CHAPTER 1 SIMPLE STRESS AND STRAIN 1.1.Load In any engineering structure or mechanism the individual components will be subjected to external forces arising from the service conditions or environment in which the component works.If the component or member is in equilibrium,the resultant of the external forces will be zero but,nevertheless,they together place a load on the member which tends to deform that member and which must be reacted by internal forces which are set up within the material. If a cylindrical bar is subjected to a direct pull or push along its axis as shown in Fig.1.1, then it is said to be subjected to tension or compression.Typical examples of tension are the forces present in towing ropes or lifting hoists,whilst compression occurs in the legs of your chair as you sit on it or in the support pillars of buildings. Area A Stress:P/A P Tension Compression Fig.1.1.Types of direct stress. In the SI system of units load is measured in newtons,although a single newton,in engineering terms,is a very small load.In most engineering applications,therefore,loads appear in SI multiples,i.e.kilonewtons (kN)or meganewtons(MN). There are a number of different ways in which load can be applied to a member.Typical loading types are: (a)Static or dead loads,i.e.non-fluctuating loads,generally caused by gravity effects. (b)Live loads,as produced by,for example,lorries crossing a bridge. (c)Impact or shock loads caused by sudden blows. (d)Fatigue,fluctuating or alternating loads,the magnitude and sign of the load changing with time
CHAPTER 1 SIMPLE STRESS AND STRAIN 1.1. Load In any engineering structure or mechanism the individual components will be subjected to external forces arising from the service conditions or environment in which the component works. If the component or member is in equilibrium, the resultant of the external forces will be zero but, nevertheless, they together place a load on the member which tends to deform that member and which must be reacted by internal forces which are set up within the material. If a cylindrical bar is subjected to a direct pull or push along its axis as shown in Fig. 1.1, then it is said to be subjected to tension or compression. Typical examples of tension are the forces present in towing ropes or lifting hoists, whilst compression occurs in the legs of your chair as you sit on it or in the support pillars of buildings. ,Are0 A Tension Compression Fig. 1.1. Types of direct stress. In the SI system of units load is measured in newtons, although a single newton, in engineering terms, is a very small load. In most engineering applications, therefore, loads appear in SI multiples, i.e. kilonewtons (kN) or meganewtons (MN). There are a number of different ways in which load can be applied to a member. Typical loading types are: (a) Static or dead loads, i.e. non-fluctuating loads, generally caused by gravity effects. (b) Liue loads, as produced by, for example, lorries crossing a bridge. (c) Impact or shock loads caused by sudden blows. (d) Fatigue,fluctuating or alternating loads, the magnitude and sign of the load changing with time. 1
2 Mechanics of Materials §1.2 1.2.Direct or normal stress ( It has been noted above that external force applied to a body in equilibrium is reacted by internal forces set up within the material.If,therefore,a bar is subjected to a uniform tension or compression,i.e.a direct force,which is uniformly or equally applied across the cross- section,then the internal forces set up are also distributed uniformly and the bar is said to be subjected to a uniform direct or normal stress,the stress being defined as load P stress (o)= 一 area A Stress a may thus be compressive or tensile depending on the nature of the load and will be measured in units of newtons per square metre(N/m2)or multiples of this. In some cases the loading situation is such that the stress will vary across any given section, and in such cases the stress at any point is given by the limiting value of oP/6A as 6A tends to zero. 1.3.Direct strain (e) If a bar is subjected to a direct load,and hence a stress,the bar will change in length.If the bar has an original length L and changes in length by an amount oL,the strain produced is defined as follows: strain (e)-change in lengthL original length L Strain is thus a measure of the deformation of the material and is non-dimensional,i.e.it has no units;it is simply a ratio of two quantities with the same unit(Fig.1.2). Strain e=8L/L ◇ Fig.1.2. Since,in practice,the extensions of materials under load are very small,it is often convenient to measure the strains in the form of strain x 10-6,i.e.microstrain,when the symbol used becomes ue. Alternatively,strain can be expressed as a percentage strain L i.e. strain (e)= ×100% 1.4.Sign convention for direct stress and strain Tensile stresses and strains are considered POSITIVE in sense producing an increase in length.Compressive stresses and strains are considered NEGATIVE in sense producing a decrease in length
