The Structure of Metals

Transcription

1 Chpter 1 The Structure of Metls QUALITATIVE PROBLEMS 1.21 Explin your understnding of why the study of the crystl structure of metls is importnt. The study of crystl structure is importnt for number of resons. Bsiclly, the crystl structure influences mteril s performnce from both design nd mnufcturing stndpoint. For exmple, the number of slip systems in crystl hs direct bering on the bility of metl to undergo plstic deformtion without frcture. Similrly, the crystl structure hs bering on strength, ductility nd corrosion resistnce. Metls with fce-centered cubic structure, for exmple, tend to be ductile wheres hexgonl close-pcked metls tend to be brittle. The crystl structure nd size of tom determines the lrgest interstitil sites, which hs bering on the bility of tht mteril to form lloys, nd with which mterils, s interstitils or substitutionls Wht is the significnce of the fct tht some metls undergo llotropism? Allotropism (lso clled polymorphism) mens tht metl cn chnge from one crystl structure to nother. Since properties vry with crystl structures, llotropism is useful nd essentil in het treting of metls to chieve desired properties (Chpter 4). A mjor ppliction is hrdening of steel, which involves the chnge in iron from the fcc structure to the bcc structure (see Fig. 1.2 on p. 39). By heting the steel to the fcc structure nd quenching, it develops into mrtensite, which is very hrd, hence strong, structure Is it possible for two pieces of the sme metl to hve different recrystlliztion tempertures? Is it possible for recrystlliztion to tke plce in some regions of prt before it does in other regions of the sme prt? Explin. Two pieces of the sme metl cn hve different recrystlliztion tempertures if the pieces hve been cold worked to different mounts. The piece tht ws cold worked to greter 1

2 The Structure of Metls 2 extent (higher strins), will hve more internl energy (stored energy) to drive the recrystlliztion process, hence its recrystlliztion temperture will be lower. Recrystlliztion my lso occur in some regions of the prt before others if it hs been unevenly strined (since vrying mounts of cold work hve different recrystlliztion tempertures), or if the prt hs different thicknesses in vrious sections. The thinner sections will het up to the recrystlliztion temperture fster Describe your understnding of why different crystl structures exhibit different strengths nd ductilities. Different crystl structures hve different slip systems, which consist of slip plne (the closest pcked plne) nd slip direction (the close-pcked direction). The fcc structure hs 12 slip systems, bcc hs 48, nd hcp hs 3. The ductility of metl depends on how mny of the slip systems cn be opertive. In generl, fcc nd bcc structures possess higher ductility thn hcp structures, becuse they hve more slip systems. The sher strength of metl decreses for decresing b/ rtio (b is inversely proportionl to tomic density in the slip plne nd is the plne spcing), nd the b/ rtio depends on the slip system of the chemicl structure. (See Section 1.3.) 1.25 A cold-worked piece of metl hs been recrystllized. When tested, it is found to be nisotropic. Explin the probble reson. The nisotropy of the workpiece is likely due to preferred orienttion remining from the recrystlliztion process. Copper is n exmple of metl tht hs very strong preferred orienttion fter nneling. Also, it hs been shown tht below criticl mount of plstic deformtion, typiclly 5%, no recrystlliztion occurs Wht mterils nd structures cn you think of (other thn metls) tht exhibit nisotropic behvior? This is n open-ended problem nd the students should be encourged to develop their own nswers. However, some exmples of nisotropic mterils re wood, polymers tht hve been cold worked, bone, ny woven mteril (such s cloth) nd composite mterils Two prts hve been mde of the sme mteril, but one ws formed by cold working nd the other by hot working. Explin the differences you might observe between the two. There re lrge number of differences tht will be seen between the two mterils, including: () The cold worked mteril will hve higher strength thn the hot worked mteril, nd this will be more pronounced for mterils with high strin hrdening exponents. (b) Since hrdness (see Section 2.6.3) is relted to strength, the cold worked mteril will lso hve higher hrdness. (c) The cold worked mteril will hve smller grins nd the grins will be elongted. (d) The hot worked mteril will probbly hve fewer disloctions, nd they will be more evenly distributed. (e) The cold worked mteril cn hve superior surfce finish when in n s-formed condition. Also, it cn hve better tolernces.

