Lesson 1.5: Torsion of Circular Shafts

A shaft that transmits torque without breaking can still fail functionally if it twists too much: the angular deflection throws off the timing in a Geneva indexing mechanism, causes vibration in a coupling, or misaligns a sensor mounting. Designing against torsional failure means checking both the shear stress (will it yield?) and the angle of twist (will it stay accurate?). In this lesson you will analyze three real cases, a Geneva mechanism crankshaft, a power-transmitting motor shaft, and a hollow-versus-solid shaft comparison, learning the torsion formula, polar moment of inertia, and the design trade-offs that keep rotating shafts strong and stiff. #Torsion #ShearStress #ShaftDesign

Learning Objectives

By the end of this lesson, you will be able to:

1. Derive the torsional shear stress formula and explain why stress peaks at the outer surface
2. Calculate peak shear stress and angle of twist for solid and hollow circular shafts
3. Convert power and rotational speed into torque, and apply the result to a shaft design
4. Compare a hollow and a solid shaft carrying the same torque, and quantify the material saving

Real-World System Problem: Geneva Mechanism Crankshaft

The Geneva mechanism converts continuous rotation into intermittent indexing motion. It appears in film-transport mechanisms, rotary indexing tables, and automated assembly equipment. The drive crankshaft spins continuously while the Geneva wheel advances one slot at a time, and every engagement event sends a torque spike through the crankshaft.

The Torsional Challenge

Engineering Question: How do we size the crankshaft so it can transmit peak torque without yielding, and keep the angular twist small enough to preserve timing accuracy?

Both criteria matter. A shaft that survives the stress but winds up by several degrees will cause the Geneva wheel to over-travel or under-travel its slot, jamming the mechanism or marking parts at the wrong position.

Why Torsional Analysis Matters

Fundamental Theory: Torsional Mechanics

When a circular shaft carries a twisting moment (torque), every cross-section resists by developing shear stress that varies linearly from zero at the centre to a maximum at the outer surface. Three formulas govern everything in this lesson.

Torsional shear stress

Where:

• = shear stress (Pa)
• = applied torque (N·m)
• = radial distance from the shaft centre (m); maximum stress occurs at
• = polar second moment of area (m)

The stress is highest at the outer fibre because is largest there. The centre carries no shear stress at all, which is why hollow shafts waste so little capacity.

Polar second moment of area

For a solid shaft of diameter :

For a hollow shaft with outer diameter and inner diameter :

captures how the cross-section’s area is distributed about the axis. Because diameter enters as the fourth power, a modest increase in diameter raises dramatically.

Angle of twist

Where:

• = angle of twist (radians)
• = shaft length (m)
• = shear modulus of the material (Pa); for steel GPa

Angular deflection grows with torque and length, and shrinks with material stiffness and section stiffness. A shaft that passes the stress check can still fail the stiffness check if it is long or lightly proportioned.

Power and torque

A motor or gearbox rated at power (watts) running at angular speed (rad/s) delivers:

This is the gateway calculation for every driven shaft: convert the nameplate data to torque before applying the torsion formula.

Application 1: Geneva Mechanism Crankshaft

The Geneva drive crankshaft below is the original motivating problem. Three design questions are answered in sequence: is the 25 mm shaft strong enough? if not, what diameter is needed for strength? and once resized, does it meet the timing requirement?

Step 1: Polar Moment of Inertia for the 25 mm Shaft

Click to reveal the polar moment calculation

1. Apply the solid-shaft formula. Convert diameter to metres:

   ✅

2. Store this value. It feeds directly into the stress and twist calculations that follow.

Step 2: Torsional Stress Check

Click to reveal the torsional stress analysis

1. Continuous running stress:

   ✅

2. Peak engagement stress:

   ✅

   The same result follows from the compact shortcut , which is algebraically identical to for a solid shaft. Keep in metres throughout:

   ✅

3. Safety factor at peak torque:

   ✅

   The 25 mm shaft passes the strength check at SF = 3.7, above the required 3.0. ✅

Step 3: Angle of Twist Check

Click to reveal the angular deformation analysis

1. Angle of twist under peak torque ( = 200 mm = 0.2 m):

   ✅

   Converting to degrees: ✅

2. Check against the precision requirement of :

   : the shaft just exceeds the stiffness budget. A slightly larger diameter resolves this without any strength concern. ✅

3. Required diameter for rad:

   ✅

   Rounding up to the next standard size, 26 mm, meets both strength and stiffness requirements. ✅

Application 2: Sizing a Power-Transmitting Motor Shaft

A small conveyor drive unit has a 5 kW induction motor running at 960 rpm. The output shaft connects the motor to a worm gearbox over a span of 400 mm. The shaft material is steel with GPa and an allowable shear stress of 60 MPa. A 20 mm solid shaft has been proposed. Verify the shear stress and angle of twist.

