Slope-Deflection Method for Frames without Sidesway

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The slope-deflection method is a method of analysis for indeterminate structures. The slope-deflection method is the inverse of the Force Method and is a similar method to the Moment Distribution Method. The slope-deflection method is a stiffness method, meaning that it relates and solves for the unknown rotations and displacements first then the reactions of the structure are solved using equilibrium equations.[1][2]

This section will consist of slope-deflection method for a frame without sidesway. The analysis of this method for a frame with sidesway can be found here, Slope-Deflection Method for Frames with Sidesway. The slope-deflection method may also be applied to a beam. This method of analysis is summarized on this page, Slope-Deflection Method for Continuous Beams.


Figure 1: Simple Frame
A frame is a 2-dimensional (for the scope of this section) structure comprised of columns and beams fastened by pinned or fix connections. The connections are know as joints and are the rigid connections that hold frame members in place.[2]
In the study of structural analysis a member is often referred to as a beam, girder or column. The members of a frame system are the structural components that span between joints, hinges and supports.
A joint is described as a node on a frame where two or more members are linked together. A joint may also be defined as the location of a support. In this case it is possible to have only one member located at the reaction node.
Members of a frame subject to an externally applied load will cause the joints of a frame to move in a direction relative to their point of origin. This movement is known as translation. Since we are dealing with two-dimensional analysis, possible translations are in the horizontal and vertical directions.
Joints along a frame may be subjected to rotation. Since we are dealing with structures in a two-dimensional plane, rotation may only occur along one plane of rotation (clockwise and counter-clockwise rotations).
Sidesway describes the ability of the joints on a frame system to undergo translations. Since frames in this section are restrained from deforming axially, this cannot be caused by axial deformation of the frame.[1]
Degrees of Freedom
At any joint of a statically indeterminate structure, the unknown rotations and/or displacements of a joint are the degrees of freedom. If the degrees of freedom are all known or is equal to zero, the structure is statically determinate.[1][2]
Equations of Equilibrium
Equilibrium for a rigid body is expressed as the sum of all forces and moments at a particular point on the structure being equal to zero. In this section, these are of importance at the joints with degrees of freedom. [3]
Flexural Rigidity
Flexural Rigidity (EI) is the product of the moment of inertia and the modulus of elasticity of the material. The more rigid something is, the more force it can resist before it starts to deform.[2]

Sidesway of a Frame

Figure 2: Frame with-out and with sidesway
Figure 3: Frame and its deflected shape without sidesway

For frames, axial deformations are much smaller than the bending deformations and are neglected in the analysis. With this assumption the frame shown in Figure 3 exhibits no sidesway. This means that the frames joints will not translate to the left or right. For example, in Figure 3, joint B and joint E cannot translate in either horizontal or vertical direction due to the fixities A, C and D.[4][2]

Sidesway Degrees of Freedom

To distinguish whether a frame exhibits sidesway for analysis, the following equation can be used to find the sidesways degrees of freedom, $ ss $;

$ ss = 2j - [2(f + h) + r + m] $ [1] ~ Equation 1
Where $ j $ is the number of joints, $ f $ is the number of fixed ends, $ h $ is the number of hinged supports, $ r $ is the number of roller supports and $ m $ is the number of members of the frame.[1]

If the sidesway degrees of freedom, $ ss $, is zero, the frame exhibits zero or no sidesway.[1] If the sidesway degrees of freedom is greater than zero, then the frame exhibits sidesway.[1] The value of the sidesway degrees of freedom will always be an integer and will describe the number of directions the joints of the frame are allowed to translate.[1]

The sidesway degrees of freedom will allow insight for the proper method to use for the analysis of the slope deflection method of a frame.

Symetric Frames with Symmetric Loading

A symmetric frame with symmetric loading will appear to be a sidesway frame from the calculation of the sidesway degrees of freedom.[1][4][2] For this special case, due to the symmetry of the frame and loading, the joints do not translate. Since the joints of the frame do not translate the frame is considered to exhibit zero or no sidesway.[1][4][2]

Frames which are symmetric and have parallel loading to its columns are an exception to the sidesway degrees of freedom equation and are always without sidesway.

