Introduction to Materials and Manufacturing
3.5 Properties of Solid Materials
3.5.2 Modulus of Elasticity, Poisson’s Ratio, and Shear Modulus
Themodulus of elasticity(orYoung’s modulus) is defined as the slope of the linear-elastic part of the stress-strain curve. In Fig. 3.6, the linear portion of the stress-strain curve is between the origin and point P, or at stresses lower than the propor- tional limit stress. The modulus of elasticity can be written as
E= σ
. (3.3)
Since strain has no dimension, the modulus of elasticity has the same units as stress, or N/m2. Figure 3.12 and Table 3.1
Properties of Solid Materials 61 Table 3.1: Typical physical properties of common engineering materials.
Thermal Thermal Specific Elastic Poisson’s Density, conductivity, expansion heat,
modulus, ratio, , Kt coefficient,a¯ Cp
Material E, (GPa) ν (kg/m3) (W/m◦C) (µm/m◦C) (J/kg◦C)
Metals
0 0 9 3
2 9
0 2 0
0 7 2 3
3 . 0 2
6 m
u n i m u l
AAluminum alloys 70 0.33 2630-2820 221-239 24 900
0 6 9 4
2 0
8 1 0
0 7 2 3
3 . 0 3
6 n
it m u n i m u l
ABabbitt, lead-based 29 0.33 7530 24 20 150
Babbitt, tin-based 52 0.33 7340 56 23 210
0 9 3 9
1 0
2 1 0 4 9 8 - 0 7 4 7 3 3 . 0 0
0 1 s
e s s a r
BBronze, aluminum 117 0.33 8940 50 18 380
Bronze, leaded 97 0.33 9100 47 18 380
Bronze, phosphor 110 0.33 8500 50 18 380
Bronze, porous 60 0.22 8040 30 18 380
0 8 3 8
1 0
7 1 0
7 9 8 3
3 . 0 4
2 1 r
e p p o
Cron,graycast 109 0.26 7860 50 11 420
Iron,malleablecast 170 0.26 7860 11 420
IIron, spheroidal graphite 159 0.26 7860 30 11 420
0 6 4 2
1 8
2 0
6 4 7 0
2 . 0 0
8 s
u o r o p , n o r
Iron,wrought 170 0.30 7860 70 12 460
IMagnesium
alloys 41 0.33 1770 110 27 1000
0 4 4 5
1 3
6 - 2 5 0
5 8 8 1
3 . 0 1
2 2 s
y o ll a l e k c i
NSteel, low alloys 196 0.30 7800 35 11 450
Steel, medium and high alloys 200 0.30 7850 30 11 450
Steel, stainless 193 0.30 8030 15 17 500
Steel, high speed 212 0.30 7860 30 11 450
Titanium alloys 110 0.32 4510 8-12 8.4 520
0 0 4 7
2 0
1 1 5
3 1 7 7
2 . 0 0
5 s
y o ll a c n i PolymersZ
Acetal (polyformaldehyde) 2.7 0.35 1400 0.24 90 1460
Nylons (polyamides) 1.9 0.40 1140 0.25 100 1700
Polyethylene, high density 0.9 0.35 940 0.5 126 1800
Phenol formaldehyde 7.0 0.35 1362 0.17 25-40 1600
Rubber, natural 0.004 0.50 930 1.6 80-120 2000
Ceramics
Alumina (Al2O3) 390 0.28 961 25 5.0 880
0 4 8 0
. 4 - 4 . 1 5
2 1 0
0 4 2 1
3 . 0 7
2 e
ti h p a r
GSilicon carbide (SiC) 450 0.19 3210 15 4.3 750
Silicon nitride (Si2N4) 314 0.26 3290 30 3.2 710
give data for the elastic modulus for various metals, poly- mers, and ceramics at room temperature. The elastic moduli for metals and ceramics are high and quite similar, but those for polymers are considerably lower.
The elastic moduli of most materials depend on two fac- tors: bond strength and bond density per unit area. A bond is like a spring: it has a spring rate,k(in newtons per meter).
The modulus of elasticity,E, is roughly E= k
ro. (3.4)
whererois the atom size (this can be obtained from the mean atomic volume4πr3o/3, which is generally known). The wide range of modulus of elasticity in Fig. 3.12 and Table 3.1 is largely caused by the range ofkin materials. The covalent bond is stiff (k = 20 to 200 N/m), while the metallic and ionic bonds are somewhat less stiff (k= 15 to 100 N/m). Dia- mond, although not shown in Fig. 3.12 or Table 3.1, has a very high modulus of elasticity because the carbon atom is small (giving a high bond density) and its atoms are linked by ex- tremely strong bonds (200 N/m). Metals have a high modu- lus of elasticity because close packing gives a high bond den- sity and the bonds are strong, although not as strong as those of diamond. Polymers contain both strong diamond-like co- valent bonds and weak hydrogen (or van der Waals) bonds (k= 0.5 to 2 N/m); these weak bonds stretch when a polymer is deformed, giving a lower modulus of elasticity. Elastomers have a low modulus of elasticity because they have only an extremely weak restoring force, which is associated with tan- gled, long-chain molecules when the material is loaded.
