# Haghighi 1 1 t p x tlij n tlij 0 p y 0 1p

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Unformatted text preview: ⎫ b = ∫ [N] ⎨ ⎬dV = ⎢ ⎮ 0 N ⎥ ⎨Y ⎬dV = ⎮ ⎢ N Y ⎥dV ⎩Y ⎭ j ⎥⎩ ⎭ V ⎮⎢ j ⎥ ⎮⎢ ⎮ ⎢Nk X ⎥ ⎮ ⎢Nk 0 ⎥ ⎮ ⎢N Y ⎥ ⎮⎢ 0 N ⎥ ⌡⎣ k ⎦ ⌡⎣ k⎦ V V Chapter 23 Page 17 K. Haghighi Using area coord. ⌠ ⎡L1X ⎤ ⎡X⎤ ⎢Y⎥ ⎮ ⎢L Y ⎥ ⎢⎥ ⎮⎢ 1 ⎥ tA ⎢ X ⎥ ⎮ ⎢L2X ⎥ = t⎮ ⎢ ⎥dA = ⎢ ⎥ 3 ⎢Y⎥ ⎢L2Y ⎥ ⎮ ⎢X⎥ ⎮ ⎢L3X ⎥ ⎢⎥ ⎮ ⎢L Y ⎥ ⎣Y⎦ ⌡⎣ 3 ⎦ A So, the integral is dividing each body force comp. such as (t A X) or (t A Y) equally among the 3 nodes of the ∆ . Chapter 23 Page 18 K. Haghighi T ⎧p x ⎫ c = [N] ⎨ ⎬dΓ ∫ ⎩p y ⎭ Γ so involves surface stresses and must be evaluated along the edge of the elem. and also dΓ = t d 2 1 c = tL⌠ [N]T ⎧p x ⎫d ⎨p ⎬ 2 has 3 values for 3 sides ⎮ of ∆ ⌡0 ⎩ y⎭ length of a side Chapter 23 Page 19 K. Haghighi For the side ij 1 ⌠ ⎡ Ni 0 ⎤ ⎮⎢ 0 N ⎥ i⎥ ⎮⎢ 1 ⌠ T ⎧p x ⎫ ⎮ ⎢ Nj 0 ⎥ ⎧p x ⎫ tLij ⎮ [N] ⎨ ⎬ = tLij ⎮ ⎢ ⎥ ⎨p ⎬d 2 ⌡0 ⎩p y ⎭ ⎢ 0 Nj ⎥ ⎩ y ⎭ ⎮ ⎮ ⎢Nk 0 ⎥ ⎮⎢ 0 N ⎥ ⌡⎣ k⎦ 0 But Nk = 0 &amp; using area coord., Chapter 23 Page 20 K. Haghighi 1 ⌠⎡ ⎮⎢ ⎮⎢ 1 ⌠ T ⎧p x ⎫ ⎮⎢ tLij ⎮ [N] ⎨ ⎬ = tLij ⎮ ⎢ ⌡0 ⎩p y ⎭ ⎮⎢ ⎮⎢ ⎮⎢ ⌡⎣ 0 1p x ⎤ py ⎥ 1⎥ 2p x ⎥ ⎥d 2 2p y ⎥ 0⎥ ⎡p x ⎤ ⎥ 0⎦ ⎢p ⎥ ⎢ y⎥ tLij ⎢p x ⎥ = ⎢⎥ 2 ⎢p y ⎥ ⎢0⎥ ⎢⎥ ⎣0⎦ Chapter 23 Page 21 K. Haghighi or (p x tLij ) &amp; (p y tLij )are force comp. acting on side ij and integral has allotted 1/2 of each force comp. to each node on side ij. ⎧p x ⎫ ⎧0⎫ Similarly for other 2 sides: ⎪ ⎪ ⎪p ⎪ 0 ⎪ y⎪ ⎪⎪ tLjk ⎪p x ⎪ tLik ⎪ 0 ⎪ ⎪⎪ T ⎧p x ⎫ [N] ⎨ ⎬dΓ = ⎪ ⎪; ⎨0⎬ ⎨p ⎬ 2 ⎪ y⎪ 2⎪ ⎪ ⎩p y ⎭ Γ ⎪p x ⎪ ⎪p x ⎪ ⎪⎪ ⎪⎪ ⎪p y ⎪ ⎪p y ⎪ ⎩⎭ ⎩⎭ ∫ for side jk for side ik Chapter 23 Page 22 K. Haghighi Note: p x &amp; p y are (+) when they act in the (+) coord. direction. py y x py px px y x The direction of positive surface stresses. Chapter 23 Page 23 K. Haghighi Example: A plane s...
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## This document was uploaded on 01/29/2014.

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