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

12.2 Stress at a General point in an Arbitrarily
Loaded Body
z
FIGURE 12.1 Solid body in
equilibrium.
z
P 3
P 3
P 4
P 4
y
y
P 2
P 5
P 1
FIGURE 12.2a Resultant forces
on area ∆A.
y
P 2
ΔV x
Q
P 5
P 2
ΔF x
ΔA
x
x
In Chapter 1, the concept of stress was introduced by considering the internal force distribution
required to satisfy equilibrium in a portion of a bar under axial load. The nature
of the force distribution led to uniformly distributed normal and shear stresses on transverse
planes through the bar. (See Section 1.5.) In more complicated structural members
or machine components, the stress distributions will not be uniform on arbitrary internal
planes; therefore, a more general concept of the state of stress at a point is needed.
Consider a body of arbitrary shape that is in equilibrium under the action of a system
of several applied loads P 1 , P 2 , and so on (Figure 12.1). The nature of the stresses
created at an arbitrary interior point Q can be studied by cutting a section through the
body at Q, where the cutting plane is parallel to the y–z plane, as shown in Figure 12.2a.
This free body is subjected to some of the original loads (P 1 , P 2 , etc.), as well as to
normal and shearing forces, distributed on the exposed plane surface. We will focus on
a small portion ∆A of the exposed plane surface. The resultant force acting on ∆A can
be resolved into components that act perpendicular and parallel, respectively, to the
surface. The perpendicular component is a normal force ∆F x , and the parallel component
is a shear force ∆V x . The subscript x is used to indicate that these forces act on a
plane whose normal is in the x direction (termed the x plane).
Although the direction of the normal force ∆F x is well defined, the shear force ∆V x
could be oriented in any direction on the x plane. Therefore, ∆V x will be resolved into
two component forces, ∆V xy and ∆V xz , where the second subscript indicates that the
shear forces on the x plane act in the y and z directions, respectively. The x, y, and z
components of the normal and shear forces acting on ∆A are shown in Figure 12.2b.
If each force component is divided by the area ∆A, an average force per unit area
is obtained. As ∆A is made smaller and smaller, three stress components are defined at
point Q (Figure 12.3):
∆Fx
∆Vxy
σx
= lim τxy
= lim τ xz = lim
∆A→ ∆A
∆A→ ∆A
∆A→
0 0 0
∆Vxz
∆A
(12.1)
P 3
ΔV xy
ΔV xz
Q
ΔF x
To reiterate, the first subscript on stresses σ x , τ xy , and τ xz indicates that these stresses
act on a plane whose normal is in the x direction. The second subscript on τ xy and τ xz
indicates the direction in which the shear stress acts on the x plane.
z
P 4
P 5
FIGURE 12.2b Resultant forces
on area ∆A resolved into x, y, and z
components.
x
Next, suppose that a cutting plane parallel to the x–z plane is passed through the
original body (from Figure 12.1). This cutting plane exposes a surface whose normal is
in the y direction (Figure 12.4). According to the previous reasoning, three stresses are
obtained on the y plane at Q: a normal stress σ y acting in the y direction, a shear stress
τ yx acting on the y plane in the x direction, and a shear stress τ yz acting on the y plane in
the z direction.
Finally, a cutting plane parallel to the x–y plane is passed through the original body
to expose a surface whose normal is in the z direction (Figure 12.5). Again, three stresses
are obtained on the z plane at Q: a normal stress σ z acting in the z direction, a shear
stress τ zx acting on the z plane in the x direction, and a shear stress τ zy acting on the
z plane in the y direction.
If a different set of coordinate axes (say, x′–y′–z′) had been chosen in the previous
discussion, then the stresses found at point Q would be different from those determined
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