Giao trinh bai tap chapter 4 hoạch định chiến lược tiếp thị giữa các tổ chức

M Vable Mechanics of Materials: Chapter Stress Transformation • Transforming stress components from one coordinate system to another at a given point • Relating stresses on different planes that pass through a point σxx T T P P τθx σxx τxθ Cast Iron Aluminum Learning Objectives • Learn the equations and procedures of relating stresses (on different planes) in different coordinate system at a point Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Develop the ability to visualize planes passing through a point on which stresses are given or are being found, particularly the planes of maximum normal stress and maximum shear stress August 2012 8-1 M Vable Mechanics of Materials: Chapter Wedge Method • The fixed reference coordinate system in which the entire problem is described is called the global coordinate system • A coordinate system that can be fixed at any point on the body and has an orientation that is defined with respect to the global coordinate system is called the local coordinate system Vertical plane y y t t ␪ in cl In pl x e an ␪ n ␪ ed n Outward normal to inclined plane Horizontal plane z x z (a) (b) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Plane stress problem: We will consider only those inclined planes that can be obtained by rotation about the z-axis August 2012 8-2 M Vable Mechanics of Materials: Chapter C8.1 In the following problems one could say that the normal stress on the incline AA is in tension, compression or can’t be determined by inspection Similarly we could say that the shear stress on the incline AA is positive, negative or can’t be determined by inspection Choose the correct answers for normal and shear stress on the incline AA by inspection Assume coordinate z is perpendicular to this page and towards you (a) σ y (b) y τ A A A 300 A 600 x Class Problem y τ (c) (d) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm A 300 August 2012 A x 8-3 x M Vable Mechanics of Materials: Chapter General Procedure for Wedge Method Step 1: A stress cube with the plane on which stresses are to be found, or are given, is constructed Step 2: A wedge made from the following three planes is constructed: • a vertical plane that has an outward normal in the x-direction, • a horizontal plane that has an outward normal in the y-direction, and • the specified inclined plane on which we either seek stresses or the stresses are given Establish a local n-t-z coordinate system using the outward normal of the inclined plane as the n-direction All the known and the unknown stresses are shown on the wedge The diagram so constructed will be called a stress wedge Step 3: Multiply the stress components by the area of the planes on which the stress components are acting, to obtain forces acting on that plane The wedge with the forces drawn will be referred to as the force wedge Step 4: Balance forces in any two directions to determine the unknown stresses Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Step 5: Check the answer intuitively August 2012 8-4 M Vable C8.2 Mechanics of Materials: Chapter Determine the normal and shear stress on plane AA A 300 ksi A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 10 ksi August 2012 8-5 M Vable Mechanics of Materials: Chapter Stress Transformation By Method of Equations Stress Cube Stress Wedge σyy y τyx B t τxy σxx Δy σxx C τxy θ O τyx τxy x A Δt Δx τyx σyy Force Wedge σnn θ θ σxx τnt n σyy Force Transformation τnt (Δt Δz) y σnn(Δt Δz) θ t θ n Fy θ sθ co Fx θ θ Fy τxy (Δt cosθ Δz) θ s co sin Fx σxx (Δt cosθ Δz) Fy sin Fx τyx (Δt sinθ Δz) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm