Question 4.SP.10: A couple of magnitude M0 = 1.5 kN⋅m acting in a vertical pla......
A couple of magnitude M_0 = 1.5 kN⋅m acting in a vertical plane is applied to a beam having the Z-shaped cross section shown. Determine (a) the stress at point A and (b) the angle that the neutral axis forms with the horizontal plane. The moments and product of inertia of the section with respect to the y and z axes have been computed and are
\begin{aligned} & I_y=3.25 \times 10^{-6} m ^4 \\ & I_z=4.18 \times 10^{-6} m ^4 \\ & I_{y z}=2.87 \times 10^{-6} m ^4 \end{aligned}
STRATEGY: The Z-shaped cross section does not have an axis of symmetry, so it is first necessary to determine the orientation of the principal axes and the corresponding moments of inertia. The applied load is then resolved into components along the principal axes. The stresses due to the axial load and the two couples are then superposed to determine the stress at point A. The angle between the neutral axis and horizontal plane is then found using Eq. (4.57).
\tan \phi=\frac{I_z}{I_y} \tan \theta (4.57)
MODELING and ANALYSIS:
Principal Axes. We draw Mohr’s circle and determine the orientation of the principal axes and the corresponding principal moments of inertia (Fig. 1).^†
\begin{aligned} & \tan 2 \theta_m=\frac{F Z}{E F}=\frac{2.87}{0.465} \quad 2 \theta_m=80.8^{\circ} \quad \theta_m=40.4^{\circ} \\ \\ & R^2=(E F)^2+(F Z)^2=(0.465)^2+(2.87)^2 \quad R=2.91 \times 10^{-6} m ^4 \\ \\ & I_u=I_{\min }=O U=I_{\text {ave }}-R=3.72-2.91=0.810 \times 10^{-6} m ^4 \\ \\ & I_ν=I_{\max }=O V=I_{\text {ave }}+R=3.72+2.91=6.63 \times 10^{-6} m ^4 \end{aligned}
Loading. As shown in Fig. 2, the applied couple M _0 is resolved into components parallel to the principal axes.
\begin{aligned} & M_u=M_0 \sin \theta_m=1500 \sin 40.4^{\circ}=972 N \cdot m \\ & M_ν=M_0 \cos \theta_m=1500 \cos 40.4^{\circ}=1142 N \cdot m \end{aligned}
a. Stress at A. The perpendicular distances from each principal axis to point A shown in Fig. 3 are
\begin{aligned} & u_A=y_A \cos \theta_m+z_A \sin \theta_m=50 \cos 40.4^{\circ}+74 \sin 40.4^{\circ}=86.0 mm \\ & ν_A=-y_A \sin \theta_m+z_A \cos \theta_m=-50 \sin 40.4^{\circ}+74 \cos 40.4^{\circ}=23.9 mm \end{aligned}
Considering separately the bending about each principal axis, note that M _u produces a tensile stress at point A while M _ν produces a compressive stress at the same point.
\begin{aligned} \sigma_A & =+\frac{M_u ν_A}{I_u}-\frac{M_ν u_A}{I_ν}=+\frac{(972 N \cdot m )(0.0239 m )}{0.810 \times 10^{-6} m ^4}-\frac{(1142 N \cdot m )(0.0860 m )}{6.63 \times 10^{-6} m ^4} \\ & =+(28.68 MPa )-(14.81 MPa ) \quad \sigma_A=+13.87 MPa \end{aligned}
b. Neutral Axis. As shown in Fig. 4, we find the angle \phi that the neutral axis forms with the ν axis.
\tan \phi=\frac{I_ν}{I_u} \tan \theta_m=\frac{6.63}{0.810} \tan 40.4^{\circ} \quad \phi=81.8^{\circ}
The angle β formed by the neutral axis and the horizontal is
\beta=\phi-\theta_m=81.8^{\circ}-40.4^{\circ}=41.4^{\circ} \quad \beta=41.4^{\circ}
^†See Ferdinand F. Beer and E. Russell Johnston, Jr., Mechanics for Engineers, 5th ed., McGraw-Hill, New York, 2008, or Vector Mechanics for Engineers–12th ed., McGraw-Hill, New York, 2019, Secs. 9.3–9.4.
