Module #3. Transformation of stresses in 3-D READING LIST. DIETER: Ch. 2, pp Ch. 3 in Roesler Ch. 2 in McClintock and Argon Ch.
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1 HOMEWORK From Dieter -3, -4, 3-7 Module #3 Transformation of stresses in 3-D READING LIST DIETER: Ch., pp Ch. 3 in Roesler Ch. in McClintock and Argon Ch. 7 in Edelglass
2 The Stress Tensor z z x O zz zy zx xz xy xx y yz yy yx x y ij yx yy xx xy xz yz zx zy zz In three dimensions the state of stress is described by the stress tensor. We can transform from one coordinate system to another in the same way that we did for two dimensions.
3 Method Lets resolve an arbitrary 3D state of stress onto an oblique plane ABC (area A). To make the problem easier, let S be parallel to the plane normal (meaning that it is a principal stress acting on a principal plane (i.e., the plane w/o shear). z z C normal yy yz xy yx xz O zx xx A S = B y x x z y y x A zy zz DIRECTION COSINES l = cos θ x m = cos θ y n = cos θ z
4 z DIRECTION COSINES l = cos θ x m = cos θ y n = cos θ z C xy xx S = yy yz yx O xz zx A B y x A zy zz z z normal x y y L* x
5 The components of S parallel to the original x-, y-, and z- axes (i.e., S x, S y, S z ) are: S x = Sl = σl To balance force we need Area COB = Al. S y = Sm = σm the areas that each stress Area AOC = Am. S z = Sn = σn components acts on Area AOB = An. z x yy A z C xy yx xz yz zy O zz zx xx A S S z S x B S y y S S S S x y z x x z y normal Recall: l = cos θ x m = cos θ y n = cos θ z y
6 All forces must balance to meet the conditions for static equilibrium (i.e., F=0): Fx Sx A SAl xxal yxam zxan Fy Sy A SAm yyam xyal zyan F S A SAn An Al Am z z zz xz yz Fx ( xx S) l yxmzxn0 Fy x yl( yy S) mzyn0 F l m ( S) n 0 z xz yz zz S S xx yx zx xy yy zy xz yz zz S l 0 m 0 n 0 When written in matrix form
7 Method cont d The solution of the determinant of the matrix on the left yields a cubic equation in terms of S. S S S 3 ( xx yy zz ) ( xx yy yyzz xxzz xy yz xz ) ( xx yyzz xy yzxz xx yz yyxz zzxy ) 0 S or I S I SI In this problem, S =. Thus, the three roots of this cubic equation represent the principal stresses, 1,, and 3.
8 Method cont d The directions in which the principal stresses act are determined by substituting 1,, and 3, each for S in: ( xx S) l yxmzxn0 xyl ( yy S) mzyn 0 l m ( S) n0 xz yz zz Then the resulting equations must be solved simultaneously for l, m, and n (using the relationship l +m +n = 1). (a) Substitute 1 for S ; solve for l, m, and n; (b) Substitute for S ; solve for l, m, and n; (c) Substitute 3 for S ; solve for l, m, and n.
9 Invariants of the Stress Tensor I I 1 xx yy zz xx xy yy yz xx xz yx yy zy zz zx zz xx yy yy zz xx zz xy yz xz I xx xy xz 3 yx yy yz xx yyzz xy yzxz xx yz yyxz zzxy zx zy zz Whenever stresses are transformed from one coordinate system to another, these three quantities remain constant.
10 Example Problem #1 Determine the principal normal stresses for the following state of stress: 0, 10, 75, xx yy zz 50, 0 xy yz xz or MPa
11 Example Problem #1 solution This problem can be solved by substituting the known state of stress into the cubic equation: where S = σ. 3 S I1S ISI3 0 This is detailed on the next page.
