UNIT 11 THIERMAL STRESSES

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1 UNIT 11 THIERMAL STRESSES Structure 11.1 Introduction Objectives Effect of Temperature on Bodies 11.3 Thermal Stresses in Bodies 11.4 Thermal Stresses in Uniform Bars Fully Restrained Bar Partially Restrarned Bar 11.5 Thermal Stresses in Stepped Bar Fully Restrained Stepped Bar Partially Restrained Stepped Bar 11.6 Thermal Stresses in Tapered Bars Fully Restriuned Tapered Bar Partially Restrained Tapered Bar 11.7 Thermal Stresses in Composite or Compound Bars and Tubes 11.8 Summary 11.9 Answers to SAQs 11. INTRODUCTION This unit introduces you to the effect of temperature on materials and completely or partially restrained bodies. It introduces the concept of thermal stresses and the method to determine them in simple and compound bars. Objectives After studying this unit, you should be able to explain the effect of temperature on bodies, understand how thermal stresses are developed in materials, and determine thermal stresses in uniform, stepped, tapered and compound bars EFFECT OF TEMPERATURE ON BODIES It is a common observation and experience that the size of a body increases as the temperature increases while it decreases as the temperature decreases. Thus, temperature rise results in expansion, i.e. increase in linear dimensions of the body, and temperature fall results in contraction, i.e. reduction in the linear dimensions of the body. The. magnitude of the changes brought in linear dimensions of the body by temperatur is defined through an intrillsic property of the material, of which the body is made VI, called coefficient of linear thermal expansions or contraction. The coefficienl of linear thermal expansion or contraction of a material is defined as the change in length per unit lenght per degree change in temperature of the body which is made of the malerial. The values of this coefficient, a of some of the common engineering materials are listed in Table From the definition, a can be written as, where, AL is the change in length over a total length L when the temperature is changcd by AT. Thus.

2 Str.. irs in Shafts & Shells and 'I'hennal Stresses Material Low carbon (mild) steel Cast h n : Gray Malleable Nickle-chrome steel ' Aluminium Brass Bronze Magnesium Table 11.1 a, Coeflicient of Linear Thermal Expansion w') 11.7 X 12 x lo x 23 x x 18 x lod6 28.8~10-~ ' 11.3 THERMAL STRESSES IN BODIES In the earlier section we had seen that if the temperature of a body is increased by AT, there will be an expansion AZ, over a length L. However, this expansion occurs only if the body is free to expand. During this free expansion, the body is not stressed. But, if this free expansion is restrained either fully or partially, forces of restraint are generated and the body develops internal strain resulting in internal stresses. These stresses are called lfhermal stresses. A similar situation arises if the body is restrained against contraction Ind the temperature is reduced. When expansion is restrained compressive stresses will lie generated because the restraint is equivalent to applying a compressive force. Similarly, when contraction is restrained tensile stresses will be generated THERMAL STRESSES IN UNIFORM BARS 11: lli~s section we shall discuss the problem of thermal stresses in bars of uniform ~ross-section when they are partially or totally restrained against change in length due to temperature changes. Since the area of cross-section is the same over the entire length, the stress will be the same on every cross-section..i Fully Restrained Bar Let us consider a bar of uniform cross-sectional area A and length L and at constrained both ends w as shown in Figure 11.1 (a) and (b). AREA OF C.S:A Ngure 11.1 (a) Fipm 11.1 (b) Let us increase the temperature of the bar by AT. If the bar were free to expand, it would have increased to a length L + AL, as in Figure 11.1 (b). However, this increase in length AL is completely restrained. Hence, the restraining force P will be equivalent to a force which would have produced the same change in length. Following these arguments PL alat= - or P = AEaAT AE Therefore, AL, Thermal Strain = - = a AT L Thermal Stress in the bar = Strain x E = E a AT P which is the same as -. Since the force P is compressive, the stresss developed is also A I compressive in nature.

