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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 Kc =PL¹2, W_â t L' One path to the algebraic solution is as follows: W = LP E q= Lin 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 Kc =PL¹2, W_â t L' One path to the algebraic solution is as follows: W = LP E q= Lin L'

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Number 3. solve for all, Pm, and Wm. I’ve attached a reference image for the solution method I just need clarification
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
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
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