Phase VII · FE Civil Academy
Mechanics of Materials
How members respond internally to load: stress, strain, deflection and stability.
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Mechanics of materials — stress, strain and beam behaviour
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Engineering story
A mezzanine in a municipal maintenance building is being repurposed to store spare valve assemblies. The facilities manager asks whether the existing W12×26 floor beams can take the new 1.6 kip/ft storage load without replacement.
- What is the engineering problem?
- Strength is only half of the question. The mezzanine also supports a laser alignment bench, and deflection — not bending stress — is likely to govern the answer.
- What information is missing?
- Nobody has computed the current bending stress, the deflection under the proposed load, or the applicable serviceability limit.
- What should the engineer evaluate?
- Maximum bending moment, section modulus demand, actual bending stress against allowable, and midspan deflection against L/360.
- What decision must be made?
- Approve as-is, restrict the stored load, or add an intermediate support to halve the effective span.
- How does this lesson help?
- This lecture connects internal forces from statics to the stresses and deformations that decide whether a member is acceptable, and it makes explicit why the deflection check so often governs before the stress check does.
Why this matters
- Mechanics of materials is the bridge between the equilibrium of statics and the code checks of structural design; the FE exam samples it heavily.
- Serviceability failures — sagging floors, cracked finishes, vibrating walkways — are far more common in practice than strength failures, and they are decided here.
- The relationship between load, span and deflection is highly nonlinear, which makes engineering intuition unreliable without the calculation.
- Combined loading and Mohr's circle underpin every stress-based failure criterion you will meet later.
Learning objectives
- Compute normal, shear and bearing stress for axially and transversely loaded members.
- Relate stress and strain through Hooke's law and compute axial deformation.
- Locate the maximum bending moment and compute flexural stress using σ = Mc/I.
- Evaluate midspan deflection for standard load cases and compare against serviceability limits.
- Determine principal stresses and maximum in-plane shear stress for a plane-stress element.
ABET evidence: SO1 · SO2 · CE-PC2 · CE-PC3
Prerequisite review
Shear and moment diagrams
Shear is the integral of load; moment is the integral of shear. Maximum moment occurs where shear crosses zero.
Second moment of area
For a rectangle, I = bh³/12. For rolled shapes, read I and S directly from the section tables.
Equilibrium
Internal forces come from a cut free-body diagram — the same tool used throughout statics.
1. Stress, strain and the elastic relationship
Normal stress is force distributed over the area that resists it: σ = P/A. The subtlety is always which area. A bolted tension member resists on its net area at the hole line but on its gross area away from the connection, and the governing check is whichever produces the larger stress. A bearing plate resists on the contact area, which may be far smaller than the member cross-section.
Strain is dimensionless deformation, ε = δ/L. Within the elastic range these are linked by Hooke's law, σ = Eε, which rearranges to the axial deformation relationship δ = PL/(AE). Steel has E ≈ 29,000 ksi and normal-weight concrete roughly 3,600 ksi at 4 ksi strength; that eight-to-one ratio explains why composite behaviour and cracked-section analysis matter so much in reinforced concrete.
Thermal strain adds to mechanical strain and is independent of load: ε_T = αΔT. In a restrained member this strain has nowhere to go and converts directly into stress, σ_T = EαΔT. A 60 °F temperature rise in a fully restrained steel member generates roughly 11 ksi — enough to buckle a member that was never analysed for it. This is why expansion joints exist.
σ = P/A ε = δ/L σ = Eε δ = PL/(AE)
- P = axial force
- A = resisting area
- E = modulus of elasticity
- L = member length
σ_T = E α ΔT (fully restrained)
- α ≈ 6.5 × 10⁻⁶ /°F for steel
- ΔT = temperature change
Avoid this trap. Using gross area where the net area governs, or vice versa. Decide which cross-section actually resists the force at the location you are checking before selecting A.
2. Flexure: from moment to stress
Bending stress follows σ = Mc/I, or equivalently σ = M/S where S = I/c is the elastic section modulus. The equation says something physically important: stress depends on how far material sits from the neutral axis, squared through I and linearly through c. Doubling the depth of a rectangular beam multiplies its moment capacity by four while only doubling its weight, which is why beams are deep and why wide-flange shapes concentrate area in the flanges.
Before applying the equation, locate the maximum moment. For a simply supported beam under uniform load w, M_max = wL²/8 at midspan. For a single concentrated load P at midspan, M_max = PL/4. For a cantilever with uniform load, M_max = wL²/2 at the fixed end. These three cases cover the majority of FE flexure questions, and recognising them saves the time of constructing a full moment diagram.
Transverse shear stress in a beam follows τ = VQ/(Ib) and peaks at the neutral axis, not at the extreme fibre where bending stress peaks. In rolled steel sections a conservative and widely used approximation takes the web as carrying all the shear: τ ≈ V/(d × t_w). Shear rarely governs in long-span flexural members but frequently governs in short, heavily loaded spans and at coped connections.
σ = Mc/I = M/S
- c = distance to extreme fibre
- I = moment of inertia about the bending axis
- S = elastic section modulus
τ = VQ/(Ib)
- Q = first moment of the area above the level considered
- b = width at that level
Δ_max = 5wL⁴/(384EI) (uniform, simply supported)
- Δ in same length units as L
- Watch unit consistency: convert kip/ft and ft to kip/in and in
Avoid this trap. Deflection scales with L⁴. A 20 percent span increase raises deflection by 107 percent while raising moment by only 44 percent — this is precisely why serviceability so often governs before strength.
