A timed, course-specific sprint through stress state, failure theories, manufacturing, and beam deflection. Learn the decision paths, then prove them under pressure.
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TOTAL120MINUTES
20 stress
35 failure
15 mfg.
35 deflection
15 drill
THE PLAN
Five passes. One objective.
00–20Orient stresstensor → principal → Mohr
20–55Predict failurematerial → theory → n
55–70Lock in manufacturingprocess → property
70–105Control deflectionload → integrate → BCs
105–120Pressure test16 questions / no notes
How to use it: read the amber “exam move,” work the checkpoint before revealing it, then mark the module complete. Do not chase derivations beyond the ones shown.
01
00:00–00:20 · FOUNDATION
Stress state at a point
Translate components into principal stresses—the language every failure theory uses.
CORE IDEA
The stress tensor is symmetric
σxxτxyτxzτyxσyyτyzτzxτzyσzz
→
τxy = τyxτxz = τzxτyz = τzy
Moment equilibrium leaves six independent components. Change orientation and the components change, but the physical state does not.
EXAM MOVE
Always reduce first
Write the plane-stress state: σx, σy, τxy.
Find σA, σB with Mohr’s circle.
Add the out-of-plane principal stress 0.
Order: σ1 ≥ σ2 ≥ σ3.
INTERACTIVE MOHR LAB
See the invariants move
Adjust a plane stress state. The circle gives the principal stresses and maximum in-plane shear.
center C30.0radius R = τmax14.1σA44.1σB15.9
C = (σx + σy)/2 · R = √[(( σx − σy)/2)² + τxy²]
PRINCIPAL STRESSESσA,B = C ± R
MAX IN-PLANE SHEARτmax = R
PRINCIPAL PLANEshear = 0
SIGNtension + / compression −
LOAD → LOCAL STRESS → FAILURE
The complete stress-analysis path
Failure theories never start with the applied force. They start with the stress state at the most critical material point.
1Cut the memberUse equilibrium to find internal N, V, M, and T at the likely critical section.
→
2Choose a pointMaximum bending is at the outer fiber; torsional shear is maximum at the outside radius.
→
3Build componentsAdd normal stresses with signs and add shear components on the same faces.
→
4TransformFind principal stresses, include the plane-stress zero, then order σ1 ≥ σ2 ≥ σ3.
→
5Test failureSelect MSS/DE for ductile or CM/MM for brittle, then calculate n.
AXIALσN = N/A
BENDINGσb = −My/I
TORSIONτT = Tρ/J
TRANSVERSE SHEARτV = VQ/(Ib)
Course shortcut: for a slender member with l/d > 8, transverse shear from V is often negligible at the critical outer surface. Check bending + torsion first, but state the assumption.
TRUE vs ENGINEERING STRESS
Which area is in the denominator?
ENGINEERINGσeng = F/A0
Uses the original area. This is what the standard stress-strain curve reports and what tabulated Sy and Su are normally based on.
TRUEσtrue = F/A
Uses the instantaneous area. As a tensile specimen thins, true stress is higher than engineering stress for the same force.
Before necking: σtrue = σeng(1 + εeng)
TRUE STRAIN
Add the tiny increments
εtrue = ∫ dL/L = ln(L/L0) = ln(1 + εeng). If plastic deformation preserves volume, A0L0 ≈ AL, so:
εtrue = ln(A0/A) = ln[1/(1−R)]
After necking: deformation is no longer uniform. Use the local minimum neck area for true stress; the simple conversion from engineering strain is no longer reliable.
QUICK CONVERSION · BEFORE NECKING ONLY
Example: 20% engineering strain at 50 ksi
GIVENεeng = 0.20σeng = 50 ksi
→
TRUE STRAINln(1.20) = 0.182
True strain is smaller than engineering strain at positive elongation because it accumulates each increment relative to the growing length.
→
TRUE STRESS50(1.20) = 60 ksi
True stress is larger because the instantaneous area is smaller than A0.
Do not mix definitions: compare tabulated engineering strengths with engineering stress unless the problem explicitly supplies a true-stress/true-strain material law.
02
00:20–00:55 · HIGHEST YIELD
Static failure theories
Choose the theory from material behavior, then compare equivalent stress with strength.
THEORY SELECTOR
What material are you holding?
