13.16 The computed theoretical density is 3.54 g/cm3 and the measured density is 3.51
g/cm3
V
-22
3
S 6.43 x 10
cm
13.23 APF = V =
-22
3 = 0.842
C 7.64 x 10
cm
13.40 (a)  = 78.3 MPa (10,660 psi)
fs
-2
-3
(b) y =7.0 x 10 mm (2.5 x 10 in.)
13.42 F =10,100 N (2165 lb )
f
f
14.7 (a) Below is shown the logarithm viscosity versus reciprocal of temperature plot for the
borosilicate glass, using the data in Figure.
40
ln Viscosity (Pa-s)
30
20
10
8.00e-4
9.00e-4
1.00e-3
1.10e-3
1.20e-3
1.30e-3
1/T (1/K)
(b) Q
vis
=362,000 J/mol
14.9 (a) Residual thermal stresses are introduced into a glass piece when it is cooled because
surface and interior regions cool at different rates, and, therefore, contract different amounts;
since the material will experience very little, if any deformation, stresses are established.
(b) Yes, thermal stresses will be introduced because of thermal expansion upon heating for
the same reason as for thermal contraction upon cooling.
(c)
The thinner the thickness of a glass ware the lower the thermal stresses that are
introduced when it is either heated or cooled. The reason for this is that the difference in
temperature across the cross-section of the ware, and, therefore, the difference in the
degree of expansion or contraction will decrease with a decrease in thickness.
_
n
= x M = 73,800 g/mol
i i
w
= w M = 81,450 g/mol
i i
15.7 (a) M
_
(b) M
_
_
M
(c) Due to m = n
w
=104.42 g/mol, polystyrene is the material since its mer molecular
w
weight is closest to that calculated above.
_
M
n
(d) n = m = 707
n
15.8 Such a homopolymer is possible since the calculated n
w
is 585.
3
3
15.27 (a)  = 2.000 g/cm ;  = 2.301 g/cm
a
c
(b) % crystallinity = 87.9%
3
3
15.28 (a)  = 1.091 g/cm ;  = 1.242 g/cm
a
c
3
(b)  = 1.170 g/cm
s
16.6 TS = 44 MPa
_
16.7 M
n
=213,000 g/mol
17.12 Since V and V are equal, the proposed composite is possible.
fl
ft
17.13 (a) For of the glass-reinforced epoxy material: 486 MPa
For the carbon-fiber reinforced epoxy: 775 MPa
For the aramid-fiber reinforced epoxy: 986 MPa
For the 440A tempered martensitic steel: 229 MPa
For the normalized 1020 plain carbon steel: 56 MPa
For the 2024-T3 aluminum alloy: 175 MPa
For the C36000 brass (cold worked): 47 MPa
For the AZ31B (rolled) magnesium alloy: 164 MPa
For the annealed Ti-6Al-4V titanium alloy: 203 MPa
(b) The longitudinal specific modulus is just the longitudinal tensile modulus-specific gravity ratio.
For the glass-fiber reinforced epoxy, this ratio is: 21.4 GPa
For the carbon-fiber reinforced epoxy: 90.6 GPa
And, for the aramid-fiber reinforced epoxy: 54.3 GPa
The specific moduli for the metal alloys (Tables B.1 and B.2) are as follows:
For the 440A tempered martensitic steel: 25.6 GPa
For the normalized 1020 plain-carbon steel: 26.4 GPa
For the 2024-T3 aluminum alloy: 26.1 GPa
For the C36000 brass: 11.4 GPa
For the AZ31B magnesium alloy: 25.4 GPa
For the Ti-6Al-4V titanium alloy: 25.7 GPa
F
f
17.17 (a) F =23.4 Or, F = 23.4F
f
m
m
(b) F = 1824 N (410 lb ) and F = 42,676 N (9590 lb )
m
f
f
f
(c)  = 445 MPa (63,930 psi) and  = 8.14 MPa (1170 psi)
f
m
-3
(d)  = = 3.39 x 10
m
f
17.19 * =  ' (1 - V ) +*V = 633 MPa (91,700 psi)
cl
m
f
f
f
17.20 V = 0.357.
f
17.28 (a) E = E V + E V + E V , in which the subscripts f1 and f2 refer to the two types
cl
m m
f1 f1
f2 f2
of fibers.
6
(b) E = 69.1 GPa (10.0 x 10 psi)
cl
17.D1 Possible candidate: carbon (PAN standard-modulus), the aramid
17.D4 * = 4495 MPa (650,000 psi).
f
3.11 If cost is unimportant, CFRP is the clear choice. 7075Al is a good compromise considering
cost, and is resistant to corrosion and rot.
3.12 (a) The best choice is the CFRP (cost is unimportant and light weight is considered).
