Chapter 10: AXISYMMETRIC ISO-P ELEMENTS 10–14

Homework Exercises for Chapter 10

Solutions

EXERCISE 10.1
Use the centroidal rule for the first one:

ζi d A(e) = A F( 13 , 13 , 13 ) = A( 13 ) = A/3. (E10.28)
A(e)

Use the midpoint rule (simplest exact rule) for the second one:

ζi ζ j d A(e) = (A/3)[F( 12 , 12 , 0) + F(0, 12 , 12 ) + F( 12 , 0, 12 )] = (A/3)[( 12 )2 + ( 12 )2 δi j ] = A(1 + δi j )/12.
A(e)
(E10.29)
For (E8.7) we need the 7-point rule. Three cases will be distinguished. For i = j = k,

ζi ζ j ζk d A(e) = A[w0 ( 13 )3 + 3w1 α1 β12 + 3w1 α2 β22 ] = A/60. (E10.30)
A(e)

For i = j = k,

ζi2 ζk d A(e) = A[w0 ( 13 )3 + 3w1 β1 (α12 + α1 β1 + β12 ) + 3w2 β2 (α22 + α2 β2 + β22 )] = A/30. (E10.31)
A(e)

For i = j = k,

ζi3 d A(e) = A[w0 ( 13 )3 + 3w1 (α13 + 2β13 ) + 3w2 (α23 + 2β23 )] = A/10. (E10.32)
A(e)

Combining these 3 cases the result is γi jk A/60 as stated.

For (E8.8) we use (E8.5):
 
(e)
r dA = (r1 ζ1 + r2 ζ2 + r3 ζ3 ) d A(e) = A(r1 + r2 + r3 )/3. (E10.33)
A(e) A(e)

For (E8.9) we use (E8.6):

 
(e)
ζi r d A = ζi (r1 ζ1 + r2 ζ2 + r3 ζ3 ) d A(e) = (A/12)r j (1 + δi j ) = 1
12
A(r1 + r2 + r3 + ri ). (E10.34)
A(e) A(e)

For (E8.10) we use (E8.6) again, but more steps are needed:
 
(e)
2
r dA = (r1 ζ1 + r2 ζ2 + r3 ζ3 )2 d A(e)
A(e) A(e)
    
ζ1 ζ1 ζ1 ζ2 ζ1 ζ3 r1
= [ r1 r2 r3 ] ζ2 ζ2 ζ2 ζ3 d A(e) r2
A(e)
symm ζ3 ζ3 r3
    
A 2 1 1 r1
(E10.35)
= [ r1 r2 r3 ] 1 2 1 d A(e) r2
A(e) 12 1 1 2 r3
  1 1 1   1 0 0   
r1
A
= [ r1 r2 r3 ] 1 1 1 + 0 1 0 d A(e) r2
12 A(e) 1 1 1 0 0 1 r3
 
= 1
12
A (r1 + r2 + r3 )2 + r12 + r22 + r32 .

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10–15 Solutions to Exercises

EXERCISE 10.2
It’s an easy one using the summation convention to write r 2 = r j rk ζ j ζk :
  
(e) (e)
ζi r d A
2
= ζi ζ j ζk r j rk d A = r j rk ζi ζ j ζk d A(e) = r j rk γi jk /60. (E10.36)
A(e) A(e) A(e)

EXERCISE 10.3
For (E8.12):
Rule Case (a) Case (b)
1 point 1.5000A/a 0.5000A
3 points (E8.2) 1.8000A/a 0.5222A
3 points (E8.3) 1.6667A/a 0.5222A
7 points 1.8800A/a 0.5232A
Exact (to 5 digits) 2.0000A/a 0.52324A

For case (a) the 7-point rule is 6% off; next best result is given by the 3-interior-point rule.
For (E8.13):

Rule ζ1 /r ζ2 /r = ζ3 /r
1 point 0.5000A/a 0.5000A/a
3 points (E8.2) 0.8000A/a 0.5000A/a
3 points (E8.3) 0.6667A/a 0.5000A/a
7 points 0.8800A/a 0.5000A/a
Exact 1.0000A/a 0.5000A/a

