Determine the equation of the deflection curve for a cantilever beam AB supporting a triangularly distributed load of maximum intensity $q_{0}$ (Fig. 9-14a).
Also, determine the deflection $\delta_{B}$ and angle of rotation $\theta_{B}$ at the free end (Fig. 9-14b). Use the fourth-order differential equation of the deflection curve (the load equation). (Note: The beam has length L and constant flexural rigidity EI.)

Question

Determine the equation of the deflection curve for a cantilever beam AB supporting a triangularly distributed load of maximum intensity [latex]q_{0}[/latex] (Fig. 9-14a).
Also, determine the deflection [latex]\delta_{B}[/latex] and angle of rotation [latex] heta_{B}[/latex] at the free end (Fig. 9-14b). Use the fourth-order differential equation of the deflection curve (the load equation). (Note: The beam has length L and constant flexural rigidity EI.)

Accepted Answer

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

[latex]q=\frac{q_{0}(L-x)}{L}[/latex] (9-40)

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

[latex]EI u^{\prime\prime\prime\prime}=-q=-\frac{q_{0}(L-x)}{L}[/latex] (a)

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

[latex]EI u^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}+C_{1}[/latex] (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:

[latex]EI u^{\prime\prime\prime}=V[/latex] (9-12b)

ν′′′(L) = 0

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

[latex]EI u^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}[/latex] (c)

and the shear force in the beam is

[latex]V=EI u^{\prime\prime\prime}=\frac{q_{0}}{2L}(L-x)^{2}[/latex] (9-41)

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

[latex]EI u^{\prime\prime}=-\frac{q_{0}}{6L}(L-x)^{3}+C_{2}[/latex] (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:

[latex]EI u^{\prime\prime}=M[/latex] (9-12a)

ν″(L) = 0

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

[latex]M=EI u^{\prime\prime}=-\frac{q_{0}}{6L}(L-x)^{3}[/latex] (9-42)

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

[latex]EI u^{\prime}=\frac{q_{0}}{24L}(L-x)^{4}+C_{3}[/latex] (e)

[latex]EI u=-\frac{q_{0}}{120L}(L-x)^{5}+C_{3}x+C_{4}[/latex] (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

[latex]C_{3}=-\frac{q_{0}L^{3}}{24}[/latex] [latex]C_{4}=\frac{q_{0}L^{4}}{120}[/latex]

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

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

[latex] u=-\frac{q_{0}x^{2}}{120LEI}(10L^{3}-10L^{2}x+5Lx^{2}-x^{3})[/latex] (9-44)

Angle of rotation and deflection at the free end of the beam. The angle of rotation [latex] heta_{B}[/latex] and deflection [latex]\delta_{B}[/latex] 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

[latex] heta_{B}=- u^{\prime}(L)=\frac{q_{0}L^{3}}{24EI}[/latex] [latex]\delta_{B}=- u(L)=\frac{q_{0}L^{4}}{30EI}[/latex] (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.

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