First course in the finite element method si version 5th edition logan solutions manual

First Course in the Finite Element Method SI Version 5th Edition Logan Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/first-course-in-the-finite-element-method-si-version-5th-edition-logan-solutionsmanual/

Chapter 9 9.1 (a) m

ii

jj

Figure 9.1a [k] = 2 π r A [ B ] T [D] [ B ] ri = 0, zi = 0, rj = 2, zj = 0, rm = 0, zm = 2

αi = rj zm – zj rm = 2.2 – 0 = 4 cm2 αj = rm zi – zm ri = 0.0 – 2.0 = 0 cm2 αm = ri zj – zi rj = 0.0 – 0.2 = 0 cm2 βi = zj – zm = 0 – 2 = – 2 cm βj = zm – zi = 2 – 0 = 2 cm βm = zi – zj = 0 – 0 = 0 cm

r =

γ i = rm – rj = 0 – 2 = – 2 cm

1 2 (2) = cm 3 3 z =

γ j = ri – rm = 0 – 0 = 0 cm

A=

1 2 (2) = cm 3 3

1 (2) (2) = 2 cm 2

γ m = rj – ri = 2 – 0 = 2 cm  −2

2 0 0 10  0 −2 0 0 0 [B] = 4 2 0 2 0 2   −2 −2 0 2 2 2 

0

0  2 0  0

0.75 0.25 0.25 0    210 × 10 0  0.25 0.75 0.25 [D] = (1 + 0.25)(1 − 0.5) 0.25 0.25 0.75 0     0 0 0 0.25 9

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0 1 – 0.5  –1  – 0.5 –1.5 – 0.5 – 0.5   1 2 0  210 × 10 11  2 T [ B ] [D] =   0 0 0.5  2.5  0  0.5 0.5 1.5 0.5    1.5 0.5 0   0.5

 5   1  0 [k] = 175.84 × 107  −1  1   0

1

0 −1

1

4 −2 −1 −2 −2 8 0 4 −1 0 1 1 −2 4 1 4 −3 2 0 1

0  −3 2  0 1  3

(b) m

j

i

Figure 9.1b ri = 0, zi = 0, rj = 2, zj = 0, rm = 2, zm = 2

αi = 4, αj = 0, αm = 0 βi = – 2, βj = 2, βm = 0 γ i = 0, γ j = – 2, γ m = 2 r = 2×

2 2 = 1.333 cm, z = cm, A = 2 cm2 3 3  −2

2 

[B] =

10  0 4 1  0

0 0  −2 0 2  1 0 1 0  −2 −2 2 2 0  0 0 0

2 0

0

0.75 0.25 0.25 0    210 × 109 0  0.25 0.75 0.25 [D] =  (1 + 0.25) (1 − 0.5 0  2 ) 0.25 0.25 0.75    0 0 0 0.25

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0   –1.25 – 0.25 0.25  0 0 0 – 0.5   0.75 1.25 – 0.5 210 × 1011  1.75 T [ B ] [D] =   2.5  – 0.5 –1.5 – 0.5 0.5   0.25 0.25 0.75 0.5    1.5 0.5 0   0.5 0 – 2.25 0.5 0.25 –0.5  2.75  0 1 1 –1 –1 0    1 5.75 – 2.5 0.25 1.5  7  – 2.25 [k] = 151.59 × 10   –1 – 2.5 4 0.5 – 3   0.5  0.25 –1 0.25 0.5 1.75 0.5    1.5 –3 0.5 3   – 0.5 0 (c) E = 210 × 109

N

v = 0.25 (Mathcad used here) m2 Triangle coordinate definition  0 cm  r = 0 i=  This defines an array variable   0 cm  z = 1 x coordinate is the top y coordinate is the bottom 1  2 cm  base × height Area of triangle = j=   2  0 cm  1  1 cm  (jr – ir) (mz – iz) A = 2 cm2 m=  A=  2  2 cm 

