University of Waterloo. ME Mechanical Design 1. Partial notes Part 1

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1 University of Waterloo Department of Mechanical Engineering ME 3 - Mechanical Design 1 Partial notes Part 1 G. Glinka Fall 005 1

2 Forces and stresses Stresses and Stress Tensor Two basic types of forces act on a body to produce stresses: - Forces of the first type are called surface forces and act on the surface of the body. Surface forces are generally exerted when one body comes in contact with another. - Forces of the second type are called body forces since they act on each particle of the body. Body forces are commonly produced by centrifugal, gravitational, or other force fields. A solid body of arbitrary size, shape, and material acted upon by an equilibrium system of surface and /or body forces will respond so as to develop a system of internal forces also in equilibrium at any point within the body. If the body were cut by an arbitrary imaginary plane, these internal forces would in general be distributed continuously over the cut surface and would vary across the surface in both direction and intensity. Furthermore, the internal force distribution would also be a function of the orientation of the plane chosen for investigation. Stress is the term used to define the intensity and direction of the internal forces acting at given point on a particular plane. Stress has a magnitude and a sense, and is also associated with an area (or planeover which it acts. Such mathematical objects are known as TENSORS and are of higher order than vectors. However, the components of the stress on the same plane (area are vectors. In other words the Tensor is made of Vectors. The intensity of force (vector normal to the plane (section is called the normal stress. The intensity of force acting parallel to the plane (section is called shear stress τ.

3 The meaning of stress tensor F zz F z F zy F x F zx F y ij Fij Fij = lim Akl 0 A x x kl k l 3

4 Fn normal force F = lim = lim A 0 A A 0 surface area A Ft shear force Ft τ = lim = lim A 0 A A 0surface area A n Consider first a surface (plane whose outer normal is the positive direction of axis z. A complete description of the magnitudes and directions of stresses on all possible planes through a point constitute the stress state at a point. As illustrated in the Figure below, it is possible to resolve force F into two components: one F n normal to the surface and the other F t tangential to the surface. The resultant force F associated with a particular plane (surface can be further resolved into three Cartesian components along the x, y, z axes i.e., F x, F y, F z. Subsequently, the Cartesian stress components on this plane can be defined as: Fz zz = lim A 0 A Fx τ zx = lim A 0 A Fy τ zy = lim A 0 A All the components are associated with the plane having the outer normal parallel to the z axis. If the same procedure is followed using surfaces whose outer normals are in the positive x and y directions, two more sets of Cartesian components xx, τ xy, τ xz and yy, τ yx, τ yz respectively can be obtained. 4

5 S T R E S S E S xx xy xz = zx zy zz ij yx yy yz 5

6 S T R A I N S ε xx εxy εxz εij = εyx εyy εyz ε zx εzy ε zz Note! γ ij εij = for i ε = ε for i = ij ij j j 6

7 Strains in terms of stresses ( Hooke' s law 1 εxx = xx ν ( yy zz E + 1 εyy = yy ( zz xx E ν + 1 εzz = zz ν ( xx yy E + 1 ( + ν 1 ( + ν 1 ( + ν γ = τ γ = τ γ = τ E E E xy xy yz yz zx zx Stresses in terms of strains ( Hooke' s law E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν xx xx yy zz E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν yy yy zz xx E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν zz zz xx yy E E E τ = γ τ = γ τ = γ ( + ν ( + ν ( + ν xy xy yz yz zx zx 7

8 Uni-axial stress state, 11 = 33 = 1 = 3 = 31 =0 and 0 ν ν ε11 = ; ε = ; ε33 = ; E E E ε = 0; ε = 0; ε = 0; ε ε = 0 ε 0 ; = 0 0 ij ij 0 0 ε Thin plate under uniaxial tension or bending 8

9 Plane stress state in principal axes, 33 = 1 = 3 = 31 =0 and 11 0, 0 ( + 11 ν ν ν 11 ε11 = ; ε = ; ε33 = E E E 11 ε = 0; ε = 0; ε = 0; ε ε = ε 0 ; = 0 ; ij ij 0 0 ε Plane stress state in principal axes, 33 = 1 = 3 = 31 =0 and 11 0, 0 Thin rotating disk ω 11 9

10 General plane strain state, 3 = 31 =0, ε 33 =0 and 11 0, 0, 33 0, 1 0 ( ( ( 11 ν + 33 ν ν 11 + ε11 = ; ε = ; ε33 = = 0 E E E ε = 0; ε = 0; ε = 0; then ( 33 ν = = ν + E ( ( 1 ( 1+ ( 1 ( ν ν ν ν ν11 ν ε11 = ; ε = ; ε33 = 0 E E ε = 0; ε = 0; ε = 0; or ε ε = 0 ε 0 ; = 0 0 ij ij ε = 0 0 ; ij = 0 0 εij ε ν + ( 11 10

