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One of the most challenging problems in design is that of relating the strength of a mechanical element to the internal stresses which are produced by the external loads. Usually we have only a single value for the strength, such as the yield strength, but several stress components. One of the problems to be faced in a future chapter is how to take a stress element like Figure 02-01b and relate it to a single strength magnitude to achieve safety. This section is the first step in the solution.
Suppose the element of Figure 02-01b is cut by an oblique plane at angle phi to the x axis as shown if Figure 02-02. This section is concerned with the stresses sigma and tau which act upon this oblique plane. By summing the forces caused by all the stress components to zero, the stresses sigma and tau are found to be
sigma = (sig_x + sig_y)/2 + (sig_x - sig_y) *
cos(2*phi) + tau_xy * sin(2*phi) [eq2-2]
tau = -((sig_x - sig_y)/2) * sin(2*phi) +
tau_xy * cos(2*phi) [eq2-3]
Differentiating the first equation with respect to phi and setting the result equal to zero gives
tan(2*phi) = (2*tau_xy)/(sig_x-sig_y) [eq2-4]
Equation [2-4] defines two particular values for the angle 2*phi, one of which defines the maximum normal stress, sig_1 and the other, the minimum normal stress sig_2. These two stresses are called the principal stresses, and their corresponding directions, the principal directions. The angle phi, between the principal directions is 90 degrees. In a similar manner, we differentiate equation [2-3], set the result equal to zero, and obtain
tan(2*phi) = -(sig_x-sig_y)/(2*tau_xy) [eq2-5]
Equation [2-5] difines the two values of 2*phi at which the shear stress tau is maximum. It is interesting to note that equation [2-4] can be written in the form
2*tau_xy*cos(2*phi) = (sig_x-sig_y)*sin(2*phi)
or
sin(2*phi) = (2*tau_xy*cos(2*phi))/(sig_x-sig_y) [a]
Now substitute equation [a] for sin(2*phi) in equation [2-3]. We obtain
tau = -(sig_x-sig_y)/2*(2*tau_xy*cos(2*phi))/
(sig_x-sig_y)+tau_xy*cos(2*phi)
= 0 [2-6]
Equation [2-6] states that the shear stress associated with both principal directions is zero. Solving equation [2-5] for sin(2*phi), in a similar manner, and substituting the result in equation [2-2] yields
sigma = (sig_x-sig_y)/2 [2-7]
Equation [2-7] tells us that the two normal stresses associated with the directions of the two maximum shear stresses are equal.
Formulas for the two principal stresses can be obtained by substituting the angle 2*phi from equation [2-4] into equation [2-2]. The result is
sig_1 = (sig_x+sig_y)/2 +
sqrt(((sig_x-sig_y)/2)^2+tau_xy^2) and
sig_2 = (sig_x+sig_y)/2 -
sqrt(((sig_x-sig_y)/2)^2+tau_xy^2) [2-8]
In a similar manner the two maximum shear stresses are found to be
tau_max = sqrt(((sig_x-sig_y)/2)^2+tau_xy^2) and
tau_max = -sqrt(((sig_x-sig_y)/2)^2+tau_xy^2) [2-9]
But this may not be the maximum shear stress when the z direction is considered too. See Section 2-03.
A graphical method ... UNDER CONSTRUCTION...
