Lesson 1.2: Strain, Material Properties, and Shear in Actuator Systems

A CNC actuator shaft that stretches even a few micrometres under load throws off positioning accuracy, and choosing a material by cost alone risks either over-designing (heavy, expensive) or under-designing (flexible, imprecise). Selecting the right material requires comparing elastic modulus, yield strength, and elongation at failure across candidates, then computing the actual deformation under your expected loads. In this lesson you will perform that analysis on a CNC actuator shaft, read a steel tensile test curve to extract yield strength, toughness, and ductility, and compute shear deflection in a bonded elastomer mount. #AxialStress #MaterialSelection #CNCDesign

Learning Objectives

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

1. Analyze axial stress and strain in a precision actuator shaft and select between steel and aluminum based on strength and stiffness criteria
2. Interpret a tensile test stress-strain curve to identify proportional limit, 0.2% offset yield strength, UTS, fracture strain, modulus of resilience, and modulus of toughness
3. Apply the shear modulus relation to calculate shear stress, shear strain, and lateral deflection in an elastomer component

Real-World System Problem: CNC Z-Axis Actuator Shaft

In a CNC machine the Z-axis actuator controls vertical tool movement. The shaft must transmit precise forces while maintaining dimensional accuracy under varying loads. A shaft that deflects too far under the cutting force moves the tool out of the programmed path, producing scrap.

The Shaft Selection Problem

Engineering Question: How do you select the right material and shaft diameter to ensure the CNC system maintains a deformation limit of 0.05 mm under a maximum axial load of 5,000 N?

Answering that requires two independent checks: a strength check (the stress must stay below the allowable strength) and a stiffness check (the elongation must stay within the positional tolerance). Either check can set the minimum diameter, and in precision systems the stiffness check almost always wins.

Why Material Selection Matters

Fundamental Theory: Strain, Hooke’s Law, and the Stress-Strain Curve

The previous lesson defined stress. Here we build the quantities needed to predict deformation and to read a material test curve.

Axial Strain and Deformation

Axial Strain and Deflection

Where:

• = axial strain (dimensionless)
• = change in length (mm or m)
• = original length
• = axial force (N)
• = cross-sectional area (mm² or m²)
• = Young’s modulus (GPa)

The second form is the most-used relation in this lesson: it links a known load and geometry to the actual elongation.

The Stress-Strain Curve

A tensile test pulls a standard coupon to fracture while recording force and extension. The resulting curve divides material behaviour into regions:

Key Points on the Curve

• Proportional limit: the highest stress at which stress and strain remain strictly proportional (the curve is still a straight line)
• Yield strength (0.2% offset): found by drawing a line with slope starting at ; where it intersects the curve is the yield point
• Ultimate tensile strength (UTS): the peak stress the material ever carries
• Fracture point: where the coupon breaks; the strain here gives the percent elongation (a ductility measure)

Modulus of Resilience and Modulus of Toughness

Energy Absorption Measures

Resilience measures how much elastic energy the material stores per unit volume before yielding. Toughness measures the total energy per unit volume absorbed before fracture, including all the plastic work.

Shear Modulus

Shear Modulus and Its Relation to E and Poisson's Ratio

Where:

• = shear modulus (Pa or MPa)
• = shear stress (Pa or MPa)
• = shear strain (dimensionless, in radians)
• = thickness of the sheared layer

For metals, falls between and . For elastomers with , , much smaller than for metals.

Application 1: CNC Actuator Shaft, Steel vs Aluminum

The CNC Z-axis shaft must transmit a maximum axial force of 5,000 N while keeping elongation within 0.05 mm over its 400 mm length. Both a strength check and a stiffness check are required.

Run the steel and the aluminum shaft as the A and B states in the simulator and compare the deflection directly:

Step 1: Minimum Diameter from the Strength Criterion

Click to reveal the strength calculations

1. Allowable stress for each material with SF = 3:

   Aluminum: ✅

   Steel: ✅

2. Minimum area from :

   Aluminum: ✅

   Steel: ✅

3. Minimum diameter from :

   Aluminum: ✅

   Steel: ✅

   Strength alone would permit quite a thin shaft for either material. ✅

Step 2: Minimum Diameter from the Stiffness Criterion

Click to reveal the stiffness calculations

1. Required area from mm, rearranged to :

   Aluminum: ✅

   Steel: ✅

   (Here is in MPa = N/mm², so in mm and in mm are consistent.)

2. Required diameter from :

   Aluminum: ✅

   Steel: ✅

3. Compare the two criteria:

   For aluminum: strength needs 8.4 mm, stiffness needs 27.0 mm. Stiffness governs by a factor of 3.2. ✅

   For steel: strength needs 6.0 mm, stiffness needs 16.0 mm. Stiffness governs by a factor of 2.7. ✅

Step 3: Verify the Final Design and Compare Materials

Click to reveal the design comparison

1. Confirm deformation for the stiffness-governed diameters. For the 27 mm aluminum shaft:

   ✅

   (just within 0.05 mm) ✅

   For the 16 mm steel shaft:

   ✅

   (just within 0.05 mm) ✅

2. Estimate shaft mass using :

   Aluminum ( kg/m³): kg ✅

   Steel ( kg/m³): kg ✅

   The masses are nearly equal because steel’s higher density is almost exactly offset by its smaller cross-section. ✅

3. Design decision: Steel 1045 at 16 mm diameter meets both criteria, uses a 41% smaller diameter (16 mm versus 27 mm, which fits tighter machine envelopes), and has a similar mass to the aluminum option. ✅

Application 2: Interpreting a Tensile Test of a Steel Coupon

A tensile test is the standard way to measure the mechanical properties of a material before it goes into a design. Understanding how to read the stress-strain curve lets you extract the numbers that feed every other analysis.