11 Strains in terms of stresses ( Hooke' s law 1 εxx = xx ν ( yy zz E + 1 εyy = yy ( zz xx E ν + 1 εzz = zz ν ( xx yy E + 1 ( + ν 1 ( + ν 1 ( + ν γ = τ γ = τ γ = τ E E E xy xy yz yz zx zx Stresses in terms of strains ( Hooke' s law E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν xx xx yy zz E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν yy yy zz xx E = + ν ε + ν ε + ε ( ( ( 1 ( 1+ ν 1 ν zz zz xx yy E E E τ = γ τ = γ τ = γ ( + ν ( + ν ( + ν xy xy yz yz zx zx 11

12 Transformation of the stress tensor y yy ϕ τ xy xx x The stress components xx, yy, and τ xy can be calculated on planes x and y, i.e. in x-y co-ordinates. However we need to know stress components xx, yy, and τ xy on planes x and y being at angle ϕ with respect to the original system of co-ordinates x and y., yy y ϕ, τ xy, xx x The stress components xx, yy, and τ xy can be calculated on planes x and y, i.e. in x-y co-ordinates. It is possible to determine stress components xx, yy, and τ xy by transformation of the original stress components xx, yy, and τ xy. 1

13 Transformation of the stress tensor the mathematics y yy, yy y τ xy xx x, τ xy, xx ϕ x τ, xx, yy, xy xx + yy xx yy = + cos ϕ + τxy sin ϕ xx + yy xx yy = cos ϕ τxy sin ϕ xx yy = sin ϕ + τxy cos ϕ 13

14 The Mohr circle B( yy, τ yx y B ( yy, τ yx yy A( xx, τ xy y, yy xx τ xy x, τ xy, xx ϕ A ( xx, τ xy x xx yy B ( yy, τ yx A( xx, τ xy τ yx 0 τ yx R O ϕ α τ xy τ xy B( yy, τ yx yy ( xx + yy / A ( xx, τ xy τ xx 14

15 The Mohr circle - interpretation xx yy B ( yy, τ yx τ yx A( xx, τ xy O α τ xy 0 τ yx B( yy, τ yx R ϕ τ xy τ yy ( xx + yy / xx A ( xx, τ xy Position of the Mohr circle center: O [ ( xx + yy /, 0] Radius of the Mohr circle: R xx yy = + ( τ xy 15

16 The Mohr circle - interpretation xx yy B ( yy, τ yx A( xx, τ xy 0 τ yx τ yx R O ϕ α τ xy τ xy B( yy, τ yx yy ( xx + yy / A ( xx, τ xy τ τ xy sinα = ( / xx xx yy α τ, xx, yy, xy xx + yy = + Rcos( ϕ α xx + yy = Rcos( ϕ α = Rsin( ϕ α 16

17 The Mohr circle Principal stresses xx yy B ( yy, τ yx A( xx, τ xy 0 τ yx τ yx R O ϕ α 1 τ xy τ xy B( yy, τ yx yy ( xx + yy / A ( xx, τ xy τ xx xx + yy xx + yy xx yy 1 = + R = + + xx + yy xx + yy xx yy 1 = R = + τ xy sin α = α; ( / xx yy ( τxy ( τxy angle of principal axes 17

18 Measurement of basic mechanical properties of engineering materials Smooth laboratory specimens used for the determination of the ε ε ε 1 1 ε ε 11 = ε 33 = νε 6-8 mm ij = ε ij = ε ε ε 33 Stress and strain state in specimens used for the determination of material properties 18

19 uts ys Engineering stress Engineering strain Typical engineering stress-strain behavior to fracture point F. The ultimate tensile strength uts is indicated at point M. The inserts show the geometry of the deformed specimen at various points along the curve. 19

20 Typical stress-strain behavior of engineering materials f uts ys Ductile material f ys 0. Brittle material 0

21 Macroscopic behavior of engineering material under uni-axial tension u u u Stress, Elastic brittle material Y Stress, Mild steel Stress, High strength steel ε f Strain, ε ε f Strain, ε Strain, ε ε f M O D E L S Stress, Ideal elastic material Stress, Y ε = E ε = E for < Y Ideal elasticplastic material Stress, E E E ε 1 n' = + ' E K Strain hardening material Strain, ε Strain, ε Strain, ε Y Specimen Stress, ε = < A 0 for Y Ideal rigidplastic material Strain, ε F = F A F 1

