ERRATA for "Introduction to Composite Materials Design—Second Edition"

Ever J. Barbero, CRC, 2007, ISBN 978-1420079159

Updated April 27, 2011

Note: In addition to corrections for typos, this file includes comments that attempt to clarify and enhance the existing content in the book as well as highlight new code and examples as they become available in the Website.

Page

Where it reads

Correction

xxxiii

xxxiv , insert

 

nT Transverse Poisson’s ratio

xxxv , insert

 

Fmc Apparent matrix tensile strength

xxxv , insert

 

Fms Apparent matrix shear strength

 

 

 

12, footnote

[27]

[242]

16, last line before Ex. 1.6

[27]

[242]

24, Table 1.3

Transverse Poisson’s ratio n12

Transverse Poisson’s ratio n23

25, Table 1.4, footnote

G23=2.8—5.9…

c G23=2.8—5.9…

38, 3rd paragraph in Sect. 2.4

…and a fiberglass backing…

…and sometimes a fiberglass backing…

40, line 6 in Sect. 2.4.2

…the textile tread,…

…the textile thread…

66, Table 2.7

above 3

-

66, Table 2.7

above 4.5

-

97, end of sect. 4.1.5

…orthotropic material.

…orthotropic material: E1, E2, E3, G12, G13, G23, n12, n13, n23.

97, end of sect. 4.1.6

five constants.

five constants: E1, E2, G12, n12, n23.

103, Sect. 4.2.4, above (4.34)

The strength of material approach…

The mechanics of materials approach…

112, (4.69) 3rd and 4th eq.

˝ C*44

1/4  C*44

112, (4.69)

Add a 6th equation

C44=(C22-C23)/2

114, Ex. 4.4

technique

techniques

114, Solution Ex. 4.4

a=60 10-6

a=30 10-6

116, end of 3rd line in Sect. 4.4

…from bulk properties.

…from bulk properties (see Table 2.5 and Sect. 2.2.1).

121, second line from top

Since 2.9 does not list…

Since Table 2.9 does not list…

130, penultimate line

…to be a=0.8/4=0.2 mm.

…to be a0=0.8/4=0.2 mm.

152, Ex. 5.1

The out of plane shear modulus G23=Gm=0.385 GPa.

n23=0.385.

152, Solution Ex. 5.1

Q44*=0.385 GPa

Q44*=4.112 GPa

179, 1st line after 1st eq.

and at q=55o

and at q=-55o

179

Example 5.1, p. 152

Example 5.1, p. 152, with n23=0.385

179

Q44*=0.385 GPa

Q44*=4.122 GPa

 

Q44*(1)=2.669

Q45*(1)=1.599

Q55*(1)=1.505

Q44*(1)=3.90

Q45*(1)=-0.15

Q55*(1)=4.01

179

Q44*(2)=2.669

Q45*(2)=-1.599

Q55*(2)=1.505

Q44*(2)=3.90

Q45*(2)=0.15

Q55*(2)=4.01

179

H44=2.82

H55=1.59

H45=0

H44=4.13

H55=4.24

H45=0

184, Sect. 6.3.4, above (6.26)

…material; and for each 0-lamina there is a complementary 90-lamina also of the same thickness and material. By…

…material. By…

209, below the Table, RHS of 1st eq. (forces)

10-3

N/mm

209, below the Table, RHS of 2nd eq. (moments)

10-3

N

209, 3rd text line

Middles surface

Middle surface

210, 2nd paragraph, 7th line

…angle is 55. Therefore…

…angle is 55o. Therefore…

212, above the 3rd eq. from bottom

With the load having a COV of 20%,…

With the load having a COV of 30%,…

241, Sol. Ex. 7.6

For Glass—Epoxy, form Table 1.3, F2t=40.

For Glass—Epoxy, form Table 1.3, F2t=40 MPa.

242, Ex. 7.7, 3rd paragraph, 4th line

N={3500,7000,0} has and a…

N={3500,7000,0} has a…

258, 2nd paragraph in Sect. 7.5

The theoretical SCF relates…

The theoretical SCF, Kt, relates…

259, after (7.61)

The notch sensitivity…

For isotropic materials, the notch sensitivity…

259, end of paragraph after (7.61)

…laminate type.

…laminate type, and it can be larger than one (see Fig. 7.18).

261, line above (7.69)

That is sy=F0 at…

That is, sx=F0 at…

261, line after (7.7)

a0 is the characteristic

d0 is the characteristic

262, line above (7.74)

Using (7.63) and…

Using (7.73) and…

270, line 7

…onto the broken fiber (8.2).

…onto the broken fiber (Figure 8.2).

271, (8.7)

(m e F1tm)-1

[m e (F1t)m]-1

283, after (8.36)

delta Dirac

Kronecker delta

303, (9.24)

All zero subscripts,

,must be superscripts as in (9.23)

304, (9.26)

All zero subscripts,

,must be superscripts as in (9.23)

326, Sect. 9.9

(CSM, Figure 2.3).

(CSM, Figure 2.2).

327, (9.64)

 

Note that for CSM,  for any rotation angle q, so Q22=Q11, explaining why Q22 is not listed in (9.64)

341, Sect. 10.1.1, line 2

Carpet plot of Ex (Figure 6.8).

Carpet plot of Ex (Figure 6.8).1

Footnote 1. Not Exb because the flanges are subjected to predominantly tensile or compressive stress resultants.

 

 

 

 

 

 

References for this Errata

[*1] Barbero, E. J., Finite Element Analysis of Composite Materials (2007), CRC Press, Boca Raton FL http://www.mae.wvu.edu/barbero/feacm/index.html

 


Please e-mail additional corrections to http://www.mae.wvu.edu/barbero/


Contributors to this page & Users of the Book

Corrections have been made thanks to the helpful contributions from:

·         Joaquin Gutierrez, West Virginia University

·         Fernando Cosso, West Virginia University

·         Any Others??