Lecture mechanics of materials chapter 9 strain transformation

M Vable Mechanics of Materials: Chapter Strain Transformation • Ideas, definitions, and equations in strain transformation are very similar to those in stress transformation But there are also several differences Learning objective Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Learn the equations and procedures of relating strains at a point in different coordinate systems August 2012 9-1 M Vable Mechanics of Materials: Chapter Line Method Plane Strain Global coordinate system is x,y, and z Local coordinate system is n, t, and z We assume εxx, εyy, and γxy are known at a point Objective is to find εnn, εtt, and γnt Procedure Step View the ‘n’ and ‘t’ directions as two separate lines and determine the deformation and rotation of each line as described in steps below Step Construct a rectangle with a diagonal in direction of the line Step Relate the length of the diagonal to the lengths of the rectangle’s sides Step Calculate deformation due to the given strain component and draw the deformed shape Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Step Find the deformation and rotation of the diagonal using small strain approximations Step Calculate normal strains by dividing the deformation by the length of the diagonal Step Calculate the change of angle from the rotation of the lines in the ‘n’ and ‘t’ directions August 2012 9-2 M Vable Mechanics of Materials: Chapter C9.1 At a point, the only non-zero strain component is ε xx = – 400μ Determine the strain components in ‘n’ and ‘t’ coordinate system shown y t x 30o Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm n August 2012 9-3 M Vable Mechanics of Materials: Chapter Visualizing Principal Strain Directions • Principal coordinates directions are the coordinate axes in which the shear strain is zero • The angles the principal axes makes with the global coordinate system are called the principal angles • Normal strains in principal directions are called principal strains • The greatest principal strain is called principal strain one (ε1) Observations • Principal strains are the maximum and the minimum normal strain at a point • A circle in undeformed state will become an ellipse during deformation with major axis as principal axis and minor axis as principal axis Visualizing Procedure Step Visualize or draw a square with a circle drawn inside it Step Visualize or draw the deformed shape of the square due to just normal strains Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Step Visualize or draw the deformed shape of the rectangle due to the shear strain Step Using the eight 45o sectors shown report the orientation of principal direction Also report principal direction as two sectors counterclockwise from the sector reported for principal direction y August 2012 9-4 x M Vable Mechanics of Materials: Chapter C9.2 The state of strain at a point in plane strain is as given in each problem Estimate the orientation of the principal directions and report your results using sectors shown ε xx = – 400 μ ε yy = 600 μ γ xy = – 500 μ y x Class Problem Estimate the orientation of the principal directions and report your results using sectors shown ε xx = 400 μ ε yy = – 600 μ γ xy = – 500 μ ε yy = – 600 μ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ε xx = 400 μ August 2012 9-5 γ xy = 500 μ M Vable Mechanics of Materials: Chapter Method of Equations Calculations for εxx acting alone n y Pn n φ n1 φ2 t t1 P Δn Δy Δ x = ( Δ n ) cos θ (1) (1) ε nn = ε xx cos θ ε tt P1 φ1 θ o θ x A ε xx Δ x = ε xx cos ( θ + 90 ) = ε xx sin θ φ = ε xx sin θ cos θ φ = ε xx sin ( θ + 90 ) cos ( θ + 90 ) = ε xx sin θ cos θ (1) γ nt = – ( φ + φ ) = – 2ε xx sin θ cos θ Calculations for εyy acting alone (2) (2) ε nn = ε yy sin θ ε tt (2) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Calculations for γxy acting along (3) (3) ε tt (3) = – γ xy sin θ cos θ 2 γ nt = ( φ – φ ) = γ xy ( cos θ – sin θ ) Total Strains: August 2012 (1) = ε yy sin ( θ + 90 ) = ε yy cos θ γ nt = ( φ + φ ) = 2ε yy sin θ cos θ ε nn = γ xy sin θ cos θ (2) (3) ε nn = ε nn + ε nn + ε nn 9-6 n1 M Vable Mechanics of Materials: Chapter Strain Transformation Equations 2 ε nn = ε xx cos θ + ε yy sin θ + γ xy sin θ cos θ 2 ε tt = ε xx