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Mechanics of Materials · Chapter 9 of 10 · Advanced

Deflection of Beams

A beam can be strong enough yet sag too much. Deflection comes from integrating the moment twice, EI y'' = M, and for everyday cases a table of formulas and superposition do the work.

Readiness check

This chapter integrates the bending moment to get deflection. Tick only what you can do closed-notes.

Find M(x) along a beam.
Integrate polynomials and apply constants.
Use boundary conditions (zero deflection at supports).
Recall the flexural rigidity EI.
Add quantities by superposition.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsReview moment diagrams in Chapter 5.

3 or more weak itemsRefresh integration in integrals first.

The core idea

The curvature of a loaded beam is proportional to the bending moment, EI y'' = M(x); integrating twice gives the slope and deflection.

EI y″ = M(x)δ_max = PL³/48EI (SS, central P)superpose: δ = Σ δ_i

Bending bends the beam into an elastic curve whose curvature at each point equals M/EI. Integrating once gives the slope, again gives the deflection, with the constants set by the supports (zero deflection at a pin, zero slope at a fixed end). Because the relation is linear, deflections from several loads add, so most problems are solved by looking up standard cases and superposing them rather than integrating from scratch. Stiffness, the flexural rigidity EI, controls deflection just as it controlled axial and torsional deformation.

The skill works when: you integrate M/EI with correct boundary conditions, or superpose tabulated cases.

The skill breaks down when: deflection is judged from strength alone, or constants of integration are dropped.

The concept. The loaded beam deflects into an elastic curve. Its curvature is the moment over the flexural rigidity, M/EI; integrating that twice, with the support conditions, gives the deflection.

The skills, taught in order

Deflection is integration of the moment, made practical by tables and superposition. Five skills cover the curve, integration, the formulas, superposition, and indeterminate beams.

9.1 The elastic curve

The deflected shape y(x) is the elastic curve. For small deflections its curvature is y″ = M/EI, so EI y″ = M(x). The flexural rigidity EI combines material stiffness E and section shape I; a deeper or stiffer beam deflects less.

9.2 Double integration

Integrate EI y″ = M(x) once for the slope EI y′ and again for the deflection EI y. Two constants of integration appear; fix them with boundary conditions: zero deflection at each support, and zero slope at a fixed end. This gives the deflection anywhere.

9.3 Standard deflection formulas

Common cases are tabulated, so you rarely integrate by hand.

Beam and load	Maximum deflection
Simply supported, central point P	PL³/48EI
Simply supported, UDL w	5wL⁴/384EI
Cantilever, end point P	PL³/3EI
Cantilever, UDL w	wL⁴/8EI

9.4 Superposition of deflections

Because the governing equation is linear, the deflection under several loads is the sum of the deflections each would cause alone. Combine tabulated cases at the point of interest, the fastest route for most real beams.

9.5 Statically indeterminate beams

When a beam has more supports than equilibrium can resolve (a propped cantilever, a fixed-fixed beam), use a deflection compatibility condition, such as zero deflection at the extra support, to find the redundant reaction. Superposition of standard cases makes this tractable.

Engineering connection: floor joists, machine beds, shafts, and aircraft wings are often governed by a deflection limit (such as L/360), not by strength; EI is the lever to control it.

Worked example 1: deflection under a central load

A simply supported steel beam spans 4 m and carries a 20 kN load at mid-span. With E = 200 GPa and I = 40×10⁶ mm⁴, find the maximum deflection.

Figure 1. The central point load deflects the beam most at mid-span. The deflection scales with the cube of the span, so length matters far more than load.

ProblemFind the maximum deflection of the beam in Figure 1.
Given / findL = 4 m, P = 20 kN, E = 200 GPa, I = 40×10⁶ mm⁴, central load. Find δ_max.
AssumptionsSmall deflection, linear-elastic, prismatic beam.
ModelUse the standard simply supported, central-point-load deflection.
Equationsδ_max = PL³/48EI
Solveδ_max = (20 000 × 4000³)/(48 × 200 000 × 40×10⁶) = (20 000 × 6.4×10¹⁰)/(3.84×10¹⁴) = 3.33 mm.
CheckAs a fraction of span, 3.33/4000 = L/1200, well within a typical L/360 limit, so the beam is stiff enough. Note δ ∝ L³: doubling the span would multiply the deflection eightfold.
ConclusionThe standard formula gives the deflection in one line. The cube-of-span dependence is why long beams deflect so much and why depth (through I) is the cure.

Result. Maximum deflection δ = 3.33 mm (about L/1200).

Worked example 2: superposing two loads

The same beam (L = 4 m, E = 200 GPa, I = 40×10⁶ mm⁴) now carries both the 20 kN central load and a uniformly distributed load of 6 kN/m. Find the total mid-span deflection by superposition.

Figure 2. The point-load and UDL deflections are found separately from the table, then added. Superposition replaces re-integrating the combined load.

