Solutions manual for first course in the finite element method si edition 6th edition by logan ibsn

Solutions Manual for First Course in the Finite Element Method SI Edition 6th Edition by Logan IBSN 9781305637344 Full clear download (no formatting errors) at: http://downloadlink.org/p/solutions-manual-for-first-course-in-the-finiteelement-method-si-edition-6th-edition-by-logan-ibsn-9781305637344/

Chapter 2 2.1 (a)

[k(1)] =

[k(2)] =

[k 3(3)] =

k1

0 – k1

0

0

0

0

0

– k1 0

0 0

k1 0

0 0

0 0 0 0

0 0

0 0

0 0 k2 0 0 – k2 0 0

0 k3

– k2 k2

0 0 0 – k3

0 0 0 – k3

0 0

0 k3

[K] = [k(1)] + [k(2)] + [k(3)]

[K] =

k1

0

– k1

0

0

k3

0

– k3

– k1 0

0 – k3

k1 k2 – k2

– k2 k2 k3

(b) Nodes 1 and 2 are fixed so u1 = 0 and u2 = 0 and [K] becomes [K] =

k1

k2

– k2

– k2 k2 k3

{F} = [K] {d} F3 x F4 x 0 P

=

=

k1 k2 – k2

– k2 k2 k3

k1

– k2 k2 k3

k2

– k2

{F} = [K] {d}

u3 u4 u3 u4

[K] –1 {F} = [K ]–1 [K] {d}

K] –1 {F} = {d} Using the adjoint method to find [K –1] C11 = k2 + k3

C21 = (– 1)3 (– k2)

C12 = (– 1)1 + 2 (– k2) = k2

C22 = k1 + k2

[C] =

k3

k2

k2

k2

k1

and CT =

k2

k3

k2 k2

k2 k1

k2

det [K] = | [K] | = (k1 + k2) (k2 + k3) – ( – k2) (– k2) | [K] | = (k1 + k2) (k2 + k3) – k22 [K –1] =

[K –1] =

[C T ] det K k2 k k3

k2 k k3 k k2 k2 k k2 k2 2 1 = k22) ( k2 1k3 ) – k2 2 k1 k2 k1 k3 k2 k3

(k1

k3

k2 u3 u4

=

u3 = u4 =

k2 k1 k2

0 k1 k2 P k1 k3 k2 k3

k2 P k1 k3

k1 k2 (k1 k1 k2

k2

k2 k3

k2 ) P k1 k3 k2 k3

(c) In order to find the reaction forces we go back to the global matrix F = [K] {d} F1x F2 x F3 x

=

F4 x

k1 0

0 k3

k1 0

0 k3

k1 0

0 k3

k1 k2 k2

k2 k 2 k3

F1x = – k1 u3 = – k1 F1x =

k1 k2

k1 k2 k1 k2 P k1 k3 k2 k3

F2x = – k3 u4 = – k3 F2x = 2.2

(k1 k1 k2

k1 k2k3 (k k11 k3k2 )k2P k3

k2 P k1 k3 k2 k3

k2 ) P k1 k3 k2 k3

u1 u2 u3 u4

k1 = k2 = k3 = 100

N N = 1000 cm mm

(1) (2) (2) (3) k (1) k k k (2) ; [k(2)] = k k (2) k k (3)

[k(1)] =

By the method of superposition the global stiffness matrix is constructed.

[K] = Node 1 is fixed

(1)

(2)

(3)

k k 0

k k k

0 (1) k (2) k (3)

k

[K] =

k k 0

k 2k k

u1 = 0 and u3 =

{F} = [K] {d} F1x F2 x F3 x

?