σyy (Δt sinθ Δz) 2 σ nn = σ xx cos θ + σ yy sin θ + 2τ xy sin θ cos θ 2 τ nt = – σ xx cos θ sin θ + σ yy sin θ cos θ + τ xy ( cos θ – sin θ ) Trigonometric identities August 2012 8-6 x M Vable Mechanics of Materials: Chapter cos θ = ( + cos 2θ ) ⁄ 2 cos θ – sin θ = cos 2θ sin θ = ( – cos 2θ ) ⁄ cos θ sin θ = ( sin 2θ ) ⁄ ( σ xx + σ yy ) ( σ xx – σ yy ) σ nn = + cos 2θ + τ xy sin 2θ 2 ( σ xx – σ yy ) τ nt = – sin 2θ + τ xy cos 2θ Maximum normal stress 2τ xy ⎛ dσ nn ⎞ = 0⎟ ⇒ tan 2θ p = -⎜ ( σ xx – σ yy ) d θ ⎝ ⎠ θ = θp −τxy -(σxx − σyy)/2 2θp τxy R 2θp (σxx − σyy)/2 θ1 = θp θ2 =90 + θp Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm R σ xx – σ yy ( σ xx + σ yy ) ⎛ τ 1, = σ 1, = ± -⎞ + τ xy ⎝ ⎠ 2 • Planes on which the shear stresses are zero are called the principal planes • The normal direction to the principal planes is referred to as the principal direction or the principal axis • The angles the principal axis makes with the global coordinate system are called the principal angles • Normal stress on a principal plane is called the principal stress • The greatest principal stress is called principal stress one • Only θ defining principal axis one is reported in describing the principal coordinate system in two-dimensional problems Counterclockwise rotation from the x axis is defined as positive August 2012 8-7 M Vable Mechanics of Materials: Chapter σ nn + σ tt = σ xx + σ yy = σ + σ • The sum of the normal stresses is invariant with the coordinate transformation ⎧0 σ = σ zz = ⎨ ⎩ ν ( σ xx + σ yy ) = ν ( σ + σ ) Plane Stress Plane Strain In-Plane Maximum Shear Stress Vertical plane y t in cl In pl x e an ␪ n ␪ ed n Outward normal to inclined plane Horizontal plane x z • The maximum shear stress on a plane that can be obtained by rotating about the z axis is called the in-plane maximum shear stress Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm ( σ xx – σ yy ) τ nt = – sin 2θ + τ xy cos 2θ – ( σ xx – σ yy )⎞ ⎛ dτ nt ⎞ ⎛ σ1 – σ2 τ p = -= 0⎟ ⇒ ⎜ tan 2θ s = -⎟ ⎜ 2τ xy ⎝ d θ θ = θs ⎠ ⎝ ⎠ • maximum in-plane shear stress exists on two planes, each of which are 45o away from the principal planes Maximum Shear Stress • The maximum shear stress at a point is the absolute maximum shear stress that acts on any plane passing through the point August 2012 8-8 M Vable Mechanics of Materials: Chapter Planes of Maximum Shear Stress p2 p2 rotation about principal axis p1 p1 σ2 – σ3 τ 23 = – τ 32 = -2 p3 p3 p2 p2 rotation about principal axis p1 p1 σ3 – σ1 τ 31 = – τ 13 = -2 p3 p3 p2 p2 rotation about principal axis (In-plane) p1 p1 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm σ1 – σ2 τ 21 = – τ 12 = -2 p3 p3 σ1 – σ2 σ2 – σ3 σ3 – σ1 ⎞ , , -τ max = max ⎛ -⎝ ⎠ 2 Plane Stress ⎧0 σ = σ zz = ⎨ Plane Strain ⎩ ν ( σ xx + σ yy ) = ν ( σ + σ ) • The maximum shear stress value may be different in plane stress and in plane strain August 2012 8-9 M Vable Mechanics of Materials: Chapter C8.3 Determine the normal and shear stress on plane AA using the method of equations (resolving problem C8.2) A 300 ksi A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 10 ksi August 2012 8-10 M Vable Mechanics of Materials: Chapter Stress Transformation By Mohr’s Circle ( σ xx + σ yy ) ( σ xx – σ yy ) σ nn = + cos 2θ + τ xy sin 2θ 2 ( σ xx – σ yy ) τ nt = – sin 2θ + τ xy cos 2θ Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm σ xx + σ yy σ xx – σ yy ⎛ σ – ⎞ ⎛ 2 - + τ nt = -⎞ + τ xy 8.1 ⎝ nn ⎠ ⎝ ⎠ 2 • Each point on the Mohr’s circle represents a unique plane that passes through the point at which the stresses are specified • The coordinates of the point on Mohr’s Circle are the normal and shear