact.

Limitations:

(a) often costlier;(b) requires assembly;(c) weaker in shear; and(d) more likely to loosen (bolted connections).

12.3 What are the similarities and differences be-tween consumable and nonconsumable elec-trodes?

By the student. Review Sections 12.3 and 12.4.Comment, for example, on factors such as therole of the electrodes, the circuitry involved, theelectrode materials, and the manner in whichthey are used.

12.4 What determines whether a certain weldingprocess can be used for workpieces in horizon-tal, vertical, or upside-down positions, or for alltypes of positions? Explain, giving appropriateexamples.

By the student. Note, for example, that somewelding operations (see Table 12.2 on p. 734)cannot take place under any conditions ex-cept horizontal, such as submerged arc weld-ing, where a granular flux must be placed onthe workpiece. If a process requires a shieldinggas, it can be used in vertical or upside-downpositions. Oxyacetylene welding would be dif-ficult upside-down because the flux may dripaway from the surface instead of penetratingthe joint.

12.5 Comment on your observations regardingFig. 12.5.

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By the student. The students are encouragedto develop their answers considering the signif-icance of the layered weld beads and the qual-ity of their interfaces. For example, there maybe concerns regarding the weld strength sincethe interfaces between adjacent beads may havesome slag or surface contaminants that have notbeen removed. The heat-affected zone and fa-tigue implications of such welds are also a sig-nificant concern.

12.6 Discuss the need for and role of fixtures in hold-ing workpieces in the welding operations de-scribed in this chapter.

By the student. The reasons for using fixturesare basically to assure proper alignment of thecomponents to be joined, reduce warpage, andhelp develop good joint strength. The fixturescan also be a part of the electrical circuit in arcwelding, where a high clamping force reducesthe contact resistance. See also Section 14.11.1.

12.7 Describe the factors that influence the size ofthe two weld beads in Fig. 12.13.

The reason why electron-beam weld beads arenarrower than those obtained by arc welding isthat the energy source in the former is muchmore intense, confined, and controllable, allow-ing the heating and the weld bead to be morelocalized. Other factors that influence the sizeof the weld bead are workpiece thickness, mate-rial properties, such as melting point and ther-mal conductivity. See also pp. 749-751.

12.8 Why is the quality of welds produced by sub-merged arc welding very good?

Submerged arc welding (see Fig. 12.6) hasvery good quality because oxygen in the atmo-sphere cannot penetrate the weld zone wherethe shielding flux protects the weld metal. Also,there are no sparks, spatter, or fumes as inshielded metal arc and some other welding pro-cess.

12.9 Explain the factors involved in electrode selec-tion in arc welding processes.

By the student. Refer to Section 12.3.8. Elec-trode selection is guided by many factors, in-cluding the process used and the metals to bewelded.

12.10 Explain why the electroslag welding process issuitable for thick plates and heavy structuralsections.

Electroslag welding (see Fig. 12.8) can be per-formed with large plates because the tempera-tures attainable through electric arcs are veryhigh. A continuous and stable arc can beachieved and held long enough to melt thickplates.

12.11 What are the similarities and differences be-tween consumable and nonconsumable elec-trode arc welding processes?

By the student. Similarities: both require anelectric power source, arcing for heating, and anelectrically-conductive workpiece. Differences:the electrode is the source of the weld metal inconsumable-arc welding, whereas a weld metalmust be provided in nonconsumable-arc weld-ing processes.

12.12 In Table 12.2, there is a column on the distor-tion of welded components, ranging from lowestto highest. Explain why the degree of distortionvaries among different welding processes.

By the student. Refer to Table 12.2 on p. 734.The distortion of parts is mainly due to ther-mal warping because of temperature gradientsdeveloped within the component. Note thatthe lowest distortions are in electron beam andlaser beam processes, where the heat is highlyconcentrated in narrow regions and with deeperpenetration. This is unlike most other processeswhere the weld zones are large and distortioncan be extensive.

12.13 Explain why the grains in Fig. 12.16 grow inthe particular directions shown.

The grains grow in the directions shown inFig. 12.16 because of the same reasons grainsgrow away from the wall in casting process so-lidification, described in Section 5.2. Heat fluxis in the opposite direction as grain growth,meaning a temperature gradient exists in thatdirection, so only grains oriented in the direc-tion perpendicular from the solid-metal sub-strate will grow.

12.14 Prepare a table listing the processes describedin this chapter and providing, for each process,

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the range of welding speeds as a function ofworkpiece material and thickness.

By the student. This is a good assignment forstudents, although it can be rather demandingbecause such extensive data is rarely available,except with a wide range or only for a particu-lar group of materials. Consequently, it can bedifficult to compare the processes; they never-theless should be encouraged to develop such alist as best they can.

12.15 Explain what is meant by solid state welding.

As descried briefly on p. 733, in solid state weld-ing, the metals to be joined do not melt; there isno liquid state in the interface. Note that thereare six processes listed under this category.

12.16 Describe your observations concerningFigs. 12.19 through 12.21.

By the student. This is a challenging questionand students are encouraged to develop and listas many answers as they can. For example,they can consider the crack locations, developan ability to identify them through inspection,describe the causes of the defects, and the ef-fects of different workpiece materials and pro-cessing conditions.

12.17 What advantages does friction welding haveover the other joining methods described in thischapter?

By the student. As described in Section 12.9,the main advantages of friction welding are thatthe entire cross-sectional area can be welded, in-stead of a mere bead along the periphery, andis suitable for a wide variety of materials. Also,with proper process control, the weld zone canbe very small and thin, so that thermal distor-tions will be minimal.

12.18 Why is diffusion bonding, when combined withsuperplastic forming of sheet metals, an attrac-tive fabrication process? Does it have any lim-itations?

By the student. As shown in Fig. 12.41,diffusion bonding combined with superplasticforming can produce lightweight, rigid, andstrong aerospace structures with high stiffness-to-weight ratios. The main drawback is the long

production time and the high costs involved,which may be justified for many aerospace ap-plications. The students are encouraged to findother examples of applications for this impor-tant process.

12.19 Can roll bonding be applied to various part con-figurations? Explain.

Roll bonding (Fig. 12.28) is mainly used in flatrolling, although other applications may be pos-sible. The important consideration is that thepressure (normal stress) between the sheets tobe joined be sufficiently high and the interfacesare clean and free of oxide layers. To meet thiscondition for shapes other than flat is likely tobe a difficult task and involve complex tooling.Also, any significant variation in pressure dur-ing rolling can make the bonded structure be-come not uniform and unreliable. The studentis encouraged to search the literature and at-tempt to find examples of such applications.

12.20 Comment on your observations concerningFig. 12.40.

By the student. The explosion welding oper-ation results in wavy interfaces (as shown inthe figures) due to the very high interfacialvelocities and pressures involved. The ripplesobserved are actually due to stress waves inthe interface, and help improve joint strengthby mechanical interlocking of the mating sur-faces. Some students may wish to investigateand elaborate further as to how these wavesare developed and how they affect interfacialstrength.

12.21 If electrical components are to be attached toboth sides of a circuit board, what solderingprocess(es) would you use? Explain.

A challenging problem arises when a printedcircuit board (see Section 13.13) has bothsurface-mount and in-line circuits on the sameboard and it is desired to solder all the jointsvia a reliable automated process. An importantpoint is that all of the in-line circuits shouldbe restricted to insertion from one side of theboard. Indeed, there is no performance require-ment which would dictate otherwise, but thisrestriction greatly simplifies manufacturing.

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The basic steps in soldering the connections onsuch a board are as follows:

(a) Apply solder paste to one side.

(b) Place the surface-mount packages ontothe board; also, insert in-line packagesthrough the primary side of the board.

(c) Reflow the solder (see bottom of p. 777).

(d) Apply adhesive to the secondary side ofthe board.