12 MPa ( xx yy zz ) ( xx yy yyzz xxzz xy yz xz ) ( xx yy zz xy yz xz xx yz yy xz zz xy ) ( 75 ) [(0 10) (10 75) (0 75) ( 50) (0) (0) ] [0 ( 1075) ( 50 00) (00 ) (10 0 ) ( 7550 )] 0 3 ( 65) [ 350] [187500] 0 I 1 I I
13 You can determine the principal stresses by plotting this equation OR you can solve it using more traditional means (10 6 MPa) = -75 = = (MPa)
14 Example Problem #1 solution This problem is easier than most because there are no shear stresses along the z-axis. It should have been obvious toyou that one of the principal stresses is = -75 MPa (since zx = xz = 0 and zy = yz = 0). Can you determine the directions in which the principal stresses act? (I RECOMMEND THAT YOU TRY IT)
15 Resources on the Web There are many useful eigenvalue calculators on the world wide web. Here are a few: /MohrCircles-3D/Applets/applet.htm H*
16 Example Problem # p. 1/11 Determine (a) the principal stresses, (b) maximum shear stress, and (c) the orientations of the principal planes for the state of stress provided below: MPa ( xx yy zz ) ( xx yy yyzz xxzz xy yz xz ) ( xx yy zz xy yz xz xx yz yy xz zz xy ) ) 3 ( 0 0) [(80 40) ( 40 60) (80 60) (0) (30) (50 [( ) ( ) (80 30 ) ( ) (60 0 )] 0 ] ( 100) [ 4600] [ 48000]
17 p. / (10 6 MPa) = 11 MPa; = 36 MPa; 3 =-57 MPa = 36 1 = 11 3 = (MPa)
18 MPa ( ) l m n 0 xx yx zx l ( ) m n 0 xy yy zy l m( ) n 0 xz yz zz (80 ) l 0m50n0 0 l ( 40 ) m30n0 50l 30 m(60 ) n0 substitute,, and in place of and solve simultaneous equations p. 3/ (80 ) l 0m50n0 41l 0m50n l ( 40 ) m30n 0 0l 161m30n 0 50l 30 m(60 ) n0 50l 30m61n 0 3 (3 [1]) + (5 []) yields: 13l 60m150n0 100l 805m150n0 3l 745m0n0; m0.0309l
19 Sample Problem # cont d p. 4/11 substitute,, and in place of and solve simultaneous equations (80 ) l 0m50n0 41l 0m50n l ( 40 ) m30n 0 0l 161m30n 0 50l 30 m(60 ) n0 50l 30m61n (3 [1]) + (- [3]) yields: 13l 60m150n0 100l 60m1n0 3l 0m8n0; n0.81l
20 Sample Problem # cont d p. 5/11 l m n 1 substitute expresions for m & n l 0.031l 0.81l 1.673l l m0.031l 0.04 n0.81l Orientations of principal planes associated with 1
21 Sample Problem # cont d p. 6/11 substitute,, and in place of and solve simultaneous equations (80 ) l 0m50n 0 44l 0m50n0 [4] 0 l ( 40 ) m30n 0 0l 76m30n0 [5] 50l 30 m(60 ) n 0 50l 30m4n0 [6] 5 (3 [4]) + (5 [5]) yields: 13l 60m150n0 100l 380m150n0 3l 30m0n0; m0.75l
22 Sample Problem # cont d p. 7/11 substitute,, and in place of and solve simultaneous equations (80 ) l 0m50n 0 44l 0m50n0 [4] 0 l ( 40 ) m30n 0 0l 76m30n0 [5] 50l 30 m(60 ) n 0 50l 30m4n0 [6] 6 (3 [4]) + (- [6]) yields: 13l 60m150n0 100l 60m48n0 3l 0m198n0; n1.17l
23 Sample Problem # cont d p. 8/11 l m n 1 substitute expresions for m & n l 0.75l 1.17l.899l 7 1 l m0.75l 0.46 n1.17l Orientations of principal planes associated with
24 Sample Problem # cont d p. 9/11 substitute,, and in place of and solve simultaneous equations (80 ) l 0m50n0 137l 0m50n l ( 40 ) m30n 0 0l 17m30n l 30 m(60 ) n 0 50l 30m117n (3 [7]) + (5 [8]) yields: 411l 60m150n0 100l 85m150n0 511l 145m0n0; m3.54l
25 Sample Problem # cont d p. 10/11 substitute,, and in place of and solve simultaneous equations (80 ) l 0m50n0 137l 0m50n l ( 40 ) m30n 0 0l 17m30n l 30 m(60 ) n 0 50l 30m117n (3 [7]) + (- [6]) yields: 411l 60m150n0 100l 60m34n0 511l 0m384n0; n1.331l
26 Sample Problem # cont d p. 11/11 l m n 1 substitute expresions for m & n l 3.54l 1.331l l 10 1 l m3.54l n1.331l 0.34 Orientations of principal planes associated with max 11 ( 57) 89 MPa
27 5 minute break
28 General Method for Triaxial States of Stress [p in Dieter] In previous example/method, we assumed that the stress on the inclined plane was a principal stress. What if the stress on the new plane is not a principal stress? The math is nearly the same. S S S S n s x yy A yz x y z C yx zy O zz z xy xz zx xx S s n B x y x z z y normal y
29 Triaxial Stress States cont d From summation of forces parallel to the x, y, z axes, we find the components S x, S y, S z : S l m n x xx yx zx S l m n y xy yy zy S l m n z xz yz zz The normal stress on the oblique plane equals the sum of the components S x, S y, S z parallel to the plane normal. SlSmSn n x y z l m n lm mn nl xx yy zz xy yz zx
30 Triaxial Stress States cont d From the expression = σ + τ S n s, the shear stress can be obtained. When written in terms of principal axes, it becomes: s 1 lm 1 3 ln 3 mn The maximum shear stress occurs when: max 1 3 max min
31 Mohr s s Circle in 3-D3 We can use a 3-D Mohr s circle to visualize the state of stress and to determine principal stresses. Essentially three -D Mohr s circles corresponding to the x-y, x-z, and y-z faces of the elemental cubic element. z zz zx zy yz xz yy xy yx xx y x
32 1 max Uniaxial Tension = = 0 3 = max 1 max = = 3 3 max = 1 Biaxial Tension = 3 max = = = = 0 σ 3 1 Uniaxial Compression = = 3 Triaxial Tension (unequal) Adapted from G.E. Dieter, Mechanical Metallurgy, 3 rd ed., McGraw-Hill (1986) p. 37 σ = - = - 3 = 3 1 Uniaxial Tension plus Biaxial Compression
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