| 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.
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