(b) Pine is a good choice
3.13
(a) If both light weight and cost are important, 7075Al or GFRP are possibilities. Other
possibilities exist, such as a steel if weight is not important, or CFRP or Ti-6Al-4V, if weight
is critical.
(b) the vessel thickness required is:
For 1020 steel: 8.08mm;
For 4340 steel: 1.90mm;
For 7075 Al: 4.48mm;
For Ti-6-4: 1.77mm;
For PC: 33.87mm;
For Pine: 23.86mm;
For GFRP: 5.53mm;
For CFRP: 2.26mm.
3.14. (a) All but 1020 steel would give a lighter component, but Ti-6-4 and CFRP would be very
expensive.
(b) For 1020 steel, h=5.04mm and X=0.352.
(c) All but Ti-6-4, GFRP, and CFRP fail due to h too large or X<1.4.
5.5 eat t = 60s.
5.17
PD
dv
=
(5-2)
4tE
V
5.18
dv 3PD
=
(1-)
V
4tE
5.21 (a) Yes, 
(b) E1 =
z
E
-
v x
(1+) and    X (1+ )
E
E
1  v  v 2 (1   )
(c) λ = -1, E1 = 0.77E; λ = 0, E1 = 1.10E; λ = +1, E1 = 1.92E. The effect of λ on E1 for v=0.3
is substantial.
5.23 (a) For MgO, ΔTdown = -30.4 oC, ΔTup = 182 oC;
For Al2O3, ΔTdown = -69 oC, ΔTup = 687 oC;
For ZrO2, ΔTdown = -48 oC, ΔTup = 686 oC;
For SiC, ΔTdown = -135 oC, ΔTup = 1103 oC;
For Si3N4, ΔTdown = -365oC, ΔTup = 2801 oC;
6.8 (a) σMax = 159.2MPa; Max = 79.6 MPa
(b) σMax = 86.5 MPa; Max = 43.3 MPa
3
6.15 (a) max =
(b)
3
3 Pr1 r2
3
3
4 R 3 (r2  r1 )
σ1 = σ2 =339.5 MPa; σ3 = σr = -300MPa; 1= 0; 2 = 3 = 319.8 MPa.
6.17 max = 271.3 MPa, at R=r1.
6.22 σmax =
Pd
Pd
Pd
; max =
; h =
2t
4t
2 6t
7.2 X0 = 2.99 (maximum shear stress criterion); X0 = 3.45 (Octahedral shear stress criterion)
7.8 X0 = 1.78 (maximum shear stress criterion); X0 = 1.95 (Octahedral shear stress criterion)
7.9 (a) X =
0
4 2   2
32 X
2
M 2  3T )1/3
(b) d = (
4
 0
7.11 σz = σ0, No effect of confinement or  if use maximum shear stress criterion
Z 
o
1   2
, modest effect is predicted that depends on  if octahedral shear stress
criterion
8.2 For -150 oC, KIc = 42 MPa
m , σ0 = 700 MPa, S =
749
MPa;
a
For +10 oC, KIc = 160 MPa
m , σ0 = 480 MPa, S =
2855
MPa.
a
8.9 Q = 1.630; X = 1.26, X is inadequate to safely operate.
8.13 (a) No, Xa = 2.99.
(b) Yes, Xa = 24.7.
(c) KIc = 113 MPa
m
(d) Xk = 3.
8.22 (a) Yes, Cc =156 mm, Cc ≥ t.
(b) X0 =4.51.
(c) Tmin = -20 oC.
(d) Reduced pressure is 5.15 MPa. Xa=3 and Xk=1.73 and are not very large, and an even
more severe pressure limit might be advisable, around P = 3 MPa.
9.4 (a) A = 850, B = -0.0867.
(b) A = 881MPa, B = -0.09161.
(c) σf’ = 939 MPa; b = B = -0.0916.
9.12 (a) Nf = 5.48*104 cycles.
(b) Nf = 1.31*104 cycles.
(c) Nf = 1.99*105 cycles.
9.13 (a) σa = 838*(2 Nf)-0.0648.
(b) σa = 938*(2 Nf)-0.0648.
(c) σa = 1038*(2 Nf)-0.0648.
The lines are parallel on a log-log plot, and lower for higher σm.
9.20 N3 = 3930 cycles.
9.22 Bf = 4450 repetitions.
11.2 m = 24, c = 3.54*10-19
mm cycle
. The very high value of m= 24 indicates that the
( MPa m ) m
growth rates in this material are extremely sensitive to stress level and crack size.
11.3 (a) m = 3.10, c= 1.09*10-8
mm cycle
.
( MPa m ) m
(b) b = -8.06, c = 8.71*10-9, m = 3.22.
11.20 Nif = 47,400 cycles.
11.22 ai = 0.475 mm.