Principal Stresses, Yielding Criteria, wall structures

Transcription

1 Principal Stresses, Yielding Criteria, St i thi Stresses in thin wall structures

2 Introduction The most general state of stress at a point may be represented by 6 components, x, y, z τ xy, τ yz, τ zx normal stresses shearing stresses (Note : τ xy = τ yx, τ yz = τ zy, τ zx = τ Same state of stress is represented by a different set of components if axes are rotated. The first part of the chapter is concerned with how the components of stress are transformed under a rotation of the coordinate axes. The second part of the chapter is devoted dto a similar analysis of the transformation of the components of strain. xz ) 7 -

3 Introduction Plane Stress state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by,, τ and τ = τ 0. x y xy z = zx zy = 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 any point of the surface not subjected to an external force. 7-3

4 Transformation of Plane Stress Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x, y, and x axes. Fx = 0 = x A x( Acosθ ) cosθ τ xy( Acosθ ) sinθ y ( Asinθ ) sinθ τ xy( Asinθ ) cosθ F y = 0 = τ x y A + x ( Acosθ ) sinθ τ xy ( Acosθ ) cosθ ( Asinθ ) cosθ + τ ( Asinθ ) sinθ y The equations may be rewritten to yield x + y x y x = + cosθ + τ xy sin θ x + y x y y = cosθ τ xy sin θ x y τ x y = sin θ + τ xy cosθ xy 7-4

5 Principal Stresses The previous equations are combined to yield parametric equations for a circle, ( ) + τ = R x ave + x y where x + y x y ave = R = + τ xy Principal stresses occur on the principal planes of stress with zero shearing stresses. 7-5

6 Maximum Shearing Stress Maximum shearing stress occurs for = x ave 7-6

7 Example 1 For the state of plane stress shown, determine (a) the principal planes, (b) the principal stresses, (c) the maximum shearing stress and the corresponding normal stress. SOLUTION: Find the element orientation for the principal stresses from τ xy tan θ p = x Determine the principal stresses from y x + y x y max,min = ± + τ Calculate the maximum shearing stress with x y τ max = + +τ xy xy x + y = 7-7

8 Example 1 x = + 50MPa τ xy = 10MPa x = + 40MPa SOLUTION: Find the element orientation for the principal stresses from τ xy ( + 40) tan θ p = = = θ p θ p x = 53.1, y 33.1 = 6.6, ( ) Determine the principal stresses from x + y x y max,min = ± + τ xy ( 30) ( 40) = 0 ± + max = 70MPa min = 30 MPa 7-8

9 Example 1 Calculate the maximum shearing stress with τ x y max = + τ xy = ( 30) + ( 40) x x = + 50MPa τ xy = 10MPa = + 40MPa τ max = 50MPa θ s = θ p 45 θ = 18.4, θ s The corresponding normal stress is = ave = x + y = = 0MPa 7-9

10 Sample Problem 1 A single horizontal force P of 600 N 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 x and y axes, (b) the principal planes and principal stresses at the point H. SOLUTION: Determine an equivalent force-couple system 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. 7-10

11 Sample Problem 1 SOLUTION: Determine an equivalent force-couple system at the center of the transverse section passing through H. Evaluate the normal and shearing stresses at H. 7-11

12 Sample Problem 1 Determine the principal planes and calculate the principal stresses. 7-1

13 Mohr s Circle for Plane Stress With the physical significance of Mohr s circle for plane stress established, it may be applied with simple geometric considerations. Critical values are estimated t graphically or calculated. l For a known state of plane stress x, y, plot the points X and Y and construct the circle centered at C. x + y x y ave = R = + τ The principal stresses are obtained at A and B. max,min tan θ p = = τ x ave xy ± R y The direction of rotation of Ox to Oa is the same as CX to CA. xy 7-13 τ xy

14 Mohr s Circle for Plane Stress With Mohr s circle uniquely defined, the state of stress at other axes orientations may be depicted. For the state of stress at an angle θ with respect to the xy axes, 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. 7-14

15 Mohr s Circle for Plane Stress Mohr s circle for centric axial loading: x P, y = τ = 0 A = xy x = = τ y xy = P A Mohr s circle for torsional loading: Tc Tc x = y = 0 τ xy = x = y = τ xy = 0 J J 7-15

16 Example For the state of plane stress shown, (a) construct Mohr s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress. SOLUTION: Construction o of Mohr s circlec ( 50) + ( 10) x + y ave = = CF = 50 0 = 30MPa R = CX = = 0MPa FX = 40MPa ( 30) + ( 40) = 50MPa 7-16

17 Example Principal planes and stresses max = OA = OC + CA = max = 70MPa min = OB = OC BC min = 30 MPa = 0 50 tan θ p θ p θ p FX = = CP = =