2 Mechanics of Materials $1.2 1.2. Direct or normal stress (a) It has been noted above that external force applied to a body in equilibrium is reacted by internal forces set up within the material. If, therefore, a bar is subjected to a uniform tension or compression, i.e. a direct force, which is uniformly or equally applied across the crosssection, then the internal forces set up are also distributed uniformly and the bar is said to be subjected to a uniform direct or normal stress, the stress being defined as load P stress (a) = - = - area A Stress CT may thus be compressive or tensile depending on the nature of the load and will be measured in units of newtons per square metre (N/mZ) or multiples of this. In some cases the loading situation is such that the stress will vary across any given section, and in such cases the stress at any point is given by the limiting value of 6P/6A as 6A tends to zero. 1.3. Direct strain (E) If a bar is subjected to a direct load, and hence a stress, the bar will change in length. If the bar has an original length L and changes in length by an amount 6L, the strain produced is defined as follows: change in length 6L strain (E) = =- original length L Strain is thus a measure of the deformation of the material and is non-dimensional, Le. it has no units; it is simply a ratio of two quantities with the same unit (Fig. 1.2). Strain C=GL/L Fig. 1.2. Since, in practice, the extensions of materials under load are very small, it is often convenient to measure the strains in the form of strain x i.e. microstrain, when the symbol used becomes /ALE. Alternatively, strain can be expressed as a percentage strain 6L L i.e. strain (E) = - x 100% 1.4. Sign convention for direct stress and strain Tensile stresses and strains are considered POSITIVE in sense producing an increase in length. Compressive stresses and strains are considered NEGATIVE in sense producing a decrease in length
s1.5 Simple Stress and Strain 3 1.5.Elastic materials-Hooke's law A material is said to be elastic if it returns to its original,unloaded dimensions when load is removed.A particular form of elasticity which applies to a large range of engineering materials,at least over part of their load range,produces deformations which are proportional to the loads producing them.Since loads are proportional to the stresses they produce and deformations are proportional to the strains,this also implies that,whilst materials are elastic,stress is proportional to strain.Hooke's law,in its simplest form*, therefore states that stress (o)oc strain (8) stress i.e. constant* strain It will be seen in later sections that this law is obeyed within certain limits by most ferrous alloys and it can even be assumed to apply to other engineering materials such as concrete, timber and non-ferrous alloys with reasonable accuracy.Whilst a material is elastic the deformation produced by any load will be completely recovered when the load is removed; there is no permanent deformation. Other classifications of materials with which the reader should be acquainted are as follows: A material which has a uniform structure throughout without any flaws or discontinuities is termed a homogeneous material.Non-homogeneous or inhomogeneous materials such as concrete and poor-quality cast iron will thus have a structure which varies from point to point depending on its constituents and the presence of casting flaws or impurities. If a material exhibits uniform properties throughout in all directions it is said to be isotropic;conversely one which does not exhibit this uniform behaviour is said to be non- isotropic or anisotropic. An orthotropic material is one which has different properties in different planes.A typical example of such a material is wood,although some composites which contain systematically orientated "inhomogeneities"may also be considered to fall into this category. 1.6.Modulus of elasticity-Young's modulus Within the elastic limits of materials,i.e.within the limits in which Hooke's law applies,it has been shown that stress constant strain This constant is given the symbol E and termed the modulus ofelasticity or Young's modulus. Thus stress o E= strain (1.1) P L PL FA÷D=A6L (1.2) Readers should be warned that in more complex stress cases this simple form of Hooke's law will not apply and mis-application could prove dangerous;see $14.1,page 361
$1.5 Simple Stress and Strain 3 1.5. Elastic materials - Hooke’s law A material is said to be elastic if it returns to its original, unloaded dimensions when load is removed. A particular form of elasticity which applies to a large range of engineering materials, at least over part of their load range, produces deformations which are proportional to the loads producing them. Since loads are proportional to the stresses they produce and deformations are proportional to the strains, this also implies that, whilst materials are elastic, stress is proportional to strain. Hooke’s law, in its simplest form*, therefore states that stress (a) a strain (E) i.e. stress strain -- - constant* It will be seen in later sections that this law is obeyed within certain limits by most ferrous alloys and it can even be assumed to apply to other engineering materials such as concrete, timber and non-ferrous alloys with reasonable accuracy. Whilst a material is elastic the deformation produced by any load will be completely recovered