Mathematics Chapter 6 For maximum displacement of the bottom mass

Solutions — Chapter 6

6.1.1. (a)K= 3−2

−2 3 !; (b)u=0

@

18

5

17

51

A= 3.6

3.4!; (c) the first mass has moved the

farthest; (d)e=“18

5,−1

5,−17

5”T= ( 3.6,−.2,−3.4 )T, so the first spring has stretched

the most, while the third spring experiences the most compression.

6.1.3. 6.1.1: (a)K= 3−2

−2 2 !; (b)u= 7

17

2!= 7.0

8.5!; (c) the second mass has

moved the farthest; (d)e=“7,3

2”T= ( 7.0,1.5 )T, so the first spring has stretched the

most.

2,3”T= ( 3.5,3.)T, so the first spring has stretched slightly farther.

6.1.4.

(a)u= ( 1,3,3,1 )T,e= ( 1,2,0,−2,−1 )T. The solution is unique since Kis invertible.

(b) Now u= ( 2,6,7.5,7.5 )T,e= ( 2,4,1.5,0 )T. The masses have all moved farther, and

the springs are elongated more; in this case, no springs are compressed.

6.1.5.

(a) Since e1=u1,ej=uj−uj+1, for 2 ≤j≤n, while en+1 =−un, so

e1+···+en+1 =u1+ (u2−u1)+(u2−u1)+···+ (un−un−1)−un= 0.

Alternatively, note that z= ( 1,1,…,1 )T∈coker Aand hence z·e=e1+···+en+1 = 0

♣6.1.7. Top and bottom support; constant force:

20 40 60 80 100

2

4

6

8

10

12

20 40 60 80 100

-0.4

-0.2

0.2

0.4

Top and bottom support; linear force:

20 40 60 80 100

10

20

30

40

50

20 40 60 80 100

0.2

0.4

0.6

0.8

1

Top support only; linear force:

160

6.1.8.

(a) For maximum displacement of the bottom mass, the springs should be arranged from

weakest at the top to strongest at the bottom, so c1=c= 1, c2=c0= 2, c3=c00 = 3.

(b) In this case, the order giving maximum displacement of the bottom mass is c1=c= 2,

c2=c0= 3, c3=c00 = 1.

♣6.1.9.

(a) When the bottom end is free, for maximum displacement of the bottom mass, the springs

6.1.10.

The sub-diagonal entries of Lare li,i−1=−1

i, while the diagonal entries of Dare dii =i+ 1

i.

♥6.1.11.

(a) Since y=Au, we have y∈rng A= corng AT. Thus, according to Theorem 5.59, yhas

minimal Euclidean norm among all solutions to ATy=f.

6.1.12. Regular Gaussian Elimination reduces them to 0

B

@

2−1 0

03

21

0 0 4

3

1

C

A,0

B

@

2−1 0

03

21

0 0 1

3

1

C

A, respec-

tively. Since all three pivots are positive, the matrices are positive definite.

161

6.1.14.

(a)p(u) = 1

2(u1u2) 3−2

−2 3 ! u1

u2!−(u1u2) 4

3!

=3

2u2

1−2u1u2+3

2u2

2−4u1−3u2,so p(u?) = p“3.6,3.4”=−12.3.

(b) For instance, p(1,0) = −2.5, p(0,1) = −1.5, p(3,3) = −12.

6.1.16.

(a) Two masses, both ends fixed, c1= 2, c2= 4, c3= 2, f= ( −1,3 )T;

equilibrium: u?= ( .3, .7 )T.

6.1.17. In both cases, the homogeneous system Au=0requires 0 = u1=u2=··· =un, and so

ker A={0}, proving linear independence of its columns.

6.1.18. This is an immediate consequence of Exercise 4.2.9.

♥6.1.19.

(a) When only the top end is supported, the potential energy is lowest when the springs are

6.1.20. True. The potential energy function (6.16) uniquely determines the symmetric stiﬀness

matrix Kand the external force vector f. According to (6.12), (6.15), the oﬀ-diagonal en-

6.2.1. (a) (b)

(c) (d) (e)

ﬂowing through.

