Question 9.1: Determine the equation of the deflection curve for a simple ...

Bending moment in the beam. The bending moment at a cross section distance x from the left-hand support is obtained from the free-body diagram of Fig. 9-9. Since the reaction at the support is qL/2, the equation for the bending the moment is

M = \frac{qL}{2}\left(x\right)-qx\left(\frac{x}{2} \right) = \frac{qLx}{2}-\frac{qx^{2}}{2}          (9-14)

Differential equation of the deflection curve. By substituting the expression for the bending moment (Eq. 9-14) into the differential equation (Eq. 9-12a), we obtain

EIv″ = M     (9-12a)

EIv″ = \frac{qLx}{2}-\frac{qx^{2}}{2}     (9-15)

This equation can now be integrated to obtain the slope and deflection of the beam.
Slope of the beam. Multiplying both sides of the differential equation by dx, we get the following equation:

EIv″ dx = \frac{qLx}{2}dx-\frac{qx^{2}}{2}dx

Integrating each term, we obtain

EI\int{ν″ dx } = \int{\frac{qLx}{2}dx}-\int{\frac{qx^{2}}{2}dx}

or

EIv′ = \frac{qLx^{2}}{4}-\frac{qx^{3}}{6}+C_{1}     (a)

in which C_{1} is a constant of integration.
To evaluate the constant C_{1}, we observe from the symmetry of the beam and its load that the slope of the deflection curve at midspan is equal to zero. Thus, we have the following symmetry condition:

ν′ = 0 when x = \frac{L}{2}

This condition may be expressed more succinctly as

ν′\left(\frac{L}{2}\right) = 0

Applying this condition to Eq. (a) gives

0 = \frac{qL}{4}\left(\frac{L}{2} \right)^{2}-\frac{q}{6}\left(\frac{L}{2}^{3} \right)+C_{1}   or   C_{1}=-\frac{qL^{3}}{24}

The equation for the slope of the beam (Eq. a) then becomes

EIν′ = \frac{qLx^{2}}{4}-\frac{qx^{3}}{6}+-\frac{qL^{3}}{24}     (b)

or    ν′ = –\frac{q}{24EI} \left(L^{3}-Lx^{2}+4x^{3}\right)     (9-16)

As expected, the slope is negative (i.e., clockwise) at the left-hand end of the beam (x = 0), positive at the right-hand end (x = L), and equal to zero at the midpoint (x = L/2).
Deflection of the beam. The deflection is obtained by integrating the equation for the slope. Thus, upon multiplying both sides of Eq. (b) by dx and integrating, we obtain

EIν = \frac{qLx^{3}}{12}-\frac{qx^{4}}{24}+-\frac{qL^{3}x}{24}+C_{2}     (c)

The constant of integration C_{2} may be evaluated from the condition that the deflection of the beam at the left-hand support is equal to zero; that is, ν = 0 when x = 0, or

ν(0) = 0

Applying this condition to Eq. (c) yields C_{2} = 0; hence the equation for the deflection curve is

EIν = \frac{qLx^{3}}{12}-\frac{qx^{4}}{24}+-\frac{qL^{3}x}{24}     (d)

or    ν′ = –\frac{qx}{24EI} \left(L^{3}-2Lx^{2}+x^{3}\right)     (9-17)

This equation gives the deflection at any point along the axis of the beam. Note that the deflection is zero at both ends of the beam (x = 0 and x = L) and negative elsewhere (recall that downward deflections are negative).
Maximum deflection. From symmetry we know that the maximum deflection occurs at the midpoint of the span (Fig. 9-8b). Thus, setting x equal to L/2 in Eq. (9-17), we obtain

ν\left(\frac{L}{2} \right) = -\frac{5qL^{4}}{384EI}

in which the negative sign means that the deflection is downward (as expected). Since \delta_{max} represents the magnitude of this deflection, we obtain

\delta_{max} = \left|ν\left(\frac{L}{2} \right)\right| = \frac{5qL^{4}}{384EI}     (9-18)

Angles of rotation. The maximum angles of rotation occur at the supports
of the beam. At the left-hand end of the beam, the angle \theta_{A}, which is a clockwise angle (Fig. 9-8b), is equal to the negative of the slope ν′. Thus, by substituting x = 0 into Eq. (9-16), we find

\theta_{A} = -ν′(0) = \frac{qL^{3}}{24EI}      (9-19)

In a similar manner, we can obtain the angle of rotation \theta_{B} at the right-hand end of the beam. Since \theta_{B} is a counterclockwise angle, it is equal to the slope at the end:

\theta_{B} = ν′(L) = \frac{qL^{3}}{24EI}      (9-20)

Because the beam and loading are symmetric about the midpoint, the angles of rotation at the ends are equal.
This example illustrates the process of setting up and solving the differential equation of the deflection curve. It also illustrates the process of finding slopes and deflections at selected points along the axis of a beam.
Note: Now that we have derived formulas for the maximum deflection and
maximum angles of rotation (see Eqs. 9-18, 9-19, and 9-20), we can evaluate
those quantities numerically and observe that the deflections and angles are indeed small, as the theory requires.
Consider a steel beam on simple supports with a span length L = 6 ft. The cross section is rectangular with width b = 3 in. and height h = 6 in. The intensity of uniform load is q = 8000 lb/ft, which is relatively large because it produces a stress in the beam of 24,000 psi. (Thus, the deflections and slopes are larger than would normally be expected.)
Substituting into Eq. (9-18), and using E = 30 × 10^{6} psi, we find that the maximum deflection is \delta_{max} = 0.144 in., which is only 1/500 of the span length. Also, from Eq. (9-19), we find that the maximum angle of rotation is \theta_{A} = 0.0064 radians, or 0.37° , which is a very small angle.
Thus, our assumption that the slopes and deflections are small is validated.