3 Partially Restrained Bar Thennal st-- Now, let us assume that one of the supports of the bar in Figure 11.1 (a) can move by a distance ALf(AL' c AL) while the bar is expanding. In this case the bar is free to expand by an amount hl' but is restrained in expanding further by rest of the amount (hl- hl'). Thus, the force of restraint P is now a force which would produce a change in length (AL-M) in the bar.,' Thus,, -, Strain = dat-hl' = faat-g) [ Y'; Thermal Stress = Strain x E = E a AT - - Example 11.1 Two parallel walls are stayed together by a steel rod of 5 cm diameter passing through metal plates and nuts at both ends. The nuts are tightened, when the rod is at 150 C, to keep the walls 10 m apart. Determine the stresses in the rod when the temperature falls down to 50 C, if (a) the ends do not yield, and (b) the ends yield by 1 cm. so111 t'ion Given Length of the rod, L = 10 m = lo4 mm Diameter of the rod, d = 5 cm = 50 cm Change in temperature, AT = = 100' C a = 12x lod~'-' (a) When the ends do not yield (let the stress be ol) Thermal stress, 01, in the rod = E a AT (b) When the ends yield by 1 cm (let the stress be 02) r Thermal Stress, 02 = a AT -- x E Y51 = 2x105x12x10dx100 = 240~1mm~ SAQ 1 Two parallel walls h m apart, are slayed together by a steel rod 20 mm diameter, passlng through ~netal plales and nuts at each end. The nuts are tightened, when the rod is at a temperature of 100OC. Determine the stress in the rod, when the lcnlperaturc falls down to 20 C, if (a) thc ends do not yield. and (b) the ends yield by 1 mm. 4-1 ~ake1:'=2~10%/mm~ and a = 12x10 K.

4 Stn?sser Rnd The~?nd Stresses 11.5 THERMAL STRESSES IN STEPPED BARS In this section we shall discuss the determination of thermal stresses in bars whose cross-sectional area changes in steps over the length Fully Restrained Stepped Bar Consider a bar of length L which has a uniform cross-sectional area Al over length L1 while in the rest of the length Lq the cross-sectional area is A2. Let the Young's modulus and the coefficient of linear expansion for the two parts be ElVal and E2,a2 respectively. The ends of the bar are fully restrained. The bar is shown in Figure Let the temperature of the bar be raised by AT. Rgure 11.2 If the bar were free to expand, it would have extended by a length AL given by AL = L1 al AT+L2a2AT Since the bar is fully restrained, P, the compressive force developed would have to produce a contraction equal to AL. Let this force produce a stress crl in part 1 and a;! in part 2 (these are the thermal stresses). (Jl a1 will produce a strain equal to - and a change in length AL1 equal to -. El 0lLl El Similarly, q will produce a change in length equal to *. Thus. the total change in E2 length due to the application of force P will be equal to (AL1 + a), we get, But this should be equal to 4L. Therefore, ~ll1 02L = AL = (Llal + L2a2) AT El E2 But, Hence,

5 Herein, we have assumed that the two parts of the bar are made of two different materials. Instead, if the entire bar is of a single material of Young's rnodulus E and coefficient of linear expansion a, El = E2 = E and al = a2 = a Thennal Stresses Then al and 0 2 reduce to ATEA2aL 01 = A1 Lz+A2L1 and Further, if the cross-section is uniform throughout (Al = A2 = A), both the above expressions reduce to ol = 6 2 = 0 = EaAT a result which we have aiready derived for uniform bars Partially Restrained Stepped Bar Let us now consider the case of a stepped bar as in Figure 11.2, but which is free to extend by an amount AL' (AL' < AL) but restrained thereafter. In this case, during the free expansion of AL' the bar remains unstressed but thereafter develops the stresses 01 and 0 2 whose values will now be different than those of Section By a similar argument as in the previous section, we obiain & 02k! El + - = (AL - AL') = [AT (Llal + L2a2) - AL'] E2 From the above two expressions, the thermal stresses a1 and 02 can now be written as E1E2A2 [AT (Llal + ha2) - AL'] a1 = ' AlElL2 + A2E2L1 In the case of a single material bar 02 = (31 = 02 = EIE2Al [AT (Llal + L2a2) - AL'] A lellz + A2E2L1 E A2 [AT 05 - AL'] A1L2 + A2Ll EAI [AT al - AL'] AIL2 + A2L1 For a uniform bar the stress expression reduces to a result already observed. Example 11.2 A stepped bar made of aluminium is held between two supports. The bar is 900 mm in length at 3g0C, 600 mm of which is having a diameter of 50 mm, while the rest is of 25 mm diameter. Determine the L9ermal stress in the bar at a temperature of 21 O C, if (a) the supports are unyielding, and (b) when the supports move towards each other by 0.:.nm.

