Compiled by: John Wachira
Continuous beams are common elements in building structures where multiple spans are supported by three or more supports. Determining moments in these beams is crucial for designing safe and economical structural elements in Kenya’s construction industry. The three-moment equation is a powerful tool used to analyze continuous beams by relating moments at three consecutive supports. This chapter focuses on developing the three-moment expressions to enable accurate moment calculations for continuous beams in building technology.
Continuous beams transfer loads over multiple supports, making their analysis more complex than simply supported beams. The three-moment equation provides a relationship between the bending moments at three consecutive supports, accounting for beam span lengths and applied loads. Mastering this method equips building technologists to design beams that meet safety and serviceability requirements in Kenya’s diverse construction projects.
The three-moment equation relates the bending moments at three consecutive supports of a continuous beam resting on elastic supports. It is derived from the principles of equilibrium, compatibility, and the moment-curvature relationship for elastic beams. The equation is expressed as:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Where:
For beams on rigid supports with no settlement, \(\Delta_1 = \Delta_2 = 0\), simplifying the equation to:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Example 1: A continuous beam has two spans: \(L_1 = 4\,m\) and \(L_2 = 5\,m\). The moments at supports 1 and 3 are zero (\(M_1 = M_3 = 0\)). Calculate \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 4\,m\),
\(L_2 = 5\,m\),
\(\Delta_1 = \Delta_2 = 0\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Substitution:$$ 0 \times 4 + 2 M_2 (4 + 5) + 0 \times 5 = 0 $$$$ 2 M_2 \times 9 = 0 $$$$ 18 M_2 = 0 $$$$ M_2 = 0\,kNm $$
Answer: \(M_2 = 0\,kNm\)
Example 2: For the same beam with \(M_1 = 0\,kNm\), \(M_3 = 0\,kNm\), but with support settlements \(\Delta_1 = 2\,mm\) downwards and \(\Delta_2 = 0\), calculate \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 4\,m\),
\(L_2 = 5\,m\),
\(\Delta_1 = 2\,mm = 0.002\,m\),
\(\Delta_2 = 0\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Substitution:$$ 0 \times 4 + 2 M_2 (4 + 5) + 0 \times 5 = -6 ( \frac{0.002}{4} + 0 ) $$$$ 2 M_2 \times 9 = -6 \times 0.0005 $$$$ 18 M_2 = -0.003 $$$$ M_2 = \frac{-0.003}{18} = -0.0001667\,kNm $$
Answer: \(M_2 = -0.00017\,kNm\) (hogging moment)
Example 3: A continuous beam has three supports with spans \(L_1 = 6\,m\), \(L_2 = 8\,m\). The supports do not settle. Moments at supports 1 and 3 are \(M_1 = 10\,kNm\) and \(M_3 = 15\,kNm\). Find \(M_2\).
Given:
\(M_1 = 10\,kNm\),
\(M_3 = 15\,kNm\),
\(L_1 = 6\,m\),
\(L_2 = 8\,m\),
\(\Delta_1 = \Delta_2 = 0\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Substitution:$$ 10 \times 6 + 2 M_2 (6 + 8) + 15 \times 8 = 0 $$$$ 60 + 2 M_2 \times 14 + 120 = 0 $$$$ 2 M_2 \times 14 = -180 $$$$ 28 M_2 = -180 $$$$ M_2 = \frac{-180}{28} = -6.43\,kNm $$
Answer: \(M_2 = -6.43\,kNm\)
When continuous beams are subjected to uniformly distributed loads, the moment expressions can be simplified by calculating the fixed-end moments and using the three-moment equation to find the unknown moments at supports. This approach is essential in building designs where slabs or beams carry uniform loads such as floor finishes or roofing.
