# Equations - AE 321 Equation Sheet – Final Exam N 0 1 2 3...

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Unformatted text preview: AE 321 - Equation Sheet – Final Exam N 0 1 2 3 4 order zero (scalar) one (vector) two three four transformation law A=A Ai = ij A j Aij = ik jl Akl Aijk = il jm Aijkl = im jn kn Almn kp lq Amnpq Tensor operations: rr a • b = aibi [ r a r b ij = aib j [ rr ab ij = ijk a j bk ( • A ) i = Aij , j ( A ) ij = Ti = ij , j ipq A jq , p ( A )ijk = Aij,k Cauchy relation: ij nj and ij Equilibrium equations: + fi = 0 k ij = ji ( Principal stresses and directions: ij 3 k ) n (jk ) = 0 k Q1 33 2 k ii + Q2 = tr( 13 33 Q3 = 0 Q1 = Invariants of stress: 11 + 22 12 + + 1 6 = ) 22 23 23 33 Q2 = 11 12 11 13 + kl = 22 1 ( 2 ii jj ij ij ) Q3 = det ( oct norm )= ii mik njl ij mn = 1 3 Octahedral stresses: oct sh 1 = 3 [( 1 2 ) 2 +( max 2 3 ) 2 +( 3 1 ) 1 22 Absolute maximum shear stress: max sh = = 1 [( 2 1 3 )] 1 Stress Analysis in 2-D: + 2 + ' ( ) = 11 22 2 ' 11 ( )= 11 22 + 11 22 2 11 22 cos 2 + cos 2 12 12 sin 2 sin 2 22 2 11 22 12 ' 12 ( )= + 2 + 2 * 22 2 + 11 sin 2 + 2 22 cos 2 1 = = 11 2 2 11 22 + + 2 12 11 22 2 2 2 12 r n (1) = cos r n (2) = sin r r e1 + sin *e2 r with tan 2 *r e1 + cos *e2 11 * = 2 11 12 22 2-D equilibrium: x1 12 + + 12 x2 22 + f1 = 0 + f2 = 0 x1 Deviator and mean: x2 [ ]=[ [ M M + [ D] m 0 m = 0 0 11 0 0 m 12 22 23 m 33 13 23 m 0 m 12 13 where m = 1 3 ii [ D] = 2 Strain displacement relations: 1 (ui, j + u j ,i + uk,i uk, j ) 2 1 eij = ( ui, j + u j ,i uk,i uk , j ) 2 1 (ui, j + u j ,i ) ij = 2 E ij = Physical significance: dS = 1 + 2 E ii no sum ds 2 E ij sin = 1 + 2 E ii 1 + 2 E jj 11, 22 22 , 33 33,11 no sum + + + + + + ii 22 ,11 33, 22 11, 33 13,12 21, 23 32 , 31 2 2 2 12 ,12 23, 23 31, 31 =0 =0 =0 11, 23 Strain compatibility equations: 12 ,13 23, 21 31, 32 23,11 31, 22 12 , 33 =0 =0 =0 22 , 31 33,12 Volume dilatation: = Strain deviator and mean normal strain: Same expressions as for stress given above Material behavior: General anisotropic: ij = Cijkl kl Isotropic: ij = 2μ ij + ij ij kk = E (1 + ) ij + E (1 + )(1 2 ) ij kk ij = (1 + ) E E 1 2 ij kk Strain energy density: W = ij ij Bulk modulus: p = k V V0 with k = 2μ + 3 E = 3 3 (1 2 ) 3 Failure criteria: Tresca: max sh = 1 2 y, Von Mises: 1 oct sh = 2 3 y Maximum principal stress: Miner’s rule for fatigue failure: = y, where y is the uniaxial yield stress. i ni =1 Ni Known stress solutions Extension F Bending x A F [ P00 ]= 0 0 0 000 with P = F A y M Torsion x M [ My I ]= 0 0 00 00 00 T r z T [ 0 ]= 0 μr 0 μr 0 0 0 0 0 0 0 0 = [ ]= Prandtl stress function for torsion 2 μ x2 μ x1 x1 μ x2 μ x1 0 T μJ ( x1, x2 ) 2 such that = 2μ 13 = x2 