0 = ?

k k 0

k 2k k

0 F3 x

2k k

k k

=

k u2 = = 2k

0 k k

uu1 0 2 ? u3

u2

0 F3x

20 mm = 2 2

2k u2

k

k u2 k u2 = 10 mm = 1 cm

F3x = – k (1 cm) + k (2 cm) F3x = (– 1000

N N ) (1 cm) + (1000 ) (2 cm) cm cm

F3x = 1000 N Internal forces Element (1) f1x (1) f2 x

( 2)

=

k k

k u1 0 k u2 1 cm

f1x (1) = (– 1000 f 2 x (1) = (1000 Element (2)



N ) (1 cm) cm

N ) (1 cm) cm

f1x (1) = – 1000 N f 2 x (1) = 1000 N

0 k k

f 2 x ( 2)

=

f 3x ( 2)

k k

k u2 1 cm k u3 2 cm

f 2 x (2) – 1000 N



f3x (2) 1000 N

2.3

k k k k By the method of superposition we construct the global [K] and knowing {F} = [K] {d} we have k 0 0 F1x ? u12 0 k 0 F2 x 0 k 2k k 0 0 u 0 k 2k k 0 u3 F3 x P =

(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =

F54x

0 0

?0

0 (b)

P = 2kk 0 0

0 0

k 0

k 2k

0 k

u2 u3

k

2k

4

uu u u 2 = 3 ; u4 = 3 2 2

2k k 0 P

k u54 0 k 2ku2 ku3 ku2 2ku3 ku3 2ku4

ku4

(1) (2) (3)

0

Substituting in the second equation above P = – k u2 + 2k u3 – k u4 P= –k

u3 2

+ 2k u3 – k

u3 2

P = k u3 u3 =

P k

u2 =

P P ; u4 = 2k 2k

(c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation {F} = [K] {d}

Check

F1x = – k u2 = –k

P 2k

F1x =

P 2

F5x = – k u4 = –k

P 2k

F5x =

P 2

Fx = 0

F1x + F5x + P = 0 P + 2

P 2

+P=0

0=0 2.4

(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =

k k

k k

By the method of superposition the global [K] is constructed.

Also

{F} = [K] {d} and u1 = 0 and u5 =

F1x F2 x F3 x

?

0 0 =

F4 x

0

F5 x

?

k 2k k 0 0

k k 0 0 0

0 k 2k k 0

0 0 k 2k k

(b) 0 = 2k u2 – k u3

0 0 0 k k

u1 0 u2 ? u3 ? u4 ? u5 (1)

0 = – ku2 + 2k u3 – k u4

(2)

0 = – k u3 + 2k u4 – k

(3)

From (2) u3 = 2 u2 From (3) 2 u2 2

u4 =

Substituting in Equation (2) – k (u2) + 2k (2 u2) – k – u2 + 4 u2 – u2 – u3 = 2 u4 =

u3 =

4 2 2

4

2

=0

2 u4 =

3 4

2 u 2  2  u2 = 4

(c) Going back to the global equation {F} = [K] {d} F1x = – k u2 =

k

k 4

F1x =

4

F5x = – k u4 + k = – k k F5x = 4

3 4

+k

2.5

[k (1)] =

[k (3)] =

d1 d2 200 200  ; [k (2)] = 200 200

d2 400 400

d2 d4 600 600  ; [k (4)] = 600 600 

d2 d4 800 800 800 800

d4 d3 1000 1000 [k (5)] = 1000 1000

d4 400 400



Assembling global [K] using direct stiffness method 200 200 0 0 200 200 400 600 800 0 400 600 800 [K] = 0 0 1000 1000 0 400 600 800 1000 400 600 800 1000 Simplifying 200 200 0 200 2000 0 [K] = 0 0 1000 0 1800 1000

0 1800 N 1000 mm 2800 

2.6 Now apply + 10,000 N at node 2 in spring assemblage of P 2.5.

F2x = 10,000 N



[K]{d} = {F} [K] from P 2.5 200 200 200 2000 0 0 0 1800

0 0 1000 1000

0   1800 1000  2800

u1 u2 u3

0 0

F1 =

u4

10, 000   F3 0 

where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed.

