STRESSSTRAIN RELATIONS


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1 STRSSSTRAIN RLATIONS Strain Strain is related to change in dimensions and shape of a material. The most elementar definition of strain is when the deformation is along one ais: change in length strain (. original length When a material is stretched, the change in length and the strain are positive. When it is compressed, the change in length and strain are negative. This conforms with the signs of the stresses which would accompan these strains, tensile stresses being positive and compressive stresses negative. This definition refers to what are termed normal strains, which change the dimensions of a material but not its shape; in other words, angles do not change. In general, there are normal strains along three mutuall perpendicular aes. B contrast, strains which involve no length changes but which do change angles are known as shear strains. The measure of strain is the change in the angle. Normal and shear strains are illustrated below.
2 Strained shapes are the broken lines. The shear strain is the total change in the angle φ. For normal strains the usual smbol is. The same sstem of subscripts is used as for stresses. With respect to (,, aes we have strains, and, with, as with stresses, the first subscript giving the normal to the plane which is being displaced, and the second the direction of the displacement. Since the subscripts are alwas repeated, we sometimes just use a single subscript. Conventionall, the smbol for shear stress is γ and the same subscripting sstem gives rise to strains such as γ, γ, and γ. Other quantities that are associated with strain are displacements. These are simpl the distance moved b an point of material. Usuall, the displacements in the (,, directions are denoted respectivel b (u,v,w. When a strain is present, the displacements must var from point to point. It is eas to see that normal strains are given b u v (2. w This is a more rigorous definition of normal strain than ( above. In this module, strains are alwas small. This enables us to use simple theor. In this contet, small means that the square of the strain is negligible in comparison to the strain itself, e.g. 2 <<. In general, the action of stresses will cause material to change in volume. Onl the normal stresses are associated with volume strain. We define volumetric strain e as: change in volume e (3. original volume The volumetric strain is simpl related to the normal strains. Consider the rectangular solid illustrated. The original shape is on the left and the deformed shape on the right. The original volume is given b Vo. For the deformed bod on the right, (+ (+ (+ 2
3 the volume is given b V ( + ( + + ( ( We now recall that the strains are small, so that we can neglect all the high order terms so that V ( V ( Now, we can use the definition of volumetric strain (3 which becomes e V V + + (4. V This is the simple relationship we seek, which applies at small strains onl. Stressstrain relations Materials undergo strain when the are subject to stress. The relationship between stress and strain is different for different materials, and can be appreciated b plotting stress against strain. Suppose a material specimen is subject to a unaial tensile load. It will deform in a manner characteristic of the material. amples of possible behaviour are shown here. In this module, we assume that materials obe the first tpe of behaviour, linear elastic. + 3
4 This means that stresses and strains are assumed to be related b Hooke s Law. In its simplest form, Hooke s Law can be stated as: stress constant (5 strain where the constant is known as an elastic modulus or simpl a modulus. Normal strains and stresses For normal stress along the direction onl, (5 becomes σ or σ (6 where the constant is known as Young s modulus. ample Determine the total elongation of the member ABC if AB is steel ( 2Gpa and BC is aluminium ( 7 Gpa. ample A solid of revolution is of radius R at one end and tapers so that its radius r at an aial distance from the end is r( Re α. It is subject to an aial force F. If its initial length is L, and its modulus, what is its stretched length? 4
5 Suppose the stress is tensile, and the specimen of material is stretched along. Then, it will get thinner across the direction of stretching, in the and directions, as in (a. The strains will be negative along and. The are related to the strain along b a positive constant ν: ν. ν is known as Poisson s ratio. Using (6 we can see that the stress along produces transverse strains: σ ν. This is true when we onl have stress along. In general, we have stress along and also. These other stresses will cause transverse strains also. The transverse strains along the same ais arising from different stresses simpl add together. When three stresses act, the general result is ( ( σ ν σ + σ σ ν σ + σ σ ν σ + σ ( ( ( ( (7. This is threedimensional Hooke s Law for normal stresses. 5
6 Shear strains and stresses These relationships are less comple than those for normal strains. quation (5 still applies, and each strain component is proportional to the corresponding stress component. We can write three equations similar to (6: γ γ γ G G G τ τ τ (8. The constant G is the shear modulus. quations (7 and (8 constitute the full statement of Hooke' Law in three dimensions. Volumetric strain and hdrostatic stress. These two quantities are linearl related. The version of equation (5 is e σ (9 K where e is the volumetric strain, σ the hdrostatic stress and the constant K the bulk modulus. Summar The quantities, ν, G and K introduced above are elastic constants. The are not all independent. The first two are sufficient to full define an linearl elastic material. Plane strain and plane stress quations (7 and (8 give the full threedimensional form of Hooke s Law. Man problems can be approimated b a twodimensional approach, b assuming that stress or strain varies onl in a single plane, usuall the  plane. There are two was to create such an approimation: b assuming that the force normal to the plane is ero the plane stress assumption; or b assuming that the strain normal to the plane is ero the plane strain assumption. The two approimations We shall outline the plane stress tpe of analsis first. We simpl take equations (7 and put σ. Then we have ( σ νσ ( σ νσ (. ν ( σ + σ 6
7 Finall, the ero force in the direction implies that γ γ. Therefore onl the first of equations (8 is nontrivial: γ τ (. G quations ( and ( define the plane stress formulation. The plane strain analsis is a little more comple. Putting in the last equation (7 gives an epression for σ : σ ν σ + σ. ( This can be substituted into the first two equations (7 to give epressions involving σ and σ onl. The first equation becomes ( σ ν( σ + ν( σ + σ 2 ( σ ( ν ν( σ ( + ν (2. 2 ν ν σ σ ν We can define a plane strain modulus, such that, and a quantit ν such 2 ν ν that ν. Then we can write the above equation as ν ( σ ν σ (3 which resembles the first of equations (. Similarl, for the direction ( σ ν σ (4. We complete the formulation with what we alread know: (5 and note that equation ( still applies in plane strain. quations ( and (3 (5 complete the plane strain relations. 7
8 The states of plane stress and plane strain, for material stretched along, are illustrated. ample The illustrated device stretches a thin sheet of material while holding the width constant. The sheet is mm thick, and is linearl elastic with Poisson s ratio ν.3. if the pulling force along is 5 kn, what are the forces in the members AC and BD? 8
9 Thermal strains A rise in temperature causes materials to epand. In simple cases, the increase in length is simpl proportional to the temperature rise, via a constant of proportionalit α, the coefficient of epansion. The temperature rise multiplied b α gives the increase in strain caused b the temperature the thermal strain. For an initial temperature T o and an operating temperature T, the thermal strain is α(t T o. These strains are simpl added to the strains caused b the stresses. For varing temperature, equations (7 are revised as ( σ ν( σ + σ ( σ ν( σ + σ + α(t T + α(t T ( σ ν( σ + σ + α(t T (9. ample Two bars are joined together and attached at supports as shown. The left bar is brass for which 9 Gpa, α 26 C  and the right bar is aluminium for which. 7 Gpa, α 256 C . The cross sectional area of the brass bar is 5 mm 2 and that of the aluminium bar is 75 mm 2. The sstem is initiall stress free and the temperature then drops b 2 C. The bars have negligible weight. (a If the supports are unielding, find the normal stress in each bar. (b If the right supports ields.mm, find the normal stress in each bar. Ans. (a brass 4 Mpa, aluminium Mpa; (b brass 28.4 Mpa, aluminium 9 Mpa. 9
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