Question 9.4: Determine the equation of the deflection curve for a cantile...

Differential equation of the deflection curve. The intensity of the distributed load is given by the following equation (see Fig. 9-14a):

q=\frac{q_{0}(L-x)}{L}              (9-40)

Consequently, the fourth-order differential equation (Eq. 9-12c) becomes

EI\nu^{\prime\prime\prime\prime}=-q=-\frac{q_{0}(L-x)}{L}                  (a)

Shear force in the beam. The first integration of Eq. (a) gives

EI\nu^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}+C_{1}              (b)

The right-hand side of this equation represents the shear force V (see Eq. 9-12b). Because the shear force is zero at x = L, we have the following boundary condition:

EI\nu^{\prime\prime\prime}=V        (9-12b)

ν′′′(L) = 0

Using this condition with Eq. (b), we get C_{1} = 0. Therefore, Eq. (b) simplifies to

EI\nu^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}                (c)

and the shear force in the beam is

V=EI\nu^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}                  (9-41)

Bending moment in the beam. Integrating a second time, we obtain the following equation from Eq. (c):

EI\nu^{\prime\prime}=-\frac{q_{0}}{6L}(L-x)^{3}+C_{2}                (d)

This equation is equal to the bending moment M (see Eq. 9-12a). Since the bending moment is zero at the free end of the beam, we have the following boundary condition:

EI\nu^{\prime\prime}=M                (9-12a)

ν″(L) = 0

Applying this condition to Eq. (d), we obtain C_{2} = 0, and therefore the bending moment is

M=EI\nu^{\prime\prime}=-\frac{q_{0}}{6L}(L-x)^{3}                (9-42)

Slope and deflection of the beam. The third and fourth integrations yield

EI\nu^{\prime}=\frac{q_{0}}{24L}(L-x)^{4}+C_{3}                        (e)

EI\nu=-\frac{q_{0}}{120L}(L-x)^{5}+C_{3}x+C_{4}                      (f)

The boundary conditions at the fixed support, where both the slope and deflection equal zero, are

ν′(0) = 0            ν(0) = 0

Applying these conditions to Eqs. (e) and (f), respectively, we find

C_{3}=-\frac{q_{0}L^{3}}{24}                C_{4}=\frac{q_{0}L^{4}}{120}

Substituting these expressions for the constants into Eqs. (e) and (f ), we obtain the following equations for the slope and deflection of the beam:

\nu^{\prime}=-\frac{q_{0}x}{24LEI}(4L^{3}-6L^{2}x+4Lx^{2}-x^{3})              (9-43)

\nu=-\frac{q_{0}x^{2}}{120LEI}(10L^{3}-10L^{2}x+5Lx^{2}-x^{3})          (9-44)

Angle of rotation and deflection at the free end of the beam. The angle of rotation \theta_{B} and deflection \delta_{B} at the free end of the beam (Fig. 9-14b) are obtained from Eqs. (9-43) and (9-44), respectively, by substituting x = L. The results are

\theta_{B}=-\nu^{\prime}(L)=\frac{q_{0}L^{3}}{24EI}                  \delta_{B}=-\nu(L)=\frac{q_{0}L^{4}}{30EI}          (9-45a,b)

Thus, we have determined the required slopes and deflections of the beam by solving the fourth-order differential equation of the deflection curve.