Strength of materials
Method of three moments
- Introduction to the method of three moments
- Definition and alternative name (Clapeyron method)
- Application for continuous beams
- Assumptions of the method
- Advantages of the method
- Algorithm for solution
- Establishing the static scheme of the beam
- Determining geometric parameters
- Drawing moment diagrams
- Solving secondary beams
- Formulating the system of equations
- Solving the system of equations
- Calculating support reactions
- Determining internal force diagrams
- Equation of three moments
- For beams with constant stiffness
- For beams with variable stiffness
- Alternative forms of the equation
- Legend for formulas
- Coefficients and parameters
- Example 1 - Two-span beam
- Reduction of the determinable part
- Determining boundary moments
- Formulating the equation of three moments
- Graphical integration of figures
- Simpson's formula
- Determining support reactions
- Shear force diagram
- Calculating the extremum of moments
- Example 2 - Beam with fixed support
- Different stiffness of spans
- Span of zero length
- Formulating the system of equations
- Substituting data
- Determining support moments
- Calculating reactions
- Equilibrium of spans
Introduction to the three-moment method
The three-moment method, also known as Clapeyron's method, is a technique used to solve continuous beams (multi-span beams) that are statically indeterminate. This method is based on the equations of equilibrium of bending moments formulated for three consecutive supports in a continuous beam.
The fundamental assumption of the three-moment method is the use of geometric and static relationships between the bending moments at three consecutive supports and the loads acting on the adjacent spans of the beam. This method allows for the determination of the values of support moments without the need for detailed analysis of all reaction forces.
Using the three-moment method, equations are formulated that describe the conditions of continuity of deflections and rotation angles at intermediate supports. For a beam with multiple spans, a system of linear equations is created, where the unknowns are the support moments. Solving this system allows for further determination of the distribution of moments, shear forces, and deflections throughout the beam.
The advantage of the three-moment method is its relatively simple and systematic computational procedure, particularly effective for multi-span beams with any number of spans and various loading schemes.
Algorithm for solving
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Establishing the static scheme of the beam:
- Determine the number of supports and spans.
- Identify the supports (end, intermediate) and assign them appropriate numbers or symbols.
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Determining the geometric parameters of the beam:
- Determine the lengths of the consecutive spans \(l_n, l_{n+1}, \ldots\).
- Determine the stiffness of individual spans of the beam (constant or variable stiffness EI).
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Drawing moment diagrams for the given load:
- For each loaded span, determine the moment diagram.
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Solving secondary beams:
- Here, two approaches can be taken:
- 1) Calculate the reactions at the intermediate support for two adjacent beams \(R_{N1}, R_{Np}\) - the reaction at support "n" from the left side and the reaction at support "n" from the right side. The load on the secondary beam for which we determine the reactions is the moment diagram from the primary beam converted into a continuous load.
- 2) Calculate the rotation angle at the intermediate support for both adjacent beams - that is, integrate the moment diagrams that were created in the primary state and in the state from unit moments at the intermediate support. This integration can be performed, for example, by calculating the area under the diagrams on both beams, i.e.: \(\Omega_n, \Omega_{n+1}\) and the positions of their centers of gravity \(a_n, b_{n+1}\)
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Formulating the system of equations for the three-moment method:
- Apply the formulas of the three-moment method (Clapeyron), compiling them for the next three supports.
- For a beam with \(n\) supports \((n-2)\) linear equations with unknown support moments.
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Solving the system of equations:
- Solve the obtained system of linear equations for the unknown support moments \(M_{n-1}, M_n, M_{n+1}\).
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Calculating support reactions and shear forces:
- Using the obtained support moments, determine the reactions at the supports and the shear forces in the individual spans of the beam.
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Determining the diagrams of bending moments and shear forces:
- Prepare the bending moment diagram using the calculated support moments and the influence of loads in the spans.
- Determine the shear force diagram based on the previously determined support reactions.
