Question 9.2: Determine the equation of the deflection curve for a cantile......

Use a four-step problem-solving approach.
1. Conceptualize: The beam is statically determinate. Begin by finding reaction force R_A and reaction moment M_A. The left hand free-body diagram in Fig. 9-11 is then used to obtain an expression for internal moment M(x).
Bending moment in the beam: The bending moment at distance x from the fixed support is obtained from the free-body diagram of Fig. 9-11. 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
\quad\quad\quad\quad M=-\frac{q{L}^{2}}{2}+q{L}x-\frac{q x^{2}}{2}\quad\quad(9-27)
2. Categorize: Use the expression for internal moment M(x) in the bending moment equation (Eq. 9-14a) to find expressions for slopes and deflections for this beam.
Differential equation of the deflection curve: When the preceding expression for the bending moment is substituted into the differential equation (Eq. 9-16a), the following equation is obtained:

\quad\quad\quad\quad E I{\frac{d^{2}\nu}{d\ x^{2}}}=M\quad\quad(9-14a)

\quad\quad\quad\quad EIv^{\prime\prime}=M\quad\quad(9-16a)

\quad\quad\quad\quad EI\nu^{\prime\prime}=-{\frac{q L^{2}}{2}}+{{q}}{L}x-{\frac{q x^{2}}{2}}\quad\quad(9-28)
3. Analyze: Now integrate both sides of this equation to obtain the slopes and deflections.
Slope of the beam: The first integration of Eq. (9-28) gives the following equation for the slope:
\quad\quad\quad\quad {E}I\nu^{\prime}=-\frac{q{L}^{2}x}{2}+\frac{q{L}x^{2}}{2}-\frac{q x^{3}}{6}+C_{1}\quad\quad(a)
The constant of integration C_1 can be found from the boundary condition that the slope of the beam is zero at the support; which is expressed as
\quad\quad\quad\quad v^\prime(0) = 0
When this condition is applied to Eq. (a), the result is C_1 = 0. Therefore, Eq. (a) becomes
\quad\quad\quad\quad E I{\nu}{'} = -{\frac{q L^{2}x}{2}}+{\frac{q L x^{2}}{2}}-\,{\frac{q x^{3}}{6}}\quad\quad(b)
and the slope is
\quad\quad\quad\quad \nu^{\prime}=-{\frac{q x}{6E I}}(3L^{2}\,-\,3L x+x^{2})\quad\quad(9-29)
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. (b)] yields
\quad\quad\quad\quad E I\nu = -{\frac{q L^{2}x^{2}}{4}}+{\frac{q L x^{3}}{6}}-\,{\frac{q x^{4}}{24}}+\,C_{2}\quad\quad(c)
The constant C_2 is found from the boundary condition that the deflection of the beam is zero at the support:
\quad\quad\quad\quad {v\,(0)}= 0
When this condition is applied to Eq. (c), the result is C_2 = 0. Therefore, the equation for the deflection v is
\quad\quad\quad\quad \displaystyle{v\,=-{\frac{q x^{2}}{24E I}}(6L^{2}\,-\,4\,{L}x\,+\,x^{2}\,)}\quad\quad(9-30)
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 θ_B at end B of the beam (Fig. 9-10b) is equal to the negative of the slope at that point. Thus, use Eq. (9-29) to get
\quad\quad\quad\quad \theta_{B}=-{\nu}^{\prime}({L})={\frac{q L^{3}}{6{E}I}}\quad\quad(9-31)
This angle is the maximum angle of rotation for the beam.
Deflection at the free end of the beam: Since the deflection δ_B is downward (Fig. 9-10b), it is equal to the negative of the deflection obtained from Eq. (9-30):
\quad\quad\quad\quad \delta_{B}=-\nu({L})=\frac{q{L}^{4}}{8EI}\quad\quad(9-32)
This deflection is the maximum deflection of the beam.
4. Finalize: Equations (9-29) to (9-32) are listed as Case 1 in Table H-1, Appendix H.