Page 217 - Mechanical Behavior of Materials

218 Chapter 5 Stress–Strain Relationships and Behavior

Table 5.3 Elastic Constants and Density for Fiber-Reinforced Epoxy with 60%
Unidirectional Fibers by Volume
(a) Reinforcement (b) Composite, V r = 0.60
Type E r ν r E X E Y G XY ν XY ρ
3
3
GPa (10 ksi) GPa (10 ksi) g/cm 3
E-glass 72.3 0.22 45 12 4.4 0.25 1.94
(10.5) (6.5) (1.7) (0.64)
Kevlar 49 124 0.35 76 5.5 2.1 0.34 1.30
(18.0) (11.0) (0.8) (0.3)
Graphite 218 0.20 132 10.3 6.5 0.25 1.47
(T-300) (31.6) (19.2) (1.5) (0.95)
Graphite 531 0.20 320 5.5 4.1 0.25 1.61
(GY-70) (77.0) (46.4) (0.8) (0.6)
Note: For approximate matrix properties, use E m = 3.5GPa (510 ksi) and ν m = 0.33.
Sources: Data in [ASM 87] pp. 175–178, and [Kelly 94] p. 285.

Example 5.4
A plate of the epoxy reinforced with unidirectional Kevlar 49 fibers in Table 5.3 is subject to
stresses as follows: σ X = 400, σ Y = 12, and τ XY = 15 MPa, where the coordinate system is that
of Fig. 5.15(a). Determine the in-plane strains ε X , ε Y , and γ XY .
Solution Equation 5.47 applies directly.

σ X ν YX ν XY σ Y τ XY
ε X = − σ Y , ε Y =− σ X + , γ XY =
E X E Y E X E Y G XY
Since ν YX is not given in the table, it is convenient to employ Eq. 5.48.

ν YX ν XY 0.34 −6
= = = 4.474 × 10 1/MPa
E Y E X 76,000
Substituting this quantity and the given stresses, with E X , E Y , and G XY from Table 5.3,
converted to MPa, gives the strains:

400 −6
ε X = − 4.474 × 10 (12) = 0.00521 Ans.
76,000
12
−6
ε Y =− 4.474 × 10 (400) + = 0.00039 Ans.
5500
15
γ XY = = 0.00714 Ans.
2100