Home / Courses / Finite Element Methods / Bar Elements

Finite Element Methods · Chapter 3 of 10 · Intermediate

Bar and Truss Elements

A bar is a spring with a stiffness AE/L and a stress to recover. Tilt it into a plane and a transformation turns the same element into a truss member, the workhorse of structural FEA.

01

Readiness check

This chapter builds the bar and truss element. Tick only what you can do closed-notes.

  • Recall axial stress σ = P/A and δ = PL/AE.
  • Use the spring assembly from Chapter 2.
  • Recall sine and cosine of an angle.
  • Multiply a vector by a row of direction cosines.
  • Resolve a force into x and y components.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsRevisit assembly in Chapter 2.
3 or more weak itemsReview truss analysis in Statics: Structural Analysis.
02

The core idea

A bar element has stiffness AE/L along its axis. To place it in a plane truss, a transformation built from the bar's direction cosines maps that axial stiffness into global x and y.

local: (AE/L)[1 −1; −1 1]global block: (AE/L)[c² cs; cs s²]σ = E(u₂ − u₁)/L

Along its own axis a bar behaves exactly like a spring of stiffness AE/L, so stepped bars assemble and solve just as springs did, with the bonus that an axial stress σ = E(u₂ − u₁)/L can be recovered from the displacements. A truss member, though, points in an arbitrary direction, so its two axial degrees of freedom must be expressed in the global x and y of the structure. The transformation uses the direction cosines c = cos θ and s = sin θ of the member, turning the simple axial stiffness into a global element matrix whose blocks scale by c², cs, and s². Assembling these global element matrices and solving gives nodal displacements, from which each member's axial force and stress follow.

The skill works when: you use AE/L axially and transform with direction cosines for inclined members.
The skill breaks down when: a tilted member is assembled with its local stiffness, ignoring its orientation.
The concept. A truss member at angle θ in the global frame. Its axial stiffness AE/L is mapped into global x and y through the direction cosines c and s, producing the global element stiffness.
03

The skills, taught in order

Five skills cover the bar stiffness, stepped bars, stress recovery, transformation, and member forces.

3.1 The bar element stiffness

Along its axis a bar of area A, modulus E, and length L has stiffness AE/L and the element matrix (AE/L)[1, −1; −1, 1], identical in form to a spring. This is the local stiffness, valid in the bar's own coordinate.

3.2 Stepped bars and assembly

A bar of changing section is modelled as several elements in series, each with its own AE/L. Assembly and solution proceed exactly as for springs, and because the elements carry the same axial force in series, the thinner sections see the higher stress.

3.3 Axial stress recovery

Once displacements are known, the strain in an element is (u₂ − u₁)/L and the stress is σ = E(u₂ − u₁)/L. Recovering and checking these stresses against σ = P/A is the hand calculation that validates an axial model.

3.4 Coordinate transformation

A member at angle θ is transformed into global coordinates using c = cos θ and s = sin θ. The global element stiffness is built from blocks scaled by c², cs, and s², so a horizontal member loads only x and a vertical one only y, with inclined members coupling both.

Member angle θc²css²
0° (horizontal)100
30°0.750.4330.25
45°0.50.50.5
90° (vertical)001

3.5 Truss member forces

The axial force in a member is (AE/L) times its axial elongation, found by projecting the global nodal displacements onto the member direction. Positive is tension, negative compression, and these member forces must satisfy equilibrium at every joint.

Engineering connection: the transformation idea here generalises to beams and frames, where rotations join translations as degrees of freedom.

04

Worked example 1: a stepped bar

A stepped steel bar (E = 200 GPa) has two segments in series: element 1 with A₁ = 400 mm², L₁ = 500 mm, and element 2 with A₂ = 200 mm², L₂ = 500 mm. The bar is fixed at one end and loaded by 20 kN at the free end. Find the total elongation and the stress in each segment.

Figure 1. Two bar elements in series carry the same axial force, so the thinner segment 2 has twice the stress of segment 1.
  1. ProblemFind the elongation and segment stresses for the stepped bar in Figure 1.
  2. Given / findE = 200 000 MPa, A₁ = 400, L₁ = 500, A₂ = 200, L₂ = 500, P = 20 kN. Find δ and σ₁, σ₂.
  3. AssumptionsLinear elastic; each segment is one uniform bar element; series load path.
  4. ModelElement stiffnesses k = AE/L; elongation from the series stiffness; stress σ = P/A in each (equal force).
  5. Equationsk₁ = A₁E/L₁ = 160 000, k₂ = A₂E/L₂ = 80 000 N/mmσ = P/A
  6. SolveSeries stiffness = 1/(1/160 000 + 1/80 000) = 53 333 N/mm, so δ = 20 000/53 333 = 0.375 mm. σ₁ = 20 000/400 = 50 MPa; σ₂ = 20 000/200 = 100 MPa.
  7. CheckBoth elements carry the full 20 kN in series, so the thinner one has the higher stress, as the result shows. The elongation equals the sum of segment stretches, 0.125 + 0.25 mm.
  8. ConclusionStepped bars are just springs in series with stresses recovered. The critical section is the smallest area, where stress peaks.
Result. δ = 0.375 mm; σ₁ = 50 MPa, σ₂ = 100 MPa.
05

Worked example 2: transforming an inclined member

A truss member (E = 200 GPa, A = 600 mm², L = 2 m) is oriented at 30° to the horizontal. Find its axial stiffness AE/L and the global element stiffness terms scaled by c², cs, and s².

