Question 16.15: Find the support reactions in the three-span continuous beam...

The beam in Fig. 16.33 is the beam that was solved using the flexibility method in Ex. 16.7, so that this example provides a comparison between the two methods.

Initially we consider the beam as comprising three separate fixed beams \mathrm{AB}, \mathrm{BC} and \mathrm{CD} and calculate the values of the FEMs, M_{\mathrm{AB}}^{\mathrm{F}}, M_{\mathrm{BA}}^{\mathrm{F}}, M_{\mathrm{BC}}^{\mathrm{F}}, etc. Thus, using the results of Exs 13.20 and 13.22 and remembering that clockwise moments are positive and anticlockwise moments negative

\begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-M_{\mathrm{BA}}^{\mathrm{F}}=-\frac{6 \times 1.0}{8}=-0.75  \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{BC}}^{\mathrm{F}}=-M_{\mathrm{CB}}^{\mathrm{F}}=-\frac{10 \times 1.0}{8}=-1.25  \mathrm{kN} \mathrm{m} \\ & M_{\mathrm{CD}}^{\mathrm{F}}=-M_{\mathrm{DC}}^{\mathrm{F}}=-\frac{12 \times 1.0^{2}}{12}=-1.0  \mathrm{kN} \mathrm{m} \end{aligned}

In the beam of Fig. 16.33 the vertical displacements at all the supports are zero, i.e. v_{\mathrm{A}}, v_{\mathrm{B}}, v_{\mathrm{C}} and v_{\mathrm{D}} are zero. Therefore, from Eqs (16.28) and (16.30)

M_{\mathrm{AB}} =-\frac{2 E I}{L}\left[2 \theta_{\mathrm{A}}+\theta_{\mathrm{B}}+\frac{3}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+M_{\mathrm{AB}}^{\mathrm{F}} \qquad (16.28)

M_{\mathrm{BA}} =-\frac{2 E I}{L}\left[2 \theta_{\mathrm{B}}+\theta_{\mathrm{A}}+\frac{3}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+M_{\mathrm{BA}}^{\mathrm{F}} \qquad (16.30)

we have

\begin{aligned} & M_{\mathrm{AB}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{A}}+\theta_{\mathrm{B}}\right)-0.75 &(\text{i}) \\ & M_{\mathrm{BA}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{B}}+\theta_{\mathrm{A}}\right)+0.75 &(\text{ii}) \\ & M_{\mathrm{BC}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{B}}+\theta_{\mathrm{C}}\right)-1.25 & (\text{iii}) \\ & M_{\mathrm{CB}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{C}}+\theta_{\mathrm{B}}\right)+1.25 & (\text{iv})\\ & M_{\mathrm{CD}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{C}}+\theta_{\mathrm{D}}\right)-1.0 & (\text{v})\\ & M_{\mathrm{DC}}=-\frac{2 E I}{1.0}\left(2 \theta_{\mathrm{D}}+\theta_{\mathrm{C}}\right)+1.0 & (\text{vi}) \end{aligned}

From the equilibrium of moments at the supports

M_{\mathrm{AB}}=0 \quad M_{\mathrm{BA}}+M_{\mathrm{BC}}=0 \quad M_{\mathrm{CB}}+M_{\mathrm{CD}}=0 \quad M_{\mathrm{DC}}=0

Substituting for M_{\mathrm{AB}}, etc., from Eqs (i)-(vi) in these expressions we obtain

\begin{aligned} 4 E I \theta_{\mathrm{A}}+2 E I \theta_{\mathrm{B}}+0.75 & =0 & (\text{vii}) \\ 2 E I \theta_{\mathrm{A}}+8 E I \theta_{\mathrm{B}}+2 E I \theta_{\mathrm{C}}+0.5 & =0 & (\text{viii})\\ 2 E I \theta_{\mathrm{B}}+8 E I \theta_{\mathrm{C}}+2 E I \theta_{\mathrm{D}}-0.25 & =0 & (\text{ix})\\ 4 E I \theta_{\mathrm{D}}+2 E I \theta_{\mathrm{C}}-1.0 & =0 & (\text{x}) \end{aligned}

