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W=LP₁ (KIC) = PL¹/²₁ 9 = Qin/2, N = 2 E' 7 √F

W=LP₁ (KIC) = PL¹/²₁ 9 = Qin/2, N = 2 E' 7 √F

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter14: Discrete Dynamical Systems
Section14.3: Determining Stability
Problem 13E: Repeat the instruction of Exercise 11 for the function. f(x)=x3+x For part d, use i. a1=0.1 ii...
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Number 1 and 2. SOLVE for Lm, Pm, and Wm. I attached a reference for the solving method.
Po TS | W=LP₁ (KIC) =PL¹/₂²₁ 9 = Qin/2, N = E' c 7 + IL - W = LP₁ (KIC) = PL¹/²₂ 9 = Q₁n/2, 19 L E ان VF 3VS 1 c' W=LD W €²₁ 9=W³P 2 - Qin q= 1 1 q E' + VF M'L L 1 W=LP 9 = C₁ q = W³P, 9 = Qin q 드 E VF M'L L
Transcribed Image Text:Po TS | W=LP₁ (KIC) =PL¹/₂²₁ 9 = Qin/2, N = E' c 7 + IL - W = LP₁ (KIC) = PL¹/²₂ 9 = Q₁n/2, 19 L E ان VF 3VS 1 c' W=LD W €²₁ 9=W³P 2 - Qin q= 1 1 q E' + VF M'L L 1 W=LP 9 = C₁ q = W³P, 9 = Qin q 드 E VF M'L L
With these in mind, the approximation equations are written in brackets next to the governing equations. The solution for the toughness dominated regime can be found from simultaneously satisfying Elasticity, Continuity, Propagation, and the Inlet Boundary Condition, that is K₁=PL¹2, W = â= One path to the algebraic solution is as follows: W LP E W = O.. L LP E W t W ĝ t L W g t L' Lin à.. W Lt Ľ² K₁=PL¹/¹², LP LP Qi E tE Ľ² Q = →W= :: t Ľ QEt P= → K₁=PL¹² :: Ke Ľ²³ K. - QK¹ - L-(K) QiE't = [5/2 QEt LP L₁, P₁ →W=W= E 2 = ⇒P= 2/5 QEt Ľ â = Lin QE't Ľ -L¹/2 = QE't LS/2 q= 2/5 QE't L₁₂ = ⇒P= - (CE₁) +P = OF ² = P = QE ( 2 E₁)" :-:-(5) P = QEt Ľ Et EQt K 1 K₁ EQEt EQ) 1/5 L :.W₁ = KQt E'A
Transcribed Image Text:With these in mind, the approximation equations are written in brackets next to the governing equations. The solution for the toughness dominated regime can be found from simultaneously satisfying Elasticity, Continuity, Propagation, and the Inlet Boundary Condition, that is K₁=PL¹2, W = â= One path to the algebraic solution is as follows: W LP E W = O.. L LP E W t W ĝ t L W g t L' Lin à.. W Lt Ľ² K₁=PL¹/¹², LP LP Qi E tE Ľ² Q = →W= :: t Ľ QEt P= → K₁=PL¹² :: Ke Ľ²³ K. - QK¹ - L-(K) QiE't = [5/2 QEt LP L₁, P₁ →W=W= E 2 = ⇒P= 2/5 QEt Ľ â = Lin QE't Ľ -L¹/2 = QE't LS/2 q= 2/5 QE't L₁₂ = ⇒P= - (CE₁) +P = OF ² = P = QE ( 2 E₁)" :-:-(5) P = QEt Ľ Et EQt K 1 K₁ EQEt EQ) 1/5 L :.W₁ = KQt E'A
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