Question 10.1: A propped cantilever beam AB of length L supports a uniform ...

Because the load on this beam acts in the vertical direction (Fig. 10-6), we conclude that there is no horizontal reaction at the fixed support. Therefore, the beam has three unknown reactions \left(M_{A} , R_{A} , \text { and } R_{B}\right). Only two equations of equilibrium are available for determining these reactions, and therefore, the beam is statically indeterminate to the first degree.

Since we will be analyzing this beam by solving the bending-moment equation, we must begin with a general expression for the moment. This expression will be in terms of both the load and the selected redundant.

Redundant reaction. Let us choose the reaction R_{B} at the simple support as the redundant. Then, by considering the equilibrium of the entire beam, we can express the other two reactions in terms of R_{B}:

R_{A}=q L-R_{B} \quad M_{A}=\frac{q L^{2}}{2}-R_{B} L    (a,b)

Bending moment. The bending moment M at distance x from the fixed support can be expressed in terms of the reactions as

M=R_{A} x-M_{A}-\frac{q x^{2}}{2}    (c)

This equation can be obtained by the customary technique of constructing a free-body diagram of part of the beam and solving an equation of equilibrium.
Substituting into Eq. (c) from Eqs. (a) and (b), we obtain the bending moment in terms of the load and the redundant reaction:

M=q L x-R_{B} x-\frac{q L^{2}}{2}+R_{B} L-\frac{q x^{2}}{2}    (d)

Differential equation. The second-order differential equation of the deflection curve [Eq. (9-16a) EIv^{\prime \prime}=M] now becomes

EIv^{\prime \prime}=M=q L x-R_{B} x-\frac{q L^{2}}{2}+R_{B} L-\frac{q x^{2}}{2}  (e)

After two successive integrations, we obtain the following equations for the slopes and deflections of the beam:

EIv^{\prime}=\frac{q L x^{2}}{2}-\frac{R_{B} x^{2}}{2}-\frac{q L^{2} x}{2}+R_{B} L_{x}-\frac{q x^{3}}{6}+C_{1}  (f)

EIv=\frac{q L x^{3}}{6}-\frac{R_{B} x^{3}}{6}-\frac{q L^{2} x^{2}}{4}+\frac{R_{B} L x^{2}}{2}-\frac{q x^{4}}{24}+C_{1} x+C_{2}  (g)

These equations contain three unknown quantities \left(C_{1}, C_{2}, \text { and } R_{B}\right).
Boundary conditions. Three boundary conditions pertaining to the deflections and slopes of the beam are apparent from an inspection of Fig. 10-6. These conditions are as follows: (1) the deflection at the fixed sup-port is zero, (2) the slope at the fixed support is zero, and (3) the deflection at the simple support is zero. Thus

v(0) = 0     v′(0) = 0       v(L) = 0

Applying these conditions to the equations for slopes and deflections given in Eqs. (f) and (g), we find C_{1} = 0 , C_{2} = 0 , and

R_{B}=\frac{3 q L}{8}    (10-1)

Thus, the redundant reaction R_{B} is now known.
Reactions. With the value of the redundant established, we can find the remaining reactions from Eqs. (a) and (b). The results are

R_{A}=\frac{5 q L}{8} \quad M_{A}=\frac{q L^{2}}{8}    (10-2a,b)

Knowing these reactions, we can find the shear forces and bending moments in the beam.
Shear forces and bending moments. These quantities can be obtained by the usual techniques involving free-body diagrams and equations of equilibrium. The results are

V=R_{A}-q x=\frac{5 q L}{8}-q x  (10-3)

M=R_{A} x-M_{A}-\frac{q x^{2}}{2}=\frac{5 q L x}{8}-\frac{q L^{2}}{8}-\frac{q x^{2}}{2}    (10-4)

Shear-force and bending-moment diagrams for the beam can be drawn with the aid of these equations (see Fig. 10-7).
From the diagrams, we see that the maximum shear force occurs at the fixed support and is equal to

V_{\max }=\frac{5 q L}{8}    (10-5)

Also, the maximum positive and negative bending moments are

M_{\text {pos }}=\frac{9 q L^{2}}{128} \quad M_{\text {neg }}=-\frac{q L^{2}}{8}    (10-6 a,b)

Finally, we note that the bending moment is equal to zero at distance x = L/4 from the fixed support.
Slopes and deflections of the beam. Returning to Eqs. (f) and (g) for the slopes and deflections, we now substitute the values of the constants of  integration ( C_{1} = 0 and C_{2} = 0) as well as the expression for the redundant
R_{B} [Eq. (10-1)] and obtain

v^{\prime}=\frac{q x}{48 F I}\left(-6 L^{2}+15 L x-8 x^{2}\right)    (10-7)

v=-\frac{q x^{2}}{48 E I}\left(3 L^{2}-5 L x+2 x^{2}\right)    (10-8)

The deflected shape of the beam as obtained from Eq. (10-8) is shown in Fig. 10-8.
To determine the maximum deflection of the beam, we set the slope [Eq. (10-7)] equal to zero and solve for the distance x_{1} to the point where this deflection occurs:

from which

x_{1}=\frac{15-\sqrt{33}}{16} L=0.5785 L    (10-9)

Substituting this value of x into the equation for the deflection [Eq. (10-8)] and also changing the sign, we get the maximum deflection:

\begin{aligned}\delta_{\max } &=-(v)_{x=x_{1}}=\frac{q L^{4}}{65,536 E I}(39+55 \sqrt{33}) \\&=\frac{q L^{4}}{184.6 E I}=0.005416 \frac{q L^{4}}{E I}\end{aligned}  (10-10)

The point of inflection is located where the bending moment is equal to zero, that is, where x = L/4 . The corresponding deflection \delta_{0} of the beam [from Eq. (10-8)] is

\delta_{0}=-(v)_{x=L 4}=\frac{5 q L^{4}}{2048 EI}=0.002441 \frac{q L^{4}}{EI}    (10-11)

Note that when x < L/4 , both the curvature and the bending moment are negative, and when x > L/4 , the curvature and bending moment are positive.
To determine the angle of rotation \theta_{B} at the simply supported end of the beam, we use Eq. (10-7), as

\theta_{B}=\left(v^{\prime}\right)_{x-L}=\frac{q L^{3}}{48 E_{j}}    (10-12)

Slopes and deflections at other points along the axis of the beam can be obtained by similar procedures.

Note: In this example, we analyzed the beam by taking the reaction R_{B} (Fig. 10-6) as the redundant reaction. An alternative approach is to take the reactive moment M_{A} as the redundant. Then we can express the bending moment M in terms of M_{A}, substitute the resulting expression into the second-order differential equation, and solve as before. Still another approach is to begin with the fourth-order differential equation, as illustrated in the next example.