6 Stresses in Shafts & Shells and Thennal Stresses Given E for aluminium = 74 kn/rnm2 a for aluminium = 23.4 x 1 o4 K-' Solution When the Supports are Unyielding We have, L1 = 600 mm; diameter dl = 50 mm Thus, ;.Al =- n (50)~-2 4, Also, & = 300 mm; diameter, d2 = 25 mm, Therefore, and AT = = 17OC Therefore, free contraction AT = L a AT = 900 x 23.4 x x 17 = mm Let 01 be the stress in 50 rnm I$ part and 02 be the stress in 25 mm I$ part Then, These stresses are tensile in nature. When the Supports Move rowards Each Other by 0.1 mm Here, AL' = 0.1 rnm. Both these stresses are tensile. SAQ 2 A hi-metallic rod c~f Icngrh 450 mm is mounted horizontally between rigid abutments. The rod has a uniform circular cross-sectio11 and is made up of a 150 p ~n length of steel and a 300 mm lcilgth of copper coaxial with each other. It the rod is initially stress-free, determine the stress in the rod caused by a ttb,mperature rise of 100 K. Given E, copper = 105 GN~-" a, copper = I x x 10-' K-'

7 11.6 THERMAL STRESSES IN TAPERED BARS Tkennd Stresses In this section we shall study the problem of thermal stresses in gradually tapering bars. Because the cross-sectional area is varying along the length of the bar, so does the stress. Herein, we shall derive expression for thermal stress in bars with circular cross-section. For other cross.-sections,'like square etc., expressions for stresses can be derived in a similar fashion Fully Restrained Tapered Bar Consider the bar shown in Figure 11.3, which is circular in cross-section but tapering from a diameter d2 to dl over the length L. The bar is completely restrained at its ends. Now, let the temperature of the bar increased by AT. Figure 113 If the bar were not restrained but free to expand it would have extended by an amount AL, given by AL = alat Due to the restraint, a compressive force P would have developed in the bar whose effect is to produce a contraction equal to AL. Under this force, a cross-section at a distance x P from the larger end would have developed a stress ax equal to - where Ax is the are2 of Ax that-cross-section. The strain at that cross-section, ex can be w'ritten as \ 1 Under this strain, a small element of the bar of length dx would have changed its length by d(al) given by d(al) = ex dx = 4Pdx The total change in length AL can then be obtained by integrating the above expression.

8 Stresses in Shafts & Shells and Thermal Stresses Thus, dt Now, on writing (a - bx) = t, we get dx = - - b Substituting these, So the compressive force generated in the bar due to restraining the free expansion for an increase in temperature AT is P = ned 1d2a AT 4 Hence, the thermal stress a, at a cross-section with an area of A, is given as where x is measured from the end with diameter dl which is the larger end. The maximum stress am,, in the bar occurs at the smaller end with diameter d2. For cross-sections o&& than circular, the derivation can be proceeded in a similar way Partially ~estrained Tapered Bar Let us consider the bar of Figure 11.3, but now let it be free to extend by an amount AL'(AL' < AL). By following similar arguments as in the previous section, now the compressive load P developed in the bar will be given by %Ed id2 (AL- AL') P = 4L and hence, Ox = Edld2 (La AT- AL') (dl - d2) d - x) 2

9 and the maximum stress, Omax = Ed (kt AT - AL') L d2 Thermal Stresses Example 11.3 A circular bar rigidly fixed at both ends is 1 m long and tapers uniformly from 20 cm diameter at one end to 10 cm diameter at the other. Find the maximum stress in the bar, if its temperature is raised through 50 C. E = 2 x 10' N/mrn2 and a = 12 x IO-~K-~. Solution Here, dl = 20cm = 200mm d;! = 10cm = 100 mm = 240 N/mm2 SiIQ 3 A straight bar has a circular cross-section, the radius of which varies linearly from 30 mm at one end A to 15 mrn at the other end B. The bar is 1 m lorlg and is fixed rigidly at A, but longitudinal movement is possible at B against a spring which opposes movement with a constant sttffness of 20 kn/mn. Initially, Lhere is 110 longitudinal stress in the bar. The temperature of the bar then falls by 100K. Determine the change in the bar length if E = 69-.G~/n1~, and 11.7 THERMAL STRESSES IN COMPOSITE OR COMPOUND BARS AND TUBES In Section 11.5 on stepped bars, we have seen one type of composite bar where bars of different materials are joined serially. In this section, we shall consider another lype of compound bar, a bar constructed from two different materials rigidly joined together a shown in Figure 11.4 (a), or two bars of different materials but of equal length held between two rigid plates as in Figure 11.4 (b). Figure 11.4 (a) Figure 11.4 (b) If the individual bars were free to expand (or contract) due to temperature changes, they would do so to different amounts (for the same change in temperature), as in