(e) Attach the surface mount devices on thesecondary side, using the adhesive.

(f) Cure the adhesive.

(g) Perform a wave-soldering operation(p. 778) on the secondary side to pro-duce electrical attachment of the surfacemounts and the in-line circuits to theboard.

12.22 Discuss the factors that influence the strengthof (a) a diffusion bonded component and (b) acold welded component.

Diffusion bonded strength (Section 12.12) is in-fluenced by temperature (the higher the tem-perature, the more the diffusion), pressure,time, and the materials being joined. Thecleanliness of the surfaces is also important tomake sure no lubricants, oxides, or other con-taminants interfere with the diffusion process.For this reason, these joints are commonly pre-pared by solvent cleaning and/or pickling to re-move oxides. Cold welded components (Section12.7) involve similar considerations except thattemperature is not a relevant parameter.

12.23 Describe the difficulties you might encounter inapplying explosion welding in a factory environ-ment.

By the student. Explosives are very dangerous;after all, they are generally used for destructivepurposes. There are safety concerns such ashearing loss, damage resulting from explosions,and fires. The administrative burden is highbecause there are many federal, state, and mu-nicipal regulations regarding the handling anduse of explosives and the registration involvedin using explosives.

12.24 Inspect the edges of a U.S. quarter, and com-ment on your observations. Is the cross-section,i.e., the thickness of individual layers, symmet-rical? Explain.

By the student. This is an interesting assign-ment to demonstrate the significance of coldwelding. The side view of a U.S. quarter isshown below. The center of the coin is a copperalloy and the outer layers are a nickel-based al-loy. (Note that pennies and nickels are typicallymade of one material.) The following observa-tions may be made about the coins:

• The core is used to obtain the properweight and feel, as well as sound.

• The strength of roll-bonded joints is veryhigh, as confirmed by the fact that onenever encounters coins that have peeledapart (although during their developmentsuch separation did occur).

• The outer layers, which are made of themore expensive alloy, are thin for cost re-duction.

• The thicknesses of the two outer layers isnot the same. This is due to the smearingaction that occurs around the peripheryduring blanking of the coins, as can be re-called from Section 7.3.

12.25 What advantages do resistance welding pro-cesses have over others described in this chap-ter?

By the student. Recall that resistance weldingis a cleaner process for which electrodes, flux, orshielding environment are not needed; the met-als to be welded provide all of these inherently.The process is easily automated and productionrate is high.

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12.26 What does the strength of a weld nugget in re-sistance spot welding depend on?

By the student. The students are encour-aged to search the literature and collect pho-tographs and more details on weld nuggets (seeFig. 12.33b). This question can be answeredfrom different viewpoints. Thus, for exam-ple, one may consider this question as a stress-analysis problem, whereby the joint strengthdepends on the size of the nugget, its relation-ship to the surrounding bodies, and the typesof materials welded and their mechanical andphysical properties. Other factors to be con-sidered are the role of process parameters suchas current, pressure, time, and the nature ofthe faying surfaces. It would also be interestingand instructional to find weld nuggets that arepoorly made.

12.27 Explain the significance of the magnitude of thepressure applied through the electrodes duringresistance welding operations.

As can be seen in Fig. 12.33, the pressure is ap-plied after sufficient heat is generated. Pressureis maintained until the current is shut off. Thehigher the pressure, the higher the strength ofthe joint (although too high a pressure will ex-cessively indent the surfaces and cause damage;see Fig. 12.27). Lower pressures produce weakjoints. It should be remembered, however, thatthe higher the force the lower the resistance,hence the lower the resistance heating. Conse-quently, proper control of pressure is importantin resistance welding.

12.28 Which materials can be friction stir welded, andwhich cannot? Explain your answer.

Friction stir welding (p. 764) has been com-monly applied to aluminum and copper alloys,and some research is being conducted to extendthe process to others as well as thermoplas-tics and reinforced thermoplastics. The mainrequirements are that the workpiece be suffi-ciently soft and have a low melting point. Theformer requirement ensures that the rotatingtool (Fig. 12.32) will have appropriate strengthfor the operation being performed, and the lat-ter to ensure that the power requirements arereasonable.

12.29 List the joining methods that would be suitablefor a joint that will encounter high stresses andwill need to be disassembled several times dur-ing the product life, and rank the methods.

By the student. Refer also to Table 12.1 onp. 733. Disassembly can be a difficult featureto assess when selecting joining methods. If thepart has to be disassembled often, bolted con-nections are likely to be the best solution, orelse a quick-disconnect clamp or similar devicesshould be used. If the number of disassembliesover the lifetime of the part is limited (such asautomobile dashboards), integrated snap fas-teners (see Fig. 12.55) and even soldering orbrazing can be options. However, soldering andbrazing are only suitable if the filler metal canbe melted without damaging the joint, and ifthe joint can be resoldered.

12.30 Inspect Fig. 12.31, and explain why the par-ticular fusion-zone shapes are developed as afunction of pressure and speed. Comment onthe influence of the properties of the material.

By the student. Inspecting the fusion zonesin Fig. 12.31, it is obvious that higher forcesand speeds both result in more pronounced fu-sion zones. The relevant material properties arestrength at elevated temperatures and physi-cal properties such as thermal conductivity andspecific heat. Because all materials soften atelevated temperatures, the hotter the interface,the more pronounced the fusion zone. Note alsothat a uniform (optimum) zone can be obtainedwith proper control of the relevant parameters.

12.31 Which applications could be suitable for theroll spot welding process shown in Fig. 12.35c?Give specific examples.

By the student. The roll spot-welding opera-tion, shown in Fig. 12.35, is commonly used tofabricate all types of containers and sheet-metalproducts. They can be leak proof provided thatthe spacing of the weld nuggets are sufficientlyclose.

12.32 Give several examples concerning the bulleteditems listed at the beginning of Section 12.1.

By the student. A visit to various stores andobserving the products displayed, as well as

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equipment and appliances found in homes, of-fices, and factories will give ample opportunityfor students to respond comprehensively to thisquestion.

12.33 Could the projection welded parts shown inFig. 12.36 be made by any of the processes de-scribed in other parts of this text? Explain.

By the student. The projection-welded partsshown could possibly be made through resis-tance spot welding (although it would requireseveral strokes) and resistance projection weld-ing. Various other processes may be able to pro-duce the parts shown, but the joint strength de-veloped or the economics of the processes maynot be as favorable. The shape can also beachieved through arc or gas welding processes(followed by finishing such as grinding, if neces-sary), as well as brazing or soldering (see Sec-tion 12.13). With a modified interface, mechan-ical fastening and adhesive bonding also couldbe suitable processes.

12.34 Describe the factors that influence flattening ofthe interface after resistance projection weldingtakes place.

Review Fig. 12.36 and note that:

(a) The projections provide localized areas ofheating, so the material in the projectionsoften and undergo diffusion.

(b) The normal force between the parts flat-tens these softened projections by plasticdeformation.

(c) Important factors are the nature of themating surfaces, the materials involved,the shape of the projections, the tempera-tures developed, the magnitude of the nor-mal force, and length of time.

12.35 What factors influence the shape of the upsetjoint in flash welding, as shown in Fig. 12.37b?

The important factors are the amount of heatgenerated (if too little heat, the material willnot deform to the required extent), the natureof the contracting surfaces (oxide layers, con-taminants, etc.), the force applied (the higherthe force, the greater the upset volume), the ex-posed length between the pieces and the clamps

(if too long, the part may buckle instead of be-ing upset), thermal conductivity (the lower theconductivity, the smaller the upset length), andthe rate at which the force is applied (the higherthe rate, the greater the force required for up-setting, due to strain-rate sensitivity of the ma-terial at elevated temperatures).

12.36 Explain how you would fabricate the structuresshown in Fig. 12.41b with methods other thandiffusion bonding and superplastic forming.

