Unit IV State of stress in Three Dimensions

Transcription

1 Unit IV State of stress in Three Dimensions

2 State of stress in Three Dimensions References Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi, Rajput R.K., "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi,

3 State of stress in Three Dimensions Contents: Determination of principal stresses and principal planes Volumetric strain Theories of failure Principal stress - Principal strain shear stress Strain energy and distortion energy theories Application in analysis of stress, load carrying capacity.

4 State of stress in Three Dimensions Determination of principal stresses and principal planes Volumetric strain 4

5 State of Stress in Three Dimensions Stress tensor at a point: The Cartesian coordinates of state of stress at a point can be represented by the following nine components of stress known as the stress tensor. τ ij = τ xx τ xy τ xz τ yx τ zx τ yy τ zy τ yz τ zz = σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z 5

6 State of Stress in Three Dimensions Stress tensor at a point: y σ y τ yz τ yx σ z τ xy σ x τ zy τ zx τ xz σ x x σ z τ yx τ yz z σ y 6

7 State of Stress in Three Dimensions Stress tensor at a point: τ xx τ xy τ xz τ ij = τ yx τ yy τ yz = τ zx τ zy τ zz σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z Spherical part of tensor τ ij = αi = 1 3 tr τ ij σ x +σ y +σ z σ x +σ y +σ z σ x +σ y +σ z 3 I = σ x+σ y +σ z = 7

8 State of Stress in Three Dimensions Stress tensor at a point: τ xx τ xy τ xz τ ij = τ yx τ yy τ yz = τ zx τ zy τ zz σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z Deviatoric part of tensor τ ij = τ ij αi 8

9 Example: α = 1 3 tr(τ ij)= = 1 State of Stress in Three Dimensions τ ij = Spherical part of τ ij =αi= Deviatoric part of tensor τ ij = τ ij αi =

10 Stress invariants: Invariant is a one whose value doesn t change when the frame of reference is changed. The invariants are the coefficients of σ 2, σ and the last term in the cubic equation obtained with the following relation τ ij σi = 0 σ x σ τ xy τ xz τ yx σ y σ τ yz τ zx τ zy σ z σ = 0 σ 3 (σ x + σ y + σ z )σ 2 State of Stress in Three Dimensions + σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 σ (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 )=0 10

11 Stress invariants: Invariant is a one whose value doesn t change when the frame of reference is changed. The invariants are the coefficients of σ 2, σ and the last term in the cubic equation obtained with the following relation τ ij σi = 0 σ x σ τ xy τ xz τ yx σ y σ τ yz τ zx τ zy σ z σ = 0 σ 3 (σ x + σ y + σ z )σ 2 State of Stress in Three Dimensions + σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 σ (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 )=0 σ 3 I 1 σ 2 + I 2 σ I 3 = 0 11

12 σ 3 (σ x + σ y + σ z )σ 2 + σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 σ (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 )=0 Where invariants are, State of Stress in Three Dimensions σ 3 I 1 σ 2 + I 2 σ I 3 = 0 I 1 = σ x + σ y + σ z I 2 = σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 I 3 = (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 ) The quantities I 1, I 2 and I 3 are known as the first, second and third invariants of stress respectively. In terms of principal stresses, I 1 = σ 1 + σ 2 + σ 3 ; I 2 = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 and I 3 = σ 1 σ 2 σ 3 12

13 Problem 1: With respect to the frame of reference oxyz, the following state of stress exists. Determine the principal stresses and their associated directions. Also check on the invariences of I 1, I 2 and I 3. τ ij =

14 Solution: For this state, σ x τ xy τ xz τ ij = τ yx σ y τ yz = τ zx τ zy σ z Calculation of invariants: I 1 = σ x + σ y + σ z = = 3 I 2 = σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 I 2 = = 3 I 3 = (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 ) I 3 = = = 1 14

15 I 1 = 3; I 2 = 3 and I 3 = 1 Using the relation τ ij σi = 0, we will get a form σ 3 I 1 σ 2 + I 2 σ I 3 = 0 Now by substituting the I 1, I 2 and I 3 values the above equation will be modified as, σ 3 3σ 2 3σ + 1 = 0 σ σ σ + 1 = 0 σ σ σ + 1 3σ σ + 1 = 0 σ + 1 (σ + 1) 2 6σ = 0 σ + 1 σ σ 6σ = 0 σ + 1 σ 2 4σ + 1 = 0 Hence one solution is σ = 1. The other two solutions are obtained from the quadratic equation, σ 2 4σ + 1 = 0 15

