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Basic triangular truss element: ABC
Additional truss element : AD & BD
Classification of Coplanar trusses
Simple truss - are constructed by starting with basic triangular element and
then connecting member/s to form an additional element.
Basic triangular truss element : ABC
Additional truss element : AD , CD , DE & BE
Note that simple truss need not have to consist of entirely
of triangles to be stable.
Simple truss : ABC and CDE
Connection : Joint C and member BE
Compound truss are formed by connecting two or more simple
trusses together. This type of trusses are quite often used to support loads
acting over a large span since it is cheaper to construct a somewhat
lighter compound truss than to use a heavier single simple truss.
Types of compound truss;
Type 1 : Simple truss are connected by a common joint and bar.
Simple truss : ABC and DEF
Connection : member EC , AD , and BF
Type 2 : Simple truss are connected by bar/s.
Type 3 : Simple truss are connected where bars of a large simple truss,
called the Main Truss, have been substituted by simple truss, called
Secondary trusses.
Main simple truss : truss ABCD
Secondary simple truss: truss AB , BC , & CD
Complex truss is a peculiarly arranged truss which, whilst statically
determinate, has a set of simultaneous equilibrium equations which
can not be readily uncoupled. Truss which that cannot be classified as
either simple or compound truss.
Method of Substitute member:
Procedure for Analysis:
1. Solve for support reactions.
2. Reduction to stable simple truss ( substitute member ).
3. External loadings on simple truss.
4. Removal of external loadings on simple truss and placing of unit load on
the member substituted.
5. Solving for the unknown force (x).
Si = S’i + xsi
Example: Determine the force in each member of the complex truss shown
in figure a. Assume joints B, F , and D are on the same horizontal line.
State whether the members are in tension or compression.
By inspection , each joint has three unknown member forces. Thus can’t be
solved by methods of joints only.
SOLUTION:
1. Support reactions:
Taking FBD of whole structure we have;
Ax = 5 KN
Ay = 4.38 KN
Ey = 4.38 KN
2. Substitute member:
By removing member CF and having a substitute member BD. The
truss is reduced to a stable simple truss. As shown in Figure.
3. External loadings on simple truss:
With the support reactions on the truss have been computed. By using
methods of joints, the stress at each member is computed
and recorded in column 2 of the table below designated as ( S’i ) .
Member S'i si xsi Si Remarks
BC 3.54
CD -3.54
AF 0
EF 0
BE 0
DE -4.38
AD 5.34
BD -2.50
BA 2.50
4. Removal of external loadings on simple truss;
Placed a 1 unit load on the truss ( members that are deducted ) .
These equal but opposite forces create no external reactions on the truss.
The stresses are computed in the same sequence as in step3. The results
are designated as ( si ) force analysis and are recorded at column 3 of the
table shown below.
Note: member analysis by method of joints excluding the
support reactions. The I unit load thus not create any support reactions.
Member S'i si xsi Si Remarks
BC 3.54 -0.707
CD -3.54 -0.707
AF 0 0.833
EF 0 0.833
BE 0 -0.712
DE -4.38 -0.250
AD 5.34 -0.712
BD -2.50 1.167
BA 2.50 -0.250
5. Solving for the unknown force ( x ).
In particular , for the substituted member BD , The force SBD = S’BD + xsBD.
Since member BD does not actually exist on the original truss, set x to have
a magnitude such that it yields zero-force in BD.
S’BD + xsBD = 0
Solving for x ; -2.50 + x(1.167) = 0
x = 2.142
Substitute value of x in equation 1.0 to solve for the stresses of each
member. Stress are listed @ column5 of the table taking negative as
compression and positive in tension.
Member S'i si xsi Si Remarks
BC 3.54 -0.707 -1.51 2.02 Tension
CD -3.54 -0.707 -1.51 -5.05 Compression
AF 0 0.833 1.78 1.78 Tension
EF 0 0.833 1.78 1.78 Tension
BE 0 -0.712 -1.53 -1.53 Compression
DE -4.38 -0.250 -0.536 -4.91 Compression
AD 5.34 -0.712 -1.52 3.81 Tension
BD -2.50 1.167 2.50 0
BA 2.50 -0.250 -0.535 1.96 Tension