CHAPTER 2 COMPOUND BARS Summary When a compound bar is constructed from members of different materials,lengths and areas and is subjected to an external tensile or compressive load W the load carried by any single member is given by E1A1 W F,二EA Σ where sufixIreferstothesinge member and EA is the sum of all such quantities for all the members. Where the bars have a common length the compound bar can be reduced to a single equivalent bar with an equivalent Young's modulus,termed a combined E. ΣEA Combined E= ZA The free expansion of a bar under a temperature change from Ti to T is a(T2-T)L where a is the coefficient of linear expansion and L is the length of the bar. If this expansion is prevented a stress will be induced in the bar given by a(T-TE To determine the stresses in a compound bar composed of two members of different free lengths two principles are used: (1)The tensile force applied to the short member by the long member is equal in magnitude to the compressive force applied to the long member by the short member. (2)The extension of the short member plus the contraction of the long member equals the difference in free lengths. This difference in free lengths may result from the tightening of a nut or from a temperature change in two members of different material (i.e.different coefficients of expansion)but of equal length initially. If such a bar is then subjected to an additional external load the resultant stresses may be obtained by using the principle of superposition.With this method the stresses in the members 27
CHAPTER 2 COMPOUND BARS Summary When a compound bar is constructed from members of different materials, lengths and areas and is subjected to an external tensile or compressive load W the load carried by any single member is given by ElAl F1 =- L1 w EA c- L EA L where suffix 1 refers to the single member and X - is the sum of all such quantities for all the members. Where the bars have a common length the compound bar can be reduced to a single equivalent bar with an equivalent Young’s modulus, termed a combined E. C EA Combined E = - CA The free expansion of a bar under a temperature change from Tl to T2 is a(T2 -T1)L where a is the coefficient of linear expansion and L is the length of the bar. If this expansion is prevented a stress will be induced in the bar given by a(T2 -T# To determine the stresses in a compound bar composed of two members of different free lengths two principles are used: (1) The tensile force applied to the short member by the long member is equal in magnitude (2) The extension of the short member plus the contraction of the long member equals the This difference in free lengths may result from the tightening of a nut or from a temperature change in two members of different material (Le. different coefficients of expansion) but of equal length initially. If such a bar is then subjected to an additional external load the resultant stresses may be obtained by using the principle ofsuperposition. With this method the stresses in the members 27 to the compressive force applied to the long member by the short member. difference in free lengths
28 Mechanics of Materials $2.1 arising from the separate effects are obtained and the results added,taking account of sign,to give the resultant stresses. N.B.:Discussion in this chapter is concerned with compound bars which are symmetri- cally proportioned such that no bending results. 2.1.Compound bars subjected to external load In certain applications it is necessary to use a combination of elements or bars made from different materials,each material performing a different function.In overhead electric cables, for example,it is often convenient to carry the current in a set of copper wires surrounding steel wires,the latter being designed to support the weight of the cable over large spans.Such combinations of materials are generally termed compound bars.Discussion in this chapter is concerned with compound bars which are symmetrically proportioned such that no bending results. When an external load is applied to such a compound bar it is shared between the individual component materials in proportions depending on their respective lengths,areas and Young's