when the load is removed; there is no permanent deformation. Other classifications of materials with which the reader should be acquainted are as follows: A material which has a uniform structure throughout without any flaws or discontinuities is termed a homogeneous material. Non-homogeneous or inhomogeneous materials such as concrete and poor-quality cast iron will thus have a structure which varies from point to point depending on its constituents and the presence of casting flaws or impurities. If a material exhibits uniform properties throughout in all directions it is said to be isotropic; conversely one which does not exhibit this uniform behaviour is said to be nonisotropic or anisotropic. An orthotropic material is one which has different properties in different planes. A typical example of such a material is wood, although some composites which contain systematically orientated “inhomogeneities” may also be considered to fall into this category. 1.6. Modulus of elasticity - Young’s modulus Within the elastic limits of materials, i.e. within the limits in which Hooke’s law applies, it has been shown that stress strain -- - constant This constant is given the symbol E and termed the modulus of elasticity or Young’s modulus. Thus stress 0 strain E E=-- _- P 6L PL A’ L ASL =--I-- -- * Readers should be warned that in more complex stress cases this simple form of Hooke’s law will not apply and misapplication could prove dangerous; see 814.1, page 361
4 Mechanics of Materials s1.7 Young's modulus E is generally assumed to be the same in tension or compression and for most engineering materials has a high numerical value.Typically,E=200 x 109 N/m2 for steel,so that it will be observed from (1.1)that strains are normally very small since 0 (1.3) In most common engineering applications strains do not often exceed 0.003 or 0.3%so that the assumption used later in the text that deformations are small in relation to original dimensions is generally well founded. The actual value of Young's modulus for any material is normally determined by carrying out a standard tensile test on a specimen of the material as described below. 1.7.Tensile test In order to compare the strengths of various materials it is necessary to carry out some standard form of test to establish their relative properties.One such test is the standard tensile test in which a circular bar of uniform cross-section is subjected to a gradually increasing tensile load until failure occurs.Measurements of the change in length of a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers and a graph of load against extension or stress against strain is produced as shown in Fig.1.3; this shows a typical result for a test on a mild (low carbon)steel bar;other materials will exhibit different graphs but of a similar general form see Figs 1.5 to 1.7. Elastic Partially piastic Test E specimen Circular cross-section 6 Gauge length Extension or strain Fig.1.3.Typical tensile test curve for mild steel. For the first part of the test it will be observed that Hooke's law is obeyed,i.e.the material behaves elastically and stress is proportional to strain,giving the straight-line graph indicated.Some point A is eventually reached,however,when the linear nature of the graph ceases and this point is termed the limit of proportionality. For a short period beyond this point the material may still be elastic in the sense that deformations are completely recovered when load is removed (i.e.strain returns to zero)but
4 Mechanics of Materials $1.7 Young’s modulus E is generally assumed to be the same in tension or compression and for most engineering materials has a high numerical value. Typically, E = 200 x lo9 N/m2 for steel, so that it will be observed from (1.1) that strains are normally very small since 0 E=- E In most common engineering applications strains do not often exceed 0.003 or 0.3 % so that the assumption used later in the text that deformations are small in relation to original dimensions is generally well founded. The actual value of Young’s modulus for any material is normally determined by carrying out a standard tensile test on a specimen of the material as described below. 1.7. Tensile test In order to compare the strengths of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross-section is subjected to a gradually increasing tensile load until failure occurs. Measurements of the change in length of a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers and a graph of load against extension or stress against strain is produced as shown in Fig. 1.3; this shows a typical result for a test on a mild (low carbon) steel bar; other materials will exhibit different graphs but of a similar general form see Figs 1.5 to 1.7. Elastic Partially plastic tP Extension or strain Fig. 1.3. Typical tensile test curve for mild steel. For the first part of the test it will be observed that Hooke’s law is obeyed, Le. the material behaves elastically and stress is proportional to strain, giving the straight-line graph indicated. Some point A is eventually reached, however, when the linear nature of the graph ceases and this point is termed the limit of proportionality. For a short period beyond this point the material may still be elastic in the sense that deformations are completely recovered when load is removed (i.e. strain returns to zero) but
