STRESSES ON AN OBLIQUE PLANE
A prism ABCD in Fig. 2Oa has the principal stresses of 6300- and 2400-lb/in^2 (43,438.5-
and 16,548.0-kPa) tension. Applying both the analytical and graphical methods, deter-
mine the normal and shearing stress on plane AE.
Calculation Procedure:
- Compute the stresses, using the analytical method
A principal stress is a normal stress not accompanied by a shearing stress. The plane on
which the principal stress exists is termed a principal plane. For a condition of plane
stress, there are two principal planes through every point in a stressed body and these
planes are mutually perpendicular. Moreover, one principal stress is the maximum normal
stress existing at that point; the other is the minimum normal stress.
Let sx and sy = the principal stress in the x and y direction, respectively; Sn = normal
stress on the plane making an angle 6 with the y axis; ss = shearing stress on this plane.
All stresses are expressed in pounds per square inch (kilopascals) and all angles in de-
grees. Tensile stresses are positive; compressive stresses are negative.
Applying the usual stress equations yields sn = sy + (sx - sy) cos^2 0; ss = l/2(sx - sy)
sin 20. Substituting gives sn = 2400 + (6300 - 2400)0.766^2 = 4690-lb/in^2 (32,337.6-kPa)
tension, and ss = '/2(63OO - 2400)0.985 = 1920 lb/in^2 (13,238.4 kPa). - Apply the graphical method of solution
Construct, in Fig. 2Qb, Mohr's circle of stress thus: Using a suitable scale, draw OA = sy9
and OB = sx. Draw a circle having AB as its diameter. Draw the radius CD making an an-
gle of 26 = 80° with AB. Through D 9 drop a perpendicular DE to AB. Then OE = Sn and
ED = ss. Scale OE and ED to obtain the normal and shearing stresses on plane AE.
Related Calculations: The normal stress may also be computed from sn =
(sx + 5^)0.5 + (sx - Sy)0.5 cos 29.
(a) Stresses on prism (b) Mohr's circle of stress
FIGURE 20