Using Equations (1) and (3) of (A) 2000 1800

1800 u2  = 2800 u4 0

10, 000  

Solving u2 = 11.86 mm, u4 = 7.63 mm 2.7

f1x = C, f2x = – C f = – k = – k(u2 – u1) f1x = – k(u2 – u1) f2x = – (– k) (u2 – u1) u1 f1x k –k = u2 f2 x –k k k –k same as for [K] = –k k tensile element 2.8

k1 = 105 So

1 1

1 ; k2 = 105 1

1 1

1 1

(A)

[K] = 10

1 2 1

1 1 0

5

0 1 1

{F} = [K] {d} 

F1 ?

F2 0

1 1

= 10

1 0 2 1

5

F3 2500

0 1



1

u1

0

u2 u3

? ?

105 u2 – 105 u3

0= 2

(1)

2500 = – 10 u2 + 10 u3 5

5

(2)

From (1) 105

u2 =

105

2

u3

u2 = 0.5 u3

(3)

Substituting (3) into (2) 2500 = – 105 (0.5 u3) + 105 u3 105 u3

2500 = 0.5

u3 = 0.05 m = 5 cm u2 = (0.5) (5 cm)

u2 = 2.5 cm

Element 1–2 f1x (1) f2 x

(1)

= 105 1

1

1 0  1 0.025 m



(1)

f1x f2 x

(1)



2500 N 2500 N

Element 2–3 f 2 x ( 2) f 3x ( 2)

= 105

F1x =

105

1

(2) 2500 N 1 0.025 m f 2 x (2) 1 1 0.05 m  f3x  2500 N

0  [1 – 1 0] 0.025 m F1x 2500 N  0.05 m

2.9

[k(1)] = 103

(1) 200 200

(2) 200 200

[k(2)] = 103

[k(3)] = 103

[K] = 103

F1x ?

F2 x 5000   = F3x 0 F4 x 20, 000 u1 = u2 = u3 = u4 =

200 200 0 0

(2) (3) 200 200 200 200 (3) (4) 200 200 200 200 (1) (2) (3) 200 200 0 200 400 200 0 200 400 0 0 200 200 400 200 0

0 200 400 200

(4) 0 0 200 200 0 0 200 200

u1

0

u2 u3 u4

0 0.075 m = 7.5 cm 0.175 m = 17.5 cm 0.275 m = 27.5 cm

Reactions F1x = 103 [200 – 200 0 0]

u1 0  0.075 2u  u3  u4 

F1x = – 15000 N

0.175 0.275

Element forces Element (1) f1x (1) (1)

f2 x Element (2) f 2 x ( 2) ( 2)

f 3x Element (3) f3x (3) f4 x 2.10

(3)

f3x (3) f4 x

(3)

= 103

= 103

= 103

 200 200 0  200 200 0.075 

f1x (1) 15, 000 N f (1)  15, 000 N

200 200 0.075  200 200 0.175 

f 2 x (2) 20, 000 N

200 200 0.175  200 200 0.275 

f3x (3) 20, 000 f (3)  20, 000

2x

f3x (2) 

4x

20, 000N

[k(1)] =103

200 200 200 200 

[k(2)] =103

100 100 100 100 

[k(3)] =103

100 100 100 100  

{F} = [K] {d} F1x ?

F2 x – 20, 000 F3x ?

F4 x ?

 = 103

u2 =

200 200 0 0

20, 000 400 103

200 400 100 100

0 100 100 0

u1 0

0

100 u 2 ?  0 u3 0 100 u4 0

= – 0.05 m = 5 cm

Reactions F1x F2 x F3 x F4 x F1x F2 x

= 103

=

0 200 200 200 400 100 0 100 100 0 100 0

10, 000  20, 000   5000 5000

F3 x 4x F



0 100 0 100

0 0.05 0  0

N

Element (1) f1x (1) f2 x

(1)

200 200 0  200 200 – 0.05

f1x (1) = f 2 x (1)

10, 000 N 10, 000

103

100 100 0.05 100 100 0 

f 2 x ( 2) = f 3x ( 2)

– 5000 N 5000

= 103

100 100 0.05 100 100 0 

f 2 x (3)

5000 N 5000

= 103

Element (2) f 2 x ( 2) f 3x

( 2)

=

Element (3) f 2 x (3) f4 x 2.11

(3)

f4 x

(3)

[k(1)] =

1000 1000  ; [k(2)] = 1000 1000 

3000 3000

3000 3000

{F} = [K] {d} F1x

?

