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=PLV2, LP Wa = E t L' One path to the algebraic solution is as follows: W à = Lin L

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Number 2. solve for all, Pm, and Wm. I’ve attached a reference image for the solution method I just need clarification
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=
1/5
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)
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= 1/5 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) L :.W₁ = KQt E'A
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