EXERCISE 10.4
With the constitutive matrix  
1 0 0 0
0 1 0 0
E= E , (E10.37)
0 0 1 0
1
0 0 0 2
we get 
Kr 1r 1 = E (qr 1 qr 1 + qθ1 qθ1 + 12 qz1 qz1 ) d A(e) , (E10.38)
A(e)

Kr 1z1 = E 1
q q
2 r 1 z1
d A(e) , (E10.39)
A(e)

Kz1z1 = E (qz1 qz1 + 12 qr 1 qr 1 ) d A(e) . (E10.40)
A(e)

From the isoparametric definition,

∂ζ1 1 ∂ζ1 ζ1 1 1
qr 1 = =− , qz1 = = 0, qθ1 = = − , (E10.41)
∂r a ∂z r r a

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Chapter 10: AXISYMMETRIC ISO-P ELEMENTS 10–16

Using the midpoint rule:

  
1 ζ12 E A 2 1  1 2 1 
K r 1r 1 = E + r d A(e) = 3
+ 3 ( 2 ) / 2 + ( 12 )2 / 12 = E A/a = 12 Eb. (E10.42)
A(e) a2 r2 a

(Using other rules would change this value, but this is enough for comparison with the computer work.)
Similarly,
K r 1z1 = 0. (E10.43)

K z1z1 = E A/(3a) = 16 Eb. (E10.44)

EXERCISE 10.5
Note br = ρω2 (ζ1 r1 + ζ2 r2 + ζ3 r3 ) = ωa(ζ2 + ζ3 ). For the displacement ordering u r 1 , u r 2 , u r 3 , u z1 , u z2 , u z3
we get

 ωa 2 ζ (ζ + ζ )2   1/10 
1 2 3

  ωa 2 ζ2 (ζ2 + ζ3 )2   1/5 
2 2
 ωa ζ (ζ + ζ )  (e)  
(e)
f =ρ  3 2 3  d A = ρ Aω a  1/5  .
2 2
(E10.45)
 0   0 
A(e)    
0 0
0 0

EXERCISE 10.6
 0
 0
 
  0   0 
  (e)  
f(e) =ρ  0  d A = −ρga A  0  . (E10.46)
 −ga ζ1 (ζ2 + ζ3 )   
A(e)    1/6 
−ga ζ2 (ζ2 + ζ3 ) 1/4
−ga ζ3 (ζ2 + ζ3 ) 1/4

EXERCISE 10.7
 ζ /r − 1/a   1/3

1

   ζ2 /r + 1/a   1 
 ζ3 /r  (e)  
(e)
f = (e)
B α T r d A = α T
T   d A = α T A  1/r . (E10.47)
 0   0 
A(e) A(e)    
−r/b −2a/(3b)
r/b 2a/(3b)

EXERCISE 10.8

EXERCISE 10.9

EXERCISE 10.10

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10–17 Solutions to Exercises

EXERCISE 10.11

(a) The displacement field u = u(r, z) is a function of r and z only (no theta dependence). Thus, it can be
modeled axisymmetrically by a FE mesh on the (r, z) plane. But in this case the FE displacement field
has the form
u r = u z = 0, u θ = u θ (r, z) (E10.48)
so that there is only one degree of freedom per node. This DOF is the displacement normal to the plane
of the mesh.
(b) The nontrivial parts of the strain-displacement and constitutive equations are
 ∂ 1
  −     
γr θ σr θ γr θ
=  ∂r r u , G 0
θ = . (E10.49)
γzθ ∂ σzθ 0 G γzθ
∂z
Consequently
 (e)
  
N1,r − N1(e) /r (e)
N2,r − N2(e) /r ... (e)
Nn,r − Nn(e) /r G 0
B= (e) (e) (e)
, E= . (E10.50)
N1,z N2,z ... Nn,z 0 G

10–17