moduli. Consider,therefore,a compound bar consisting of n members,each having a different length and cross-sectional area and each being of a different material;this is shown diagrammatically in Fig.2.1.Let all members have a common extension x,i.e.the load is positioned to produce the same extension in each member. nmember First member Length Ln Length Li Area An Areo A Modulus En Modulus E Load Fn Load F ommon extension x H Fig.2.1.Diagrammatic representation of a compound bar formed of different materials with different lengths,cross-sectional areas and Young's moduli. For the nth member, stress strain =En= FnLn Anxn EnAnx F= (2.10 Ln where F is the force in the nth member and A,and L are its cross-sectional area and length
28 Mechanics of Materials $2.1 arising from the separate effects are obtained and the results added, taking account of sign, to give the resultant stresses. N.B.: Discussion in this chapter is concerned with compound bars which are symmetrically proportioned such that no bending results. 2.1. Compound bars subjected to external load In certain applications it is necessary to use a combination of elements or bars made from different materials, each material performing a different function. In overhead electric cables, for example, it is often convenient to carry the current in a set of copper wires surrounding steel wires, the latter being designed to support the weight of the cable over large spans. Such combinations of materials are generally termed compound burs. Discussion in this chapter is concerned with compound bars which are symmetrically proportioned such that no bending results. When an external load is applied to such a compound bar it is shared between the individual component materials in proportions depending on their respective lengths, areas and Young’s moduli. Consider, therefore, a compound bar consisting of n members, each having a different length and cross-sectional area and each being of a different material; this is shown diagrammatically in Fig. 2.1. Let all members have a common extension x, i.e. the load is positioned to produce the same extension in each member. First member Modulus E, Load 7 n’’mmernber Length L, Area An Modulus E, T’ Load F, Frnmon extension x W Fig. 2.1. Diagrammatic representation of a compound bar formed of different materials with different lengths, cross-sectional areas and Young’s moduli. For the nth member, f‘n Ln E, = - strain Anxn - stress -- where F, is the force in the nth member and A, and Ln are its cross-sectional area and length
§2.2 Compound Bars 29 The total load carried will be the sum of all such loads for all the members i.e. w=-2 (2.2) Now from egn.(2.1)the force in member 1 is given by E1A1x LI But,from eqn.(2.2), W X=- EmAn Ln E1A F1= (2.3) EA i.e.each member carries a portion of the total load W proportional to its EA/L value. If the wires are all of equal length the above equation reduces to EAW F1=ZEA (2.4) The stress in member 1 is then given by F 01= (2.5) A 2.2.Compound bars-“equivalent'or“combined'”modulus In order to determine the common extension of a compound bar it is convenient to consider it as a single bar of an imaginary material with an equivalent or combined modulus E..Here it is necessary to assume that both the extension and the original lengths of the individual members of the compound bar are the same;the strains in all members will then be equal. Now total load on compound bar F:+F2+F3+...+F where F1,F2,etc.,are the loads in members 1,2,etc. But force stress x area G(A1+A2+.,+An)=O1A1+02A2+·+0nAn where o is the stress in the equivalent single bar. Dividing through by the common strain 8, (41+A2+…+A)=2A1+2A2+…+A & i.e. E(A1+A2+.··+An)=E1A1+E2A2+,··+EmAn where E.is the equivalent or combined E of the single bar
