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00 The McGraw-Hill Companies, Inc. All rights reserved. T Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Teas Tech Universit Transformations of Stress and Strain

00 The McGraw-Hill Companies, Inc. All rights reserved. 7 - Transformations of Stress and Strain Introduction Transformation of Plane Stress Principal Stresses Maimum Shearing Stress Eample 7.01 Sample Problem 7.1 Mohr s Circle for Plane Stress Eample 7.0 Sample Problem 7. General State of Stress Application of Mohr s Circle to the Three- Dimensional Analsis of Stress Yield Criteria for Ductile Materials Under Plane Stress Fracture Criteria for Brittle Materials Under Plane Stress Stresses in Thin-Walled Pressure Vessels

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-3 Introduction The most general state of stress at a point ma be represented b 6 components,,, τ, τ z z, τ z normal stresses shearing stresses (Note : τ τ, τ z τ z, τ z τ Same state of stress is represented b a different set of components if aes are rotated. The first part of the chapter is concerned with how the components of stress are transformed under a rotation of the coordinate aes. The second part of the chapter is devoted to a similar analsis of the transformation of the components of strain. z )

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-4 Introduction Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated eample, the state of stress is defined b,, τ and τ τ 0. z z z State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate. State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at an point of the surface not subjected to an eternal force.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-5 Transformation of Plane Stress F F 0 0 τ A Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the,, and aes. ( Acosθ ) cosθ τ ( Acosθ ) ( Asinθ ) sinθ τ ( Asinθ ) cosθ A + ( Acosθ ) sinθ τ ( Acosθ ) ( Asinθ ) cosθ + τ ( Asinθ ) sinθ The equations ma be rewritten to ield sinθ cosθ τ sin θ cosθ cosθ τ τ τ cosθ sin θ sin θ

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-6 Principal Stresses The previous equations are combined to ield parametric equations for a circle, ( ) ave + τ where R ave + R + τ Principal stresses occur on the principal planes of stress with zero shearing stresses. ma,min tan θ p τ + ± Note : defines two angles separated b 90 + τ o

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-7 Maimum Shearing Stress Maimum shearing stress occurs for ave τ ma R tan θ s τ Note : defines two angles separated b 90 offset fromθ ave p + b 45 + τ o o and

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-8 Eample 7.01 For the state of plane stress shown, determine (a) the principal panes, (b) the principal stresses, (c) the maimum shearing stress and the corresponding normal stress. SOLUTION: Find the element orientation for the principal stresses from τ tan θ p Determine the principal stresses from + ma,min ± + τ Calculate the maimum shearing stress with τ ma + τ +

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-9 Eample 7.01 + 50MPa 10MPa τ + 40MPa SOLUTION: Find the element orientation for the principal stresses from τ ( + 40) tan θ p 1.333 50 10 θ p θ p 53.1, 33.1 6.6, 116. 6 ( ) Determine the principal stresses from + ma,min ± + τ ma min 0 ± 70MPa 30MPa ( 30) + ( 40)

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-10 Eample 7.01 Calculate the maimum shearing stress with τ ma + τ ( 30) + ( 40) τ ma 50MPa + 50MPa 10MPa τ + 40MPa θ s θ p θ s 45 18.4, 71. 6 The corresponding normal stress is ave + 50 10 0MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-11 Sample Problem 7.1 SOLUTION: Determine an equivalent force-couple sstem at the center of the transverse section passing through H. Evaluate the normal and shearing stresses at H. Determine the principal planes and calculate the principal stresses. A single horizontal force P of 150 lb magnitude is applied to end D of lever ABD. Determine (a) the normal and shearing stresses on an element at point H having sides parallel to the and aes, (b) the principal planes and principal stresses at the point H.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-1 Sample Problem 7.1 SOLUTION: Determine an equivalent force-couple sstem at the center of the transverse section passing through H. P 150lb T M ( 150lb)( 18in).7 kip ( 150lb)( 10in) 1.5kip in Evaluate the normal and shearing stresses at H. Mc ( 1.5kip in)( 0.6in) + + I 1 4 π 0.6in τ Tc + J + in ( ) 4 (.7 kip in)( 0.6in) 1π ( 0.6in) 4 0 + 8.84 ksi τ + 7.96 ksi

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-13 Sample Problem 7.1 Determine the principal planes and calculate the principal stresses. tan θ θ p p θ p τ 61.0,119 30.5, 59. 5 ( 7.96) 0 8.84 1.8 ma,min ma min + ± + τ 0 + 8.84 ± + 13.5 ksi 4.68ksi 0 8.84 + ( 7.96)

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-14 Mohr s Circle for Plane Stress With the phsical significance of Mohr s circle for plane stress established, it ma be applied with simple geometric considerations. Critical values are estimated graphicall or calculated. For a known state of plane stress,, plot the points X and Y and construct the circle centered at C. + ave R + τ The principal stresses are obtained at A and B. ma,min tan θ p τ ave ± R The direction of rotation of O to Oa is the same as CX to CA. τ

