Question 10.5: A fixed-end beam AB supports a uniform load of intensity q a......

We can find the reactions of this beam by using superposition together with the results obtained in the preceding example (Example 10-4). In that example we found the reactions of a fixed-end beam subjected to a concentrated load P acting at distance a from the left-hand end (see Fig. 10-15a and Eqs. 10-25 and 10-26). In order to apply those results to the uniform load of Fig. 10-17a. we will treat an element of the uniform load as a concentrated load of magnitude q dx acting at distance x from the left-hand end (Fig. 10-17b). Then, using the formulas derived in Example 10-4. we can obtain the reactions caused by this element of load. Finally, by integrating over the length a of the uniform load, we can obtain the reactions due to the entire uniform load.
Let us begin with the moment reactions, which are given by Eqs. (10-25a and b). To obtain the moments caused by the element q dx of the uniform load (compare Fig. 10-17b with Fig. 10-15a), we replace P with q dx, a with x, and b with L – x. Thus, the fixed-end moments due to the element of load are

M_A=\frac{Pab^2}{L^2}\quad \quad M_B=\frac{Pa^2b}{L^2}                                    (10-25a,b)

R_A=\frac{Pb^2}{L^3}(L+2a)\quad \quad R_B=\frac{Pa^2}{L^3}(L+2b)                           (10-26a,b)

dM_A=\frac{qx(L\ -\ x)^2dx}{L^2}\quad \quad dM_B=\frac{qx^2(L\ -\ x)dx}{L^2}

Integrating over the loaded part of the beam, we get the fixed-end moments due to the entire uniform load:

M_A=\int{dM_A}=\frac{q}{L^2}\int_{0}^{a}{x(L\ -\ x)^2dx}=\frac{qa^2}{12L^2}(6L^2\ -\ 8aL+3a^2)                 (10-30a)

M_B=\int{dM_B}=\frac{q}{L^2}\int_{0}^{a}{x^2(L\ -\ x)dx}=\frac{qa^3}{12L^2}(4L\ -\ 3a)         (10-30b)

Proceeding in a similar manner but using Eqs. (10-26a and b), we obtain the following expressions for the fixed-end forces due to the element of load:

dR_A=\frac{q(L\ -\ x)^2(L+2x)dx}{L^3}\quad \quad dR_B=\frac{qx^2(3L\ -\ 2x)dx}{L^3}

Integration gives

R_A=\int{dR_A}=\frac{q}{L^3}\int_{0}^{a}{(L\ -\ x)^2(L+2x)dx}=\frac{qa}{2L^3}(2L^3\ -\ 2a^2L+a^3)                      (10-31a)

R_B=\int{dR_B}=\frac{q}{L^3}\int_{0}^{a}{x^2(3L\ -\ 2x)dx}=\frac{qa^3}{2L^3}(2L\ -\ a)                         (10-31b)

Thus, all reactions (fixed-end moments and fixed-end forces) have been found.
Uniform load acting over the entire length of the beam (Fig. 10-18). When the load acts over the entire span, we can obtain the reactions by substituting a = L into the preceding equations, yielding

M_A=M_B=\frac{qL^2}{12}\quad \quad R_A=R_B=\frac{qL}{2}                          (10-32a,b)

The deflection at the midpoint of a uniformly loaded beam is also of interest. The simplest procedure for obtaining this deflection is to use the method of superposition. The first step is to remove the moment restraints at the supports and obtain a released structure in the form of a simple beam. The downward deflection at the midpoint of a simple beam due to a uniform load (from Case 1, Table G-2) is

(δ_C)_1=\frac{5qL^4}{384EI}                                 (u)

and the upward deflection at the midpoint due to the end moments (from Case 10, Table G-2) is

(δ_C)_2=\frac{M_AL^2}{8EI}=\frac{(qL^2/12)L^2}{8EI}=\frac{qL^4}{96EI}                                 (v)

Thus, the final downward deflection of the original fixed-end beam (Fig. 10-18) is

δ_C=(δ_C)_1\ -\ (δ_C)_2

Substituting for the deflections from Eqs. (u) and (v), we get

δ_C=\frac{qL^4}{384EI}                            (10-33)

This deflection is one-fifth of the deflection at the midpoint of a simple beam with a uniform load, again illustrating the stiffening effect of fixity at the ends of the beam.