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. The reaction at the support is qL/2, and therefore the equation for the bending moment is

M=\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

Elv″=M                                 (9-12a)

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

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

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:

v^′ = 0        when         x=\frac{L}{2}

This condition may be expressed more succinctly as

v^′\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}\right)+C_1       or      C_1=-\frac{qL^3}{24}

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

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

or                                       v^′=-\frac{q}{24EI}(L^3\ -\ 6Lx^2+4x^3)              (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

EIv=\frac{qLx^3}{12}\ -\ \frac{qx^4}{24}\ -\ \frac{qL^3x}{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, v = 0 when x = 0, or

v(0) = 0

Applying this condition to Eq. (c) yields C_2 = 0; hence the equation for the deflection curve is

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

or                                    v=-\frac{qx}{24EI}(L^3\ -\ 2Lx^2+x^3)                   (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. The maximum deflection δ_{max} occurs at the midpoint of the span (Fig. 9-8b) and is obtained by setting x equal to L/2 in Eq. (9-17). However, since δ_{max} represents the magnitude of the maximum deflection, and since the deflection v is negative when downward, we must insert a minus sign in the equation, as follows:

δ_{max}=\left|v\left(\frac{L}{2}\right)\right|=-v\left(\frac{L}{2}\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 θ_A, which is a clockwise angle (Fig. 9-8b), is equal to the negative of the slope v′. Thus, by substituting x = 0 into Eq. (9-16), we find

θ_A=-v^′(0)=\frac{qL^3}{24EI}                           (9-19)

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

θ_B=v^′(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 δ_{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 θ_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.