Question 7.S-P.2: For the state of plane stress shown, determine (a) the princ...
For the state of plane stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30°.
Construction of Mohr’s Circle. We note that on a face perpendicular to the x axis, the normal stress is tensile and the shearing stress tends to rotate the element clockwise; thus, we plot X at a point 100 units to the right of the vertical axis and 48 units above the horizontal axis. In a similar fashion, we examine the stress components on the upper face and plot point Y(60, -48). Joining points X and Y by a straight line, we define the center C of Mohr’s circle. The abscissa of C, which represents s_{\text {ave }}, and the radius R of the circle can be measured directly or calculated as follows:
s _{ ave }=O C=\frac{1}{2}\left( s _{x}+ s _{y}\right)=\frac{1}{2}(100+60)=80 MPaR=\sqrt{(C F)^{2}+(F X)^{2}}=\sqrt{(20)^{2}+(48)^{2}}=52 MPa
a. Principal Planes and Principal Stresses. We rotate the diameter XY clockwise through 2 u _{p} until it coincides with the diameter AB. We have
\tan 2 u _{p}=\frac{X F}{C F}=\frac{48}{20}=2.4 2 u _{p}=67.4^{\circ} i u_{p}=33.7^{\circ} i
The principal stresses are represented by the abscissas of points A and B:
s _{\max }=O A=O C+C A=80+52 s_{\max }=+132 MPa
s _{\min }=O B=O C-B C=80-52 s_{\min }=+28 MPa
Since the rotation that brings XY into AB is clockwise, the rotation that brings Ox into the axis Oa corresponding to s_{\max } is also clockwise; we obtain the orientation shown for the principal planes.
b. Stress Components on Element Rotated 30° l. Points X^{\prime} and Y^{\prime} on Mohr’s circle that correspond to the stress components on the rotated element are obtained by rotating XY counterclockwise through 2u = 60°. We find
f = 180° – 60° – 67.4° f = 52.6°
s _{x^{\prime}}=O K=O C-K C=80-52 \cos 52.6^{\circ} s _{x^{\prime}}=+48.4 MPa
s _{y^{\prime}}=O L=O C+C L=80+52 \cos 52.6^{\circ} s _{y^{\prime}}=+111.6 MPa
t _{x^{\prime} y^{\prime}}=K X^{\prime}=52 \sin 52.6^{\circ} t _{x^{\prime} y^{\prime}}=41.3 MPa
Since X^{\prime} is located above the horizontal axis, the shearing stress on the face perpendicular to O x^{\prime} tends to rotate the element clockwise.