52.2 Compound Bars 29 The total load carried will be the sum of all such loads for all the members i.e. Now from eqn. (2.1) the force in member 1 is given by But, from eqn. (2.2), .. F1=- L1 w (2.3) EA c- L i.e. each member carries a portion of the total load W proportional to its EAIL value. If the wires are all of equal length the above equation reduces to The stress in member 1 is then given by 2.2. Compound bars - “equivalent” or “combined” modulus In order to determine the common extension of a compound bar it is convenient to consider it as a single bar of an imaginary material with an equivalent or combined modulus E,. Here it is necessary to assume that both the extension and the original lengths of the individual members of the compound bar are the same; the strains in all members will then be equal. Now total load on compound bar = F1 + Fz + F3 + . . . + F, where F1, F2, etc., are the loads in members 1, 2, etc. But force = stress x area .. a(Al+Az+ ... +An)=~lAl+~zAz+ ... +a,A, where 6 is the stress in the equivalent single bar. Dividing through by the common strain E, d 61 on E E E -(Al + Az + . . . +An) = -Al + :AZ + . . . + -A, i.e. Ec(Ai+Az+ ... +An)=EIA1+EzAz+ ... +E,An where E, is the equivalent or combined E of the single bar
30 Mechanics of Materials §2.3 combined E=E4+E2++EA A1十A2+·..+An ΣEA i.e. Ee= ZA (2.6 With an external load W applied, stress in the equivalent bar= W ΣA and W strain in the equivalent bar= E,ΣA=D ..since stress =E strain WL common extension x= E∑A (2.7) extension of single bar 2.3.Compound bars subjected to temperature change When a material is subjected to a change in temperature its length will change by an amount aLt where a is the coefficient of linear expansion for the material,L is the original length and t the temperature change.(An increase in temperature produces an increase in length and a decrease in temperature a decrease in length except in very special cases of materials with zero or negative coefficients of expansion which need not be considered here.) If,however,the free expansion of the material is prevented by some external force,then a stress is set up in the material.This stress is equal in magnitude to that which would be produced in the bar by initially allowing the free change of length and then applying sufficient force to return the bar to its original length. Now change in length =aLt aLt strain=ot Therefore,the stress created in the material by the application of sufficient force to remove this strain =strain x E =Eat Consider now a compound bar constructed from two different materials rigidly joined together as shown in Fig.2.2 and Fig.2.3(a).For simplicity of description consider that the materials in this case are steel and brass
30 Mechanics of Materials 42.3 .. i.e. EiAi+E,AZ+ . . . +E,An Al+A2+ . . . +An combined E = XEA EA E, = - With an external load W applied, W stress in the equivalent bar = ~ XA and wx strain in the equivalent bar = - = - EJA L .’. since stress strain -- -E WL common extension x = - E,XA = extension of single bar 2.3. Compound bars subjected to temperature change When a material is subjected to a change in temperature its length will change by an amount aLt where a is the coefficient of linear expansion for the material, L is the original length and t the temperature change. (An increase in temperature produces an increase in length and a decrease in temperature a decrease in length except in very special cases of materials with zero or negative coefficients of expansion which need not be considered here.) If, however, the free expansion of the material is prevented by some external force, then a stress is set up in the material. This stress is equal in magnitude to that which would be produced in the bar by initially allowing the free change of length and then applying sufficient force to return the bar to its original length. Now change in length = aLt . aLt .. strain = - = at L Therefore, the stress created in the material by the application of sufficient force to remove this strain = strain x E = Eat Consider now a compound bar constructed from two different materials rigidly joined together as shown in Fig. 2.2 and Fig. 2.3(a). For simplicity of description consider that the materials in this case are steel and brass