1000

F2 x F3 x

0 ?

= 1000 0

1000 4000 3000

0 3000  3000

u1 0 u2 ? u3 0.02 m

u2 = 0.015 m Reactions Element (1)

F1x = (– 1000) (0.015)

F1x = – 15 N

f1x  = f2 x 

0  0.015

1000 1000  1000 1000 



Element (2)

f 2 x  3000 –3000 0.015  = f3 x  –3000 3000  0.02

f 1x f2 x 

f 2 x 15 = N f3 x  15

2.12

[k(1)] = [k(3)] = 10000

1 1

1 1

3 3  3 3

[k(2)] = 10000 {F} = [K] {d} F1x ?  F2 x 450 N   = 10000  F3 x 0 F4 x ?

1 1 0 0 1 4 3 0 0 3 4 1 0 0 1 1

15 N 15

u1 u2 u3 u4

0 ? ?

0

0 = – 3 u2 + 4 u3

u2 =

4 u3 3

u2 = 1.33 u3

450 N = 40000 (1.33 u3) – 30000 u3 450 N = (23200

N ) u3 m 10–2)

u2 = 1.5 (1.94

10–2 m

u3 = 1.93

10–2 m

u2 = 2.57

Element (1) f1x  = 10000 f2 x 



2.57 102 

f1x (1) 257 N f (1)  257 N

1 1

2.57 102 

f2 x

1.93 102

f3x (2) 193 N

1 1

1.93 102  0

f3x (3) 193 N

1 1

1 1

1 1

1 1

0

2x

Element (2) f2 x  = 30000 f3 x 

(2)

 193 N

Element (3) f3x f3x  = 10000 f4 x f4 x  Reactions {F1x} = (10000

0 N ) [1 – 1]  m 2.57 10

{F4x} = (10000

N 1.93 102  ) [–1 1] m 0

f4 x  2

2.13

1 1

1 1

{F} = [K] {d} F1x ?  F2 x 0 F3 x 5 kN = 60 F4 x 0   F5 x ?

1 1

1 2

0 1

0 0

0 0

0 0 0

1 0 0

2 1 0

1 2 1

0 1 1

u1 u2 u3 u4 u5

193 N

F1x = – 257 N

F4x = – 193 N

[k(1)] = [k(2)] = [k(3)] = [k(4)] = 60

(3)

0 ? ? ?

0

0 2 u2 – u3 u2 0.5 u3 0 – u3 2 u4 u4 0.5 u3

u2 = u 4

5 kN = – 60 u2 + 120 (2 u2) – 60 u2 5 = 120 u2

u2 = 0.042 m

u4 = 0.042 m u3 = 2(0.042)

u3 = 0.084 m

Element (1)

f1x  = 60 f2 x 

1 1

1 1

0  0.042

f1x (1) 2.5 kN

1 1

1 0.042 1 0.084 

f 2 x (2) 2.5 kN

f2 x

(1)

2.5 kN

Element (2) f 2x 1 1 = 60 f 3 x 1 1

f3x (2) 2.5 kN

Element (3) f 3x 1 1 = 60 f4 x 1 1

f 3x f4 x

(3) f 3x(3) 1  0.208.54kN f3x (3) 2.5 kN f4(3) f 14 x  0.042 x

1 1

2.5 kN 2.5 kN

Element (4) f4 x f  f5 x f

f14 x (4)0.024.52 kN f 4 x (4) 2.5 kN f15 x (4)02.5 kN f5 x (4) 2.5 kN 0  F1x = 60 [1 –1] F1x = – 2.5 kN 0.042

1 1 = 60 1 1 5x  4x

1 1

F 5x = 60 [–1 1]

0.042 0

F5x = – 2.5 kN

2.14

1 1

1 1

[k(1)] = [k(2)] = 4000 {F} = [K]d} F1x ? = 4000 F2 x 100 F3 x 200

1 1 0

1 2 1

0 1 1

100 = 8000 u2 – 4000 u3 – 200 = – 4000 u2 + 4000 u3

u1 0 u2 ? u3 ?

