τ = 2T/(π·d²·b)

Cylindrical bonded joint torque capacity

As with the flat lap joint, this is a nominal shear stress evenly spread over the bond circumference; real loading concentrates at the ends of the bond line.

MINTSI
01

Inputs

Approximate transmissible torque of the cylindrical bond line.

Diameter of the cylindrical bonded surface between shaft and hub.

Axial length of the cylindrical bond line.

Bond strength divided by safety factor; typical adhesive shear strengths are roughly 5 to 20 MPa.

02

Result

Select a target and calculate.

Calculation

T = τzul · π · d² · b / 2

As with the flat lap joint, this is a nominal shear stress evenly spread over the bond circumference; real loading concentrates at the ends of the bond line.

Understand the inputs
  • Transmissible torque TApproximate transmissible torque of the cylindrical bond line.
  • Bond diameter dDiameter of the cylindrical bonded surface between shaft and hub.
  • Bond width (engagement length) bAxial length of the cylindrical bond line.
  • Allowable shear stress τzulBond strength divided by safety factor; typical adhesive shear strengths are roughly 5 to 20 MPa.
Example

A 20 mm bond diameter, 30 mm engagement length and an 8 MPa allowable shear stress give T = 8 MPa · π · (20 mm)² · 30 mm / 2 ≈ 150.8 N·m.

Assumptions and limits

Nominal shear stress evenly distributed over circumference and engagement length; real bonded joints under torsion show an uneven stress distribution, additionally influenced by joint clearance, bond-line thickness and surface preparation.

Technical article

Understand Cylindrical bonded joint torque capacity

This calculator finds the approximately transmissible torque capacity of a cylindrical bonded shaft-hub joint from bond diameter, engagement length and allowable shear stress.

What does this quantity describe?

In a cylindrical bonded joint under torsion, shear stress acts tangentially over the entire circumference of the bonded surface. Rearranging the nominal, evenly distributed shear stress τ = 2T/(π·d²·b) gives the approximately transmissible torque capacity T = τallow·π·d²·b/2.

Picture the cylindrical bond line as many small lever arms acting in parallel, all at distance d/2 from the rotation axis: each area element contributes its share to the total moment, and summing over the full circumference and engagement length gives the transmissible torque.

Formula and variables

T = τzul · π · d² · b / 2

  • T = τallow·π·d²·b/2
  • τallow = 2T/(π·d²·b)
  • d = √(2T/(τallow·π·b))
Symbol / inputMeaning
Transmissible torque TApproximate transmissible torque of the cylindrical bond line.
Bond diameter dDiameter of the cylindrical bonded surface between shaft and hub.
Bond width (engagement length) bAxial length of the cylindrical bond line.
Allowable shear stress τzulBond strength divided by safety factor; typical adhesive shear strengths are roughly 5 to 20 MPa.

Choose the inputs correctly

Bond diameter d, axial engagement length b and allowable shear stress τallow set the transmissible torque. τallow should come from the adhesive's bond strength, typically roughly 5 to 20 MPa, divided by an appropriate safety factor.

How to use the calculator

Enter bond diameter, engagement length and allowable shear stress to compute transmissible torque. For a sizing question — what diameter or engagement length a target torque needs — choose d or b as the target.

Worked example

A 20 mm bond diameter, 30 mm engagement length and 8 MPa allowable shear stress give T = 8 MPa · π · (20 mm)² · 30 mm / 2 ≈ 150.8 N·m.

Understand the result and units

A transmissible torque of 150.8 N·m at just 20 mm diameter shows the basic suitability of cylindrical bonded joints for compact shaft-hub connections — diameter enters capacity with the square, while engagement length enters only linearly.

Diameter and engagement length are given in mm, stress in MPa, and the resulting torque in N·m.

Why diameter enters with the square but length only linearly

In the formula T = τallow·π·d²·b/2, diameter appears twice: once through the lever arm d/2 that converts a circumferential force into a moment, and once through the circumference π·d that helps set the available bonded area. Both effects multiply, so doubling diameter at the same engagement length quadruples transmissible torque. Engagement length b, by contrast, enters only once through bonded area and therefore acts only linearly on torque — for the same added space, a larger diameter is usually more effective than a correspondingly longer engagement.

Typical applications

The calculation is used for pre-sizing bonded shaft-hub joints, for example sensor or coupling attachments where a positive-fit or friction-fit connection would be unfavorable for space or weight reasons; the same nominal stress model applies analogously to cylindrical brazed joints with a correspondingly higher allowable stress.

Assumptions, limits and common mistakes

The model assumes shear stress evenly distributed over circumference and engagement length. Volkersen-type analyses show that actual stress distribution in cylindrical bonded joints under torsion is strongly uneven and concentrates at the axial ends of the joint; joint clearance, bond-line thickness and surface preparation additionally affect real capacity substantially.

Common mistake: A common mistake is arbitrarily increasing engagement length to reach a higher torque, even though real capacity grows more slowly than the nominal formula predicts because of the uneven stress distribution. It is also easy to carry over a τallow value determined for flat lap joints unchanged to the cylindrical geometry, even though joint clearance and cure conditions can differ for a round bonded joint.

Frequently asked questions

Why does diameter matter more than engagement length?

Because diameter sets both the lever arm and, via circumference, the bonded area, while engagement length enters only through area; see the section above.

What allowable shear stresses are typical for adhesives?

Depending on the adhesive system and safety margin applied, allowable values often fall around 5 to 20 MPa; the specific adhesive's datasheet governs.

Does the formula also apply to brazed joints?

Yes, the nominal stress model is identical; brazed joints simply use a markedly higher allowable shear stress.

What does Volkersen's analysis show about real stress distribution?

It shows that shear stress in a cylindrical bond line under torsion concentrates at the axial ends of the joint rather than distributing evenly, similar to the flat lap joint case.

How do I choose diameter or engagement length for a target torque?

Choose d or b as the target and enter the desired torque along with the other dimension and allowable stress, keeping in mind that diameter acts with the square.