Enter the motor shaft in the simulator and read the peak shear stress and the angle of twist over its length:

Step 1: Torque from Power and Speed

Click to reveal the torque calculation

1. Convert rotational speed to rad/s:

   ✅

2. Compute the transmitted torque:

   ✅

Step 2: Polar Moment and Peak Shear Stress

Click to reveal the stress calculation

1. Polar second moment for the 20 mm solid shaft:

   ✅

2. Peak shear stress at the outer surface ( = 10 mm = 0.010 m):

   ✅

3. Safety factor:

   ✅

   The shaft is safe. A factor of 1.89 is acceptable for a well-characterised steady load; if the drive sees shock loading or fatigue, a larger diameter would be chosen. ✅

Step 3: Angle of Twist

Click to reveal the angle of twist calculation

1. Apply the twist formula ( = 400 mm = 0.4 m):

   ✅

2. Convert to degrees:

   ✅

   Over 400 mm the shaft twists by just under one degree. For a conveyor drive this is entirely acceptable. For a precision indexing shaft it would require a larger diameter to stiffen the section. ✅

Application 3: Hollow Versus Solid Shaft for the Same Torque

A drive shaft must transmit = 200 N·m. The allowable shear stress is 60 MPa. Two designs are compared: a minimum solid shaft sized to just meet the allowable, and a hollow shaft with the same outer diameter as the next standard size up. The goal is to show how much cross-sectional area (and therefore material) the hollow design saves.

Step 1: Size the Solid Shaft

Click to reveal the solid shaft sizing

1. Minimum diameter from the allowable stress. Rearranging :

   Round up to the standard size: = 26 mm. ✅

2. Polar moment for the 26 mm solid shaft:

   ✅

3. Shear stress check:

   ✅

   This is just below the 60 MPa allowable, confirming the sizing is correct. ✅

4. Cross-sectional area:

   ✅

Step 2: Analyse the Hollow Shaft

Click to reveal the hollow shaft analysis

1. Polar moment for the hollow shaft ( = 32 mm, = 26 mm):

   Working out the fourth powers:

   • m ✅
   • m ✅
   • Difference: m ✅

   ✅

2. Peak shear stress at the outer surface ( = 16 mm = 0.016 m):

   ✅

   The hollow shaft has a slightly lower peak stress than the solid shaft (55.1 MPa vs 58.0 MPa) despite having a larger outer diameter, because the additional outer material adds more than it adds to the stress numerator. ✅

3. Cross-sectional area:

   ✅

Step 3: Compare and Quantify the Saving

Click to reveal the comparison

1. Area saving (which equals the mass-per-metre saving for the same material):

   ✅

2. Summary side by side:

   Quantity	Solid, = 26 mm	Hollow, = 32/26 mm
   (m)
   Peak (MPa)	58.0	55.1
   Area (mm)	530.9	273.3
   Mass saving	baseline	48.5% lighter

Design Guidelines for Torsional Loading

Summary and Next Steps

Key Concepts Mastered

1. Torsional shear stress is , peaks at the outer surface, and is zero at the centre; the formula holds for any circular shaft, solid or hollow.
2. Polar moment for a solid shaft and for hollow; diameter enters as the fourth power, making it the most effective design variable.
3. Angle of twist is ; stiffness governs precision mechanisms just as strength governs high-stress shafts, and both checks are mandatory.
4. Power conversion is the gateway to every motor-shaft calculation; always carry the full torque value into the torsion formula.
5. Hollow shafts remove the low-stress core, saving roughly half the cross-sectional area for a modest increase in outer diameter, as shown in Application 3.

Results at a Glance

Application	Key inputs	Peak (MPa)	Angle of twist	Notes
Geneva crankshaft	= 150 N·m, = 25 mm	48.9	0.56°	Strength OK; stiffness marginally exceeded; use 26 mm
Motor shaft	= 5 kW, 960 rpm, = 20 mm	31.7	0.92° over 400 mm	SF = 1.89; adequate for steady drive
Solid shaft	= 200 N·m, = 26 mm	58.0		= 530.9 mm
Hollow shaft	= 200 N·m, = 32/26 mm	55.1		= 273.3 mm; 48.5% lighter

A Note on Tools

Every result in this lesson came from four formulas (, , , ) and a calculator. No finite-element package is needed for circular shafts under uniform torque; the hand calculation is the design, and simulation only confirms it for complex geometry such as stepped or keyway-notched shafts.

Next, Thin-Walled Pressure Vessels applies a similar two-stress approach to cylindrical shells, where internal pressure simultaneously creates both hoop (circumferential) and longitudinal normal stresses in the wall.

Comments

Loading comments...