Question 9.2: Determine the equation of the deflection curve for a cantile......

Bending moment in the beam. The bending moment at distance x from the fixed support is obtained from the free-body diagram of Fig. 9-11. Note that the vertical reaction at the support is equal to qL and the moment reaction is equal to qL²/2. Consequently, the expression for the bending moment M is

M=-\frac{qL^2}{2}+qLx\ -\ \frac{qx^2}{2}                            (9-21)

Differential equation of the deflection curve. When the preceding expression for the bending moment is substituted into the differential equation (Eq. 9-12a), we obtain

EIv″=M                      (9-12a)

EIv″=-\frac{qL^2}{2}+qLx\ -\ \frac{qx^2}{2}                              (9-22)

We now integrate both sides of this equation to obtain the slopes and deflections.
Slope of the beam. The first integration of Eq. (9-22) gives the following equation for the slope:

EIv^′=-\frac{qL^2x}{2}+\frac{qLx^2}{2}\ -\ \frac{qx^3}{6}+C_1                              (e)

The constant of integration C| can be found from the boundary condition that the slope of the beam is zero at the support; thus, we have the following condition:

v′(0) = 0

When this condition is applied to Eq. (e) we get C_1 = 0. Therefore. Eq. (e) becomes

EIv^′=-\frac{qL^2x}{2}+\frac{qLx^2}{2}\ -\ \frac{qx^3}{6}                              (f)

and the slope is

v^′=-\frac{qx}{6EI}(3L^2\ -\ 3Lx+x^2)                              (9-23)

As expected, the slope is zero at the support (x = 0) and negative (i.e.. clockwise) throughout the length of the beam.
Deflection of the beam. Integration of the slope equation (Eq. f) yields

EIv=-\frac{qL^2x^2}{4}+\frac{qLx^3}{6}\ -\ \frac{qx^4}{24}+C_2                              (g)

The constant C_2 is found from the boundary condition that the deflection of the beam is zero at the support:

v(0) = 0

When this condition is applied to Eq. (g). we see immediately that C_2 = 0. Therefore, the equation for the deflection v is

v=-\frac{qx^2}{24EI}(6L^2\ -\ 4Lx+x^2)                        (9-24)

As expected, the deflection v is zero at the support (x = 0) and negative (that is, downward) elsewhere.
Angle of rotation at the free end of the beam. The clockwise angle of rotation θ_B at end B of the beam (Fig. 9-10b) is equal to the negative of the slope at that point. Thus, using Eq. (9-23). we get

θ_B=-v^′(L)=\frac{qL^3}{6EI}                                  (9-25)

This angle is the maximum angle of rotation for the beam.
Deflection at the free end of the beam. Since the deflection δ_B is downward (Fig. 9-10b), it is equal to the negative of the deflection obtained from Eq. (9-24):

δ_B=-v(L)=\frac{qL^4}{8EI}                           (9-26)

This deflection is the maximum deflection of the beam.