§2.3 Compound Bars 31 Steel Brass Step. Fig.2.2. In general,the coefficients of expansion of the two materials forming the compound bar will be different so that as the temperature rises each material will attempt to expand by different amounts.Figure 2.3b shows the positions to which the individual materials will extend if they are completely free to expand(i.e.not joined rigidly together as a compound bar).The extension of any length L is given by aLt Steel (a)Original bar 6ra53 Steel Difference in free lengths (b)Expanded position; members free to expand independenty Extension of steel- Compression of br055 (c)Compound bar Steel expanded position Brass SteeL Fig.2.3.Thermal expansion of compound bar. Thus the difference of "free"expansion lengths or so-called free lengths agLt-asLt=(aB-as)Lt since in this case the coefficient of expansion of the brassis greater than that for the steels. The initial lengths L of the two materials are assumed equal. If the two materials are now rigidly joined as a compound bar and subjected to the same temperature rise,each material will attempt to expand to its free length position but each will be affected by the movement of the other.The higher coefficient of expansion material (brass) will therefore seek to pull the steel up to its free length position and conversely the lower
§2.3 Compound Bars 31 Fig. 2.2. In general, the coefficients of expansion of the two materials forming the compound bar will be different so that as the temperature rises each material will attempt to expand by different amounts. Figure 2.3b shows the positions to which the individual materials will extend if they are completely free to expand (i.e. not joined rigidly together as a compound bar). The extension of any length L is given by cxLt Fig. 2.3. Thermal expansion of compound bar. Thus the difference of "free" expansion lengths or so-called free lengths = IXBLt-IXsLt = (IXB-IXs)Lt since in this case the coefficient of expansion of the brass IXBis greater than that for the steellXs' The initial lengths L of the two materials are assumed equal. If the two materials are now rigidly joined as a compound bar and subjected to the same temperature rise, each material will attempt to expand to its free length position but each will be affected by the movement of the other, The higher coefficient of expansion material (brass) will therefore seek to pull the steel up to its free length position and conversely the lower
32 Mechanics of Materials §2.4 coefficient of expansion material(steel)will try to hold the brass back to the steel"free length" position.In practice a compromise is reached,the compound bar extending to the position shown in Fig.2.3c,resulting in an effective compression of the brass from its free length position and an effective extension of the steel from its free length position.From the diagram it will be seen that the following rule holds. Rule 1. Extension of steel+compression of brass=difference in"free"lengths. Referring to the bars in their free expanded positions the rule may be written as Extension of"short"member compression of"long"member difference in free lengths. Applying Newton's law of equal action and reaction the following second rule also applies. Rule 2. The tensile force applied to the short member by the long member is equal in magnitude to the compressive force applied to the long member by the short member. Thus,in this case, tensile force in steel compressive force in brass Now,from the definition of Young's modulus E= stress strainδ/L where o is the change in length. 6-L E Also force stress x area =A where A is the cross-sectional area. Therefore Rule 1 becomes asLL-(as-as)Lt (2.8) Es Ea and Rule 2 becomes OsAs=GBAB (2.9) We thus have two equations with two unknowns os and a and it is possible to evaluate the magnitudes of these stresses (see Example 2.2). 2.4.Compound bar (tube and rod) Consider now the case of a hollow tube with washers or endplates at each end and a central threaded rod as shown in Fig.2.4.At first sight there would seem to be no connection with the work of the previous section,yet,in fact,the method of solution to determine the stresses set up in the tube and rod when one nut is tightened is identical to that described in $2.3. The compound bar which is formed after assembly of the tube and rod,i.e.with the nuts tightened,is shown in Fig.2.4c,the rod being in a state of tension and the tube in compression.Once again Rule 2 applies,i.e. compressive force in tube tensile force in rod