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-15 Mohr s Circle for Plane Stress With Mohr s circle uniquel defined, the state of stress at other aes orientations ma be depicted. For the state of stress at an angle θ with respect to the aes, construct a new diameter X Y at an angle θ with respect to XY. Normal and shear stresses are obtained from the coordinates X Y.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-16 Mohr s Circle for Plane Stress Mohr s circle for centric aial loading: P, τ 0 A τ P A Mohr s circle for torsional loading: Tc Tc 0 τ τ 0 J J

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-17 Eample 7.0 For the state of plane stress shown, (a) construct Mohr s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maimum shearing stress and the corresponding normal stress. SOLUTION: Construction of Mohr s circle ( 50) + ( 10) + ave CF 50 0 30MPa R CX FX 40MPa ( 30) + ( 40) 50MPa 0MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-18 Eample 7.0 Principal planes and stresses ma OA OC + CA 0 + 50 ma 70MPa OB OC BC 0 50 ma ma 30 MPa tan θ θ p p θ p FX CP 53.1 6. 6 40 30

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-19 Eample 7.0 Maimum shear stress θ s θ p + 45 τ ma R ave θs 71. 6 τ ma 50 MPa 0 MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-0 Sample Problem 7. For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components eerted on the element obtained b rotating the given element counterclockwise through 30 degrees. SOLUTION: Construct Mohr s circle + 100 + ave R 60 80MPa ( CF ) + ( FX ) ( 0) + ( 48) 5MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-1 Sample Problem 7. Principal planes and stresses tan θ θ p p θ p XF CF 67.4 33.7 48 0.4 clockwise ma OA OC + CA 80 + 5 ma +13 MPa ma OA OC BC 80 5 min +8 MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7 - Sample Problem 7. Stress components after rotation b 30 o Points X and Y on Mohr s circle that correspond to stress components on the rotated element are obtained b rotating XY counterclockwise through θ 60 φ 180 60 67.4 5.6 τ OK OC KC 80 5cos5.6 OL OC + CL 80 + 5cos5.6 KX 5sin 5.6 τ + 48.4 MPa + 111.6 MPa 41.3MPa

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-3 General State of Stress Consider the general 3D state of stress at a point and the transformation of stress from element rotation State of stress at Q defined b:,,, τ, τ, τ z z z Consider tetrahedron with face perpendicular to the line QN with direction cosines: λ, λ, λ The requirement n λ + + τ F n 0 λ + z λ λ + τ λ z z leads to, λ λ + τ Form of equation guarantees that an element orientation can be found such that a b λ + λ + n a b c These are the principal aes and principal planes and the normal stresses are the principal stresses. λ c z z λ z z λ

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-4 Application of Mohr s Circle to the Three- Dimensional Analsis of Stress Transformation of stress for an element rotated around a principal ais ma be represented b Mohr s circle. Points A, B, and C represent the principal stresses on the principal planes (shearing stress is zero) The three circles represent the normal and shearing stresses for rotation around each principal ais. Radius of the largest circle ields the maimum shearing stress. 1 τ ma ma min

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-5 Application of Mohr s Circle to the Three- Dimensional Analsis of Stress In the case of plane stress, the ais perpendicular to the plane of stress is a principal ais (shearing stress equal zero). If the points A and B (representing the principal planes) are on opposite sides of the origin, then a) the corresponding principal stresses are the maimum and minimum normal stresses for the element b) the maimum shearing stress for the element is equal to the maimum inplane shearing stress c) planes of maimum shearing stress are at 45 o to the principal planes.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-6 Application of Mohr s Circle to the Three- Dimensional Analsis of Stress If A and B are on the same side of the origin (i.e., have the same sign), then a) the circle defining ma, min, and τ ma for the element is not the circle corresponding to transformations within the plane of stress b) maimum shearing stress for the element is equal to half of the maimum stress c) planes of maimum shearing stress are at 45 degrees to the plane of stress

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-7 Yield Criteria for Ductile Materials Under Plane Stress Failure of a machine component subjected to uniaial stress is directl predicted from an equivalent tensile test Failure of a machine component subjected to plane stress cannot be directl predicted from the uniaial state of stress in a tensile test specimen It is convenient to determine the principal stresses and to base the failure criteria on the corresponding biaial stress state Failure criteria are based on the mechanism of failure. Allows comparison of the failure conditions for a uniaial stress test and biaial component loading

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-8 Yield Criteria for Ductile Materials Under Plane Stress Maimum shearing stress criteria: Structural component is safe as long as the maimum shearing stress is less than the maimum shearing stress in a tensile test specimen at ield, i.e., τ Y ma < τy For a and b with the same sign, a b τ ma or < For a and b with opposite signs, Y a b τ ma < Y

