Slope-Deflection Method for Frames with Sidesway

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Structures can can be analyzed under different loading conditions depending on if they're determinate or indeterminate structures. Many methods have been developed to analyze indeterminate structures. However, certain methods could be more applicable depending on the characteristics of the structure or truss. Slope deflection method is more specific to indeterminate structures with low degrees of freedom.


Figure 1: Simple Case of Sidesway

Here, we discuss how to use the Slope-Deflection method for indeterminate frames with sidesway. A brief introduction on the Slope-Deflection method, can be found under Slope-Deflection Method for Continuous Beams. However in this section, the method will analyze frames specifically with sidesway. The term sidesway refers to the possibility of joint translation in the x or y direction of a structure.[1]

Using the slope-deflection equations as well as equilibrium conditions, the rotation angles of the sidesway and members are determined. These angles are then substituted back into our original slope-deflection equations to solve for the moments. Once the moments are determined the truss is easily solvable using equilibrium conditions. Also, axial forces have a negligible effect in the moment equilibrium equation so they are ignored in the analysis.[2]


The Slope-Deflection method goes back to the 19th century (1915) when it was proposed by G.A.Maney. It was originally a method that was widely used to analyze rigid joint beams and frames until the publication of the moment distribution method in 1930 which became much more popular.[3] [4]The Slope-Deflection method is an alternative formulation of the displacement method which is also known as the stiffness method, the slope deflection method deals only with the bending deformations of a structure.[5]

Equations used for the Analysis

Equation 1

Standard slope-deflection equation (Fixed-Fixed):[1]

$ M_n_f = \frac{2EI}{L}(2\theta_n + \theta_f - 3\psi) + FEM_n_f $

Where $ FEM_n_f $ represents the Fixed-End Momment of the near end with respect to far end.

Equation 2

Slope-deflection equation (Fixed-Pinned):[1]

$ M_r_h = \frac{3EI}{L}(\theta_r - \psi) + (FEM_r_h - \frac{FEM_h_r}{2}) $

Where $ r $ represents the rigid end and $ h $ represents the hinged, or pinned, end.

Equation 3

Chord rotation:[5]

$ \psi =\frac{ A}{L} $

Where $ A $ represents the displacement of the member in the horizontal direction.

Analysis of Frames with Sidesway Using Slope-Deflection Method

Degrees of Freedom

The concept degrees of freedom refers to the ability of a joint to rotate or displace. The number of degrees of freedom are found at the supports, joints and free ends of a member.[6] The Degrees of freedom play an important role in determining whether a frame has the ability to sway or not. The following equation is used to calculate the degree of freedom: $ SS = 2j - [2(f + h) + r + m] $ [5]

Where $ SS $ is the degrees of freedom, $ f $ is the number of fixed ends, $ h $ is the number of hinged supports, $ r $ is the number of rollers, $ m $ is the number of frame members, and $ j $ is the number of joints[5].

Figure 2. Beam Convention

If $ SS=0 $ than there is no sidesway occurring but if SS is greater than zero then the frame has the ability to sway. A frame with the ability to sway may not undergo sway if:[6]

Figure 3. Conventional Sign Convention

1. The geometry of the frame loadings are symmetrical.

2. The frame is physically restricted to move in the direction of sway.

Sign Convention

The sign convention for the slope-deflection method no longer abides by the beam convention, where the forces acting on either end of a beam both inflict a positive moment and shear, although acting in opposite directions, Refer to Figure 2. Instead, it follows the conventional sign conventions[1], as seen in Figure 2. This means that that member end moments, chord rotations & deformations are all assumed to be positive in the counter clockwise direction.[5]

Direction is critical when it comes to solving frames using Slope-Deflection method. A simple error in one section of the method will relay to other parts of the solution and the moment equilibrium at joints will not work out.


