Fracture mechanics — stress intensity factor K, toughness K_IC: Question bank

Before you start, we lock down every symbol this page uses so no trap hides behind undefined notation.

Recall The two symbols this page keeps testing

$K_{I}$ is the stress intensity factor: a number you compute from your load and crack ( $K_{I} = σ π a Y$ ). $K_{I C}$ is the fracture toughness: a number the material owns, measured in a lab. Fracture happens when the computed one reaches the owned one: $K_{I} \geq K_{I C}$ .

Geometry	$Y$ (approx.)	Why
Centre crack, infinite plate	$1.00$	The ideal reference case
Single edge crack, wide plate	$1.12$	A free edge lets the crack open more easily
Edge crack, finite width (crack fills much of the section)	$> 1.2$ , rising	Less material left to carry load
Embedded penny-shaped crack	$\frac{2}{π} \approx 0.64$	Surrounding material restrains opening

True or false — justify

TF1 — "

K_{I C}

is a property of the crack, so a longer crack has a higher

K_{I C}

False.

K_{I C}

is a material property, fixed for a given alloy and temperature; crack length changes the applied

K_{I}

, not the material's toughness

K_{I C}

TF2 — "If the applied stress is below the yield strength, the part cannot fracture."

False. Yield strength governs plastic flow of the bulk; fracture is governed by

K_{I}

K_{I C}

at the crack tip, where stress is amplified by

1/ r

and can break bonds while the bulk is still elastic.

TF3 — "Doubling the crack length doubles

K_{I}

False.

K_{I} \propto a

, so doubling

a

multiplies

K_{I}

2 \approx 1.41

, not by 2 — the square root is the whole point of the formula.

TF4 — "A tougher material (higher

K_{I C}

) always has higher yield strength too."

False. They often trade off: high-strength alloys are frequently less tough, which is exactly why fracture mechanics exists as a separate check.

TF5 — "

K_{I}

having units of

MPa m

means it is a kind of stress."

False. Real stress has units of MPa alone; the extra

m

marks

K_{I}

as a field-scaling quantity that sets how strong the

1/ r

singularity is, not the stress at any single point.

TF6 — "The crack-tip opening stress

σ_{y y} = K_{I} / 2 π r

is infinite, so every cracked part fails instantly."

False. The infinity is a mathematical idealisation of a perfectly sharp tip; real materials blunt the tip by plastic flow, and failure is decided by the whole field's scale

K_{I}

reaching

K_{I C}

, not by a literal infinite point.

TF7 — "For a centre crack of total length

2 a

you plug the full

2 a

into

K_{I} = σ π a Y

False. You use the half-length

a

; the formula is written in terms of

a

, so a centre crack of total length 8 mm uses

a = 4

mm.

TF8 — "The geometry factor

Y = 1

for every crack."

False.

Y = 1

only for the ideal centre crack in an infinite plate; an edge crack has

Y \approx 1.12

, and finite-width or curved geometries push

Y

higher still (see the

Y

table above).

TF9 — "Increasing the safety factor

n

increases the largest crack you are allowed to tolerate."

False. Since

a_{m a x} = \frac{1}{π} (\frac{K _{I C}}{Y σ n})^{2}

(derived above), a larger

n

shrinks

a_{m a x}

— you demand detection of smaller flaws to be safer.

TF10 — "Mode I is the only mode that can cause failure."

False. Modes II and III (sliding, tearing) also drive cracks; mixed loading combines them via

K_{eff} = K_{I}^{2} + K_{I I}^{2} + K_{I I I}^{2}

, but Mode I dominates most spacecraft cases (see the three-mode figure below).

Spot the error

SE1 — "Aluminium 2024-T3 has

K_{I C} = 35

MPa

m

, and my part sees

σ = 35

MPa, so I'm exactly at the fracture limit."

The error is comparing a stress (35 MPa) to a toughness (35 MPa

m

) — different units. You must first build

K_{I} = σ π a Y

and compare that to

K_{I C}

SE2 — "There's no visible crack, so

K_{I} = 0

and the part is perfectly safe."

The error is assuming zero flaw size. Damage tolerance assumes an undetectable crack exists at the NDT detection limit; you compute

K_{I}

for that hidden crack, not for a=0.

SE3 — "The angular part

cos \frac{θ}{2} (1 - sin \frac{θ}{2} sin \frac{3 θ}{2})

shows stress is largest straight ahead at

θ = 0

, which equals

K_{I} / 2 π r

; therefore the crack turns sideways."

The value at

θ = 0

is correctly

K_{I} / 2 π r

and is the peak opening stress — and

θ = 0

points straight ahead (see the polar-coordinate figure), so it drives the crack forward, not sideways; the reasoning misreads the geometry.

SE4 — "

Y

makes cracks safer because it's a 'correction', so a larger

Y

gives a larger allowable crack."

