Question 7.14: A simple beam AB is subjected to couples Mo and 2Mo acting a...
A simple beam AB is subjected to couples M_o \text { and } 2 M_o acting as shown in Figure 7.25. Calculate the angles of rotation \theta_A \text { and } \theta_B and maximum deflection, \delta_{\max } . Assume EI to be constant.
We show the free-body diagram of the beam as shown in Figure 7.26 to determine the support reactions.
Clearly,
\sum F_y=0 \Rightarrow R_{ A }=R_{ B } \quad \text { and } \quad \sum M_{ A }=0 \Rightarrow R_{ A }=R_{ B }=\frac{3 M_{ o }}{L}
Therefore, bending moment expression at any section at a distance x from end A of the beam is given by
M_x=R_{ A } x-M_{ o }\left\langle x-\frac{L}{3}\right\rangle^0-2 M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle^0
Putting expression of R_{ A } , we get
M_x=\left\lgroup \frac{3 M_{ o }}{L}\right\rgroup x-M_{ o }\left\langle x-\frac{L}{3}\right\rangle^0-2 M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle^0
From the flexure equation, by substituting the expression of M_x , we get
(E I) \frac{ d ^2 y}{ d x^2}=-M_x=\left\lgroup -\frac{3 M_{ o }}{L} \right\rgroup x+M_{ o }\left\langle x-\frac{L}{3}\right\rangle^0+2 M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle^0
Integrating successively,
(E I) \frac{ d y}{ d x}=\left\lgroup -\frac{3 M_{ o }}{2 L} \right\rgroup x^2+M_{ o }\left\langle x-\frac{L}{3}\right\rangle+2 M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle+C_1 (1)
and (E I) y=\left\lgroup -\frac{M_{ o }}{2 L} \right\rgroup x^3+\frac{M_{ o }}{2}\left\langle x-\frac{L}{3}\right\rangle^2+M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle^2+C_1 x+C_2 (2)
The boundary conditions are: (a) at x = 0, y = 0 and (b) at x = L, y = 0. Putting these in Eq. (2) and noting Macaulay’s function values, we get
From condition (a), C_2=0 and
From condition (b),
-\frac{M_{ o } L^2}{2}+\frac{M_{ o }}{2}\left\lgroup\frac{2 L}{3} \right\rgroup^2+M_{ o }\left\lgroup\frac{L}{3} \right\rgroup^2+C_1 L=0
\begin{aligned} & \Rightarrow C_1=\frac{M_{ o } L}{2}-\frac{2 M_{ o } L}{9}-\frac{M_{ o } L}{9}=\frac{9-4-2}{18} M_{ o } L \\ & \Rightarrow C_1=\frac{1}{6} M_{ o } L \end{aligned}
Thus, Eqs. (1) and (2) now become
(E I) \frac{ d y}{ d x}=\left\lgroup -\frac{3 M_{ o }}{2 L} \right\rgroup x^2+M_{ o }\left\langle x-\frac{L}{3}\right\rangle+2 M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle+\frac{1}{6} M_{ o } L (3)
and (E I) y=\left\lgroup -\frac{M_{ o }}{2 L} \right\rgroup x^3+\frac{M_{ o }}{2}\left\langle x-\frac{L}{3}\right\rangle^2+M_{ o }\left\langle x-\frac{2 L}{3}\right\rangle^2+\left(\frac{1}{6} M_{ o } L\right) x (4)
Now, \theta_{ A } \text { and } \theta_{ B } can be obtained by putting x = 0 and x = L in Eq. (3) above, respectively, as
\theta_{ A } \approx \tan \theta_{ A }=\left.\left\lgroup\frac{1}{E I} \right\rgroup \frac{ d y}{ d x}\right|_{x=0}=\frac{1}{6} \frac{M_{ o } L}{E I}
or \theta_{ A }=\frac{1}{6} \frac{M_{ o } L}{E I} ⦪
and
\theta_{ B } \approx \tan \theta_{ B }=\left.\left(\frac{1}{E I}\right) \frac{ d y}{ d x}\right|_{x=L}=\frac{L}{E I}\left[-\frac{3 M_{ o } L}{2}+\frac{2 M_{ o } L}{3}+\frac{2 M_{ o } L}{3}+\frac{1}{6} M_{ o } L\right]
or \theta_{ B }=0
To obtain \delta_{\max } let us assume deflection occurs within [0, L/3], that is, 0 ≤ x ≤ L/3. For \delta_{\max } we must have dy/dx = 0. Therefore assuming 0 ≤ x ≤ L/3 and setting dy/dx = 0, we get from Eq. (4)
\begin{aligned} & -\frac{3 M_{ o }}{2 L} x^2+M_{ o }\left\lgroup x-\frac{L}{3} \right\rgroup +\frac{M_{ o } L}{6}=0 \\ & \Rightarrow \quad\left\lgroup \frac{3}{2 L} \right\rgroup x^2-M_{ o } x+\frac{L}{6}= \\ & \Rightarrow 9 x^2-6 L x+L^2=0 \Rightarrow x=\frac{L}{3} \end{aligned}
which lies within the domain of 0 ≤ x ≤ L/3.
Thus, putting x = L/3 in Eq. (4), we obtain:
(E I) \delta_{\max }=-\frac{M_{ o }}{2 L}\left\lgroup \frac{L^3}{27} \right\rgroup+\frac{M_{ o } L}{6} \cdot \frac{L}{3}=\frac{M_{ o } L^2}{18}-\frac{M_{ o } L^2}{54}=\frac{M_{ o } L}{27}
Therefore,
\delta_{\max }=\frac{M_{ o } L}{27 E I}(\downarrow)