18 Example Maximum shear stress θ s = θ p + 45 τ max = R = ave θs = τ max = 500 MPa = 0 MPa 7-18

19 Sample Problem For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30 degrees. SOLUTION: Construct Mohr s circle x + y = = ave 60 = 80MPa ( CF ) + ( ) = ( 0) + ( 48) = 5MPa R = FX 7-19

20 Sample Problem Principal planes and stresses tan θ θ p p θ p XF = = CF = = =.4 clockwise max = OA = OC + CA = max = OA = OC BC = 80 5 max = +13 MPa min = +8 MPa 7-0

21 Sample Problem Stress components after rotation by 30 o Points X and Y on Mohr s circle that correspond to stress components on the rotated element are obtained by rotating XY counterclockwise through θ = 60 φ = = 5.6 x y = OK = OC KC = 80 5cos5.6 = OL = OC + CL = cos5.6 = KX = 5sin 5. τ x y 6 τ x y x y = MPa = MPa = 41.3MPa 7-1

22 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 by:,,, τ, τ, τ x y z xy yz zx Consider tetrahedron with face perpendicular to the line QN with direction cosines: λ λ, λ The requirement n = x x λ + xy y F n = 0 y λ + x y z λ yz z x, y leads to, + τ λ λ + τ λ λ + τ Form of equation guarantees that an element orientation can be found such that a b b = λ + λ + n a c These are the principal axes and principal planes and the normal stresses are the principal stresses. λ c y z zx λ z z λ x 7 -

23 Application of Mohr s Circle to the Three- Dimensional Analysis of Stress Transformation of stress for an element rotated around a principal axis may be represented by Mohr s circle. Points A, B, and C represent the principal p stresses on the principal p planes (shearing stress is zero). The three circles represent the normal and shearing stresses for rotation around each principal axis. Radius of the largest circle yields the maximum shearing stress. 1 τ max = max min 7-3

24 Application of Mohr s Circle to the Three- Dimensional Analysis of Stress In the case of plane stress, the axis perpendicular p to the plane of stress is a principal axis (shearing stress equals 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 maximum and minimum i normal stresses for the element. b) the maximum shearing stress for the element is equal to the maximum inplane shearing stress. c) planes of maximum shearing stress are at 45 o to the principal planes. 7-4

25 Application of Mohr s Circle to the Three- Dimensional Analysis 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 max, min, and τ max for the element is not the circle corresponding to transformations ti within the plane of stress. b) maximum shearing stress for the element is equal to half of the maximum stress. c) planes of maximum shearing stress are at 45 degrees to the plane of stress. 7-5

26 Yield Criteria for Ductile Materials Under Plane Stress 7-6 Fil Failure of a machine component subjected to uniaxial stress is directly predicted from an equivalent tensile test. Failure of a machine component subjected to plane stress cannot be directly predicted from the uniaxial 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 biaxial stress state. Fil Failure criteria i are based on the mechanism of failure. Allows comparison of the failure conditions for a uniaxial stress test and biaxial component loading.

27 Yield Criteria for Ductile Materials Under Plane Stress Maximum shearing stress criteria: Structural component is safe as long as the maximum shearing stress is less than the maximum shearing stress in a tensile test specimen at yield, i.e., τ < τ = max τ Y Y For a and b with the same sign, a b τ max = or < For a and b with opposite signs, a b τ max = < Y Y 7-7

28 Yield Criteria for Ductile Materials Under Plane Stress Maximum distortion energy criteria: Structural component is safe as long as the distortion energy per unit volume is less distortion energy per unit volume is less than that occurring in a tensile test specimen at yield. ( ) ( ) Y Y b b a a Y d G G u u + < + < Y b b a a < + 7-8

29 Fracture Criteria for Brittle Materials Under Plane Stress Brittle materials fail suddenly through rupture or fracture in a tensile test. The failure condition is characterized by the ultimate strength U. Maximum normal stress criteria: Structural component is safe as long as the maximum normal stress is less than the ultimate strength of a tensile test specimen. < a U < b U 7-9

30 Stresses in Thin-Walled Pressure Vessels Cylindrical vessel with principal stresses: 1 = hoop stress = longitudinal stress Hoop stress: F z = 0 = pr 1 = t Longitudinal stress: ( t x) p( r x)

31 Stresses in Thin-Walled Pressure Vessels Points A and B correspond to hoop stress, 1, and longitudinal stress,. Maximum in-plane shearing stress: τ 1 max( in plane) = = pr 4tt Maximum out-of-plane shearing stress corresponds to a 45 o rotation of the plane stress element around a longitudinal axis τ max = = pr tt 7-31

32 Stresses in Thin-Walled Pressure Vessels Spherical pressure vessel: 1 1 = = pr t Mohr s circle for in-plane transformations reduces to a point = = τ 1 max(in-plane) = constant = 0 Maximum out-of-plane shearing stress pr τ 1 max = 1 = 4t 7-3