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Overview

Truss elements are 2D structural members that carry only axial forces (tension or compression). They are ideal for modeling:
  • Truss bridges
  • Roof trusses
  • Space frames (2D projection)
  • Bracing systems
  • Tension/compression members
Truss elements have 2 degrees of freedom per node (translations in x and y). Moments and shear forces are not transmitted through truss connections.

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Adding Truss Elements

model.add_truss(
    id=1,
    node_i_id=1,
    node_j_id=2,
    section_name='L4x4x1/2'
)
Parameters:
  • id (int) → Truss member ID
  • node_i_id (int) → Initial node ID
  • node_j_id (int) → Final node ID
  • section_name (str) → Section name (typically defined by area)

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Degrees of Freedom

Each node has only 2 translational DOF: 
Node i: [Uxi, Uyi]
Node j: [Uxj, Uyj]
Where:
  • Ux = Displacement in x direction
  • Uy = Displacement in y direction
  • No rotational DOF (pinned connections assumed)
The absence of rotational DOF means truss elements automatically have pinned connections at both ends.

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Element Formulation

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Local Stiffness Matrix

The local stiffness matrix for a truss element is: 
K_local = (E * A / L) * [[ 1, -1],
                         [-1,  1]]
Where:
  • E = Young’s modulus
  • A = Cross-sectional area
  • L = Element length

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Transformation Matrix

The transformation from local to global coordinates: 
L = sqrt((xj - xi)² + (yj - yi)²)
cx = (xj - xi) / L  # Direction cosine x
cy = (yj - yi) / L  # Direction cosine y

T = [[cx, cy,  0,  0],
     [ 0,  0, cx, cy]]

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Global Stiffness Matrix

K_global = T^T * K_local * T
This results in a 4×4 global stiffness matrix relating the 4 DOF.

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Distributed Loads

You can apply distributed axial loads along truss members: 
from milcapy.loads.load import DistributedLoad

truss.set_current_load_pattern('Wind')
truss.set_distributed_load(DistributedLoad(
    p_i=5.0,  # Axial load at node i (force/length)
    p_j=5.0   # Axial load at node j (force/length)
))
The equivalent nodal forces are computed as: 
F_i = (2*p_i + p_j) * L / 6
F_j = (2*p_j + p_i) * L / 6
Distributed loads are automatically converted to equivalent nodal forces during assembly.

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Example: Simple Truss

import milcapy as mx

# Create model
model = mx.Model()

# Define nodes for a simple truss
model.add_node(1, 0, 0)     # Left support
model.add_node(2, 4, 0)     # Right support  
model.add_node(3, 2, 3)     # Top node

# Define material (only E and A needed for trusses)
model.add_material('Steel', E=200e9, v=0.3, rho=7850)

# Define section (area is the key property)
model.add_section('Rod', material='Steel', A=0.001)  # 1000 mm²

# Add truss members
model.add_truss(id=1, node_i_id=1, node_j_id=3, section_name='Rod')
model.add_truss(id=2, node_i_id=2, node_j_id=3, section_name='Rod')
model.add_truss(id=3, node_i_id=1, node_j_id=2, section_name='Rod')

# Apply boundary conditions
model.add_support(node_id=1, ux=True, uy=True)  # Pin support
model.add_support(node_id=2, ux=False, uy=True) # Roller support

# Apply loads
model.add_load_pattern('Live')
model.add_nodal_load(node_id=3, fx=0, fy=-10000)  # 10 kN downward

# Solve
model.solve('Live')

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Example: Bridge Truss

import milcapy as mx

# Create a Warren truss bridge
model = mx.Model()

# Bottom chord nodes
for i in range(6):
    model.add_node(i+1, i*2, 0)

# Top chord nodes  
for i in range(5):
    model.add_node(i+7, i*2+1, 2)

# Material and sections
model.add_material('Steel', E=200e9, v=0.3, rho=7850)
model.add_section('Chord', material='Steel', A=0.003)
model.add_section('Web', material='Steel', A=0.002)

# Bottom chord
for i in range(5):
    model.add_truss(i+1, i+1, i+2, 'Chord')

# Top chord
for i in range(4):
    model.add_truss(i+6, i+7, i+8, 'Chord')

# Web members (diagonal and vertical)
model.add_truss(11, 1, 7, 'Web')   # Diagonal
model.add_truss(12, 7, 2, 'Web')   # Diagonal
model.add_truss(13, 2, 8, 'Web')   # Diagonal
# ... continue pattern

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Analysis Results

After solving, you can extract:

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Axial Force

# Get member displacements
u = model.get_member_displacements(member_id=1)

# Calculate axial force
E = truss.section.E()
A = truss.section.A()
L = truss.length()

# Local displacement
u_local = truss.transformation_matrix() @ u
axial_force = (E * A / L) * (u_local[1] - u_local[0])

print(f"Axial force: {axial_force:.2f} N")
print(f"Stress: {axial_force/A:.2f} Pa")

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Member Stress State

if axial_force > 0:
    print("Member in TENSION")
else:
    print("Member in COMPRESSION")
Important: Truss members in compression must be checked for buckling. Milcapy computes axial forces only; stability checks must be performed separately.

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Best Practices

Node Ordering

Node order affects force sign convention. Positive force = tension from node i to node j.

Section Properties

Only E (modulus) and A (area) affect truss behavior. Other properties are ignored.

Stability

Check compression members for buckling using Euler’s formula or code provisions.

Geometry

Avoid nearly collinear members which can cause numerical instability.

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Comparison with Beam Elements

FeatureTruss ElementBeam Element
DOF per node2 (Ux, Uy)3 (Ux, Uy, Rz)
Forces transmittedAxial onlyAxial, shear, moment
End connectionsAlways pinnedCan be fixed/pinned/released
Use caseTension/compressionBending members
Stiffness matrix4×46×6
For members that may carry bending moments, use beam elements instead of trusses.

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