Mechanics of Materials · Chapter 2

Strain

How deformation is measured: normal strain, shear strain, and the small-strain assumption.

Chapter Objectives

Describe deformation as the change in shape and size of a body under load or temperature change.
Define normal strain and shear strain, and the six Cartesian strain components at a point.
Apply small-strain analysis to compute strains from the geometry of deformed bodies.

Physical meaning

Interpret the result in terms of transmitted load and deformation rather than formula symbols alone.

Problem-solving strategy

Draw the system.
State assumptions.
Select a relation.
Compute consistently.
Audit the result.

2.1Deformation

Deformation

Any change in a body's shape or size produced by an applied force or by a change in temperature. Deformation may be dramatic (a stretched rubber band) or imperceptible (a building frame under occupancy loads); thermal expansion and contraction of a roof with the weather is a familiar temperature-driven example.

In general, deformation is not uniform throughout a body. The change in geometry of a line segment drawn in the material can vary along its length, so deformation is studied using very short line segments in the neighborhood of a point. The result also depends on the orientation of the segment at that point: under the same loading, a segment in one direction may elongate while a segment in another direction contracts, and an inclined segment may both change length and rotate.

2.2Strain

Strain expresses deformation in terms of changes in length of line segments and changes in angle between them. Strain is the geometric quantity that is actually measured by experiment; the next chapter relates measured strain to the stress in the body through material properties.

Average Normal Strain (Axial Member)

\[ \epsilon_{avg} = \frac{L - L_0}{L_0} = \frac{\delta}{L_0} \]

\(L_0\) = original length, \(L\) = final length, \(\delta = L - L_0\) = change in length
Positive when the line elongates, negative when it contracts.

Normal Strain at a Point

\[ \epsilon = \lim_{\Delta s \to 0} \frac{\Delta s' - \Delta s}{\Delta s} \]

\(\Delta s\) = original length of a tiny line segment at the point; \(\Delta s'\) = its length after deformation.

Units of Normal Strain

Strain is a ratio of two lengths, so it is dimensionless. Because engineering strains are small, they are commonly reported as micrometres per metre (\(\mu\text{m/m}\)), in./in., as a percent, or simply in "micros": a strain of \(480 \times 10^{-6}\) may be written 480 μm/m, 0.0480 %, or 480 μ.

Shear Strain

\[ \gamma = \frac{\pi}{2} - \theta \]

\(\gamma\) (gamma) = change in angle between two line segments that were originally perpendicular; \(\theta\) = angle between them after deformation.
Measured in radians. Positive if \(\theta < \pi/2\) (angle closes), negative if \(\theta > \pi/2\) (angle opens).

Normal strain changes the lengths (and volume) of an element; shear strain changes the angles between its sides (its shape).

FIG. 2-1 (original diagram)

Cartesian Strain Components

A tiny rectangular element at a point deforms into a parallelepiped. The normal strains change its side lengths to \((1+\epsilon_x)\Delta x\), \((1+\epsilon_y)\Delta y\), \((1+\epsilon_z)\Delta z\) — a change of volume — while the shear strains change the angles between the sides to \(\tfrac{\pi}{2}-\gamma_{xy}\), \(\tfrac{\pi}{2}-\gamma_{yz}\), \(\tfrac{\pi}{2}-\gamma_{xz}\) — a change of shape. The state of strain at a point is therefore characterized by six components: \(\epsilon_x, \epsilon_y, \epsilon_z\) and \(\gamma_{xy}, \gamma_{yz}, \gamma_{xz}\), all of which depend on the original orientation and location of the segments.

Small Strain Analysis

Most engineering design permits only small deformations, so normal strains are assumed very small compared to 1 (\(\epsilon \ll 1\)). For small angle changes \(\Delta\theta\) this allows the first-order approximations \(\sin\Delta\theta \approx \Delta\theta\), \(\cos\Delta\theta \approx 1\), \(\tan\Delta\theta \approx \Delta\theta\), which greatly simplify strain calculations.

EXAMPLE 1Gauge-length normal strain

A 200 mm extensometer gauge on a tensile coupon lengthens by 0.246 mm. Determine average normal strain, express it in microstrain, and state whether the gauge segment elongates or contracts.

Recognition strategy

A measured change in a specified original gauge length directly calls for engineering average normal strain; stress cannot be inferred without a material constitutive relation.

Given information

Original gauge length is 200 mm and measured extension is positive 0.246 mm.

Model and interpretation

Use the original, not final, gauge length as the reference. Assume grip slip and thermal drift have been removed from the extensometer reading.

Step-by-step solution

1. Plan. Set the reference configuration as \(L_0=200\ \text{mm}\) and assign positive displacement to extension.

2. Formulate. Use the definition \(\epsilon_{avg}=\delta/L_0\), which distinguishes a dimensionless strain from the measured displacement.