10 stresses Shdh Sheas Figure 11.4 (c) and Figure 11.4 (d), due to the difference in the coefficients of linear and Thermal Stresses expansion of the two materials. Dl FFERENCE IN r FREE EX%NSION Figure 11.4 (c) DIFFERENCE IN r FREE EXPANSION Fig (d) Fig (e) Figure 11.4 (f) However, since the two materials are 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 will try to pull the lower expansion material to its free length, but will be held back by the latter to its own free length position. In practice, a compromise will be reached with both extending to a common position in between the individual free length positions. This, in effect, is equivalent to a contraction in bar 2 from its free length position and an expansion of bar 1 from its free position. Thus, the higher coefficient of expansion material develops compressive stress and the lower coefficient of expansion material develops tensile stresses, when the temperature of the compound bar increases. It will be vice-versa when the temperature decreases. From Figure 11.4 (c) to 11.4 (f) it is clear that extension of bar 1 + contraction of bar 2 = difference in free lengths Let the stresses in bars 1 and 2 be a1 and 02 due to the temperature change. Then the above nlle can be written as Since there are no external forces acting on the compound bar, for equilibrium, the compressive force in bar 2 should be equal to the tensile force in bar 1. This means that

11 From the above two expressions, a1 and 02 can be written as Thermal Stresses 61 = A2ElE2 (a2- a1) AT. and AlEl+ A2E2 The extension of the compound bar, i.e., AL is given as A2E2 (a;! - al) L AT = LalAT+ A 1El+ A2E2 Example 11.4 A compound bar is constructed from three bars 50 mm wide by 12 rnm 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/rn2 and the outside bars are of brass with E = 100 GN/m2. If the bars are initially fastened at 18OC and the.temperature of the whole assembly is then raised to 50 C, determine the stresses sel up in the brass and the aluminium. For computation purposes, take following values : Solution adumiium = 22 x K-' Let the stress in aluminium bar be a, and that in each brass bar be ab. Then for equilibrium Force in brass = Force in aluminium 0~~50~12x2 = 0 ~~12x50 or 2 a b = a, and from extension considerations or ab = 3.32 ~ /mm~ (tensile) and. Example 11.5 a, = 6.64 N/mm2 (compressive) A hollow steel cylinder of cross-sectional area 2000 mm2 concentrically surrounds a solid aluminium cylinder of cross-sectional area 6000 mm2. Both cylinders have the same length of 500 mm before a rigid block weighing 200 kn is applied at 20 C as shown in Figures11.5. Determine (a) the load carried by each cylinder at 60 C. (b) the temperature rise required for the entire load to be carried by the aluminium cylinder alone.

12 stresses in sh& & shells For computation purposes, take following values : and Thermal Stresses 2 Estml = 210 GNIm and Edumiiu, = 70 G N / ~ ~ = 12 x 10 K and adumi~,, = 23 x 10 K 500 rnrn Solution Figure 11s Figure 11.6 shows the free thermal expansions A, and A, together with the common expansion A under the load of 200 kn (the subscripts a and s standing for aluminium and steel respectively). For a temperature rise of AT K. 3 r-7 Aa Figure 11.6 we have, A, = 500 x 23 x x AT = 11.5 x AT mm 'A, = 500~12~10-~x~~ = 6 ~10-~~Tmm Under load, the strains are Ea A,-A = - and / 500 and the corresponding stresses are as follows : 6, = OX id 500 (A, - A) = 140 (A, - A) N For equilibrium of vertical forces, oax6000+o,x2000 = 200x10~~. Substituting for o,, o,, A, and A, we get, Hence, 5 A = 8.75 x AT

