Question 9.13: A beam ABCDE on simple supports is constructed from a wide-f......

Differential equations of the deflection curve. In this example we will determine the slopes and deflections of the beam by integrating the bending-moment equation, that is, the second-order differential equation of the deflection curve (Eq. 9-12a). Since the reaction at each support is P/2, the bending moment throughout the left-hand half of the beam is

EIv″ = M                                  (9-12a)

M=\frac{Px}{2}\quad \quad \quad \left(0≤x≤\frac{L}{2}\right)                                   (a)

Therefore, the differential equations for the left-hand half of the beam are

EIv″=\frac{Px}{2}\quad \quad \quad \left(0≤x≤\frac{L}{2}\right)                                  (b)

E(2I)v″=\frac{Px}{2}\quad \quad \quad \left(\frac{L}{4}≤x≤\frac{L}{2}\right)                                  (c)

Each of these equations can be integrated twice to obtain expressions for the slopes and deflections in their respective regions. These integrations produce four constants of integration that can be found from the following four conditions:

(1) Boundary condition: At support A (x = 0), the deflection is zero (v = 0).
(2) Symmetry condition: At point C (x = L/2), the slope is zero (v′ = 0).
(3) Continuity condition: At point B (x = L/4), the slope obtained from part
AB of the beam is equal to the slope obtained from part BC of the beam.
(4) Continuity condition: At point B (x = L/4), the deflection obtained from part AB of the beam is equal to the deflection obtained from part BC of the beam.

Slopes of the beam. Integrating each of the differential equations (Eqs. b and c), we obtain the following equations for the slopes in the left-hand half of the beam:

v^′=\frac{Px^2}{4EI}+C_1\quad \quad \quad \left(0≤x≤\frac{L}{4}\right)                                  (d)

v^′=\frac{Px^2}{8EI}C_2\quad \quad \quad \left(\frac{L}{4}≤x≤\frac{L}{2}\right)                                  (e)

Applying the symmetry condition (2) to Eq. (e), we obtain the constant C_2:

C_2=-\frac{PL^2}{32EI}

Therefore, the slope of the beam between points B and C (from Eq. e) is

v^′=\frac{P}{32EI}(L^2\ -\ 4x^2)\quad \quad \quad \left(\frac{L}{4}≤x≤\frac{L}{2}\right)                                  (9-70)

From this equation we can find the slope of the deflection curve at point B where the moment of inertia changes from I to 21:

v^′\left(\frac{L}{4}\right)=-\frac{3PL^2}{128EI}                     (f)

Because the deflection curve is continuous at point B, we can use the continuity condition (3) and equate the slope at point B as obtained from Eq. (d) to the slope at the same point given by Eq. (f). In this manner we find the constant C_1:

\frac{P}{4EI}\left(\frac{L}{4}\right)^2+C_1=-\frac{3PL^2}{128EI}\quad \quad or\quad \quad C_1=-\frac{5PL^2}{128EI}

Therefore, the slope between points A and B (see Eq. d) is

v^′=-\frac{P}{128EI}(5L^2\ -\ 32x^2)\quad \quad \left(0≤x≤\frac{L}{4}\right)                    (9-71)

At support A, where x = 0, the angle of rotation (Fig. 9-28c) is

θ_A=-v^′(0)=\frac{5PL^2}{128EI}                            (9-72)

Deflections of the beam. Integrating the equations for the slopes (Eqs. 9-71 and 9-70). we get

v=-\frac{P}{128EI}\left(5L^2x\ -\ \frac{32x^3}{3}\right)+C_3\quad \quad \left(0≤x≤\frac{L}{4}\right)                    (g)

v=-\frac{P}{32EI}\left(L^2x\ -\ \frac{4x^3}{3}\right)+C_4\quad \quad \left(\frac{L}{4}≤x≤\frac{L}{2}\right)                    (h)

Applying the boundary condition at the support (condition 1) to Eq. (g), we get C_3 = 0. Therefore, the deflection between points A and B (from Eq. g) is

v=-\frac{Px}{384EI}(15L^2\ -\ 32x^2)\quad \quad \left(0≤x≤\frac{L}{4}\right)                    (9-73)

From this equation we can find the deflection at point B:

v\left(\frac{L}{4}\right)=-\frac{13PL^3}{1536EI}                       (i)

Since the deflection curve is continuous at point B. we can use the continuity condition (4) and equate the deflection at point B as obtained from Eq. (h) to the deflection given by Eq. (i):

-\frac{P}{32EI}\left[L^2\left(\frac{L}{4}\right)\ -\ \frac{4}{3}\left(\frac{L}{4}\right)^3\right]+C_4=-\frac{13PL^3}{1536EI}

from which

C_4=-\frac{PL^3}{768EI}

Therefore, the deflection between points B and C (from Eq. h) is

v=-\frac{P}{768EI}(L^3+24L^2x\ -\ 32x^3)\quad \quad \left(\frac{L}{4}≤x≤\frac{L}{2}\right)                    (9-74)

Thus, we have obtained the equations of the deflection curve for the left-hand half of the beam. (The deflections in the right-hand half of the beam can be obtained from symmetry.)
Finally, we obtain the deflection at the midpoint C by substituting x = L/2 into Eq. (9-74):

δ_C=-v\left(\frac{L}{2}\right)=\frac{3PL^3}{256EI}                        (9-75)

All required quantities have now been found, and the analysis of the nonprismatic beam is completed.
Notes: Using the differential equation for finding deflections is practical only if the number of equations to be solved is limited to one or two and only if the integrations are easily performed, as in this example. In the case of a tapered beam (Fig. 9-27), it may be difficult to solve the differential equation analytically because the moment of inertia is a continuous function of .x. In such a case, the differential equation has variable coefficients instead of constant coefficients.

When a beam has abrupt changes in cross-sectional dimensions, as in this example, there are stress concentrations at the points where changes occur. However. because the stress concentrations affect only a small region of the beam, they have no noticeable effect on the deflections.