DifferentialEquations_02_Strain_Disp_Eqns

DifferentialEquations_02_Strain_Disp_Eqns - Section 1.2 1.2...

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Section 1.2 Solid Mechanics Part II Kelly 9 1.2 The Strain-Displacement Relations The strain was introduced in Part I: §3.6. Expressions which relate the displacements of material particles to the strains for a continuously varying strain field are derived in what follows. 1.2.1 The Strain-Displacement Relations Normal Strain Consider a line element of length x Δ emanating from position ) , ( y x and lying in the x - direction, denoted by AB in Fig. 1.2.1. After deformation the line element occupies B A , having undergone a translation, extension and rotation. Figure 1.2.1: deformation of a line element The particle that was originally at x has undergone a displacement ) , ( y x u x and the other end of the line element has undergone a displacement ) , ( y x x u x Δ + . By the definition of (small) normal strain, x y x u y x x u AB AB B A x x xx Δ Δ + = = ) , ( ) , ( * ε (1.2.1) In the limit 0 Δ x one has x u x xx = (1.2.2) This partial derivative is a displacement gradient , a measure of how rapid the displacement changes through the material, and is the strain at ) , ( y x . Physically, it represents the (approximate) unit change in length of a line element, as indicated in Fig. 1.2.2. x A B A B ) , ( y x u x ) , ( y x x u x Δ + x x x Δ + y * B
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Section 1.2 Solid Mechanics Part II Kelly 10 Figure 1.2.2: unit change in length of a line element Similarly, by considering a line element initially lying in the y direction, the strain in the y direction can be expressed as y u y yy = ε (1.2.3) Shear Strain The particles A and B in Fig. 1.2.1 also undergo displacements in the y direction and this is shown in Fig. 1.2.3. In this case, one has x x u B B y Δ = * (1.2.4) Figure 1.2.3: deformation of a line element A similar relation can be derived by considering a line element initially lying in the y direction. A summary is given in Fig. 1.2.4. Form the figure, x u x u x u y x y + = / 1 / tan θ provided that (i) is small and (ii) the displacement gradient x u x / is small. A similar expression for the angle λ can be derived, and hence the shear strain can be written in terms of displacement gradients.
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This note was uploaded on 01/20/2012 for the course ENGINEERIN 2 taught by Professor Staff during the Fall '11 term at Auckland.

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DifferentialEquations_02_Strain_Disp_Eqns - Section 1.2 1.2...

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