3 The Structure of Metls 3 (f) A cold worked mteril will hve lower recrystlliztion temperture thn hot worked mteril Do you think it might be importnt to know whether rw mteril to be used in mnufcturing process hs nisotropic properties? Wht bout nisotropy in the finished product? Explin. Anisotropy is importnt in cold-working processes, especilly sheet-metl forming where the mteril s properties should preferbly be uniform in the plne of the sheet nd stronger in the thickness direction. As shown in Section 16.7, these chrcteristics llow for deep drwing of prts (like beverge cns) without ering, tering, or crcking in the forming opertions involved. In finished prt, nisotropy is importnt so tht the strongest direction of the prt cn be designed to support the lrgest lod in service. Also, the efficiency of trnsformers cn be improved by using sheet steel with nisotropy tht cn reduce mgnetic hysteresis losses. Hysteresis is well known in ferromgnetic mterils. When n externl mgnetic field is pplied to ferromgnet, the ferromgnet bsorbs some of the externl field. When sheet steel is highly nisotropic, it contins smll grins nd crystllogrphic orienttion tht is fr more uniform thn for isotropic mterils, nd this orienttion will reduce mgnetic hysteresis losses Explin why the strength of polycrystlline metl t room temperture decreses s its grin size increses. Strength increses s more entnglements of disloctions occur with grin boundries (Section on p. 45). Metls with lrger grins hve less grin-boundry re per unit volume, nd hence will not be s ble to generte s mny entnglements t grin boundries, thus the strength will be lower Describe the technique you would use to reduce the ornge-peel effect on the surfce of workpieces. Ornge peel is surfce roughening induced by plstic strin. There re number of wys of reducing the ornge peel effect, including: Performing ll forming opertions without lubricnt, or else very thin lubricnt film (smller thn the desired roughness) nd very smooth tooling. The gol is to hve the surfce roughness of the tooling imprted onto the workpiece. As discussed on pge 46, lrge grins excerbte ornge peel, so the use of smll grined mterils would reduce ornge peel. If deformtion processes cn be designed so tht the surfces see no deformtion, then there would be no ornge peel. For exmple, upsetting beneth flt dies cn led to reduction in thickness with very little surfce strins beneth the plten (see Fig on p. 340). Finishing opertions cn remove ornge peel effects Wht is the significnce of the fct tht such metls s led nd tin hve recrystlliztion temperture tht is bout room temperture?

4 The Structure of Metls 4 Recrystlliztion round room temperture prevents these metls from work hrdening when cold worked. This chrcteristic prevents their strengthening nd hrdening, thus requiring recrystlliztion cycle to restore their ductility. This behvior is lso useful in experimentl verifiction of nlyticl results concerning force nd energy requirements in metlworking processes (see Prt III of the text) It ws stted in this chpter tht twinning usully occurs in hcp mterils, but Fig. 1.6b shows twinning in rectngulr rry of toms. Cn you explin the discrepncy? The hcp unit cell shown in Fig. 1.5 on p. 41 hs hexgon on the top nd bottom surfces. However, n intersecting plne tht is verticl in this figure would intersect toms in rectngulr rry s depicted in Fig. 1.6b on p. 44. Thus, twinning occurs in hcp mterils, but not in the hexgonl (close pcked) plne such s in the top of the unit cell It hs been noted tht the more metl hs been cold worked, the less it strin hrdens. Explin why. This phenomenon cn be observed in stress-strin curves, such s those shown in Figs. 2.2 nd 2.5. Recll tht the min effects of cold working re tht grins become elongted nd tht the verge grin size becomes smller (s grins brek down) with strin. Strin hrdening occurs when disloctions interfere with ech other nd with grin boundries. When metl is nneled, the grins re lrge, nd smll strin results in grins moving reltively esily t first, but they incresingly interfere with ech other s strin increses. This explins tht there is strin hrdening for nneled mterils t low strin. To understnd why there is less strin hrdening t higher levels of cold work, consider the extreme cse of very highly cold-worked mteril, with very smll grins nd very mny disloctions tht lredy interfere with ech other. For this highly cold-worked mteril, the stress cnnot be incresed much more with strin, becuse the disloctions hve nowhere else to go - they lredy interfere with ech other nd re pinned t grin boundries Is it possible to cold work metl t tempertures bove the boiling point of wter? Explin. The metllurgicl distinction between cold nd hot working is ssocited with the homologous temperture s discussed on p. 50. Cold working is ssocited with plstic deformtion of metl when it is below one-third of its melting temperture on n bsolute scle. At the boiling point of wter, the temperture is 100 C, or 373 Kelvin. If this vlue is one-third the melting temperture, then metl would hve to hve melting temperture of 1119K, or 846 C. As cn be seen in Tble 3.1 on p. 89, there re mny such metls Comment on your observtions regrding Fig This is n open-ended problem with mny potentil nswers. Students my choose to ddress this problem by focusing on the shpe of individul curves or their reltion to ech other. The instructor my wish to focus the students on curve or two, or sk if the figure would give the sme trends for mteril tht is quickly heted, held t tht temperture for few secongs, nd then quenched, or lterntively for one tht is mintined t the tempertures for very long times.

