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 , DC , CE , BE , BF & EF

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