errata geere-goodno 8 si.pdf

Thursday, June 18, 2015

Errata Log Gere/Goodno Mechanics of Material, 8th Edition SI (ISBN: 9781111577742)

(Corrections for 1st & 2nd reprint)

Page

Number

Description of Correction Print

Runs

Affected

10 - Entry in Table 1-1: Delete blue rectangle in Table 1-1 graphic

- Photo labels are reversed in (2)Pin support of Table 1-1. The entire captions with credits need should be switched around. (Lower photo

is by J. Kerkhoff, upper photo by Goodno)

1st & 2nd

11 Figure in middle column, Item (4): fig. with weld symbol should

show steel not concrete hatch pattern. Replace this figure

> with this figure

1st & 2nd

30 Change equation numbers from (1-1) to (1-6) and (1-2) to (1-7) as

shown below

1st & 2nd

62 Paragraph at bottom of page: example numbers are incorrect 1st & 2nd


65 Delete one occurrence of z in Fig. 1-52e as noted below:

1st & 2nd

84 Prob. 1.2-7: change given in feet to given in meters 1st

84 Prob. 1.2-10:

modification to

figure

Vertical dashed

line from z axis

should connect to

point B as shown

by red line at right

1st & 2nd

86 Figure for Prob. 1.2-15: delete arrow & text 1900 N 1st

97 Prob. 1.6-6: change peformed to performed 1st

97 Prob. 1.6-8: change numerical date to numerical data 1st

99 Prob. 1.7-4: (1) change self weight to self-weight; (2) change which is attached to that is attached

1st

99 In Fig. 1.7-4 at bottom right, change q = 36 N/m to q = 40 N/m

1st & 2nd

100 Prob. 1.7-5: change show in the figure to shown in the figure 1st

104 Use this revised problem statement for problem 1.7-16:

1st


A removable sign post on a hurricane evacuation route (see figure part

a) consists of an upper pole with a slotted base plate which is bolted to

a short post anchored in the ground. The lower post is capped with a

separate conventional base plate having four holes of diameter db. The

upper slotted base plate has slots (or cut outs) at bolt locations 1 to 4

and is bolted to the conventional lower base plate at these four points

(see figure part b). Each of the four bolts has a diameter of db and a

washer with diameter of dw. The bolts are arranged in a rectangular

pattern (b h). Consider only wind force Wy applied in the y direction

at the center of pressure of the sign structure at height z = L above the

base. Neglect the weight of the sign and post, and also neglect friction

between the upper and lower base plates. Assume that the lower

conventional base plate and short anchored post are rigid.

(a) Find the average shear stress (MPa) at bolt 1 (see figure part c) due to the wind force Wy; repeat for bolt 4.

(b) Find the average bearing stress b (MPa) between the bolt and the upper slotted base plate (thickness t) at bolt 1; repeat for bolt 4.

(c) Find the average bearing stress b (MPa) between the upper base plate and washer at bolt 4 due to the wind force Wy (assume the

initial bolt pretension is zero).

(d) Find the average shear stress (MPa) through the upper base plate at bolt 4 due to the wind force Wy.

(e) Find an expression for the normal stress in bolt 3 due to the wind force Wy.

[See Prob. 1.8-15 for additional discussion of wind on a sign, and the

resulting forces acting on a conventional base plate.]

105 Figure for Prob. 1.7-16(b):

- Change (b) to (c) - outer plate dimensions in should be uppercase H - Change Square base plate to Square upper base plate - Change (H H) to lowercase: (h h) at upper right of

figure part (c)

- Change Slot in base to Slot in upper base

1st

105 Photo for Prob. 1.7-16: photo deleted in 2nd printing and replaced with

new figure 1.7-16(b): 1st


105 Prob. 1.7-16(c): outer plate dimensions in fig. (c) should be uppercase

H

1st & 2nd

107 Prob. 1.8-1 in problem statement, change 0.08 mm to 0.8 mm 1st & 2nd

108 Prob. 1.8-8: insert of after The thickness of each 1st

110 Prob. 1.8-12

replace L102 76 6.4 with L130 65 8

replace Table F-5(b) with Table E-5

replace A = 2190 mm2 with A = 3018 mm2

in Fig. (b), replace 6.4 mm with 8 mm

1st

117 Prob. 1.9-15, in 1st line of problem statement: Change AB to AC [should read Two bars AC and BC of the ]

1st

143 2nd = sign in Eq. 2-12 should be a + sign 1st & 2nd


174 Last line in 3rd par.: change Fig. 1-10 to Fig. 1-28

1st & 2nd

237 Prob. 2.5-3 in problem statement, change 32 mm to 3.2 mm 1st & 2nd

258 Change load in both statement & fig. for Prob. A-2.5 (8e-US) & Prob.

R-2.5 (8e-SI); change P at joint 2 to 2P as shown below:

1st & 2nd

260 Prob. A-2.11:

Change 100 MPa to 50 MPa and 48 MPa to 24 MPa 1st & 2nd

319 Change k1 to k2 in 4 locations as shown below:

1st & 2nd

339 Prob. 3.3-3 in problem statement, change 255 mm to 225 mm 1st & 2nd

342 Prob. 3.3-14: add Assume that G = 28 GPa. at the end of part (b) statement.

1st & 2nd


344 Problem 3.4-9: in problem statements, change 276 to 27 and change Pa to GPa So change G = 276 Pa to G = 27 GPa

1st & 2nd

353 In the statement for Prob. #3.8-12, change T= 2500 kNm to T = 2500 Nm

1st & 2nd

379 In Fig. 4-14 (a, b, c): delete B in each figure as shown below

1st & 2nd

379 In the equation just above Eq. (k), delete (-) sign in first term as follows:

1st & 2nd

397 Example 4-6:

In Eq. (b), delete last part of equ. (i.e., delete last phrase

0

( )

x

q x dx )

1st

430 Equation 5-19b: change =3

6 to =

2

6 1

st & 2nd


443 In Eq. 5-29(b), change 1 to I as shown below:

512 Prob. 5.11-5(a): spacing Smax should be spacing smax (i.e., lowercase s in smax)

1st

520 R-5.3, change 6m to 4 m in problem statement as shown below.

1st & 2nd

526 Above Eq. (6-1), change (5-6) to (5-5) as shown below:

1st & 2nd

539 Example 6-3: Answer for 1C should be 1.7 MPa (not 1.13 MPa) 1st & 2nd

551 Typo in subscript in unnumbered equ. which follows Eq. 6-32;

correction is as follows:

1st

558 Example 6-8, last two lines on page:

Just right of = sign: Iz/Iy, not Iz/Iz [i.e., y subscript in denom.] = 78.6 [not 89.1]

1st

588 Problem 6.2-4: Change all instances of steel to titanium and brass to copper.

Change all instances of S to Ti and all instances of B to Cu

Inside the figure, change S to Ti and B to Cu.

1st & 2nd

590 Problem 6.3-3(SI) change 5.5 mm long to 5.5 m long in problem statement in text.

1st & 2nd

615 Change 3-30a and 3-30b to 3-29a and 3-29b as shown below:

1st & 2nd


625 &

repeated

on 626

Example 7-3, Fig. 7-14(d): Direction of one arrow on top face of

element should be reversed (i.e., arrow circled below)

1st

658 Change 7-67 to 7-76 as shown below:

1st & 2nd

673 Prob. 7.2-5: Direction of arrows on top and left faces of element

should be reversed (i.e., arrows circled below).

1st

736 Problem 8.2-3:

Part (a), change 190 mm to 350 mm Part (c), change 1.85 MPa to 3 MPa

1st & 2nd

742 Change Problem Statements as follows:

1st & 2nd


775 Below Eq. (9-52b), change 9-16b to 9-15b as shown below:

1st & 2nd

907 See below for changes:

Chapter 11 Columns874

Example 11-1

Structure 1

Structure 2

L

L/2

L/2

L/2

L/2

Initialposition

Initialposition

C = DL

B = 0

A

C = DL

C = 2D

A

B = D(3L/2)

(a) (b)

D

D

R

C C

L

D

B B

A A

D

P PFig. 11-5Example 11-1: Buckled positions of two idealizedstructures, (a) one supportedlaterally by a translationalspring, and (b) the other supported by a rotational elastic connection

Two idealized columns are shown in Fig. 11-5. Both columns are initiallystraight and vertical. The first column (Structure 1, Fig. 11-5a) consists ofa single rigid bar ABCD which is pinned at D and laterally supported at Bby a spring with translational stiffness . The second column (Structure 2,Fig. 11-5b) is comprised of rigid bars ABC and CD that are joined at C byan elastic connection with rotational stiffness . Structure 2 ispinned at D and has a roller support at B. Find an expression for criticalload Pcr for each column.

