The Direct Stiffness Method
for Truss Analysis

Geometry Definition

Material Properties and Cross-Sections

For this analysis, we'll assign member properties by group:

Table 3: Member group properties including Area (A) and Elastic Modulus (E)

Where:

• A = Cross-sectional area of the member
• E = Young's modulus (elastic modulus) of the material

Applied Loads

External forces are applied to the nodes as follows:

Table 4: Applied nodal loads in x and y directions

Step 1: Calculate Member Properties

For each member in the truss, we need to determine:

Member starting node

Member ending node

Member rotation angle from horizontal

Member direction

Figure 2: Generic member showing local coordinate system, orientation angle θ, length L, and global x-y coordinates

Member length (L):

\[ L = \sqrt{(x_E - x_S)^2 + (y_E - y_S)^2} \]

Where \( (x_S, y_S) \) and \( (x_E, y_E) \) are the coordinates of the start and end nodes, respectively.

Member orientation angle (θ):

\[ \theta = \arctan{\left( \frac{y_E - y_S}{x_E - x_S} \right)} \]

This angle is measured counterclockwise from the positive x-axis.

Direction cosines:

\[ c = \cos(\theta) = \frac{x_E - x_S}{L} \]

\[ s = \sin(\theta) = \frac{y_E - y_S}{L} \]

These values will be used repeatedly in forming the member stiffness matrices.

For example, for member 0:

\[ \theta_0 = \arctan{\left( \frac{2.5 - 0}{3 - 0} \right)} = 39.81^\circ \]

\[ c_0 = \cos(39.81^\circ) = 0.768 \]

\[ s_0 = \sin(39.81^\circ) = 0.640 \]

Step 2: Form Individual Member Stiffness Matrices

Each truss member has a stiffness matrix that relates the forces at its ends to the displacements at its ends. In the global coordinate system, a 2D truss member has 4 degrees of freedom (DOF): horizontal and vertical displacements at each end node.

The member stiffness matrix in global coordinates is:

Where:

• k_i = Stiffness matrix for member i
• A = Cross-sectional area
• E = Elastic modulus
• L = Member length
• c = cos(θ)
• s = sin(θ)

The rows and columns correspond to DOFs in the order of horizontal and vertical stiffness components at the start node (S) and end node (E): \( [ S_x, S_y, E_x, E_y ] \).

For example, the stiffness matrix for member 0 is:

Step 3: Assemble the Global Structure Stiffness Matrix

The global stiffness matrix K combines all member stiffness matrices by summing contributions at each degree of freedom. For a truss with N nodes, K is a (2N × 2N) matrix.

Assembly process:

1. Initialize K as a (2N × 2N) zero matrix
2. For each member:
   • Identify the global DOF indices for its start and end nodes
   • Add the member's stiffness contributions to the appropriate positions in K

The global DOF numbering typically follows this pattern:

• Node 0: DOFs 0 (x-direction), 1 (y-direction)
• Node 1: DOFs 2 (x-direction), 3 (y-direction)
• Node i: DOFs 2i (x-direction), 2i+1 (y-direction)

For this example:

Only member 0 and member 4 connect to node 0, so the top left corner of the global stiffness matrix receives contributions from just these two members. We can use the above method to calculate the member stiffness matrices as:

Now, assembling these two members into an empty global stiffness matrix gives:

Global Stiffness Matrix K with only members 0 and 4 assembled

As we continue for all members and all nodes, we will obtain the full global stiffness matrix for this truss:

Global Structure Stiffness Matrix K

Step 4: Apply Boundary Conditions (Reduce the Stiffness Matrix)

Support conditions constrain certain DOFs to zero displacement. To solve for unknown displacements, we remove rows and columns corresponding to these constrained DOFs, creating the reduced stiffness matrix K_R.

For example:

• Pin support: Both x and y displacements = 0 (remove 2 DOFs)
• Roller support (vertical): y displacement = 0 (remove 1 DOF)
• Roller support (horizontal): x displacement = 0 (remove 1 DOF)

In this example, we know that Node 0 is a pin support and Node 6 is a roller support in the vertical direction.

Therefore, both columns and rows can be removed from the global stiffness matrix K corresponding to \( N_{0,x} \), \( N_{0,y} \), and \( N_{6,y} \) to form the reduced stiffness matrix:

Reduced Stiffness Matrix K_R

Step 5: Form the Reduced Load Vector

Similarly, create a load vector Q containing all applied forces at each DOF:

Complete Load Vector Q with DOF labels

Then form the reduced load vector Q_R by removing entries corresponding to constrained DOFs, just like we did for the stiffness matrix.

Step 6: Solve for Unknown Displacements

The fundamental stiffness equation relates forces to displacements:

Where D_R is the vector of unknown displacements.

Now, solving for displacements:

This requires inverting (or more commonly, performing LU decomposition or Cholesky decomposition on) the reduced stiffness matrix. Although this is computationally intensive when done by hand, it's straightforward to program on a computer using Matlab, Python, or any other language suited for matrix algebra.

Alternatively, the computation can be reduced further for hand calculations by partitioning the stiffness, force, and displacement matrices.

Displacement results:

The calculation of the reduced displacement vector results in displacements for all of the unknown degrees of freedom:

By filling in the known 0-displacements at supports, we now know all node displacements in the system.

Table 5: Nodal displacements in x and y directions

Step 7: Calculate Member Axial Forces

Once nodal displacements are known, individual member axial forces can be calculated using the transformation:

Where:

• q_i = Axial force in member i (positive = tension, negative = compression)
• \( Δ_{S,x} \), \( Δ_{S,y} \) = Displacements at start node
• \( Δ_{E,x} \), \( Δ_{E,y} \) = Displacements at end node

For this example, the member axial force calculation for member 0 would be:

Complete member force results:

Repeating this calculation for all members gives us the axial forces in each member:

Table 6: Member axial forces for all members

Figure 3: Truss with members colored by force magnitude (scale in kips, tension positive)

Step 8: Calculate Support Reactions

Support reactions are found by multiplying the full stiffness matrix by the complete displacement vector (including zero displacements at supports):

\[ Q = K \cdot D \]

The reaction forces are the values of Q at the constrained DOFs. If there were applied loads at supports, subtract these to get the net reactions.

Support reaction results:

Table 7: Reaction forces at each support

Static Equilibrium Check

Sum all forces (applied loads and reactions) in each direction:

\[ ΣF_x = 0 \]

\[ ΣF_y = 0 \]

\[ ΣM = 0 \]

Where M is the moment about any point in the structure.

If these don't sum to zero (within numerical tolerance), there's an error in the analysis.

Comparison with Hand Calculations

For simple, statically determinate trusses, compare member forces with results from the method of joints or method of sections.

Physical Reasonableness

• Do compression members make sense given the loading?
• Are displacement magnitudes reasonable?
• Do reactions point in expected directions?

Sign Conventions

• Member forces: Positive = tension, negative = compression
• Displacements: Positive in the positive x or y direction
• Reactions: Report as positive when acting in the positive coordinate direction

Numerical Considerations

• Matrix conditioning: Very stiff members relative to flexible ones can cause numerical issues
• Units consistency: Ensure all units are consistent (e.g., all forces in kN, lengths in m, stress in MPa)
• Precision: Use sufficient numerical precision to avoid round-off errors in matrix operations

Modeling Tips

• Number nodes and members systematically
• Define member orientation consistently (though the method handles any orientation)
• Leverage symmetry when possible to reduce problem size
• Start with simplified models before adding complexity