stress on the plane represented by the point • On Mohr’s circle, plane are separated by twice the actual angle between the planes August 2012 8-11 M Vable Mechanics of Materials: Chapter Construction of Mohr’s Circle Step Show the stresses σxx, σyy, and τxy on a stress cube and label the vertical plane as V and the horizontal plane as H Step Write the coordinates of points V and H as V (σxx , τxy )and H (σyy , τyx ) The rotation arrow next to the shear stresses corresponds to the rotation of the cube caused by the set of shear stress on planes V and H y σyy τyx τxy H σxx V τxy V τ H τyx σxx R E σyy (C) x H τyx (CW) σxx C D R τxy σ (T) V σyy (CCW) σ xx + σ yy σ xx – σ yy -2 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Step Draw the horizontal axis with the tensile normal stress to the right and the compressive normal stress to the left Draw the vertical axis with clockwise direction of shear stress up and counterclockwise direction of rotation down 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 8-12 M Vable Mechanics of Materials: Chapter Principal Stresses & Maximum In-Plane Shear Stress from Mohr’s Circle τ S1 (CW) H τ12 R P3 σ3 (C) P2 2θp C E D 2θp τ21 V σ1 (CCW) (T) τxy R σ2 σ P1 σ xx + σ yy -2 S2 σ xx – σ 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 Stress τ (CW) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm τ23 (C) τ32 τ12 P3 σ (T) P1 P2 τ21 (CCW) August 2012 τ13 8-13 τ31 M Vable Mechanics of Materials: Chapter Stresses on an Inclined Plane σyy y τ τyx H σxx V A V β H θA A τxy τyx (CW) τxy σxx A H 2θA 2βA F D C (C) E x V σyy σA (CCW) Sign of shear stress on incline: Coordinates of point A: ( σ , τ ) A A t V A n τA A V n τA Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm H August 2012 H 8-14 t τA (T) σ M Vable Mechanics of Materials: Chapter Principal Stress Element τ S1 (CW) P2 p2 R P2 (C) C E y τ12 D 2θ1 σ2 P1 H V V σ1 P2 (T) τ21 R S1 σ P1 p1 P1 S2 H V θ1 H S2 x (CCW) y σav τ21 P1 σ1 τ21 y σav S1 P2 σ2 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm x Cast Iron P1 σxx S1 P2 H S2 August 2012 P1 θ1 x P σxx S2 P2 σ2 θ1 P σ1 σ xx τ max = V P1 σ = σ xx S2 P1 P2 Aluminum S2 8-15 M Vable Mechanics of Materials: Chapter C8.4 In a thin body (plane stress) the stresses in the x-y plane are as shown on each stress element (a) Determine the normal and shear stresses on plane A (b) Determine the principal stresses at the point (c) Determine the maximum shear stress at the point (d) Draw the principal element 20 ksi 30 ksi A 42o Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C8.4 August 2012 8-16 10 ksi M Vable Mechanics of Materials: Chapter Class Problem Associate the stress cubes with the appropriate Mohr’s circle for stress 40 ksi 40ksi ksi ksi 40 ksi cube cube cube ksi circle B circle A circle C circle D Class Problem Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Determine the two possible values of principal angle one (θ1) in each question y V H H V V H August 2012 24o 24o x V 8-17 H M Vable Mechanics of Materials: Chapter Class Problem Explain the failure surfaces in cast iron and aluminum due to torsion by drawing the principal stress element T T τθx τxθ Cast Iron Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Aluminum August 2012 8-18 M Vable Mechanics of Materials: Chapter C8.5 A broken in x in wooden bar was glued together as shown Determine the normal and shear stress in the glue F = 12 kips F in Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 60o August 2012 8-19 M Vable Mechanics of Materials: Chapter C8.6 If the applied force P = 1.8 kN, determine the principal stresses and maximum shear stress at points A, B, and C which are on the surface of the beam P A B C 30 mm 15 mm 30 mm 0.4 m Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 0.4 m August 2012 8-20 mm mm [...]