5 The Structure of Metls Is it possible for metl to be completely isotropic? Explin. This nswer cn be nswered only if isotropy is defined within limits. For exmple: A single crystl of metl hs n inherent n unvoidble nisotropy. Thus, t length scle tht is on the order of mteril s grin size, metl will lwys be nisotropic. A metl with elongted grins will hve lower strength nd hrdness in one direction thn in others, nd this is unvoidble. However, metl tht contins lrge number of smll nd equixed grins will hve the first two effects essentilly mde very smll; the metl my be isotropic within mesurement limits. Anneling cn led to equixed grins, nd depending on the mesurement limits, this cn essentilly result in n isotropic metl. A metl with very smll grin size (i.e., metl glss) cn hve no pprent crystl structure or slip systems, nd cn be essentilly isotropic. QUANTITATIVE PROBLEMS 1.37 How mny toms re there in single repeting cell of n fcc crystl structure? How mny in repeting cell of n hcp structure? For n fcc structure, refer to Fig. 1.4 on p. 41. The toms t ech corner re shred by eight unit cells, nd there re eight of these toms. Therefore, the corners contribute one totl tom. The toms on the fces re ech shred by two cells, nd there re six of these toms. Therefore, the toms on the fces contribute totl of three toms to the unit cell. Therefore, the totl number of toms in n fcc unit cell is four toms. For the hcp, refer to Fig. 1.4 on p. 41. The toms on the periphery of the top nd bottom re ech shred by six cells, nd there re 12 of these toms (on top nd bottom), for contribution of two toms. The toms in the center of the hexgon re shred by two cells, nd there re two of these toms, for net contribution of one tom. There re lso three toms fully contined in the unit cell. Therefore, there re six toms in n hcp unit cell The tomic weight of copper is 63.55, mening tht toms weigh g. The density of copper is 8970 kg/m 3, nd pure copper forms fcc crystls. Estimte the dimeter of copper tom. Consider the fce of the fcc unit cell, which consists of right tringle with side length nd hypoteneuse of 4r. From the Pythgoren theorem, = 4r/ 2. Therefore, the volume of the unit cell is ( ) 4r 3 V = 3 = = 22.63r 3 2 Ech fcc unit cell hs four toms (see Prob. 1.34), nd ech tom hs mss of Mss = g = g

6 The Structure of Metls 6 So tht the density inside fcc unit cell is ρ = Mss V = 8970 kg/m 3 = 4( ) kg 22.63r 3 Solving for r yields r = m, or Å. Note tht the ccepted vlue is 2.5 Å; the difference is ttributble to number of fctors, including impurities in crystl structure nd concentrtion of mss in the nucleus of the tom Plot the dt given in Tble 1.1 in terms of grins/mm 2 versus grins/mm 3, nd discuss your observtions. The plot is shown below. It cn be seen tht the grins per cubic millimeter increses fster thn the grins per squre millimeter. This reltionship is to be expected since the volume of n equixed grin depends on the dimeter cubed, wheres its re depends on the dimeter squred. 3.5 x x 10 4 Grins/mm x x x x x x x x 10 6 Grins/mm A strip of metl is reduced from 30 mm in thickness to 20 mm by cold working; similr strip is reduced from 40 to 30 mm. Which of these cold-worked strips will recrystllize t lower temperture? Why? The metl tht is reduced to 20 mm by cold working will recrystllize t lower temperture. on p. 50, the more the cold work the lower the temperture required for recrystlliztion. This is becuse the number of disloctions nd energy stored in the mteril increses with cold work. Thus, when recrystllizing more highly cold worked mteril, this energy cn be recovered nd less energy needs to be imprted to the mteril The bll of bllpoint pen is 1 mm in dimeter nd hs n ASTM grin size of 10. How mny grins re there in the bll? From Tble 1.1 on p. 46, we find tht metl with n ASTM grin size of 10 hs bout 520,000 grins/mm 3. The volume of the bll is V = 4 3 πr3 = 4 3 π(1)3 = mm 3