Figure 2. The member's axial stiffness is distributed into global x and y by the direction cosines. The horizontal coupling (c²) is largest at this shallow angle.
  1. ProblemFind AE/L and the global stiffness terms for the member in Figure 2.
  2. Given / findE = 200 000 MPa, A = 600 mm², L = 2000 mm, θ = 30°. Find AE/L and the c², cs, s² terms.
  3. AssumptionsTwo-dimensional truss member; direction cosines from the member angle.
  4. ModelCompute AE/L, then multiply by c² = cos²θ, cs = cosθ sinθ, and s² = sin²θ.
  5. EquationsAE/L = (600 × 200 000)/2000c = cos 30° = 0.866, s = sin 30° = 0.5
  6. SolveAE/L = 60 000 N/mm. c² = 0.75 → 45 000; cs = 0.433 → 25 981; s² = 0.25 → 15 000 N/mm.
  7. CheckThe three terms sum with the symmetry the global matrix needs, and at 30° the horizontal stiffness (c²) exceeds the vertical (s²), as a shallow member should. A horizontal member would give c² = 1 and s² = 0.
  8. ConclusionThe transformation spreads one axial stiffness across the global directions. Assembling these global element matrices builds the truss stiffness.
Result. AE/L = 60 000 N/mm; global terms 45 000, 25 981, 15 000 N/mm.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Local stiffness for a tilted memberWrong truss displacements"Is this member horizontal?"Transform with direction cosines for any inclined member.
Stress from the wrong areaSegment stress under-predicted"Which area carries this force?"Use σ = P/A with each segment's own area.
Sign of member forceTension and compression confused"Did the member lengthen or shorten?"Positive elongation is tension, negative is compression.
Wrong angle referenceDirection cosines swapped"Is θ measured from the global x-axis?"Take θ consistently from the positive x-axis to the member.
07

Practice ladder

Level 1 · Direct skill

A bar has A = 250 mm², L = 800 mm, E = 200 GPa. Find AE/L.

Show answer

AE/L = (250 × 200 000)/800 = 62 500 N/mm.

Level 2 · Mixed concept

For the Worked Example 1 bar, what is the elongation of element 2 alone under the 20 kN load?

Show answer

δ₂ = PL₂/(A₂E) = 20 000 × 500/(200 × 200 000) = 0.25 mm, the larger share because element 2 is thinner.

Level 3 · Independent problem

A member at 45° has A = 500 mm², L = 1.5 m, E = 200 GPa. Find AE/L and the c² and cs global terms.

Show answer

AE/L = (500 × 200 000)/1500 = 66 667 N/mm. At 45°, c² = cs = 0.5, so both terms are 33 333 N/mm.

Level 4 · Transfer to real engineering

For a real truss member, identify its orientation, compute its global stiffness contribution, and state how you would recover and check its axial force.

What good work looks like

Direction cosines from the geometry, a global stiffness with c², cs, s² terms, and a member force recovered as AE/L times elongation, checked against joint equilibrium.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I transformed the inclined member with the right direction cosines."
"Give me three member angles; I will write the c², cs, s² terms."
"Solve this truss." Building the global stiffness yourself is the skill.
"What is the member force?" Projecting displacements onto the axis is the point.

Portfolio task

Model a small truss: transform each member to global coordinates, assemble and solve, then recover member forces and stresses with a joint-equilibrium check.

Must include: transformed element stiffnesses, a solved displacement, and equilibrium-checked member forces.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the bar element stiffness.

(AE/L)[1, −1; −1, 1] along the bar axis.

2. How is axial stress recovered?

σ = E(u₂ − u₁)/L from the element's nodal displacements.

3. What does the transformation use?

The direction cosines c = cos θ and s = sin θ of the member.

4. Give the global stiffness blocks.

Scaled by c², cs, and s² times AE/L.

5. How is a member force found?

AE/L times the axial elongation; positive is tension.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the stepped bar and transformation from a blank page.
+3 daysTransform two new members and assemble.
+7 daysCarry transformation into beam elements, Chapter 4.
+30 daysReuse the bar element when validating any axial FEA.
10

Textbook mapping

ItemMapping
Primary sourceLogan, A First Course in the Finite Element Method, Chapter 3 (Development of Truss Equations)
Cross-referenceHutton, Ch. 3 · Statics: Structural Analysis
Core topics3.1 Bar stiffness · 3.2 Stepped bars · 3.3 Stress recovery · 3.4 Transformation · 3.5 Member forces
Engineering connectionThe transformation generalises to beams and frames.
Read nextChapter 4: Beam Elements.