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Machine Elements · Chapter 3 of 10 · Intermediate

Deflection and Stiffness

A part can be strong enough and still fail its job by flexing too much. Stiffness, the load needed per unit deflection, is a design target in its own right.

01

Readiness check

This chapter is about how much things move under load. Tick only what you can do closed-notes.

  • Recall the second moment of area I for a round section.
  • Use Hooke's law and the modulus E.
  • State a standard beam deflection formula.
  • Add fractions to combine reciprocals.
  • Relate force, deflection, and a spring constant.
0 or 1 weak itemsContinue with this chapter.
2 weak itemsReview beam deflection in Mechanics of Materials, Chapter 9.
3 or more weak itemsRevisit axial stiffness and Hooke's law first.
02

The core idea

Every elastic part behaves like a spring: the load is proportional to the deflection, and the constant of proportionality is the stiffness, k = F/δ. Geometry and modulus set it.

k = F/δδ = FL³/(3EI) (cantilever tip)kaxial = AE/L

Hooke's law makes deflection proportional to load, so any part has a spring rate k = F/δ. For an axial member k = AE/L; for a cantilever loaded at the tip k = 3EI/L³. Stiffness rises with modulus and section, and falls steeply with length, especially in bending where it scales with 1/L³. Combining parts changes the rate: springs sharing a load add (parallel), while springs carrying the same load in turn add as reciprocals (series).

The skill works when: you treat each part as a spring and combine rates by how the parts share load and motion.
The skill breaks down when: series and parallel are swapped, or stiffness is assumed to track strength.
The concept. A loaded cantilever flexes by δ. The ratio of load to deflection is its stiffness; for a beam this falls with the cube of the length, so small length changes matter a lot.
03

The skills, taught in order

Five skills define the spring rate, list the standard stiffness formulas, and combine springs the way real assemblies do.

3.1 The spring rate

Any elastic element obeys F = kδ, where k = F/δ is the spring rate or stiffness, in newtons per millimetre. It captures how hard the part resists motion, and it is the natural language for tying machine parts together as a system of springs.

3.2 Axial, torsional, and bending stiffness

Each loading has its own rate. An axial bar has k = AE/L; a shaft in torsion has an angular rate kt = GJ/L; a beam's rate depends on how it is supported and loaded. Bending is the softest, because it scales with 1/L³.

ElementDeflectionSpring rate k
Axial barδ = FL/AEAE/L
Cantilever, end loadδ = FL³/3EI3EI/L³
Simply supported, centre loadδ = FL³/48EI48EI/L³
Helical compression springδ = 8FD³N/d⁴Gd⁴G/8D³N

3.3 Standard beam deflections

Most shaft and bracket deflections come from a short table of standard cases, superposed when several loads act. Knowing the cantilever and simply supported results, and that deflection scales with FL³/EI, covers most machine geometry without re-integrating the beam equation each time.

3.4 Springs in series and parallel

Parts that share the load and deflect together are in parallel, and their rates add: k = k1 + k2. Parts that carry the same load in turn, with deflections that add, are in series, and their reciprocals add: 1/k = 1/k1 + 1/k2. Series is always softer than either spring; parallel is always stiffer.

3.5 Stiffness as a design target

Stiffness and strength are independent. A shaft may be strong yet deflect enough to misalign gears or bearings, so machine specifications often set a deflection or slope limit that sizes the part before stress does.

Engineering connection: the spring rate returns directly in the helical-spring design of Chapter 8 and in shaft critical speed in Chapter 6.

04

Worked example 1: deflection and stiffness of a cantilever

A round steel cantilever 300 mm long and 20 mm in diameter carries a 500 N load at its free end. With E = 207 GPa, find the tip deflection and the spring rate.

Figure 1. The cantilever bends under the tip load. Its deflection follows the standard FL³/3EI result, and load over deflection gives the spring rate.
  1. ProblemFind the tip deflection and stiffness of the cantilever in Figure 1.
  2. Given / findL = 300 mm, d = 20 mm, F = 500 N, E = 207 000 MPa. Find δ and k.
  3. AssumptionsLinear elastic; small deflection; bending only (shear deflection negligible).
  4. ModelUse I = πd⁴/64, the cantilever deflection δ = FL³/3EI, then k = F/δ.
  5. EquationsI = πd⁴/64δ = FL³/(3EI)k = F/δ
  6. SolveI = π(20)⁴/64 = 7854 mm⁴. δ = 500(300)³/(3·207 000·7854) = 1.35×10¹⁰/4.88×10⁹ = 2.77 mm. k = 500/2.77 = 181 N/mm.
  7. CheckUnits: N·mm³/(MPa·mm⁴) = mm. A few millimetres of tip droop on a 300 mm steel cantilever is reasonable; doubling the length would increase deflection eightfold.
  8. ConclusionDeflection scales with L³, so length is the strongest lever on stiffness. The spring rate is the part's behaviour in one number.
Result. δ = 2.77 mm and k = 181 N/mm.
05

Worked example 2: springs in series and parallel

Two springs of rate k1 = 180 N/mm and k2 = 120 N/mm support a 600 N load. Find the combined rate and the deflection when they are arranged in series and when they are in parallel.