16 σ 2 4σ + 1 = 0 σ = 4 ± = 4 ± 16 4 = 2 ± σ 1 = 1, σ 2 = = 3.73, σ 3 = 2 3 = Check on the invariance: With the set of axes chosen along the principal axes, the stress matrix will have the form τ ij = I 1 = = 3 I 2 = σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 = 1(2 + 3) + (2 + 3) ( 1)(2 3) = = 3 I 3 = σ 1 σ 2 σ 3 = ( 1)(2 + 3)(2-3) = 4 3 = 1 16

17 Direction of principal axes or direction cosines: (i) For σ 1 = 1 τ ij σ 1 I n = n x n y n z = n x n y n z = 0 2n x +2n y + n z = 0 (1) 2n x +2n y + n z = 0 (2) n x +n y + 2n z = 0 (3) 17

18 Direction of principal axes or direction cosines: together with n x 2 + n y 2 + n z 2 = 1 4 From (2) and (3 ) equations above, n z = 0 and using this in (3) we will obtain n x = n y. Now by substituting n z = 0 and n x = n y in (4), n y 2 + n y 2 = 1 n y = ± 1 2 n x = 1 2 Hence σ 1 = 1 is in the direction + 1 2, It should be noted that the plus and minus signs associated with n x, n y and n z represent the same line. 18

19 Direction of principal axes or direction cosines: (ii) For σ 2 = τ ij σ 2 I n = n x n y n z = ( 1 3)n x +2n y + n z = 0 (5) 2n x +( 1 n x +n y + ( 1 3)n y + n z = 0 (6) 3)n z = 0 (7) n x n y n z = 0 19

20 Direction of principal axes or direction cosines: together with n x 2 + n y 2 + n z 2 = 1 4 Solving, we get 2n x +( 1 3)n y + n z = 0 (6) 2n x +2n y + ( 2 2 3)n z = 0 (7) 2 ( 1 3 2) n y + ( )n z = 0 ( 3 3) n y = ( ) n z ( ) n y = (3 + 3) n z n y = 1.366n z From (7), n x n z + ( 1 3)n z = 0 n x = 1.366n z 20

21 Direction of principal axes or direction cosines: From (4), (1.366n z ) 2 + (1.366n z ) 2 + n z 2 = 1 n z = n x = 1.366n z = = n x = n y = 1.366n z = = n y =

22 Direction of principal axes or direction cosines: (iii) For σ 3 = τ ij σ 3 I n = n x n y n z = ( 1 + 3)n x +2n y + n z = 0 (8) 2n x +( 1 + n x +n y + ( 1 + 3)n y + n z = 0 (9) 3)n z = 0 (10) n x n y n z = 0 22

23 Direction of principal axes or direction cosines: together with n x 2 + n y 2 + n z 2 = 1 4 Solving, we get 2n x +( 1 + 3)n y + n z = 0 (9) 2n x +2n y + ( )n z = 0 (10) 2 ( ) n y + ( )n z = 0 ( 3 + 3) n y = (3 2 3) n z (3 2 3) n y = (3 3) n z n y = 0.366n z From (10), n x 0.366n z + ( 1 + 3)n z = 0 n x = 0.366n z 23

24 Direction of principal axes or direction cosines: From (4), ( 0.366n z ) 2 + ( 0.366n z ) 2 + n z 2 = 1 n z = n x = 0.366n z = = n x = n y = 0.366n z = = n y =

25 Problem 2 The rectangular stress components of a point in three dimensional stress system are defined as σ x = 1; σ y = 2, σ z = 4, τ xy = 2, τ yz = 3 and τ zx = 1. All in units of kpa. Find principal stresses and their direction cosines. Note: Tutorial class 25

26 Problem 3: The rectangular stress components of a point in three dimensional stress system are defined as σ x = 20 MPa; σ y = 40 MPa, σ z = 80 Mpa, τ xy = 40 Mpa, τ yz = 60 MPa a nd τ zx = 20 MPa. Determine the principal stresses and principal planes. Find also the maximum shear stress (April/May 2011). 26

27 Solution: For this state, τ ij = σ x τ xy τ xz τ yx τ zx σ y τ zy τ yz σ z = Calculation of invariants: I 1 = σ x + σ y + σ z = = 60 I 2 = σ x σ y + σ y σ z + σ z σ x τ 2 xy τ 2 yz τ 2 zx I 2 = = 8000 I 3 = (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 ) I 3 = =

28 σ 3 I 1 σ 2 + I 2 σ I 3 = 0 σ 3 60σ σ = 0 Now cos 3θ = 4 cos 3 θ 3 cos θ cos 3 θ 3 4 cos θ 1 cos 3θ = 0 (a) 4 Put σ = r cos θ + I 1 = r cos θ The cubic equation becomes, r cos θ r cos θ r cos θ = 0 r 3 cos 3 θ + 60r 2 cos 2 θ + 120r cos θ r 2 cos 2 θ r cos θ 8000r cos θ = 0 r 3 cos 3 θ 9200r cos θ = 0 28

