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

12.3 Equilibrium of the Stress Element
Keep in mind that the square
shown in Figure 12.7 is simply a
two-dimensional projection of
the cube shown in Figure 12.6.
In other words, we are seeing
only one side of the infinitesimally
small cube in Figure 12.7,
but the stress element we are
talking about is still a cube.
Although the shear stress and
shear force arrows in Figure 12.7
are shown slightly offset from
the faces of the stress element, it
should be understood that the
shear stresses and the shear
forces act directly on the face.
The arrows are shown offset
from the faces of the stress
element for clarity.
Figure 12.7a shows a two-dimensional projection of a stress element having width dx and
height dy. The thickness of the stress element perpendicular to the x–y plane is dz. The
stress element represents an infinitesimally small portion of a physical object. If an object
is in equilibrium, then any portion of the object that one chooses to examine must also be
in equilibrium, no matter how small that portion may be. Consequently, the stress element
must be in equilibrium.
Equilibrium involves forces, not stresses. To consider equilibrium of the stress element
in Figure 12.7a, the forces produced by the stresses that act on each face must be
found by multiplying the stress acting on each face by the area of the face. These forces can
then be considered on a free-body diagram of the element.
Since the stress element is infinitesimally small, we can assert that the normal
stresses σ x and σ y acting on opposite faces of the stress element are equal in magnitude
and aligned collinearly in pairs. Hence, the forces arising from normal stresses counteract
each other, and equilibrium is assured with respect to both translation (ΣF = 0) and
rotation (ΣM = 0).
Next, consider the shear stresses acting on the x and y faces of the stress element (Figure
12.7b). Suppose that a positive shear stress τ xy acts on the positive x face of the stress
element. Then, the shear force produced on the x face in the y direction by this stress is V xy =
τ xy (dy dz) (where dz is the out-of-plane thickness of the element). To satisfy equilibrium
in the y direction (ΣF y = 0), the shear stress on the −x face must act in the −y direction.
Similarly, a positive shear stress τ yx acting on the positive y face of the stress element
produces a shear force V yx = τ yx (dx dz) in the x direction. To satisfy equilibrium in the
x direction (ΣF x = 0), the shear stress on the −y face must act in the −x direction. Therefore,
the shear stresses shown in Figure 12.7 satisfy equilibrium in the x and y directions.
The moments created by the shear stresses must also satisfy equilibrium. Consider the
moments produced about point O, located at the lower left corner of the stress element. The
lines of action of the shear forces acting on the −x and −y faces pass through point O; therefore,
these forces do not produce moments. The shear force V yx acting on the +y face
(a distance dy from point O) produces a clockwise moment V yx dy. The shear force V xy
acting on the +x face (a distance dx from point O) produces a counterclockwise moment
equal to V xy dx. Application of the equation ΣM O = 0 yields
∑ M = V dx − V dy = τ ( dydzdx ) − τ ( dxdzdy ) = 0
O xy yx xy yx
which reduces to
τ
yx
= τ
(12.2)
xy
σ
x
τ
xy
y
dy
dx
τ
yx
σ
y
σ
τ
y
yx
τ
xy
σ
x
x
y
V yx = τ yx dx dz
dy V xy = τxy dy dz
dx
x
O
dz = out-of-plane thickness
FIGURE 12.7a
FIGURE 12.7b
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