By the student. These structures can be madethrough a combination of sheet-metal formingprocesses (Chapter 7) and resistance welding,brazing, mechanical joining, or adhesive bond-ing. Note, however, that such complex partsand interfaces may not allow easy implementa-tion of these various operations without exten-sive tooling.

12.37 Make a survey of metal containers used forhousehold products and foods and beverages.Identify those that have utilized any of the pro-cesses described in this chapter. Describe yourobservations.

By the student. This is an interesting projectfor students. It will be noted that some foodand beverage containers are three-piece cans,with a welded seam along the length of the can;others may be soldered or seamed (see, for ex-ample, Fig. 12.53). These containers are typ-ically used for shaving cream, laundry starchsprays, and various spray cans for paints andother products.

12.38 Which process uses a solder paste? What arethe advantages to this process?

Solder paste is used in reflow soldering, de-scribed in Section 12.13.3, which is also used forsoldering integrated circuits onto printed circuitboards (Section 13.13).

12.39 Explain why some joints may have to be pre-heated prior to welding.

Some joints may have to be preheated prior towelding in order to:

(a) control and reduce the cooling rate, espe-cially for metals with high thermal con-ductivity, such as aluminum and copper,

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(b) control and reduce residual stresses devel-oped in the joint, and

(c) for more effective wave soldering (p. 778).

12.40 What are the similarities and differences be-tween casting of metals (Chapter 5) and fusionwelding?

By the student. Casting and fusion weldingprocesses are similar in that they both involvemolten metals that are allowed to recrystallize,cool, and solidify. The mechanisms are similarin that solidification begins with the formationof columnar grains (Section 5,3). The cooledstructure is essentially identical to a cast struc-ture with coarse grains. However, the weld joint(Fig. 12.15) is different in that selection of fillersand heat treatment (after welding) influence thejoint’s properties.

12.41 Explain the role of the excessive restraint (stiff-ness) of various components to be welded onweld defects.

Refer to Section 12.6.1. The effect of stiffnesson weld defects is primarily through the stressesdeveloped during heating and cooling of theweld joint. Note, for example, that not allowingfor contraction (such as due to a very stiff sys-tem) will cause cracks in the joint due to highthermal stresses (see Fig. 12.22).

12.42 Discuss the weldability of several metals, andexplain why some metals are easier to weld thanothers.

By the student. This is a challenging assign-ment and will require considerable effort. Re-view Section 12.62 and note that, as expected,weldability depends on many factors. See alsoTable 3.8 and the Bibliography at the end ofthis chapter.

12.43 Must the filler metal be of the same composi-tion as that of the base metal to be welded?Explain.

It is not necessary for the filler metal, rod, orwire to be the same as the base metal to bewelded. Filler metals are generally chosen forthe favorable alloying properties that they im-part to the weld zone. The only function thefiller metal must fulfill is to fill in the gaps in

the joint. The filler metal is typically an al-loy of the same metal, due to the fact that theworkpiece and the filler should melt at reason-ably close temperatures. To visualize why thisis the case, consider a copper filler used with amaterial with a much higher melting tempera-ture, such as steel. When the copper melts, thesteel workpiece is still in a solid state, and theinterface will be one of adhesion, with no sig-nificant diffusion between the copper and thesteel. (See also bottom of p. 743 and p. 773.)

12.44 Describe the factors that contribute to the dif-ference in properties across a welded joint.

By the student. An appropriate response willrequire the students to carefully review Sec-tion12.6.

12.45 How does the weldability of steel change as thesteel’s carbon content increases? Why?

By the student. Review Section 12.6.2. As thecarbon content increases, weldability decreasesbecause of martensite formation, which is hardand brittle (see p. 238).

12.46 Are there common factors among the weldabil-ity, solderability, castability, formability, andmachinability of metals? Explain, with appro-priate examples.

By the student. This is an interesting, but verychallenging, assignment and appropriate for astudent paper. As to be expected, the rela-tionships are complex, as can also be seen byreviewing Table 3.8 on p. 117. Note that forsome aluminum alloys, for example, machin-ability and weldability are opposite (i.e., D-Cvs. A ratings). The students should analyze thecontents of the following: Weldability - Section12.6.2; solderability - p. 777; castability - Sec-tions 5.4.2 and 5.6; formability - Sections 6.2.6and 7.7; machinability - Section 8.5.

12.47 Assume that you are asked to inspect a weldfor a critical application. Describe the proce-dure you would follow. If you find a flaw duringyour inspection, how would you go about deter-mining whether or not this flaw is important forthe particular application?

By the student. This is a challenging task, re-quiring a careful review of Section 12.6.1. Note,

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for example, that visual examination can detectsome defects, such as undercuts and toe cracks;however, underbead cracks or incomplete fusioncannot be detected visually. There are nonde-structive techniques (Section 4.8) for evaluat-ing a weld, acoustic and X-ray techniques be-ing the most common for determining porosityand large inclusions. Proof stressing a weld isa destructive approach, but it certainly can besuitable since defective welds cannot be placedin service safely.

Some analysis on flaw behavior and crack prop-agation in metal structures can be attempted,probably with finite-element methods or by us-ing advanced concepts for crack propagation.An understanding of the loads and the result-ing stresses often determines whether or not aflaw is important. For example, if the defect ina weld in a beam is at the neutral axis in bend-ing, the flaw is not likely to be critical. Onthe other hand, a defect in a highly loaded areaor in a stress concentration would raise seriousconcerns.

12.48 Do you think it is acceptable to differentiatebrazing and soldering arbitrarily by tempera-ture of application? Comment.

By the student. The definition is somewhat ar-bitrary. The temperature classification differ-entiates between the filler metals that can beused in thhe two processes. Note also that, withsoldering, thermal distortions are not as criticalbecause of the lower temperatures involved.

12.49 LoctiteR© is an adhesive used to keep metal boltsfrom vibrating loose; it basically glues the boltto the nut once the bolt is inserted in the nut.Explain how this adhesive works.

LoctiteR© is an anaerobic adhesive (see Table12.6 on p. 782), meaning that it cures in theabsence of oxygen, hence it does not solidify inair. Such a situation exists in the interfaces be-tween threaded fasteners and their nuts, as wellas pins and sleeves, so that the adhesive can beapplied to the threaded fastener and it does notcure until assembled. The students are encour-aged to also review the company literature.

12.50 List the joining methods that would be suitablefor a joint that will encounter high stresses and

cyclic (fatigue) loading, and rank the methodsin order of preference.

By the student. This is a challenging topicwhere the answers will depend on the work-piece materials that are being considered (seealso Table 12.1 on p. 733). Students should notbe limited to the answers given here, but shouldbe encouraged to rely upon their experience andtraining. However, some of the suitable meth-ods for such loadings are:

(a) Riveting is well-suited for such applica-tions, since the rivet can expand uponheading and apply compression to thehole; this can help arrest fatigue cracks.

(b) Bolts can be used for such applications;the use of a preload on a nut can lead tostiff joints with good fatigue resistance.

(c) Welding can be suitable, so long as theweld and the members are properly sized;fatigue crack propogation through theheat-affected zone is a concern.

(d) Brazing can be suitable for such applica-tions, depending on the materials to bebrazed.

(e) Adhesive bonding can also be suitable, aslong as the joints are properly designed(see Fig. 12.60 on p. 793). The mechanicalproperties of the adhesive is an importantconsideration, as well as the strength ofbond with the workpiece.

(f) Combinations of these methods are alsosuitable, such as combining adhesion withriveting as shown in Fig. 12..60d on p. 793.

12.51 Why is surface preparation important in adhe-sive bonding?

By the student. See Section 12.4.2. Surfacepreparation is important because the adhesivestrength depends greatly on its ability to prop-erly bond to a surface (see also Section 4.5). If,for example, there are lubricant residues on asurface, this ability is greatly hindered. As anexample, try sticking masking tape on a dusty,moist or greasy surface, or to your finger coatedwith a very thin layer of oil or grease.