In a tension test, there will be axial deformation, but there will also be dimensional changes in the transverse di- rection, for as a bar extends axially, it contracts transversely.
The transverse strain,t, is related to the axial strain,a, by Poisson’s ratio,ν, such that
t=−νa. (3.5)
The negative sign simply means that the transverse deforma- tion will be in the opposite sense to the axial deformation.
Poisson’s ratio is dimensionless. Table 3.1 gives quantitative values of Poisson’s ratio for various metals, polymers, and ce- ramics at room temperature. The highest Poisson’s ratio ap- proaches 0.5 for rubber, and the lowest is 0.19 for silicon car- bide and cemented carbides (although it approaches zero for cork or foams in compression). Poisson’s ratio cannot be less than zero (or else the second law of thermodynamics would be violated), nor can it exceed 0.5 (or else a material’s volume would increase when compressed).
Shear stress and strain are proportional to each other;
that is
τ=Gγ, (3.6)
whereGis theshear modulusormodulus of rigidity. This relation is only true for the linear-elastic region of the shear stress-strain curve (from 0 toP in Fig. 3.6). Ghas units of stress (N/m2).
The three material propertiesE,G, andνare related by the following equation:
G= E
2(1 +ν). (3.7)
Lead Copper
Aluminum tin Aluminum
Magnesium Steels Cast iron Zinc alloys Sintered iron
Alumina Silicon nitride Silicon carbide
Graphite Ceramics Metals Polymers
Silicone rubber
Natural rubber Polyethylene Acetal
Phenol formal- dehyde Nylon
8 x 102 103 104
Density, ρ, kg/m3
Figure 3.11: Density for various metals, polymers, and ceram- ics at room temperature (20◦C).
Thus, when two parameters are known, the third can easily be determined from Eq. (3.7). That is, if for a particular mate- rial, the modulus of elasticity and Poisson’s ratio are obtained from Table 3.1, then the shear modulus can be obtained from Eq. (3.7).
The material presented thus far is valid for metals, poly- mers, or ceramics. To establish the modulus of elasticity for a unidirectional fiber-reinforced composite in the direction of the fibers, it is assumed that the fiber-matrix interfacial bond is good, so that deformation of both matrix and fibers is the same. Under these conditions, the total load sustained by the composite,Pc, is equal to the loads carried by the matrix,Pm, and the fiber,Pf, or
Pc=Pm+Pf, (3.8)
where subscriptsc,m, andfrefer to composite, matrix, and fiber, respectively. Substituting Eq. (3.8) into the definition of stress given by Eq. (2.7) results in
σc=σmAm
Ac +σfAf
Ac. (3.9)
If the composite, matrix, and fiber lengths are equal, Eq. (3.9) becomes
σc=σmvm+σfvf, (3.10) wherevmis the volume fraction of the matrix andvf is the volume fraction of the fiber in the composite material. Be- cause the same deformation of matrix and fibers was as- sumed and the composite consists of only matrix and fibers
Phenol formal- dehyde Steels
Cast iron Brass, bronze Aluminum Zinc alloys Magnesium alloys Babbitts
Carbides Alumina
Graphite Ceramics
Metals Polymers
Acetal Nylon Polyethylene
Natural rubber
Modulus of elasticity, E, Pa
106 107 108 109 1010 1011 1012
Figure 3.12: Modulus of elasticity for various metals, poly- mers, and ceramics at room temperature (20◦C).
(i.e.,vm+vf = 1), Eq. (3.10) can be rewritten in terms of the moduli of elasticity as
Ec=Emvm+Efvf (3.11) or
Ec=Em(1−vf) +Efvf. (3.12) Thus, Eq. (3.12) enables the composite’s modulus of elasticity to be determined when the elastic moduli of the matrix and the fiber and the volume fractions of each are known. It can also be shown that the ratio of the load carried by the fibers to that carried by the matrix is
Pf
Pm
= Efvf
Emvm
. (3.13)
Recall that Eqs. (3.8) to (3.13) are only applicable to unidirec- tional, fiber-reinforced composites.
Properties of Solid Materials 63
Example 3.6: Stiffness of a Fiber Reinforced Polymer
Given: A fiber-reinforced plastic contains 10 vol% glass fibers (E= 70 GPa,Su= 0.7 GPa).
Find:Calculate how this fiber percentage has to be changed to give the same elastic properties if the glass fibers are changed to carbon fibers (E= 150 GPa,Su= 1 GPa). Assume the matrix material hasEm= 2 GPa.
Solution: According to Eq. (3.11), the modulus of elasticity for a fiber composite is
Ec=Emvm+Efvf. Therefore, for the glass fiber-reinforced material,
Ec= 2
×109
(0.9) + 70
×109
(0.1) = 8.8GPa.
This composite modulus of elasticity should be maintained for the carbon-reinforced plastic:
8.8×109= 2
×109
x+ 150
×109 (1−x), which is solved asx= 0.954or1−x= 0.046. Thus, the plas- tic should contain 4.6 vol% carbon fibers to get the same elas- tic properties as the glass-fiber-reinforced plastic with 10%
glass fibers.