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Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Number 1 and 2. SOLVE for Lm, Pm, and Wm. I attached a reference for the solving method.

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

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
Expert Solution
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