Problem 9.1. Moments of multispan beam

Maximum negative moment, M^–_max [kNm]=

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Positive moment at the middle of the beam, M⁺ [kNm]=

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Step 1.  Define a primary structure. Choose the redundants.

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Step 2.  Draw moment diagrams of the primary structure from the original load and from the unit redundants.

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Step 3.  Determine redundants from the compatibility condition.

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Step 4.  Apply superposition to give the moment diagram of the statically indeterminate structure.

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The degree of indeterminancy is four. Four hinges are introduced above the intermediate supports, thus the primary structure is a set of simply supported girders. The four redundants are the moment couples, X₁, X₂, X₃, X₄ shown in the Figure.

Moment diagrams of the primary structure from the original load and from the unit redundants are given below.

Redundants are determined from the compatibility condition.The relative rotations above the intermediate supports must be zero. To calculate the rotations Castigliano’s theorem can be applied.

Rotations at the supports from the uniformly distributed load is

$a_{i0}={\int }_0^l\frac{M_0{M}_i}{EI}dx=–\frac{p{L}^3}{24EI}$

Relative rotations above the hinges from the unit redudants are

$$\begin{gathered} a_{ii}={\int }_0^l\frac{M_i{M}_i}{EI}dx=2A_i\frac{2}{3}=2\times \frac{1\times l}{2}\frac{2}{3}=\frac{2L}{3EI} \\ {a}_{i,i+1}={\int }_0^l\frac{M_i{M}_{i+1}}{EI}dx=A_i\frac{1}{3}=\frac{1\times l}{2}\frac{1}{3}=\frac{L}{6EI} \\ {a}_{i–1,i}={\int }_0^l\frac{M_{i–1}{M}_i}{EI}dx=A_i\frac{1}{3}=\frac{1\times l}{2}\frac{1}{3}=\frac{L}{6EI} \end{gathered}$$

where $a_{i,i+1}$ is the relative rotation above the i+1-th support from the i-th unit redundant.

According to the recipocal theorem, we may write:

$a_{i–1,i}=a_{i,i–1}$

Compatibilty conditions written for all the four intermediate supports result in a linear equation system, from which the redundants can be calculated:

$$\begin{gathered} \begin{array}{c}{a}_{10}+a_{11}{X}_1+a_{21}{X}_2=0\\ a_{20}+a_{12}{X}_1+a_{22}{X}_2+a_{23}{X}_3=0\\ a_{30}+a_{23}{X}_2+a_{33}{X}_3+a_{34}{X}_4=0\\ a_{20}+a_{34}{X}_3+a_{44}{X}_4=0\end{array} \to \left\{\begin{array}{c}{a}_{10}\\ a_{20}\\ a_{30}\\ a_{40}\end{array}\right\}+\left[\begin{array}{cccc}{a}_{11}& a_{21}& 0& 0\\ a_{12}& a_{22}& a_{32}& 0\\ 0& a_{23}& a_{33}& a_{43}\\ & & a_{34}& a_{44}\end{array}\right]\left\{\begin{array}{c}{X}_1\\ X_2\\ X_3\\ X_4\end{array}\right\} \\ \to –\frac{p{l}^3}{24EI} \left\{\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right\}+\frac{l}{6EI}\left[\begin{array}{cccc}4& 1& 0& 0\\ 1& 4& 1& 0\\ 0& 1& 4& 1\\ 0& 0& 1& 4\end{array}\right]\left\{\begin{array}{c}{X}_1\\ X_2\\ X_3\\ X_4\end{array}\right\}=0 \\ \to \left\{\begin{array}{c}{X}_1\\ X_2\\ X_3\\ X_4\end{array}\right\}=\frac{p{l}^2}{4}{\left[\begin{array}{cccc}4& 1& 0& 0\\ 1& 4& 1& 0\\ 0& 1& 4& 1\\ 0& 0& 1& 4\end{array}\right]}^{–1}\left\{\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right\}=p{l}^2\left\{\begin{array}{c}5.26\\ 3.95\\ 3.95\\ 5.26\end{array}\right\}{10}^{–2}=4.0\times 3.0^2\left\{\begin{array}{c}5.26\\ 3.95\\ 3.95\\ 5.26\end{array}\right\}{10}^{–2}=\left\{\begin{array}{c}1.89\\ 1.42\\ 1.42\\ 1.89\end{array}\right\}\mathrm{kNm} \end{gathered}$$

Moment diagram of the statically indeterminate structure is obtained by superposition.

$M=M_0+M_1{X}_1+M_2{X}_2+M_3{X}_3+M_4{X}_4$

The maximum negative moment is

${\mathit{M}}_{\mathbf{max}}^{\mathbf{–}}=X_1\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{89}\mathbf{ }\mathbf{kNm}$

Positive moment at the middle of the girder is

${\mathit{M}}_{}^{\mathbf{+}}=M_0+M_2{X}_2+M_3{X}_3=\frac{p{l}^2}{8}–\frac{1}{2}{X}_2–\frac{1}{2}{X}_2=\frac{4.0\times 3.0^2}{8}–2\frac{1.42}{2}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{08}\mathbf{ }\mathbf{kNm}$

The results are compared to the moments with infinite number of spans. Increasing the number of spans, n  all the redundant would result in

$X=X_i=\frac{p{l}^2}{24}$

and the moment diagram would be