BIZOMAT MAKEOS · Springs
Torsion Bar Spring Design (Diameter for Stress and Twist)
What diameter must this torsion bar be to carry the torque within the allowable shear stress and twist, and what rate and stroke will it give?
Design of a solid round torsion bar spring: torque from load and arm, the least diameter for the allowable shear stress and the least for the largest twist, the governing diameter, and with the diameter you choose the shear stress, twist, torsional rate, arm-tip travel (as the torsion bar tool) and energy stored.
Open the calculatorRuns in your browser. No account needed.
How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Load at the arm | N | Load applied at the end of the arm. |
| Arm length | mm | Length of the lever arm. |
| Active bar length | mm | Active length of the bar between the fixed end and the arm. |
| Shear modulus | GPa | Shear modulus of the bar. |
| Allowable shear stress | MPa | Allowable shear stress for the duty (your value; presetting raises it). |
| Largest twist allowed | ° | Largest twist you allow (for travel or stops). |
| Bar diameter chosen | mm | Diameter you intend to use. |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Load at the arm
- What it is
- The force applied at the free end of the lever arm, square to the arm, which twists the bar.
- Units and format
- Enter a number in N: newtons. The form accepts 0.000001 N to 10000000 N. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 1000 N as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Torque, stress and twist all rise in proportion to the load.
- Get it right
- Use the force at the arm tip, not the torque; the tool multiplies by the arm length.
What this means: Arm length
- What it is
- The lever arm length from the bar axis to the point where the load acts.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 5000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 200 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Torque is load × arm, so a longer arm raises stress and twist and increases tip travel.
- Get it right
- Measure from the centre of the bar to the line of the load, not to the end of the arm casting.
What this means: Active bar length
- What it is
- The working length of the bar that actually twists, between the fixed anchor and the arm, excluding the splined or hexagon ends.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 10000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 800 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Twist grows in proportion to length and rate falls; shear stress is unchanged.
- Get it right
- Do not include the end fittings; they barely twist.
What this means: Shear modulus
- What it is
- The shear modulus of the bar material, its stiffness in twisting. Take it from the material data sheet or a materials reference.
- Units and format
- Enter a number in GPa: gigapascals (1 GPa = 1000 MPa). The form accepts 0.000001 GPa to 100 GPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 79.3 GPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher shear modulus reduces twist and raises the rate.
- Get it right
- Enter GPa, not MPa: the tool multiplies by 1000.
What this means: Allowable shear stress
- What it is
- The highest shear stress you will allow in the bar for this duty, from your design code, material data and fatigue requirements. It sets the least diameter for strength.
- Units and format
- Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 2000 MPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 600 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A higher allowable lowers the diameter needed for stress.
- Get it right
- A fatigue duty needs a lower allowable than a static one; use your design basis.
What this means: Largest twist allowed
- What it is
- The largest angle the bar may twist under the load, set by the travel available or the stops. It sets the least diameter for stiffness.
- Units and format
- Enter a number in °: degrees of angle. The form accepts 0.1° to 90°. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 10° as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Allowing more twist lowers the diameter needed for twist.
- Get it right
- Enter degrees of twist of the bar, not the arm-tip travel in millimetres.
What this means: Bar diameter chosen
- What it is
- The diameter of the solid round bar you intend to use. The tool reports its stress, twist, rate, tip travel and stored energy.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 500 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 20 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- Shear stress falls with the cube of diameter and twist with the fourth power.
- Get it right
- Use the diameter of the active length, not of the end fittings.
Method and formulas
- Torque T = F × arm.
- Diameter for stress d = (16T ÷ π τ_allow)^⅓; for twist d = (32 T L ÷ π G θ_max)^¼.
- τ = 16T ÷ π d³; θ = T L ÷ G J with J = π d⁴ ÷ 32; rate = G J ÷ L; tip travel = arm × θ (as the torsion bar tool); energy = T θ ÷ 2.
- Example: 1000 N on a 200 mm arm needs 18.524 mm; a 20 mm bar twists 7.3595° at 127.32 MPa.
Limitations
- Solid round bar; the splined or hexagon ends need a larger diameter and their own stress check.
- Static stress; fatigue governs repeated loading.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size a torsion bar for a counterbalance or suspension
- Check a bar for a higher load
- Trade length against diameter
Who it is for
- Design engineers
- Vehicle engineers
- Machine designers
What you get
- Diameter needed for stress and for twist
- Stress, twist, rate and tip travel with the chosen diameter
- Energy stored
Interpreting the result
- A longer bar twists more for the same stress: use length, not diameter, to soften the rate.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 295–303, 344. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · BIZOMAT original book
Torsion bars and flat springs
What it covers: How to size a torsion bar for stress and twist and balance a heavy panel with it, and how to check flat cantilever springs, with their limits at the ends and in large deflection.
BIZOMAT tools are screening and decision-support utilities. Verify important financial, engineering, safety, legal or operational decisions using appropriate professional standards and your organisation’s approved data.
Open the Torsion Bar Spring Design (Diameter for Stress and Twist) calculator