§1.7 Simple Stress and Strain 5 Hooke's law does not apply.The limiting point B for this condition is termed the elastic limit. For most practical purposes it can often be assumed that points A and B are coincident. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will thus be some permanent deformation or permanent set when load is removed. After the points C,termed the upper yield point,and D,the lower yield point,relatively rapid increases in strain occur without correspondingly high increases in load or stress.The graph thus becomes much more shallow and covers a much greater portion of the strain axis than does the elastic range of the material.The capacity of a material to allow these large plastic deformations is a measure of the so-called ductility of the material,and this will be discussed in greater detail below. For certain materials,for example,high carbon steels and non-ferrous metals,it is not possible to detect any difference between the upper and lower yield points and in some cases no yield point exists at all.In such cases a proof stress is used to indicate the onset of plastic strain or as a comparison of the relative properties with another similar material.This involves a measure of the permanent deformation produced by a loading cycle;the 0.1% proof stress,for example,is that stress which,when removed,produces a permanent strain or "set"of 0.1%of the original gauge length-see Fig.1.4(a). 6 1b) Proof stress 6 Stron e Stra1n∈ 01% Permanent 'set' Fig.1.4.(a)Determination of 0.1%proof stress. Fig.1.4.(b)Permanent deformation or "set"after straining beyond the yield point. The 0.1proof stress value may be determined from the tensile test curve for the material in question as follows: Mark the point P on the strain axis which is equivalent to 0.1%strain.From P draw a line parallel with the initial straight line portion of the tensile test curve to cut the curve in N.The stress corresponding to Nis then the 0.1%proofstress.A material is considered to satisfy its specification if the permanent set is no more than 0.1%after the proof stress has been applied for 15 seconds and removed. Beyond the yield point some increase in load is required to take the strain to point E on the graph.Between D and E the material is said to be in the elastic-plastic state,some of the section remaining elastic and hence contributing to recovery of the original dimensions if load is removed,the remainder being plastic.Beyond E the cross-sectional area of the bar
51.7 Simple Stress and Strain 5 Hooke’s law does not apply. The limiting point B for this condition is termed the elastic limit. For most practical purposes it can often be assumed that points A and B are coincident. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will thus be some permanent deformation or permanent set when load is removed. After the points C, termed the upper yield point, and D, the lower yield point, relatively rapid increases in strain occur without correspondingly high increases in load or stress. The graph thus becomes much more shallow and covers a much greater portion of the strain axis than does the elastic range of the material. The capacity of a material to allow these large plastic deformations is a measure of the so-called ductility of the material, and this will be discussed in greater detail below. For certain materials, for example, high carbon steels and non-ferrous metals, it is not possible to detect any difference between the upper and lower yield points and in some cases no yield point exists at all. In such cases a proof stress is used to indicate the onset of plastic strain or as a comparison of the relative properties with another similar material. This involves a measure of the permanent deformation produced by a loading cycle; the 0.1 % proof stress, for example, is that stress which, when removed, produces a permanent strain or “set” of 0.1 % of the original gauge length-see Fig. 1.4(a). b ”7 E i5 c r-=F I’ \ I 0 I % 1 Permanent ‘set’ Fig. 1.4. (a) Determination of 0.1 % proof stress. Fig. 1.4. (b) Permanent deformation or “set” after straining beyond the yield point. The 0.1 % proof stress value may be determined from the tensile test curve for the material in question as follows: Mark the point P on the strain axis which is equivalent to 0.1 % strain. From P draw a line parallel with the initial straight line portion of the tensile test curve to cut the curve in N. The stress corresponding to Nis then the 0.1 %proof stress. A material is considered to satisfy its specification if the permanent set is no more than 0.1 %after the proof stress has been applied for 15 seconds and removed. Beyond the yield point some increase in load is required to take the strain to point E on the graph. Between D and E the material is said to be in the elastic-plastic state, some of the section remaining elastic and hence contributing to recovery of the original dimensions if load is removed, the remainder being plastic. Beyond E the cross-sectional area of the bar
