Question 4.CA.9: A vertical 4.80-kN load is applied as shown on a wooden post......
A vertical 4.80-kN load is applied as shown on a wooden post of rectangular cross section, 80 by 120 mm (Fig. 4.57a). (a) Determine the stress at points A, B, C, and D. (b) Locate the neutral axis of the cross section.
a. Stresses. The given eccentric load is replaced by an equivalent system consisting of a centric load P and two couples M _x \text { and } M _z represented by vectors directed along the principal centroidal axes of the section (Fig. 4.57b). Thus
\begin{aligned} & M_x=(4.80 kN )(40 mm )=192 N \cdot m \\ & M_z=(4.80 kN )(60 mm -35 mm )=120 N \cdot m \end{aligned}
Compute the area and the centroidal moments of inertia of the cross section:
\begin{aligned} & A=(0.080 m )(0.120 m )=9.60 \times 10^{-3} m ^2 \\ & I_x=\frac{1}{12}(0.120 m )(0.080 m )^3=5.12 \times 10^{-6} m ^4 \\ & I_z=\frac{1}{12}(0.080 m )(0.120 m )^3=11.52 \times 10^{-6} m ^4 \end{aligned}
The stress \sigma_0 due to the centric load P is negative and uniform across the section:
\sigma_0=\frac{P}{A}=\frac{-4.80 kN }{9.60 \times 10^{-3} m ^2}=-0.5 MPa
The stresses due to the bending couples M _x \text { and } M _z are linearly distributed across the section with maximum values equal to
\begin{aligned} & \sigma_1=\frac{M_x z_{\max }}{I_x}=\frac{(192 N \cdot m )(40 mm )}{5.12 \times 10^{-6} m ^4}=1.5 MPa \\ & \sigma_2=\frac{M_z x_{\max }}{I_z}=\frac{(120 N \cdot m )(60 mm )}{11.52 \times 10^{-6} m ^4}=0.625 MPa \end{aligned}
The stresses at the corners of the section are
\sigma_y=\sigma_0 \pm \sigma_1 \pm \sigma_2
where the signs must be determined from Fig. 4.57b. Noting that the stresses due to M _x are positive at C and D and negative at A and B, and the stresses due to M _z are positive at B and C and negative at A and D, we obtain
\begin{aligned} & \sigma_A=-0.5-1.5-0.625=-2.625 MPa \\ & \sigma_B=-0.5-1.5+0.625=-1.375 MPa \\ & \sigma_C=-0.5+1.5+0.625=+1.625 MPa \\ & \sigma_D=-0.5+1.5-0.625=+0.375 MPa \end{aligned}
b. Neutral Axis. The stress will be zero at a point G between B and C, and at a point H between D and A (Fig. 4.57c). Since the stress distribution is linear,
\begin{array}{ll} \frac{B G}{80 mm }=\frac{1.375}{1.625+1.375} & B G=36.7 mm \\ \frac{H A}{80 mm }=\frac{2.625}{2.625+0.375} & H A=70 mm \end{array}
The neutral axis can be drawn through points G and H (Fig. 4.57d).
The distribution of the stresses across the section is shown in Fig. 4.57e.
