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, we conclude that there is no horizontal reaction at the fixed support. Therefore, the beam has three unknown reactions (M_A, R_A, and R_B). 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 bending moment. This expression will be in terms of both the load and the redundant reaction.
Redundant reaction. Let us choose the reaction R_B at the simple support as the redundant. Then, by considering the equihbrium of the entire beam, we can express the other two reactions in terms of R_B:

R_A=qL\ -\ R_B\quad \quad M_A=\frac{qL^2}{2}\ -\ R_BL                                     (a,b)

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

M=R_Ax\ -\ M_A\ -\ \frac{qx^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=qLx\ -\ R_Bx\ -\ \frac{qL^2}{2}+R_BL\ -\ \frac{qx^2}{2}                                 (d)

Differential equation. The second-order differential equation of the deflection curve (Eq. 9-12a) now becomes

EIv″=M=qLx\ -\ R_Bx\ – \ \frac{qL^2}{2}+R_BL\ -\ \frac{qx^2}{2}                                (e)

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

EIv^′=\frac{qLx^2}{2}\ -\ \frac{R_Bx^2}{2}\ -\ \frac{qL^2x}{2}+R_BLx\ -\ \frac{qx^3}{6}+C_1                      (f)

EIv=\frac{qLx^3}{6}\ -\ \frac{R_Bx^3}{6}\ -\ \frac{qL^2x^2}{4}+\frac{R_BLx^2}{2}\ -\ \frac{qx^4}{24}+C_1x+C_2                      (g)

These equations contain three unknown quantities (C_1, C_2, and R_B).
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 support 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 (Eqs. f and g), we find C_1 = 0, C_2 = 0, and

R_B=\frac{3qL}{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{5qL}{8}\quad \quad M_A=\frac{qL^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\ -\ qx=\frac{5qL}{8}\ -\ qx                    (10-3)

M=R_Ax\ -\ M_A\ -\ \frac{qx^2}{2}=\frac{5qLx}{8}\ -\ \frac{qL^2}{8}\ -\ \frac{qx^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{5qL}{8}                            (10-5)

Also, the maximum positive and negative bending moments are

M_{pos}=\frac{9qL^2}{128}\quad \quad M_{neg}=-\frac{qL^2}{8}                                (10-6a,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 = and C_2 = 0) as well as the expression for the redundant (Eq. 10-1) and obtain

v^′=\frac{qx}{48EI}(-6L^2+15Lx\ -\ 8x^2)                         (10-7)

v=-\frac{qx^2}{48EI}(3L^2\ -\ 15Lx\ -\ 2x^2)                         (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 δ_{max} of the beam, we set the slope (Eq. 10-7) equal to zero and solve for the distance x_1 to the point of maximum deflection:

v′ = 0         or           -6L² + 15Lx – 8x² = 0

from which

x_1=\frac{15\ -\ \sqrt{33}}{16}L = 0.5785L                      (10-9)

Substituting this value of x into the equation for the deflection (Eq. 10-8), we get

δ_{max}=-(v)_{x=x_1}=\frac{qL^4}{65,536EI}(39+55\sqrt{33})\\ =\frac{qL^4}{184.6EI}=0.005416\frac{qL^4}{EI}                           (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 of the beam (from Eq. 10-8) is

v_0=(v)_{x=L/4}=-\frac{5qL^4}{2048EI}=-0.002441\frac{qL^4}{EI}                       (10-11)

The negative sign means that the deflection is downward, as expected. Also, we note that when x < L/A, both the curvature and the bending moment are negative, and when x > L/A, the curvature and bending moment are positive.
To determine the angle of rotation θ_B at the simply supported end of the beam, we use Eq. (10-7), as follows:

θ_B=(v^′)_{x=L}=\frac{qL^3}{48EI}                         (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.