Simple Truss Problems Linear Algebraic Systems
Post on 20-Feb-2022
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Definition of a truss
● A truss is a rigid frame consisting of slendermembers connected at their endpoints.
truss
A
D
C
B
E
Simple Trusses
The simplest configuration for a stable truss is atriangle as shown in red above.
A
D
C
B
E
Simple Trusses
The simplest configuration for a stable truss is atriangle as shown in red above.
The Triangular Truss
45°
2 m
A
B
C
Ay=500 N
500 N
2 m
Ax=500 N
Cy=500 N
500 N
FBA
FBC
Free Body Diagram (FBD)500 N
2 m
0 N
Static Equilibrium
500 N
FBA
FBC
B● The joint B is not moving and is therfore
said to be in static equilibrium.
Static Equilibrium
500 N
FBA
FBC
B● The joint B is not moving and is therfore
said to be in static equilibrium.
● Physically speaking, this means that thereare no unbalanced forces so if we add allof the forces acting in the x-direction,their sum should be zero.
Static Equilibrium
500 N
FBA
FBC
B● The joint B is not moving and is therfore
said to be in static equilibrium.
● Physically speaking, this means that thereare no unbalanced forces so if we add allof the forces acting in the x-direction,their sum should be zero.
● The same is true for forces acting in the y-direction
Unique vs. Singular Systems
● Some systems of equations do not haveunique solutions.
y
x
unique solution
singular system
Statically Indeterminate Example
● Previously, we showed a system of twoequations that had two unknowns. Now,adding member BD and CD:
A C
B
Statically Indeterminate Example
● Previously, we showed a system of twoequations that had two unknowns. Now,adding member BD and CD:
A C
B D
C
Statically Indeterminate Example
● Previously, we showed a system of twoequations that had two unknowns. Now,adding member BD and CD:
A C
B D
C
0500 =+−=∑ BDxBCx FFF
At pin B:
0=−=∑ yy BABCy FFF
Statically Indeterminate Example
● Previously, we showed a system of twoequations that had two unknowns. Now,adding member BD and CD:
A C
B D
C
0500 =+−=∑ BDxBCx FFF
At pin B:
0=−=∑ yy BABCy FFF
Hence, we have onlytwo equations forthree unknowns.
Expressing Sets of Linear AlgebraicEquations in Matrix Form
● Summation of the forces in the x and y directionscan be written as:
0)1(45cos
500)0(45sin
=−°−=+°−
BABC
BABC
FF
FF
● These two equations can be equivalently expressedin matrix form as...
Generalized System of LinearEquations
mnmnmm
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
=+++
=+++=+++
...
...
...
2211
22222121
11212111
MMMM
Generalized MatrixRepresentation of Linear System
=
mnmm
n
n
aaa
aaa
aaa
L
MLMM
L
L
21
22221
11211
A
=
nx
x
x
M2
1
x
=
mb
b
b
M2
1
b
In matrix A, m is the index that identifies the row and n is theindex that identifies the column. Thus, the requirement that thenumber of unknowns must equal the number of equations in orderfor a unique solution to exist, is at the root of matrixmultiplication. i.e. m must be equal to n
Solution Techniques
● One method of solving involves successiveelimination of variables until only oneequation and one unknown variable remains.Gauss Elimination
● Cramer’s Method is based on finding matrixdeterminants for the system
● Another technique particularly suited toMATLAB is based on the matrix inversemethod
Solution of Linear System UsingMATLAB
Start script file
Ask user fornumber of equations
Call function build.mto create matrix A
Call function build.mto create matrix B
Send A and B to functionlinsolve1.m
Display results
end file
Script matalg.m
● Calls the function build.m twice• build.m performs a dedicated task to inout data
● Calls a separate function, linsolve1.m to dothe dedicated task of computing the solution
● Displays the answer
Function linsolve1.m
● This function introduces use of the MATLAB
backslash( \ ), matrix operator to solvelinear systems of the general form:
bAx =
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