R (2/5)L2

L/2

C = DL

A

MC

C

B

A

L/2

P

P

C = 2D

(c)

L/2

L/2

Initialposition

B

A

C = DL

(d)

D

C

D

R

C

B

L

D

P

A

L/2

MC = R (C D)

CL

C

HBC

(e)

B

A

L/2

P

P

HB = B

77735_11_ch11_p868-953.qxd:77735_11_ch11_p868-953 10/19/11 4:00 PM Page 874

bg6Rectangle

bg6Text Boxuse updated fig. (c)at left (add force HB in 2 locations as shown)

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akaupertText Box906

11.2 Buckling and Stability 875

SolutionStructure 1. We begin by considering the equilibrium of Structure 1 in a dis-turbed position caused by some external load and defined by small rotationangle D (Fig. 11-5a). Summing moments about D, we get the followingequilibrium equation:

(a)

where (b)

and (c)

Since the angle D is small, lateral displacement A is obtained using Eq. (b).The force HB in the translational spring at B is the product of spring constant and small horizontal displacement B. Substituting the expression for Afrom Eq. (b) and the expression for HB from Eq (c) into Eq. (a), and solvingfor P, we find that the critical load Pcr for Structure 1 is

(d)

The buckled mode shape for Structure 1 is the disturbed position shown inFig. 11-5a.

Structure 2. The translational spring at B is now replaced by a roller sup-port, and the structure is assembled using two rigid bars (ABC and CD) joinedby a rotational spring having stiffness R. If we sum moments about D forthe undisturbed structure, we conclude that horizontal reaction HB is zero.Next, we consider the equilibrium of Structure 2 in a disturbed position, onceagain defined by small rotation angle D (Fig. 11-5b). Using a free-body dia-gram of the upper bar ABC (Fig. 11-5c) and noting that the moment Mc isequal to rotational stiffness R times the total relative rotation of the spring,we have

(e)

We see that equilibrium of bar ABC requires that

(f)

Substituting expressions for Mc, A, and c into Eq. (f), we obtain

So the critical load Pcr for Structure 2 is

(g)

The buckled mode shape for Structure 2 is the disturbed position shown inFig. 11-5b.

Combined Model and Analysis. We can create a more advanced or complex structure model by combining the features of Structure 1 andStructure 2 into a single structure, as shown in Fig. 11-5d. This idealized

HB B cDa3L2 b d

A DaL 2 L2 b D(2L)

MD 0 PA HBa3L2 b

Pcr 3R2L

or Pcr 32La2

5L2b 3

5L

Pcr MC

A C

R(3D)

Ca L2 b D(L)

R(3D)

D(2L)

MC 0 MC P(A C) 0

MC R(C D) R(2D D) R(3D)

Pcr HBAa3L

2b

Da3L2 bD(2L)

a3L2b 9

8L

Continues


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bg6Text Boxreplace this entiresection on Structure 2> see p. 5 below

akaupertText Box907

Chapter 11 Columns876

structure is shown in its disturbed position and now has both translationalspring at B and rotational elastic connection R at joint C where rigid barsABC and CD are joined. Note that two rotation angles, C and D, are nowrequired to uniquely describe any arbitrary position of the disturbed structure (alternatively, we could use translations B and C, for example,instead of C and D). We will refer to position angles C and D as degreesof freedom. Hence, the combined structure has two degrees of freedomand, therefore, has two possible buckled mode shapes and two differentcritical loads, each of which causes the associated buckling mode. In con-trast, we see now that Structures 1 and 2 are single degree of freedom struc-tures, because only D is needed (or alternatively, C) to define the buckledshape of each structure depicted in Figs. 11-5a and b.

We can now observe that if rotational spring R becomes infinitely stiffin the combined structure (Fig. 11-5d) (but remains finite), the two degreeof freedom (2DOF) combined model reduces to the single degree of free-dom (SDOF) model of Fig. 11-5a. Similarly, if translational spring becomesinfinitely stiff in Fig. 11-5d (while R remains finite), the elastic support at Bbecomes a roller support. We conclude that the solutions for Pcr forStructures 1 and 2 in Eqs. (d) and (g) are simply two special-case solutions ofthe general combined model in Fig. 11-5d.

