Module 2 Stresses in machine elements. Version 2 ME, IIT Kharagpur

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1 Module Stresses in machine elements Version M, IIT Kharagpur

2 Lesson 3 Strain analsis Version M, IIT Kharagpur

3 Instructional Objectives At the end of this lesson, the student should learn Normal and shear strains. 3-D strain matri. Constitutive equation; generalied Hooke s law Relation between elastic, shear and bulk moduli,, K). Stress- strain relation considering thermal effects..3. Introduction No matter what stresses are imposed on an elastic bod, provided the material does not rupture, displacement at an point can have onl one value. Therefore the displacement at an point can be completel given b the three single valued components u, v and w along the three co-ordinate aes, and respectivel. The normal and shear strains ma be derived in terms of these displacements..3. Normal strains Consider an element AB of length δ figure-.3..). If displacement of end A is u, that of end B is therefore the strain in -direction is v and w u+ δ. This gives an increase in length of u+ δ-u) and.similarl, strains in and directions are.therefore, we ma write the three normal strain components as v w, and. Version M, IIT Kharagpur

4 u+ δ A δ B u A ' B '.3.3 Shear strain.3..f- Change in length of an infinitesimal element. In the same wa we ma define the shear strains. For this purpose consider an element ABCD in - plane and let the displaced position of the element be A B C D Figure-.3.3.). This gives shear strain in plane as where α is the angle made b the displaced line B C with the vertical and β is the angle made b the displaced line A D with the horiontal. This gives δ v δ v α and β δ δ α +β u + δ v v + δ v B A δ B' α A' u C D β D' C' v v + δ u + δ.3.3.f- Shear strain associated with the distortion of an infinitesimal element. Version M, IIT Kharagpur

5 We ma therefore write the three shear strain components as u v v w w u +, + and + Therefore, the complete strain matri can be written as v u 0 w.3.4 Constitutive equation The state of strain at a point can be completel described b the si strain components and the strain components in their turns can be completel defined b the displacement components u, v, and w. The constitutive equations relate stresses and strains and in linear elasticit we simpl have σ where is σ modulus of elasticit. It is also known that σ produces a strain of in - direction, νσ in -direction and νσ in -direction. Therefore we ma write the generalied Hooke s law as ), ) and σ ν σ +σ σ ν σ +σ σ ν σ +σ ) It is also known that the shear stress γ, where is the shear modulus and γ is shear strain. We ma thus write the three strain components as γ γ γ, and In general each strain is dependent on each stress and we ma write Version M, IIT Kharagpur

6 K K K3 K4 K5 K6 σ K K K K K K σ K K K K K K σ γ K4 K4 K43 K44 K45 K46 γ K5 K5 K53 K54 K55 K 56 γ K K K K K K For isotropic material K K K33 ν K K3 K K3 K3 K3 K44 K55 K66 Rest of the elements in K matri are ero. On substitution, this reduces the general constitutive equation to equations for isotropic materials as given b the generalied Hooke s law. Since the principal stress and strains aes coincide, we ma write the principal strains in terms of principal stresses as σ ν σ +σ3 σ ν σ 3+σ 3 σ3 ν σ +σ [ )] [ )] [ )] From the point of view of volume change or dilatation resulting from hdrostatic pressure we also have σ KΔ where σ σ +σ +σ ) σ +σ +σ3) and Δ + + ) + + 3) 3 3 Version M, IIT Kharagpur