The general form of the three-moment equation for uniform loads \(w_1\) and \(w_2\) on spans \(L_1\) and \(L_2\) is:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4} $$
Example 1: A continuous beam with spans \(L_1 = 5\,m\), \(L_2 = 7\,m\) carries uniform loads \(w_1 = 4\,kN/m\) and \(w_2 = 3\,kN/m\). Moments at supports 1 and 3 are zero. Find \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 5\,m\),
\(L_2 = 7\,m\),
\(w_1 = 4\,kN/m\),
\(w_2 = 3\,kN/m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4} $$
Substitution:$$ 0 \times 5 + 2 M_2 (5 + 7) + 0 \times 7 = -\frac{4 \times 5^3}{4} - \frac{3 \times 7^3}{4} $$$$ 2 M_2 \times 12 = -\frac{4 \times 125}{4} - \frac{3 \times 343}{4} $$$$ 24 M_2 = -125 - 257.25 $$$$ 24 M_2 = -382.25 $$$$ M_2 = \frac{-382.25}{24} = -15.93\,kNm $$
Answer: \(M_2 = -15.93\,kNm\)
Example 2: A continuous beam with spans \(L_1 = 6\,m\), \(L_2 = 6\,m\) carries uniform loads \(w_1 = 5\,kN/m\) and \(w_2 = 5\,kN/m\). Given moments \(M_1 = 10\,kNm\), \(M_3 = 15\,kNm\), find \(M_2\).
Given:
\(M_1 = 10\,kNm\),
\(M_3 = 15\,kNm\),
\(L_1 = 6\,m\),
\(L_2 = 6\,m\),
\(w_1 = 5\,kN/m\),
\(w_2 = 5\,kN/m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4} $$
Substitution:$$ 10 \times 6 + 2 M_2 (6 + 6) + 15 \times 6 = -\frac{5 \times 6^3}{4} - \frac{5 \times 6^3}{4} $$$$ 60 + 2 M_2 \times 12 + 90 = -\frac{5 \times 216}{4} - \frac{5 \times 216}{4} $$$$ 60 + 24 M_2 + 90 = -270 - 270 $$$$ 24 M_2 + 150 = -540 $$$$ 24 M_2 = -690 $$$$ M_2 = \frac{-690}{24} = -28.75\,kNm $$
Answer: \(M_2 = -28.75\,kNm\)
Example 3: For a continuous beam with spans \(L_1 = 8\,m\), \(L_2 = 5\,m\), uniform loads \(w_1 = 6\,kN/m\), \(w_2 = 4\,kN/m\), and no moments at supports 1 and 3, find \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 8\,m\),
\(L_2 = 5\,m\),
\(w_1 = 6\,kN/m\),
\(w_2 = 4\,kN/m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4} $$
Substitution:$$ 0 \times 8 + 2 M_2 (8 + 5) + 0 \times 5 = -\frac{6 \times 8^3}{4} - \frac{4 \times 5^3}{4} $$$$ 2 M_2 \times 13 = -\frac{6 \times 512}{4} - \frac{4 \times 125}{4} $$$$ 26 M_2 = -768 - 125 $$$$ 26 M_2 = -893 $$$$ M_2 = \frac{-893}{26} = -34.35\,kNm $$
Answer: \(M_2 = -34.35\,kNm\)
In building structures, beams often have unequal span lengths due to architectural or site constraints. The three-moment equation accommodates unequal spans by incorporating span lengths \(L_1\) and \(L_2\) explicitly, allowing for accurate moment calculations. This is essential for Kenyan building projects where irregular spans are common.
The general three-moment equation remains:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
When no settlements occur, it simplifies to:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Example 1: A continuous beam has spans \(L_1 = 3\,m\), \(L_2 = 6\,m\) with zero moments at supports 1 and 3. Find \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 3\,m\),
\(L_2 = 6\,m\),
\(\Delta_1 = \Delta_2 = 0\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Substitution:$$ 0 \times 3 + 2 M_2 (3 + 6) + 0 \times 6 = 0 $$$$ 2 M_2 \times 9 = 0 $$$$ 18 M_2 = 0 $$$$ M_2 = 0\,kNm $$
Answer: \(M_2 = 0\,kNm\)
Example 2: For the same beam, \(M_1 = 5\,kNm\), \(M_3 = 10\,kNm\), find \(M_2\).
Given:
\(M_1 = 5\,kNm\),
\(M_3 = 10\,kNm\),
\(L_1 = 3\,m\),
\(L_2 = 6\,m\),
\(\Delta_1 = \Delta_2 = 0\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = 0 $$
Substitution:$$ 5 \times 3 + 2 M_2 (3 + 6) + 10 \times 6 = 0 $$$$ 15 + 2 M_2 \times 9 + 60 = 0 $$$$ 18 M_2 + 75 = 0 $$$$ 18 M_2 = -75 $$$$ M_2 = \frac{-75}{18} = -4.17\,kNm $$
Answer: \(M_2 = -4.17\,kNm\)
Example 3: A continuous beam with spans \(L_1 = 4\,m\), \(L_2 = 7\,m\) carries uniform loads \(w_1 = 3\,kN/m\), \(w_2 = 5\,kN/m\), with zero moments at supports 1 and 3. Find \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 4\,m\),
\(L_2 = 7\,m\),
\(w_1 = 3\,kN/m\),
\(w_2 = 5\,kN/m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -\frac{w_1 L_1^3}{4} - \frac{w_2 L_2^3}{4} $$
Substitution:$$ 0 \times 4 + 2 M_2 (4 + 7) + 0 \times 7 = -\frac{3 \times 4^3}{4} - \frac{5 \times 7^3}{4} $$$$ 2 M_2 \times 11 = -\frac{3 \times 64}{4} - \frac{5 \times 343}{4} $$$$ 22 M_2 = -48 - 428.75 $$$$ 22 M_2 = -476.75 $$$$ M_2 = \frac{-476.75}{22} = -21.67\,kNm $$
Answer: \(M_2 = -21.67\,kNm\)
In real construction projects, supports may settle or experience movements affecting beam moments. The three-moment equation can incorporate these effects through the terms \(\Delta_1\) and \(\Delta_2\), representing vertical displacements at supports 1 and 3 relative to support 2. Additionally, known moments at supports can be included to solve for unknown moments.
The complete three-moment equation is:
$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Example 1: A continuous beam with spans \(L_1 = 5\,m\), \(L_2 = 5\,m\) has support settlements \(\Delta_1 = 3\,mm\) downwards, \(\Delta_2 = 1\,mm\) upwards. Moments at supports 1 and 3 are zero. Find \(M_2\).
Given:
\(M_1 = 0\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 5\,m\),
\(L_2 = 5\,m\),
\(\Delta_1 = 3\,mm = 0.003\,m\),
\(\Delta_2 = -1\,mm = -0.001\,m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Substitution:$$ 0 \times 5 + 2 M_2 (5 + 5) + 0 \times 5 = -6 ( \frac{0.003}{5} + \frac{-0.001}{5} ) $$$$ 2 M_2 \times 10 = -6 \times (0.0006 - 0.0002) $$$$ 20 M_2 = -6 \times 0.0004 $$$$ 20 M_2 = -0.0024 $$$$ M_2 = \frac{-0.0024}{20} = -0.00012\,kNm $$
Answer: \(M_2 = -0.00012\,kNm\) (negligible moment)
Example 2: For the same beam, if \(M_1 = 8\,kNm\), \(M_3 = 12\,kNm\), find \(M_2\).
Given:
\(M_1 = 8\,kNm\),
\(M_3 = 12\,kNm\),
\(L_1 = 5\,m\),
\(L_2 = 5\,m\),
\(\Delta_1 = 0.003\,m\),
\(\Delta_2 = -0.001\,m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Substitution:$$ 8 \times 5 + 2 M_2 (5 + 5) + 12 \times 5 = -6 \times (0.0006 - 0.0002) $$$$ 40 + 20 M_2 + 60 = -0.0024 $$$$ 20 M_2 + 100 = -0.0024 $$$$ 20 M_2 = -100.0024 $$$$ M_2 = \frac{-100.0024}{20} = -5.00012\,kNm $$
Answer: \(M_2 = -5.00\,kNm\)
Example 3: A beam with spans \(L_1 = 7\,m\), \(L_2 = 4\,m\) has settlements \(\Delta_1 = 0\), \(\Delta_2 = 5\,mm\) downward. Moments at supports 1 and 3 are \(M_1 = 7\,kNm\), \(M_3 = 0\). Find \(M_2\).
