STRESS AND STRAINS - ikbooks.com

1 CHAPTER2 STRESS AND STRAINSS tressStress can be classified broadly in three types as described STRESS : It is illustrated in Fig. a tensile loadWis applied to a uni-form rod fixed at one STRESS , =WCross-sectional area of rod=WAunit is N/mm2or MN/m2, (Greek letter sigma). STRESS : As shown in Fig. loadWtends to compress a rodof cross-section areaA, then compressivestress = STRESS : If two plates are joinedtogether with rivet as shown in Fig. STRESS in rivet is known as shear STRESS ,it is denoted by (Greek letter tau), shearstress in rivet, = Strength of MaterialsIt may be noted that pointAis the limit of proportionality andBis the elastic limit. Between pointAandBit is a curve thus not linear relationship. Therefore, actuallyEis constant within the limitof proportionality, though in Hooke s law, we had mentioned, within elastic limit , becauseAandBare very close to each other.

2 Let us name the important points on the graph:A: Limit of proportionalityB: Elastic limit: It may be noted that on removal of load up to elastic limits, specimen comesback to its original : Higher yield point: This is the point where yielding of the material : Lower yield point: The STRESS associated with the lower yield point is known as yield : Maximum STRESS : Here the STRESS is maximum because due to plastic behaviour of the mate-rial, area of cross section is very : Point of fracture: At this point waisting occurs as shown in Fig. the material is loaded beyond the elastic limit and then load is removed, a permanent extensionremains, calledpermanent STRESS : For engineering purposes it is desirable to know the STRESS to which a highlyductile material such as aluminium can be loaded safely before a permanent extension takes STRESS is known as theproof STRESS or offset stressand is defined as the STRESS at whicha specified permanent extension has taken place in the tensile test.

3 Proof STRESS is found from thestress- strain curve as given in Fig. The extension specified is usually , or per cent ofgauge and STRAINS 7 The proof STRESS here is found on the basis of per cent : Draw a line parallel to the initial slope of the curve. The STRESS at the point where thisline cuts the curve is the STRESS . The per cent proof STRESS is also found in the : Though we define Hooke s law to be taken without elastic limit, but strictly speaking it isapplicable up to the point of proportionalityBin Fig. Materials: Fig. shows that the STRESS - strain graph for brittle materials such as cast metal is almost elastic and up to fracture but does not obey Hooke s law. A material such asthis which has little plasticity or ductility and does not neck down before fracture is termed brittle.

4 The modulus of elasticity for cast iron is not constant but depends on the portion of the curve fromwhich it is iron in tensionThe following table is very useful for mechanical properties:PercentageYield proof stressUltimate tensile stressMaterialelongationMN/m2MN/m2MN/m2 Copper annealed60 60220 Copper hard4 320400 Aluminium soft35 3090 Aluminium hard5 140150 Black mild steel25 26 Bright mild steel14 17 Structural steel20220 250 430 500 Cast ironSpheroidalGraphiteCast iron (annealed) 280 340 Stainless steel60230 6008 Strength of MaterialsNowE=stressstrainLetF,L,A,dlbe the force, length, area of cross section, and extension or contractionrespectively,thenE= :A bar of mild steel has an overall length of m. The diameter up to 700 mmlength is 56 mm, the diameter of the remaining m is 35 mm. Calculate the extension of the bardue to a tensile load of 55 : Remember 1 GN/m2=1 kN/mm2 E=200 kN/mm2We knowE=FlAdl dl= , for portion of 700 mm,the extensiondl1=55000 700 7 4200000 22 56 56=564mm= mmNowdl2for 1400 mm length of 35 mm dia,dl2=55000 1400 7 4200000 22 35 35= mmTotal extension=dl1+dl2= + bars: When two or more materials (members) are rigidly fixed together so that theyshare the same load and extend or compress by same amount, the two members form compoundbar.