6 Mechanics of Materials $1.7 begins to reduce rapidly over a relatively small length of the bar and the bar is said to neck. This necking takes place whilst the load reduces,and fracture of the bar finally occurs at point F. The nominal stress at failure,termed the maximum or ultimate tensile stress,is given by the load at E divided by the original cross-sectional area of the bar.(This is also known as the tensile strength of the material of the bar.)Owing to the large reduction in area produced by the necking process the actual stress at fracture is often greater than the above value.Since, however,designers are interested in maximum loads which can be carried by the complete cross-section,the stress at fracture is seldom of any practical value. If load is removed from the test specimen after the yield point C has been passed,e.g.to some position S,Fig.1.4(b),the unloading line STcan,for most practical purposes,be taken to be linear.Thus,despite the fact that loading to S comprises both elastic (OC)and partially plastic (CS)portions,the unloading procedure is totally elastic.A second load cycle, commencing with the permanent elongation associated with the strain OT,would then follow the lineT'S and continue along the previous curve to failure at F.It will be observed,however, that the repeated load cycle has the effect of increasing the elastic range of the material,i.e. raising the effective yield point from C to S,while the tensile strength is unaltered.The procedure could be repeated along the line Po,etc.,and the material is said to have been work hardened. In fact,careful observation shows that the material will no longer exhibit true elasticity since the unloading and reloading lines will form a small hysteresis loop,neither being precisely linear.Repeated loading and unloading will produce a yield point approaching the ultimate stress value but the elongation or strain to failure will be much reduced. Typical stress-strain curves resulting from tensile tests on other engineering materials are shown in Figs 1.5 to 1.7. Nickel chrome steel 1500 Medium carbon steel-heat treoted 1200 Cold rolled steel 900 Medum corbon steel- onnealed Hard bronze -Low corbon steel Soft bross 00 20 30 40 50 Stroin, Fig.1.5.Tensile test curves for various metals
6 Mechanics of Materials 01.7 begins to reduce rapidly over a relatively small length of the bar and the bar is said to neck. This necking takes place whilst the load reduces, and fracture of the bar finally occurs at point F. The nominal stress at failure, termed the maximum or ultimate tensile stress, is given by the load at E divided by the original cross-sectional area of the bar. (This is also known as the tensile strength of the material of the bar.) Owing to the large reduction in area produced by the necking process the actual stress at fracture is often greater than the above value. Since, however, designers are interested in maximum loads which can be carried by the complete cross-section, the stress at fracture is seldom of any practical value. If load is removed from the test specimen after the yield point C has been passed, e.g. to some position S, Fig. 1.4(b), the unloading line STcan, for most practical purposes, be taken to be linear. Thus, despite the fact that loading to S comprises both elastic (OC) and partially plastic (CS) portions, the unloading procedure is totally elastic. A second load cycle, commencing with the permanent elongation associated with the strain OT, would then follow the line TS and continue along the previous curve to failure at F. It will be observed, however, that the repeated load cycle has the effect of increasing the elastic range of the material, i.e. raising the effective yield point from C to S, while the tensile strength is unaltered. The procedure could be repeated along the line PQ, etc., and the material is said to have been work hardened. In fact, careful observation shows that the material will no longer exhibit true elasticity since the unloading and reloading lines will form a small hysteresis loop, neither being precisely linear. Repeated loading and unloading will produce a yield point approaching the ultimate stress value but the elongation or strain to failure will be much reduced. Typical stress-strain curves resulting from tensile tests on other engineering materials are shown in Figs 1.5 to 1.7. 1500tr /Nickel chrome steel u- Medium carbon steel - &i ?! 600 Y I I I I I 0 0 20 30 40 50 Strain. % Fig. 1.5. Tensile test curves for various metals
$1.7 Simple Stress and Strain 7 600 Mild steel 450 E Copper 300 Alumnium 150 2 6 Strain, Fig.1.6.Typical stress-strain curves for hard drawn wire materials-note large reduction in strain values from those of Fig.1.5. 00 Gloss reinforced polycarbonate -Nylon-dry type 60 Unreinforced polycorbonate 40 Nyion-wet type 20 20 30 40 50 Stroin, Fig.1.7.Typical tension test results for various types of nylon and polycarbonate. After completing the standard tensile test it is usually necessary to refer to some"British Standard Specification"or"Code of Practice"to ensure that the material tested satisfies the requirements,for example: BS4360 British Standard Specification for Weldable Structural Steels BS970 British Standard Specification for Wrought Steels. BS153 British Standard Specification for Steel Girder Bridges. BS449 British Standard Specification for the use of Structural Steel in Building,etc
