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 qL²/2. 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 (Eq. 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^2x}{2}+\frac{qLx^2}{2}- \frac{qx^3}{6}+ C_1 (e)
The constant of integration C_l can be found from the boundary condition that the slope 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^2x}{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^2x^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.
