Math 272 - Tower Truss Project
Math 272 - Tower Truss Project - Background
Goal
To design a determinate truss that will support the strongest or most distant loads.
Background
You are to design and test (in simulation) a tower of height H meters, constructed out of a truss. The tower will be supported
at its base by one joint at a fixed pin, and another joint at a roller.
For this task, you will be given a supply of tubular steel, with the following properties:
• The material has modulus of elasticity of E = 2.1 × 1011 Pascals
• The tubular steel has exterior diameter 3 cm, and interior diameter 2 cm, giving cross-sectional area A = 3.9 × 10−4
m2 , moment of inertia I = 3.2 × 10−8 m4 , and linear density of ρ = 3.1 kg/m of length.
The beams can fail in the following ways:
• to buckle in compression, the compressive force on a beam is at the Euler load:
Fbuckling =
EIπ 2
L2
This is approximately 60,000 N for a 1 m long beam.
This failure mode does depend on the length of the beam, with longer beams failing at lower forces.
• to crush in compression requires 280 MPa, or 280 × 106 · (Area) ≈ 110, 000 N. This failure mode does not depend on
the length of the beam.
• to yield in tension requires a stress of 250 MPa, or 250 × 106 · (Area) ≈ 98, 000 N. This failure mode does not depend
on the length of the beam.
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Truss Representation
A flexible format for representing the truss is made by specifying two matrices and two scalars:
• joints - A list of (x, y) coordinates representing the joint locations, and
• beams - A list of (ji , jj ) pairs representing a beam joining joint i and joint j.
• pinned_joint - the number of the joint attached to the pin
• roller_joint - the number of the joint attached to the roller
Example:
This truss can be described by the matrices below:
joints
2,
0,
2,
0,
2,
0,
2,
= [0 , 0;
0;
3;
3;
6;
6;
9;
9];
beams = [ 1 , 2 ; % j o i n t numbers
1 , 3;
2 , 3;
2 , 4;
3 , 4;
3 , 5;
4 , 5;
4 , 6;
5 , 6;
5 , 7;
6 , 7;
6 , 8;
7, 8];
% x/y l o c a t i o n s , in m
pinned joint = 1;
r o l l e r j o i n t = 2;
Using these matrices builds the truss and, implicitly, assigns the joint and beam numbering as shown below:
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Member Forces
To handle various design possibilities, we will need to compute the x and y components of the forces of each beam on a
joint, given the (x, y) location of the joints. The diagram below outlines how that calculation can be performed using vector
components, based on the joint locations at each end of a beam. Note that no explicit angle calculations are used in this
method, which (trust me!) simplifies the programming substantially.
Joint Configuration
Beam Vector = ~v1→2 = [x2 − x1 , y2 − y1 ]
1
~v1→2
Unit Beam Vector = ~u =
||~v1→2 ||
Force Vector = ~uFi , where Fi is the resultant force in the beam i under the applied load
(x2 − x1 )
Fx =
×Fi
||~v1→2 ||
|
{z
}
Coeff of Fi in x/horiz’l equation for second joint
(y2 − y1 )
×Fi
Fy =
||~v1→2 ||
|
{z
}
Coeff of Fi in y/vert’l equation for second joint
Beam Vector = ~v2→1 = [x1 − x2 , y1 − y2 ]
1
~v2→1
Unit Beam Vector = ~u =
||~v2→1 ||
Force Vector = ~uFi
(x1 − x2 )
×Fi
Fx =
||~v2→1 ||
|
{z
}
Coeff of Fi in x/horiz’l equation for second joint
(y1 − y2 )
Fy =
×Fi
||~v2→1 ||
|
{z
}
Coeff of Fi in y/vert’l equation for second joint
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Notes:
• All beam forces are assumed to be positive which represents tension. That is why we point the force vectors away
from the joint we are studying.
• With this construction, the signs of the forces all work themselves out directly through the calculated values (coefficients
are negative when the force is applied left or down, positive otherwise). By working with the vector forms, you do not
need to worry about different cases for different geometric arrangements.
• The entries in the A matrix of A F = B are just the coefficients indicated in the diagram above. Multiplying A by F
produces the full term, (coeff) ×Fi .
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