– 100 = 4000 u2

u2 = – 0.025 m

100 = 8000 (– 0.025) – 4000 u3

u3 = – 0.075 m

Element (1) f1x  = 4000 f2 x  Element (2)

1 1

1 1

f 2 x f2 x  = 4000 f3x f3x 

1 –1 –1 1

0  0.025

f1x (1) 100 N f2 x

(1)

100 N

f 2 x (2) 200 N

0.025 0.075 

f3x

(2)

200 N

Reaction {F1x} = 4000 [1 –1]

0 0.025

F1x = 100 N

2.15

[k(1)] = F1x ?    F2 x ? = F3 x 4 kN  F4 x ?

500 500 500 0 500 0

500 ; [k(2)] = 500 0 500 500 0

500 500 2000 1000

500 500 0 0 1000 1000

500 ; [k(3)] = 500 u1 u3

0 0 ?

u4

0

u2

u3 = 0.002 m Reactions F1x = (– 500) (0.002) F2x = (– 500) (0.002) F4x = (– 1000) (0.002) Element (1)

f1x  = f3 x 

500 500

500 500

F1x = – 1.0 kN F2x = – 1.0 kN F4x = – 2.0 kN 0  0.002

f1x  = f3 x 

1.0 kN 1.0 kN

Element (2) 500f 2 x 500 500 500 0 1f.02 xkN 1.0 kN = = 1 0.002 .f03 xkN 1.0 kN 500 f3 x  500 500 500 Element (3)  1000 0.001 f3 x  f 3 x 2.0 kN 10000.002  = = f4 x  f4 x  0 0  2.0 kN 1000 1000

1000 1000

1000 1000

2.16 F1x  20 200   500 20 20 20  = 500 0 20 F4 x 0 0 500  40 20 3 = 10 500  20 40  u2 = 8.33 u3 =

8.33

10

3

10

m 3

0 20 20 20 20 u2 u3

0 0



20  20

0 u2 u3 0

8.33 mm m

8.33 mm

2.17

F1x 0

u1

?

1000 N

= 500 500

F4 x

500 300 –500 400 300

u2 0 300 400)

(

–300 – 300

0 = 1500 u2 – 600 u3 0 –300 – 300 300 ? 1000 = – 600 u2 + 1000 u3 0 – 400 – 400 15 0 0 u3 = u2 = 2.5 u2 60 0 1000 = – 600 u2 + 1000 (2.5 u2) 1000 = 1900 u2 u2 =

1 1000 = mm = 0.526 mm 1900 1.9

u3 = 2.5

1 1.9

0

mm = 1.316 mm

0 – 400 – 400 400 400

u3 u4

0

F1x = – 500

1 = – 263.16 N 1.9

F4x = – 400

1 – 400 1.9

2.5

1 1.9

1 2.5 = –736.84 N 1.9 1.9 Fx = – 263.16 + 1000 – 736.84 = 0 = – 400

2.18 (a)

As in Example 2.4 p= U + 1 k x2, = – Fx U= 2 Set up table 1 (2000) x2 – 5000 x = 1000 x2 – 5000 x p= 2 Deformation x, cm – 7.5 – 5.0 – 2.5 0.0 0.5 2.5 7.5 p

x

= 2000 x – 5000 = 0

p,

Ncm

93,750 50,000 18,750 0 6250 0 18,750 x = 2.5 cm yields minimum p as table verifies.

