Question 16.15: A uniform beam of arbitrary, unsymmetrical cross-section and......
A uniform beam of arbitrary, unsymmetrical cross-section and length 2l is built in at one end and simply supported in the vertical direction at a point halfway along its length. This support, however, allows the beam to deflect freely in the horizontal x direction (Fig. P.16.15). For a vertical load W applied at the free end of the beam, calculate and draw the bending moment diagram, putting in the principal values.
Answer: M_{\mathrm{C}}=0,\,M_{\mathrm{B}}=W l,\,M_{\mathrm{A}}=-W l/2;
The beam is allowed to deflect in the horizontal direction at B so that the support reaction, R_{B}, at B is vertical. Then the total complementary energy, C, of the beam is given by
C=\int_{L}^{}{\int_{0}^{M}{\mathrm{d}\theta\mathrm{d}M-R_{\mathrm{B}}\Delta_{\mathrm{B}}-W\Delta_{\mathrm{C}}} } (i)
From the principle of the stationary value of the total complementary energy of the beam and noting that \Delta_{\mathrm{B}}=0,
\frac{\partial C}{\partial R_{B}} =\int_{L}^{}{\mathrm{d}\theta{\frac{\partial M}{\partial R_{\mathrm{B}}}}=0}Thus,
\frac{\partial C}{\partial R_{B}} =\int_{L}^{}{{\frac{M\ \partial M}{E I\partial R_{\mathrm{B}}}}\mathrm{d}z=0} (ii)
In CB,
M=W(2l-z)~~\mathrm{and}~~\partial M/\partial R_{\mathrm{B}}=0In BA,
M=W(2l-z)-R_{\mathrm{B}}(l-z)\;\;\mathrm{and}\;\;\partial M/\partial R_{\mathrm{B}}=-(l-z)Substituting in Eq. (ii),
\int_{0}^{l}{[w(2l-z)-R_{\mathrm{B}}(l-z)](l-z)\mathrm{d}z=0}from which
R_{\mathrm{B}}={\frac{5W}{2}}Then
M_{\mathrm{C}}=0\,\,\,M_{\mathrm{B}}=W l\,\,\,M_{\mathrm{A}}=-W l/2and the bending moment diagram is as shown in Fig. S.16.15.