7 The Structure of Metls 7 Multiplying the volume by the grins per cubic millimeter gives the number of grins in the pper clip s bout 2.18 million How mny grins re there on the surfce of the hed of pin? Assume tht the hed of pin is sphericl with 1-mm dimeter nd hs n ASTM grin size of 12. Note tht the surfce re of sphere is given by A = 4πr 2. Therefore, 1 mm dimeter hed hs surfce re of A = 4πr 2 = 4π(0.5) 2 = π mm 2 From Eq. (1.2), the number of grins per re is N = 2 11 = 2048 This is the number of grins per mm 2 of ctul re; therefore the number of grins on the surfce is N g = 2048 = 99, 750 grins (π) 1.43 The unit cells shown in Figs cn be represented by tennis blls rrnged in vrious configurtions in box. In such n rrngement, the tomic pcking fctor (APF) is defined s the rtio of the sum of the volumes of the toms to the volume of the unit cell. Show tht the APF is 0.68 for the bcc structure nd 0.74 for the fcc structure. Note tht the bcc unit cell in Fig. 1.3 on p. 41 hs 2 toms inside of it; one inside the unit cell nd eight toms tht hve one-eighth of their volume inside the unit cell. Therefore the volume of toms inside the cell is 8πr 3 /3, since the volume of sphere is 4πr 3 /3. Note tht the digonl of fce of unit cell hs length of 2, which cn be esily determined from the Pythgoren theorem. Using tht digonl nd the height of results in the determintion of the digonl of the cube s 3. Since there re four rdii cross tht digonl, it cn be deduced tht 3 = 4r = 4r 3 The volume of the unit cell is 3, so V = 3 = Therefore, the tomic pcking fctor is APF bcc = ( ) 4r 3 ( ) 4 3 = 3 r πr3 ( 4 3 ) 3 r 3 = 0.68 For the fcc cell, there re four toms in the cell, so the volume of toms inside the fcc unit cell is 16πr 3 /3. On fce of the fcc cell, it cn be shown from the Pythgoren theorem tht the hypotenuse is 2. Also, there re four rdii cross the dimeter, so tht 2 = 4r = 2r 2

8 The Structure of Metls 8 Therefore, the volume of the unit cell is so tht the tomic pcking fctor is V = 3 = APF fcc = ( 2r ) πr3 ( 2 2 ) 3 r 3 = Show tht the lttice constnt in Fig. 1.4 is relted to the tomic rdius by the formul = 2 2R, where R is the rdius of the tom s depicted by the tennis-bll model. For fce centered cubic unit cell s shown in Fig. 1.3, the Pythgoren theorem yields = (4r) 2 Therefore, 2 2 = 16r 2 = 2 2r 1.45 Show tht, for the fcc unit cell, the rdius r of the lrgest hole is given by r = 0.414R. Determine the size of the lrgest hole for the iron toms in the fcc structure. The lrgest hole is shown in the sketch below. Note tht this hole occurs in other loctions, in fct in three other loctions of this sketch. R R Lrgest hole dimeter = 2r For fce centered cubic unit cell s shown in Fig. 1.3, the Pythgoren theorem yields Therefore, Also, the side dimension is Therefore, substituting for, = (4R) = 16R 2 = 2 2R = 2R + 2r 2 2R = 2R + 2r r = ( 2 1 ) R = 0.414R