5 Let us say that in Fig. we have to find STRESS in each material and amount of Aof ESMaterial Bof EBFigure the outer tube of materialAhas outsidedia asd1and inside dia asd2and inner tube ofmaterialBhas outside dia asd3and inside diaasd4. Both ends are joined rigidly to make com-pound bar of Strength of MaterialsHence,d2=5630 mm2ord=75 :A steel bar of 20 mm diameter and 400 mm long is placed concentrically insidea gunmetal tube (Fig. ). The tube has inside diameter 22 mm and thickness 4 mm. The length ofthe tube exceeds the length of the steel bar by mm. Rigid plates are placed on the compoundassembly. Find: a) the load which will just make tube and bar of same length and b) the stresses inthe steel and gunmetal when a load of 50 kN is steel=213 GN/m2,Efor gunmetal=100 mmPFigure of gunmetal tube,Ag= 4( )= m2 Area of steel barAs= 4( )2= m2a) For tube to compress mm: strain = ,Let 1be the STRESS in the tube 1Eg= , 1100= 1= 100= GN/m2=30000 kN/m2 Hence, load=30000 kNb) Load available to compress bar and tube as a compound bar is given by, let 2be the addi-tional STRESS produced in the gunmetal tube due to this load and sbe the corresponding stressin the steel bar, thenLoad on compound bar=50 kNP= 2Ag+ 2 + 3 (i) STRESS and STRAINS 11 Also 2Eg= sEs, 2=1002100 s(ii)From Eqns.

6 (i) and (ii) 2=40,600 kN/m2= MN/m2 s=85300 kN/m2= MN/m2 Final STRESS in gunmetal= 1+ 2=40600+30000=70,600 kN/m2= MN/m2 Deformation of a Body Due to Self WeightBAlxdxFigure us consider a barABwhich is hanging freely under itsown weight (see Fig. )Letw=specific weight of the bar materialNow consider a small sectiondxat a of the bar for a length=w volume=wAx(A is cross section of the bar)Now elongation of the elementry lengthdxdue to weight of the bar for lengthx,(wAx)= (wAx) elongation=l 0wxdxE=wEl [x22]l0 elongation,dl=wl22 EBecause total weight of bar,W= elongationdlcan be written ,dl=Wl2AE12 Strength of MaterialsThis result also proves that the extension due to own weight is half if same weight is applied atthe end (of course neglecting extension due to self weight). :A steel barABC18 m long is having cross-sectional area 4 mm2weighs N(Refer Fig.)

7 If modulus of elasticity of wire is 210 GN/m2, find the deflections m9 mCFigure atCdue to self weight of wireAC=dlcdlc=Wl2AE= 180002 4 210000= mmDeflection at B:Now deflection atBis due to two reasons: i) due to selfweight ofABand ii) due to weight l/22AE+W/2 dlB=W/2 l/2AE(12+1)= 90002 4 210000( )= mmSometimes a machine member is a acted upon by a number of forces, some acting at outer edgeswhile some are acting inside the body. In such cases in order to find out the total extension orcontraction, the principle of superposition is applied. This has been very well made clear by thefollowing :A steel barABCof 400 mm length and 20 mm diameter is subjected to a pointload as shown in Fig. Determine the total change in the length of bar. TakeE=200 kN40 kNCB20 kN200 mm200 mmFigure and STRAINS 13 SOLUTION:For simplification split it into two parts as under:AC400 mm40 kN40 kNACB200 mm20 kN20 kN = PlAEA= 4(20)2=314 mm2 AC=40 103 400314 200000= mm AB=20 103 200314 200000= mmTotal = + :A copper rodABCDof 800 mm2cross-sectional area and m long is subjectedto forces as shown in Fig.