ProblemFind the total mid-span deflection of the beam in Figure 2.
Given / findL = 4 m, P = 20 kN central, w = 6 kN/m, E = 200 GPa, I = 40×10⁶ mm⁴. Find δ_total.
AssumptionsLinear-elastic (so superposition holds), small deflection.
ModelAdd the standard central-point-load and UDL deflections.
Equationsδ_P = PL³/48EI δ_w = 5wL⁴/384EI δ = δ_P + δ_w
Solveδ_P = 3.33 mm (from Example 1). δ_w = 5(6)(4000)⁴/(384 × 200 000 × 40×10⁶) = 2.50 mm. Total δ = 3.33 + 2.50 = 5.83 mm.
CheckBoth deflections are downward, so they add. The total (5.83 mm ≈ L/686) is still within an L/360 limit. Superposition gives in two lookups what double integration would take a page to do.
ConclusionSuperposition is the workhorse of deflection analysis: decompose the loading into tabulated cases, add the deflections. The same idea finds redundant reactions in indeterminate beams.

Result. Total mid-span deflection δ = 5.83 mm (3.33 from the point load, 2.50 from the UDL).

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Strength implies stiffness	Strong beam sags too much	"Is there a deflection limit?"	Check δ against a serviceability limit (such as L/360) separately.
Dropping integration constants	Wrong slope or deflection	"Did I apply the boundary conditions?"	Use the supports to fix both constants.
Wrong standard formula	Off by a numeric factor	"Cantilever or simply supported? Point or UDL?"	Match the case exactly before using a tabulated δ.
Adding deflections of unlike sign	Total too small or large	"Do the loads deflect the same way?"	Superpose with sign; opposing loads partly cancel.

Practice ladder

Level 1 · Direct skill

A cantilever 2 m long (EI = 200 GPa × 8×10⁶ mm⁴) carries a 5 kN end load. Find the tip deflection.

Show answer

δ = PL³/3EI = 5000 × 2000³/(3 × 200 000 × 8×10⁶) = 4×10¹³/4.8×10¹² = 8.33 mm. Cantilevers deflect far more than simply supported beams of the same span.

Level 2 · Mixed concept

For the Worked Example 1 beam, what I is needed to keep the deflection under an L/360 limit?

Show answer

Limit δ = 4000/360 = 11.1 mm. Since δ ∝ 1/I, the current 3.33 mm at I = 40×10⁶ is already well under, so almost any practical I works; I_min = 40×10⁶ × 3.33/11.1 = 12×10⁶ mm⁴ would just meet it.

Level 3 · Independent problem

A propped cantilever (fixed at one end, simply supported at the other) carries a UDL. Outline how to find the prop reaction.

Show answer

It is statically indeterminate. Remove the prop and find the UDL tip deflection of the cantilever (wL⁴/8EI); then find the upward deflection from an unknown prop force R (RL³/3EI). Set their sum to zero (the prop allows no deflection): R = 3wL/8. Compatibility supplies the missing equation.

Level 4 · Transfer to real engineering

Find a real beam with a deflection concern (a bookshelf, a diving board, a floor joist). Estimate its deflection from a standard formula and compare with a sensible limit.

What good work looks like

The correct standard case identified, EI estimated, δ computed, and a comparison with a serviceability limit such as L/360.

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I matched the right standard case and applied the boundary conditions."

"Give me five beams; I will pick the tabulated deflection formula for each."

"Find the deflection." Selecting the case and superposing is the skill.

"Is it stiff enough?" Comparing δ with the limit is the point.

Portfolio task

Analyse one beam's deflection: use a standard formula (or superpose two), and check the result against a serviceability limit.

Must include: the standard case(s), EI, the deflection, and a comparison with a limit such as L/360.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the governing equation for the elastic curve.

EI y″ = M(x); integrate twice for slope and deflection.

2. What sets the constants of integration?

Boundary conditions: zero deflection at supports, zero slope at a fixed end.

3. Give the max deflection for a simply supported beam under a central point load.

δ = PL³/48EI.

4. Why does superposition work for deflection?

The governing equation is linear, so deflections from separate loads add.

5. How is a propped cantilever solved?

By a compatibility condition (zero deflection at the prop) to find the redundant reaction.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the deflection and superposition from a blank page.

+3 daysOne single-case and one superposition problem.

+7 daysFinish with columns and buckling, Chapter 10.

+30 daysCheck a real design against a deflection limit.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 9 (Deflection of Beams)
Cross-reference	Hibbeler, Ch. 12 · Gere and Goodno, Ch. 9
Core topics	9.1 Elastic curve · 9.2 Double integration · 9.3 Standard formulas · 9.4 Superposition · 9.5 Indeterminate beams
Engineering connection	Serviceability limits on joists, machine beds, shafts, and wings.
Read next	Chapter 10: Columns and Buckling.