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

σ
σ
Modulus of
toughness
Fracture
719
wORk ANd STRAIN ENERgy
σ Y
Modulus of
resilience
ε
ε
(a)
(b)
FIGURE 17.8 Geometric interpretation of (a) modulus of resilience and (b) modulus of toughness.
Strain-energy density has units of energy per unit volume. In the SI, an appropriate
unit for strain-energy density is joules per cubic meter (J/m 3 ). In the U.S. customary
system, units of lb ⋅ ft/ft 3 or lb ⋅ in./in. 3 are suitable. However, notice that all of these units
reduce to stress units; therefore, strain-energy density can also be expressed in pascals (Pa)
or pounds per square inch (psi).
The area under the straight-line portion of the stress–strain curve (Figure 17.8a),
evaluated from zero to the proportional limit, represents a material property known as
the modulus of resilience: the maximum strain-energy density that a material can
store or absorb without exhibiting permanent deformations. In practice, the yield stress
σ Y , rather than the proportional limit, is generally used to determine the modulus of
resilience.
The area under the entire stress–strain curve from zero to fracture (Figure 17.8b) gives
a property known as the modulus of toughness. This modulus denotes the strain-energy
density necessary to rupture the material. From the figure, it is evident that the modulus of
toughness depends greatly on both the strength and the ductility of the material. A high
modulus of toughness is particularly important when materials are subjected to dynamic or
impact loads.
The total strain energy associated with uniaxial normal stress can be found by integrating
the strain-energy density [Equation (17.4)] over the volume of the member:
U
2
σ x
= ∫
2E dV
(17.6)
V
Equation (17.6) can be used to determine strain energy for both axially loaded bars and
beams in pure bending.
Strain-Energy Density for Shear Stress
Next, consider an elemental volume dV subjected to a shear stress τ xy = τ yx (Figure 17.9).
Notice that the shear stress on the upper face displaces the upper face of the element relative
to the lower face. The vertical faces of the element do not displace relative to each
other—they only rotate. Therefore, only the shear force acting on the upper face performs
work as the element deforms. The shear force acting on the y face is dF = τ xy dx dz, and this
force displaces through a horizontal distance γ xy dy relative to the bottom face. The work
done by dF—and hence, the strain energy stored by the element—is
1
dU = 2
( τ dx dz)
γ dy
xy
xy
dy
y
z
x τ yx = τ xy
γ xy
dx
dz
τ xy
FIGURE 17.9 Volume element
subjected to pure shear stress
τ xy = τ yx .