The equation of three moments
For beams with constant stiffness
\[ M_{n-1} l_{n}+2 M_{n}\left(l_{n}+l_{n+1}\right)+M_{n+1} l_{n+1}=-6\left(R_{N l}+R_{N p}\right) \]For beams with variable stiffness
\[ c M_{n-1} l_{n}+2 M_{n}\left(c l_{n}+l_{n+1}\right)+M_{n+1} l_{n+1}=-6\left(c R_{N l}+R_{N p}\right) \]One may also encounter the notation in the form:
\[ c M_{n-1} l_{n}+2 M_{n}\left(c l_{n}+l_{n+1}\right)+M_{n+1} l_{n+1}=-6 c \frac{\Omega_{n} a_{n}}{l_{n}}-6 \frac{\Omega_{n+1} b_{n+1}}{l_{n+1}} \]Legend for the formulas
- \(M_{n-1}, M_n, M_{n+1}\) - bending moments at the consecutive supports (previous, current, and next)
- \(l_n, l_{n+1}\) - lengths of the consecutive spans of the beam
- \(R_N, R_{N p}\) - quantities dependent on the type and distribution of loads on the spans (equivalent moments resulting from the loads)
- \(c\) - coefficient accounting for the variability of the beam's stiffness: \[c = \frac{(EI)_{n+1}}{(EI)_n}\]
- \(\Omega_n, \Omega_{n+1}\) - areas under the moment diagrams from the loads on the individual spans
- \(a_n, b_{n+1}\) - distances of the centers of gravity of the moment areas from the respective supports
Alternative notation of the equation
One may also encounter the notation in the form below:
\[ \mathrm{x}_{\mathrm{k}-1} \cdot \mathrm{l}_{\mathrm{k}^{\prime}}+2 \cdot \mathrm{x}_{\mathrm{k}} \cdot\left(\mathrm{l}_{\mathrm{k}^{\prime}}+\mathrm{l}_{\mathrm{k}+1^{\prime}}\right)+\mathrm{x}_{\mathrm{k}+1^{-1}} \mathrm{l}_{\mathrm{k}+1^{\prime}}=\mathrm{N}_{\mathrm{k}} \mathrm{P} \]Legend:
- \(\mathrm{x}_{\mathrm{k}-1}, \mathrm{x}_{\mathrm{k}}, \mathrm{x}_{\mathrm{k}+1^{-1}}\) – bending moments at the consecutive supports (previous, current, next)
- \(\mathrm{l}_{\mathrm{k}'}\) – reduced (comparative) length of the span between supports \(k-1\) and \(k\)
- \(\mathrm{l}_{\mathrm{k}+1'}\) – reduced (comparative) length of the span between supports \(k\) and \(k+1\)
- \(\mathrm{N}_{\mathrm{k}}\) = \(-6 \mathrm{EI} \cdot\left(\varphi_{k \mathrm{~L}}+\varphi_{k \mathrm{P}}\right)\)
where: \(I_C\) - comparative moment of inertia (the moment of inertia of one of the spans is chosen as comparative)
Example 1
Content
Solve the given beam using the three-moment method. Draw the final diagrams of bending moments and shear forces.
Solution
We will show the solution here using the notation from the last variant of the formula we provided in the introduction.
Reduction of the determinable part, determination of boundary moments
If \(I = \text{const}\), then \(l_k' = l_k\)
Data:
- \(l_1 = l_1' = 6\,\text{m}\)
- \(l_2 = l_2' = 5\,\text{m}\)
The equation of three moments
\[ x_{k-1} \cdot l_k' + 2 \cdot x_k \cdot (l_k' + l_{k+1}') + x_{k+1} \cdot l_{k+1}' = N_k P \] \[ x_0 \cdot l_1' + 2 \cdot x_1 \cdot (l_1' + l_2') + x_2 \cdot l_2' = N_1 P \] \[ -3 \cdot 6 + 2 \cdot x_1 \cdot (6 + 5) + 0 \cdot 5 = N_1 P \] \[ 22 \cdot x_1 = N_1 P + 18 \]Calculating the right side of the equation of three moments
Approach with graphical integration of figures
For the integration of the above examples, the Simpson's rule
was used. \[ N_{1P} = -6EI \cdot (\varphi_{1L} + \varphi_{1P}) = -6EI \cdot \left( \frac{13{,}5}{EI} + \frac{125}{12EI} \right) \] \[ N_{1P} = -143{,}5 \]Solution of the 3M equation
\[ 22x_1 = N_{1P} + 18 = -143{,}5 + 18 \] \[ x_1 = -5{,}7\ \text{kNm} \] \[ M_{\text{ost}} = M_1 \cdot x_1 + M_P \]
