Instructor Solution Manual To Accompany Fracture Mechanics (Solutions)

67 Pages·2011·0.967 MB·English

by  C. T. (Chin-Teh ) Sun, Z. (Zhihe) Jin

Solution Manual

for

Fracture Mechanics

by

C.T. Sun and Z.-H. Jin

Updated April 15, 2013

1
Chapter 2

Prob 2.1:
a)
2
P 1 P
P 1 3 P
P 3 P
4

No load is carried by Part 2 and Part 4 . There is no strain energy stored in them.

Constant Force Boundary Condition
The total strain energy per unit width stored in Part 1 and Part 3 is
U =∫L P2 dx+∫a P2 dx= P2(L- a)+ P2a
a 2AE 0 2A E 2AE 2A E
1 3 1 3
A =h, A =h/3
1 3
Thus the total strain energy is
3aP2 P2(L- a)
U = +
2Eh 2Eh
Finally, the energy release rate is
1 ¶ U P2
G = =
2 ¶ a 2Eh

Fixed End Boundary Condition
dW = 0 since dδ = 0, then energy released due to the crack extension is
e
1
dW =- dU= P2dS
s 2

2da
dS =
hE
P2da
⇒dW =
s Eh

Finally, the energy release rate is
1 dW P2
G = s =
2 da 2hE

2
b)
P/2 P/2
2 P/2
2, 4
P 1 3 P
P P
3
4 P/2

No load is carried by Part 1. There is no strain energy stored in it.

Constant Force Boundary Condition
The total strain energy per unit width stored in Part 1 and Part 3 is
U =∫a P2 +2(P 2)2 dx= P2a +2 P2a
0 2AE 2AE 2AE 8AE
A=h/3
Thus the total strain energy is
3aP2 3aP2 9P2a
U = +2 =
2Eh 8Eh 4Eh
Finally, the energy release rate is
1 ¶ U 9P2
G = =
2 ¶ a 8Eh

Fixed End Boundary Condition

Based on the fixed end boundary condition, we can have dW = 0 ⇒ dW = -dU
e s
Thus the energy released due to the crack extension is
1 P 1
dW =- dU= 2 ( )2dS+ P2dS
s 2 2 1 2 2
- 3da 3da
dS = ,dS =
1 2hE 2 hE
3P2da
⇒dW =
s 4EA
Eventually, the energy release rate is
1 dW 9P2
G= s =
2 da 8Eh

3
Prob 2.2:
a) Symmetric case
P
P

The strain energies stored in each part are
U =U = ∫a M2 dx = ∫a (Px)2 dx = P2a3 , I = th3
1 2 0 2EI 0 2EI 6EI 1 12
So the total strain energy is
P2a3 4P2a3
U =U +U +U = =
1 2 3 3EI Eth3
The energy release rate is given by
1dU 12P2a2
G = =
I t da Eh3t2
b) Anti-symmetric case

P
P

The strain energies stored in each part are
U =U = ∫a (Px)2 dx = P2a3 , I = th3
1 2 0 2EI 6EI 1 12
1 1
U = ∫L(2Px)2 dx = 2P2 (L3 - a3), I = 2th3
3 a 2EI 3EI 3 3
3 3
So the total strain energy is
P2
U =U +U +U = (L3 +3a3)
1 2 3 12EI
The energy release rate is given by
1dU 9P2a2
G = =
II t da Et2h3

4
Prob 2.3:
Since the structure is symmetric (both loading and geometry), we can represent the given
cracked beam as two cracked beams as shown in Figure 1.

M is the moment needed to ensure zero slope at the location where we made the free
body cut.

Assuming that the beam marked "1" is a cantilever beam with a tip load P/2, the slope (cid:1)
at the tip is given by
( )
P 2 a2
q =
P 2EI
The slope at the tip of a cantilever subjected to a tip moment is given by
Ma
q =-
M EI
To ensure that the system shown in Figure 1 accurately represents the actual center
cracked structure with a mid-section load, we need to satisfy
q +q =0
p M
( )
P 2 a2 Ma Pa
- = 0 ⇒ M =
2EI EI 4

The moment distribution along the length of beam 1 is
P Pa
M(x)= x- (x=0 being the tip of the cantilever beam).
2 4
The strain energies can be written as
a M(x)2dx 1 aPx Pa2 P2a3
U =U = ∫ = ∫ -  dx=
1 2 2EI 2EI  2 4  96EI
0 0
and U = 0. The total strain energy is therefore,
3
P2a3
U =U +U +U =
1 2 3 48EI
Strain energy release rate can be calculated as
5
1dU P2a2
G = =
t da 16EIt
The moment of inertia for the specified dimensions is, I =4.7· 10- 10m4
The crack will propagate if G‡ G . Hence, we can calculate the minimum load P
c min
required to propagate the crack as given below.
P2 a2
G =G = min
c 16EIt

16EItG
⇒ P = c =216N
min a2

6
Prob 2.4:

Let us assume that the crack extends by da as shown above
Total distance d moved by the loading point is given by
dd =Elasticextension+peeledoff length
= e (L+da)-e L+ da
=da(1+e )
 P 
=da1+  ;assuming unit width
 At
Work done by the external force is therefore,
 P 
dW = Pdd = P1+ da
e  Et
Change in strain energy is given by

dU =U - U
L+da L
= 1P P (L+da)A- 1 P P (L)A
2 AEA 2 AEA
P2da
=
2Et
Hence,
dW =dW - dU
s e
 P 
= P1+ da
 2Et
Strain energy release rate is given by
7
dW  P 
G = s = P1+ 
da  2Et

Note that in this particular problem, dW „ dU
s

8
Prob 2.5:

dd ddd

Let the crack extend by da as shown above. The incremental displacement can be written
as
Pda
dd = ,taking unit width
Et
Work done by external force,
P2da
dW = Pdd =
e Et
Change in strain energy is given by
dU =U - U
L+da L
= 1P P (L+da)A- 1 P P (L)A
2 AEA 2 AEA
P2da
=
2Et
Hence,
P2da
dW =dW - dU=
s e 2Et
Strain energy release rate is
dW P2
G = s = .
da 2Et
In this case, dW =dU .
s

9
Comparing Prob 2.4 and 2.5

As seen in the two problems above, the strain energy release rate and the strain energy
gained by the film are not equal for prob 2.4 and are equal for prob 2.5. This is due to
nature of the P-δ curves shown in Figure 4. In prob 2.4, the nature of the loading makes
the system non-conservative. It should also be noted that for the same load P, prob 2.4
has a higher G and hence it is easier to propagate the crack in prob 2.4 than in prob 2.5.

P P

Pda
Pda
da Et
Et

d dd d d dd d
PL P(L+da) PL P(L+da)
+da
Et Et Et Et

Prob 2.4 Prob 2.5

10