12.52 Why have mechanical joining and fasteningmethods been developed? Give several specificexamples of their applications.

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By the student. Mechanical joining methods,described in Section 12.15, date back to 3000-2000 B.C., as shown in Table 1.1 on p. 3. Thesemethods have been developed mainly becausethey impart design flexibility to products, theygreatly ease assembly (especially disassembly,thus simplifying repair and part replacement),and have economic advantages.

12.53 Explain why hole preparation may be impor-tant in mechanical joining.

By the student. See Section 12.15.1. Note,for example, that if a hole has large burrs(see Fig. 7.5) it can adversely affect joint qual-ity, and also possibly causing crevice corrosion(p. 109). If the hole is significantly larger thanthe rivet, no compressive stress will be devel-oped on its cylindrical surface when the rivet isupset.

12.54 What precautions should be taken in mechani-cal joining of dissimilar metals?

By the student. In joining dissimilar metals,one must be careful about their possible chemi-cal interaction. Often, two dissimilar metals re-act in a cathodic process, causing galvanic cor-rosion and corrosive wear (see Section 3.9.7).This is especially a concern in marine applica-tions, where sea salt can cause major degrada-tion, as well as in chemical industries.

12.55 What difficulties are involved in joining plas-tics? What about in joining ceramics? Why?

By the student. See Section 12.16. Plastics canbe difficult to join. The thermal conductivity isso low that, if melted, plastics will flow beforethey resolidify; thermosets will not melt, butwill degrade as temperature is increased. Ther-moplastics are generally soft and thus cannotbe compressed very much in threaded connec-tions, so the bonds with these processes will notbe very strong. Thermoplastics are usually as-sembled with snap fasteners when strength isnot a key concern, or with adhesives. Ceramicscan be joined by adhesive bonding, and also bymechanical means in which the brittleness andnotch sensitivity of these materials are impor-tant concerns.

12.56 Comment on your observations concerning the

numerous joints shown in the figures in Section12.17.

By the student. The students may respond tothis question in different ways. For example,they can compare and contrast adhesive bondedjoints with those of welded and mechanicallyassembled joints. Note also the projected areaof the joints, the type of materials used, theirgeometric features, and the locations and direc-tions of the forces applied.

12.57 How different is adhesive bonding from otherjoining methods? What limitations does ithave?

By the student. Review Section 12.14. Adhe-sive bonding is significantly different from otherjoining methods in that the workpiece materi-als are of various types, there is no penetrationof the workpiece surfaces, and bonding is doneat room temperature. Its main limitations arethe necessity for clean surfaces, tight clearances,and the longer times required.

12.58 Soldering is generally applied to thinner com-ponents. Why?

Solders have much lower strength than brazefillers or weld beads. Therefore, in joining mem-bers to be subjected to significant loads, whichis typical of members with large thickness, onewould normally consider brazing or welding,but not soldering. A benefit of soldering whenjoining thin components is that it takes place atmuch lower temperatures than brazing or weld-ing, so that one does not have to be concernedabout the workpiece melting due to localizedheating, or significant warping in the joint area.

12.59 Explain why adhesively bonded joints tend tobe weak in peeling.

Adhesives are weak in peeling because there isa concentrated, high tensile stress at the tip ofthe joint when being peeled (see Fig. 12.50);consequently, their low tensile strength reducesthe peeling forces. (Recall that this situationis somewhat analogous to crack initiation andpropagation in metals under tensile stresses; seeFig. 3.30.) Note, however, that tougher adhe-sives can require considerable force and energyto peel, as can be appreciated when trying topeel off some adhesive tapes.

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12.60 Inspect various household products, and de-scribe how they are joined and assembled. Ex-plain why those particular processes were used.

By the student. Metallic food containers aregenerally seamed from sheet. Knife blades areoften riveted/bonded to their handles. Somepots and pans have a number of cold-weldedlayers of sheet, which are then deep drawn andformed to desired shapes. The reason pots havea number of layers different materials is to com-bine their desirable qualities, such as high ther-mal conductivity of copper with the strengthand ease of cleaning of stainless steel. Handleson pots and pans are typically spot welded, riv-eted, or assembled with threaded fasteners. Allof these processes meet functional, technologi-cal, economic, or aesthetic requirements.

12.61 Name several products that have been assem-bled by (a) seaming, (b) stitching, and (c) sol-dering.

By the student. Note, for example:

(a) Products assembled by seaming are foodcontainers and tops of beverage cans.

(b) Products made through stitching arecardboard and wood boxes, insulationand other construction materials, andfootwear.

(c) Soldered parts include electrical compo-nents such as diodes attached to circuitboards, pipe fittings, and electrical termi-nals.

12.62 Suggest methods of attaching a round bar madeof thermosetting plastic perpendicularly to aflat metal plate.

By the student. Consider, for example, the fol-lowing methods:

(a) Threading the end of the rod, drilling andtapping a hole into the plate, and screwingthe rod in, using a sealer if necessary.

(b) Press fit.

(c) Riveting the rod in place.

(d) Fittings can be employed.

12.63 Describe the tooling and equipment that arenecessary to perform the double-lock seaming

operation shown in Fig. 12.53, starting with flatsheet. (See also Fig. 7.23.)

By the student. With some search of the techni-cal literature and the Bibliography given at theend of Chapter 7, the students should be ableto describe designs and equipment required forperforming this operation.

12.64 What joining methods would be suitable toassemble a thermoplastic cover over a metalframe? Assume that the cover has to be re-moved periodically.

By the student. Because the cover has to beremoved periodically, the most feasible joiningmethod is simply snapping the lid on, as is doneon numerous food products (such as polypropy-lene lids on shortening or coffee cans) which, af-ter opening, can easily be resealed. The sealingis due to the elastic recovery of the lid after it isstretched over the edge of the container. Note,however, that at low temperatures (even in therefrigerator) the lid may crack due to lack ofsufficient ductility and severe notch sensitivityof the plastic. The students are encouraged toelaborate further.

12.65 Repeat Question 12.64, but for a cover made of(a) a thermosetting plastic, (b) metal, and (c)ceramic. Describe the factors involved in yourselection of methods.

By the student. Consider the following sugges-tions:

(a) For part (a):

i. A method similar to Answer 12.64above, since thermosetting plasticsalso have some small elastic recovery.

ii. Some mechanical means.iii. Methods would include snap fits.iv. Threaded interfaces.

(b) For (b):

i. Similar to (a) above, especiallythreaded interfaces, such as screwcaps on bottles.

(c) For (c):

i. The generally low ductility of ceram-ics would be a significant concern asthe cover may crack under repeated

198




tensile hoop stresses involved in theiruse. The students are encouraged torespond to the question as to why aceramic cover may even be necessaryif the container is made of metal.

12.66 Do you think the strength of an adhesivelybonded structure is as high as that obtainedby diffusion bonding? Explain.

By the student. Because they rely on bondstrength, the joint strength in adhesivelybonded joints is usually not as high as thatachieved through diffusion bonding. Diffusionbonding (Section 12.12) is exceptional in thatthe two components, typically metals, are dif-fused into each other, making the joint verystrong. Adhesives are generally not as strongas the material they bond (see Table 12.6 onp. 782), unless the materials are inherentlyweak, such as paper, cardboard, and some plas-tics. (See also Section 4.4.)

12.67 Comment on workpiece size limitations, if any,for each of the processes described in this chap-ter.

By the student. This is a good topic for aproject. Basically, large parts can be accom-modated in these processes by appropriate fix-turing. Small parts, on the other hand, maybe delicate and thin, hence will require carefulhandling. Leads for electronic components aregenerally soldered; the wires are typically muchsmaller than 1-mm in diameter. (See also Table12.1 on p. 733.)