6.2.3. The reduced incidence matrix is A?=0

B

@

1−1

1 0

0 1

1

C

A

, and the equilibrium equations are

3−1

−1 3 !u= 3

0!, with solution u=0

@

9

8

3

A= 1.125

.375 !; the resulting currents are

6.2.4. (a)A=

B

@

100−1 0

010−1 0

0 0 1 0 −1

0 0 0 1 −1

C

A

; (b)u=0

B

@

35

19

35

16

35

C

A

=0

B

@

.5429

.4571

C

A;

y=“11

35 ,4

35 ,3

7,9

35 ,1

5,19

35 ,16

35 ”T= ( .3143, .1143, .4286, .2571, .2000, .5429, .4571 )T;

(c) wire 6.

163

♠6.2.7. There is no current on the two wires connecting the same poles of the batteries (positive-

positive and negative-negative) and 2.5 amps along all the other wires.

♠6.2.8.

(a) The potentials remain the same, but the currents are all twice as large.

♣6.2.10.

(a) For n= 2, the potentials are

0

B

@

1

16

1

8

1

16

1

8

3

8

1

8

1

C

A=0

B

@

.0625 .125 .0625

.125 .375 .125

.0625 .125 .0625 1

C

A.

where all wires are oriented from left to right, so the currents are all going away from

the center. The currents in the vertical wires are given by the transpose of the matrix.

For n= 4, the potentials are

0

B

.0165 .0331 .0478 .0551 .0478 .0331 .0165

.0331 .0680 .1029 .125 .1029 .0680 .0331

.0478 .1029 .1710 .2390 .1710 .1029 .0478

1

C

164

The currents along the horizontal wires are

0

B

−.0165 −.0165 −.0147 −.0074 .0074 .0147 .0165 .0165

−.0331 −.0349 −.0349 −.0221 .0221 .0349 .0349 .0331

1

C

As n→ ∞ the potentials approach a limit, which is, in fact, the fundamental so-

lution to the Dirichlet boundary value problem for Laplace’s equation on the square,

[47,59]. The horizontal and vertical currents tend to the gradient of the fundamental

solution. But this, of course, is the result of a more advanced analysis beyond the scope

of this text. Here are graphs of the potentials and horizontal currents for n= 2,3,4,10:

6.2.11. This is an immediate consequence of Theorem 5.59, which states that the minimum

norm solution to ATy=fis characterized by the condition y= corng AT= rng A. But,

solving the system ATAu=fresults in y=Au∈rng A.

6.2.12.

(a) (i)u= ( 2,1,1,0 )T,y= ( 1,0,1 )T; (ii )u= ( 3,2,1,1,0 )T,y= ( 1,1,0,1 )T;

6.2.13.

(i)u=“3

2,1

2,0,0”T,y=“1,1

2,1

2”T;

165

For the exclusive use of adopters of the book Applied Linear Algebra, by Peter J. Olver and Cheri Shakiban. ISBN 0-13-147382-4.

6.2.14. According to Exercise 2.6.8(b), a tree with nnodes has n−1 edges. Thus, the reduced

incidence matrix A?is square, of size (n−1) ×(n−1), and is nonsingular since the tree is

connected.

6.2.15.

6.2.16. In general, if v1, . . . , vmare the rows of the (reduced) incidence matrix A, then the re-

sistivity matrix is K=ATC A =

m

X

civT

6.2.17.

(a) If fare the current sources at the nodes and bthe battery terms, then the nodal volt-

age potentials satisfy ATC Au=f−ATCb.

(b) By linearity, the combined potentials (currents) are obtained by adding the potentials

(currents) due to the batteries and those resulting from the current sources.

♦6.2.18. The resistivity matrix K?is symmetric, and so is its inverse. The (i, j) entry of (K?)−1

is the ith entry of uj= (K?)−1ej, which is the potential at the ith node due to a unit cur-

6.2.19. If the graph has kconnected subgraphs, then there are kindependent compatibility

6.3.2. The bar will be stress-free provided the vertical force is 1.5 times the horizontal force.

6.3.3.