(b)

p

=

1 2 kx – Fx = 2500 x2 – 5000 x 2 x, cm – 3.0 – 2.0 – 1.0 0 1.0 2.0 3.0

p,

N cm 37,5000 20,000 7500 0 – 2500 0 7500

p

= 5000 x – 5000 = 0 x x = 1.0 in. yields p minimum (c)

p

=

1 (2000) x2 – 3924 x = 1000 x2 – 3924 x 2

p

= 2000 x – 3924 = 0

x

x = 1.962 mm yields p minimum p min

1 (2000) (1.962)2 – 3924 (1.962) 2

= – 3849.45 N mm

p min

(d)

=

1 (400) x2 – 981 x 2

=

p

p

= 400 x – 981 = 0

x

x = 2.4525 mm yields p minimum p min = p min =

1 (400) (2.4525)2 – 981 (2.4525) 2

– 1202.95 N mm

2.19 1 2 kx – Fx 2

=

p

p

1 (125) x2 – 5000 x 2

= (62.5)x2 – 5000 x

p p

x

=

= (125)x – 5000 = 0

x = 40.0 mm 2.20

F = k2 dU = F dx

(x = )

U=

0 (kx2) dx

U=

k x3 3

x

= – Fx 1 3 p= kx – 2500 x 3 p

= 0 = kx2 – 2500

x

0 = 2500 x2 – 2500 x = 1.0 cm (equilibrium value of displacement) 1 p min = (2500) (1.0)3 (2500) (7.0) 3 p min

= – 1666.7 N cm

2.21 Solve Problem 2.10 using P.E. approach 3 p

=

p e 1

(e) =

1 1 1 k1 (u2 – u1)2 + k2 (u3 – u2)2 + k3 (u4 – u2)2 2 2 2

– f1x(1) u1 – f2x(1) u2 – f2x(2) u2 – f3x(2) u3 – f2x(3) u2 – f4x(3) u4 p

u1

= – k1 u2 + k1 u1 – f1x(1) = 0

(1)

p

= k1 u2 – k1 u1 – k2 u3 + k2 u2 – k3 u4 u2 + k3 u2 – f2x(1) – f2x(2) – f2x(3) = 0 p

u3

= k2 u3 – k2 u2 – f3x(2) = 0

(2) (3)

p

u4

= k3 u4 – k3 u2 – f4x(3) = 0

(4)

In matrix form (1) through (4) become k1 k1

k1 k1 k2

0 0

k3

k2 k3

0 k2

0 k3

u1 u2

k2 0

0 k3

u3

f1x (1) =

f 2 x (1)

u4

f 2 x ( 2) f3 x

( 2)

f4 x

(3)

f 2 x (3)

(5)

or using numerical values 200 200 0 0  200 400 200 –200  103 0 200 200 0   0 200 0 200 

u1 u2 u3 u4

0

F1x 20, 000  = F3 x 0 F4 x 0

Solution now follows as in Problem 2.10 Solve 2nd of Equations (6) for u2 = – 0.05 m = 5 cm For reactions and element forces, see solution to Problem 2.10 2.22 Solve Problem 2.15 by P.E. approach 3 p=

p e 1

(e)

=

1 1 k1 (u3 – u1)2 + k2 (u3 – u2)2 2 2 +

1 2

k (u – u )2 – f 3

4

3

u 1x

(1)

1

– f3x(1) u3 – f2x(2) u2 – f3x(2) u3 – f3x(3) u3 – f3x(4) u4 p

u1 p

u2 p

u3

= 0 = – k1 u3 + k1 u1 – f1x(1) = 0 = – k2 u3 + k2 u2 – f2x(2) = 0 = k1 u3 + k2 u3 – k2 u2 – k3 u4 + k3 u3 – f3x(2) – f3x(3) – f3x(1) – k1 u1

(6)

p

= 0 = k3 u4 – k3 u3 – f3x(4)

u In matrix form 4

k1

0

0 k1 0

kk1

k2 k2 0

k1 k2 2 k3 k3

0

u1

F1x

0 k3 k3

u2 u3 u4

F2 x = 4 kNF3 x F4 x

For rest of solution, see solution of Problem 2.15.

2.23 I = a1 + a2 x I (0) = a1 = I1 I (L) = a1 + a2 L = I2 a2 =

I2

I1 L I = I1 +

I 2 I1 x L

Now V = IR V = – V1 = R (I2 – I1) V = V2 = R (I2 – I1) V1 V2

=R

1 1

1 1

I1 I2

23

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