Figure 4. Determination of Psi

The first step in solving an indeterminate structure with sidesway using the slope deflection method is to find the degrees of freedom in each support, joints and free end. The number of degrees of freedom will determine the number of equations of equilibrium that will be needed in order to solve the indeterminate structure.[5] Referring to Figure 1. We can see that the number of degrees of freedom is three. The translation in the x direction from the top member, and the rotation of the two vertical members.

Finding our fixed end moments ($ FEM $) is the next step. In order to solve for the fixed end moments of each member, the free body diagrams are considered. For the purpose of solving for fixed end moments, we need to treat each member as if it were fixed on both ends. From there we solve for the moments at each end of the member by using the fixed end moment equations located in the back of Aslam Kassimali's textbook: Structural Analysis 4th ed.[5]

Next, determine the rotation $ \psi $ of the beams that are involved in the sway of the structure. Rotation in the clockwise direction is negative, while rotation in the counterclockwise direction is positive as stated in Sign Convention. Using the assumption that the bending creates a linear chord[1] we can use right angle triangle rules to determine and equation for $ \psi $. Refer to Figure 4. and Equation 3. If the lengths of each of the vertical members are different, it will result in a different rotation angle.[5]

Using the slope deflection equation, Equation 1, form equations for all the end moments of each member. This gives us one set of equations but we need two more, both of which are found using equilibrium conditions.[5]

The next set of equations comes from the use of moment equilibrium of the joints. Summing of the moments at each joint will result in equations with only internal moments as the variables.

Our final set of equations are determined by looking at the free body diagram for each member individually. A sum of moments from the top of each vertical member will result in two equations with the variables being the support reactions in the direction of interest since the values for all moments have already been accounted for with equations.

By substituting our equations into each other we can solve for the rotation and deflections.

Now that the displacements and rotations have been found, they are subbed back into our initial moments equations obtained from Equation 1 to solve for each moment individually.

After all moments have been solved for, the shear and axial forces of our now determinate structure can be found using equilibrium equations. From there, the shear, axial, and moment diagrams can be easily drawn and the structure is hence forth fully determined.[1]

Summary Of Method

Step 1. Find the degrees of freedom.

Step 2. Determine the Fixed End Moments for each member.

Step 3. Determine the equations for the chord rotation values.

Step 4. Use slope deflection equations to form equations for the moment at either ends of the members.

Step 5. Use moment equilibrium at joints to form a second set of equations.

Step 6. Using free body diagrams for each vertical member, use sum of moments at the top to obtain third set of equations.

Step 7. Solve rotations and displacements by substitution of equations.

Step 8. Use newly found rotations and displacements to solve for numerical values of member moments.

Step 9. Structure is now determinate and remaining shear and axial forces can be found through equilibrium.

Example Problem

Let us consider an indeterminate structure, shown in Figure 5. We will solve this indeterminate structure using the slope defelection method for indeterminate structures with sidesway.

Figure 5: FBD and Deflected Shape of the Frame

1. Find the Degrees of Freedom

Considering the deflected shape due to the applied forces, as shown in Figure 5. It is seen that there are 3 degrees of freedom: $ \theta_B $, $ \theta_C $ and $ \Delta $.

2. Determine Fixed End Moments

Starting with member AB, the member is assumed to be fixed at both ends (Figure 5.), and the equation for a fixed end moment due to a point load: $ \ FEM_A_B=\frac{Pab^2}{L^2} $ [5]. is used to solve.

$ \ FEM_A_B=5.625KNm $

$ \ FEM_B_A=-16.875KNm $

Figure 6: Beams AB and BC shown with fixed ends.

The same is then done to member BC (Figure 6.), and the equation for a fixed end moment due to a uniform load: $ \ FEM_B_C=\frac{wL^2}{12} $ [5].

$ \ FEM_B_C=-FEM_C_B=25KNm $

Due to the absence of external forces, both $ \ FEM_C_D $ and $ \ FEM_D_C $ are equal to 0.

