2.8. Stresses at inclined sections under tension (compression) in one direction
It is necessary to be able to determine the stresses acting at any inclined section of a tensed (compressed) element. To conclude about the material strength (Fig. 2.6).
а) b)
c)
Fig.2.6.
Consider
the normal stresses at the bar
cross section to be known (for example
).
Let us determine the stresses arising at the inclined BC section with the normal to it under the angle of the direction. Accept the direction which is opposite to the clock motion as positive to count off the angle.
Denote: A is the cross-area perpendicular to the bar axis; Аα is the inclined cross area herewith
(2.1)
normal
and
shear
stresses act at the inclined section in general case. We determine
their values from the equilibrium condition, for example of the cut
low part (Fig. 2.6 b). Project the forces on the
direction:
.
(2.2)
Using the relation (2.1) we find
(2.3)
Projecting the forces on the direction, we get
then
(2.4)
With value positive, (i.e. tension) and at 0 ≤ ≤ 90° we receive the positive value for . It means that the shearing stress will be as pictured in Fig. 2.6 b.
The given direction of the shearing stress is characterized by the necessity to rotate the external n normal to the section area clockwise to coincide with the shearing stress.
The shearing stresses of this direction are considered positive. If it is necessary to rotate the normal to the area against the clock to coincide with the shearing stress, this stress is then considered negative (Fig. 2.6 c).
From the formula (2.3) it follows that the maximum values of the normal stresses are set up under =0 i.e. at the section perpendicular to the bar axis.
From the formula (2.4) it follows that under = 0 the shearing stress is equal to zero.
From the formulae (2.3) and (2.4) it is evident that under = 90° we have = 0 and = 0.
Thus, the longitudinal sections have neither normal nor shearing stresses.
The areas having no shearing stresses are called principal areas and the normal stresses acting along them are the principal stresses. Consequently, the normal stress at the cross section for the tensed or compressed bar is the principal one.
It is also evident from the formula (2.4) that the maximum shearing stress takes place at the section under = 45° and it is equal to half principal stress:
(2.5)
2.9. Law of the shearing stresses couple
Let us determine normal and shearing stresses at two mutually perpendicular planes.
For the inclined under angle plane by the formulae (2.3) and (2.4) we have:
For the
mutually perpendicular plane under
angle value, we can determine the normal and shearing stresses either
from the equilibrium condition of the upper or lower bar parts
(Fig.2.6 c) or by the formulae (2.3) and (2.4) replacing
by
.
To study
the material mechanical properties irrespective of the specimen
dimensions a diagram in the coordinates “stress - deformation”
is applied. The diagrams in question differ only by scales.
Applying the formulae (2.3) and (2.4), we get
Analyzing the obtained results we can see that, firstly,
i.e. the sum of the normal stresses in two mutually perpendicular areas is constant and equal to the principal stress, secondly,
So, at two mutually perpendicular areas shearing stresses (law of the shearing stresses couple) equal by value and reverse at their signs act.
The law of shearing stresses couple (reciprocation) is valid not only for the monoaxial, but for any other tensed condition: biaxial and volumetric.