Question 7.2: For the state of plane stress already considered in Example ...
For the state of plane stress already considered in Example 7.01, (a) construct Mohr’s circle, (b) determine the principal stresses, (c) determine the maximum shearing stress and the corresponding normal stress.
(a) Construction of Mohr’s Circle. We note from Fig. 7.21a that the normal stress exerted on the face oriented toward the x axis is tensile (positive) and that the shearing stress exerted on that face tends to rotate the element counterclockwise. Point X of Mohr’s circle, therefore, will be plotted to the right of the vertical axis and below the horizontal axis (Fig. 7.21b). A similar inspection of the normal stress and shearing stress exerted on the upper face of the element shows that point Y should be plotted to the left of the vertical axis and above the horizontal axis. Drawing the line XY, we obtain the center C of Mohr’s circle; its abscissa is
s _{\text {ave }}=\frac{ s _{x}+ s _{y}}{2}=\frac{50+(-10)}{2}=20 MPa
Since the sides of the shaded triangle are
CF = 50 – 20 = 30 MPa and FX = 40 MPa
the radius of the circle is
R=C X=\sqrt{(30)^{2}+(40)^{2}}=50 MPa
(b) Principal Planes and Principal Stresses. The principal stresses are
s _{\max }=O A=O C+C A=20+50=70 MPas _{\min }=O B=O C-B C=20-50=-30 MPa
Recalling that the angle ACX represents 2 u _{p} (Fig. 7.21b), we write
\tan 2 u _{p}=\frac{F X}{C F}=\frac{40}{30}
2 u _{p}=53.1^{\circ} u_{p}=26.6^{\circ}
Since the rotation which brings CX into CA in Fig. 7.22b is counterclockwise, the rotation that brings Ox into the axis Oa corresponding to s_{\max } in Fig. 7.22a is also counterclockwise.
(c) Maximum Shearing Stress. Since a further rotation of 90° counterclockwise brings CA into CD in Fig. 7.22b, a further rotation of 45° counterclockwise will bring the axis Oa into the axis Od corresponding to the maximum shearing stress in Fig. 7.22a. We note from Fig. 7.22b that t _{\max } = R = 50 MPa and that the corresponding normal stress is s ^{\prime}= s _{\text {ave }}= 20 MPa. Since point D is located above the s axis in Fig. 7.22b, the shearing stresses exerted on the faces perpendicular to Od in Fig. 7.22a must be directed so that they will tend to rotate the element clockwise.
