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 qL² / 2. Consequently, the expression for the bending moment M is

M=-\frac{q L^{2}}{2}+q L x-\frac{q x^{2}}{2}             (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.

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

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:

E I v^{\prime \prime}=M              (9-16a)

E I v^{\prime \prime}=-\frac{q L^{2}}{2}+q L x-\frac{q x^{2}}{2}           (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:

E I v^{\prime}=-\frac{q L^{2} x}{2}+\frac{q L x^{2}}{2}-\frac{q x^{3}}{6}+C_{1}           (a)

The constant of integration C_{I} can be found from the boundary condition that the slope of the beam is zero at the support; which is expressed as

When this condition is applied to Eq. (a), the result is C_{1} = 0. Therefore, Eq. (a) becomes

E I v^{\prime}=-\frac{q L^{2} x}{2}+\frac{q L x^{2}}{2}-\frac{q x^{3}}{6}             (b)

and the slope is

v^{\prime}=-\frac{q x}{6 E I}\left(3 L^{2}-3 L x+x^{2}\right)            (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

E I v=-\frac{q L^{2} x^{2}}{4}+\frac{q L x^{3}}{6}-\frac{q x^{4}}{24}+C_{2}               (c)

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. (c), the result is C_{2} = 0. Therefore, the equation for the deflection v is

v=-\frac{q x^{2}}{24 E I}\left(6 L^{2}-4 L x+x^{2}\right)          (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 \theta_{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

\theta_{B}=-v^{\prime}(L)=\frac{q L^{3}}{6 E I}            (9-31)

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. 9-10b), it is equal to the negative of the deflection obtained from Eq. (9-30):

\delta_{B}=-v(L)=\frac{q L^{4}}{8 E I}            (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.

Table H-1
Deflections and Slopes of Cantilever Beams
	Notation:
	v = deflection in the y direction (positive upward)
	v′ = dv/dx = slope of the deflection curve
	\delta_{B}=-v(L)= deflection at end B of the beam (positive downward)
	\theta_{B}=-v^{\prime}(L)= angle of rotation at end B of the beam (positive clockwise)
	EI = constant
	v=-\frac{q x^{2}}{24 E I}\left(6 L^{2}-4 L x+x^{2}\right)   \quad v^{\prime}=\frac{q x}{6 E I}\left(3 L^{2}-3 L x+x^{2}\right)
	\delta_{B}=\frac{q L^{4}}{8 E I}   \quad \theta_{B}=\frac{q L^{3}}{6 E I}
	v=-\frac{q x^{2}}{24 E I}\left(6 a^{2}-4 a x+x^{2}\right)   \quad(0 \leq x \leq a)
	v^{\prime}=-\frac{q x}{6 E I}\left(3 a^{2}-3 a x+x^{2}\right)   \quad(0 \leq x \leq a)
	v=-\frac{q a^{3}}{24 E I}(4 x-a) \quad v^{\prime}=-\frac{q a^{3}}{6 E I}   \quad(a \leq x \leq L)
	At  x=a: v=-\frac{q a^{4}}{8 E I}   \quad v^{\prime}=-\frac{q a^{3}}{6 E I}
	\delta_{B}=\frac{q a^{3}}{24 E I}(4 L-a)   \quad \theta_{B}=\frac{q a^{3}}{6 E I}
	v=-\frac{q b x^{2}}{12 E I}(3 L+3 a-2 x)          (0 \leq x \leq a)
	v^{\prime}=-\frac{q b x}{2 E I}(L+a-x)            (0 \leq x \leq a)
	v=-\frac{q}{24 E I}\left(x^{4}-4 L x^{3}+6 L^{2} x^{2}-4 a^{3} x+a^{4}\right)   \quad(a \leq x \leq L)
	v^{\prime}=-\frac{q}{6 E I}\left(x^{3}-3 L x^{2}+3 L^{2} x-a^{3}\right)  \quad(a \leq x \leq L)
	At x=a:   v=-\frac{q a^{2} b}{12 E I}(3 L+a)  \quad v^{\prime}=-\frac{q a b L}{2 E I}
	\delta_{B}=\frac{q}{24 E I}\left(3 L^{4}-4 a^{3} L+a^{4}\right)  \quad \theta_{B}=\frac{q}{6 E I}\left(L^{3}-a^{3}\right)
	v=-\frac{P x^{2}}{6 E I}(3 L-x)  \quad v^{\prime}=-\frac{P x}{2 E I}(2 L-x)
	\delta_{B}=\frac{P L^{3}}{3 E I}  \quad \theta_{B}=\frac{P L^{2}}{2 E I}
	v=-\frac{P x^{2}}{6 E I}(3 a-x) \quad v^{\prime}=-\frac{P x}{2 E I}(2 a-x)  \quad(0 \leq x \leq a)
	v=-\frac{P a^{2}}{6 E I}(3 x-a)  \quad v^{\prime}=-\frac{P a^{2}}{2 E I}  \quad(a \leq x \leq L)
	At x=a:  \quad v=-\frac{P a^{3}}{3 E I}  \quad v^{\prime}=-\frac{P a^{2}}{2 E I}
	\delta_{B}=\frac{P a^{2}}{6 E I}(3 L-a)  \quad \theta_{B}=\frac{P a^{2}}{2 E I}
	v=-\frac{M_{0} x^{2}}{2 E I} \quad v^{\prime}=-\frac{M_{0} x}{E I}
	\delta_{B}=\frac{M_{0} L^{2}}{2 E I}  \quad \theta_{B}=\frac{M_{0} L}{E I}
	v=-\frac{M_{0} x^{2}}{2 E I}  \quad v^{\prime}=-\frac{M_{0} x}{E I} \quad(0 \leq x \leq a)
	v=-\frac{M_{0} a}{2 E I}(2 x-a)  \quad v^{\prime}=-\frac{M_{0} a}{E I} \quad(a \leq x \leq L)
	At x=a:  \quad v=-\frac{M_{0} a^{2}}{2 E I}  \quad v^{\prime}=-\frac{M_{0} a}{E I}
	\delta_{B}=\frac{M_{0} a}{2 E I}(2 L-a)  \quad \theta_{B}=\frac{M_{0} a}{E I}
	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)
	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)
	\delta_{B}=\frac{q_{0} L^{4}}{30 E I} \quad \theta_{B}=\frac{q_{0} L^{3}}{24 E I}
	v=-\frac{q_{0} x^{2}}{120 L E I}\left(20 L^{3}-10 L^{2} x+x^{3}\right)
	v^{\prime}=-\frac{q_{0} x}{24 L E I}\left(8 L^{3}-6 L^{2} x+x^{3}\right)
	\delta_{B}=\frac{11 q_{0} L^{4}}{120 E I}  \quad \theta_{B}=\frac{q_{0} L^{3}}{8 E I}
	v=-\frac{q_{0} L}{3 \pi^{4} E I}\left(48 L^{3} \cos \frac{\pi x}{2 L}-48 L^{3}+3 \pi^{3} L x^{2}-\pi^{3} x^{3}\right)
	v^{\prime}=-\frac{q_{0} L}{\pi^{3} E I}\left(2 \pi^{2} L x-\pi^{2} x^{2}-8L^{2} \sin \frac{\pi x}{2 L}\right)
	\delta_{B}=\frac{2 q_{0} L^{4}}{3 \pi^{4} E I}\left(\pi^{3}-24\right)   \quad\theta_{B}=\frac{q_{0} L^{3}}{\pi^{3} E I}\left(\pi^{2}-8\right)