8 Find the total elongation of the bar. TakeE=100 GPaA40 kN50 kNCDB20 kN30 mSOLUTION:Splitting into three figures as shown m40 kN40 m20 kN20 kNDB4 m10 kN10 kNFigure and STRAINS 15 Hence, cross-sectional area at distancexfrom larger endA = d 24= 4(d1 kx)2 STRESS at this section, =PA =4P (d1 kx)2 strain = = E=4P E(d1 kx)2 Extension of elementary lengthdx= dx=4P dx E(d1 kx)2 Total extention of the bar= =4P El 0dx(d1 kx)2=4P E[(dl kx) 1 1 k]lo=4P Ek[1d1 kx]l0=4P Ek{1d1 kl 1d1}butk=d1 d2l =4P E(d1 d2)[1d1 d1+d2 1d1]=4Pl E(d1 d2)(1d2 1d1)=4Pl E(d1 d2) d1 d2d1d2 =4Pl Ed1d2If both the diameters are equal =4Pl :A round steel rod of different cross-sections is loaded as shown in Fig. Findthe maximum STRESS induced in the rod and its deformations. TakeE=210 and STRAINS 17 Extension of Tapered Rectangular StripPPbtxxxxdxaFigure any sectionx xdistantxfrom the bigger endWidth of the section=t Area of the section=Pt(a kx) Extension of an elemental lengthdx=Pdxt(a kx)E Total extension of the rod= =PtEl 0dxa kx= PtE 1k loge[(a kx)]l0= PtE[loge(a kl) logea]=PtkE[aa k]Butk=a bl = (a b)logeab18 Strength of :A straight bar of steel rectangular in section is 3 m long and of thickness of 12 width of rod varies uniformly from 110 mm or one end to 35 mm at the other end.

9 If the rod issubjected to an axial load (tensile) of 25 kN, find the extension of the rod. TakeE=200000 of the rod, =PlEt(a b)logeabP=25000 N,l=3000 mm,t=12 mma=110 mm,ab=35 mm &E=200000 N/mm2 =25000 30002 105 12(110 35)loge11035=25000 30002 105 12 75 :A rigid barABis attached to two vertical rods as shown in Fig. is horizontalbefore the load is applied. Determine the vertical movement ofPif it is of magnitude 60 aluminiumL = 3 mA = 500 mm2E = 75 GPaFor steelL = 4 mA = 300 mm2E = 210 GPaCB60 mFigure and STRAINS 19 SOLUTION:ForAl, MB=0,6 PAl= 60 PA= 606=25 kN=25000 N Al= 3000500 75000=2 mmFor steel MA=0 givesPst= 60 Pst= 606=35 kN ST=35000 4 1000300 200000= mm2 mmYAC1A1C2B2B1 BFigure from similar trianglesA1C1C2andA1,B1,B2YA1C1=B1B2A1B1 ; 26 Y= mmNow vertical movement ofP=CC2=CC1+Y=2+ mmAns20 Strength of MaterialsBar of Uniform StrengthAs we have seen earlier that the STRESS due to self weight is not constant.

10 It increases with theincrease of distance from the lower wish to find the shape of the bar of which the self weight is considered and is having uniformstress on all sections when subjected to an axialP. Figure shows such a bar of uniform stressin which the area increases from the lower end to the upper A1 Area A1 (A+dA) A+wAdxArea A2P(a)(b)LFigure the length of bar, having areaA1, and areaA2be cross-sectional areas of the bar at topand bottom, the specific weight of the bar material ( weight per unit volume of the bar).The forces acting on the elementary stripe are:i) Weight of the strip acting downward and is equal tow volume of ) Force on sectionABdue to uniform STRESS is equal to A. This is acting of elementary ) Force on sectionCDdue to uniform( )is equal to (A+dA). This is acting force acting upwards=Total force acting downwards (A+dA)= A+wA dx A+ dA= A+ dx22 Strength of MaterialsUsing equation,A1=A2ewL A1= 10 4A1= :A steel rod of 25 mm dia passes centrally through a copper tube of 30 mm insidediameter and 40 mm outside diameter.