DE
DUCTILE · εf > 0.05
Distortion Energy (von Mises)
More accurate for ductile metals. Yielding begins when distortion energy reaches the uniaxial-yield benchmark.
n = Sy / σ′
MSSDUCTILE · CONSERVATIVE
Maximum shear stress
σ1 − σ3 = Sy
Yielding starts when the largest 3D Mohr-circle radius reaches the uniaxial-test value Sy/2. Only the largest and smallest ordered principal stresses are needed.
n = Sy / (σ1 − σ3)
DEDUCTILE · ACCURATE
Distortion energy
σ′ = √{[(σ1−σ2)² + (σ2−σ3)² + (σ3−σ1)²]/2}
Von Mises stress measures shape-changing energy. Equal pressure added to all three principal stresses cancels because only differences appear.
plane: σ′ = √(σx² − σxσy + σy² + 3τxy²)
CMBRITTLE · CONSERVATIVE
Coulomb–Mohr
mixed signs: σ1/Sut − σ3/Suc = 1/n
A straight line joins the tensile failure point +Sut to the compressive failure point −Suc. Different branches apply when both principal stresses have the same sign.
Use ultimate strengths as positive magnitudes: Sut > 0 and Suc > 0.
MMBRITTLE · ACCURATE
Modified Mohr
piecewise: tensile cutoff + modified mixed-sign line
Modified Mohr follows brittle-test behavior more closely. When tension controls, compression does not immediately reduce the tensile capacity; the modified branch begins only after compression dominates.
Select the branch using signs and the ratio |σ3|/σ1.
BEFORE ANY THEORY
One setup shared by MSS, DE, CM, and MM
1
Find the local stress components. At the critical point, combine axial and bending normal stress; combine torsional and transverse shear only when they act on compatible faces.
2
Find principal stresses. For plane stress calculate σA, σB, then include 0.
3
Order them. Rewrite the set as σ1 ≥ σ2 ≥ σ3. Never assume zero is the middle value.
Controlled heating and cooling changes microstructure, phase, and properties.
slow / softAnnealNormalizeQuenchfast / hard
COLD WORK vs HOT WORK
COLD ROLL / DRAWroom temperature
strain hardening
strength ↑
ductility ↓
VS
HOT ROLLT > 0.5 Tm
removes strain hardening
lower strength
higher ductility
STEEL NUMBER DECODER
4140
alloyfamily≈ 0.40 wt% C
Read from the right first: in a basic four-digit SAE/AISI grade, the last two digits give nominal carbon content in hundredths of a weight percent. The leading digits identify the alloy family.
41xxchromium–molybdenum4140 is the high-frequency exam example
43xxnickel–chromium–molybdenum4340 ≈ 0.40%C
51xxchromium steel5160 ≈ 0.60%C
86xxnickel–chromium–molybdenumlow-alloy family distinct from 43xx
Limit of the shortcut: the number is a grade identifier, not a complete chemical recipe. Letter prefixes/suffixes, five-digit grades, stainless grades, and tool steels use additional conventions.
WROUGHT6061major alloy familyoriginal / modificationalloy ID, not wt%
1xxxnearly pure aluminum1050 means at least 99.50% Al
2xxxcopperhigh strength; often heat treatable
3xxxmanganesemoderate strength, good formability
4xxxsiliconlower melting range; filler alloys
5xxxmagnesiumstrain hardenable; good corrosion resistance
6xxxmagnesium + silicon6061 is heat treatable and widely used
7xxxzincvery high-strength families
8xxxother elementsspecial-purpose alloy families
EXAMPLE6061-T6
6xxx = Mg + Si; 0 = original alloy; 61 identifies the specific alloy; T6 = solution heat-treated and artificially aged.
COMMON TEMPERS
F as fabricated · O annealed · H strain hardened · W solution heat-treated, unstable · T thermally treated to a stable condition.
Important distinction: for wrought 2xxx–8xxx aluminum, the final two digits identify an alloy—they are not a composition percentage. Cast aluminum instead uses three digits plus a decimal; .0 denotes a casting and .1/.2 denotes ingot.
CHECKPOINT Quenching makes steel strong but brittle. What repairs ductility? Reveal
Tempering. Reheat quenched steel to a controlled lower temperature: ductility rises while strength falls. The tempering temperature controls the tradeoff.
04
01:10–01:45 · METHOD
Beam deflection
Keep load, internal action, slope, and deflection in the right order.
externalq(x)load
d/dx
internalV(x)shear
d/dx
internalM(x)moment
÷ EI
shapeθ(x)slope
∫ dx
shapey(x)deflection
Course convention: q/EI = y⁗, V/EI = y‴, M/EI = y″, θ = y′. Integrating q → V → M → EIθ → EIy introduces constants only at the slope and deflection steps.
METHOD A
Singularity functions
Best for a straight beam with complex loading when E and I are constant.
