Question 8.4: 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. 8-14a):

q=\frac{q_0(L-x)}{L}                    (8-40)

Consequently, the fourth-order differential equation (Eq. 8-12c) becomes

EIv''''=-q              (8-12c)

=-\frac{q_0(L-x)}{L}                      (a)

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

EIv'''=\frac{q_0}{2L} (L-x)^2+C_1                  (b)

The right-hand side of this equation represents the shear force V (see Eq. 8-12b). Because the shear force is zero at x = L, we have the following boundary condition:

EIv''' = V               (8-12b)

v'''(L) = 0

Using this condition with Eq. (b), we get C_1 = 0. Therefore, Eq. (b) simplifies to

EIv'''=\frac{q_0}{2L} (L-x)^2                    (c)

and the shear force in the beam is

V=EIv'''=\frac{q_0}{2L} (L-x)^2                      (8-41)

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

EIv''=-\frac{q_0}{6L} (L-x)^3+C_2                    (d)

This equation is equal to the bending moment M (see Eq. 8-12a). Since the bending moment is zero at the free end of the beam, we have the following boundary condition:

EIv'' = M                 (8-12a)

v''(L) = 0

Applying this condition to Eq. (d), we obtain C_2 = 0, and therefore the bending moment is

M=EIv''=-\frac{q_0}{6L} (L-x)^3                  (8-42)

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

EIv'=\frac{q_0}{24L} (L-x)^4+C_3                  (e)

EIv=-\frac{q_0}{120L} (L-x)^5+C_3x+C_4              (f)

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

v'(0) = 0               v(0) = 0

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

C_3=-\frac{q_0L^3}{24}                       C_4=\frac{q_0L^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:

v'=-\frac{q_0x}{24LEI} (4L^3-6L^2x+4Lx^2-x^3)                  (8-43)

v=-\frac{q_0x^2}{120LEI} (10L^3-10L^2x+5Lx^2-x^3)                (8-44)

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. 8-14b) are obtained from Eqs. (8-43) and (8-44), respectively, by substituting x = L. The results are

\theta _B=-v'(L)=\frac{q_0L^3}{24EI}                        \delta _B=-v(L)=\frac{q_0L^4}{30EI}                  (8-45 a,b)

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