Slope-Deflection Analysis

The slope-deflection analysis finds the unknown displacements through the help of the slope-deflection equation and the equations of equilibrium at joints with degrees of freedom.[1] Once these displacements are found, they are back substituted into the equations to find the moments, shears and reactions of the structure.[1]

Slope-Deflection Equation

Figure 4: Slope Deflection Reactions

The following is the general form of the slope-deflection equation for an end moment at end n of member nf, $ M_{nf} $:

$ M_{nf} = \frac{2EI}{L}(2\theta_n + \theta_f - 3\psi) + FEM_{nf} $[1] ~ Equation 2
Where $ n $ is the near end and $ f $ is the far end of the member, $ M_{nf} $ is the end moment of member nf, $ E $ is Youngs modulus of elasticity, $ I $ Moment of inertia of the member, $ \theta_n $ and $ \theta_f $ are the rotations of the joints $ n $ and $ f $ respectively, $ \theta $ is the rotation of the chord formed by the ends of the deformed shape of the member and $ FEM_{nf} $ is the fixed end moment at $ n $ of member nf.[1]

If we assume small rotations, the chord rotation, $ \psi $ can be represented as;

$ \psi = \frac{\theta}{L} $[1] ~ Equation 3
Where $ \theta $ is the relative translation between the ends of the member and $ L $ is the length of the member.[1]

Sign Convention

In this section, the sign convention for the moment, shear and axial forces are absolute. Moments are taken to be positive if they are in the counter-clockwise direction and are negative if in the clockwise direction. Shear forces are taken as positive if in the upward direction and negative if in the downwards direction.

Fixed-end Moments

To find the fixed-end moment of each member in the frame under analysis, a fixity is added at both ends of the member and the new moment at this fixed end is called the fixed-end moment.[1] Table 1 shows simplified equations which can be used to find the fixed-end moments for the following four common cases.

Table 1: Common Fixed-end Moment Cases and Formulas:[1]

Fixed-end Moment Diagram
Fixed-end Case 1 $ M_{cd} = M_{dc} = \frac{PL}{8} $ FEM1.png
Fixed-end Case 2 $ M_{cd} = \frac{Pab^{2}}{L^{2}} $, $ M_{dc} = \frac{Pa^{2}b}{L^{2}} $ FEM2.png
Fixed-end Case 3 $ M_{cd} = \frac{Mb}{L^{2}}(b - 2a) $, $ M_{dc} = \frac{Ma}{L^{2}}(2b - a) $ FEM3.png
Fixed-end Case 4 $ M_{cd} = M_{dc} = \frac{wL^{2}}{12} $ FEM4.png

Fixed-end moments which do not follow the previous four cases in table 1 can be found through the equilibrium equations and solving a system of equations.

General Analysis of Frames without Sidesway

The following is an adaptation of a procedure of the analysis of slope deflection method for frames developed by Kassimali.[1]

Step 1: Determine whether the frame has sidesway or not using Equation 1. If the frame does not have sidesway continue on with this procedure, if the frame does have sidesway follow this procedure.
Step 2: Find the number of degrees of freedom in the frame and its location(s).
Step 3: Evaluate the fixed-end moments for each member by using the equations listed in Table 1 or solve the system of equations if the case does not coincide with those from Table 1.
Step 4: Apply the slope-deflection equation (Equation 2) for each member in the frame. This will give you the number of necessary equation - DOF.
Step 5: Do an equilibrium analysis at each joint with degrees of freedom to find the equilibrium equation. The number of equations are now sufficient to be able to solve for the unknown translations and rotations in the frame.
Step 6: Substitute the slope deflection equations into the equilibrium equations and solve the system of equations in order to find the translations and rotations of the joints.
Step 7: Now with the previously unknown displacements known, back substitute the joint displacements in order to find the end-moments of each member of the frame.
Step 8: Break the frame into its member components for analysis. With the end-moments of each member known, continue to find the end shears, axial forces and reactions at the support across the entire frame.
Step 9: Everything is now known about the beam. If required, draw the axial force diagram, shear diagram and moment diagram of the frame.

Example Problem

Determine the end moments and reactions for the frame shown in Figure 5 and draw the axial, shear and bending moment diagrams of this frame. (EI is constant and equals 120)

Figure 5: Example Problem

Step 1: Check for sidesway:
$ ss = 2j - [2(f + h) + r + m] $ [1] ~Eq.1 (Where j=5, f=2, h=1, r=0 and m=4)
$ ss = 0 $ ($ \therefore $ Frame has no sidesway)

Step 2: Find the degrees of freedom of each of the 5 joints to find the total D.O.F.
Joint A: $ D.O.F. = 0 $ (Joint A cannot displace or rotate)
Joint B: $ D.O.F. = 0 $ (Joint B cannot displace or rotate)
Joint C: $ D.O.F. = 1 $ (Joint C can rotate but not displace)
Joint D: $ D.O.F. = 1 $ (Joint D can rotate but not displace)
Joint E: $ D.O.F. = 1 $ (Joint E can rotate but not displace)
$ \therefore $ Total $ D.O.F. = 3 $ (Three equations are needed to solve)