$1.7 Simple Stress and Strain 7 Strain, % Fig. 1.6. Typical stressstrain curves for hard drawn wire material-note large reduction in strain values from those of Fig. 1.5. Glass remforced polycarbonate eo ? ’ I I / ,Unreinforced pdycnrbonote c, Y I I I I I 0 10 20 30 40 50 Strain, % Fig. 1.7. Typical tension test results for various types of nylon and polycarbonate. After completing the standard tensile test it is usually necessary to refer to some “British Standard Specification” or “Code of Practice” to ensure that the material tested satisfies the requirements, for example: BS 4360 BS 970 BS 153 BS 449 British Standard Specification for Weldable Structural Steels. British Standard Specification for Wrought Steels. British Standard Specification for Steel Girder Bridges. British Standard Specification for the use of Structural Steel in Building, etc
8 Mechanics of Materials §1.8 1.8.Ductile materials It has been observed above that the partially plastic range of the graph of Fig.1.3 covers a much wider part of the strain axis than does the elastic range.Thus the extension of the material over this range is considerably in excess of that associated with elastic loading.The capacity of a material to allow these large extensions,i.e.the ability to be drawn out plastically,is termed its ductility.Materials with high ductility are termed ductile materials, members with low ductility are termed brittle materials.A quantitative value of the ductility is obtained by measurements of the percentage elongation or percentage reduction in area,both being defined below. Percentage elongation=increase in gauge length to fracture ×100 original gauge length Perntage reduction indctinn rosciorea f ecked portio -×100 original area The latter value,being independent of any selected gauge length,is generally taken to be the more useful measure of ductility for reference purposes. A property closely related to ductility is malleability,which defines a material's ability to be hammered out into thin sheets.A typical example of a malleable material is lead.This is used extensively in the plumbing trade where it is hammered or beaten into corners or joints to provide a weatherproof seal.Malleability thus represents the ability of a material to allow permanent extensions in all lateral directions under compressive loadings. 1.9.Brittle materials A brittle material is one which exhibits relatively small extensions to fracture so that the partially plastic region of the tensile test graph is much reduced(Fig.1.8).Whilst Fig.1.3 referred to a low carbon steel,Fig.1.8 could well refer to a much higher strength steel with a higher carbon content.There is little or no necking at fracture for brittle materials. Fig.1.8.Typical tensile test curve for a brittle material. Typical variations of mechanical properties of steel with carbon content are shown in Fig.1.9
8 Mechanics of Materials 51.8 1.8. Ductile materials It has been observed above that the partially plastic range of the graph of Fig. 1.3 covers a much wider part of the strain axis than does the elastic range. Thus the extension of the material over this range is considerably in excess of that associated with elastic loading. The capacity of a material to allow these large extensions, i.e. the ability to be drawn out plastically, is termed its ductility. Materials with high ductility are termed ductile materials, members with low ductility are termed brittle materials. A quantitative value of the ductility is obtained by measurements of the percentage elongation or percentage reduction in area, both being defined below. increase in gauge length to fracture original gauge length Percentage elongation = x loo reduction in cross-sectional area of necked portion original area Percentage reduction in area = x 100 The latter value, being independent of any selected gauge length, is generally taken to be the more useful measure of ductility for reference purposes. A property closely related to ductility is malleability, which defines a material's ability to be hammered out into thin sheets. A typical example of a malleable material is lead. This is used extensively in the plumbing trade where it is hammered or beaten into corners or joints to provide a weatherproof seal. Malleability thus represents the ability of a material to allow permanent extensions in all lateral directions under compressive loadings. 1.9. Brittle materials A brittle material is one which exhibits relatively small extensions to fracture so that the partially plastic region of the tensile test graph is much reduced (Fig. 1.8). Whilst Fig. 1.3 referred to a low carbon steel, Fig. 1.8 could well refer to a much higher strength steel with a higher carbon content. There is little or no necking at fracture for brittle materials. E Fig. 1.8. Typical tensile test curve for a brittle material. Typical variations of mechanical properties of steel with carbon content are shown in Fig. 1.9