22 MONOTONIC STRESS-STRAIN BEHAVIOUR Basic Definitions. A monotonic tension test of a smooth specimen is usually used to determine the engineering stress-strain behaviour of a material where S = engineering stress = e engineering strain P A o = = = o l l l l l o o where: P = applied load L o = original length d o = original diameter A o = original area l = instantaneous length d = instantaneous diameter A = instantaneous area Due to changes in cross-sectional area during deformation the true stress in tension is larger than the engineering stress, = true stress = P A True or natural strain, based on an instantaneous gage length l, is defined as dl l ε = true strain = = ln l l l l 0 o

23 A 0 d 0 l 0 F A F - applied load; l 0 - original length d 0 - original diameter; A 0 - original area l - instantaneous length d - instantaneous diameter A 0 - instantaneous area d l S ( 1 ε ln( 1 = S + e and = + e e = engineering stress = engineering strain 0 = = = F = true stress = F A dl l ε = true strain = = ln l l l l 0 F A 0 l l l l l

24 . True stress and strain can be related to engineering stress and strain. ε = ln(1 +e = S(1 + e The equation above is only valid up to necking. At necking the strain is no longer uniform throughout the gage length. The total true strain ε in a tension test can be separated into elastic and plastic components: Linear elastic strain: that portion of the strain, which is recovered upon unloading, ε e. Plastic strain (nonlinear: that portion which cannot be recovered on unloading, ε p. Stated in equation form, ε = ε + ε e p For most metals a log-log plot of true stress versus true plastic strain is modelled as a straight line. Consequently, this curve can be expressed using a power function. ε p = K 1/n where: K is the strength coefficient and n is the strain hardening exponent. 4

25 At fracture two important quantities can be defined. These are true fracture strength and true fracture ductility. True fracture strength, f, is the true stress at final fracture. = f P A f f where: A f is the area at fracture and P f is the load at fracture. True fracture ductility, ε f, is the true strain at final fracture. This value can be defined in terms of the initial cross-sectional area and the area at fracture. Ao 1 ε f = ln = ln A 1 RA f A A = = A o f RA reduction in area o The strength coefficient, K, can be defined in terms of the true stress at fracture, f, and the true strain at fracture, ε f. f K = ε n f We can also define plastic strain in terms of these quantities. ε p ε = f f 1/n The total strain can be expressed as: ε = ε + ε e p 5

26 Static Strength Theories The material properties such as the yield limit, S ys ( ys, and the ultimate strength, S ut ( ut, are obtained in simple uni-axial tensile tests. However the analyst or designing engineer must asses the critical loading conditions for machine and structural components which are most often subjected to multiaxial stress states. Therefore a theory is needed (or failure criterion allowing to translate the mutiaxial stress state into an equivalent uniaxial stresses state making possible the use of uniaxial material properties. Maximum Normal Stress Theory Failure is predicted to occur in the multiaxial stress state of stress when the maximum principal normal stress becomes equal to or exceeds the maximum normal stress at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e., failure is predicted by the maximum normal stress theory to occur if: ; ; ; or 1 f t f t 3 f t ; ; ; 1 f c f c 3 f c Where: f-t is the uni-axial tensile strength of the material f-c is the uni-axial compressive strength of the material The maximum normal stress theory provides good results for brittle materials. 6

27 Graphical representation of the maximum normal stress failure criterion General bi-axial plane stress state, 33 = 1 = 31 =0 and 11 0, 0, 3 0 reduces two two non-zero principal stresses: 1 0, 0 and 3 =0 f-t f-c 0 f-t 1 f-c principal stresses: + = + + ( ( = + = 0 7

28 Maximum Shearing Stress Theory (Tresca Failure is predicted to occur in the multiaxial stress state of stress when the maximum shearing stress becomes equal to or exceeds the maximum shearing stress magnitude at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e., failure is predicted by the maximum normal stress theory to occur if: τ τ ; τ τ ; τ τ ; 1 f f 3 f Where: τ f - is the shear strength of the material obtained in uni-axial tension test. τ f = The maximum normal stress theory provides good results for ductile materials. The Tresca theory in principal stresses: General bi-axial plane stress state, 33 = 1 = 31 =0 and 11 0, 0, 3 0 reduces two two non-zero principal stresses: 1 0, 0 and 3 =0 1 ys τ1 τ f; ; 1 0 ys τ τ f; ; 0 1 ys τ3 τ f; ; 1 ys ys ; ys ys 8

29 Graphical representation of the Maximum Shearing Stress Theory (Tresca 1 = 0 = ys 1 = ys = ys 1 = - ys / = ys / 1 =- ys = 0 safe region = ys = 0 1 = ys / = - ys / 1 = - ys = - ys 1 = 0 = - ys 9