12.68 Describe part shapes that cannot be joined bythe processes described in this chapter. Givesspecific examples.

By the student. A review of the various figuresand illustrations in this chapter will clearly in-dicate that part shape is not a significant diffi-culty in joining processes. The basic reason isthat there is such a very wide variety of pro-cesses and possibilities available. The studentis encouraged to think of specific illustrationsof parts that may negate this statement. Inrare cases, if a part shape, as designed, is notsuitable for joining with other components, itsshape could indeed be modified to enable its as-

sembly with other components (see also designfor assembly, Section 14.11).

12.69 Give several applications of electrically con-ducting adhesives.

By the student. See also Section 12.14.4 whereseveral examples are given.

12.70 Give several applications for fasteners in vari-ous household products, and explain why otherjoining methods have not been used instead.

By the student. The students are encour-aged to carefully inspect the variety of productsavailable and to review Sections 12.15, 12.17.4,and 14.10. Note that fasteners are commonlyused in many household products, such as cof-fee makers, electric irons, appliances, furniture,which greatly facilitate assembly, as well as dis-assembly.

12.71 Comment on workpiece shape limitations, ifany, for each of the processes described in thischapter.

By the student. See Question 12.67 and notethat it pertained to size limitations, whereasthis question concerns shapes. Refer also todesign variability in Table 12.1 on p. 733 andwelding position in Table 12.2. Although thereare some limitations, these are often associatedwith fixturing requirements. Consider the fol-lowing: Roll bonding is generally used withsheet metals, so parts that do not involve thinlayers are difficult to roll bond. Ultrasonic weld-ing is typically restricted to thin foils. Frictionwelding requires parts be mounted into chucksor similar fixtures in order to be able to rotateone of the comments to be joined. Spot weld-ing operations can handle complex shapes byappropriate design of electrode holders. Diffu-sion bonding can produce complex shapes, ascan brazing and mechanical fastening.

12.72 List and explain the rules that must be fol-lowed to avoid cracks in welded joints, such ashot tearing, hydrogen-induced cracking, lamel-lar tearing, etc.

By the student. See Section 12.6.1 where allrelevant parameters are discussed.

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12.73 If a built-up weld is to be constructed (seeFig. 12.5), should all of it be done at once, orshould it be done a little at a time, with suffi-cient time allowed for cooling between beads?

With proper welding techniques (see also slaginclusions in Section 12.6.1) and care, the weldjoint can be built continuously, as this proce-dure will prevent excessive oxidation betweenbead interfaces, as well as reducing weld timeand thus making the process economical.

12.74 Describe the reasons that fatigue failure gener-ally occurs in the heat-affected zone of weldsinstead of through the weld bead itself.

Fatigue failure and crack propagation (see Sec-tions 2.7 and 3.8) are complex phenomena. Re-call that the base metal of the workpiece is of-ten a wrought product with varying degrees ofcold work. Thus, the base metal usually hasgood fatigue resistance. The weld zone itselfis highly alloyed, with high strength and alsogood fatigue resistance. However, the heat af-fected zone adjacent to the weld does not havethe advantageous metallurgy of the weld northe microstructure of the worked base metal;it has a large grained, equiaxed strucure (seeFig. 12.15 on p. 749). In addition, there is astress concentration associated with the weld,and the heat affected zone is generally in a vol-ume that is highly stressed. Thus, it is not sur-prising that the heat affected zone is the usualsite of fatigue failure.

12.75 If the parts to be welded are preheated, is thelikelihood that porosity will form increased ordecreased? Explain.

Weld porosity arises from a number of sources,including micropores (similar to those foundin castings; see Section 5.12.1), entrained orevolved gases, and bridging and cracking. Ifthe part is preheated, bridging and cracking arereduced and the cooling rate is lower, there-fore large shrinkage pores are less likely. How-ever, since cooling is slower with preheat, sol-uble gases may be more likely to be entrainedunless effective shielding gases are used.

12.76 What is the advantage of electron-beam andlaser-beam welding, as compared to arc weld-ing?

The main advantages of these processes are as-sociated with the very small weld zone, and thelocalized energy input and small heat-affectedzone. Weld failures, especially by fatigue, oc-cur in the heat-affected zone; thus, minimiz-ing this volume reduces the likelihood of largeflaws and rapid crack growth. Also, the low en-ergy input means that thermal distortions andwarping associated with these processes is muchlower than with arc welding.

12.77 Describe the common types of discontinuities inwelds, explain the methods by which they canbe avoided.

By the student. Note that discontinuities inwelds are discussed in Section 12.6. Some ofthe common defects are porosity, inclusions, in-complete fusion/penetration, underfilling, un-dercutting, overlaps, and cracks. The methodsby which they can be avoided are discussed inSection 12.6.1.

12.78 What are the sources of weld spatter? How canspatter be controlled?

Weld spatter arises from a number of sources.If the filler metal is a powder, errant parti-cles can strike the surface and loosely adhereto the surface, similar to the thermal spray-ing process (pp. 156-157). Even a continuouselectrode will spatter, as a violently evolvingor pumped shielding gas can cause the moltenmetal to emit droplets, which then adhere tothe workpiece surface near the weld zone.

12.79 Describe the functions and characteristics ofelectrodes. What functions do coatings have?How are electrodes classified?

By the student. The functions of electrodes in-clude:

(a) Serve as part of the electrical circuit deliv-ering the power required for welding.

(b) Melt and provide a filler metal.

(c) Have a coating or core that provides ashielding gas and flux.

(d) Help stabilize the arc and make the pro-cess more robust.

There are many characteristics of electrodesand the student is encouraged to develop an

200




appropriate list, noting the two classes of elec-trodes in arc welding processes: Consumableand nonconsumable. Students will need to per-form a literature search to determine classifi-cation of electrodes. For example, the follow-ing is taken from Hamrock, Schmid, and Ja-cobson, Fundamentals of Machine Elements, 2ded., McGraw-Hill, 2004.

Ultimatetensile Yield Elonga-

Electrode strength, strength, tion, ek,number Su, ksi Sy, ksi percentE60XX 62 50 17-25E70XX 70 57 22E80XX 80 67 19E90XX 90 77 14-17E100XX 100 87 13-16E120XX 120 107 14

12.80 Describe the advantages and limitations of ex-plosion welding.

Explosion welding is discussed in Section 12.11.The main advantage is that very dissimilar ma-terials can be bonded, producing high jointstrength, as well as specialized applications.The basic limitation is that it is a basically verydangerous operation.

12.81 Explain the difference between resistance seamwelding and resistance spot welding.

By the student. The difference between resis-tance seam welding and resistance spot weldingis in the spacing of the weld nuggets (see Sec-tions 12.10.1 and 12.10.2). If the nuggets over-lap, it is a seam weld; if they do not overlap, itis a spot weld.

12.82 Could you use any of the processes described inthis chapter to make a large bolt by welding thehead to the shank? (See Fig. 6.17.) Explain theadvantages and limitations of this approach

By the student. Note that processes such asarc welding and gas welding, as well as frictionwelding, can be used to join the two compo-nents. However, the advantage of the latter isthat the weld is over the entire contact area be-tween the two joined components, instead of asmall bead along the periphery of the contactlocation. Brazing is another method of join-ing the two components. The advantages are

that unique designs can be incorporated andusing different materials; the process is econom-ical for relatively few parts. The limitations arethe higher production times required, includingsubsequent finishing operations, as compared toheading operations, which is a common processfor making bolt heads.

12.83 Describe wave soldering. What are the advan-tages and disadvantages to this process?

Wave soldering, described on p. 778, involvesmoving a circuit board with inserted compo-nents over a stationary wave of solder, as shownin Fig. 12.48. The basic advantage to thisprocess is that it can simultaneously producea number of high-quality joints inexpensively.The main drawback is that it places restrictionson the layout of integrated circuit packages ona circuit board.

