Question 10.7: The continuous beam shown in Fig. 10-23a has three spans of ......

This beam rests on four simple supports (supports 1, 2, 3, and 4), and therefore it is statically indeterminate to the second degree. The bending moments M_2 and M_3 at supports 2 and 3. respectively, are the redundant quantities.
Since the moments of inertia and span lengths are the same for all three spans, we can use the three-moment equation in the form given by Eq. (10-39):

M_A+4M_B+M_C=-\frac{6EI}{L}(θ_{BA})_1\ -\ \frac{6EI}{L}(θ_{BC})_1                     (10-39)

M_A+4M_B+M_C=-\frac{6EI}{L}(θ_{BC})_1                     (a)

Now we will write this equation at supports 2 and 3.
Three-moment equation at support 2. The adjoining spans are 1-2 and 2-3, and therefore the moments M_A, M_B, and M_C in Eq. (a) become M_1,M_2, and M_3, respectively. Since M_1 = 0. only the moments M_2 and M_3 will appear in the equation.
The angle of rotation (θ_{BA})_1 is the rotation at the right-hand end of member 1-2 when it is treated as a simple beam. Thus, from Eq. (10-40a). we get

(θ_{BA})_1=\frac{qL_A^3}{24EI_A}                          (10-40a)

(θ_{BA})_1=(θ_{21})_1=\frac{qL^3}{24EI}                          (b)

Similarly, the angle (θ_{BC})_1 is the simple-beam rotation at the left-hand end of member 2-3. Since there is no load on member 2-3, we obtain

(θ_{BC})_1=(θ_{23})_1 = 0                               (c)

Thus, the three-moment equation for support 2 (see Eq. a) becomes

4M_2+M_3=-\frac{6EI}{L}\left(\frac{qL^3}{24EI}\right)=-\frac{qL^2}{4}                         (d)

Three-moment equation at support 3. At support 3 the moments M_A, M_B, and M_C in Eq. (a) become M_2, M_3, and M_4, respectively. Since M_4 = 0, only the moments M_2 and M_3 will appear in the equation.
The angle of rotation (θ_{BA})_1 is the angle at the right-hand end of member 2-3 when it is treated as a simple beam. Thus,

(θ_{BA})_1=(θ_{32})_1 = 0                                     (e)

The angle (θ_{BC})_1 is the rotation at the left-hand end of member 3-4. Since the load on member 3-4 is a concentrated load, we determine the angle of rotation from Case 5, Table G-2. Using the formula for the angle θ_A in that table, we obtain

(θ_{BC})_1=(θ_{34})_1=\frac{Pab(L+b)}{6LEI}=\frac{P(3L/4)(L/4)(L+L/4)}{6LEI}=\frac{5PL^2}{128EI}

Since the load P is equal to qL, this equation becomes

(θ_{BC})_1=(θ_{34})_1=\frac{5qL^3}{128EI}                             (f)

Thus, the three-moment equation for support 3 (Eq. a) becomes

M_2+4M_3=-\frac{6EI}{L}\left(\frac{5qL^3}{128EI}\right)=-\frac{15qL^2}{64}                        (g)

Solution of equations. To find the redundant bending moments, we solve simultaneously the three-moment equations (Eqs. d and g). The results are

M_2=-\frac{49qL^2}{960}\quad \quad M_3=-\frac{11qL^2}{240}                        (h,i)

Reactions. The reactions at each support can now be obtained by using free-body diagrams and equations of equilibrium, or by substituting directly into Eq. (10-42).
At support 1, the terms in Eq. (10-42) are as follows: (1) The simple-beam reaction for member AB is zero because span AB is nonexistent; (2) the simple-beam reaction for member BC, which is now member 1-2, is qL/2; (3) the moment M_A is zero because it is nonexistent; (4) the moment M_B is zero because it is the bending moment at support 1; and (5) the moment M_C is the bending moment M_2, which is given by Eq. (h). Thus, substituting term-by-term into Eq. (10-42), we get

R_B=\text{Simple-beam reaction of member AB due to the loads}\\ \quad +\text{Simple-beam reaction of member BC due to the loads}\\ \quad +\frac{M_A}{L_A}\ -\ \frac{M_B}{L_A}\ -\ \frac{M_B}{L_B}+\frac{M_C}{L_B}                                      (10-42)

R_1=0+\frac{qL}{2}+0\ -\ 0\ -\ 0\ -\ \frac{49qL^2}{960L}=\frac{431qL}{960}

At support 2, the terms in Eq. (10-42) are as follows: (1) The simple-beam reaction for member AB, which is now member 1-2, is qL/2; (2) the simple-beam reaction for member BC, which is now member 2-3. is zero; (3) the moment M_A is the bending moment at support 1 and is equal to zero; (4) the moment M_B is the bending moment M_B, which is given by Eq. (h); and (5) the moment M_2 is the bending moment M_C which is given by Eq. (i). Thus, substituting term-by-term into Eq. (10-42), we get

R_2=\frac{qL}{2}+0+0+\frac{49L^2}{960L}+\frac{49qL^2}{960L}\ -\ \frac{11qL^2}{240L}=\frac{89qL}{160}

Proceeding in this manner for the remaining two reactions, we are able to obtain all reactions of the beam from Eq. (10-42). As an alternative to using this equation, we can simply draw free-body diagrams of each member and solve the equations of equilibrium.

The final results are

R_1=\frac{431qL}{960} = 0.449qL           R_2=\frac{89qL}{160} = 0.556qL         (j,k)

R_3=\frac{93qL}{320} = 0.291qL           R_4=\frac{169qL}{240} = 0.704qL         (l,m)

Shear-force and bending-moment diagrams. Now that the reactions have been found, we can isolate parts of the beam as free bodies and use equations of equilibrium to determine the shear forces and bending moments throughout the beam. The resulting diagrams are shown in Figs. 10-23b and c.