7 These equations allow the principal strain components to be defined in terms of principal stresses. For isotropic and homogeneous materials onl two constants vi. and ν are sufficient to relate the stresses and strains. The strain transformation follows the same set of rules as those used in stress transformation ecept that the shear strains are halved wherever the appear..3.5 Relations between, and K The largest maimum shear strain and shear stress can be given b σ σ 3 ma γ ma 3 and ma and since γ ma we have σ σ 3 3 σ ν σ +σ σ ν σ +σ ) ) 3 and this gives + ν) Considering now the hdrostatic state of stress and strain we ma write ) 3 K 3) 3 σ+σ +σ + +. Substituting, and 3 in terms of σ, σ and σ3 we ma write 3 ) K [ ) )] σ +σ +σ σ +σ +σ ν σ +σ +σ and this gives K 3 ν )..3.6 lementar thermoelasticit So far the state of strain at a point was considered entirel due to applied forces. Changes in temperature ma also cause stresses if a thermal gradient or some eternal constraints eist. Provided that the materials remain linearl elastic, stress pattern due to thermal effect ma be superimposed upon that due to applied forces and we ma write Version M, IIT Kharagpur

8 ) σ ν σ +σ +αt ) σ ν σ +σ +αt σ ν σ +σ ) +α T and It is important to note that the shear strains are not affected directl b temperature changes. It is sometimes convenient to epress stresses in terms of strains. This ma be done using the relation Δ + +. Substituting the above epressions for, and we have, and substituting Δ ) ) 3 T ν σ +σ +σ + α K we have 3 ν) Δ σ +σ +σ ) + 3α T. 3K Combining this with ) σ ν σ +σ +α T we have 3νK Δ 3αT) αt +ν +ν +ν σ + 3νK Substituting and λ we ma write the normal and shear +ν) + ν stresses as σ +λδ 3KαT σ +λδ 3KαT σ +λδ 3KαT These equations are considered to be suitable in thermoelastic situations. Version M, IIT Kharagpur

9 .3.7 Problems with Answers Q.: A rectangular plate of 0mm thickness is subjected to uniforml distributed load along its edges as shown in figure Find the change in thickness due to the loading. 00 Pa, ν 0.3 KN /mm 50mm 4KN/mm 00mm.3.7.F A.: Here σ 400 MPa, σ 00 MPa and σ 0 ν σ +σ This gives ) Now, Δt where, t is the thickness and Δt is the change in thickness. t Therefore, the change in thickness 7.5 μm. Q.: At a point in a loaded member, a state of plane stress eists and the strains are , and If the elastic constants, ν and are 00 Pa, 0.3 and 84 Pa respectivel, determine the normal stresses σ and σ and the shear stress at the point. Version M, IIT Kharagpur

10 A.: σ νσ σ νσ This gives σ ν σ ν +ν +ν Substituting values, we get σ -.75 MPa, σ -.53 MPa and 9.3 MPa. Q.3: A rod 50 mm in diameter and 50 mm long is compressed aiall b an uniforml distributed load of 50 KN. Find the change in diameter of the rod if 00 Pa and ν0.3. A.3: Aial stress 50 σ 7.3MPa π 0.05) 4 Aial strain, Lateral strain ν.90 4 Δ Now, lateral strain, L and this gives D Δ 9.5 μm. Q.4: If a steel rod of 50 mm diameter and m long is constrained at the ends and heated to 00 o C from an initial temperature of 0 o C, what would be the aial load developed? Will the rod buckle? Take the coefficient of thermal epansion, α0-6 per o C and 00 Pa. Version M, IIT Kharagpur

11 A.4: Thermal strain, t αδ T.60 3 In the absence of an applied load, the force developed due to thermal epansion, F ta 848KN For buckling to occur the critical load is given b F π I KN. l cr Therefore, the rod will buckle when heated to 00 o C..3.8 Summar of this Lesson Normal and shear strains along with the 3-D strain matri have been defined. eneralied Hooke s law and elementar thermo-elasticit are discussed..3.9 Reference for Module- ) Mechanics of materials b.p.popov, Prentice hall of India, 989. ) Mechanics of materials b Ferdinand P. Boer,. Russel Johnson, J.T Dewolf, Tata Mcraw Hill, ) Advanced strength and applied stress analsis b Richard. Budens, Mcraw Hill, ) Mechanical engineering design b Joseph. Shigle, Mcraw Hill, 986. Version M, IIT Kharagpur