Question 4.SP.10: A couple of magnitude M0 = 1.5 kN • m acting in a vertical p...
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, (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 as follows:
I_{y} = 3.25 \times 10^{-6} m^{4} \\ I_{z} = 4.18 \times 10^{-6} m^{4} \\ I_{yz} = 2.87 \times 10^{-6} m^{4}
Principal Axes. We draw Mohr’s circle and determine the orientation of the principal axes and the corresponding principal moments of inertia.\dagger
\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_{ave }-R=3.72-2.91=0.810 \times 10^{-6} m ^{4} \\ I_{v} =I_{\max }=O V=I_{ave }+R=3.72+2.91=6.63 \times 10^{-6} m ^{4}
Loading. The applied couple \mathbf{M} _{0} is resolved into components parallel to the principal axes.
M_{u}=M_{0} \sin \theta_{m}=1500 \sin 40.4^{\circ}=972 N \cdot m \\ M_{v}=M_{0} \cos \theta_{m}=1500 \cos 40.4^{\circ}=1142 N \cdot m
a. Stress at A. The perpendicular distances from each principal axis to point A are
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 \\ v_{A}=-y_{A} \sin \theta_{m}+z_{A} \cos \theta_{m}=-50 \sin 40.4^{\circ}+74 \cos 40.4^{\circ}=23.9 mm
Considering separately the bending about each principal axis, we note that
\mathbf{M} _{u} produces a tensile stress at point A while \mathbf{M} _{v} produces a compressive stress at the same point.
\sigma_{ A }=+\frac{M_{u} v_{A}}{I_{u}}-\frac{M_{v} u_{A}}{I_{v}}=+\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) \sigma_{A} = +13.87 MPa
b. Neutral Axis. Using Eq. (4.57), we find the angle Φ that the neutral axis forms with the v axis.
\tan \phi=\frac{I_{v}}{I_{u}} \tan \theta_{m}=\frac{6.63}{0.810} \tan 40.4^{\circ} \quad \phi=81.8^{\circ}
The angle \beta formed by the neutral axis and the horizontal is
\beta=\phi-\theta_{m}=81.8^{\circ}-40.4^{\circ}=41.4^{\circ} \quad \quad \quad \quad \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–9th ed., McGraw-Hill, New York, 2010, Secs. 9.8–9.10.