13 The loads taken by the aluminium and the steel are therefore, Pa = oa x 6000 N Thennal Stresses These equations will be valid as long as A is less th& A,. The load will be completely carried by aluminium when A, becomes equal to A. (a) at 60 C, AT = = 40K = kn P, = = 7.6 kn (b) The load will be carried completely by aluminium when 5 6 x x AT = 8.75 x low3 x AT Example 11.6 i.e., at a temperature of ( ) = 63.3 OC. A steel bolt of diameter 12 mm and length 175 mm is used to clamp a brass sleeve of length 150 mm to a rigid base plate as in Figure The sleeve has an internal diameter of 25 mm and a wall thickness of 3 mm. The thickness of the base plate is 25 mm. Initially, the nut is tightened until there is tensile force of 5 kn in the bolt. The temperature is now increased by 100' C. Determine the final stresses in the bolt and the sleeve. For computation purposes, take following values : Solution Figure 11.7 Let the free thermal expansions of steel and brass be As and Ab and A be the common expansion. Then

14 . Stresses in Shaffs & Shells and Thennal Stresaes If the initial stresses in steel and brass due to the 5 W load are asl and abl asl = + lo3 = N/mm2 (Tensile) x (1212 abl = - lo3 = Nhnm2 (Compressive) n ( ) Equilibrium of thermal stresses an and ab2 requires that or 3as2+7ab2 = 0 The thermal strains are given by A-As (A- Ab) ES = - and Eb = Thus, the thermal stresses are as follows : and 105 = - (A - Ab) X ld N/-~ 150 Substituting these in the equilibrium equation, 105 3xm(~-0.21)~ 175 ~o~+~x--(a-o.~o)x 150 lo3 = 0 Hence, A = Thus, os2 = ( ) x id = N/mm2 (tensile) ( ) x lo3 = N/mrn2 (compressive) Ob2 = 150 Total stressess are therefore as follows : 2 as = o,l + 6,2 = Nlmm 2 ab = abl.t 662 = N/mm (tensile) (compressive) SAQ 4 A steel rod of cross-sectional yea 600 mm2 and a coaxial copper tube of clc?ss-sectional area 1OOO mm- are firmly attached at their ends to form a compound bar. Determine the stress in the steel and in copper when the tenlperamre of the bar is raised by 80 C and an axial tensile force of 60 kn is applied. For steel E = 200 G N and ~ a ~ = 11 x K-' 4-1 For copper E = 100 G N and ~ a = ~ 16.5 x 10. K

15 ;,\a) 5,t ~teci rot1 of 20 n1n.r il~mit.tcr passe.; cerltially through a tight fitting copper bhe 1.t :xltrnal d~amctt'r 40 nnn. The tube 1s closed wilh the help of rigid washcrs of iiatgl~plh!? thickness ar~tl nuts thteadcltf on the rod.?'he nuts art1 tightened tiil the rvc )mpr?\\lve load on the tube 1s 50 kn Lhterniine the stresses in the rctrj imd the!?rb~. whcbil the temperature of tho assemb!) ldls by 50 C. For copper I.: GNIT-? i~nd n - 1 X x 1 O4 K-' SAQ tb..i. z-ic.;.l.,,,i :;l;i!, \vt;ighing 600 kn i:: placed ilpon twr! bronze rcid and one steel rod each iil Oj! c~i.~ CTOSS-S~I:~~~I;~: area si rr temperature of 15 C The bronze rods are _'5 cr:] \vhlli. lhc. sleel rot! is 3s) cn~ long. 8~i.0~~. the slab was placed, the top of all riii: rhrce rods are ievcl. t-'i~,i! the tcmporarme, 31 which the stress in the steel roti will hc zero. E,~,1 steel z: 'z3f)(> ~ ~ n a~rt l ' R of bronze = 80 GNI~-' 11.8 SUMMARY In this unit, we have studied (i) the effect of temperature on ixx~terials and bodies, (is) the concept of thermalktresses, and (iii) the method of determination of thermal stresses in uniform, stepped, tapered and compound bars ANSWERS TO SAQs SAQ 1 (a) N/mm2 (b) N/mm2 SAQ MPa SAQ mrn SAQ N/mm2, 3.3 ~ /mm~ SAQ 5 o, = ~/mm~ * SAQ OC

CHAPTER 2 COMPOUND BARS. Summary. F1 =- L1 w. c- L

CHAPTER 2 COMPOUND BARS. Summary. F1 =- L1 w. c- L 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