zm12_m13

# zm12_m13 - Symmetry of Stress Tensor Consider moment...

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Symmetry of Stress Tensor Consider moment equilibrium of differential element: Taking moments about x 1 axis (i.e point C): È Â M 1 = 0 : 2 s 23 d x 3 x 1 ) 2 Area of È ( x 2 ˘ - 2 32 ( x 2 x 1 ) x 3 ˘ = 0 Î˚ Î 2 ˚ Moment f 23 = face arm 32 Thus, in general mn = nm Stress tensor is symmetric. Six independent components of the stress tensor. 11 12 Ê= 21 ˆ 22 23 Á Á = 32 ˜ ˜ 33 31 Ë = 13 ¯ Note a positive (tensile) component of stress acts on a face with a positive normal in a positive direction. Thus a stress acting on a negative normal face, in a negative direction is also positive. If the stresses do not vary over the infinitesimal element, mn acts on opposite faces, in opposite directions but with equal magnitude.

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But what happens if stress varies with position? Stress Equations of Equilibrium Consider equilibrium in x 1 direction. Let stresses vary across cube. Also allow there to be a body force (per unit volume) f 1 acting on the cube in the x1 direction: x 3 s 11 d x 2 ∂s 31 d x 3 s + x 3 21 x 2 + x 2 s 21 d x 3 x 2 d x 1 s 31 x 1 11 x 1 f 1 + x 1 f n is a “body force”, e.g. due to the weight of the element ( r g), centrifugal acceleration ( r r w 2 ), electromagnetic fields etc. Taking equilibrium of forces in the x 1 direction gives: Ê ˆˆ Á 11 x 1 ˜ ( x 2 x 3 ) - 11 ( x 2 x 3 ) + Á Ê 21 + 21 x 2 ˜ ( x 1 x 3 ) - 21 ( x 1 x 3 ) Ë 11 + x 1 ¯Ë x 2 ¯ ˆ + Ê Á 31 + 31 x 3 ˜ ( x 1 x 2 ) - 31 ( x 1 x 2 ) + f 1 ( x 1 x 2 x 3 ) = 0 Ë x 3 ¯
which simplifies to: ∂s 11 + 21 + 31 + f 1 = 0 x 1 x 2 x 3 similarly, equilibrium in the x 2 and x 3 directions yields: 23 12 + 22 + 32 + f 2 = 0 and 13 + + 33 + f 3 = 0 x 1 x 2 x 3 x 1 x 2 x 3 Canceling out terms and dividing through by the common dx 1 , dx 2 , dx 3 .

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## This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.

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zm12_m13 - Symmetry of Stress Tensor Consider moment...

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