Step 3: Find fixed end moments for each member using table 1 (positive moment = anticlockwise)
$ FEM_A_D = \frac{60\times3} {8} = 22.5 kN\cdot m $ (fixed-end case 1)
$ FEM_D_A = -FEM_A_D = -22.5 kN\cdot m $ (fixed-end case 1)
$ FEM_B_E = 0 $
$ FEM_E_B = 0 $
$ FEM_C_D = 0 $
$ FEM_D_C = 0 $
$ FEM_D_E = \frac{25\times3^2}{12} = 18.75 kN\cdot m $ (fixed-end case 4)
$ FEM_E_D = -FEM_D_E = -18.75 kN\cdot m $ (fixed-end case 4)

Step 4: Apply the FEM's from step 3 to the slope-deflection equation $ M_{nf} = \frac{2EI}{L}(2\theta_n + \theta_f - 3\psi) + FEM_{nf} $[1] ~Eq.2
$ M_A_D = \frac{2\times(EI)}{3}\times \theta_D -22.5 $
$ M_D_A = \frac {2\times(EI)}{3}\times(2\theta_D) +22.5 $
$ M_B_E = \frac{2\times(EI)}{3}\times\theta_E $
$ M_E_B = \frac{2\times(EI)}{3}\times(2\theta_E) $
$ M_C_D = \frac{2\times(EI)}{3}\times(2\theta_C+\theta_D) $
$ M_D_C = \frac{2\times(EI)}{3}\times(2\theta_D+\theta_C) $
$ M_D_E = \frac{2\times(EI)}{3}\times(2\theta_D+\theta_E) +18.75 $
$ M_E_D = \frac{2\times(EI)}{3}\times(2\theta_E+\theta_D) - 18.75 $

Step 5: Doing equilibrium analysis at joints C,D and E yields:
$ M_E_B+M_E_D = 0 $ (from joint E)
$ M_D_E + M_D_C + M_D_A = 0 $ (from joint D)
$ M_C_D = 0 $ (from joint C)

Step 6: Apply equations from step 4 into equations from step 5 and solve for the three unknowns
$ [\frac{2\times(EI)}{3}\times(2\theta_E)]+[\frac{2\times(EI)}{3}\times(2\theta_E+\theta_D)-18.75]=0 $
$ [\frac{2\times(EI)}{3}\times(2\theta_D+\theta_E)+18.75]+[\frac{2\times(EI)}{3}\times(2\theta_D+\theta_C)]+[\frac{2\times(EI)}{3}\times(2\theta_D)+22.5]=0 $
$ \frac{2\times(EI)}{3}\times(2\theta_C+\theta_D)=0 $
Solving this system of equations yields:
$ \theta_C=\frac{6.67}{EI} = 0.0556 rad $
$ \theta_E=\frac{10.2}{EI} = 0.085 rad $
$ \theta_D=\frac{-13.35}{EI} = -0.1113 rad $

Step 7 Input the new $ \theta $ values into the slope-deflection equations from step 4 and solve for the end moments of each member
$ M_A_D = \frac{2\times(EI)}{3}\times \theta_D -22.5 = -31.4kN\cdot m $
$ M_D_A = \frac {2\times(EI)}{3}\times(2\theta_D) +22.5 = 4.7kN\cdot m $
$ M_B_E = \frac{2\times(EI)}{3}\times\theta_E = 6.8kN\cdot m $
$ M_E_B = \frac{2\times(EI)}{3}\times(2\theta_E) = 14kN\cdot m $
$ M_C_D = \frac{2\times(EI)}{3}\times(2\theta_C+\theta_D) = 0 kN\cdot m $
$ M_D_C = \frac{2\times(EI)}{3}\times(2\theta_D+\theta_C) = -13.4kN\cdot m $
$ M_D_E = \frac{2\times(EI)}{3}\times(2\theta_D+\theta_E) +18.75 = 7.8kN\cdot m $
$ M_E_D = \frac{2\times(EI)}{3}\times(2\theta_E+\theta_D) - 18.75 = -14kN\cdot m $

Step 8: Using the end moments of each member break the frame down into its members and solve for the axial, shear and moment forces to find the support reactions shown in the FBD below. (Figure 6)

Figure 6: Free Body Diagram

Step 9: Using prior knowledge of structural analysis and now knowing all of the forces and reactions of the members the; axial force, shear force, bending moment and deflected shape diagrams can be constructed as shown in the figures below. (Figures 7-10)

Figure 7: Axial Force Diagram

Figure 8: Shear Force Diagram

Figure 9: Bending Moment Diagram

Figure 10: Deflection Diagram


  1. 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning..
  2. 2.0 2.1 2.2 2.3 2.4 2.5 2.6 Hibbeler, R.C. (2012). Structural Analysis (8th ed). Upper Saddle River, NJ: Prentice Hall.
  3. Hibbeler, R.C. (2010). Statics & Dynamics (12th ed). Upper Saddle River, NJ: Prentice Hall.
  4. 4.0 4.1 4.2