Question 4.10: A state of plane stress consists of a tensile stress σo = 8 ...
A state of plane stress consists of a tensile stress \sigma_o = 8 ksi exerted on vertical surfaces and unknown shear stresses \textrm{τ}_o (Figure 4.49). Determine (a) the magnitude of the shear stress \textrm{τ}_o for which the maximum normal stress is 10 ksi, and (b) the corresponding maximum shear stress.
Given: Partial plane stress state.
Find: Shear stress \textrm{τ}_o ; maximum shear stress.
Assume: Plane stress.
We assume a sense (sign) for the unknown shear stress and can construct Mohr’s circle. The shearing stress \textrm{τ}_o on faces normal to the x axis tends to rotate the element clockwise, so we plot point X, whose coordinates are ( \sigma_o , \textrm{τ}_o ), above the σ axis. We see that in our initial state \sigma_y is zero and that on faces normal to the y axis \textrm{τ}_o tends to rotate the element counterclockwise; thus, we plot point Y (0, \textrm{τ}_o ), below the σ axis (Figure 4.50).
Line XY passes through the center of our circle, at
\sigma _{ave} = ½ (\sigma _x + \sigma _y) = ½ (8 + 0) = 4 ksi.
We determine the radius R of the circle by observing that the maximum normal stress, given as 10 ksi, appears a distance R to the right of the circle’s center (Figure 4.51):
\sigma _1 =\sigma _{ave} +R .
R= \sigma _1 – \sigma _{ave}.
R= 10 \textrm{ksi}- 4 \textrm{ksi}= 6 \textrm{ksi}.
Now we have Mohr’s circle to work with. We see that the rotation required to get from our initial stress state (point X) to the principal stress state at (10 ksi, 0) is either clockwise 2 \textrm{θ}_N as shown or counterclockwise 360 – 2 \textrm{θ}_N . We choose to work with the more manageable clockwise rotation, and consider the right triangle CFX.
\cos 2\theta _N=\frac{CF}{CX}=\frac{CF}{R}=\frac{4 \textrm{ksi}}{6 \textrm{ksi}}\theta _N=-24.1°
This rotation, again, is clockwise, as reflected by the negative sign. The right triangle CFX also allows us to compute the unknown shear stress, \tau_o , which is experienced at point X:
\textrm{τ}_o = FX = R \sin 2θ_N = (6 \textrm{ksi})·\sin 48.2˚ = 4.47 ksi.
The maximum shear stress is also apparent from Mohr’s circle. It is simply the radius of the circle, R:
\textrm{τ}_{\max} =R= 6 ksi.
The corresponding normal stress at this stress state is \sigma_{ave} = 4 ksi. Mohr’s circle indicates that to get from the initial stress state to the state of maximum shear stress, we must rotate the circle diameter XY counterclockwise by 2 \textrm{θ}_S or to rotate the element itself by \textrm{θ}_S . It is clear from the circle that 2 \textrm{θ}_S + \left|2 \textrm{θ}_N\right| = 90º. Hence,
2 \textrm{θ}_S =90º – \left|2 \textrm{θ}_N\right| = 90º – 48.2º = 41.8º.
With all this information in hand, we can draw properly oriented elements in each of the identified stress states (Figure 4.52).
Note: If we had originally assumed the opposite sense of the unknown \textrm{τ}_o , we would have obtained the same numerical answers, but the orientation of the elements would be as shown in Figure 4.53.