Develop stiffness matrix

αi = jr mz – jz mr αj = mr iz – mz ir αm = ir jz – iz jr

αi = 4 cm2 βi = jz – mz αj = 0 cm2 βj = mz – iz αm = 0 cm2 βm = iz – jz

βi = – 2 cm γ i = mr – jr γ i = – 1 cm βj = 2 cm γ j = ir – mr γ j = – 1 cm βm = 0 cm γm = jr – ir γm = 2 cm

Evaluate [B] at centroid of element ir + jr + mr 3

zbar =

βi   0 1  [ Bi ] = αi γ z 2 A  r + βi + ri bar bar  bar  γi

0 γi  0  βi 

rbar =

iz + j z + mz 3

rbar = 1 cm

βj   0 1  [ Bj ] = αj γ z  2 A r + β j + rj bar bar  bar  γ j

βm   0 1  [Bm] = αm γ z 2 A  r + βm + mr bar bar  bar  γm

zbar = 0.667 cm

0 γj   0  β  j

0

γm  0  βm 

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B = augment( Bi , B j , Bm )

Gradient matrix at centroid of element  −0.5 0 0.5 0 0 0    0 0 0 0.5  1 −0.25 −0.25 [ B] = (10) 2 ×   0.3333 0 0.3333 0 0.3333 0  m    −0.25 −0.5 −0.25 0.5 0.5 0  1 − v v v  v 1− v v E  [D] = (1 + v )(1 − 2v )  v v 1− v   0 0 0

0  Axisymmetric 0   stress 0  1 − 2 v  constitutive matrix 2 

 2.52 0.84 0.84 0    0.84 2.52 0.84 0  N [D] = (1011 ) ×  2  0.84 0.84 2.52 0  m    0 0 0 0.84  [k] = 2π rbar A [B]T [D] [B] Axisymmetric element stiffness matrix  0.682   0.140  −0.298 [k] = 127.75 × 106   −0.070  0.035   −0.070

0.140 0.368 −0.070 −0.053 −0.280 −0.315

−0.070  −0.070 −0.053 −0.280 −0.315 1.242 −0.280 0.315 0.035  N  −0.280 0.368 0.14 −0.315 m 0.315 0.140 0.490 0.140   0.350 −0.315 0.140 0.630  −0.298 −0.070

0.035

9.2 z

(b, h) p0

m

j

i (0, 0)

r

(b, 0)

 N1  0 p   {fs} = ∫ [ N s ]T  r  ds = ∫s  N 2 s  p2   0  N3  0

0 N1   p z 0   h0     2π rdz N2   0  0  Evaluated @ r=b N3  z = z

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Now

Ni =

1 1 (αi + βir + γ iz), Nj = (αj + βjr + γ jz) 2A 2A

Nm =

1 (αm + βmr + γ mz) 2A

ri = 0, rj = b, rm = b, zi = 0, zj = 0, zm = h

αi = rj zm – zj rm = bh – 0b = bh, αj = rm zi – zm ri = 0 βi = zj – zm = 0 – h = – h, βj = zm – zi = h – 0 = h γ i = rm – rj = b – b = 0, γ j = ri – γ m = 0 – b = – b αm = ri zj – zi rj = 0, βm = zi – zj = 0 – 0 = 0 γm = rj – ri = b – 0 = b So the shape functions evaluated at r = b and z = z

Ni =

1 (bh + (– h) b + 0 z) = 0 bh

Nj =

1 1 (0 + h b + (– b)z) = (hb – bz) bh bh

Nm =

1 1 (0 + 0 b + bz) = (bz) bh bh

 0  0  1  ( hb − bz ) h bh {fs} = ∫  0  0  1  bh (bz )   0

  0    p0 z  0   h  2π dz 1 ( hb − bz )     bh  0  0  1 (bz )  bh 0

    0 0     0 0      p bz – p0bz 2   p0bz 2 − p0bz 3  2π b h  0 2π  2 h  3h  =   dz =   h bh ∫0  0 0        p0bz 2 p0bz 3     h 3h     0 0

 0   0     f s1x     0   f s1 y   0  2    p0 bh   p0 h  2π  6   f s 2 x   6  =   ⇒  f  = 2π b   h  0   s2 y   0   p0 h   p0 bh 2   f s 3 x   3   3     0     f s3 y   0  359 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9.3