(a) For a unit horizontal force on the two nodes, the displacement vector is

6.3.4. The swing set cannot support a uniform horizontal force, since f1=f2=f,

g1=g2=h1=h2= 0 does not satisfy the constraint for equilibrium. Thus, the swing

set will collapse. For the reinforced version, under a horizontal force of magnitude f, the

displacements of the two free nodes are ( 14.5f, 0,−3f)Tand ( 14.5f, 0,3f)Trespectively,

♥6.3.5. (a)A=

0

B

@

0100

−1010

0001

0 0 1

√2

1

√2

−1

√2

1

√20 0

1

C

A

; (b)

3

2u1−1

2v1−u2= 0,

−1

2u1+3

2v1= 0,

−u1+3

2u2+1

2v2= 0,

1

2u2+3

2v2= 0.

(c) Stable, statically indeterminate. (d) Write down f=Ke1, so f1= 3

2

0!.

Joined version:

u= ( .0909, .8182,−.0909, .8182,0, .3636 )T,

e= ( .8182,−.1818, .8182, .2571, .2571, .2571, .2571 )T;

Thus, joining the nodes causes a larger vertical displacement, but smaller horizontal dis-

6.3.7.

B

0 1 0 0 0 0

−1

√2−1

√2

1

√2

1

√20 0

1

C

♥6.3.8.

(a)A=

0

B

0 1 0 0 0 0

−3

√10 −1

√10

3

√10

1

√10 0 0

−1 0 0 0 1 0

1

C

0 0 0 0 0 1

(b) One instability: the mechanism of simultaneous horizontal motion of the three nodes.

(c) No net horizontal force: f1+f2+f3= 0. For example, if f1=f2=f3= ( 0,1 )T, then

e=„3

2,q5

2,−3

2,q5

2,3

2«T

= ( 1.5,1.5811,−1.5,1.5811,1.5 )T, so the compressed di-

agonal bars have slightly more stress than the compressed vertical bars or the elongated

♣6.3.9. Two-dimensional house:

0

B

0 1 0 0 0 0 0 0 0 0

0−1 0 1 0 0 0 0 0 0

−1 0 0 0 0 0 0 0 1 0

1

C

(c) Numbering the five movable nodes in order starting at the middle left, the correspond-

ing forces f1= ( f1, g1)T,…,f5= ( f5, g5)Tmust satisfy: f1+f5= 0, f2+f3+

f4= 0, f2+1

2f3+g3= 0. When f1=f2=f4=f5= ( 0,1 )T,f3=0,

(d) To stabilize the structure, you need to add in at least three more bars.

(e) Suppose we add in an upper horizontal bar and two diagonal bars going from lower left

Three-dimensional house:

(a)

0

B

0 0 1 0 0 0 0 0 0

0−1

√2−1

√201

√2

1

√20 0 0

0−1 0 0 0 0 0 1 0

0 0 0 0 −1

√2

1

√201

√2−1

√2

0 0 0 0 0 0 0 0 1

−1 0 0 0 0 0 0 0 0

(b) 5 mechanisms: horizontal motion of (i) the two topmost nodes in the direction of the

bar connecting them; (ii ) the two right side nodes in the direction of the bar connect-

ing them; (iii ) the two left side nodes in the direction of the bar connecting them;

169

(e) To stabilize, you need to add in at least five more bars, e.g., two diagonal bars across

the front and back walls and a bar from a fixed node to the opposite topmost node.

In all cases, if a minimal number of reinforcing bars are added, the stresses remain

the same on the old bars, while the reinforcing bars experience no stress. See Exercise

6.3.21 for the general result.

♥6.3.10.

(a) Letting widenote the vertical displacement and hithe vertical component of the force

on the ith mass, 2w1−w2=h1,−w1+ 2w2−w3=h2,−w2+w3=h3.The system is

statically determinate and stable.

♥6.3.11.

(a) The incidence matrix is

A=

0

B

@

1−1

−1 1

......

1

C

A

170

the ring is zero.

(c) For instance, if c1=c2=c3=c4= 1, and f= ( 1,−1,0,0 )T, then the solution is

u=“1

4,−1

2,−1

4,0”T+t( 1,1,1,1 )Tfor any t. Nonuniqueness is telling us that the

masses can all be moved by the same amount, i.e., the entire ring is rotated, without

aﬀecting the force balance at equilibrium.

♣6.3.12.

(a)

(c)v1,v2,v3correspond to translations in, respectively, the x, y, z directions;

(d)v4,v5,v6correspond to rotations around, respectively, the x, y, z coordinate axes;

0

B

1 0 0 −100000000

0 1 0 0 0 0 0 −1 0 0 0 0

00100000000−1

−1 0 0 2 −1

2−1

2

1

20−1

201

2

0 0 0 −1

1

C

171

(f) For fi= ( fi, gi, hi)Twe require f1+f2+f3+f4= 0, g1+g2+g3+g4= 0,

♣6.3.13.

