2.10. Determination of stresses at the inclined sections in tension (compression) in two directions
Let us
consider a general case of the plane stress state, when two principal
stresses (
in
Fig. 2.7 a) differ from zero.
The indices
for denoting the principal stresses are put so as the
inequality is observed. We shall count off the positive
angle between the
direction and the normal to the arbitrary section against the clock.
The angle
between the
stress direction and the plane is equal to
The
and
stresses at the arbitrary inclined section can be determined from the
equilibrium condition of the ABC trihedral prism (Fig. 2.7 b) or
calculated by the formulae (2.3) and (2.4) by summing the stresses
caused by
and the stresses caused by
.
As a result we get
F
a) b)
c) d)
wherefrom
(2.8)
further,
wherefrom
(2.9)
From the formula (2.9) it is evident that the maximum shearing stresses are equal to half the difference of the principal stresses:
. (2.10)
Special cases.
The
first case.
Let us consider the stress state, under which
(Fig. 2.7 c).
In this
case for all sections passing through the point under study the
shearing stress
is equal to zero, and the normal stress has the same value
.
That kind of the stress state is called the
uniform plane tension
(or compression).
The
second case.
Let us consider the stress state as presented in Fig. 2.7 d,
characterized by the
principal stresses
and
.
Then,
.
Determine the stresses at the sections having the same slope to the directions and i.e. at = 45° and = 135°.
We
get
;
by
the formulae (2.8) and (2.9). This stress state is called the
pure shear.
2.11. Determining the principal stresses and the principal planes position
Let us
consider the reverse problem. The normal and shearing stresses acting
on the element plane are given (Fig. 2.8 a). It is necessary to
determine the
principal planes
positions and the principal stress values. Let us consider the
trihedral prism with the ABC base (Fig. 2.8 b). Take
.
The angle
will be counted off from the larger stress direction until the
normal to the plane. We will take the against the clock direction as
positive. The inclined plane area will be denoted by dA.
Then, the vertical plane area will be
and the horizontal one will be -
.
Projecting all forces on the direction , we get
Now we will project all forces on the direction:
Having cancelled by dA introducing the function of double angles, we get
(2.11)
(2.12)
a) b)
Fig. 2.8.
The
value is continuously changing with the angle change of the
plane incline. To find the principal planes position we have to equal
the
derivative to zero, then we get:
After the transposition we have
(2.13)
We get the formulae of the principal stresses determination when to use double angle functions are applied to transpose:
(2.
14)
The shearing stresses on the principal plane are always equal to zero.