Figure 2. In series the springs stack and deflections add, giving a softer combined rate; in parallel they share the load and rates add, giving a stiffer one.
  1. ProblemFind the combined rate and deflection for both arrangements in Figure 2.
  2. Given / findk1 = 180 N/mm, k2 = 120 N/mm, F = 600 N. Find k and δ for series and parallel.
  3. AssumptionsLinear springs; the load is shared (parallel) or carried in turn (series) ideally.
  4. ModelSeries: 1/k = 1/k1 + 1/k2. Parallel: k = k1 + k2. Then δ = F/k.
  5. Equations1/kseries = 1/k1 + 1/k2kparallel = k1 + k2
  6. SolveSeries: 1/k = 1/180 + 1/120 = 0.01389, so k = 72 N/mm and δ = 600/72 = 8.33 mm. Parallel: k = 180 + 120 = 300 N/mm and δ = 600/300 = 2.0 mm.
  7. CheckSeries is softer than either spring (72 < 120); parallel is stiffer than either (300 > 180). Both bounds hold, as they must.
  8. ConclusionArrangement, not just the springs, sets the stiffness. Series softens and parallel stiffens, a rule reused for bolted joints in Chapter 7.
Result. Series: k = 72 N/mm, δ = 8.33 mm. Parallel: k = 300 N/mm, δ = 2.0 mm.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Swapping series and parallelCombined rate on the wrong side of the inputs"Do the parts share the load or carry it in turn?"Shared load is parallel (rates add); same load in turn is series (reciprocals add).
Equating stiffness with strengthA strong part assumed to be stiff"Is the limit a stress or a deflection?"Stiffness and strength are independent; check deflection separately.
Ignoring the L³ effectLength change underestimated"How does length enter the deflection?"Bending deflection scales with L³; small length changes matter a lot.
Wrong I for the sectionDeflection off by a large factor"Is I = πd⁴/64 for this round section?"Use the correct second moment of area for the cross-section.
07

Practice ladder

Level 1 · Direct skill

A steel bar 500 mm long with a 200 mm² section carries 8 kN axially. With E = 207 GPa, find its stretch.

Show answer

δ = FL/AE = 8000·500/(200·207 000) = 4×10⁶/4.14×10⁷ = 0.0966 mm. Axial members are very stiff compared with beams.

Level 2 · Mixed concept

The cantilever of Worked Example 1 is shortened to 200 mm. What is the new deflection?

Show answer

δ scales with L³: δ = 2.77 × (200/300)³ = 2.77 × 0.296 = 0.82 mm. The shorter beam is much stiffer.

Level 3 · Independent problem

Three identical springs of 90 N/mm are arranged all in parallel, then all in series. Find each combined rate.

Show answer

Parallel: k = 3 × 90 = 270 N/mm. Series: 1/k = 3/90, so k = 30 N/mm. Parallel multiplies stiffness by the count; series divides it.

Level 4 · Transfer to real engineering

Find a product where stiffness, not strength, is the design driver (a machine-tool frame, a camera tripod, a bookshelf). Explain what deflection limit governs it.

What good work looks like

A named deflection or slope limit (accuracy, alignment, feel), a recognition that the part is nowhere near yielding, and a link from geometry to stiffness.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check whether I treated these springs as series or parallel correctly."
"Give me three assemblies; I will identify series versus parallel stiffness."
"What is the deflection?" Choosing the right formula and computing it is the skill.
"Combine these springs for me." Deciding the arrangement is the point.

Portfolio task

Model a real assembly as a set of springs, classify each connection as series or parallel, compute the system stiffness, and state the deflection under a representative load.

Must include: a spring model, a correct series or parallel classification, a system stiffness, and a deflection.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Define the spring rate.

k = F/δ, the load per unit deflection, in N/mm.

2. Give the cantilever tip deflection.

δ = FL³/3EI, so the rate is 3EI/L³.

3. How do parallel springs combine?

Rates add: k = k1 + k2, giving a stiffer result.

4. How do series springs combine?

Reciprocals add: 1/k = 1/k1 + 1/k2, giving a softer result.

5. Are stiffness and strength the same?

No; a part can be strong yet flexible. Check deflection limits separately.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive the cantilever and series/parallel results from a blank page.
+3 daysCombine three new spring arrangements.
+7 daysCarry stiffness into static failure, Chapter 4.
+30 daysReuse the spring rate in helical springs and critical speed.
10

Textbook mapping

ItemMapping
Primary sourceBudynas and Nisbett, Shigley's Mechanical Engineering Design, Chapter 4 (Deflection and Stiffness)
Cross-referenceNorton, Ch. 4 · Mechanics of Materials, Ch. 9
Core topics3.1 Spring rate · 3.2 Axial, torsional, bending stiffness · 3.3 Standard beam deflections · 3.4 Series and parallel · 3.5 Stiffness as a target
Engineering connectionThe spring rate underlies helical springs and shaft critical speed.
Read nextChapter 4: Failure from Static Loading.