Determining the final reactions and shear forces
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Left span: \[ \sum M_0 = 0 \quad \curvearrowright \] \[ -3 + 6 \cdot 3 + 5{,}7 - R_{1L} \cdot 6 = 0 \] \[ R_{1L} = 3{,}45\ \text{kN} \] \[ \sum Y = 0 \] \[ R_0 - 6 + R_{1L} = 0 \] \[ R_0 = 2{,}55\ \text{kN} \] |
Right span: \[ \sum M_1 = 0 \] \[ -5{,}7 + 2{,}5 \cdot 2{,}5 - R_2 \cdot 5 = 0 \] \[ R_2 = 3{,}86\ \text{kN} \] \[ \sum Y = 0 \] \[ R_{1P} - 2{,}5 + R_2 = 0 \] \[ R_{1P} = 6{,}14\ \text{kN} \] |
Verification
\[ R_1=R_{1L}+R_{1P}=9,59 \ kN \] \[ \sum y=0 \ \ R_0+R_1+R_2-6-2\cdot 5=0 \] \[ 0=0 \]Final shear force diagram
Calculating the extremum on the moment diagram
\[ \frac{3,86}{x} = \frac{6,14}{5-x} \] \[ 6,14x = 3,86 \cdot (5-x) \] \[ 6,14x = 19,3 - 3,86x \] \[ 10x = 19,3 \quad (:10) \] \[ x = 1,93\text{m} \] \[ M_{ex} = R_z \cdot x - 2 \cdot x \cdot \frac{x}{2} = 3,72\text{ kNm} \]Example 2
Content
Let's look at another partially solved example - to show how to treat a fixed support in the three-moment method, as well as how to account for different stiffnesses of spans.
Using the equation of three moments, prepare the shear force and bending moment diagrams for the beam shown in Fig. 8.
Data: \(q,\ l,\ c = \frac{2EI}{EI} = 2\)
Solution
The beam is doubly statically indeterminate. The missing two equations are obtained by applying the three-moment method for spans 1 and 2 and 2 and 3 (span 3 with zero length!).
The equations of three moments
\[ cM_A l_1 + 2M_B(c l_1 + l_2) + M_C l_2 = -6c \cdot \frac{\Omega_1 a_1}{l_1} - 6 \cdot \frac{\Omega_2 b_2}{l_2} \] \[ M_B l_2 + 2M_C(l_2 + l_3) + M_0 l_3 = -6 \cdot \frac{\Omega_2 a_2}{l_2} - 6 \cdot \frac{\Omega_3 b_3}{l_3} \]Substituting data
The equations should be substituted with:
- \(l_1 = l_2 = l\)
- \(l_3 = 0\)
- \(c = 2\)
- \(M_A = 0\)
- \(\Omega_1 = 0\)
- \(\Omega_2 = \frac{2}{3} \cdot \frac{1}{8} q l^2 \cdot l = \frac{1}{12} q l^3\)
- \(\Omega_3 = 0\)
- \(a_2 = \frac{1}{2} l\)
After transforming the equations of three moments, we obtain the system of equations:
\[ \begin{cases} 6M_B + M_C = -\frac{1}{4} ql^2 \\ M_B + 2M_C = -\frac{1}{4} ql^2 \end{cases} \]Solving this system of equations, we obtain the values of moments at the supports:
\[ M_B = -\frac{1}{44} ql^2, \quad M_C = -\frac{5}{44} ql^2 \]Determining support reactions
To calculate the support reactions, we consider the equilibrium of individual spans. From the moment equation about point B (for span AB), we obtain:
\[ R_A = \frac{M_B}{l} = -\frac{1}{44} ql \]Balancing the forces acting on span AB, we obtain the value of the reaction force from the span side:
\[ R'_B = -R_A = \frac{1}{44} ql \]For span BC, using the moment equilibrium equation about point B:
\[ R_C = \frac{M_B - M_C}{l} + \frac{ql}{2} = \frac{13}{22} ql \]Next, we determine the force transmitted from the right side of the span to point B:
\[ R''_B = ql - R_C = \frac{9}{22} ql \]The total reaction at support B is the sum of the reactions from the left and right sides:
\[ R_B = R'_B + R''_B = \frac{1}{44} ql + \frac{9}{22} ql = \frac{19}{44} ql \]Having the calculated moments and reactions, we can now prepare the shear force and bending moment diagrams for the entire beam.
Source: task 10.15 (p. 379) from:
M. Banasiak, K. Grossman, M. Trombski, “Set of problems in material strength”, Scientific Publishing House PWN, Warsaw 2012.