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

also be added together to produce the same resultant force F R . In other words, the resultant
force F R could be expressed either as the sum of components F x and F y in an x–y coordinate
system or as the sum of components F n and F t in an n–t coordinate system. The components
are different in the two coordinate systems, but both sets of components represent the
same resultant force.
In this vector addition illustration, the transformation of forces from one coordinate
system (i.e., the x–y coordinate system) to a rotated n–t coordinate system must take into
account the magnitude and direction of each force component. The transformation of
stress components, however, is more complicated than vector addition. In considering
stresses, the transformation must account for not only the magnitude and direction of
each stress component, but also the orientation of the area upon which the stress component
acts.
A more general approach to stress transformations will be developed in Section 12.7;
however, at the outset, it is instructive to use equilibrium considerations to determine normal
and shear stresses that act on an arbitrary plane. The solution method used here is
similar to that developed in Section 1.5 for stresses on inclined sections of axial members.
Example 12.2 illustrates this method for plane stress conditions.
489
EQuILIbRIuM METHOd
FOR PLANE STRESS
TRANSFORMATIONS
ExAmpLE 12.2
At a given point in a machine component, the following stresses were determined:
150 MPa (T) on a vertical plane, 30 MPa (C) on a horizontal plane, and zero shear
stress. Determine the stresses at this point on a plane having a slope of 3 vertical
to 4 horizontal.
y
30 MPa
Plan the Solution
A free-body diagram of a wedge-shaped portion of the stress element will be investigated.
Forces acting on vertical and horizontal planes will be derived from the
given stresses and the areas of the wedge faces. Since the wedge-shaped portion
of the stress element must satisfy equilibrium, the normal and shear stresses acting
on the inclined surface can be determined.
4
3
150 MPa
x
SolutioN
Sketch a free-body diagram of the wedge-shaped portion of the stress element. From the
3:4 slope of the inclined surface, the angle between the vertical face and the inclined
surface is 53.13°. The area of the inclined surface will be designated dA. Accordingly, the
area of the vertical face can be expressed as dA cos 53.13°, and the area of the horizontal
face can be expressed as dA sin 53.13°. The forces acting on these areas are found from
the product of the given stresses and the areas.
dA
σn
dA
(150 MPa)(dA cos 53.13°)
53.13°
τnt
dA
(30 MPa)(dA sin 53.13°)