Questions: Stress Intensity Factor

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A steel component has a crack of length 4 mm and is operating such that K_I = 0.8 K_c. The crack grows to 16 mm while the applied stress and geometry factor Y remain unchanged. What is the approximate new value of K_I?

AK_I = 0.8 K_c — applied stress hasn't changed, so K_I doesn't change
BK_I = 3.2 K_c — crack length quadrupled, so K_I quadruples
CK_I = 1.6 K_c — K_I scales as √a, so it doubles when crack length quadruples; fracture occurs
DK_I = 1.13 K_c — K_I scales as the fourth root of crack length
Question 2 Multiple Choice

How does the stress intensity factor K differ fundamentally from the stress concentration factor K_t?

AK applies only to Mode I loading; K_t applies to all three fracture modes
BK has units of MPa√m and predicts fracture by comparison to K_c; K_t is dimensionless and describes local stress amplification from elasticity theory without predicting fracture
CK_t is the material property; K is the applied quantity — they are two halves of the same fracture criterion
DK measures crack tip displacement; K_t measures the stress ratio between tip and nominal stress
Question 3 True / False

According to K = Yσ√(πa), doubling the applied stress increases K by the same multiplicative factor as doubling the crack length.

TTrue
FFalse
Question 4 True / False

A component can fracture even when the average applied stress is well below the material's yield strength, if a crack is present and K_I reaches K_c.

TTrue
FFalse
Question 5 Short Answer

Explain why detecting a crack at 1 mm is disproportionately more valuable than detecting a crack at 16 mm, even if both are below the critical size at current stress levels.

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