Chap2 slides (2)

chap2 slides fm M Vable Mechanics of Materials Chapter 2 Pr in te d fr om h ttp // w w w m e m tu e du /~ m av ab le /M oM 2n d ht m Strain • Relating strains to displacements is a problem in geometry[.]

M Vable Mechanics of Materials: Chapter Strain • Relating strains to displacements is a problem in geometry Kinematics Learning objectives • Understand the concept of strain Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Understand the use of approximate deformed shape for calculating strains from displacements August 2012 2-1 M Vable Mechanics of Materials: Chapter Preliminary Definitions • The total movement of a point with respect to a fixed reference coordinates is called displacement • The relative movement of a point with respect to another point on the body is called deformation • Lagrangian strain is computed from deformation by using the original undeformed geometry as the reference geometry Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Eulerian strain is computed from deformation by using the final deformed geometry as the reference geometry August 2012 2-2 M Vable Mechanics of Materials: Chapter Average Normal Strain B A A Lo Lf – Lo δ ε av = - = -Lo Lo B Lf • Elongations (Lf > Lo) result in positive normal strains Contractions (Lf < Lo) result in negative normal strains A xA Lo B Lf A1 (xA+uA) x xB L0 = xB – xA B1 x (xB+uB) Lf = ( xB + uB ) – ( xA + uA ) = Lo + ( uB – u uB – uA ε av = -xB – xA Units of average normal strain Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • To differentiate average strain from strain at a point • in/in, or cm/cm, or m/m • percentage 0.5% is equal to a strain of 0.005 • prefix: μ = 10-6 1000 μ in / in is equal to a strain 0.001 in / in August 2012 2-3 M Vable Mechanics of Materials: Chapter C2.1 Due to the application of the forces in Fig C2.1, the displacement of the rigid plates in the x direction were observed as given below Determine the axial strains in rods in sections AB, BC, and CD u B = – 1.8 mm u C = 0.7 mm u D = 3.7 mm x A 1.5 m F1 F2 B C F1 2.5 m Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C2.1 August 2012 2-4 D F2 2m F3 F3 M Vable Mechanics of Materials: Chapter Average shear strain Undeformed grid Deformed grid (b) (a) Wooden Bar with Masking Tape A Wooden Bar with Masking Tape A B π/2 C A1 γ B α Wooden Bar with Masking Tape C Wooden Bar with Masking Tape π γ av = - – α • Decreases in the angle (α < π / 2) result in positive shear strain Increase in the angle (α > π / 2) result in negative shear strain Units of average shear strain • To differentiate average strain from strain at a point • rad Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • prefix: μ = 10-6 1000 μ rad is equal to a strain 0.001 rad August 2012 2-5 M Vable Mechanics of Materials: Chapter Small Strain Approximation P1 Lf = Lf 2 L o + D + 2L o D cos θ D ␪ A P2 P D D L f = L o + ⎛⎝ -⎞⎠ + ⎛⎝ -⎞⎠ cos θ Lo Lo L0 Lf – Lo ε = - = Lo D D + ⎛ ⎞ + ⎛ ⎞ cos θ – ⎝L ⎠ ⎝L ⎠ o o 2.5 D cos θ ε small = -Lo Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm εsmall Eq 2.6 ε Eq 2.5 2.6 % error 1.0 1.23607 19.1 0.5 0.58114 14.0 0.1 0.10454 4.3 0.05 0.005119 2.32 0.01 0.01005 0.49 0.005 0.00501 0.25 • Small-strain approximation may be used for strains less than 0.01 • Small normal strains are calculated by using the deformation component in the original direction of the line element regardless of the orientation of the deformed line element • In small shear strain (γ) calculations the following approximation may sin γ ≈ γ cos γ ≈ be used for the trigonometric functions: tan γ ≈ γ • Small-strain calculations result in linear deformation analysis • Drawing approximate deformed shape is very important in analysis of small strains August 2012 2-6 M Vable Mechanics of Materials: Chapter C2.2 A thin triangular plate ABC forms a right angle at point A During deformation, point A moves vertically down by δA Determine the average shear strain at point A δ A = 0.008 in in A δA Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C2.2 August 2012 2-7 3i n C B M Vable Mechanics of Materials: Chapter C2.3 A roller at P slides in a slot as shown Determine the deformation in bar AP and bar BP by using small strain approximation ␦P ⫽ 0.02 in P A 40⬚ 110⬚ B Fig C2.3 Class Problem Draw an approximate exaggerated deformed shape Using small strain approximation write equations relating δAP and δBP to δP B 30° 75° A P Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ␦P ⫽ 0.25 mm August 2012 2-8 M Vable Mechanics of Materials: Chapter Strain Components y Δv Δu ε xx = Δx Δy Δv ε yy = Δy Δx x Δu Δw ε zz = -Δz Δz Δw z y Δu Δu Δv γ xy = - + Δy Δx Δy ⎛π ⎞ ⎝ 2- – γ xy⎠ z Δv x Δx Δv Δu γ yx = - + - = γ xy Δx Δy y Δv Δw γ yz = - + -Δz Δy Δw Δy ⎛ π - – γ ⎞ ⎝ yz⎠ Δz Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm z Δw Δv γ zy = + - = γ yz Δy Δz x Δv y ⎛π ⎞ ⎝ 2- – γ zx⎠ Δz z August 2012 Δx Δw Δw Δu γ zx = + Δx Δz x Δu Δw γ xz = - + = γ zx Δz Δx Δu 2-9 M Vable Mechanics of Materials: Chapter Engineering Strain ε xx γ xy γ xz γ yx ε yy γ yz γ zx γ zy ε zz ε xx γ xy γ yx ε yy 0 0 Engineering strain matrix Plane strain matrix Strain at a point ε xx = Δu lim ⎛ -⎞ Δx → 0⎝ Δx⎠ = ∂u ∂x Δu Δv ∂u ∂v γ xy = γ yx = lim ⎛ - + -⎞ = + ⎝ ⎠ Δy Δx ∂ y ∂x Δx → Δy → ∂v ε yy = ∂y Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ∂w ε zz = ∂z ∂v ∂w + γ yz = γ zy = ∂z ∂y ∂w ∂u + γ zx = γ xz = ∂x ∂z • tensor normal strains = engineering normal strains • tensor shear strains = (engineering shear strains)/ Strain at a Point on a Line ε xx = August 2012 2-10 du ( x ) dx M Vable Mechanics of Materials: Chapter C2.4 Displacements u and v in the x and y directions respectively were measured by Moire Interferometry method at many points on a body Displacements of four points on a body are given below Determine the average values of strain components εxx, ε yy, and γxy at point A shown in Fig C2.4 0.0005 mm y C D A B x uA = vA = u B = 0.625μmm v B = – 0.3125μmm u C = – 0.500 μmm v C = – 0.5625 μmm u D = 0.250μmm v D = – 1.125 μmm 0.0005 mm Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C2.4 August 2012 2-11 M Vable Mechanics of Materials: Chapter C2.5 The axial displacement in a quadratic one-dimensional finite element is as given below u1 u2 u3 u ( x ) = - ( x – a ) ( x – 2a ) – - ( x ) ( x – 2a ) + - ( x ) ( x – a ) 2 2a a 2a Determine the strain at Node Node Node x2= a x1= Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm x August 2012 2-12 Node x3= 2a