29 r 3 cos 3 θ 9200r cos θ = 0 cos 3 θ cos θ r2 r 3 Hence, equations (a) and (b) are identical if 9200 r 2 = 3 4 r 2 = r = r 3 = 1 cos 3θ 4 cos 3θ = cos 3θ = = 0 b 29

30 cos 3θ = θ 1 = , θ 2 = , θ 3 = Since θ 2 = 120 θ 1 and θ 3 = θ 1 r cos θ 1 = cos 19.6 = r cos θ 2 = cos = r cos θ 3 = cos = σ 1 = r cos θ = = MPa σ 2 = r cos θ = = 0 MPa σ 3 = r cos θ = = MPa Therefore principal stresses are Mpa, 0 and Mpa with the Principal planes with , and Maximum shear stress, τ max = σ 1 σ 3 2 = ( 64.33) 2 = MPa. 30

31 Therefore principal stresses are Mpa, 0 and Mpa with the Principal planes with , and Maximum shear stress, τ max = σ 1 σ 3 2 = ( 64.33) 2 = MPa. 31

32 Problem 4: With respect to the frame of reference oxyz, the following state of stress exists. Determine the principal stresses and orientation of principal planes. τ ij =

33 Solution: τ ij = σ x τ xy τ xz τ yx τ zx σ y τ zy τ yz σ z = Calculation of invariants: I 1 = σ x + σ y + σ z = = 3 I 2 = σ x σ y + σ y σ z + σ z σ x τ xy 2 τ yz 2 τ zx 2 I 2 = = 3 I 3 = (σ x σ y σ z + 2τ xy τ yz τ zx σ x τ yz 2 σ y τ zx 2 σ z τ xy 2 ) I 3 = = = 1 33

34 I 1 = 3; I 2 = 3 and I 3 = 1 Using the relation τ ij σi = 0, we will get a form σ 3 I 1 σ 2 + I 2 σ I 3 = 0 Now by substituting the I 1, I 2 and I 3 values the above equation will be modified as, σ 3 3σ 2 3σ + 1 = 0 Now cos 3θ = 4 cos 3 θ 3 cos θ Put σ = r cos θ + I 1 3 cos 3 θ 3 4 cos θ 1 cos 3θ = 0 (a) 4 = r cos θ + 1 The cubic equation becomes, r cos θ r cos θ r cos θ = 0 34

35 r cos θ r cos θ r cos θ = 0 r 3 cos 3 θ r 2 cos 2 θ + 3r cos θ 3r 2 cos 2 θ 3 6r cos θ 3r cos θ 2 = 0 r 3 cos 3 θ 6r cos θ 4 = 0 cos 3 θ 6 r 2 cos θ 4 r 3 = 0 b Hence, equations (a) and (b) are identical if 6 r 2 = 3 4 r 2 = 8 r = r 3 = 1 cos 3θ 4 35

36 cos 3θ = θ 1 = 15 0, θ 2 = 105 0, θ 3 = Since θ 2 = 120 θ 1 and θ 3 = θ 1 r cos θ 1 = 2 2 cos 15 = 2.73 r cos θ 2 = 2 2 cos 105 = 0.73 r cos θ 3 = 2 2 cos 135 = 2 σ 1 = r cos θ = = 3.73 σ 2 = r cos θ = = σ 3 = r cos θ = = 1 Therefore principal stresses are 3.73, and -1 with the Principal planes with 15 0, and Maximum shear stress, τ max = σ 1 σ 3 2 = 3.73 ( 1) 2 =

37 Problem1: A.U. Question paper problems The Principal tensile stresses at a point across two perpendicular planes are 120 MN/m 2 and 60 MN/m 2. Find (i) the normal and tangential stress and the resultant stress and its obliquity on a plane at 20 0 with major principal plane. (ii) the intensity of stress which acting alone can produce the same maximum strain. Take Poisson s ratio=1/4. (May/June 2013) 37

38 Problem 2: The state of stress (Cartesian components of stress) at a point are σ xx = 7 MPa, σ yy = 6 MPa, σ zz = 5 MPa, τ xy = 2 Mpa, τ yz = 2 Mpa, τ xz = 0 Mpa. Determine the values of principal stresses (May/June 2012). A.U. Question paper problems 38

39 2 marks Questions and Answers 1. Define stress tensor. 2. Define principal stress and principal strain? 3. What are the various stress invariants for three dimensional state of stress. 39