Ch_23_annotated

# 515 18750 n force2 pna pntl 5000 25 2

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Unformatted text preview: tress elem. is shown. The elem. experiences a 10 C temp. difference. y t = .5 cm k(1,3) (e ) E = 15(10)6 N / cm2 [k ]= ? (e ) ] = ? [f Solution: µ = .25 α = 6(10−6 )/ C i(0,0) j(2,0) x [k(e )]= tA[B]T [ D][ B] Chapter 23 Page 24 K. Haghighi but ⎡bi 0 b j 0 bk 1⎢ [B] = ⎢ 0 ci 0 c j 0 2A ⎢ ci b i c j b j ck ⎣ where 0⎤ ⎥ ck ⎥ bk ⎥ ⎦ A=3(2)/2=3 cm2 . Find all b’s & c’s, subst. to get: 0 3 ⎡− 3 1⎢ [B] = ⎢ 0 − 1 0 6 ⎢−1 − 3 −1 ⎣ 0 0 0⎤ − 1 0 2⎥ ⎥ 3 2 0⎥ ⎦ Chapter 23 Page 25 K. Haghighi Also ⎡ ⎤ 0⎥ 0⎤ ⎢1 µ 6 ⎡1 14 E⎢ ⎥ = 15 10 ⎢1 4 1 [D] = 0 µ1 0⎥ ⎥ ⎥ 1 − .252 ⎢ 1 − µ2 ⎢ 1−µ ⎢0 0 3 8⎥ ⎢0 0 ⎥ ⎣ ⎦ ⎣ 2⎦ () or ⎡16 4 0⎤ [D] = 106 ⎢ 4 16 0⎥ ⎢ ⎥ ⎢ 0 0 6⎥ ⎣ ⎦ Chapter 23 Page 26 K. Haghighi then evaluate [B]T [D] and finally ⎡ ____ ⎤ ⎢ ____ ⎥ (see the ⎥ book!) k (e ) = tA[B]T [D][B] = 41667 ⎢ ⎢ ____ ⎥ ⎥ ⎢ ⎣ ____ ⎦ 6×6 The thermal force vector, [f (e )]= [B]T [D]{εT}tA Chapter 23 Page 27 K. Haghighi where ⎧αδT⎫ ⎧60⎫ ⎪ ⎪ −6 ⎪ ⎪ {ε T } = ⎨αδT⎬ = 10 ⎨60⎬ ⎪0⎪ ⎪0⎪ ⎩ ⎭ ⎩⎭ ⎧ − 600 ⎫ and finally ⎪ − 200 ⎪ ⎪ ⎪ (e ) = ⎪ 600 ⎪ f ⎨ ⎬ ⎪ − 200 ⎪ ⎪0⎪ ⎪ ⎪ ⎩ 400⎭ {} Chapter 23 Page 28 K. Haghighi Example: Calculate the equivalent set of conc. forces acting @ nodes 10, 12 & 18. pn = 5000 N / cm2 18 1.5 2 cm Surface stress and elements along an exterior boundary. (1) 12 t = 2.50 cm ( 2) 10 60 Chapter 23 Page 29 K. Haghighi Solution: 1 ⌠ T ⎧p x ⎫ Use ⎮ [N] ⎨ ⎬dΓ ⎩p y ⎭ ⌡ Γ ⎧0⎫ ⎪0⎪ ⎪⎪ tL jk ⎪p x ⎪ ⎪⎪ = ⎨p ⎬ 2 ⎪ y⎪ ⎪p x ⎪ ⎪⎪ ⎪p y ⎪ ⎩⎭ Chapter 23 Page 30 K. Haghighi 2 Use physical interpretation of 1 (easier & faster!) ∴ calc. total force acting on the side of the elem., then divide bet. the 2 connecting nodes equally. or and force(1) = PnA = PntL = 5000( 2.5)(1.5) = 18750 N force(2 ) = PnA = PntL = 5000( 2.5)( 2) = 25000 N Chapter 23 Page 31 K. Haghighi dividing these forces between the nodes; 9375 9375 18 4688 12 8119 10938 18 12 9375 12500 18 12500 10 (a) 21875 12 12500 1...
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## This document was uploaded on 01/29/2014.

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