and 23 = Compatibility equation: x12 + 2 x2 on cross-section , with = 0 along . Relation to applied torque T = 2 ( x1, x2 ) d 4 Plane problems Plane strain: Plane strain: 3i = i3 =0 33 = ( E 11 + 11 22 ) 22 3i = i3 =0 33 = ( + ) for pland = 1 for pl- Constitutive relations: E 11 = 11 22 = 22 with E = 11 2 μ 12 = 12 1 Airy stress function – Cartesian coordinates 22 2 11 = 2 x2 2 22 E E for plE for pl2 +V +V 2 = = x12 with V = x1 V = x2 f1 and 4 4 4 4 = f2 x14 +2 2 x12 x2 + 4 x2 =0 x1 x2 Airy stress function – Polar coordinates 1 12 + +V rr = 2 r r r2 2 12 = r2 +V with 1 r= rr General solution in polar coordinates: ( r, ) = a0 log r + b0 r 2 + c0 r 2 log r + d0 r 2 + a0 + + n=2 V =R r 1V = r 4 =0 2 and 2 = r2 + 1 1 +2 rr r 2 2 a1 r sin + b1r 3 + a1r 1 + b1r log r cos 2 ( ) c1 r cos + d1r 3 + c1r 1 + d1r log r sin 2 ( ) (a r n n + bn r n + 2 + an r n + bn r n+2 ) cos n + n=2 (c r n n + dn r n + 2 + cn r n + dn r n+2 ) sin n Known solutions Cylinder of inner radius a and outer radius b with internal pressure Pi and outer pressure Po : rr = = ( P0 Pi )a 2b 2 1 (b 2 a 2 ) r 2 ( P0 Pi )a 2b 2 1 (b 2 a 2 ) r 2 ( P0b 2 (b 2 ( P0b 2 (b 2 Pi a 2 ) a2 ) Pi a 2 ) a2 ) r =0 5 CIRCULAR CYLINDRICAL COORDINATES EQULIBRIUM EQUATIONS 1 r r 1 r + r r rr + r + + rz z z + + r rr r 2 r zr r +R=0 + =0 z zz zr r + z r + z + +Z =0 STRAIN-DISPLACEMENT RELATIONS rr = r u r = v r 2 u v + rr v u + 2 rr = zz = w r w z + v z 2 rz z = = u w + z r COMPATIBILITY EQUATIONS 2 r r rr r z z 2 2r 2 rr rr 2 r rr r r 2 + r =0 2 rr (r ) + 2 r ( 2r r ) r 2 zz ( 2r ) r z 2 2 zr zr =0 =0 + r2 r r 1 rr 2 2 r r z ( 2r ) r z r 2 z 2 +r zz r + 2 + 2 z 2 zr zz 2 z r +2 zr =0 zr 2 2 1 rr 2 rr 2 z z rz =0 12 r =0 z + r zz 2 6 SPHERICAL COORDINATES EQUILIBRIUM EQUATIONS r r r + + 1 r sin 1 r sin 1 r sin r + + 1 r 1 r 1 r r + + + 2 3 3 r + +2 r +( r r cot r cos +R=0 r r r + ) cot =0 + =0 r + + r STRAIN DISPLACEMENT RELATIONS r = = = u r 1 r sin = 1 r sin + v r w r v v r + u cot + w r r = 1 r sin w = + 1w u + r r 1v r cot v r u r 1u w + r r COMPATIBILITY EQUATIONS 2 2 2sin r 2 2 2 r r cot + 1 2 2 + r r r 1 2 r cos cot r sin r ) cot =0 sin 2 r + r2 sin r r r +r cos2 sin r r =0 + r2 2 r r r r r sin 2 r sin 2 r +2 2 cos sin2 =0 =0 2r sin r r r sin r +( r + ( r sin ) + r sin r 2 r 2 r r + r 2 r 2 2 r sin 2 r sin cos =0 2 + r + sin 2 r sin r r ( + 2 r 2 ) + 2( sin 2 r ) sin + sin r r r r2 r r (r r ) sin cos r r (r ) r =0 7 ...
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## This note was uploaded on 03/07/2011 for the course AE 311 taught by Professor Dutton,j during the Spring '08 term at University of Illinois, Urbana Champaign.

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