Question 8.SP.5: Three forces are applied as shown at points A, B, and D of a......
Three forces are applied as shown at points A, B, and D of a short steel post. Knowing that the horizontal cross section of the post is a 40 × 140-mm rectangle, determine the principal stresses, principal planes, and maximum shearing stress at point H.
STRATEGY: Begin by determining the forces and couples acting on the section containing the point of interest, and then use these to calculate the normal and shearing stresses acting at the point. Using Mohr’s circle, these stresses can then be transformed to obtain the principal stresses, principal planes, and maximum shearing stress.
MODELING and ANALYSIS:
Internal Forces in Section EFG. Replace the three applied forces by an equivalent force-couple system at the center C of the rectangular section EFG (Fig. 1).
\begin{aligned} & V_x=-30 kN \quad P=50 kN \quad V_z=-75 kN \\ & M_x=(50 kN )(0.130 m )-(75 kN )(0.200 m )=-8.5 kN \cdot m \\ & M_y=0 \quad M_z=(30 kN )(0.100 m )=3 kN \cdot m \end{aligned}
Note that there is no twisting couple about the y axis. The geometric properties of the rectangular section are
\begin{aligned} & A=(0.040 m )(0.140 m )=5.6 \times 10^{-3} m ^2 \\ & I_x=\frac{1}{12}(0.040 m )(0.140 m )^3=9.15 \times 10^{-6} m ^4 \\ & I_z=\frac{1}{12}(0.140 m )(0.040 m )^3=0.747 \times 10^{-6} m ^4 \end{aligned}
Normal Stress at H. The normal stresses \sigma_y are produced by the centric force P and by the bending couples M _x \text { and } M _z . The sign of each stress is determined by carefully examining the force-couple system at C (Fig. 2).
\begin{aligned}\sigma_y & =+\frac{P}{A}+\frac{\left|M_z\right| a}{I_z}-\frac{\left|M_x\right| b}{I_x} \\ & =\frac{50 kN }{5.6 \times 10^{-3} m ^2}+\frac{(3 kN \cdot m )(0.020 m )}{0.747 \times 10^{-6} m ^4}-\frac{(8.5 kN \cdot m )(0.025 m )}{9.15 \times 10^{-6} m ^4} \\ & =8.93 MPa +80.3 MPa -23.2 MPa \end{aligned}
\sigma_y=66.0 MPa
Shearing Stress at H. Considering the shearing force V _x , we note that Q = 0 with respect to the z axis, since H is on the edge of the cross section.
Thus, V _x produces no shearing stress at H. The shearing force V _z does produce a shearing stress at H (Fig. 3).
\begin{aligned} & Q=A_1 \bar{y}_1=[(0.040 m )(0.045 m )](0.0475 m )=85.5 \times 10^{-6} m ^3 \\ & \tau_{y z}=\frac{V_z Q}{I_x t}=\frac{(75 kN )\left(85.5 \times 10^{-6} m ^3\right)}{\left(9.15 \times 10^{-6} m ^4\right)(0.040 m )} \end{aligned}
\tau_{y z}=17.52 MPa
Principal Stresses, Principal Planes, and Maximum Shearing Stress at H. Draw Mohr’s circle for the stresses at point H (Fig. 4).
\tan 2 \theta_p=\frac{17.52}{33.0} \quad 2 \theta_p=27.96^{\circ} \quad \theta_p=13.98^{\circ}
\begin{aligned} R & =\sqrt{(33.0)^2+(17.52)^2}=37.4 MPa & & \tau_{\max }=37.4 MPa \\ \sigma_{\max } & =O A=O C+R=33.0+37.4 & & \sigma_{\max }=70.4 MPa \\ \sigma_{\min } & =O B=O C-R=33.0-37.4 & & \sigma_{\min }=-7.4 MPa \end{aligned}