 RB  Z   B 2π rA  RB  Equation to be evaluated is {fB} =   3 Z B   RB    Z B  r = 4+2×

2 3

ZB = 288 × 103

⇒ r = 5.333 × 10−2 m N m3

2π × 20  2  288 × 103  −2 RB = w2ρ r =    (5.33 × 10 )  9.81 60     RB = 6863.44

N m3

2π rA 2 = π (5.333 ×10−2) (2 × 10−4) = 2.232 × 105 m3 3 3 So

fb1r = (2.232 × 10−5)6863.44 = 0.153 N fb1z = (–2.232 × 10−5) (288 × 103) = – 6.428 N fb2r = (2.232 × 10−5) (6863.44) = 0.153 N fb3z = (– 2.232 × 10−5) (288 ×103) = – 6.428 N fb3r = (2.232 × 10−5)6863.44 = 0.153 N fb3z = (– 2.232 × 10−5) (288 × 103) = – 6.428 N 9.4 (a)

Element Figure 9.4 a The equation to be evaluated is {σ} = [D] [B} {d}

ri = 0, zi = 0, rj = 2, zj = 0, rm = 1, zm = 3

αi = 6, αj = 0, αm = 0, βi = – 3, βj = 3, βm = – 3, γ i = – 1, γ j = – 1, γm = 2 r = 1, z = 1, A =

1 (3) (2) = 3 cm2 2

Figure 9.4a

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 −3 0 3 0 −3 0   2  10  0 −1 0 −1 0 2  [ B] = 6  2 0 2 0 −1 0     −1 −3 −1 3 2 −3 0.75 0.25 0.25 0    210 × 10 0  0.25 0.75 0.25 [D] = (1 + 0.25)(1 − 0.5) 0.25 0.25 0.75 0     0 0 0 0.25 9

σ r  0.75 0.25 0.25 0  σ   2 210 × 109 0.25 0.75 0.25  z 0   10      = (1.25)(0.5) 0.25 0.25 0.75 0  6  σ θ     0 τ rz  0 0 0.25  1   −3 0 3 0 −3 0  2   0 −1 0 −1 0 2  5 –6 ×   × 2.5 × 10  2 0 2 0 −1 0  6    −1 −3 −1 3 2 −3 0   0 

σ r  140.0  σ     z   0.0  =     MPa σ θ  140.0  τ rz   21.0 

(3, 0)

(b) Element Figure 9.4b

ri = 1, zi = 0, rj = 3, zj = 0, rm = 3, zm = 3

αi = 9, αj = – 3, αm = 0, βi = – 3, βj = 3, βm = 0 γi = 0, γ j = – 2, γm = 2

(2, 0) 2

r = 2.333, zm = 0.666, A = 3 cm

Figure 9.4 b

 −3 0 3 0 0 0  2  10 0 0 0 −2 2  0 [ B] = 6 0.858 0 1.14 0 0.512 0     0 3 0 −3 −2 2 0.75 0.25 0.25 0    210 × 10 0  0.25 0.75 0.25 [ D] = (1 + 0.25)(1 − 0.5) 0.25 0.25 0.75 0     0 0 0 0.25 9

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{σ} = [D] [ B ] {d} σ r  106.95  σ     z  −61.04  =   MPa σ θ   68.85  τ rz   7.00  (0, 2)

3

(c)

u1 = 0.00025 cm

w1 = 0.0005 cm

u2 = 0.00125 cm

w2 = 0.0015 cm

u3 = 0.0 cm

w3 = 0.0 cm

1 (0, 0)

2 (2, 0)

Figure 9.4c  u1  w   1 u  d=  2  w2   u3     w3 

164.49 × 103  σr    σ    z  = [D] [B] {d} =  38.49  MPa  σθ  199.47       τ rz  31.5