12.84 What are the similarities and differences be-tween a bolt and a rivet?

By the student. Bolts and rivets are very simi-lar in that two or more components are joinedby a mechanical means. Both preload the com-ponents to function in highly stressed joints.The main difference is that a bolt uses a threadand can thus be disassembled; a rivet is up-set and disassembly requires destruction of therivet.

12.85 It is common practice to tin plate electrical ter-minals to facilitate soldering. Why is tin a suit-able material?

Note in Table 12.5 on p. 777 that solders thatare suitable for general purpose and for elec-tronics applications are lead-tin alloys. Thus,the surface tension of the molten solder withthe tin plate will be very low, thus allowinggood wetting by the solder and resulting in agood joint.

12.86 Review Table 12.3 and explain why some ma-terials require more heat than others to melt agiven volume.

Refer to Section 3.9.2. Recall that the meltingpoint of a metal depends on the energy requiredto separate its atoms, thus it is a characteristicof the individual metal. For an alloy, it dependson the melting points of the individual alloyingelements. Additional factors are:

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Problems

12.87 Two flat copper sheets (each 1.5 mm thick) arebeing spot welded by the use of a current of7000 A and a current flow time of 0.3 s. Theelectrodes are 5 mm in diameter. Estimate theheat generated in the weld zone. Assume thatthe resistance is 200 µΩ.

This problem is very similar to Example 12.5 onp. 765. Note in Eq. (12.6) that the quantitiesnow are I = 7000 A and t = 0.3 s. As in theexample, the resistance is 200 µΩ. Therefore,

H = (7000)2(0.0002)(0.3) = 2940 J

As in the example, we take the weld nuggetvolume to be the projected volume below theelectrode, or

V =π

4d2t =

π

4(5)2(3) = 58.9 mm3

From Table 12.3 on p. 737, the specific en-ergy needed to melt copper is u = 6.1 J/mm3.Therefore, the heat needed is

Hmelt = (58.9)(6.1) = 359 J

The remaining heat (that is, 2940-359 J = 2581J) is dissipated into the volume of metal sur-rounding the weld nugget.

12.88 Calculate the temperature rise in Problem12.87, assuming that the heat generated is con-fined to the volume of material directly betweenthe two electrodes and that the temperaturedistribution is uniform.

The volume of metal directly under the 5-mmelectrodes is

V =π

4d2t =

π

4(5)2(3) = 58.9 mm3,

and this volume has a mass of (58.9)(0.00897)= 0.53 g = 0.00053 kg. The specific heat forcopper is 385 J/kgK. Therefore, the theoreticaltemperature rise is

∆T =2940 J

(385 J/kgK)(0.00053 kg)= 14, 400 K

Note that the melting point of copper is 1082C(1355 K), thus much more energy has been

provided than is needed for this small volume.Clearly, in practice, very little of the heat is con-centrated in this small volume. A more elabo-rate model of temperature distributions is pos-sible, but beyond the scope of this book. Textssuch as Carslaw, H.S., and Jaeger, J.C., Con-duction of Heat in Solids, Oxford UniversityPress, 1959, address such problems in detail.

12.89 Calculate the range of allowable currents forProblem 12.87, if the temperature should bebetween 0.7 and 0.85 times the melting tem-perature of copper. Repeat this problem forcarbon steel.

This problem can be interpreted as between0.7 and 0.85 times the melting temperature onan absolute (Kelvin) or a Celsius temperaturescale. This solution will use a Celsius scale,so that the final target temperature is between765 and 925 C. Using the same approach as inProblem 12.87, the allowable energy for thesecases is 100 and 121 J, respectively. With a re-sistance of 200 µΩ, the currents are 1310 and1420 A, respectively. The solution for carbonsteel is left for the student to supply, but usesthe same approach.

12.90 In Fig. 12.24, assume that most of the top por-tion of the top piece is cut horizontally with asharp saw. Thus, the residual stresses will bedisturbed, and, as described in Section 2.10, thepart will undergo shape change. For this case,how will the part distort? Explain.

Inspecting Fig. 12.24 and recalling Answer 2.25regarding Fig. 2.30, we arrive at the followingobservations and conclusions: (1) The top por-tion of the top piece is subjected to longitudinalcompressive residual stresses. (2) If we cut thisportion with a sharp saw (so that we do not in-duce further residual stresses during cutting),stresses will rearrange themselves and the partwill bend downward, i.e., it will hold water, as-suming it will not warp in the plane of the page.For details, recall the spring analogy in Prob-lem 2.25.

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12.91 The accompanying figure shows a metal sheavethat consists of two matching pieces of hot-rolled, low-carbon-steel sheets. These twopieces can be joined either by spot welding or byV-groove welding. Discuss the advantages andlimitations of each process for this application.

Spot weld

2 716 in.

0.135 in.

V-groove weld

(b) (c)

(a)

By the student. The original method of joiningthe two sheaves was by resistance spot weld-ing, with 16 welds equally spaced around theperiphery, as shown in (b). Although the weldquality was satisfactory, the welding time persheave was 1 minute. In order to increase pro-duction rate, an alternative process was chosen(gas-metal-arc welding, GMAW) with a contin-uous weld around the periphery of the sheave asshown in (c). With an automated welding pro-cess, the welding time per sheave was reducedto 40 s.

12.92 A welding operation takes place on analuminum-alloy plate. A pipe 50-mm in diam-eter with a 4-mm wall thickness and a 60-mmlength is butt-welded onto a section of 15 x 15x 5 mm angle iron. The angle iron is of an L-shape and has a length of 0.3 m. If the weldzone in a gas tungsten-arc welding process isapproximately 8 mm wide, what would be thetemperature increase of the entire structure dueto the heat input from welding only? What ifthe process were an electron-beam welding op-eration with a bead width of 6 mm? Assume

that the electrode requires 1500 J and the alu-minum alloy requires 1200 J to melt one gram.

For the first part of the problem, assume thatthe electrode is placed around the entire pipe,so that the weld length is πD = π(50 mm) =0.157 m. If the weld cross section is triangular,its volume is approximately

V =12bhL =

12(0.008 m)2(0.157 m)

or V = 5.02× 10−6 m3 = 5020 mm3. The elec-trode material should be matched to aluminum,so it will likely be an aluminum alloy in orderto approximately match melting temperaturesand compatibility. The density should thereforebe around 2700 kg/m3 (see Table 3.3 on p. 106,where it is also noted that C = 900 J/kg-K).The specific heat to melt aluminum alloys isgiven by Table 12.3 as 2.9 J/mm3. Therefore,the energy input is (2.9)(5020) = 14.5 kJ. Thetotal volume of the aluminum is

V =π

4(d2

o − d2i )L+ 2btl

=π

4(502 − 422)(60) + 2(15)(5)(300)

= 79, 683 mm3

or V = 7.968× 10−5 m3. The temperature riseis then calculated as:

E = ρV C∆T

Solving for ∆T ,

∆T =E

ρV C

=14, 500

(2700)(7.968× 10−5)(900)

Or ∆T = 75C. For the second part of the prob-lem, the change to be made is in the input en-ergy. Using the same approach as above, wehave

V =12bhL =

12(0.006 m)2(0.157 m)

or V = 2.826×10−6 m3=2826 mm3. The inputenergy is (2.9)(2826)=8.20 kJ. The temperaturerise is therefore

∆T =E

ρV C

=8200

(2700)(7.968× 10−5)(900)= 42C

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12.93 A shielded metal arc welding operation is tak-ing place on carbon steel to produce a fillet weld(see Fig. 12.21b). The desired welding speed isaround 25 mm/sec. If the power supply is 10V, what current is needed if the weld width isto be 7 mm?