3. Combined stress and principal planes
Real members rarely carry one action alone. A crane runway girder carries bending from wheel loads plus axial force from lateral bracing plus torsion from eccentric rail placement. Where the stresses are all normal stresses on the same plane they superpose directly: σ_total = P/A ± Mc/I. Sign matters, and the governing fibre is the one where the two contributions add rather than cancel.
When normal and shear stresses act together, the largest normal stress does not act on the plane you happened to draw. Principal stresses give the maximum and minimum normal stress and the planes on which they act, obtained from the plane-stress transformation. Mohr's circle is the graphical form of the same algebra and remains the fastest way to answer FE questions that ask for maximum in-plane shear stress.
The centre of Mohr's circle sits at the average normal stress (σx + σy)/2 and its radius equals the maximum in-plane shear stress. Once you have those two numbers, the principal stresses are simply centre plus radius and centre minus radius. Most exam questions are answerable from the centre and radius alone, without ever computing the angle of the principal plane.
σ_1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]
- σ_1 = maximum principal stress
- σ_2 = minimum principal stress
τ_max = √[((σx − σy)/2)² + τxy²]
- Equals the radius of Mohr's circle
Avoid this trap. Maximum in-plane shear stress is the radius of Mohr's circle, not half the difference of the applied normal stresses — the shear term τxy must be inside the square root.
Table 1. Standard beam cases — maximum moment, shear and deflection
| Case | M_max | V_max | Δ_max |
|---|---|---|---|
| Simply supported, uniform w | wL²/8 at midspan | wL/2 at supports | 5wL⁴/(384EI) |
| Simply supported, P at midspan | PL/4 at midspan | P/2 | PL³/(48EI) |
| Cantilever, uniform w | wL²/2 at fixed end | wL at fixed end | wL⁴/(8EI) |
| Cantilever, P at free end | PL at fixed end | P | PL³/(3EI) |
| Fixed–fixed, uniform w | wL²/12 at ends | wL/2 | wL⁴/(384EI) |
Professional workflow
- Establish the load case and span, and confirm the support idealisation.
- Draw shear and moment diagrams, or recall the standard case.
- Select the governing section and read I, S and c from the section table.
- Compute bending stress and compare with the allowable or design strength.
- Compute deflection and compare with the serviceability limit (commonly L/360 live, L/240 total).
- Check shear at supports and at any coped or notched location.
- State which limit state governs and by what margin.
Applicable standards
NCEES FE Reference Handbook — Mechanics of Materials
Stress and strain relations, beam formula tables, plane-stress transformation and Mohr's circle.
AISC Steel Construction Manual
Section properties, available flexural strength and serviceability guidance.
IBC Table 1604.3
Deflection limits for floor and roof members under live and total load.
Case study — Hyatt Regency walkway collapse, Kansas City (1981)
Suspended walkways in a hotel atrium collapsed during a crowded dance event, killing 114 people. It remains the deadliest structural collapse in U.S. history.
- A shop-drawing change split a single continuous hanger rod into two, doubling the load on the fourth-floor box-beam connection.
- The connection detail was never independently recalculated after the change; the load path through the bearing area was simply assumed to be unchanged.
- Even the original detail did not meet the applicable code requirement, so the as-built condition had no reserve at all.
Lesson learned. A change in geometry is a change in the load path and demands a fresh stress calculation on the actual bearing area. Reviewing engineers must recompute, not assume, when details change.
Source: NBS Building Science Series 143; Missouri Board of Architects, Professional Engineers and Land Surveyors disciplinary findings.
Practical scenario
A simply supported W-shape carries a uniform load. You must draw the shear and moment diagrams, check bending stress and estimate deflection.
Why this matters
Member sizing, serviceability and stability all come from mechanics of materials — the largest single block of structural FE content.
Concept explanation
- Shear and moment diagrams follow from equilibrium: load is the slope of shear, shear is the slope of moment.
- Flexural stress varies linearly through the depth; maximum stress occurs at the extreme fiber.
- Axial, torsional and flexural stresses superpose when they act on the same section.
- Slender columns fail by buckling well below the yield stress.
Unit guidance
- Write units on every substituted value.
- Cancel units symbolically before evaluating numbers.
- Confirm the result unit matches the quantity you were asked for.
Handbook navigation
- Open the Mechanics of Materials section of the handbook.
- Search the term "shear and moment diagrams" rather than scrolling.
- Note the equation number so you can return to it quickly.
Calculator guidance
- Confirm angle mode before any trig entry.
- Store intermediate values in memory instead of re-keying rounded numbers.
- Round only at the final answer.
FE strategy
- Sketch the shear diagram first; the maximum moment is the area under it.
- Confirm which limit state the question is actually asking about.
Conceptual example
A conceptual mechanics of materials item will ask which quantity increases, decreases, or stays the same when one input changes. Reason from the governing relationship, not from memory of a numeric answer.
Common mistakes
- Using I about the wrong axis for a rotated section.
- Forgetting to convert L to inches in a deflection formula.
- Applying Euler buckling to a stocky column that yields first.
Guided practice
- Work the first worked example with the solution visible.
- Re-solve it closed-book and compare every step.
- Explain the unit cancellation out loud.
Independent practice
- Complete a 10-question Mechanics of Materials quiz in practice mode.
- Classify every miss by error category.
- Re-test the missed subtopic within 48 hours.
Summary
- σ = Mc/I for bending, τ = VQ/Ib for shear.
- Deflection scales with L⁴ — span dominates.
- Check slenderness before applying Euler.
Capstone application: Member sizing and serviceability checks in structural design.