Step 1: Confirm the 0.2% Offset Yield Construction

Click to reveal the 0.2% offset construction

1. What “0.2% offset” means. Draw a straight line with slope equal to (200 GPa) that starts not at the origin but at on the strain axis. Where that line intersects the actual stress-strain curve is defined as the yield strength. The 0.002 offset accounts for the permanent set that is considered acceptable for a yield criterion.

2. Check the elastic slope at the yield point. At MPa the elastic strain is:

   ✅

   The total strain at the 0.2% offset yield point is , which is consistent with where the offset line meets the curve. ✅

3. Read the other landmarks:

   • Proportional limit MPa at strain (curve is still linear)
   • UTS = 520 MPa at a strain of roughly 12% (the peak of the curve)
   • Fracture at 22% elongation and 410 MPa (necking has reduced the load-bearing area)

Step 2: Modulus of Resilience

Click to reveal the resilience calculation

1. Formula. The modulus of resilience is the area under the elastic portion of the curve, approximated as a triangle:

2. Calculation:

   ✅

3. Meaning. This steel can absorb 306 kJ per cubic metre elastically before yielding. A material with higher resilience makes a better spring; higher toughness (see Step 3) makes a better crash absorber. ✅

Step 3: Estimate the Modulus of Toughness

Click to reveal the toughness estimate

1. Approach. The modulus of toughness is the total area under the stress-strain curve. A practical estimate splits the curve into the elastic triangle plus a trapezoidal plastic region:

   where is the strain at yield (approximately 0.00175) and .

2. Calculation:

   Elastic contribution: J/m³ (from Step 2) ✅

   Average stress over plastic region: MPa ✅

   Plastic strain range: ✅

   Plastic contribution: J/m³ MJ/m³ ✅

   Total: MJ/m³ ✅

3. Interpretation. The steel absorbs roughly 95 MJ per cubic metre before fracturing. Nearly all of that (99.7%) is plastic work during the large permanent elongation. This high toughness is what makes medium-carbon steel suitable for parts that must absorb impact without shattering. ✅

Step 4: Percent Elongation (Ductility)

Click to reveal the percent elongation

1. Percent elongation is defined over the original gauge length:

   ✅

   (The fracture elongation of 22% was read from the curve; it means the gauge section grew from 50 mm to 61 mm before breaking.) ✅

2. Classification. Materials with are considered ductile. At 22%, this steel has substantial ductility: it gives visual warning (necking and elongation) well before fracture, which is the property that makes ductile metals forgiving in overload situations. ✅

Application 3: Shear Modulus and a Bonded Elastomer Anti-Vibration Mount

Shear modulus governs any component that is loaded across its thickness rather than along its length. Bonded elastomer mounts are a common example: the rubber layer is glued between two metal plates, and when the structure moves horizontally the rubber is sheared. The mount’s compliance in shear is exactly what provides the vibration isolation.

Step 1: Shear Modulus from E and Poisson’s Ratio

Click to reveal the shear modulus calculation

1. Apply the elastic relation between , , and :

   ✅

2. Context. For nearly incompressible materials such as rubber ( to ), the denominator approaches , so . This is why rubber is much easier to shear than to compress: its shear modulus is only a third of its tensile modulus, while for steel the ratio is about . ✅

Step 2: Shear Stress in the Rubber Layer

Click to reveal the shear stress calculation

1. Shear stress is force over the area parallel to the load:

   ✅

2. Check against the rubber’s short-term shear strength. Natural rubber typically carries shear stresses up to 1 to 2 MPa before tearing. At 0.125 MPa the mount is well within its elastic range. ✅

Step 3: Shear Strain and Lateral Deflection

Click to reveal the shear strain and deflection

1. Shear strain from :

   ✅

2. Lateral shear deflection. Shear strain is the tangent of the shear angle, which for small angles equals :

   ✅

3. Interpret the result. A lateral deflection of about 1.9 mm under 800 N gives a lateral stiffness of N/mm. Anti-vibration mounts for light machinery typically target lateral stiffnesses in the range of 200 to 800 N/mm, so this mount is in the right design region. ✅

Design Guidelines for Strain and Material Properties

Summary and Next Steps

Key Concepts Mastered

1. Axial deflection follows . In precision systems the stiffness criterion (deformation limit) almost always sets a larger minimum section than the strength criterion, and both must be checked.
2. The tensile test curve captures stiffness (, the slope), yield strength (0.2% offset construction), UTS (peak), fracture strain (ductility), and the energy per unit volume the material can absorb (resilience and toughness).
3. Shear modulus is related to Young’s modulus and Poisson’s ratio by . Shear-loaded parts (elastomer mounts, pins, adhesive joints) use and to find the shear deflection .

Results at a Glance

Application	Key result	Governing criterion
CNC shaft: aluminum 27 mm, steel 16 mm	Al: mm, mass 0.62 kg; steel: mm, mass 0.63 kg	Stiffness governs both
Steel tensile coupon	kJ/m³, MJ/m³, ductility 22%	Yield at 350 MPa, UTS 520 MPa
Elastomer mount (80 mm x 80 mm, 20 mm thick)	MPa, MPa, rad, mm	Shear governs deflection

A Note on Tools

Every number in this lesson came from three formulas (, , ) and a calculator. Finite-element software is not needed for uniform axial or shear members; the hand calculation is the design, and simulation is for validating complex geometry after the sizing is done.

Next, Compound Bars and Composite Systems extends the axial analysis to members made of two different materials in series or parallel, which is the situation in every bolted joint, press fit, and bi-material actuator.

Comments

Loading comments...