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### Direct Displacement Method

Introduction

The direct displacement method is another technique that can be used to analyze indeterminate structures. This method can be generalized and is commonly used in structural analysis software.

In this method, all degrees of freedom of a structure are restrained, i.e. "locked", and the member fixed end forces are calculated due to any applied loads on the member. As in moment distribution, each degree of freedom is then independently released, i.e."un-locked", and the member end forces are determined due to an application of a unit displacement that corresponds to each degree of freedom. This displacement can either be a rotation or a translation. The actual member end forces are then calculated by satisfying equilibrium conditions at each of the degrees of freedom.

The steps required to complete an analysis based on the direct displacement method are illustrated in the following simplified example (more detailed examples are given below):

Given a structure:

i) Determine the kinematic degree of indeterminacy, i.e. the number of degrees of freedom.

The kinematic degree of freedom is the number of independent joint displacements (rotations and translations).

This sample problem has two degrees of kinematic indeterminacy since the structure can undergo rotations at both joints B and C.

However, the moment at Joint C is determinate. However, since the moment at C is zero, modified stiffness can be utilized for member BC. Modified stiffness is used by not including the rotation at C as an unknown reaction. This reduces the kinematic indeterminacy of the structure to only one.

The figures below show the kinematic degree of freedom associated with a variety of  two-dimensional plane structures.  For reference, the number of degrees of static indeterminacy is also listed.  Note: axial deformation is not considered.

Structure Kinematic Degree of Freedom Degree of Static Indeterminacy
2 0
1 1
2 4
4 2
a)   ** 2 2
b)   ** 1 2
2 2
3 1

** a) This problem can also be classified as twice kinematically indeterminate if the effect of the cantilever portion is considered as joint loads at the roller support on the far right.

** b) This problem can also be classified as once kinematically indeterminate if modified stiffness is used for the beam between the two rollers in conjunction with the correct fixed end moment at the intermediate roller. This accounts for the fact that the moment at the roller on the far right is known.
For more details, see the two-span cantilever example below.

ii) Determine the restraining, i.e. fixed end forces of each member.

In this example, the structure is separated into two sections;

The corresponding sections and their fixed end forces (refer to the Table of Fixed End Moments or a structural analysis text book for these values),

note: since modified stiffness is being used on member BC the moment at C is equal to 0 and the moment at B is wl2/8 rather than wl2/12.

iii) Calculate the member end forces due to the application of a unit displacement in the direction of and at the location of each degree of freedom. (clockwise rotation is positive in Direct Displacement)

The resulting member end forces on each member due to a unit rotation at B are:

and

note: the member end forces are those required to induce a unit rotation at support B. For derivations, please refer to any structural analysis text.

iv) Determine the equilibrium conditions that correspond to each degree of freedom.

The equilibrium conditions are found by using statics to sum all forces that act on a specific degree of freedom. In this example, the sum of the fixed end moments and those induced due to the unknown degrees of freedom, X1, at B.

The equation of statics at B is expressed as,

PL/8 - wL2/8 + (4EI/L + 3EI/L)*X1 = 0

Where PL/8 and wL2/8 are the fixed end forces and (4EI/L + 3EI/L) is the summation of the stiffness coefficients at Joint B, i.e., the moments due to unit rotation at Joint B. These coefficients need to be multiplied by the unknown rotation X1, so that equilibrium can be attained. The left hand side equation is set equal to zero, since there is no external moment applied at Joint B.

Repeat this step for each degree of freedom. This will yield a number of equations that equal the number of the unknown displacements.

v) Calculate the unknown displacements.

X1=(wl2/8-PL/8)*L/(7EI)

vi) Utilize the calculated unknowns to determine the member end forces.

The resulting member end forces are now found by adding a correction moment to the fixed end moments. These correction forces are equal to the member end forces corresponding to a unit displacement at the end of the member, multiplied by the associated displacement X.

In this example;

 MAB = -PL/8 + (2EI/L)*X1 YA = P/2 - ( 6(EI)/L2 ) * X1 MBA = PL/8 + (4EI/L)*X1 YB - left = P/2 + ( 6(EI)/L2 ) * X1 MBC = -wL2/8 + (3EI/L)*X1 YB - right = 5wl/8 - ( 3(EI)/L2 ) * X1 MCB = 0 YC = 3wl/8 + ( 3(EI)/L2 ) * X1

vii) Calculate remaining member end reactions.

Use the fixed end moments and any applied loads in conjunction with the static equations to calculate the remaining member end reactions.

The following detailed examples further define and illustrate this method.

Examples

Determine the joint reactions of a two-span indeterminate beam.
Determine the joint reactions of a three-span indeterminate beam.
Determine the joint reactions of a two-span indeterminate beam with a cantilever.
Determine the joint reactions of a frame.