A state of plane stress consists of a tensile stress s0 = 8 ksi exerted on vertical surfaces and of unknown shearing stresses. Determine (a) the magnitude of the shearing stress t0 for which the largest normal stress is 10 ksi, (b) the corresponding maximum shearing stress.

Question

A state of plane stress consists of a tensile stress [latex]s_0[/latex] = 8 ksi exerted on vertical surfaces and of unknown shearing stresses. Determine (a) the magnitude of the shearing stress [latex]t_0[/latex] for which the largest normal stress is 10 ksi, (b) the corresponding maximum shearing stress.

Accepted Answer

Construction of Mohr’s Circle. We assume that the shearing stresses act in the senses shown. Thus, the shearing stress [latex]t_0[/latex] 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 [latex]t_0[/latex] above the horizontal axis. Considering a horizontal face of the element, we observe that [latex]s_y = 0[/latex] and that [latex]t_0[/latex] tends to rotate the element counterclockwise; thus, we plot point Y at a distance [latex]t_0[/latex] below O.

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

[latex]s _{ ext {ave }}=\frac{1}{2}\left( s _{x}+ s _{y} ight)=\frac{1}{2}(8+0)=4 ksi[/latex]

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

[latex]s_{\max }= s_{ ext {ave }}+R[/latex]

10 ksi = 4 ksi + R R = 6 ksi

a. Shearing Stress [latex]t_0[/latex]. Considering the right triangle CFX, we find

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

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

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

[latex]t _{\max }=R=6 ksi[/latex] [latex]t _{\max }=6 ksi[/latex]

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

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 [latex]t_0[/latex] 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.

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

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