Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

σ x
y
Finally, suppose the material is subjected to shear stress as shown in Figure 13.19c.
Then the relation between shear stress and shear strain is
z
x
σ x
γ
xy
=
τ
G
xy
xy
FIGURE 13.19a Loading in x
direction.
y
σ y
where G xy is the shear modulus in the x–y plane.
Collecting these equations and using the principle of superposition, we can express
the strains in terms of the stresses as follows:
σ x
ε x = − v
E
x
yx
σ
E
y
y
z
x
ε
y
σ y σ x
= − vxy
(13.32)
E E
y
x
FIGURE 13.19b Loading in y
direction.
z
σ y
y
τ xy
x
τ xy
FIGURE 13.19c Shear stress
loading in the x–y plane.
γ
xy
τ
=
G
Solving these equations for the stresses gives the following:
Ex
σx
= ( εx + vyxε
y)
1 − v v
xy
yx
xy
xy
Ey
σ y = ( εy + vxyεx)
(13.33)
1 − v v
τ
xy
yx
= G γ
xy xy xy
Of the five material constants, only four are independent, because the major and minor
Poisson’s ratios are related reciprocally; that is,
v
E
xy
x
vyx
Ey
= , or vyx
= vxy
(13.34)
E
E
y
x
Note that if the material is isotropic, then E x = E y = E, ν xy = ν xy = ν, and G xy = G. Thus,
Equations (13.32) and (13.33) reduce to Equations (13.24), (13.25), and (13.26) for isotropic
materials subjected to plane stress.
Some typical material properties for fiber-reinforced epoxy material with 60% unidirectional
fibers by volume are shown in Table 13.3.
Table 13.3 Typical material properties for Fiber-Reinforced Epoxy materials
type Material E x GPa (ksi) E y GPa (ksi) G xy GPa (ksi) ν xy
T-300 Graphite–Epoxy 132 (19,200) 10.3 (1,500) 6.5 (950) 0.25
GY-70 Graphite–Epoxy 320 (46,400) 5.5 (800) 4.1 (600) 0.25
E-Glass Glass–Epoxy 45 (6,500) 12 (1,700) 4.4 (640) 0.25
Kevlar 49 Aramid–Epoxy 76 (11,000) 5.5 (800) 2.1 (300) 0.34
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