FEA HW#3 Solution updated
Prob. 3.2
Beam theory: by assuming no trans, shear deformation, defelection of cantilever beam due to the
applied moment M at tip
vF , vE , vD = −
ML2
EI
The moment of inertia I can be calculated as follows
t ( 2c )
2tc 3
=
I =
12
3
3
3ML2
vF = vE = vD = −
4 Etc 3
ML 3ML
0 ; uF =
uE =
c
−uD =
=
EI 2 Etc 2
(a) The strain can be calculated as below
εx 
1
 
ε y  =
γ  2 A
 xy 
 y23
0

 x32
0
x32
y23
y31
0
x13
0
x13
y31
y12
0
x21
0
x21 
y12 
 u1 
v 
 1
 u2 
 
 v2 
 u3 
 
 v3 
(3.2-3)
εx 
1
 
=
ε y 
γ  2 Lc
 xy 
 − c 0 2c 0 − c 0 


 0 −L 0 0 0 L 
 − L −c 0 2c L −c 
ε x = 0 ; ε y = 0; γ xy = −
3ML
4 Etc3
For ν = 0, the elasticity matrix becomes
1 0 0 
E = E 0 1 0 
0 0 1 / 2 
for plane stress
σ = Eε
σ x 
 
σ y 
τ 
 xy 
(b)
1 0 0   ε x 
0 1 0   ε 
E=

 y 
0 0 1 / 2  2γ xy 


 0 


 0 
 ML 
− 3 
 8tc 
0 
 0
 
 0
 
vE
0 
 
0 
εx 
 
ε y 
γ 
 xy 
c
0 c 0
 −2c 0
1 
0
0
0 − L 0 L 

2 Lc
 0 −2c − L c L c 
ε x = 0 ; ε y = 0; γ xy = −
 0 
 0
 
 u D 
 
 vD
u F 
 
 v F 
3ML
4 Etc 3
Accordingly the results will be similar to (a)
(C)
εx 
1
 
=
ε y 
γ  4ab
 xy 
− (b − y )

0
0
(b − y )


− (a − x)
− ( a + x ) 
0
0

 − ( a − x ) − ( b − y ) − ( a + x )
( b − y ) 
For the coordinate system as below
 u1 
v 
 1
 u2 
 
 v2 
 
 
 v4 
(3.4-4)
Fig. 3.4.-1
We get a =L / 2 , b = c, and then we can rewrite above Eq. as below
εx 
1
 
=
ε y 
γ  2 Lc
 xy 
 − (c − y )

0
0
(c − y )


− ( L / 2 − x)
− ( L / 2 + x ) 
0
0

 − ( L / 2 − x )
− (c − y )
− ( L / 2a + x )
( c − y ) 
3My
2 Etc 3
εy = 0
εx =
3ML
4 Etc
γ xy =
−
+
3
3M  L

+ x
3 
2 Etc  2

1 0 0 
E = E 0 1 0 
0 0 1 / 2 
for plane stress
The stresses can be calculated as below
3My
2tc 3
σy = 0
σx =
τ xy =
3M
x
4tc 3
For our coordinate system x→ (x−L/2), y → (y − c), then
 0 
 0 


 uD 


 vD 


−uD 
 vD 


 0 
 0 


3M ( y − c )
2tc3
σy = 0
σx =
=
τ xy
L
3M 
x− 
3 
4tc 
2
Prob. 3.8