32 Mechanics of Materials $2.4 coefficient of expansion material (steel) will try to hold the brass back to the steel “free length” position. In practice a compromise is reached, the compound bar extending to the position shown in Fig. 2.3c, resulting in an effective compression of the brass from its free length position and an effective extension of the steel from its free length position. From the diagram it will be seen that the following rule holds. Rule 1. Extension of steel + compression of brass = dixerence in “free” lengths. Extension of “short” member + compression of“1ong” member = dixerence in free lengths. Applying Newton’s law of equal action and reaction the following second rule also applies. Rule 2. The tensile force applied to the short member by the long member is equal in magnitude to the compressive force applied to the long member by the short member. Referring to the bars in their free expanded positions the rule may be written as Thus, in this case, Now, from the definition of Young’s modulus tensile force in steel = compressive force in brass stress o E=- -- strain 6/L - where 6 is the change in length. OL .. 6=- E Also where A is the cross-sectional area. force = stress x area = OA Therefore Rule 1 becomes a,L tl L -++=(ag-as)Lt E, E, and Rule 2 becomes @,A, = cBAB We thus have two equations with two unknowns os and oB and it is possible to evaluate the magnitudes of these stresses (see Example 2.2). 2.4. Compound bar (tube and rod) Consider now the case of a hollow tube with washers or endplates at each end and a central threaded rod as shown in Fig. 2.4. At first sight there would seem to be no connection with the work of the previous section, yet, in fact, the method of solution to determine the stresses set up in the tube and rod when one nut is tightened is identical to that described in $2.3. The compound bar which is formed after assembly of the tube and rod, i.e. with the nuts tightened, is shown in Fig. 2.4c, the rod being in a state of tension and the tube in compression. Once again Rule 2 applies, i.e. compressive force in tube = tensile force in rod
§2.4 Compound Bars 33 (a) Difference in free lengths adistance moved by nut b Compression of tube Extension of rod Fig.2.4.Equivalent "mechanical"system to that of Fig.2.3. Figure 2.4a and bshow,diagrammatically,the effective positions of the tube and rod before the nut is tightened and the two components are combined.As the nut is turned there is a simultaneous compression of the tube and tension of the rod leading to the final state shown in Fig.2.4c.As before,however,the diagram shows that Rule 1 applies: compression of tube +extension of rod difference in free lengths axial advance of nut i.e.the axial movement of the nut (number of turns n x threads per metre)is taken up by combined compression of the tube and extension of the rod. Thus,with suffix t for tube and R for rod, + -=n x threads/metre (2.10) E ER also ORAR=0:A (2.11) If the tube and rod are now subjected to a change of temperature they may be treated as a normal compound bar of $2.3 and Rules I and 2 again apply(Fig.2.5), i.e. :L ORL」 =(a:-aR)Lt (2.12) E ER Difference in free lengths G99C525502566 (o)Free independent expansion a227207z Compression of tube Extension of rod 27722 (b】Compound bar expansion 2222 3 Original length Fig.2.5
$2.4 Compound Bars 33 Difference in free lengths I 7 = distance moved by nut I Compression of tube I 4 +€Extension of rod Fig. 2.4. Equivalent “mechanical” system to that of Fig. 2.3. Figure 2.4a and b show, diagrammatically, the effective positions of the tube and rod before the nut is tightened and the two components are combined. As the nut is turned there is a simultaneous compression of the tube and tension of the rod leading to the final state shown in Fig. 2.4~. As before, however, the diagram shows that Rule 1 applies: compression of tube +extension of rod = difference in free lengths = axial advance of nut i.e. the axial movement of the nut ( = number of turns n x threads per metre) is taken up by combined compression of the tube and extension of the rod. Thus, with suffix t for tube and R for rod, OIL d L - + R = n x threads/metre El (2.10) also uRA, = a,A, (2.1 1) If the tube and rod are now subjected to a change of temperature they may be treated as a normal compound bar of $2.3 and Rules 1 and 2 again apply (Fig. 2.5), i.e. (a) Free independent expansion Difference in free lengths Compression of tube Extension of rod (b) Cornpouna bar expansoon (2.12) t--GGz Id Fig. 2.5