⟨x−a⟩−2concentrated moment
⟨x−a⟩−1concentrated force
⟨x−a⟩0distributed-load start/stop
⟨x−a⟩1ramp
Critical rule: a finite UDL must have a start term and an equal opposite stop term.
METHOD B
Superposition
Best when table solutions exist or the structure mixes bending, torsion, axial loading, materials, or geometry.
Separate the loading.
Match each case to a known solution.
Respect the segment/domain.
Add compatible displacements with signs.
3D frame: total point displacement can contain bending of multiple legs + twist-induced motion.
SINGULARITY WORKFLOW
1ReactionsΣF = 0, ΣM = 0
→
2Write q(x)include starts + stops
→
3Integrate 4×q, V, M, EIθ, EIy
→
4Apply BCssupports/fixed ends
→
5Evaluatewatch inactive brackets
WHAT THE BRACKET MEANS
One expression that turns on at x = a
⟨x−a⟩n =0 for x < a (x−a)n for x ≥ a
For ordinary powers n ≥ 0, the bracket is simply inactive to the left of a and active to the right. It lets one equation represent several beam regions without writing a separate M(x) for every interval.
At x = 3, a = 4⟨3−4⟩2 = 0the load has not started
At x = 5, a = 4⟨5−4⟩2 = 1the term is active
COURSE INTEGRATION RULES
Negative exponents are special symbols
∫⟨x−a⟩−2dx = ⟨x−a⟩−1∫⟨x−a⟩−1dx = ⟨x−a⟩0∫⟨x−a⟩ndx = ⟨x−a⟩n+1/(n+1), n ≥ 0
Do not apply the usual power-rule denominator to −2 or −1. They encode a concentrated couple and point force; the course convention advances their order directly.
FINITE UDL FROM a TO b−w⟨x−a⟩0 + w⟨x−b⟩0
The opposite stop term cancels the load after b. If the mathematical beam domain ends exactly at b, some solutions omit the stop term because no x > b is evaluated.
FULL WALKTHROUGH
Cantilever: point force + partial UDL
Fixed at x = 0 · L = 8 m · 30 kN downward at x = 4 m · 10 kN/m downward from x = 4 to 8 m
LOADq = −360⟨x⟩−2 + 70⟨x⟩−1 − 30⟨x−4⟩−1 − 10⟨x−4⟩0The reaction moment, reaction force, point force, and UDL all enter at their x-location.
SHEARV = −360⟨x⟩−1 + 70⟨x⟩0 − 30⟨x−4⟩0 − 10⟨x−4⟩1Integrate once. The moment singularity remains as an order −1 bookkeeping term.
MOMENTM = −360⟨x⟩0 + 70⟨x⟩1 − 30⟨x−4⟩1 − 5⟨x−4⟩2Integrate twice. This is the physical bending-moment expression.
SLOPEEIθ = −360⟨x⟩1 + 35⟨x⟩2 − 15⟨x−4⟩2 − (10/6)⟨x−4⟩3 + C1Use EIθ = ∫M dx for constant E and I.
DEFLECTIONEIy = −180⟨x⟩2 + (70/6)⟨x⟩3 − 5⟨x−4⟩3 − (10/24)⟨x−4⟩4 + C1x + C2At the fixed end, θ(0)=0 and y(0)=0, so C1=C2=0.
Sign convention: these signs match the course notes. If your instructor’s positive load/moment convention differs, the signs may reverse, but the bracket locations, exponents, integration steps, and boundary conditions do not.
BOUNDARY CONDITIONS
Choose conditions the support actually enforces
Fixed endy = 0 and θ = 0
Pin / rollery = 0
Symmetry planeθ = 0
Free endV = 0 and M = 0 unless end loads/couples act
Internal joinmatch y and θ when using piecewise equations
ERROR CHECK
Five mistakes that cost the whole problem
Placing a term at the wrong x-coordinate.
Forgetting the UDL stop term when the beam continues past the load.
Using the ordinary power rule on orders −2 or −1.
Evaluating ⟨x−a⟩n as a negative number when x < a; it must be zero.
Applying slope/deflection boundary conditions before reaching EIθ and EIy.
WORKED EXAMPLE · COURSE PROBLEM
Simply supported beam with overhang load
supports O and A (8 in apart) · overhang AB = 8 in · w = 100 lbf/in · E = 10×106 psi · I = 0.1 in4
Stress state notes · MSS / DE / CM / MM notes · manufacturing notes · singularity-function and superposition notes · static-failure problems and solutions · deflection problems and solutions · Quiz 2 review.