Our goal now is to find a general solution for the 2DOF model inFig. 11-5d and then show that solutions for Pcr for Structures 1 and 2 can beobtained from this general solution.

First, we consider the equilibrium of the entire 2DOF model in the dis-turbed position shown in Fig. 11-5d. Summing moments about D, we get

where

and

Combining these expressions, we obtain the following equation in terms ofthe two unknown position angles (C and D) as

(h)

We can obtain a second equation which describes the equilibrium ofthe disturbed structure from the free-body diagram of bar ABC alone(Fig. 11-5e). The moment at C is equal to rotational spring stiffness R timesthe relative rotation at C, and the spring force HB is equal to the spring con-stant times the total translational displacement at B:

(i)

and

(j)HB B aC L2 DLb

MC R(C D)

CaP 34 Lb DaP 32Lb 0

HB B aC L2 DLb

A (C D)L

MD 0 PA HBa3L2 b 0

Example 11-1 - Continued


akaupertText Box908

11.2 Buckling and Stability 877

Summing moments about C in Fig. 11-5e, we get the second equilibriumequation for the combined model as

(k)

Inserting expressions for MC using Eq. (i) and HB using Eq. (j) into Eq. (k) andsimplifying gives

(l)

We now have two algebraic equations in Eqs. (h) and (l) and twounknowns (C,D). These equations can have nonzero (i.e., nontrivial) solutions only if the determinant of the coefficients of C and D is equal tozero. Substituting the assumed expression for and then evalu -ating the determinant produces the following characteristic equation forthe system:

(m)

Solving Eq. (m) using the quadratic formula results in two possible val-ues of the critical load:

These are the eigenvalues of the combined 2DOF system. Usually, the lowervalue of the critical load is of more interest, because the structure willbuckle first at this lower load value. If we substitute Pcr1 and Pcr2 back intoEqs. (h) and (l), we can find the buckled mode shape (i.e., eigenvector) asso-ciated with each critical load.

Application of combined model to Structures 1 and 2. If the rotationalspring stiffness R goes to infinity while the translational spring stiffness remains finite, the combined model (Fig. 11-5d) reduces to Structure 1because the rotation angles C and D are equal, as shown in Fig. 11-5a.Equating C and D in Eq. (h) and solving for P results in , whichis the critical load for Structure 1 [see Eq. (d)].

If the rotational spring stiffness R remains finite while the transla-tional spring stiffness goes to infinity, the combined model (Fig. 11-5d)reduces to Structure 2. The translational spring becomes a roller support,so (i.e., ) while rotation angle (i.e., C is clockwise, so negative, as shown in Fig. 11-5b). Inserting and

into Eq. (l) gives the critical load for Structure 2 [see Eq. (g)].C 2D 0

HB 0 C 2DB 0

Pcr (9/8)L

MC 0 P(CL) MC HBL2

0

Pcr2 La41 124140 b 1.413L

Pcr1 La41 124140 b 0.637L

P2 a4120

LbP 910

(L)2 0

R (2/5L2)

CaP 14 L RLb Da

RL

12Lb 0


bg6Rectangle

bg6Calloutsee below for revised wording for this par.

bg6Text BoxIf the rotational spring stiffness R remains finite, while the translational spring stiffness goes to infinity, the combined model (Fig. d) reduces to Structure 2. The translational spring becomes a roller support so B = 0 (so HB = 0) while rotation angle C = -2D (i.e., C is clockwise so negative as shown in Fig. b). Inserting = 0 and C = -2D into Eq. l leads to the critical load for Structure 2 (see Eq. g).

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akaupertText Box909

B. Goodno, Georgia Tech p. 3 of 9 EG-11-1-new_REVISED.xmcd

Structure 2: The translational spring at B is now replaced by a roller support, and the structure is assembledusing two rigid bars (ABC and CD) joined by a rotational spring having stiffness R. We investigate theequilibrium of Structure 2 in a disturbed position, once again defined by small rotation angle D (Fig. 11-5(b)).Using a free body diagram of upper bar ABC (Fig. 11-5(c)), and noting that the moment MC is equal to rotationalstiffness R times the total relative rotation of the spring, we have

MC R C D R = R 2 D D R 3 D R 2 D D e( )We see that equilibrium of bar ABC requires that

MC 0 MC P A C HB L2 0MC P A C HB L2MC P A C HB L2 f( )where MD 0 for the entire structure gives HB

23 L P A AAA

Substituting expressions for MC, HB, A and C into Eq. (f), we obtain

PcrMC

23A C

MC =

R 3 D C

L3

D L( )

= 95 L R

So the critical load Pcr for Structure 2 is

Pcr95 L

25 L2

=1825

L g( )

The buckled mode shape for Structure 2 is the disturbed positionshown in Fig. (b).