§1.10 Simple Stress and Strain 9 190 Tensile strength 60 Elostic limit 20 Elongotion 30 Reduction in orea 0.2 04 060.8 C Fig.1.9.Variation of mechanical properties of steel with carbon content. 1.10.Poisson's ratio Consider the rectangular bar of Fig.1.10 subjected to a tensile load.Under the action of this load the bar will increase in length by an amount oL giving a longitudinal strain in the bar of L= L Fig.1.10. The bar will also exhibit,however,a reduction in dimensions laterally,i.e.its breadth and depth will both reduce.The associated lateral strains will both be equal,will be of opposite sense to the longitudinal strain,and will be given by 6bδd 8at=-6=-d Provided the load on the material is retained within the elastic range the ratio of the lateral and longitudinal strains will always be constant.This ratio is termed Poisson's ratio. i.e. Poisson's ratio(v)=】 lateral strain (-6d/d) longitudinal strainL/L (1.4) The negative sign of the lateral strain is normally ignored to leave Poisson's ratio simply as
$1.10 Simple Stress and Strain 9 i90 I I I 02 04 06 08 10 Yo c Fig. 1.9. Variation of mechanical properties of steel with carbon content. 1.10. Poisson’s ratio Consider the rectangular bar of Fig. 1.10 subjected to a tensile load. Under the action of this load the bar will increase in length by an amount 6 L giving a longitudinal strain in the bar of 6L EL= - L Fig. 1.10. The bar will also exhibit, however, a reduction in dimensions laterally, i.e. its breadth and depth will both reduce. The associated lateral strains will both be equal, will be of opposite sense to the longitudinal strain, and will be given by 6b 6d b d Provided the load on the material is retained within the elastic range the ratio of the lateral &l,t = -- - -- and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio. lateral strain (-6d/d) (1.4) 1.e. Poisson’s ratio (v) = - longitudinal strain 6LIL The negative sign of the lateral strain is normally ignored to leave Poisson’s ratio simply as
10 Mechanics of Materials §1.11 a ratio of strain magnitudes.It must be remembered,however,that the longitudinal strain induces a lateral strain of opposite sign,e.g.tensile longitudinal strain induces compressive lateral strain. For most engineering materials the value of v lies between 0.25 and 0.33. Since longitudinal stress o longitudinal strain=Young's modulus E (1.4a) Hence lateral strain y E (1.4b) 1.11.Application of Poisson's ratio to a two-dimensional stress system A two-dimensional stress system is one in which all the stresses lie within one plane such as the X-Y plane.From the work of $1.10 it will be seen that if a material is subjected to a tensile stress o on one axis producing a strain a/E and hence an extension on that axis,it will be subjected simultaneously to a lateral strain of v times o/E on any axis at right angles.This lateral strain will be compressive and will result in a compression or reduction of length on this axis. Consider,therefore,an element of material subjected to two stresses at right angles to each other and let both stresses,o and o,,be considered tensile,see Fig.1.11. Fig.1.11.Simple two-dimensional system of direct stresses. The following strains will be produced: (a)in the X direction resulting from ax =ox/E, (b)in the Y direction resulting from oy=o,/E. (c)in the X direction resulting from a=-v(o/E). (d)in the Y direction resulting from ox=-v(/E) strains (c)and (d)being the so-called Poisson's ratio strain,opposite in sign to the applied strains,i.e.compressive. The total strain in the X direction will therefore be given by: sv(ax-va,) E-v Ex= (1.5)
10 Mechanics of Materials $1.11 a ratio of strain magnitudes. It must be remembered, however, that the longitudinal strain induces a lateral strain of opposite sign, e.g. tensile longitudinal strain induces compressive lateral strain. Since For most engineering materials the value of v lies between 0.25 and 0.33. longitudinal stress a Young’s modulus E (1.4a) longitudinal strain = - -- Hence (1.4b) d lateral strain = v - E 1.11. Application of Poisson’s ratio to a two-dimensional stress system A two-dimensional stress system is one in which all the stresses lie within one plane such as the X-Y plane. From the work of $1.10 it will be seen that if a material is subjected to a tensile stress a on one axis producing a strain u/E and hence an extension on that axis, it will be subjected simultaneously to a lateral strain of v times a/E on any axis at right angles. This lateral strain will be compressive and will result in a compression or reduction of length on this axis. Consider, therefore, an element of material subjected to two stresses at right angles to each other and let both stresses, ux and cy, be considered tensile, see Fig. 1.11. Fig. 1.11. Simple twodimensional system of direct stresses. The following strains will be produced (a) in the X direction resulting from ax = a,/E, (b) in the Y direction resulting from cy = a,/E. (c) in the X direction resulting from 0, = - v(a,/E), (d) in the Y direction resulting from ax = - v(a,/E). strains (c) and (d) being the so-called Poisson’s ratio strain, opposite in sign to the applied strains, i.e. compressive. The total strain in the X direction will therefore be given by: 6” 1 & = - - voy = -(ax - va,) “E EE (1.5)