Since its width is 7 mm, the cross-sectional areaof the weld is A = 1

2 (7 in2) = 24.5 mm2 =2.45×10−5 m2. For shielded metal arc welding,we obtain from Section 12.3.1 that C = 75%.From Table 12.3, u is assigned a mean value of9.7 J/mm3. From Eq. (12.5), the weld speed istherefore calculated as

v = eV I

uA

Solving for the current, I,

I =uvA

eV=

(9.7)(25)(24.5)(0.75)(10)

or I = 792 A.

12.94 The energy applied in friction welding is givenby the formula E = IS2/C, where I is the mo-ment of inertia of the flywheel, S is the spindlespeed in rpm, and C is a constant of propor-tionality (5873, when the moment of inertia isgiven in lb-ft2). For a spindle speed of 600 rpmand an operation in which a steel tube (3.5 in.OD, 0.25 in. wall thickness) is welded to a flatframe, what is the required moment of inertiaof the flywheel if all of the energy is used toheat the weld zone (approximated as the mate-rial 0.25 in. deep and directly below the tube)?Assume that 1.4 ft-lbm is needed to melt theelectrode.

The flywheel moment of inertia can be calcu-lated as:

E =IS2

C

Solving for I,

I =EC

S2=

(1.4)(5873)(600)2

= 0.0228 lb-ft2

12.95 In oxyacetylene, arc, and laser-beam cutting,the processes basically involve melting of theworkpiece. If an 80 mm diameter hole is tobe cut from a 250 mm diameter, 12 mm thickplate, plot the mean temperature rise in the

plate as a function of kerf. Assume that one-half of the energy goes into the plate and one-half goes into the blank.

The volume melted is

V = (πD)th = π(80 mm)(12 mm)t = 30106t

where t is the kerf width in mm. The energyinput is then E = uV/2 = 1508ut, where u isthe specific energy required to melt the work-piece, as given in J/mm3 in Table 12.1. Notethat we have divided the energy by two becauseonly one-half of the energy goes into the blank.The volume of the blank is

V =π

4d2h

=π

4

[(250 mm)2 − (80 mm)2

](12 mm)

= 5.29× 10−4 m3

The temperature rise in the blank is ∆T =E/ρV Cp; substituting for the input energy,

∆T =1508ut

ρCp(5.29× 10−4)

= (2.85× 106)t(

u

ρCp

)It can be seen that the plot of temperature riseis a linear function of width, t. This is plottedbelow for selected materials.

500

400

300

200

100

0Avg

. Tem

p. I

ncre

ase

in b

lank

(°C

)

0 10 20 30Kerf width (mm)

AlCuSteelTi

12.96 Refer to the simple butt and lap joints shownin Fig. 12.1. (a) Assuming the area of the buttjoint is 3 mm × 20 mm and referring to the ad-hesive properties given in Table 12.6, estimatethe minimum and maximum tensile force thatthis joint can withstand. (b) Estimate theseforces for the lap joint assuming its area is 15mm × 15 mm.

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Referring to Table 12.6 on p. 782, note thatthe lowest adhesive strength is for epoxy orpolyurethane at 15.4 MPa, and the highesttension-shear strength is for modified acrylic at25.9 MPa. These values are used in the solutionbelow.

(a) For a butt joint, assuming there is strongadhesion between the adhesive and work-piece, the full strength of the adhesive canbe developed. In this case, we can calcu-late the required load-bearing area as

A = (3)(20) = 60 mm2 = 6.0× 10−5 m2

Consequently, we have

Fmin = (15.4× 106)(6.0× 10−5) = 924 N

and

Fmax = (25.9×106)(6.0×10−5) = 1554 N

(b) For the lap joint, we similarly obtain A =(15)(15) = 225 mm2 = 2.25 × 10−4 m2.Note that in this case, the joint is loadedin shear, and the shear strength is one-half the tensile strength, as discussed incourses on mechanics of solids. Therefore,

Fmin =12(15.4× 106

) (2.25× 10−4

)or Fmin = 1730 N. Also,

Fmax =12(25.9× 106

) (2.25× 10−4

)or Fmax = 2910 N.

12.97 As shown in Fig. 12.61, a rivet can buckle if itis too long. Using information from solid me-chanics, determine the length-to-diameter ratioof a rivet that will not buckle during riveting.

The riveting process is very similar to head-ing (see Section 6.2.4). Basically, the designrequirement is that the length-to-diameter ra-tio should be 3 or less. If the heading tool has acontrolled geometry, a longer length can be ac-commodated if the head diameter is not morethan 1.5 times the shank diameter.

12.98 Repeat Example 12.2 if the workpiece is (a)magnesium, (b) copper or (c) nickel.

(a) For the magnesium workpiece, Table 12.3gives u = 2.9 J/mm3. Therefore, fromEq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(2.9)(30)

= 34.5 mm/s.

(b) For the copper workpiece, we have u = 6.1J/mm3. Therefore, from Eq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(6.1)(30)

= 16.4 mm/s.

(c) For the nickel workpiece, we have u = 9.8J/mm3. Therefore, from Eq. (12.5),

v = eV I

uA= (0.75)

(20)(200)(9.8)(30)

= 10.2 mm/s.

12.99 A submerged arc welding operation takes placeon 10 mm thick stainless steel, producing a buttweld as shown in Fig. 12.20c. The weld geom-etry can be approximated as a trapezoid with15 mm and 10 mm as the top and bottom di-mensions, respectively. If the voltage providedis 40 V at 400 A, estimate the welding speed ifa stainless steel filler wire is used.

A sketch of the weld cross section is shown be-low.

15

10

10

The area of the trapezoid is

A = (10)(10) + 2(

12

)(2.5)(10) = 125 mm2

For submerged arc welding, it is stated in Sec-tion 12.3.1 that e = 0.90. For a stainless steelworkpiece, the unit specific energy is obtainedfrom Table 12.3 as u = 9.4 J/mm3. Therefore,from Eq. (12.5),

v = eV I

uA

= 0.9(40)(400)(9.4)(125)

= 12.2 mm/s

205




12.100 Assume that you are asked to give a quiz to stu-dents on the contents of this chapter. Preparethree quantitative problems and three qualita-tive questions, and supply the answers.

By the student. This is a challenging, open-ended question that requires considerable focusand understanding on the part of the students,and has been found to be a very valuable home-work problem.

Design

12.101 Design a machine that can perform frictionwelding of two cylindrical pieces, as well asremove the flash from the welded joint. (SeeFig. 12.30.)

By the student. Note that this machine can bevery similar to a lathe, where one-half of theworkpiece is held in a fixture attached to thetailstock, the other half is in the rotating chuck,and a cutting tool is used as in turning.

12.102 How would you modify your design in Problem12.101 if one of the pieces to be welded is non-circular?

By the student. The machine is more com-plicated, machining can become more difficult,with essentially a milling operation taking placeafter welding.

12.103 Describe product designs that cannot be joinedby friction welding processes.

By the student. Consider, for example, that ifone of the components is a very thin tube, it willnot be able to support the large axial loads in-volved in friction welding; likewise, if the othercomponent is very think and slender.

12.104 Make a comprehensive outline of joint designsrelating to the processes described in this chap-ter. Give specific examples of engineering ap-plications for each type of joint.

By the student. Refer to Section 12.17. Thisis a challenging problem, and would be suitablefor a project or a paper.

12.105 Review the two weld designs in Fig. 12.58a, and,based on the topics covered in courses on thestrength of materials, show that the design on

the right is capable of supporting a larger mo-ment, as shown.

The problem statement assumes that the failurein the part on the left will be in the weld itself;if the material strength determines the momentthat can be supported, then the weld designis irrelevant. Assuming that the weld zone isroughly square, it is better to place the welds asshown on the right because the strength arisesfrom the cube of the distance from the neutralaxis. In the design on the left, only the extremeends are fully loaded, and some material (at theneutral axis) is subjected to very little stress.

