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Truss displacement due to member temperature changes
Joint displacement of an initially stress-free, statically determinate truss caused by temperature changes ΔT in its members, with n the force in each member from a unit virtual load, α the thermal expansion coefficient and L the length.
- Ref
- CIV-35
- Rev
- v1
- Field
- Civil
- Result
- δT [m]
If members change temperature, they change length by αΔTL even with no load. The joint displacement in a statically determinate truss follows from the virtual work expression δT = Σ n α ΔT L, where n is the force in each member from a unit virtual load. No stress develops, since the truss can adjust freely. Fixed support positions are assumed.
Joint displacement of an initially stress-free, statically determinate truss caused by temperature changes ΔT in its members, with n the force in each member from a unit virtual load, α the thermal expansion coefficient and L the length. The supports are fixed in position. Up to three members here; leave unused ones at ΔT = 0.
Calculator
- Free change in length of member 1, α₁ΔT₁L₁0.00108 m
- Free change in length of member 2, α₂ΔT₂L₂4.8×10−4 m
- Free change in length of member 3, α₃ΔT₃L₃0 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The joint moves 0.84 mm from temperature alone, with no stress in the members because the truss is statically determinate.
- A statically indeterminate truss would instead develop stress, so this displacement only applies when the supports can move or the structure is determinate.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| n1 Unit-load force in member 1 | −5 to 5 | −0.00564 | 0.00516 | 0.0108 | +0.2 | +25.7 % |
| ΔT1 Temperature change of member 1 | −100 to 100 K | −0.00384 | 0.00336 | 0.0072 | +4 K | +17.1 % |
| L1 Length of member 1 | 0.5 to 20 m | −6e-5 | 0.00696 | 0.00702 | +10% | +12.9 % |
| n2 Unit-load force in member 2 | −5 to 5 | −0.00132 | 0.00348 | 0.0048 | +0.2 | +11.4 % |
| ΔT2 Temperature change of member 2 | −100 to 100 K | −0.00132 | 0.00348 | 0.0048 | +4 K | −11.4 % |
| α1 Expansion coefficient of member 1 | 1e-6 to 3e-5 1/K | −1.5e-4 | 0.00246 | 0.00261 | +10% | +12.9 % |
| L2 Length of member 2 | 0.5 to 20 m | −1.2e-4 | 0.00105 | 0.00117 | +10% | −2.86 % |
| α2 Expansion coefficient of member 2 | 1e-6 to 3e-5 1/K | 4.8e-4 | 0.00106 | 5.8e-4 | +10% | −2.86 % |
| n3 Unit-load force in member 3 | −5 to 5 | 8.4e-4 | 8.4e-4 | 0 | +0.2 | 0 % |
| α3 Expansion coefficient of member 3 | 1e-6 to 3e-5 1/K | 8.4e-4 | 8.4e-4 | 0 | +10% | 0 % |
| ΔT3 Temperature change of member 3 | −100 to 100 K | 8.4e-4 | 8.4e-4 | 0 | +4 K | 0 % |
| L3 Length of member 3 | 0.5 to 20 m | 8.4e-4 | 8.4e-4 | 0 | +10% | 0 % |
Sweep of Temperature change of member 1
| Temperature change of member 1 (K) | Joint displacement (m) |
|---|---|
| −100 | −0.00384 |
| −75 | −0.00294 |
| −50 | −0.00204 |
| −25 | −0.00114 |
| 0 | −2.4e-4 |
| 25 | 6.6e-4 |
| 50 | 0.00156 |
| 75 | 0.00246 |
| 100 | 0.00336 |
Sweep one variable
Variables and how each one contributes
Joint displacement
What the equation gives you. Everything else feeds it.
Unit-load force in member 1
The force in member 1 from the unit virtual load (tension positive). Scales that member’s thermal contribution in direct proportion. Raising it by 10 % (to 1.1) moves the result by +12.9 %, from 8.4×10−4 to 9.48×10−4 m.
Unit-load force in member 2
The force in member 2 from the unit virtual load (tension positive). Scales that member’s thermal contribution in direct proportion. Raising it by 10 % (to −0.55) moves the result by −2.86 %, from 8.4×10−4 to 8.16×10−4 m.
Unit-load force in member 3
The force in member 3 from the unit virtual load (tension positive). Scales that member’s thermal contribution in direct proportion. Raising it by 10 % (to 0) moves the result by +0 %, from 8.4×10−4 to 8.4×10−4 m.
Expansion coefficient of member 1
The thermal expansion coefficient of member 1, in 1/K. Raises the free thermal strain in direct proportion. Raising it by 10 % (to 1.32×10−5) moves the result by +12.9 %, from 8.4×10−4 to 9.48×10−4 m.
Expansion coefficient of member 2
The thermal expansion coefficient of member 2, in 1/K. Raises the free thermal strain in direct proportion. Raising it by 10 % (to 1.32×10−5) moves the result by −2.86 %, from 8.4×10−4 to 8.16×10−4 m.
Expansion coefficient of member 3
The thermal expansion coefficient of member 3, in 1/K. Raises the free thermal strain in direct proportion. Raising it by 10 % (to 1.32×10−5) moves the result by +0 %, from 8.4×10−4 to 8.4×10−4 m.
Temperature change of member 1
The uniform temperature change of member 1, in K. Raises the change in length in direct proportion. Raising it by 10 % (to 33) moves the result by +12.9 %, from 8.4×10−4 to 9.48×10−4 m.
Temperature change of member 2
The uniform temperature change of member 2, in K. Raises the change in length in direct proportion. Raising it by 10 % (to 11) moves the result by −2.86 %, from 8.4×10−4 to 8.16×10−4 m.
Temperature change of member 3
The uniform temperature change of member 3, in K. Raises the change in length in direct proportion. Raising it by 10 % (to 0) moves the result by +0 %, from 8.4×10−4 to 8.4×10−4 m.
Length of member 1
The length of member 1, in m. A longer member changes length more. Raising it by 10 % (to 3.3) moves the result by +12.9 %, from 8.4×10−4 to 9.48×10−4 m.
Length of member 2
The length of member 2, in m. A longer member changes length more. Raising it by 10 % (to 4.4) moves the result by −2.86 %, from 8.4×10−4 to 8.16×10−4 m.
Length of member 3
The length of member 3, in m. A longer member changes length more. Raising it by 10 % (to 3.3) moves the result by +0 %, from 8.4×10−4 to 8.4×10−4 m.
Worked example with the values above
- Member terms n·α·ΔT·L: member 1 = 1 × 1.2×10−5 × 30 × 3 = 0.00108; member 2 = −0.5 × 1.2×10−5 × 10 × 4 = −2.4×10−4 m.
- Sum of the terms: 8.4×10−4 m.
- δT = 8.4×10−4 m.
Assumptions and limits
- Pin joints, loads at the joints, small displacements and linear elastic members.
- Tension positive in both the real and virtual force systems.
Common mistakes
- Using compression positive in one system and tension positive in the other.
- Mixing the real force and the virtual force between members.
- Forgetting that unused members must have zero real or virtual force.
Statistics and uncertainty
Truss displacements are exact for pin-jointed bars with axial deformation only. Real joints are slightly flexible and members have length tolerances, so measured displacements are usually 5 % to 20 % higher than the pin-jointed prediction.
Where it is used
- Deflection and stiffness checks for trusses, braces and frames.
- Estimating thermal movement and strain energy of members.
Maths you need
- Virtual work: external virtual work equals internal virtual work
- Axial deformation: ΔL = NL/EA
- Strain energy: the work stored by elastic deformation