34 Mechanics of Materials $2.5 where o;and o's are the stresses in the tube and rod due to temperature change only and a,is assumed greater than a.If the latter is not the case the two terms inside the final bracket should be interchanged. Also GRAR=G:A 2.5.Compound bars subjected to external load and temperature effects In this case the principle of superposition must be applied,i.e.provided that stresses remain within the elastic limit the effects of external load and temperature change may be assessed separately as described in the previous sections and the results added,taking account of sign, to determine the resultant total effect; i.e. total strain sum of strain due to external loads and temperature strain 2.6.Compound thick cylinders subjected to temperature changes The procedure described in $2.3 has been applied to compound cylinders constructed from tubes of different materials on page 230. Examples Example 2.1 (a)A compound bar consists of four brass wires of 2.5 mm diameter and one steel wire of 1.5 mm diameter.Determine the stresses in each of the wires when the bar supports a load of 500 N.Assume all of the wires are of equal lengths. (b)Calculate the "equivalent"or "combined"modulus for the compound bar and determine its total extension if it is initially 0.75 m long.Hence check the values of the stresses obtained in part (a). For brass E 100 GN/m2 and for steel E =200 GN/m2. Solution (a)From eqn.(2.3)the force in the steel wire is given by E,AsW F,=EA 200×10°×叠×1.52×10-6 =[200×10°×毫×152×10-6+4100×109×毫×252×10-6 500 -[ax140x25]3w-62N 2×1.52
34 Mechanics of Materials §2.5 where a; and 0; are the stresses in the tube and rod due to temperature change only and a, is assumed greater than aR. If the latter is not the case the two terms inside the final bracket should be interchanged. Also ai A, = 0; A, 2.5. Compound bars subjected to external load and temperature effects In this case the principle ofsuperposition must be applied, i.e. provided that stresses remain within the elastic limit the effects of external load and temperature change may be assessed separately as described in the previous sections and the results added, taking account of sign, to determine the resultant total effect; i.e. total strain = sum of strain due to external loads and temperature strain 2.6. Compound thick cylinders subjected to temperature changes The procedure described in $2.3 has been applied to compound cylinders constructed from tubes of different materials on page 230. Examples Example 2.1 (a) A compound bar consists of four brass wires of 2.5 mm diameter and one steel wire of 1.5 mm diameter. Determine the stresses in each of the wires when the bar supports a load of 500 N. Assume all of the wires are of equal lengths. (b) Calculate the “equivalent” or “combined modulus for the compound bar and determine its total extension if it is initially 0.75 m long. Hence check the values of the stresses obtained in part (a). For brass E = 100 GN/m’ and for steel E = 200 GN/m’. Solution (a) From eqn. (2.3) the force in the steel wire is given by 200 x 109 x 2 x 1.52 x 10-6 = [ 200 x lo9 x 2 x 1.5’ x +4(1OO x lo9 x 2 x 2.5’ x 500 = 76.27 N 1 2 x 1.5’ = [ (2 x 1.5’)+ (4 x 2.5’)
Compound Bars 35 total force in brass wires 500-76.27=423.73 N load 76.27 stress in steel area 季×1.52×10-6=43.2MN/m2 load 423.73 and stress in brass area4×0×2.52x10-6=21.6MN/m2 (b)From egn.(2.6) combined E= EA_200×10°×等×1.52×10-6+4(100×109×章×2.52×10-6) ΣA (1.52+4×2.52y0-6 200×1.52+400×2.59)109=108.26GN/m2 = (1.52+4×2.52) stress Now E= strain and the stress in the equivalent bar 500 500 2A= 01.52+4×2.510-6=23.36MN/m2 stress 23.36×106 strain in the equivalent bar E=10826×10=0.216×103 common extension strain x original length =0.216×10-3×0.75=0.162×10-3 =0.162mm This is also the extension of any single bar,giving a strain in any bar 0.162×10-3 -=0.216×10-3 as above 0.75 stress in steel=strain×E,=0.216×10-3×200×109 =43.2MN/m2 and stress in brass=strain×Eg=0.216×10-3×100×10 =21.6MN/m2 These are the same values as obtained in part(a). Example 2.2 (a)A compound bar is constructed from three bars 50 mm wide by 12 mm thick fastened together to form a bar 50 mm wide by 36 mm thick.The middle bar is of aluminium alloy for which E=70 GN/m2 and the outside bars are of brass with E=100 GN/m2.If the bars are initially fastened at 18C and the temperature of the whole assembly is then raised to 50C, determine the stresses set up in the brass and the aluminium. xg=18×10-6 perC and a4=22×10-6perC