3. Determine Chord Rotations

Using equation 3 as expressed in the previous section, the following chord rotations are found:

$ \ \Psi_A_B=-\frac{\Delta}{4} $, $ \ \Psi_B_C=0 $, $ \ \Psi_C_D=-\frac{\Delta}{6} $

4. Slope Deflections Equations

The moments at the end of each member can be determined using the Slope Deflection Equation (Equation 1.) Due to the fixed ends of the frame, $ \theta_A $ and $ \theta_D $ are equal to 0 There is also no chord rotation on the horizontal member BC.

$ \ M_A_B=\frac{2EI}{4}(\theta_B+0.75\Delta)+5.625 $ (1)

$ \ M_B_A=\frac{2EI}{4}(2\theta_B+0.75\Delta)-16.875 $ (2)

$ \ M_B_C=\frac{2EI}{10}(2\theta_B+\theta_C)+25 $ (3)

$ \ M_C_B=\frac{2EI}{10}(\theta_B+2\theta_C)-25 $ (4)

$ \ M_C_D=\frac{2EI}{6}(2\theta_C+0.5\Delta) $ (5)

$ \ M_D_C=\frac{2EI}{6}(\theta_C+0.5\Delta) $ (6)

5. Equations of Equilibrium

In this problem, there are 3 degrees of freedom, therefore we must find 3 equations of equilibrium. The first two are found using the FBD's of joints B and C, shown in Figure 7. These are:

Figure 7: FBD of Joints B and C

Joint B: $ \ M_B_C+M_B_A=0 $ (7)

Joint C: $ \ M_C_B+M_C_D=0 $ (8)

The third equation comes from summing the forces in the x direction in the FBD of the entire frame. This becomes:

$ \sum{F_x}=0 $

$ \ S_A_B+S_D_C-P=0 $

These shears can be expressed by the member end moments by summing the moments at the top of each of the members AB and CD.

$ \ S_A_B=\frac{M_A_B+M_B_A+P}{4} $

$ \ S_D_C=\frac{M_C_D+M_D_C}{6} $

The equation then becomes:

$ \ 6(M_A_B+M_B_A)+4(M_C_D+M_D_C)-540=0 $ (9)

7. Sub in the Slope Deflection Equations

After subbing equations 1-6 into equation 7-9, the three equilibrium equations become:

i) $ \ 1.4EI\theta_B+0.2EI\theta_C+0.375EI\Delta=-8.125 $

ii) $ \ 0.2EI\theta_B+1.066EI\theta_C+0.166EI\Delta=25 $

Figure 8: Showing end Moments on the Frame

iii) $ \ 9EI\theta_B+4EI\theta_C+5.8333EI\Delta=607.5 $

9. Solve the System of Equations and Original End Moment Equations

$ \ EI\theta_B = -57.1145 $

$ \ EI\theta_C = 4.6115 $

$ \ EI\Delta = 189.1012 $

$ \ M_A_B=47.98KNm $

$ \ M_B_A=-3.07KNm $

$ \ M_B_C=3.07KNm $

$ \ M_C_B=-34.58KNm $

$ \ M_C_D=34.59KNm $

$ \ M_D_C=33.05KNm $

9. Solve the Newly Formed Determinate Structure

The now determinate structure can be solved using equilibrium to find the remaining shear and axial forces, and the associated shear and moment diagrams. See Figure 8.


The Slope-Deflection method seems to be a lot more efficient to use for beams with higher degrees of statically indeterminacy than the Force-Method[1]. However, this method has been expanded to also solve frames with sway/nosway. In this method slope deflection equations are derived and used to analyze a frame with sidesway, the reactions are calculated using static equilibrium equations.

See also


  1. 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Jeffery Erochko, Class Lecture,Slope-Deflection Method: "Slope Deflection Method." Carleton University, Ottawa, Ontario, Azrieli Theatre, November 13th, 2013
  2. Norris,Wilburd,Utku. (1991). Elementary Structural Analysis(4th ed.).p.355. McGraw Hill.
  5. 5.00 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 5.10 , Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.).p.655. Stamford, CT: Cengage Learning.
  6. 6.0 6.1 (