9 The Structure of Metls A technicin determines tht the grin size of certin etched specimen is 8. Upon further checking, it is found tht the mgnifiction used ws 125, insted of the 100 tht is required by the ASTM stndrds. Determine the correct grin size. If the grin size is 8, then there re 2048 grins per squre millimeter (see Tble 1.1 on p. 46). However, the mgnifiction ws too lrge, mening tht too smll of n re ws exmined. For mgnifiction of 100, the re is reduced by fctor of 1/1.82= Therefore, there relly re 632 grins per mm 2, which corresponds to grin size between 6 nd If the dimeter of the luminum tom is 0.28 nm, how mny toms re there in grin of ASTM grin size 8? If the grin size is 8, there re 65,000 grins per cubic millimeter of luminum - see Tble 1.1 on p. 46. Ech grin hs volume of 1/65, 000 = mm 3. Note tht for n fcc mteril there re four toms per unit cell (see solution to Prob. 1.43), with totl volume of 16πR 3 /3, nd tht the digonl,, of the unit cell is given by ( = 2 ) 2 R Hence, APF fcc = ( 16πR 3 /3 ) ( 2R 2 ) 3 = 0.74 Note tht s long s ll the toms in the unit cell hve the sme size, the tomic pcking fctors do not depend on the tomic rdius. Therefore, the volume of the grin which is tken up by toms is ( )(0.74) = mm 3. (Recll tht 1 mm=10 6 nm.) If the dimeter of n luminum tom is 0.5 nm, then its rdius is 0.25 nm or mm. The volume of n luminum tom is then V = 4πR 3 /3 = 4π( ) 3 /3 = mm 3 Dividing the volume of luminum in the grin by the volume of n luminum tom yields the totl number of toms in the grin s ( )/( ) = The following dt re obtined in tension tests of brss: Grin size Yield strength (µm) (MP) Does the mteril follow the Hll Petch eqution? If so, wht is the vlue of k? First, it is obvious from this tble tht the mteril becomes stronger s the grin size decreses, which is the expected result. However, it is not cler whether Eq. (3.8) on p. 92 is pplicble. It is possible to plot the yield stress s function of grin dimeter, but it is better to plot it s function of d 1/2, s follows:

10 The Structure of Metls 10 Yield strength (MP) d -1/2 The lest-squres curve fit for stright line is Y = d 1/2 with n R fctor of This suggests tht liner curve fit is proper, nd it cn be concluded tht the mteril does follow the Hll-Petch effect, with vlue of k = 458 MP- µm Assume tht you re sked to sk quntittive problem for quiz. Prepre such question, supplying the nswer. By the student. This is chllenging, open-ended question tht requires considerble focus nd understnding on the prt of the students, nd hs been found to be very vluble homework problem. To be successful, such n ssignment will need some supervision by the instructor. A strtegy tht hs been followed by the uthors hs been to promise the students tht the best problems will become problems on the next exm; students tht mke n honest effort re therefore rewrded by being in strong position to nswer the question on the exm (lthough we hve lso seen students unble to nswer their own questions!) The tomic rdius of iron is nm, while tht of crbon tom is nm. Cn crbon tom fit inside steel bcc structure without distorting the neighboring toms? 2r h g 2r Consider the sketch shown. The hypotenuse of the tringle shown is h = 2r 2 = 2.828r The smllest gp shown is g = 2.828r 2r = 0.828r

11 The Structure of Metls 11 For r = nm, this opening is g = nm. Therefore, the crbon tom fits (with little room to spre), without distorting the lttice Estimte the tomic rdius for the following mterils nd dt: () Aluminum (tomic weight = g/mol, density = 2700 kg/m 3 ); (b) tungsten (tomic weight = g/mol, density = 19,300 kg/m 3 ); nd (c) mgnesium (tomic weight = g/mol, density = 1740 kg/m 3 ). See lso the solution to Problem 1.38 for the pproch. () From p. 40, luminum is fcc mteril. From Problem 1.43, the tomic pcking fctor is therefore For ρ = 2700, nd tomic weight of 26.98, there re 100,000 moles of toms in cubic meter. Since the tomic pcking fctor is 0.74, the toms fill 0.74 m 3 of tht spce. The volume of ech tom is V = 0.74 m 3 (100, 000)( ) = m 3 For sphere, V = 4 3 πr3, so tht for n luminum tom, r is found to be nm. (b) For tungsten, which is body centered cubic mteril (p. 40), the tomic pcking fctor is 0.68 (Problem 1.43). The volume of tungsten tom is or r = nm. V = 0.68 (104, 76)( ) = m A simple cubic structure consists of toms locted t the cube corners tht re in contct with ech other long the cube edges. Mke sketch of simple cubic structure, nd clculte its tomic pcking fctor. R () (b) The sketch is shown bove. Note tht ech side of the unit cell hs length of 2r, so tht the volume of the unit cell is 8r 3. There re 8 toms prtilly in the cell, nd ech tom is shred by 8 other cells, so tht there is one tom in the unit cell. Therefore, the tomic pcking fctor is 4 3 APF = πr3 8r 3 = π 6 = Sme s Prob. 1.39, but ASTM no. versus grins/mm 3. The plot is s follows, with log scle for the x-xis.

