Question 7.S-P.3: A state of plane stress consists of a tensile stress s0 = 8 ...

Construction of Mohr’s Circle.     We assume that the shearing stresses act in the senses shown. Thus, the shearing stress t_0 on a face perpendicular to the x axis tends to rotate the element clockwise and we plot the point X of coordinates 8 ksi and t_0 above the horizontal axis. Considering a horizontal face of the element, we observe that s_y = 0 and that t_0 tends to rotate the element counterclockwise; thus, we plot point Y at a distance t_0 below O.

We note that the abscissa of the center C of Mohr’s circle is

The radius R of the circle is determined by observing that the maximum normal stress, s_{\max }=10 ksi, is represented by the abscissa of point A and writing

10 ksi = 4 ksi + R                          R = 6 ksi

a. Shearing Stress t_0.    Considering the right triangle CFX, we find

\cos 2 u _{p}=\frac{C F}{C X}=\frac{C F}{R}=\frac{4 ksi }{6 ksi }                    2 u _{p}=48.2^{\circ} i                u _{p}=24.1^{\circ} i

t _{0}=F X=R \sin 2 u _{p}=(6 ksi ) \sin 48.2^{\circ}                  t _{0}=4.47 ksi

b. Maximum Shearing Stress. The coordinates of point D of Mohr’s circle represent the maximum shearing stress and the corresponding normal stress.

t _{\max }=R=6 ksi                              t _{\max }=6 ksi

2 u _{s}=90^{\circ}-2 u_{p}=90^{\circ}-48.2^{\circ}=41.8^{\circ} l                      u _{x}=20.9^{\circ} l

The maximum shearing stress is exerted on an element that is oriented as shown in Fig. a. (The element upon which the principal stresses are exerted is also shown.)

Note. If our original assumption regarding the sense of t_0 was reversed, we would obtain the same circle and the same answers, but the orientation of the elements would be as shown in Fig. b.