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Course: EGM 5611, Fall 2009
School: Fayetteville State...
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CIRCLE MOHRS IN THREE DIMENSIONS Consider the state of stress given by T11 T12 T13 o T = T21 T22 T23 T31 T32 T33 {e ,e ,e } o1 o2 o3 I II T is symmetric. When we determine the principal value, let T , T and T III be the principal value. Let T I T II T III be the principal values. ( 1) ( 1) ( 1) \$ ( 1) Let T I be associated n = n1 , n2 , n3 , \$ ( 1) , n( 2) , n( 3) form a rectangular Cartesian coordinate...

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CIRCLE MOHRS IN THREE DIMENSIONS Consider the state of stress given by T11 T12 T13 o T = T21 T22 T23 T31 T32 T33 {e ,e ,e } o1 o2 o3 I II T is symmetric. When we determine the principal value, let T , T and T III be the principal value. Let T I T II T III be the principal values. ( 1) ( 1) ( 1) \$ ( 1) Let T I be associated n = n1 , n2 , n3 , \$ ( 1) , n( 2) , n( 3) form a rectangular Cartesian coordinate n \$ \$ o ( 1) , e ( 2 ) , e ( 3) to system. If [Q] represents the transformation from e Note that { } for T \$( ) n = { n ( ) , n ( ) , n ( ) } for T \$ ( 2) = n ( 2) , n ( 2) , n ( 2) n 1 2 3 3 3 3 3 1 2 3 { } II III { \$ ( 1) \$ ( 2) \$ ( 3) n ,n ,n } { 1 2 3 } , then T I 0 0 0 T II 0 0 T 0 = [ Q] [ T ] [ Q] T III \$ ( 1) , n( 2 ) , n( 3) . As n \$ \$ X-axis represents the eigen value with coordinate system we rotate the radius with arbitrary angles , and with old system (eigen vector system) with 2 + 2 + 2 = 1 Thus [T ] = [Q ] T [T ]eigin [Q] ( TI + TIII , TI - TIII 2 2 ) ^ n (T,) T III T II TI ^ e T 4.7 Equations of Motion Previous: Mohr's Circle in 3-D NOTES On -ve face, -ve direction, stress component is positive There is a gradient of of stress component from ve to +ve face. Stress field is continuously varying. Note that center of cube is origin P o Let T be the stress tensor at a point P. Let X be the body force/volume at that point. Note : B be the body force/mass. B = X Next: 4.7 Equations of Motion (Cnt'd) 4.7 (b) Equations of Motion (Cnt'd) Previous: Equations of Motion Summing all the forces in the x1 direction T11 dx1 dx2 dx3 - T11dx2 dx3 T11 + x1 T + T21 + 21 dx2 dx1dx3 - T21dx1dx3 x2 T31 + T31 + dx3 dx2 dx1 - T31dx2 dx1 x3 + X 1dx1dx2 dx3 = 0 Dividing by dx1dx2 dx3 we have T11 T21 T31 + + + X1 = 0 x1 x2 x3 or T ji xi + X i = 0 or T ji , j + X i = 0 Since we already know that Tij = T ji for the moment equilibrium Tij , j + X i = 0 is the equation of the equilibrium 3 equations (3+1) terms per equation Next: 4.9 Boundary Conditions 4.9 Boundary Conditions Previous: Equations of Motion (Cnt'd) Forces/Length \$ n Surface S Volume V Fixed Roller If \$ n is the unit outward normal at a given point in surface then, \$ =Tn t o\$ (1) Next: 4.10 First P-K Stress where \$ is the force vector or traction vector per unit area. Equation (1) t represents the stress or traction boundary condition 4.10 First P-K Stress Previous: 4.9 Boundary Conditions We define Tij =Cauchy stress tensor (Force/area in the deformed geometry) We know for equilibrium equation T ji xi + Xi = 0 Define two other stress tensors, first and second Piola-Kirchoff Stress tensors Three different stress measures o o T = Cauchy Stress T o o T 0 = I Piola Kirchoff Stress o o O T = II Piola Kirchoff Stress S T o We will denote as first P-K and Note Stress Cauchy I P-K II P-K Book o S as II P-K stress o T o To o T Ours Configuration Current force Current area Current force Original area Fictitious force Original area T T S Previous: 4.10 First P-K Stress 4.10 First (a) P-K Stress Previous: 4.10 First P-K Stress Undeformed Let dso Undeformed area with N \$ ds Deformed area with n dP Same force vector with same orientation (in deformed and undeformed ) Next: 4.10(b) First P-K (Cont'd) 4.10 (b) First P-K Stress Contd Previous: 4.10(a) First P-K Stress T2 T11 T12 T21 I n TI T1 I x2 x1 T22 Resolution of forces in 2-D We can develop two basis for stress relative to the undeformed area. Define the first p-k stress tensor such that it gives the actual force dP on the deformed surface dS but measured w.r.t undeformed surface dSo Next: 4.10(c) First P-K contd 4.10(c) First P-K contd Previous: 4.10(b)First P-K contd In 1-D case 11 = dP 1 dS0 where dP is the component of dP 1 in the in the x1 direction. Let dP be the component of i Recall. dP i th direction. ti = T ji n j (in the deformed) dPi = n jT ji dS (in the deformed)...(1) dPi = N j ji (in the original)......(2) dSo Thus from (1) and (2) N j ji dSo = dPi = n jT ji dS Next: 4.10(d)First P-K contd 4.10(d) First P-K Contd Previous : 4.10(c)First P-K contd Now, we know that ni dS = o N j Substituting a j xi dS0 Finally, o ak N j ij dSo = T ji N k dSo x j o a j or N j dSo ji - Tki =0 xk Tij = xi kj o ak First P-k Stress Det F def gradient Cauchy Note, the tensor is not symmetric i.e. ij symmetric ji First P-K stress is not Next: 4.10(e) First P-K contd 4.10(e) First P-K contd Previous: 4.10 (d) First P-K contd (2) II P-K Stress Sij Sij is defined in terms of fictitious force dP applied to undeformed surface dSo . This force is related to the real force dP (applied to dS) in the same way that a material vector da at a is related by the deformation to the corresponding spatial vector dx at x . The angle is the same o dP on dS and d P on dSo a T d P i = i .dPj...

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