12 The Structure of Metls SYNTHESIS, DESIGN, AND PROJECTS 1.54 By stretching thin strip of polished metl, s in tension-testing mchine, demonstrte nd comment on wht hppens to its reflectivity s the strip is being stretched. The polished surfce is initilly smooth, which llows light to be reflected uniformly cross the surfce. As the metl is stretched, the reflective surfce of the polished sheet metl will begin to become dull. The slip nd twin bnds developed t the surfce cuse roughening (see Fig. 1.6 on p. 42), which tends to sctter the reflected light Drw some nlogies to mechnicl fibering for exmple, lyers of thin dough sprinkled with flour or melted butter between ech lyer. A wide vriety of cceptble nswers re possible bsed on the student s experience nd cretivity. Some exmples of mechnicl fibering include: () food products such s lsgn, where lyers of noodles bound suce, or pstries with mny thin lyers, such s bklv; (b) log cbins, where tree trunks re oriented to construct wlls nd then seled with mtrix; nd (c) strw-reinforced mud Drw some nlogies to the phenomenon of hot shortness. Some nlogies to hot shortness include: () brick wll with deteriorting mortr between the bricks, (b) time-relesed medicine, where slowly soluble mtrix surrounds doses of quickly soluble medicine, nd (c) n Oreo cookie t room temperture compred to frozen cookie Obtin number of smll blls mde of plstic, wood, mrble, or metl, nd rrnge them with your hnds or glue them together to represent the crystl

13 The Structure of Metls 13 structures shown in Figs Comment on your observtions. By the student. There re mny possible comments, including the reltive densities of the three crystl structures (hcp is clerly densest). Also, the ingenious nd simple solid-bll models re striking when performing such demonstrtions Tke deck of plying crds, plce rubber bnd round it, nd then slip the crds ginst ech other to represent Figs. 1.6 nd 1.7. If you repet the sme experiment with more nd more rubber bnds round the sme deck, wht re you ccomplishing s fr s the behvior of the deck is concerned? By the student. With n incresed number of rubber bnds, you re physiclly incresing the friction force between ech crd. This is nlogous to incresing the mgnitude of the sher stress required to cuse slip. Furthermore, the greter the number of rubber bnds, the higher the sher or elstic modulus of the mteril (see Section 2.4 on p. 66). This problem cn be tken to very effective extreme by using smll C-clmps to highly compress the crds; the result is n object tht cts like one solid, with much higher stiffness thn the loose crds Give exmples in which nisotropy is scle dependent. For exmple, wire rope cn contin nneled wires tht re isotropic on microscopic scle, but the rope s whole is nisotropic. All mterils my behve in n nisotropic mnner when considered t tomic scles, but when tken s continuum, mny mterils re isotropic. Other exmples include: Clothing, which overll ppers to be isotropic, but clerly hs nisotropy defined by the direction of the threds in the cloth. This nisotropic behvior cn be verified by pulling smll ptches of the cloth in different directions. Wood hs directionlity (orthotropic) but it cn be ignored for mny pplictions. Humn skin: it ppers isotropic t lrge length scles, but microscopiclly it consists of cells with vrying strengths within the cell The movement of n edge disloction ws described in Section 1.4.1, by mens of n nlogy involving hump in crpet on the floor nd how the whole crpet cn eventully be moved by moving the hump forwrd. Recll tht the entnglement of disloctions ws described in terms of two humps t different ngles. Use piece of cloth plced on flt tble to demonstrte these phenomen. By the student. This cn be clerly demonstrted, especilly with cloth tht is complint (flexible) but hs high friction with flt surfce. Two methods of ensuring this is the cse re () to use cotton mteril (s found in T-shirts) nd wetting it before conducting the experiments, or (b) sprying the bottom side of the fbric with temporry dhesives, s found in most rts nd office supply stores. The experiments (single nd two lumps) cn then be conducted nd observtions mde.