Solutions Manual – Materials for Civil and Construction Engineers 4th Edition by Mamlouk

ALL 11 CHAPTERS COVERED




SOLUTIONS MANUAL

,Table of Contents
1. Materials Engineering Concepts
2. Nature of Materials
3. Steel
4. Aluminum
5. Aggregates
6. Portland Cement, Mixing Water, and Admixtures
7. Portland Cement Concrete
8. Masonry
9. Asphalt Binders and Asphalt Mixtures
10. Wood
11. Composites

,Materials for Civil and Construction Engineers, Third Edition, by Michael S. Mamlouk and John P. Zaniewski. ISBN: 0-13-611058-4
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

CHAPTER 1. MATERIALS ENGINEERING CONCEPTS

1.2. Strength at rupture = 45 ksi
Toughness = (45 x 0.003) / 2 = 0.0675 ksi

1.3. A = 0.36 in2
= 138.8889 ksi
A = 0.0035 in/in

L = -0.016667 in/in
E = 39682 ksi
= 0.21


1.4. A = 201.06 mm2
= 0.945 GPa
A = 0.002698 m/m
L = -0.000625 m/m

E = 350.3 GPa
= 0.23


1.5. A = d2/4 = 28.27 in2
-150,.27 in2 = -5.31 ksi
E= = 8000 ksi
A / E = -5.31 ksi / 8000 ksi = -0.0006631 in/in
L = A o = -0006631 in/in (12 in) = -0.00796 in
Lf = L + Lo = 12 in – 0.00796 in = 11.992 in
= - L / A = 0.35
L d / do = - A = -0.35 (-0.0006631 in/in) = 0.000232 in/in
d = L o = 0.000232 (6 in) = 0.00139 in
df = d + do = 6 in + 0.00139 in = 6.00139 in


1.6. A = d2/4 = 0.196 in2
2
= 10.18 ksi (Less than the yield strength. Within the elastic
region)
E= = 10,000 ksi
A / E = 10.18 ksi / 10,000 ksi = 0.0010186 in/in
L = A o = 0.0010186 in/in (12 in) = 0.0122 in
Lf = L + Lo = 12 in + 0.0122 in = 12.0122 in
= - L / A = 0.33
L d / do = - A = -0.33 (0.0010186 in/in) = -0.000336 in/in
d = L o = -0.000336 (0.5 in) = -0.000168 in
df = d + do = 0.5 in - 0.000168 in = 0.49998 in


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, Materials for Civil and Construction Engineers, Third Edition, by Michael S. Mamlouk and John P. Zaniewski. ISBN: 0-13-611058-4
© 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,
storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

1.7. Lx =30 mm, Ly = 60 mm, Lz = 90 mm
x= y= z= = 100 MPa
E = 70 GPa
= 0.333

x = [ x - ( y + z ) ] /E
6 6 6 9 -4
x = [100 x 10 - 0.333 (100 x 10 + 100 x 10 )] / 70 x 10 = 4.77 x 10 = y = z =
Lx = x Lx = 4.77 x 10-4 x 30 = 0.01431 mm
Ly = x Ly = 4.77 x 10-4 x 60 = 0.02862 mm
Lz = x Lz = 4.77 x 10-4 x 90 = 0.04293 mm
V = New volume - Original volume = [(Lx - Lx) (Ly - Ly) (Lz - Lz)] - Lx Ly Lz
= (30 - 0.01431) (60 - 0.02862) (90 - 0.04293)] - (30 x 60 x 90) = 161768 - 162000
= -232 mm3


1.8. Lx =4 in, Ly = 4 in, Lz = 4 in
x= y= z= = 15,000 psi
E = 1000 ksi
= 0.49

x = [ x - ( y + z ) ] /E
x = [15 - 0.49 (15 + 15)] / 1000 = 0.0003 = y = z =
Lx = x Lx = 0.0003 x 15 = 0.0045 in
Ly = x Ly = 0.0003 x 15 = 0.0045 in
Lz = x Lz = 0.0003 x 15 = 0.0045 in
V = New volume - Original volume = [(Lx - Lx) (Ly - Ly) (Lz - Lz)] - Lx Ly Lz
= (15 - 0.0045) (15 - 0.0045) (15 - 0.0045)] - (15 x 15 x 15) = 3371.963 - 3375
= -3.037 in3


1.9. = 0.3 x 10-16 3
At = 50,000 psi, = 0.3 x 10-16 (50,000)3 = 3.75 x 10-3 in./in.
50,000
Secant Modulus = 3
= 1.33 x 107 psi
3.75x10
d
0.9 x 10-16 2
d
d
At = 50,000 psi, 0.9 x 10-16 (50,000)2 = 2.25 x 10-7 in.2/lb
d
d 1
Tangent modulus = 7
= 4.44x106 psi
d 2.25x10




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