30 Distortion Energy Theory (Huber-von Mises-Hencky Failure is predicted to occur in the multiaxial stress state of stress when the distortion energy per unit volume becomes equal to or exceeds the distortion energy per unit volume at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e., failure is predicted by the Distortion Energy theory to occur if: eq Where: - for general 3-D stress state: 1 eq = ys ( ( ( 6 ( ( ( for 3-D stress state in principal stresses: 1 eq = + + ( ( ( for general plane stress state: ( ( ( = eq = ( 11 + ( 0 + ( ( 1 + ( 0 + ( 0 + ( ( 3( = for plane stress state in principal stresses: 1 eq = ( ( 0 ( 0 ( ( =

31 - for plane strain state in principal stresses: eq ( ( ( = ( ( ν( ν( ν ( = ( 1 ( ( 1 ν ν 1( 1 ν ν = Graphical representation of the Maximum Distortion Energy Theory (H-M-H 1 = 0 = ys 1 = ys = ys 1 = - ys / 3 = ys / 3 1 =- ys = 0 0 safe region 1 1 = ys = 0 1 = ys / 3 = - ys / 3 1 = - ys = - ys 1 = 0 = - ys 31

32 Comparison of the Tresca and H-M-H failure criterion of for plane stress states 1 = - ys / 3 = ys / 3 1 = 0 = ys 1 = ys = ys 1 = - ys / = ys / 1 =- ys = 0 0 safe region 1 1 = ys = 0 1 = ys / 3 = - ys / 3 1 = - ys = - ys 1 = 0 = - ys 1 = ys / = - ys / 3

33 Example: A thick-wall cylindrical pressure vessel made of carbon steel plate (ASME SA-85M, Grade C, ut = 65 ksi, ys = 30 ksi is pressurized internally. Find the maximum internal pressure, p i, at the initiation of plastic yielding in the cylinder s wall. The internal and external diameter is 8 in and 1 in respectively. Use the Tresca and H-M-H theory. Assume open ended cylinder with: a unrestrained ends, b with restrained ends D i = a = 8 in. D o = b =1 in. a b = θθ θθ p 11 = -p a Internally pressurized thick wall cylinder with unrestrained ends, 33

34 Radial and hoop stresses in an open ended internally pressurized thick wall cylinder (with un-restrained ends p a b p a b rr = 1 ; 1 θθ = + b a r b a r = 0 zz Maximum stresses at the inner surface of the cylinder: rr( at r= a zz( at r= a θθ ( at r= a then 3 rr( at r= a p a b = 1 = p; = 0 b a r p a b b + a = 1+ = p b a r b a b + a = θθ ( at r= a = p p.6 p = = b a 6 4 = 0 = = p Plastic yielding will commence according to the Tresca criterion when: ( 1 3 = ys.6 p p = 30ksi 3.6 p = 30ksi lb p = 8.777ksi p= 8777 in. 34

35 Plastic yielding will commence according to the H-M-H criterion when: eq ys eq ( ( ( = ( ( ( ( ( = p +.6 p + 0 p.6 p 0 0 = 3.18p eq = ys 3.18p = 30ksi lb p= 9.3 ksi p= 93 in 35

36 b Internally pressurized thick wall cylinder with restrained ends, Thick cylinder under internal pressure and torsion with fixed ends (no axial expansion = θθ θθ p = -p Radial and hoop and axial stresses in an open ended internally pressurized thick wall cylinder with restrained ends rr p a b p a b = 1 ; 1 ; θθ = + b a r b a r ( = ν + ; because ε = 0 zz rr θθ zz 36

37 Maximum stresses at the inner surface of the cylinder: p a b = 1 = p b a r rr ( at r= a p a b b + a = 1 + = p b a r b a θθ ( at r= a a = ν( + = ν p b a zz( at r= a rr θθ then b + a = θθ ( at r= a = p p.6 p = = b a 6 4 a 4 = zz( at r= a = ν p = 0. 3p = 0.48p b a = rr ( at r= a = p Plastic yielding will commence according to the Tresca criterion when: ( 1 3 = ys.6 p p = 30ksi 3.6 p = 30 ksi lb p = 8.777ksi p = 8777 in. 37

38 Plastic yielding will commence according to the H-M-H criterion when: eq ys eq ( ( ( = ( p + (.6 p + ( 0.48p ( p(.6 p (.6 p( 0. 48p ( 0.48 p( p = = p eq = ys p = 30 ksi lb p = 9.57 ksi p = 957 in 38

39 The effect of the mutiaxiality of the stress state on the tendency to plastic yielding The change of the maximum stress necessary for plastic deformation of a ductile material under various stress states: a uni-axial tension, b tension with transverse compression, c bi-axial tension, and d hydrostatic compression (after N. Dowling 39