The error is treating

Y

as protective. Larger

Y

raises

K_{I}

for the same crack, so it reduces the allowable size

a_{m a x} \propto 1/ Y^{2}

SE5 — "Since

K_{I} \geq K_{I C}

means failure, using

K_{I} \leq K_{I C} / n

is over-cautious double-counting."

Not double-counting.

K_{I} = K_{I C}

is the physical failure edge; dividing by

n

deliberately steps back from that edge to absorb uncertainty in loads, crack size, and material scatter.

SE6 — "Ceramics have tiny

K_{I C}

(2–5), so we should never use them on spacecraft."

The error ignores the load context. Thermal tiles carry very low mechanical stress, so even a brittle, low-

K_{I C}

ceramic can keep

K_{I}

well below

K_{I C}

; toughness only matters relative to the stress it must survive.

SE7 — "Plugging crack length in millimetres into

K_{I} = σ π a Y

is fine as long as

σ

is in MPa."

The error is unit inconsistency.

K_{I}

in MPa

m

needs

a

in metres; a 2 mm crack must enter as 0.002 m, or the answer is off by

1000 \approx 31.6

Why questions

WHY1 — Why does the crack-tip stress scale as

1/ r

and not, say,

1/ r

Because the elasticity solution with traction-free crack faces demands exactly this square-root singularity to keep the total force transmitted across the tip finite; a

1/ r

blow-up would carry infinite force.

WHY2 — Why is there a

π a

and not just

a

K_{I}

The

π

falls out of integrating the stress field around the elliptical crack front in the exact Westergaard solution; it is not free-chosen but fixed so

K_{I}

matches the real field a physical crack produces.

WHY3 — Why is

K_{I}

useful if the tip stress is technically infinite for everyone?

Because every sharp crack shares the same

1/ r

shape, so a single scaling number

K_{I}

fully characterises the field; comparing that one number to

K_{I C}

replaces an impossible point-by-point stress comparison.

WHY4 — Why does damage-tolerance design compute an

a_{m a x}

instead of just "no cracks allowed"?

Because manufacturing, impacts, and fatigue guarantee cracks exist; you cannot forbid them, so you instead set the largest size that stays safe and demand inspection catch anything bigger.

WHY5 — Why do high-strength alloys often need extra fracture-mechanics scrutiny?

Their high yield strength lets them carry large stresses without yielding, but their toughness is often lower, so

K_{I}

can reach

K_{I C}

at crack sizes that a strength check would never flag.

WHY6 — Why does the failure criterion use

\geq

(a threshold) rather than a gradual "damage accumulates" law?

Once

K_{I}

reaches

K_{I C}

, bonds break faster than plastic flow can blunt the tip, so propagation is unstable — it runs away, making the event a sharp threshold, not gradual creep.

WHY7 — Why must inspectors know

a_{m a x}

before choosing an NDT method?

The chosen method must reliably detect flaws smaller than

a_{m a x}

; if eddy-current/ultrasound finds 1–2 mm cracks and

a_{m a x}

is 5.4 mm, there is real margin — but if

a_{m a x}

were 0.5 mm, that method would be useless.

Edge cases

EC1 — What is

K_{I}

when the crack length

a = 0

(a truly flawless part)?

K_{I} = σ π \cdot 0 Y = 0

, so fracture mechanics predicts no crack-driven failure — the part is then limited by ordinary yield strength instead.

EC2 — What happens to

a_{m a x}

as the applied stress

σ \to 0

a_{m a x} = \frac{1}{π} (K_{I C} / Y σ n)^{2} \to \infty

: with essentially no load, any crack is tolerable, which matches intuition that unloaded structures don't fracture.

EC3 — What does

K_{I} = K_{I C}

exactly (equality, not

>

) represent physically?

The precise onset of unstable propagation — the crack is on the knife-edge, neither safely arrested nor yet running; design never sits here, which is why

n > 1

pushes you below it.

EC4 — For a very high-toughness material (

K_{I C} \to \infty

), what limits the structure?

Nothing on the fracture side —

a_{m a x} \to \infty

— so failure reverts to conventional yielding or buckling; toughness stops being the governing constraint.

EC5 — In pure Mode I loading, what are

K_{I I}

and

K_{I I I}

, and what does

K_{eff}

become?

Both are zero, so

K_{eff} = K_{I}^{2} + 0 + 0 = K_{I}

; the mixed-mode formula gracefully reduces to the plain Mode I case.

EC6 — Along the crack plane (

θ = 0

), what does the angular factor

cos \frac{θ}{2} (1 - sin \frac{θ}{2} sin \frac{3 θ}{2})

evaluate to, and why does that matter?

It becomes

cos 0 (1 - 0) = 1

, so

σ_{y y} = K_{I} / 2 π r

straight ahead — the cleanest place to define and interpret

K_{I}

, since the messy angular terms vanish (this is the

θ = 0

ray in the polar figure).