sin θ + ε yy cos θ – γ xy sin θ cos θ 2 γ nt = – 2ε xx sin θ cos θ + 2ε yy sin θ cos θ + γ xy ( cos θ – sin θ ) Stress Transformation equations 2 σ nn = σ xx cos θ + σ yy sin θ + 2τ xy sin θ cos θ 2 σ tt = σ xx sin θ + σ yy cos θ – 2τ xy cos θ sin θ 2 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm τ nt = – σ xx cos θ sin θ + σ yy sin θ cos θ + τ xy ( cos θ – sin θ ) • The coefficient of the shear strain term is half the coefficient of the shear stress term This difference is due to the fact that we are using engineering strain instead of tensor strain August 2012 9-7 M Vable Mechanics of Materials: Chapter Principal Strains ( ε xx + ε yy ) ε xx – ε yy γ xy ⎛ ⎞ ⎛ ε 1, = ± - + -⎞ ⎝ ⎠ ⎝ 2⎠ 2 ε nn + ε tt = ε xx + ε yy = ε + ε • The angle of principal axis one from the x-axis is only reported in describing the principal coordinate system in two dimensional problems γ xy tan 2θ p = -( ε xx – ε yy ) ⎧0 ⎪ ε3 = ⎨ ⎛ ν ⎞ ν - ( ε xx + ε yy ) = – ⎛ ⎞ ( ε + ε ) ⎪ – ⎝ ⎠ ⎝ – ν⎠ ⎩ 1–ν Plane Strain Plane Stress Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Maximum Shear strain • The maximum shear strain in coordinate systems that can be obtained by rotating about the z-axis is called the in-plane maximum shear strain γp ε1 – ε2 - = -2 • The maximum shear strain at a point is the absolute maximum shear strain that can be obtained in a coordinate system by considering rotation about all three axes γ max ε1 – ε2 ε2 – ε3 ε3 – ε1 ⎞ , , - = max ⎛ -⎝ 2 2 ⎠ • Maximum shear strain in plane stress and plane strain will be different August 2012 9-8 M Vable Mechanics of Materials: Chapter C9.3 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem Using Method of Equations determine (a) the principal strains and principal angle one (b) the maximum shear strain (c) the strain components in the n-t coordinate system shown in each problem ε xx = – 600 μ t y ε yy = – 800 μ n γ xy = 500 μ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 20o August 2012 9-9 x M Vable Mechanics of Materials: Chapter Mohr’s Circle for Strains ( ε xx + ε yy ) ( ε xx – ε yy ) γ xy ⎛ ε nn = + cos 2θ + -⎞ sin 2θ ⎝ 2⎠ 2 ( ε xx – ε yy ) γ xy γ nt ⎛ ⎞ = – sin 2θ + ⎛ -⎞ cos 2θ ⎝ 2⎠ ⎝ 2⎠ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ε xx + ε yy γ nt γ xy ε xx – ε yy ⎛ ε – ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ - + - = - + -⎞ ⎝ nn ⎠ ⎝ ⎠ ⎝ 2⎠ ⎝ 2⎠ 2 • Each point on the Mohr’s circle represents a unique direction passing through the point at which the strains are specified • The coordinates of each point on the circle are the strains (εnn , γnt/2) • On Mohr’s circle, lines are separated by twice the actual angle between the lines August 2012 9-10 M Vable Mechanics of Materials: Chapter Construction of the Mohr’s Circle for strain Step Draw a square with deformed shape due to shear strain γxy Label the intersection of the vertical plane and x-axis as V and the intersection of the horizontal plane and y-axis as H y y H H γxy > V γxy < x V x Step Write the coordinates of point V and H as: V ( ε xx, γ xy ⁄ ) H ( ε yy, γ xy ⁄ and ) γ xy > for Step Draw the horizontal axis to represent the normal strain, with extension to the right and contractions to the left Draw the vertical axis to represent half the shear strain, with clockwise rotation of a line in the upper plane and counter-clockwise rotation of a line of rotation lower plane γ /2 CW H γ yx R Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm (C) C E D R CCW ε xx + ε yy -2 γ xy ε (E) ε xx – ε yy V Step Locate points V and H and join the points by drawing a line Label the point at which the line VH intersects the horizontal axis as C Step With C as center and CV or CH as radius draw the Mohr’s circle August 2012 9-11 M Vable Mechanics of Materials: Chapter Principal Strains & Maximum In-Plane Shear Strain S1 γ /2 CW H P3 ε3 (C) γ yx R 2θp P2 D γp ⁄ C 2θp ε2 R E P1 ε (E) γ xy γp ⁄ V ε1 CCW S2 ε xx – ε yy -2 ε xx + ε yy -2 • The principal angle one θ1 is the angle between line CV and CP1 Depending upon the Mohr circle θ1 may be equal to θp or equal to (θp+ 90o) Maximum Shear Strain γ /2 CW Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm γp ⁄ P3 (C) P1 P2 γp ⁄ CCW August 2012 9-12 γ max ⁄ ε (E) γ max ⁄ M Vable Mechanics of Materials: Chapter Strains in a Specified Coordinate