Question 4.8: Shaft Deflection by the Moment-Area Method A simple shaft ca...
Shaft Deflection by the Moment-Area Method
A simple shaft carries its own weight of intensity w, as depicted in Figure 4.13a. Determine the slopes at the ends and center deflection.
Assumption: Bearings act as simple supports.
Inasmuch as the flexural rigidity EI is constant, the M/EI diagram has the same parabolic shape as the bending moment diagram (Figure 4.13b), where the area properties are taken from Table A.3. The elastic curve is depicted in Figure 4.13c, with the tangent drawn at A.
The shaft and loading are symmetric about the center C; hence, the tangent to the elastic curve at C is horizontal: \theta_c=0 \text {. Therefore, } \theta_{C A}=0-\theta_A \text { or } \theta_A=-\theta_{C A} and
A_1=\frac{2}{3}\left(\frac{L}{2}\right)\left(\frac{w L^2}{8 E I}\right)=\frac{w L^3}{24 E I}
By the first moment-area theorem, \theta_{C A}=A_1 ,
\theta_A=-\frac{w L^3}{24 E I}=-\theta_B (4.26)
The minus sign means that the end A of the beam rotates clockwise, as shown in the figure.
Through the use of the second moment-area theorem, Equation 4.25,
t_{B A}=\left[\operatorname{area~of~} \frac{M}{E I} \text { diagram between } A \text { and } B\right] \bar{x}_2 (4.25)
t_{C A}=A_1\left(\frac{3 L}{16}\right)=\frac{w L^4}{128 E I}
in which t_{C A}=C C^{\prime} \text { and } \theta_A E / 2=C^{\prime} C^{\prime \prime} (Figure 4.13c). The maximum deflection \upsilon _{max} = C C^{\prime \prime}, is
\upsilon _{\max }=-\frac{w L^3}{24 E I}\left(\frac{L}{2}\right)+\frac{w L^3}{24 E I}=-\frac{5 w L^4}{384 E I} (4.27)
The minus sign indicates that the deflection is downward.
Comment: Alternatively, the moment of area A_{1} about point A, Equation 4.24, readily gives the numerical value of \upsilon _{max}.
t_{B A}=\int\limits_{A}^{B}x_1\frac{Mdx}{EI}=\left[\operatorname{area~of~} \frac{M}{E I} \text { diagram between } A \text { and } B\right] \bar{x}_1 (4.24)