(a)

1

2√3−1

2√3

(b)v1=

0

B

1

0

1

0

1

C

,v2=

0

B

0

1

0

1

0

1

C

,v3=

0

B

0

1

0

1

0

1

C

,v4=

0

B

−√2

√6

−1

−2√2

0

1

0

1

C

,v5=

0

B

0

−2

0

√3

1

0

−√3

1

C

,v6=

0

B

−√2

−√6

−1

0

−2√2

1

C

;

172

This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced,

in any form or by any means, without permission in writing from the publisher.

For the exclusive use of adopters of the book Applied Linear Algebra, by Peter J. Olver and Cheri Shakiban. ISBN 0-13-147382-4.

(e)K=

0

B

11

60−√2

3−3

4

√3

40−3

4−√3

40−1

30√2

3

01

20√3

4−1

40−√3

4−1

40 0 0 0

−√2

302

30 0 0 0 0 0 √2

30−2

3

−3

4

√3

405

6−1

√3

1

12

1

30−2

3−1

3√2

3−1

3√2−1

3002

1

C

(f) For fi= ( fi, gi, hi)Twe require

f1+f2+f3+f4= 0,

−√2f1+√6g1−h1−2√2f2+h2= 0,

i.e., there is no net horizontal force and no net moment of force around any axis.

(g) You need to fix three nodes. Fixing only two nodes still permits a rotational motion

6.3.14. True, since stability only depends on whether the reduced incidence matrix has trivial

kernel or not, which depends only on the geometry, not on the bar stiﬀnesses.

6.3.15. (a) True. Since Ku=f, if f6=0then u6=0also. (b) False if the structure is unsta-

ble, since any u∈ker Ayields a zero elongation vector y=Au=0.

6.3.16.

173

6.3.19. Since y=e=Au∈rng A= corng AT, which, according to Theorem 5.59, is the

condition for the solution of minimal norm to the adjoint equation.

♦6.3.20.

(a) We are assuming that f∈rng K= corng A= rng AT, cf. Exercise 3.4.31. Thus, we can

write f=ATh=ATCgwhere g=C−1h.

(b) The equilibrium equations Ku=fare ATC Au=ATCgwhich are the normal equa-

tions (4.57) for the weighted least squares solution to Au=g.

♥6.3.21. Let Abe the reduced incidence matrix of the structure, so the equilibrium equations are

ATC Au=f, where we are assuming f∈rng K= rng ATC A. We use Exercise 6.3.20 to

write f=ATCgand characterize uas the weighted least squares solution to Au=g, i.e.,

the vector that minimizes the weighted norm kAu−gk2.

♥6.3.22.

(a)A?=0

B

@

1

√2

1

√20 0 0

−1 0 1 0 0

0 0 −1

√2

1

√2

1

√2

1

C

A;K?u=f?where K?=

0

B

@

3

2

1

2−1 0 0

1

2

1

2000

−1 0 3

2−1

2

0 0 −1

2

1

2

1

2

0 0 −1

2

1

2

1

2

1

C

A

.

174

♥6.3.23.

(a)A?=0

B

@

1

√2

1

√20 0 0

−1 0 1 0 0

0 0 −1

√2

1

√2−1

√2

1

C

A;

♥6.3.24.

(a) Horizontal roller: A?=0

B

@

.7071 .7071 0 0 0

−1 0 1 0 0

0 0 −.7071 .7071 .7071

−.9487 .3162 0 0 .9487

1

C

A

;

0

1

(b) Vertical roller: A?=0

B

@

.7071 .7071 0 0 0

−1 0 1 0 0

0 0 −.7071 .7071 −.7071

−.9487 .3162 0 0 −.3162

1

C

A

;

6.3.25.

(a) Yes, if the direction of the roller is perpendicular to the vector between the two nodes,

the structure admits an (infinitesimal) rotation around the fixed node.

175

For the exclusive use of adopters of the book Applied Linear Algebra, by Peter J. Olver and Cheri Shakiban. ISBN 0-13-147382-4.