PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS

Since s
x, sy, txy are all constant, then from Eqs. 9–1 and 9–2 it can be seen
that the magnitudes of sx? and tx?y? only depend on the angle of inclination u
of the planes on which these stresses act. In engineering practice it is often
important to determine the orientation that causes the normal stress to be a
maximum, and the orientation that causes the shear stress to be a maximum.
We will now consider each of these cases.
In-Plane Principal Stresses. To determine the maximum and
minimum normal stress, we must differentiate Eq. 9–1 with respect to u
and set the result equal to zero. This gives
ds
x?
du = -
s
x - sy
2
(2 sin 2u) + 2txy cos 2u = 0
Solving we obtain the orientation u = up of the planes of maximum and
minimum normal stress.
tan 2u

p =
t
xy
(sx - sy)>2 (9–4)
Orientation of Principal Planes
The solution has two roots, up1 and up2. Specifically, the values of 2up1 and
2u
p2 are 180° apart, so up1 and up2 will be 90° apart.
t
To obtain the maximum and minimum normal stress, we must
substitute these angles into Eq. 9–1. Here the necessary sine and cosine
of 2u
p1 and 2up2 can be found from the shaded triangles shown in Fig. 9–8,
which are constructed based on Eq. 9–4, assuming that txy and (sx - sy)
are both positive or both negative quantities.
After substituting and simplifying, we obtain two roots, s1 and s2.
They are
s1,2 =
s
x + sy
2
{ C¢ sx -2 sy ?2 + txy2 (9–5)
Principal Stresses
These two values, with s1 ? s2, are called the in-plane principal stresses,
and the corresponding planes on which they act are called the principal
planes of stress, Fig. 9–9. Finally, if the trigonometric relations for up1 or
u
p2 are substituted into Eq. 9–2, it will be seen that tx?y? = 0; in other
words, no shear stress acts on the principal planes, Fig. 9–9.