... Materials: Chapter 8 Class Problem 2 Associate the stress cubes with the appropriate Mohr’s circle for stress 40 ksi 40 ksi 8 ksi 8 ksi 40 ksi cube 1 cube 2 cube 3 8 ksi circle B circle A circle C circle D Class Problem 3 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Determine the two possible values of principal angle one (θ1) in each question y 1 2 V H H V V H August 2012 24o 24o x V 8-17... Materials: Chapter 8 C8 .4 In a thin body (plane stress) the stresses in the x-y plane are as shown on each stress element (a) Determine the normal and shear stresses on plane A (b) Determine the principal stresses at the point (c) Determine the maximum shear stress at the point (d) Draw the principal element 20 ksi 30 ksi A 42 o Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C8 .4 August 2012... Vable Mechanics of Materials: Chapter 8 Stresses on an Inclined Plane σyy y τ τyx H σxx V A V β H θA A τxy τyx (CW) τxy σxx A H 2θA 2βA F D C (C) E x V σyy σA (CCW) Sign of shear stress on incline: Coordinates of point A: ( σ , τ ) A A t V A n τA A V n τA Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm H August 2012 H 8- 14 t τA (T) σ M Vable Mechanics of Materials: Chapter 8 Principal Stress Element... H August 2012 24o 24o x V 8-17 H M Vable Mechanics of Materials: Chapter 8 Class Problem 4 Explain the failure surfaces in cast iron and aluminum due to torsion by drawing the principal stress element T T τθx τxθ Cast Iron Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Aluminum August 2012 8-18 M Vable Mechanics of Materials: Chapter 8 C8.5 A broken 2 in x 6 in wooden bar was glued together... http://www.me.mtu.edu/~mavable/MoM2nd.htm 60o August 2012 8-19 M Vable Mechanics of Materials: Chapter 8 C8.6 If the applied force P = 1.8 kN, determine the principal stresses and maximum shear stress at points A, B, and C which are on the surface of the beam P A B C 30 mm 15 mm 30 mm 0 .4 m Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 0 .4 m August 2012 8-20 6 mm 6 mm ... direction of shear stress up and counterclockwise direction of rotation down Step 4 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 5 With C as center and CV or CH as radius draw the Mohr’s circle August 2012 8-12 M Vable Mechanics of Materials: Chapter 8 Principal Stresses & Maximum In-Plane Shear Stress from Mohr’s Circle...M Vable Mechanics of Materials: Chapter 8 Stress Transformation By Mohr’s Circle ( σ xx + σ yy ) ( σ xx – σ yy ) σ nn = + cos 2θ + τ xy sin 2θ 2 2 ( σ xx – σ yy ) τ nt = – sin 2θ + τ xy cos... the normal and shear stress on the plane represented by the point • On Mohr’s circle, plane are separated by twice the actual angle between the planes August 2012 8-11 M Vable Mechanics of Materials: Chapter 8 Construction of Mohr’s Circle Step 1 Show the stresses σxx, σyy, and τxy on a stress cube and label the vertical plane as V and the horizontal plane as H Step 2 Write the coordinates of points ... principal element 20 ksi 30 ksi A 42 o Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C8 .4 August 2012 8-16 10 ksi M Vable Mechanics of Materials: Chapter Class Problem Associate the... Class Problem Associate the stress cubes with the appropriate Mohr’s circle for stress 40 ksi 40 ksi ksi ksi 40 ksi cube cube cube ksi circle B circle A circle C circle D Class Problem Printed from:... of principal angle one (θ1) in each question y V H H V V H August 2012 24o 24o x V 8-17 H M Vable Mechanics of Materials: Chapter Class Problem Explain the failure surfaces in cast iron and aluminum