(c)

bg6Text Boxp. 907 (8e-SI){also Fig. 11-5(c) on previous page}

bg6Text BoxAlso, replace last par. in E.g. 11-1 with the following:

bg6Text BoxIf the rotational spring stiffness R remains finite, while the translational spring stiffness goes to infinity, the combined model (Fig. d) reduces to Structure 2. The translational spring becomes a roller support so B = 0 (so HB = 0) while rotation angle C = -2D (i.e., C is clockwise so negative as shown in Fig. b). Inserting = 0 and C = -2D into Eq. l leads to the critical load for Structure 2 (see Eq. g).

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957 Figure for Problem 11.3-20: add member BE (see below)

1st & 2nd

981 Change Ic to Iy in Eq. (12-19b)

1st & 2nd

986 Above Eq. (12-28), change (12-20) to (12-27) as shown below:

1st & 2nd

1031,

8e-SI

Appendix D [8e-US], Case 2: insert = after Ip

1st & 2nd

1049

1st & 2nd

1058 Prob. 1.3-5: change Answers to the following Change 12.74 to 16.28; (b) change two instances of 383 to 489

1st & 2nd

1059 In Answers for 1.7-4(c), change bshoe = 7.36 MPa to brail = 18.4 MPa

1st & 2nd


1059 Problem 1.7-11: ANSWERS

(a) change answer for f from 143.5 to 45.7 (b) change answer for f from 124.8 to 39.7

1st & 2nd

1061 Prob. 2.3-7, Part (c): Answers

Change 0.654 to 1.0 and change 0.84 to 1.284

1st & 2nd

1062 Prob. 2.6-3: change Answer to the following: 104 kN instead of 52 kN

1st & 2nd

1068 Prob. 4.5-19: change 2nd Answer to the following: Mmax = 1200 Nm instead of 4000 Nm

1st & 2nd

1068 Prob. 4.5-24: change Vmax = P/12 to Vmax = P/6 1st & 2nd

1069 Problem 5.5-7(SI) answers

1st & 2nd

1070 Prob. 5.6-12 (b): change

dmin = 45.2 mm, area(b)/area(a) = 0.635 to dmin = 42.4 mm, area(b)/area(a) = 0.557

1st

1070 Problem 5.6-17 Answers: change last number to 0.794

1 : 1.260 : 1.408 : 0.794

1st & 2nd

1070 Problem 5.8-3(b) Answer:

Change 10.27 to 3.24

1st & 2nd

1070 Problem 5.8-8(b): Answer for Mmax should be 36.0 Nm (not 9.01) 1st & 2nd

1071 Problem figure-1(SI) answer for 96.7c MPa

1st & 2nd

1072 R5-3: change D to A 1st


1072

1st & 2nd

1073 Problem 6.10-2(a): correct typo in equation as shown: In Answers,

change 2 to r2 as shown below

1st & 2nd

1075 Problem 7.4-12: in Answers

1st & 2nd

1078 Problem 8.2-3 (SI):

Part (a), change 4.66 kN to 15.81 kN and change 1.707 MPa to 3.14 MPa

Part (b), change 7.18 mm to 13.23 mm Part (c), change 206 mm to 334 mm

1st & 2nd

1079 Prob. 8.5-3(c): correct answer is tmin = 14.11 mm 1st

1079 Change both US & SI texts ANSWERS as follows:

1st & 2nd

1080 Prob. 9.2-4, Part (b): Typo in answer: change A to 4 in denominator,

i.e., change 20 /AM q L A to 2

0 / 4AM q L

1st & 2nd


1084 Answers for Prob. 10.3-4:

1st

1086 Answers for Prob. 10.4-26:

1st

1122 Answers: Prob. 6.5-5(a) change 36.9 MPa to 43.5 MPa 1st & 2nd

errata geere-goodno 8 si.pdf

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