12.106 In the building of large ships, there is a needto weld large sections of steel together to forma hull. For this application, consider each ofthe welding operations described in this chap-ter, and list the benefits and drawbacks of thatoperation for this product. Which welding pro-cess would you select? Why?

By the student. This specialized topic is verysuitable for a student paper, requiring a searchof the technical literature in shipbuilding tech-nologies. For example, the following may besuggested:

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Process Advantages DisadvantagesOxyfuel Inexpensive;

portable.Significant jointdistortion; weld-ing of thicksections is diffi-cult.

SMAW Inexpensive;portable.

Thick sectionsneed a built-up weld (seeFig. 12.5 onp. 738), possiblycompromisingjoint strength.

SAW Good weldstrength; can beautomated

Limitedworkspace;workpiece mustbe horizontal, adifficult restric-tion for boathulls

ESW Good weldstrength, well-suited for verticalwelds.

More compli-cated equipmentrequired.

12.107 Examine various household products, and de-scribe how they are joined and assembled. Ex-plain why those particular processes are usedfor these applications.

By the student. Consider the following: Metal-lic food containers that are seamed from sheet.Knife blades that are riveted/bonded to theirhandles. Pots with a number of cold-weldedlayers of sheet, which are then deep drawn andformed to desired shapes. All of these pro-cesses are used because other processes whichare technologically feasible may lack economic,functional, or aesthetic advantages.

12.108 A major cause of erratic behavior (hardwarebugs) and failure of computer equipment is fa-tigue failure of the soldered joints, especially insurface-mount devices and devices with bondwires. (See Fig. 12.48.) Design a test fixturefor cyclic loading of a surface-mount joint forfatigue testing.

By the student. This is a very demandingproject, and can be expanded into a groupdesign project. Students can consider if thetest should duplicate the geometry of a surfacemount or if an equivalent geometry can be ana-lyzed. They can determine loading cycle dura-tions and amplitudes, as well as various othertest parameters.

12.109 Using two strips of steel 1 in. wide and 8in. long, design and fabricate a joint that givesthe highest strength in a tension test in the lon-gitudinal direction.

By the student. This is a challenging problemand an experimental project, as well; it couldalso be made into a contest among students inclass. It must be noted, however, that the thick-ness of the strips is not given in the statement ofthe problem (although the word strip generallyindicates a thin material). The thickness is afactor that students should recognize and com-ment on, and supply their answers accordingly.It can also be seen that most of the processesdescribed in Chapter 12 can be used for such ajoint. Consequently, a wide variety of processesand designs should be considered, making theresponse to this question extensive.

In using a single bolt through the two strips,for example, it should be apparent that if thebolt diameter is too large, the stresses in therest of the cross section may be too high, caus-ing the strips to fail prematurely. If, on theother hand, the bolt diameter is too small, itwill easily shear off under the applied tensileforce. Thus, there has to be an optimum tobolt size. The students are encouraged to con-sider multiple-bolt designs, as well as a host ofother processes either singly or in combination.

12.110 Make an outline of the general guidelines forsafety in welding operations. For each of theoperations described in this chapter, prepare aposter which effectively and concisely gives spe-cific instructions for safe practices in welding(or cutting). Review the various publicationsof the National Safety Council and other simi-lar organizations.

By the student. This is a valuable study by thestudents, and the preparation of a poster or aflyer is a good opportunity for students. Safetyin Welding is a standard published by theAmerican National Standards Institute (ANSIZ49.1) and describes in detail the safety pre-cautions that must be taken. Most of the stan-dards are process-specific. As an example, somesafety guidelines for shielded metal-arc weldingare:

• The operator must wear eye and skin pro-

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tection against radiation.

• Leather gloves and clothing should beworn to prevent burns from arc spatter.

• Welding should be done in properly venti-lated areas, where fresh air is available toworkers and the work area is not floodedby shielding gases.

• To prevent electric shock, the weldershould not weld while standing on a wetsurface.

• The workpiece should be positioned tominimize trauma to the back and arms.

12.111 A common practice for repairing expensive bro-ken or worn parts, such as may occur when, forexample, a fragment is broken from a forging,is to fill the area with layers of weld bead andthen to machine the part back to its originaldimensions. Make a list of the precautions thatyou would suggest to someone who uses thisapproach.

By the student. Considerations are that the in-terface between the forging and the filling maynot have sufficient strength. The weld beadwill have different properties than the substrate(the forging) and have an uneven surface, thusmachining may result in vibration and chat-ter. The weld material may cause the cuttingtools to wear more rapidly. The weld may frac-ture during machining and compromise part in-tegrity. The weld material may have insufficientductility and toughness for the application.

12.112 In the roll bonding process shown in Fig. 12.28,how would you go about ensuring that the in-terfaces are clean and free of contaminants, sothat a good bond is developed? Explain.

By the student. The students are encouragedto perform a literature search for particular ap-proaches. The basic procedure has been (a)wire brushing the surfaces, which removes ox-ide from the surfaces, and (b) solvent cleaning,which removes residues and organic films fromthe surface. (See also Section 4.5.2.)

12.113 Alclad stock is made from 5182 aluminum al-loy, and has both sides coated with a thin layerof pure aluminum. The 5182 provides high

strength, while the outside layers of pure alu-minum provide good corrosion resistance, be-cause of their stable oxide film. Alclad is com-monly used in aerospace structural applicationsfor these reasons. Investigate other commonroll bonded materials and their uses, and pre-pare a summary table.

By the student. This topic could be a chal-lenging project for students. Examples includecoinage (see also Question 12.24) and a thincoating of metals on workpieces where the coat-ing serves as a solid lubricant in metalworking(see p. 152).

12.114 Obtain a soldering iron and attempt to soldertwo wires together. First, try to apply the sol-der at the same time as you first put the solder-ing iron tip to the wires. Second, preheat thewires before applying the solder. Repeat thesame procedure for a cool surface and a heatedsurface. Record your results and explain yourfindings.

By the student. This is a valuable and inex-pensive laboratory experience, showing the im-portance of surface tension. With cold wires,molten solder has high surface tension againstthe wires, and thus the solder does not wet thesurface. At elevated temperatures, the solderhas low surface tension and the solder coats thewire surfaces very effectively. Students can beasked to examine this phenomenon further byplacing a small piece of solder of known volume(which can be measured with a precision scale)on a steel plate section. When heated, the sol-der spreads according to the surface tempera-ture of the steel. It will be noted that above athreshold value, the solder will flow freely andcoat the surface.

12.115 Perform a literature search to determine theproperties and types of adhesives used to affixartificial hips onto the human femur.

By the student. Sometimes an adhesive isused, but with some designs this is not nec-essary, as they rely upon osteointegration orbone-ingrowth to affix the implant. Usually thecement is polymethylmethacrylate, an acrylicpolymer often referred to as bone cement, orelse a hydroxyapetite polymer is used. Newmaterials are constantly being developed and

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a number of variations can be found in the lit-erature and through an Internet search. A com-mon trend is to develop cements from calciumphosphates, as these are closer matches to themineral content of bone.

12.116 Using the Internet, investigate the geometry ofthe heads of screws that are permanent fasten-ers (that is, ones that can be screwed in but notout).

By the student. These heads usually present astraight vertical surface for the screwdriver inone direction, but a curved surface in the op-posite direction, so that a screwdriver simplyslips when turned counterclockwise and is noteffective for unscrewing. The sketch on the leftwas obtained from www.k-mac-fasteners.com,while the photo on the right was obtained fromwww.storesonline.com.

12.117 Obtain an expression similar to Eq. (12.6), butfor electron beam and laser welding.

By the student. The heat input is generallygiven by

H = cIA

where c is a constant that indicates the portionof laser energy absorbed by the material, and Iis the intensity of light over the area A.

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68416 ch 12