Question 9.4: Determine the equation of the deflection curve for a cantile...
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 δ_{B} and angle of rotation θ_{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.
Use a four-step problem-solving approach.
1. Conceptualize: The cantilever beam is statically determinate. Use the load equation to find the deflection curve of the beam. An expression for load intensity q(x) can be obtained as shown in Fig. 9-14a.
Differential equation of the deflection curve: The intensity of the distributed load is given by (see Fig. 9-14a):
q(x)=\frac{q_{0}(L-x)}{L} (9-46)
2. Categorize: Consequently, the fourth-order differential equation [Eq. (9-16c)] becomes
E I v^{\prime \prime}=M \quad E I v^{\prime \prime \prime}=V \quad E I v^{\prime \prime \prime \prime}=-q (9-16a,b,c)
E I v^{\prime \prime \prime \prime}=-q(x)=-\frac{q_{0}(L-x)}{L} (a)
3. Analyze:
Shear force in the beam: The first integration of Eq. (a) gives
E I v^{\prime \prime \prime}=\frac{q_{0}}{2 L}(L-x)^{2}+C_{1} (b)
The right-hand side of this equation represents the shear force V [see Eq. (9-16b)]. Because the shear force is zero at x = L, the boundary condition is:
v^{\prime \prime \prime}(L)=0Use this condition with Eq. (b), to get C_{1} = 0. Therefore, Eq. (b) simplifies to
E I v^{\prime \prime \prime}=\frac{q_{0}}{2 L}(L-x)^{2} (c)
and the shear force in the beam is
V=E I v^{\prime \prime \prime}=\frac{q_{0}}{2 L}(L-x)^{2} (9-47)
Bending moment in the beam: Integrate a second time to obtain the following equation from Eq. (c):
E I v^{\prime \prime}=-\frac{q_{0}}{6 L}(L-x)^{3}+C_{2} (d)
This equation is equal to the bending moment M [see Eq. (9-16a)]. The bending moment is zero at the free end of the beam, so the boundary condition is
v^{\prime \prime}(L)=0Apply this condition to Eq. (d) to obtain C_{2} = 0, so the bending moment is
M=E I v^{\prime \prime}=-\frac{q_{0}}{6 L}(L-x)^{3} (9-48)
Slope and deflection of the beam: The third and fourth integrations yield
E I v^{\prime}=\frac{q_{0}}{24 L}(L-x)^{4}+C_{3} (e)
E I v=-\frac{q_{0}}{120 L}(L-x)^{5}+C_{3} x+C_{4} (f)
The boundary conditions at the fixed support, where both the slope and deflection are equal to zero, are
v^{\prime}(0)=0 \quad v(0)=0Apply these conditions to Eqs. (e) and (f), respectively, to find
C_{3}=-\frac{q_{0} L^{3}}{24} \quad C_{4}=\frac{q_{0} L^{4}}{120}Substitute these expressions for the constants into Eqs. (e) and (f) to obtain the equations for the slope and deflection of the beam:
v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(4 L^{3}-6 L^{2} x+4 L x^{2}-x^{3}\right) (9-49)
v=-\frac{q_{0} x^{2}}{120 L E I}\left(10 L^{3}-10 L^{2} x+5 L x^{2}-x^{3}\right) (9-50)
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. (9-49) and (9-50), respectively, by substituting x = L. The results are
\theta_{B}=-v^{\prime}(L)=\frac{q_{0} L^{3}}{24 E I} \quad \delta_{B}=-v(L)=\frac{q_{0} L^{4}}{30 E I} (9-51a,b)
4. Finalize: The required slopes and deflections of the beam were found by solving the fourth-order differential equation of the deflection curve. This was shown to be an efficient approach for use with the linearly varying distributed load.