System y t T γ /2 H N CW n H N θΗ θV V 2θH γ nt ⁄ x 2θV E (C) D C εtt V T CCW (E) ε εnn Sign of shear strain The coordinates of point N and T are as shown below: N ( ε nn, γ nt ⁄ ) and T ( ε tt, γ nt ⁄ ) t t1 T n N Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm n1 • Increase in angle results in negative shear strain and decrease in angle results in positive shear strains August 2012 9-13 M Vable Mechanics of Materials: Chapter C9.4 At a point in plane strain, the strain components in the x-y coordinate system are as given in each problem Using Mohr’s circle determine (a) the principal strains and principal angle one (b) the maximum shear strain (c) the strain components in the n-t coordinate system shown in each problem ε xx = – 600 μ t y ε yy = – 800 μ n γ xy = 500 μ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 20o August 2012 9-14 x M Vable Mechanics of Materials: Chapter Class Problem The Mohr’s circle corresponding to a given state of strain are shown Identify the circle you would use to find the strains in the n, t coordinate system in each question y y t n T H N 250 n x 250 T N x V V Circle A Circle B T V H t V N o 50 50o T H H N Circle C V N Circle D V 50o T 50o N Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm H T August 2012 9-15 H M Vable Mechanics of Materials: Chapter Generalized Hooke’s Law in Principal Coordinates • Generalized Hooke’s Law is valid for any orthogonal coordinate system • Principal coordinates for stresses and strains are orthogonal • For isotropic materials, the principal directions for strains are the same as principal directions for stresses ε1 = [ σ1 – ν ( σ2 + σ3 ) ] ⁄ E ε2 = [ σ2 – ν ( σ3 + σ1 ) ] ⁄ E Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ε3 = [ σ3 – ν ( σ1 + σ2 ) ] ⁄ E August 2012 9-16 M Vable Mechanics of Materials: Chapter C9.5 In a thin body (plane stress) the stresses in the x-y plane are as shown on the stress element The Modulus of Elasticity E and Poisson’s ratio ν are as given Determine: (a) the principal strains and the principal angle one at the point (b) the maximum shear strain at the point 60 MPa E = 70 GPa ν = 0.25 40 MPa Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 30 MPa August 2012 9-17 M Vable Mechanics of Materials: Chapter Strain Gages • Strain gages measure only normal strains directly • Strain gages are bonded to a free surface, i.e., the strains are in a state of plane stress and not plane strain • Strain gages measure average strain at a point 2 ε a = ε xx cos θa + ε yy sin θa + γ xy sin θ a cos θ a 2 ε b = ε xx cos θ + ε yy sin θ + γ xy sin θ b cos θ b b b 2 ε c = ε xx cos θ + ε yy sin θ + γ xy sin θ c cos θ c c c Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Strain Rosette o • The change in strain gage orientation by ± 180 makes no difference to the strain values August 2012 9-18 M Vable Mechanics of Materials: Chapter C9.6 At a point on a free surface of aluminum (E = 10,000 ksi and G =4,000 ksi) the strains recorded by the three strain gages shown in Fig C9.6 are as given Determine the stresses σxx, σyy, and τxy y ε a = – 600 μ in ⁄ in ε b = 500 μ in ⁄ in c ε c = 400 μ in ⁄ in 450 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C9.6 August 2012 9-19 b 600 a x M Vable Mechanics of Materials: Chapter C9.7 An aluminum (E = 70 GPa, and ν=0.25) beam is loaded by a force P and moment M at the free end as shown in Figure 9.7 Two strain gages at 30o to the longitudinal axis recorded the strains given Determine the applied force P and applied moment M ε a = – 386 μ m ⁄ m y 10 mm 10mm ε b = 4092 μm ⁄ m 30 mm a 30 mm b z M 0.5 m Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C9.7 August 2012 9-20 0.5m P ... 2012 9- 17 M Vable Mechanics of Materials: Chapter Strain Gages • Strain gages measure only normal strains directly • Strain gages are bonded to a free surface, i.e., the strains are in a state of. .. in negative shear strain and decrease in angle results in positive shear strains August 2012 9- 13 M Vable Mechanics of Materials: Chapter C9.4 At a point in plane strain, the strain components... 2 ⎠ • Maximum shear strain in plane stress and plane strain will be different August 2012 9- 8 M Vable Mechanics of Materials: Chapter C9.3 At a point in plane strain, the strain components in