Question 15.11: Determine the deflected shape of the beam shown in Fig. 15.2...

In this problem, an external moment M_{0} is applied to the beam at B. The support reactions are found in the normal way and are

The bending moment at any section Z between B and C is then given by

M=-R_{ A } z-M_{0}  (i)

Equation (i) is valid only for the region BC and clearly does not contain a singularity function which would cause M_{0} to vanish for z \leq b. We overcome this difficulty by writing

M=-R_{ A } z-M_{0}[z-b]^{0} \quad\left(\text { Note }:[z-b]^{0}=1\right)  (ii)

Equation (ii) has the same value as Eq. (i) but is now applicable to all sections of the beam, since [z-b]^{0} disappears when z \leq b. Substituting for M from Eq. (ii) in the second of Eqs. (15.31), we obtain

u^{\prime \prime}=-\frac{M_{y}}{E I_{y y}}, \quad v^{\prime \prime}=-\frac{M_{x}}{E I_{x x}}  (15.31)

E I v^{\prime \prime}=R_{ A } z+M_{0}[z-b]^{0}  (iii)

Integration of Eq. (iii) yields

E l v^{\prime}=R_{ A } \frac{z^{2}}{2}+M_{0}[z-b]+C_{1}  (vi)

and

E I v=R_{ A } \frac{z^{3}}{6}+\frac{M_{0}}{2}[z-b]^{2}+C_{1} z+C_{2}  (v)

where C_{1} and C_{2} are arbitrary constants. The boundary conditions are v = 0 when z = 0 and z = L. From the first of these, we have C_{2}=0, while the second gives

from which

C_{1}=-\frac{M_{0}}{6 L}\left(2 L^{2}-6 L b+3 b^{2}\right)  (vi)

The equation of the deflection curve of the beam is then

v=\frac{M_{0}}{6 E I L}\left\{z^{3}+3 L[z-b]^{2}-\left(2 L^{2}-6 L b+3 b^{2}\right) z\right\}  (vii)

The beam deflection curve equation is obtained through the following MATLAB file: