# Slope-Deflection Method for Continuous Beams

## Contents

**Introduction**

### History

First used in 1915 by George A. Maney, the Slope-Deflection Method of Structural Analysis calculates the force in a statically indeterminate structure by dealing only with the deformations of the structure.^{[1]} This method focuses on first calculating the deflections. It then uses the displacement and rotation to find the moments in the fixed-end of the beams.^{[2]}

### Uses

The slope-deflection method is ideal for doing calculations on continuous beams as the equations do not become more complicated as the number of unknowns increases. The equations remain easy to derive no matter how many degrees of indeterminacy there are in the structure.

The slope deflection method can also be used with the stiffness factor or the chord rotation of a beam. This means that the variables that need to be known vary, and the method can be used in a variety of cases.^{[3]}

This method provides a base of knowledge that is needed to preform the matrix-stiffness method, which is the method used to evaluate structures in many electronic solutions.

**Terms**

**Rotation** of the end of a beam is the amount it pivots on its support, due to external loads placed upon it. This value is found in units of degrees or radians.^{[2]}

**Displacements** of a beam segment is the distance that the structural element has moved from its original location in a unit of length.^{[2]}

**Degrees of Freedom** are the unknown values of displacement and rotation found in a beam. Sometimes called the *Degree of Kinematic Indeterminacy*.^{[1]}

**Slope Deflection Equations**

The equations found for use in the Slope Deflection Method connect the rotations and displacements of a beam to the moments that are induced at its ends through forces.^{[1]} The elastic curve of a beam under a set of arbitrary forces is used to compare the differences in rotation of the beam to the angle of the chord.^{[3]}

$ M_{nf}=\frac{2EI}{L}(2\theta_n+\theta_f-3\psi)+M'_{nf} $ ~Eq. 1^{[1]}

Where$ M_{nf} $ is the Moment acting at near end denoted as subscript *n* of member while subscript *f* represent the far/other end of the member, $ M' $ is Fixed-End Moments. $ \theta_n \text{ and } \theta_f $ are the near end and far end rotations of the member with respect to horizontal position of the member.$ \psi $ represent the rotation of member's chord under consideration. It is given as
$ \psi = \frac{\Delta}{L} $ where, $ \Delta $ is deformation assumed to be small. All the parameters are shown in figure 1. Eq. 1 is valid only for prismatic members composed of linearly elastic member and subjected to small deformations also, axial and shear forces are neglected ^{[1]}.

**Analysis of Continuous Beams**

The method for analysis of continuous beams is adapted from the textbook, Structural Analysys (4th Ed.) by Kassimali.^{[1]} [2]

**Step 1**

Evaluate the Degree of Freedom. In the case of a continuous beam, the Degree of Freedom will be the number of unknown reactions.

**Step 2**

Formulate Equilibrium Equations[3]. From Step 1, the number of Equilibrium Equations will be known as it is equal to the Degree of Freedom of the system.

**Step 3**

Calculate fixed-end moments for each member in the system.[4] In most cases, the fixed-end moments will be one of the unknown reactions of the beam.

**Step 4**

Calculate the chord rotations for the member under consideration adjacent to support with support settlement using relation $ \psi =\frac{\Delta}{L} $. Care must be taken for sign convention clockwise rotation is negative and vice versa.

**Step 5**

Using Eq.1 formulate the slope deflection equation, for each member there will be two equations.

**Step 6**

Substitute slope deflection equation from step 5 into the equilibrium equation derived in step 2 and evaluate unknown $ \theta $ joint rotations.

**Step 7**

Substitute $ \theta $ value in slope deflection equations from step 5 and calculate members end moments.

Note * Counterclockwise moments are consider positive.

**Step 8**

To cross check solution from step 7, the member end moment muse be substituted into the Equilibrium Equation from step 2.

**Step 9**

Calculate support reactions[5]

**Step 10**

Draw a free body diagram for the beam in question.

**Step 11**

Draw Shear Force and Bending Moment diagrams for the beam.[6]

**Example Problems**

*Example Question 1*

Analyze the two span continuous beam “ABC” provided below in Figure 2 by using the slope deflection method. Take EI as constant.

Step 1. Solve for the fix end moments.

Notice that the distances “a” and “b” are 4 and 2 respectively. L is the length from points A to B.

$ M'_{AB} = -\frac{Wab^2}{L^2}=-\frac{100\times4\times2^2}{6^2} =-44.44\text{ KNm} $

$ M'_{BA} = +\frac{Wa^2b}{L^2} = \frac{100\times4^2\times2}{6^2} = +88.89\text{ KNm} $

$ M'_{BC} = - \frac{wL^2}{12} = -\frac{20\times5^2}{12} = -41.67\text{ KNm} $

$ M'_{CB} = +\frac{wL^2}{12} = \frac{20\times5^2}{12} = 41.67\text{ KNm} $

By inspection it is noticed that A is a fixed support, thus the slope at A is zero.
The slopes at B and C **cannot** be equal to zero since there is a pin support.

Step 2. Use the Slope Deflection equations and sub-in the values of the fixed end moments and slopes.

**At span AB**,

$ M_{AB} = M'_{AB} + (\frac{2EI}{L}) (2\theta_{A} +\theta_{B}) $

Thus, $ M_{AB} = -44.44 + \frac{2EI}{6}{\theta_B} = -44.44 + \frac{EI\theta_{B}}{3} $ [1]

**At span BA**,

$ M_{BA} = M'_{BA} + (\frac{2EI}{L}) (2\theta_{B} +\theta_{A}) $

Thus,$ M_{BA} = 88.89 + \frac{2EI\times2\theta_B}{6} = 88.89 + \frac{2 EI\theta_B}{3} $ [2]

**At span BC**,

$ M_{BC} = M'_{BC} + (\frac{2EI}{L}) (2\theta_{B} +\theta_{C}) $

Thus,$ M_{BC} = -41.67 + (\frac{2EI}{5}) (2\theta_{B} +\theta_{C}) = -41.67+ \frac{4 EI\theta_{B}}{5} + \frac{2 EI \theta_{C}}{5} $ [3]

**At span CB**,

$ M_{CB} = M'_{CB} + (\frac{2EI}{L}) (2\theta_{C} +\theta_{B}) $

Thus,$ M_{BC} = 41.67 + (\frac{2EI}{5}) (2\theta_{C} +\theta_{B}) = 41.67+ \frac{4 EI\theta_{C}}{5} + \frac{2 EI \theta_{B}}{5} $ [4]

Step 3. By analyzing the above equations [1], [2], [3], and [4], we notice that our unknowns are$ \theta_B $ and$ \theta_C $

By using the boundary conditions, it can be seen that $ -M_{BA}-M_{BC} = 0 \text{ and } M_{BA}+M_{BC} = 0. $

Furthermore, $ M_{CB} = 0 $, since C is a simple support at the end.

Now that ,

$ M_{BA}+M_{BC} = 89.89 + \frac{2EI\theta_{B}}{3} -41.67 + \frac{4EI\theta_{B}}{5} + \frac{2EI\theta_{C}}{5} = 47.22 +\frac{22EI\theta_{B}}{15} +\frac{2EI \theta_{C}}{5} = 0 $

$ M_{CB} = +41.67 + \frac{2 EI \theta_B}{5} + \frac{4 EI\theta_C}{5} $

Solving both these equations at the same time will result in our $ EI\theta_C\text{ and } EI \theta_{B}. $

$ EI\theta_C = -41.67 $

$ EI\theta_B = -20.83 $

Substitution in the slope deflection equations will result in$ M_{AB} \text{ and } M_{BA} $

$ M_{AB} = -44.44 + \frac{-20.83}{3} = -51.38 \text{ KNm} $

$ M_{BA} = +88.99 +\frac{2\times-20.83}{3} = +75.00 \text{ KNm} $

$ M_{CB} = +41.67 + \frac{2\times-20.83}{5} + \frac{4\times-41.67}{5}= 0 \text{ KNm} $

$ M_{BC} = -41.67 + \frac{4\times-20.83}{5} + \frac{2\times-41.67}{5} =-75.00 \text{ KNm} $

*Example Question 2*

Analyze the two span continuous beam provided below in Figure 3 by using the slope deflection method.

Take EI as constant.

Step 1: Using the moment equations to find the moments from the different segments listed.

$ M'_{12} = M'_{21} = \frac{45\times 2.5}{8} =14.06 kNm $

$ M'_{23} = M'_{32} = \frac{100\times 5}{8} = 62.50 kNm $

Step 2: Using the slope deflection equations

$ M_{12} = (\frac{2\times EI}{2.5}) (\theta_2)-14.06 $

$ M_{21}= (\frac{2\times EI}{2.5}) (2\times\theta_2)+14.06 $

$ M_{23}= (\frac{6\times EI}{5})(2\times \theta_2 + \theta_3)-62.5 $

$ M_{32}= (\frac{6\times EI}{5})( \theta_2 +2\times \theta_3) +62.5 $

Step 3: Through solving for the joints and reactions is done using the equations of equilibrium.

$ M_{21} + M_{23} = 0; $

$ M_{32} = 0; $

Substituting the$ M_{21}, M_{23}, \text{ and } M_{32} $ with the above equations;

$ [(\frac{2\times EI}{2.5}) (2\times\theta_2) +14.06] + [(\frac{6\times EI}{5})(2\times \theta_2 + \theta_3) -62.5]=0 $, [1]

$ ((\frac{6\times EI}{5})( \theta_2 +2\times \theta_3) +62.5 = 0 $, [2]

Above equation can be simplified as follow

$ 4EI \theta_2 + 1.2 EI \theta_3 = 48.44 $ [3]

$ 1.2EI\theta_2+2.4EI \theta_3=-62.5 $ [4]

solving the [3] &[4] equation simultaneously we get,

$ \theta_2= \frac{23.44}{EI} $

$ \theta_3 = \frac{-37.76}{EI} $

Step 4 : Sub in the values of$ \theta_2 $ and $ \theta_3 $ in the main slope deflection equations and this will yield in the slope deflection values of all segments.

$ M_{12} = (\frac{2\times EI}{2.5}) (\theta_2)-14.06 =4.69 kNm $

$ M_{21}= (\frac{2\times EI}{2.5}) (2\times\theta_2) +14.06=51.56kNm $

$ M_{23}= (\frac{6\times EI}{5})(2\times \theta_2 + \theta_3) -62.5=-51.56 kNm $

$ M_{32}= (\frac{6\times EI}{5})( \theta_2 +2\times \theta_3) +62.5 =0 $

**References**

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}^{1.4}^{1.5}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑
^{2.0}^{2.1}^{2.2}Hibbler, R.C. (1997).*Mechanics of Materials (3rd ed.)*. Prentice Hall. - ↑
^{3.0}^{3.1}Holtz, N. (2013).*A learning resource for Course CIVE3203: Introduction to Structural Analysis - Method of Slope Deflection*.[1]