Given:
\(M_1 = 7\,kNm\),
\(M_3 = 0\,kNm\),
\(L_1 = 7\,m\),
\(L_2 = 4\,m\),
\(\Delta_1 = 0\),
\(\Delta_2 = 5\,mm = 0.005\,m\)
Formula:$$ M_1 L_1 + 2 M_2 (L_1 + L_2) + M_3 L_2 = -6 ( \frac{\Delta_1}{L_1} + \frac{\Delta_2}{L_2} ) $$
Substitution:$$ 7 \times 7 + 2 M_2 (7 + 4) + 0 \times 4 = -6 ( 0 + \frac{0.005}{4} ) $$$$ 49 + 2 M_2 \times 11 = -6 \times 0.00125 $$$$ 49 + 22 M_2 = -0.0075 $$$$ 22 M_2 = -0.0075 - 49 = -49.0075 $$$$ M_2 = \frac{-49.0075}{22} = -2.23\,kNm $$
Answer: \(M_2 = -2.23\,kNm\)
A continuous beam has spans \(L_1 = 6\,m\) and \(L_2 = 6\,m\) with no settlements. Moments at supports 1 and 3 are zero. Calculate \(M_2\) using the three-moment equation when the beam carries uniform loads of \(w_1 = 4\,kN/m\) and \(w_2 = 6\,kN/m\). (8 marks)
For a continuous beam with spans \(L_1 = 5\,m\), \(L_2 = 7\,m\), moments at supports 1 and 3 are \(M_1 = 12\,kNm\) and \(M_3 = 8\,kNm\), with no settlements. Find \(M_2\). (6 marks)
A continuous beam with spans \(L_1 = 4\,m\), \(L_2 = 5\,m\) has support settlements \(\Delta_1 = 2\,mm\) down and \(\Delta_2 = 3\,mm\) up. Moments at supports 1 and 3 are zero. Calculate \(M_2\). (8 marks)
A continuous beam with unequal spans \(L_1 = 3\,m\), \(L_2 = 7\,m\) carries uniform loads \(w_1 = 5\,kN/m\), \(w_2 = 3\,kN/m\). Moments at supports 1 and 3 are zero. Find \(M_2\). (8 marks)
For a continuous beam with spans \(L_1 = 6\,m\), \(L_2 = 6\,m\), moments at supports 1 and 3 are \(M_1 = 10\,kNm\) and \(M_3 = 20\,kNm\). Support settlements are \(\Delta_1 = 1\,mm\) down, \(\Delta_2 = 2\,mm\) down. Calculate \(M_2\). (10 marks)
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Create a free accountThis chapter focuses on determining bending moments in continuous structural members, starting with the development of three moment expressions that relate moments at supports to applied loads and spans. It then explains the application of the three moment theorem, which provides a fundamental relationship between moments at three consecutive supports of a continuous beam. Using this theorem, students learn to calculate unknown moments by considering beam geometry and loading conditions. The chapter further explores the moment distribution method, a practical iterative technique for analyzing indeterminate beams by distributing and balancing moments until equilibrium is reached. Through these methods, one can accurately determine bending moments in continuous beams, which is essential for safe and efficient structural design. The progression from theoretical expressions to practical calculation techniques equips learners with the tools needed for real-world structural analysis. Overall, the chapter integrates fundamental principles and computational methods to enable comprehensive moment analysis in continuous members.
A continuous beam spans three supports A, B, and C with spans \(L_1 = 4\,m\) and \(L_2 = 6\,m\). The beam carries uniform loads of \(w_1 = 10\,kN/m\) on span AB and \(w_2 = 15\,kN/m\) on span BC. Calculate the three moments \(M_A\), \(M_B\), and \(M_C\) using the three moment equation.
For a continuous beam with two spans each of length 5 m, carrying a point load of 20 kN at mid-span of the first span, determine the moments at supports using the three moment theorem. Assume the beam is continuous over the supports.