The solution of Eqs (vii)-(x) gives

E I \theta_{\mathrm{A}}=-0.183 \quad E I \theta_{\mathrm{B}}=-0.008 \quad E I \theta_{\mathrm{C}}=-0.033 \quad E I \theta_{\mathrm{D}}=+0.267

Substituting these values in Eqs (i)-(vi) gives

M_{\mathrm{AB}}=0 \quad M_{\mathrm{BA}}=1.15 \quad M_{\mathrm{BC}}=-1.15 \quad M_{\mathrm{CB}}=1.4 \quad M_{\mathrm{CD}}=-1.4 \quad M_{\mathrm{DC}}=0

The end moments acting on the three spans of the beam are now shown in Fig. 16.34. They produce reactions R_{\mathrm{AB}}, R_{\mathrm{BA}}, etc., at the supports; thus

\begin{gathered}R_{\mathrm{AB}}=-R_{\mathrm{BA}}=-\frac{1.15}{1.0}=-1.15  \mathrm{kN} \\R_{\mathrm{BC}}=-R_{\mathrm{CB}}=-\frac{(1.4-1.15)}{1.0}=-0.25  \mathrm{kN} \\R_{\mathrm{CD}}=-R_{\mathrm{DC}}=\frac{1.4}{1.0}=1.40  \mathrm{kN}\end{gathered}

Therefore, due to the end moments only, the support reactions are

In addition to these reactions there are the reactions due to the actual loading, which may be obtained by analysing each span as a simply supported beam (the effects of the end moments have been calculated above). In this example these reactions may be obtained by inspection. Thus

R_{\mathrm{A}, \mathrm{S}}=3.0  \mathrm{kN} \quad R_{\mathrm{B}, \mathrm{S}}=3.0+5.0=8.0  \mathrm{kN} \quad R_{\mathrm{C}, \mathrm{S}}=5.0+6.0=11.0  \mathrm{kN}

R_{\mathrm{D}, \mathrm{S}}=6.0  \mathrm{kN}

The final reactions at the supports are then

\begin{aligned} & R_{\mathrm{A}}=R_{\mathrm{A}, \mathrm{M}}+R_{\mathrm{A}, \mathrm{S}}=-1.15+3.0=1.85  \mathrm{kN} \\ & R_{\mathrm{B}}=R_{\mathrm{B}, \mathrm{M}}+R_{\mathrm{B}, \mathrm{S}}=0.9+8.0=8.9  \mathrm{kN} \\ & R_{\mathrm{C}}=R_{\mathrm{C}, \mathrm{M}}+R_{\mathrm{C}, \mathrm{S}}=1.65+11.0=12.65  \mathrm{kN} \\ & R_{\mathrm{D}}=R_{\mathrm{D}, \mathrm{M}}+R_{\mathrm{D}, \mathrm{S}}=-1.4+6.0=4.6  \mathrm{kN} \end{aligned}

Alternatively, we could have obtained these reactions by the slightly lengthier procedure of substituting for \theta_{\mathrm{A}}, \theta_{\mathrm{B}}, etc., in Eqs (16.29) and (16.31).

S_{\mathrm{AB}} =\frac{6 E I}{L^2}\left[\theta_{\mathrm{A}}+\theta_{\mathrm{B}}+\frac{2}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+S_{\mathrm{AB}}^{\mathrm{F}} \qquad (16.29)

S_{\mathrm{BA}} =-\frac{6 E I}{L^2}\left[\theta_{\mathrm{A}}+\theta_{\mathrm{B}}+\frac{2}{L}\left(v_{\mathrm{A}}-v_{\mathrm{B}}\right)\right]+S_{\mathrm{BA}}^{\mathrm{F}}\qquad (16.31)

Thus, e.g.

S_{\mathrm{AB}}=R_{\mathrm{A}}=\frac{6 E I}{L^{2}}\left(\theta_{\mathrm{A}}+\theta_{\mathrm{B}}\right)+3.0 \quad\left(v_{\mathrm{A}}=v_{\mathrm{B}}=0\right)

which gives R_{\mathrm{A}}=1.85  \mathrm{kN} as before.

Comparing the above solution with that of Ex. 16.7 we see that there are small discrepancies; these are caused by rounding-off errors.

Having obtained the support reactions, the bending moment distribution (reverting to the sagging (positive) and hogging (negative) sign convention) is obtained in the usual way and is shown in Fig. 16.35.