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

EIv″″ = -q                             (9-12c)

EIv″″=-q=-\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. 9-12b). Because the shear force is zero at x = L, we have the following boundary condition:

EIv′″ = V                             (9-12b)

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=\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):

EIv″=-\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:

EIv″ = M                             (9-12a)

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                              (9-42)

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) =              v(0) = 0

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

C_3=-\frac{q_0L^3}{24}\quad \quad 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)            (9-43)

v=-\frac{q_0x^2}{120LEI}(10L^3\ -\ 10L^2x+5Lx^2\ -\ x^3)            (9-44)

Angle of rotation and deflection at the free end of the beam. The angle of rotation θ_B and deflection δ_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

θ_B=-v^′(L)=\frac{q_0L^3}{24EI}\quad \quad δ_B=-v(L)=\frac{q_0L^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.