9.5 By Equation (9.1.35)

{fsi} = ∫

zm zj

  pr  0 1 α j + β j r j + γ j z     2π r j dz α j + β j r j + γ j z   pz  2A  0

Now

αj = rm zi – ri zm = rj zj – ri zm βj = zm – zj A= ∴

{fsj} = ∫

γj = ri – rj

Since rj = rm

zi = zm

1 (rj – ri) (zm – zj) 2

zm 2π r j   pr [rj z j − ri zm + ( zm − z j )r j + (ri − r j ) z  zj

  dz 2 A  pz [r j z j − ri zm + ( zm − z j )rj + (ri − rj ) z 

Integrating, we obtain  P (r z − r z ) ( z − z ) + r ( z − z )2 + (r − r ) ( zm − z j )  m j j m j i j 2   2π r j  r  j j i m {fsj} =   2 2 (z − z ) 2A   Pz (rj z j − ri zm ) ( zm − z j ) + rj ( zm − z j ) 2 + (ri − rj ) m 2 j     2

2

Factoring out zm – zj and simplifying − rj z j rj z j rz rz 2π rj ( zm − z j )  pr  2 − i 2m + 2 + i 2m   {fsi} =   –r z r z r z rz 2A  pz  2j j – j 2 m + j2 j + i 2m  

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1 2π rj ( zm − z j )   pr [ 2 ( zm − z j )( rj − ri )   =   1 2A  pz [ 2 ( zm − z j )( rj − ri ) 

=

{fsj} =

2π r j ( zm − z j )  pr A   2A  pz A  2π r j ( zm − z j ) 2

9.6 (a) E = 210 GPa u = 0.25 A = 1 (50)(50) 2 A = 1250 mm2 r = 16.67 mm z = 16.67 mm

αi = 50 (50) – (0) (0) = 2500 mm2 βi = – 50 mm, γ i = – 50 mm βj = 50 mm, γ j = 0 mm βm = 0 mm, γm = 50 mm −50 0 50 0 0 0    1  0 −50 0 0 0 50  1 [ B] = 2(1250)  50 0 50 0 50 0  mm   −50 −50 0 50 50 0  0.75 0.25 0.25 0    210 × 10 0.25 0.75 0.25 0  N [D] = 2 (1.25)(0.5) 0.25 0.25 0.75 0 m    0 0 0 0.25 9

 −25 9  210 10 ×  0 [D] [ B ] = 1250(1.25)  25  −12.5

−12.5 50 0 −37.5 25 0 −12.5 50 0 −12.5 0 12.5

12.5 12.5   12.5 37.5 37.5 12.5   12.5 0 

[k] = 2π r A [ B ] T [D] [ B ] =

2π (0.01667 m)(1250)(210 × 109 ) × 10−6 [ B ] T [D] [ B ] (1250)(1.25)

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Multiplying [ B ] T times [D] [ B ] , we obtain 3125 625 0 −625 625 0    2500 −1250 −625 −1250 −1875   5000 0 2500 1250  N [k] = 7.039 × 106   625 625 0 m   2500 625     Symmetry 1875 

(b)

A = 1800 mm2

βi = – 60 mm, βj = 60 mm, βm = 0 γ i = 0, γ j = – 60 mm, γ m = 60 mm [k] = 2π r A [ B ] T [D] [ B ] −60 0 60 0 0 0   1 0 0 −60 0 60 1  0 [ B] = 2(1800)  30 0 30 0 30 0  mm    0 −60 −60 60 60 0  [D] as in 9.6 (a)  −37.5

9 

[D] [ B ] =

[k] =

210 × 10  −7.5 1.25(1800)  7.5   0

0 52.5 −15 7.5 15   0 22.5 −45 7.5 45 0 37.5 −15 22.5 15   15 0 −15 −15 15

2π (0.04 m)(1800 )(210 × 109 ) T [ B ] [ D][ B ] (1.25)(1800)(2)(1800 )