Compound Bars 35 .. total force in brass wires = 500 - 76.27 = 423.73 N = 43.2 MN/m2 load 76.27 area f x 1S2 x .. stress in steel = - = and load 423.73 area stress in brass = - = = 21.6 MN/m2 4 x f x 2S2 x (b) From eqn. (2.6) CEA 200 x 109 x f x 1.52 x io-6+4(100 x 109 x 5 x 2S2 x 10-6) - f(1S2 +4 x 2.52)'0-6 combined E = - - EA stress Now E=- strain and the stress in the equivalent bar - 23.36 MN/m2 500 - 500 ZA $(1S2+4x 2.52)10-6 - =-- stress 23.36 x lo6 E 108.26 x lo9 .. strain in the equivalent bar = - = = 0.216 x .. common extension = strain x original length = 0.216 x = 0.162 mm x 0.75 = 0.162 x This is also the extension of any single bar, giving a strain in any bar 0.162 x lo-' 0.75 - = 0.216 x as above .. stress in steel = strain x E, = 0.216 x stress in brass = strain x E, = 0.216 x x 200 x lo9 x 100 x lo9 = 43.2 MN/m2 = 21.6 MN/m2 and These are the same values as obtained in part (a). Example 2.2 (a) A compound bar is constructed from three bars 50 mm wide by 12 mm thick fastened together to form a bar 50 mm wide by 36 mm thick. The middle bar is of aluminium alloy for which E = 70 GN/m2 and the outside bars are of brass with E = 100 GN/m2. If the bars are initially fastened at 18°C and the temperature of the whole assembly is then raised to 5WC, determine the stresses set up in the brass and the aluminium. ctg = 18 x per "C and u,., = 22 x per "C
36 Mechanics of Materials (b)What will be the changes in these stresses if an external compressive load of 15 kN is applied to the compound bar at the higher temperature? Solution With any problem of this type it is convenient to let the stress in one of the component members or materials,e.g.the brass,be x. Then,since force in brass force in aluminium and force stress x area x×2×50×12×10-6=04×50×12×10-6 i.e. stress in aluminium =2x Now,from egn.(2.8), extension of brass+compression of aluminium difference in free lengths =(-)(T2-T)L xL 2xL 100×10+70×10°=(2-18)10-6(50-18)L (7x+20x) 700×109 =4×10-6×32 27x=4×10-6×32×700×109 x=3.32MN/m2 The stress in the brass is thus 3.32 MN/m2(tensile)and the stress in the aluminium is 2 x 3.32 =6.64 MN/m2 (compressive). (b)With an external load of 15 kN applied each member will take a proportion of the total load given by eqn.(2.3). Force in aluminium EAAAW ΣEA 0x50x2+2x10x50x20×105]15x10 70×10°×50×12×10-6 「70 =L(70+200 15×103 =3.89kN force in brass =15-3.89 11.11 kN loa 111.11×103 stress in brass area 2×50×12×10-6 =9.26 MN/m2(compressive)
36 Mechanics of Materials (b) What will be the changes in these stresses if an external compressive load of 15 kN is applied to the compound bar at the higher temperature? Solution With any problem of this type it is convenient to let the stress in one of the component Then, since members or materials, e.g. the brass, be x. force in brass = force in aluminium and force = stress x area x x 2 x 50 x 12 x loF6 = oA x 50 x 12 x i.e. stress in aluminium oA = 2x Now, from eqn. (2.Q extension of brass + compression of aluminium = difference in free lengths = (a” -ail) (Tz -TdL XL 2xL + ___ = (22- 18)10-6(50- 18)L 100 x 109 70 x 109 700~ 109 (7x + 2ox) = 4 x 10-6 x 32 27x = 4 x lov6 x 32 x 700 x lo9 x = 3.32 MN/m2 The stress in the brass is thus 3.32 MN/m2 (tensile) and the stress in the aluminium is (b) With an external load of 15 kN applied each member will take a proportion of the total 2 x 3.32 = 6.64 MN/mz (compressive). load given by eqn. (2.3). Force in aluminium = ~ EAAA w CEA 115 x 103 70 x lo9 x 50 x 12 x = [ (70 x 50 x 12 + 2 x 100 x 50 x 12)109 x 10- 6 = 3.89 kN force in brass = 15-3.89 = 11.11 kN stress in brass = - = .. load 11.11 x 103 .. area 2 x 50 x 12 x = 9.26 MN/m2 (compressive)