A simple beam AB is subjected to couples Mo and 2Mo acting as shown in Figure 7.25. Calculate the angles of rotation θA and θB and maximum deflection, δmax. Assume EI to be constant.

Question

A simple beam AB is subjected to couples [latex] M_o ext { and } 2 M_o [/latex] acting as shown in Figure 7.25. Calculate the angles of rotation [latex] heta_A ext { and } heta_B [/latex] and maximum deflection, [latex] \delta_{\max } [/latex] . Assume EI to be constant.

Accepted Answer

We show the free-body diagram of the beam as shown in Figure 7.26 to determine the support reactions.

Clearly,

[latex] \sum F_y=0 \Rightarrow R_{ A }=R_{ B } \quad ext { and } \quad \sum M_{ A }=0 \Rightarrow R_{ A }=R_{ B }=\frac{3 M_{ o }}{L} [/latex]

Therefore, bending moment expression at any section at a distance x from end A of the beam is given by

[latex] M_x=R_{ A } x-M_{ o }\left\langle x-\frac{L}{3} ight angle^0-2 M_{ o }\left\langle x-\frac{2 L}{3} ight angle^0 [/latex]

Putting expression of [latex] R_{ A } [/latex], we get

[latex] M_x=\left\lgroup \frac{3 M_{ o }}{L} ight group x-M_{ o }\left\langle x-\frac{L}{3} ight angle^0-2 M_{ o }\left\langle x-\frac{2 L}{3} ight angle^0 [/latex]

From the flexure equation, by substituting the expression of [latex] M_x [/latex], we get

[latex] (E I) \frac{ d ^2 y}{ d x^2}=-M_x=\left\lgroup -\frac{3 M_{ o }}{L} ight group x+M_{ o }\left\langle x-\frac{L}{3} ight angle^0+2 M_{ o }\left\langle x-\frac{2 L}{3} ight angle^0 [/latex]

Integrating successively,

[latex] (E I) \frac{ d y}{ d x}=\left\lgroup -\frac{3 M_{ o }}{2 L} ight group x^2+M_{ o }\left\langle x-\frac{L}{3} ight angle+2 M_{ o }\left\langle x-\frac{2 L}{3} ight angle+C_1 [/latex] (1)

and [latex] (E I) y=\left\lgroup -\frac{M_{ o }}{2 L} ight group x^3+\frac{M_{ o }}{2}\left\langle x-\frac{L}{3} ight angle^2+M_{ o }\left\langle x-\frac{2 L}{3} ight angle^2+C_1 x+C_2 [/latex] (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), [latex] C_2=0 [/latex] and

From condition (b),

[latex] -\frac{M_{ o } L^2}{2}+\frac{M_{ o }}{2}\left\lgroup\frac{2 L}{3} ight group^2+M_{ o }\left\lgroup\frac{L}{3} ight group^2+C_1 L=0 [/latex]

[latex] \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} [/latex]

Thus, Eqs. (1) and (2) now become

[latex] (E I) \frac{ d y}{ d x}=\left\lgroup -\frac{3 M_{ o }}{2 L} ight group x^2+M_{ o }\left\langle x-\frac{L}{3} ight angle+2 M_{ o }\left\langle x-\frac{2 L}{3} ight angle+\frac{1}{6} M_{ o } L [/latex] (3)

and [latex] (E I) y=\left\lgroup -\frac{M_{ o }}{2 L} ight group x^3+\frac{M_{ o }}{2}\left\langle x-\frac{L}{3} ight angle^2+M_{ o }\left\langle x-\frac{2 L}{3} ight angle^2+\left(\frac{1}{6} M_{ o } L ight) x [/latex] (4)

Now, [latex] heta_{ A } ext { and } heta_{ B } [/latex] can be obtained by putting x = 0 and x = L in Eq. (3) above, respectively, as

[latex] heta_{ A } \approx an heta_{ A }=\left.\left\lgroup\frac{1}{E I} ight group \frac{ d y}{ d x} ight|_{x=0}=\frac{1}{6} \frac{M_{ o } L}{E I} [/latex]

or [latex] heta_{ A }=\frac{1}{6} \frac{M_{ o } L}{E I} [/latex]⦪

and

[latex] heta_{ B } \approx an heta_{ B }=\left.\left(\frac{1}{E I} ight) \frac{ d y}{ d x} ight|_{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 ight] [/latex]

or [latex] heta_{ B }=0 [/latex]

To obtain [latex] \delta_{\max } [/latex] let us assume deflection occurs within [0, L/3], that is, 0 ≤ x ≤ L/3. For [latex] \delta_{\max } [/latex] we must have dy/dx = 0. Therefore assuming 0 ≤ x ≤ L/3 and setting dy/dx = 0, we get from Eq. (4)

[latex] \begin{aligned} & -\frac{3 M_{ o }}{2 L} x^2+M_{ o }\left\lgroup x-\frac{L}{3} ight group +\frac{M_{ o } L}{6}=0 \ & \Rightarrow \quad\left\lgroup \frac{3}{2 L} ight group 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} [/latex]

which lies within the domain of 0 ≤ x ≤ L/3.

Thus, putting x = L/3 in Eq. (4), we obtain:

[latex] (E I) \delta_{\max }=-\frac{M_{ o }}{2 L}\left\lgroup \frac{L^3}{27} ight group+\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} [/latex]

Therefore,

[latex] \delta_{\max }=\frac{M_{ o } L}{27 E I}(\downarrow) [/latex]

Question 7.14: A simple beam AB is subjected to couples Mo and 2Mo acting a...

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