Unformatted text preview: First deal with XZ plane Calculate Average & Radius Avg = X + Y 0 + 0 = = 0ksi 2 2
2 2 Y 2 0  (0) 2 R= X + XY = + (5) = 5ksi 2 2 Mohr's circle is on right. Using Mohr's circle find Principal Stresses 1 = Avg + R 1 = 5ksi 2 = Avg  R 2 = 5.0ksi
Now element is in all principal Stresses Can plot three circles by looking at face of one of principal stresses, three faces to look at In the end we have. MAX = 7 ksi MIN = 5ksi INT = 5ksi ABS =
Max MAX  MIN 7ksi  (5ksi) = = 6ksi 2 2 Calculate Average & Radius X = 850 psi Y = 0 XY = 600 psi Avg = X + Y  850 psi + 0 = = 425 psi 2 2
2 2 Y 2  850  (0) 2 R= X + XY = + (600 ) = 735.3 psi 2 2 Using Mohr's circle find Principal Stresses and Orientation 1 = Avg + R = 425 psi + 735 .3 psi 1 = 310 psi 2 = Avg  R = 425 psi  735.3 psi 2 = 1160 psi tan 2 P = XY 600 = ( X  Y ) / 2 (850  0) / 2 P =  27.3o
Now find it for Max Shear = Avg = 425 psi Max = R = 735.3 psi tan 2 S = S = 17 .7 o  ( X  Y ) / 2  (850  0) / 2 = XY 600 Calculate Average & Radius X = 850(10 6 ) Y = 480(10  6 ) XY = 650(10 6 ) Avg X + Y 850(10  6 ) + 480 (10 6 ) = = = 665(10 6 ) 2 2
2 2 2 2 850(10 6 )  480(10 6 ) 650 (10 6 ) X  Y XY + = 374 (10 6 ) R= + = 2 2 2 2 Using Mohr's circle find Principal Strains and Orientation 1 = Avg + R = 665(10 6 ) + 374(10 6 ) 1 = 1039(10 6 ) 2 = Avg  R = 665(10 6 )  374(10  6 ) 2 = 291(10 6 ) XY 650 (10 6 ) tan 2 P = = ( X  Y ) (850 (10 6 )  480 (10  6 )) P = 30.2o
Now find it for Max Shear strain = Avg = 665(10 6 ) Max = 2 * R = 748(10  6 )  ( X  Y )  (850(10  6 )  480(10  6 )) tan 2 S = = XY 650(10 6 ) S = 14.8o Mohr's circle Each element Principal Strains Max Shear strain Use Strain Rosette Equations A = X cos 2 A + Y sin 2 A + XY cos A sin A B = X cos 2 B + Y sin 2 B + XY cos B sin B C = X cos 2 C + Y sin 2 C + XY cos C sin C
Where: A = 950 (10  6 ) : A = 0 o B = 380 (10  6 ) : B =  60 o C =  220 (10  6 ) : A =  120
o Substitution A = 950(10 6 ) = X B = 380(10  6 ) = X (0.25) + Y (0.75) + XY (0.433) C = 220 (10 6 ) = X (0.25) + Y (0.75) + XY (0 .433)
Take strain X results and plug in 380(10 6 ) = 950(10 6 )(0.25) + Y (0.75) + XY (0.433)  220 (10 6 ) = 950(10 6 )(0 .25) + Y (0.75) + XY (0 .433)
Add equations together and Solve for strain in Y 160(10 6 ) = 475(10 6 ) + 2 Y (0.75) Y = 210 (10  6 )
Use results and solve one of the previous equations 380(10 6 ) = 950(10 6 )(0.25) + 210(10 6 )(0.75) + XY (0.433) 300(10 6 ) = XY (0.433) XY = 692.8(10 6 )
Calculate Average & Radius X + Y 950(106 ) + ( 210(106 )) Avg = = = 370(106 ) 2 2 950(10 6 )  (210(10 6 )) 692.8(106 ) X  Y XY + = 675.6(106 R= + = 2 2 2 2 Using Mohr's circle find Principal Strains and Orientation
2 2 2 2 1 = Avg + R = 370(10 6 ) + 675 .6(10 6 ) = 1046 (10 6 ) 2 = Avg  R = 370(10 6 )  675 .6(10  6 ) = 306 (10 6 ) XY  692 .8(10 6 ) tan 2 P = = = 15.4 o (Clockwise ) 6 6 ( X  Y ) (950(10 )  (210 (10 ))) Check to ensure pressure vessel works r 1000 = = 100 > 10 t 10
The work, determine stresses, Spherical Pressure Vessel long = Hoop = Pr P(1000 mm) = = 50 P 2t 2(10 mm) Determine Strain in Gauge Max = Lat = Max = 0.012 mm = 0.6(10 3 )(mm / mm) 20 mm Using Generalized Hooke's Law 1 [ max  lat  min ] E 1 0.6(10 3 ) = [50P  (0.3)(50 P)  (0.3)(0)] 200(10 9 ) P = 3.43MPa Find the Max stress long = 50 P = 171.43MPa First Find inPlane shear stress Both stresses are the same, thus shear is zero, nothing to produce it XY =0
IN Plane Use Mohr's Circle to determine out of plane ABS =
MAX Max  Min 171.43 MPa  0 = = 85.7 MPa 2 2 Since the Material is Confined X = Y = 0
Using Generalized Hooke's Law X = 1 [ X  Y  Z ] E 0 = X  Y  Z 1 [ Y  X  Z ] E 0 = Y  X  Z Y = Solving the two equations yields the result Y = X =
Z = Z 1 Use this result in the third equation 1 [ Z  X  Y ] = 1 Z  Z  Z E E 1  1  1 Z = Z (1  )  2  2 E 1  v ( ) Z = 1 Z (1 + v)(1  2 ) E 1  v Solve for E' E' = Z 1 = E Z (1 + )(1  2 ) Increased Factor k E' 1 = E (1 + )(1  2 ) 1  0.3 k = = 1 .35 (1 + 0.3)(1  2(0.3)) k= ...
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 Spring '08
 Thouless
 Shear Stress, Stress, xy, Shear strain, Shear modulus

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