3. Calculate. Substitute \(\epsilon_{avg}=0.246/200=0.001230\) with millimetres cancelling.

4. Resolve. Convert the ratio to microstrain: \(0.001230\times10^6=1230\) microstrain; the positive sign denotes elongation.

Verification

Units: The millimetre numerator and denominator cancel, leaving dimensionless strain; microstrain is that ratio multiplied by one million.

Signs and direction: The measured extension is positive under the stated convention, so the positive strain correctly represents elongation.

Plausibility: A quarter millimetre over 200 mm is about one part in a thousand, consistent with 1230 microstrain.

Final result: \(\boxed{\epsilon_{avg}=1.23\times10^{-3}=1230\ \text{microstrain}}\)

Engineering interpretation

This is an average gauge response, not a local crack-tip strain. Gauge alignment, temperature compensation, calibration, and whether the coupon remains in the intended elastic regime affect interpretation.

EXAMPLE 2Coordinate strain in displaced wires

Two cables run from a ring to fixed anchors at equal initial 3 m horizontal and 4 m vertical offsets, giving 5 m initial length. The ring moves 12 mm right and 18 mm down. Determine average normal strain in the left and right cables.

Recognition strategy

The ring displacement must be projected through the deformed cable geometry. Each cable has a different new length, so using the same displacement as both extensions would be incorrect.

Given information

Initial horizontal and vertical projections are 3 m and 4 m; ring motion is 0.012 m right and 0.018 m down.

Model and interpretation

Keep anchors fixed. The left cable horizontal projection becomes 2.988 m, the right becomes 3.012 m, and both vertical projections become 4.018 m.

Step-by-step solution

1. Plan. Calculate each original cable length as \(L_0=\sqrt{3^2+4^2}=5.000\ \text{m}\).

2. Formulate. Apply coordinates to find \(L_L=\sqrt{2.988^2+4.018^2}=5.0072416\ \text{m}\) and \(L_R=\sqrt{3.012^2+4.018^2}=5.0216001\ \text{m}\).

3. Calculate. Use the original 5 m length in each average strain calculation rather than the final cable length.

4. Resolve. Obtain \(\epsilon_L=0.001448\) and \(\epsilon_R=0.004320\); both cables lengthen, but the right cable lengthens more.

Verification

Units: Each difference in metres divided by 5 m is dimensionless, so the reported strains are valid length ratios.

Signs and direction: Both computed final lengths exceed 5 m, giving positive strain; the rightward displacement correctly makes the right-cable value larger.

Plausibility: The right cable gains horizontal projection while the left loses it; the recomputed larger right strain is physically expected.

Final result: \(\boxed{\epsilon_L=1.45\times10^{-3},\quad\epsilon_R=4.32\times10^{-3}}\)

Engineering interpretation

These geometric strains can enter compatibility equations, but cable forces additionally depend on stiffness, pretension, sag, and whether a cable loses tension.

EXAMPLE 3Panel shear strain

A rectangular electronics panel is 320 mm high. Its top edge moves 4.8 mm right relative to the fixed bottom edge while vertical edge lengths remain unchanged. Determine engineering shear strain and the lower-left deformed angle using small-angle kinematics.

Recognition strategy

The relative lateral displacement changes an originally right angle without requiring a length change. This is simple shear, not normal extension of the panel height.

Given information

Panel height is 320 mm, top-edge relative displacement is 4.8 mm, and the original corner angle is 90 degrees.

Model and interpretation

Define positive shear as closing the lower-left right angle. For a small angle use \(\gamma\approx u/h\) and then relate angle to shear strain.

Step-by-step solution

1. Plan. Remove rigid-body motion by using the top edge displacement relative to the fixed bottom edge.

2. Formulate. Form the geometric ratio \(\tan\gamma=u/h\) and recognize that its small value permits \(\tan\gamma\approx\gamma\).

3. Calculate. Calculate \(\gamma\approx4.8/320=0.0150\ \text{rad}\), equivalent to an angle change of about 0.859 degrees.

4. Resolve. The lower-left angle closes, so \(\theta\approx90^\circ-0.859^\circ=89.14^\circ\).

Verification

Units: Millimetres cancel in the displacement-to-height ratio; radians are dimensionless angle measure used for engineering shear strain.

Signs and direction: Rightward motion of the top edge closes the lower-left angle under the declared convention, so the positive shear sign is consistent.

Plausibility: The ratio 0.015 is much smaller than one, so the small-angle approximation has a negligible geometric error here.

Final result: \(\boxed{\gamma=+0.0150\ \text{rad},\quad\theta\approx89.14^\circ}\)

Engineering interpretation

This is an average simple-shear description. Fasteners, cutouts, laminate layers, edge restraint, and local stress concentrations can produce a nonuniform strain field requiring measurement or detailed analysis.

Chapter recap

Choose the model from the physical response and assumptions.
Keep force distinct from stress and displacement distinct from strain.
Before moving on, check units, signs, applicability, and competing failure modes.