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-9 Yield Criteria for Ductile Materials Under Plane Stress Maimum distortion energ criteria: Structural component is safe as long as the distortion energ per unit volume is less than that occurring in a tensile test specimen at ield. ( ) ( ) 0 0 6 1 6 1 Y b b a a Y Y b b a a Y d G G u u < + + < + <

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-30 Fracture Criteria for Brittle Materials Under Plane Stress Brittle materials fail suddenl through rupture or fracture in a tensile test. The failure condition is characterized b the ultimate strength U. Maimum normal stress criteria: Structural component is safe as long as the maimum normal stress is less than the ultimate strength of a tensile test specimen. a b < < U U

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-31 Stresses in Thin-Walled Pressure Vessels Clindrical vessel with principal stresses 1 hoop stress longitudinal stress Hoop stress: F z 1 1 0 pr t 0 pr t ( t ) p( r ) 1 Longitudinal stress: F ( ) ( ) π rt p π r

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-3 Stresses in Thin-Walled Pressure Vessels Points A and B correspond to hoop stress, 1, and longitudinal stress, Maimum in-plane shearing stress: τ 1 ma( in plane) pr 4t Maimum out-of-plane shearing stress corresponds to a 45 o rotation of the plane stress element around a longitudinal ais τ ma pr t

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-33 Stresses in Thin-Walled Pressure Vessels Spherical pressure vessel: 1 pr t Mohr s circle for in-plane transformations reduces to a point τ 1 ma(in-plane) constant 0 Maimum out-of-plane shearing stress τ 1 ma 1 pr 4t

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-34 Transformation of Plane Strain Plane strain - deformations of the material take place in parallel planes and are the same in each of those planes. Plane strain occurs in a plate subjected along its edges to a uniforml distributed load and restrained from epanding or contracting laterall b smooth, rigid and fied supports components of ε strain : ( ε γ γ 0) ε γ z z z Eample: Consider a long bar subjected to uniforml distributed transverse loads. State of plane stress eists in an transverse section not located too close to the ends of the bar.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-35 Transformation of Plane Strain State of strain at the point Q results in different strain components with respect to the and reference frames. ε ε γ ε ε γ ( θ ) OB ε cos ε ε ε ε ( 45 ) 1 ( ε + ε + γ ) ( ε + ε ) OB + ε + ε θ + ε sin θ + γ ε ε γ sin θ + ε ε γ + cosθ + ε ε γ cosθ cosθ sinθ cosθ Appling the trigonometric relations used for the transformation of stress, sin θ sin θ

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-36 Mohr s Circle for Plane Strain The equations for the transformation of plane strain are of the same form as the equations for the transformation of plane stress - Mohr s circle techniques appl. Abscissa for the center C and radius R, ε ave ε + ε ε ε γ R + Principal aes of strain and principal strains, γ tan θ p ε ε ε ma ε ave + R ε min ε ( ε ε ) γ γ R + ma ave R Maimum in-plane shearing strain,

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-37 Three-Dimensional Analsis of Strain Previousl demonstrated that three principal aes eist such that the perpendicular element faces are free of shearing stresses. B Hooke s Law, it follows that the shearing strains are zero as well and that the principal planes of stress are also the principal planes of strain. Rotation about the principal aes ma be represented b Mohr s circles.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-38 Three-Dimensional Analsis of Strain For the case of plane strain where the and aes are in the plane of strain, - the z ais is also a principal ais - the corresponding principal normal strain is represented b the point Z 0 or the origin. If the points A and B lie on opposite sides of the origin, the maimum shearing strain is the maimum in-plane shearing strain, D and E. If the points A and B lie on the same side of the origin, the maimum shearing strain is out of the plane of strain and is represented b the points D and E.

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-39 Three-Dimensional Analsis of Strain Consider the case of plane stress, 0 a b z Corresponding normal strains, a ν ε b ε a b E ν E ν εc E a + b E ( + ) ( ε + ε ) a E b ν 1 ν Strain perpendicular to the plane of stress is not zero. If B is located between A and C on the Mohr-circle diagram, the maimum shearing strain is equal to the diameter CA. a b

00 The McGraw-Hill Companies, Inc. All rights reserved. 7-40 Measurements of Strain: Strain Rosette Strain gages indicate normal strain through changes in resistance. With a 45 o rosette, ε and ε are measured directl. γ is obtained indirectl with, OB ( ε ε ) γ ε + Normal and shearing strains ma be obtained from normal strains in an three directions, ε ε 1 cos θ + ε 1 sin θ + γ 1 sinθ cosθ 1 1 ε ε cos θ + ε sin θ + γ sinθ cosθ ε 3 ε cos θ 3 + ε sin θ 3 + γ sinθ cosθ 3 3