Question 9.5: A simple beam AB with an overhang BC supports a concentrated...

Differential equations of the deflection curve. Because reactive forces act at supports A and B, we must write separate differential equations for parts AB and BC of the beam. Therefore, we begin by finding the shear forces in each part of the beam.
The downward reaction at support A is equal to P/2, and the upward reaction at support B is equal to 3P/2 (see Fig. 9-15a). It follows that the shear forces in parts AB and BC are

V = –\frac{P}{2}     (0<x<L)     (9-46a)

V = P    (L<x<\frac{3L}{2})     (9-46b)

in which x is measured from end A of the beam (Fig. 9-12b).
The third-order differential equations for the beam now become (see Eq. 9-12b):

EIν′′′ = –\frac{P}{2}     (0<x<L)     (g)

EIν′′′ = P    (L<x<\frac{3L}{2})      (h)

Bending moments in the beam. Integration of the preceding two equations yields the bending-moment equations:

M = EIν′′′ = –\frac{Px}{2}+C_{1}     (0≤x≤L)     (i)

M = EIν′′′ =Px+C_{2}    (L<x<\frac{3L}{2})     (j)

The bending moments at points A and C are zero; hence we have the following boundary conditions:

ν″(0) = 0        ν″\left(\frac{3L}{2}\right) = 0

Using these conditions with Eqs. (i) and ( j), we get

C_{1} = 0      C_{2} = -\frac{3PL}{2}

Therefore, the bending moments are

M = EIν′′′ = –\frac{Px}{2}     (0≤x≤L)     (9-47a)

M = EIν′′′ = –\frac{P(3L-2x)}{2}    (L<x<\frac{3L}{2})     (9-47b)

These equations can be verified by determining the bending moments from freebody diagrams and equations of equilibrium.
Slopes and deflections of the beam. The next integrations yield the slopes:

EIν′ = -\frac{Px^{2}}{4}+C_{3}    (0≤x≤L)

EIν′ = –\frac{Px(3L-x)}{2}+C_{4}      (L<x<\frac{3L}{2})

The only condition on the slopes is the continuity condition at support B. According to this condition, the slope at point B as found for part AB of the beam is equal to the slope at the same point as found for part BC of the beam. Therefore, we substitute x = L into each of the two preceding equations for the slopes and obtain

This equation eliminates one constant of integration because we can express C_{4} in terms of C_{3}:

C_{4} = C_{3}+\frac{3PL^{2}}{4}     (k)

The third and last integrations give

EIν = -\frac{Px^{3}}{12}+C_{3}x+C_{5}    (0≤x≤L)     (l)

EIν = –\frac{Px^{2}(9L-2x)}{12}+C_{4}x+C_{6}      (L<x<\frac{3L}{2})     (m)

For part AB of the beam (Fig. 9-15a), we have two boundary conditions on the deflections, namely, the deflection is zero at points A and B:

v(0) = 0  and  v(L) = 0

Applying these conditions to Eq. (l), we obtain

C_{5} = 0     C_{3} = \frac{PL^{2}}{12}     (n,o)

Substituting the preceding expression for C_{3} in Eq. (k), we get

C_{4} = \frac{5PL^{2}}{6}     (p)

For part BC of the beam, the deflection is zero at point B. Therefore, the boundary condition is

v(L) = 0

Applying this condition to Eq. (m), and also substituting Eq. (p) for C_{4}, we get

C_{6} = -\frac{5PL^{3}}{4}     (q)

All constants of integration have now been evaluated.
The deflection equations are obtained by substituting the constants of integration (Eqs. n, o, p, and q) into Eqs. (l) and (m). The results are

v = \frac{Px}{12EI}(L^{2}-x^{2})    (0≤x≤L)     (9-48a)

v = -\frac{P}{12EI}(3L^{3}-10L^{2}x+9Lx^{2}-2x^{3})    (L<x<\frac{3L}{2})     (9-48b)

Note that the deflection is always positive (upward) in part AB of the beam
(Eq. 9-48a) and always negative (downward) in the overhang BC (Eq. 9-48b).

Deflection at the end of the overhang. We can find the deflection \delta_{C} at the end of the overhang (Fig. 9-15b) by substituting x = 3L/2 in Eq. (9-48b):

\delta_{C} = -v\left(\frac{3L}{2} \right) = \frac{PL^{3}}{2EI}     (9-49)

Thus, we have determined the required deflections of the overhanging beam (Eqs. 9-48 and 9-49) by solving the third-order differential equation of the deflection curve.