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2475 0 −2025 450 225 −450    −900 −900 900 900 0    5175 −2250 225 1350  [k] = 11.73 × 106   3600 450 −2700   1575 450     Symmetry 2700  (c)

n

j

i

ri = 0

zi = 0

rj = 0.002

zj = 0

rm = 0.001

zm = 0.002

A=

αi = rj zm – zj rm

αj = rm zi – zm ri

αm = ri zj – zi rj

βi = zj – zm

βj = zm – zi

βm = zi – zj

γ i = rm – rj

γ j = ri – rm

γm = rj – ri

βi   0 1  [B] = α γ z  i 2A + βi + ri  r  γ i

1 1 0.002 , v = 0.25 (rj – ri) (zm – zi), E = 210 × 109, z = × 0.002, r = 2 3 2

0

βj

0

γi

0

γj

0

βi

αj r

+ βj +

γj

γjz r

0

βj

1 − v v v  v 1 − v v E  [D] = 1− v v (1 + v) (1 − 2v)  v   0 0 0

βm 0 αm r

γ z

+ βm + mr

γm

0 γ m  0   β  m

0  0   0  1 − 2v  2 

[k] = 2π r A [B]T [D] [B]

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 8.577 × 108  8  1.759 × 10  −3.738 × 108 [k] =   −8.796 × 107   4.398 × 107   −8.796 × 107

1.759 × 108

−3.738 × 108

−8.796 × 107

4.398 × 107

4.618 × 108

−8.796 × 107

−6.597 × 107

−3.519 × 108

−8.796 × 107

1.561 × 109

−3.519 × 108

3.958 × 108

−6.597 × 107

−3.519 × 108

4.618 × 108

1.759 × 108

−3.519 × 108

3.958 × 108

1.759 × 108

6.158 × 108

−3.958 × 108

4.398 × 108

−3.958 × 108

1.759 × 108

−8.796 × 107   −3.958 × 108  4.398 × 108  −3.958 × 108   1.759 × 108   7.917 × 108 

9.7 (a)

From Problem 9.6 (a), we have [D] [ B ] ∴ {σ} = [D] [ B ] {d} σ r   −25 σ   3  z × 210 10 MPa  0   = (1250)(1.25)  25 σ θ   τ rz   −12.5

 −84    =  −84  MPa  252  −101

−12.5 50 0 −37.5 25 0 −12.5 50 0 −12.5 0 12.5

12.5 12.5 37.5 12.5

0.05   12.5   0.03  37.5 0.02    12.5  0.02   0  0     0 

(b)

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From Problem 9.6 (b), we have [D] [ B ]

σr  −37.5 0 σ  3   z ∴   = 210 × 10  −7.5 0 σθ  (1500)(1.25)  7.5 0  τrz   0 −15 (c)

52.5 22.5 37.5 −15

ui = 0.00005

wi = 0.00003

uj = 0.00002

wj = 0.00002

um = 0

wm = 0

0.05   −15 7.5 15 0.03 −103    −45 7.5 45 0.02 = −103 MPa   −15 22.5 15 0.02  112     15 15 0   0   −73     0 

 ui   wi     uj  {d} =    wj  u   m  wm 

σr  σz    = [D] [B] {d}  σθ   τ  rz

σr = – 2.87 × 109 Pa σz = – 2.45 × 109 Pa σθ = 3.57 × 109 Pa τrz = – 1.89 × 109 Pa 9.8 No, not in general, as the axisymmetric elements are rings, not plane quadrilaterals or triangles. So axisymmetric nodes are actually nodal circles whereas plane stress elements have node points.

9.9 No, the element circumferential strain is a function of r and z (see Equation (9.1.15).

9.10 Make ur = 0 for all nodes acting on the axis of symmetry.

9.11 How would you evaluate circumferential strain εθ at r = 0?

From text Equation (9.1.15)

εθ =

a z a1 + a2 + 3 r r

(1)

εr = a 2

(2) 367

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Also from text Equation (9.1.1e)

∴

εθ =

u r

(3)

εr =

∂u ∂r

(4)

u = εθ r

(5)

Substituting (1) into (5) a z a ⇒ u =  1 + a2 + 3  r = a1 + a2r + a3z r r 

(6)

Partial of (6) with reference to r ⇒

∂u = a2 ∂r

Compare to (2)

(7)

∴ εθ r = 0 = εr = a2 as stated in problem statement 9.12 What will be the stresses σr and σθ at r = 0?

From Equation (9.1.2)

σr =

a2 E  || (1 + v )(1 − 2v )  ε (1 − v ) r

=

σr =

∴ σθ =

σθ =

a2

||

||

  εθ ( v )

+ ε z (v ) +

E (az − a2 v + a6 v + a2 v ) (1 + v)(1 − 2v)

E (a2 + a6 v) (1 + v)(1 − 2v)  a2 E   (1 + v)(1 − 2v)  ε ( v ) r

=

a6

a6 

+ ε z (v ) +

a2

  εθ (1 − v ) 

E ( a2 v + a6 v + a2 – a2 v ) (1 + v)(1 − 2v) E (a2 + a6 v) (1 + v)(1 − 2v) ∴ at r = 0,

σθ = σr

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9.14

von Mises stresses (with filleted corners)

Load Case: 1 of 1 Maximum Value: 6773.52 lbf/(in^2) Minimum Value: 18.6077 lbf (in^2)

Model of a nuclear reactor

33 cm 7.5 cm 19 cm 0.4 m 1.25 m

0.6 m

5 cm

0.75 mm

The largest von Mises stress of 2079 MPa occurs at the inside corners of the pressure vessel.

9.15

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Concrete

Steel liner

400 mm

1250 mm

p

325 mm

750 mm

Note: Without the arc (inside radius), we have a 90° re-entrant corner where stress is approaching infinity. We have a singularity in the linear-elastic solution based on linear theory of elasticity.

Therefore, we need the arc as in good practice or elastic-plastic model where an upper bound on the corner stress is the yield strength of the material.

9.18

The largest von Mises stress of 200 MPa occurs near the top of the hole. This number is suspect due to the location being under a concentrated load.

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9.19

Figure 1 Thick walled open-ended cylinder

Theoretical Solution for hoop stress at inner radius q = 35 × 106 a = 1.5 b = 1.2 r = 1.2

σθ =

qb 2 a 2 + r 2 r2

a 2 – b2

σθ = 1.594 × 108 Pa Algor Results

Theoretical Results

Hoop Stress

159.5 MPa

159.4 MPa

Maximum Principal Stress

159.5 MPa

-

Minimum Principal Stress

–35 MPa

-

Deflection in y direction

0.93mm

-

The Algor results for hoop stress and the theoretical solution for hoop stress are very close which proves that the Algor model is correct. The pipe has a very minimum internal and external deflection, less than 1 mm on the inner radius. The stresses are also manageable at 159 MPa.

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9.20 A steel cylindrical pressure vessel with flat plate end caps is shown in the figure with vertical axis of symmetry. Addition of thickened sections helps to reduce stress concentrations in the corners. Analyze the design and identify the most critically stressed regions. Note that inside sharp re-entrant corners produce infinite stress concentration zones, so refining the mesh in these regions will not help you get a better answer unless you use an inelastic theory or place small fillet radii there. Recommend any design changes in your report. Let the pressure inside be 1000 kPa. 503 elements and 645 nodes. Stresses are highest at sharp corners and the middle of the top and bottom of the pressure vessel. The design is acceptable as the von Mises stresses do not reach the yield strength of the material.

9.22

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The largest von Mises stress of 242 MPa occurs at the juncture of the elliptical and cylindrical location. This model was built using brick elements instead of the axisymmetric element due to difficulties exactly modeling the elliptical shape.

The recommended head shape of the hemispherical ends versus the ellipsoidal ends would be the hemispherical ends due to a lower stress concentration at the transition between the head and the cylindrical body.

9.23 According to the von Mises stress analysis, the average stress through the glass is around 15 of the tensile strength of the glass. If the maximum force used with this syringe is 45 N, the design should be fine. However, if 45 N is the normal operating force which may increase, then it is recommended to analyze this again with a safety factor of 4 (180 N force) to make sure it will still be under 5 MPa. As for the maximum principal stresses, they are well below the tensile strength of the glass and do not appear to be an issue.

N Another analysis with a safety factor of 4 (28.64789 radian ) reveals that this syringe is still within the tensile strength of glass in all areas. With this information, it can be concluded that the syringe design is indeed safe with this material specification.

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