Question 8.2: Determine the equation of the deflection curve for a cantile...
Determine the equation of the deflection curve for a cantilever beam AB subjected to a uniform load of intensity q (Fig. 8-10a).
Also, determine the angle of rotation \theta _{B} and the deflection \delta _{B} at the free end (Fig. 8-10b). (Note: The beam has length L and constant flexural rigidity EI.)
Bending moment in the beam. The bending moment at distance x from the fixed support is obtained from the free-body diagram of Fig. 8-11. Note that the vertical reaction at the support is equal to qL and the moment reaction is equal to qL^{2} /2. Consequently, the expression for the bending moment M is
M=-\frac{qL^{2} }{2}+qLx-\frac{qx^{2} }{2} (8-21)
Differential equation of the deflection curve. When the preceding expression for the bending moment is substituted into the differential equation (Eq. 8-12a),
we obtain
EIv^{′′}=M (8-12a)
EIv^{′′}=-\frac{qL^{2}}{2} +qLx-\frac{qx^{2}}{2} (8-22)
We now integrate both sides of this equation to obtain the slopes and deflections.
Slope of the beam. The first integration of Eq. (8-22) gives the following equation for the slope:
EIv^{′}=-\frac{qL^{2}x}{2} +\frac{qLx^{2}}{2}-\frac{qx^{3}}{6}+C_{1} (e)
The constant of integration C_1 can be found from the boundary condition that the of the beam is zero at the support; thus, we have the following condition:
v^{′}(0) = 0
When this condition is applied to Eq. (e) we get C_{1}=0. Therefore, Eq. (e) becomes
EIv^{′}=-\frac{qL^{2}x}{2} +\frac{qLx^{2}}{2}-\frac{qx^{3}}{6} (f)
and the slope is
v^{′}=-\frac{qx}{6EI} (3L^{2}-3Lx+x^{2}) (8-23)
As expected, the slope obtained from this equation is zero at the support (x = 0) and negative (i.e., clockwise) throughout the length of the beam.
Deflection of the beam. Integration of the slope equation (Eq. f) yields
EIv=-\frac{qL^{2}x^{2}}{4}+\frac{qLx^{3}}{6}-\frac{qx^{4}}{24}+C_2 (g)
The constant C_2 is found from the boundary condition that the deflection of the beam is zero at the support:
v(0) = 0
When this condition is applied to Eq. (g), we see immediately that C_{2}=0.
Therefore, the equation for the deflection v is
v=-\frac{qx^{2}}{24EI}(6L^{2}-4Lx+x^{2}) (8-24)
As expected, the deflection obtained from this equation is zero at the support (x = 0) and negative (that is, downward) elsewhere.
Angle of rotation at the free end of the beam. The clockwise angle of rotation \theta_B at end B of the beam (Fig. 8-10b) is equal to the negative of the slope at that point. Thus, using Eq. (8-23), we get
\theta _{B}=-v^{′}(L)=\frac{qL^{3}}{6EI} (8-25)
This angle is the maximum angle of rotation for the beam.
Deflection at the free end of the beam. Since the deflection \delta _B is downward (Fig. 8-10b), it is equal to the negative of the deflection obtained from Eq. (8-24):
\delta _{B}=-v(L)=\frac{qL^{4}}{8EI} (8-26)
This deflection is the maximum deflection of the beam.
