NivaarExam PrepOfficial exam papers ↗

25-Nav-A4 Ship Structure and Strength of Ships · May 2016

Question 5 of 6: Risk and Reliability

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2016 — 98-Nav-A4 Ship Structure and Strength of Ships. Three-hour, closed-book exam (no notes permitted); Casio/Sharp non-programmable calculator and simple drawing equipment allowed. Format: six compulsory questions, marks indicated per sub-part, totalling 100; some formulae (beam bending relations, section modulus, shear flow, deflection/slope tables) and a Normal (cumulative) distribution table are supplied at the end of the exam and are used directly below. All six are solved in full.

Reference texts: Hughes, O.F. & Paik, J.K., Ship Structural Analysis and Design (2nd ed., SNAME, 2010) — hull-girder strength, panel/plate structure, section properties and shear flow in thin-walled hull sections; Hibbeler, R.C., Mechanics of Materials (10th ed., Pearson) — beam bending/deflection, stress–strain behaviour and fatigue basics; Muckle, W., Muckle’s Naval Architecture (2nd ed., Butterworths) — hydrostatics, Bonjean curves and structural terminology; Ang, A.H-S. & Tang, W.H., Probability Concepts in Engineering (2nd ed., Wiley) — structural reliability, load/resistance margin; IACS Common Structural Rules — steel grades, fatigue design (S–N curves, Paris Law).

Question 5: Risk and Reliability (16 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $Q\sim N(\mu_Q=120,\sigma_Q=40)\text{ kN}$; $R\sim N(\mu_R=?,\sigma_R=20)\text{ kN}$; $R_{5\%}=\gamma_T\,Q_{10\%}$ with $\gamma_T=1.20$; $Q_{10\%}$ is the upper-10%-exceedance value of Q, $R_{5\%}$ is the lower-5th-percentile value of R.

Find. $R_{5\%}$, $Q_{10\%}$; $\mu_R$; $\mu_{Margin}$, $\sigma_{Margin}$; probability of failure $P_f$.

Approach. Read the standard-normal multipliers for the 10% and 5% tail probabilities off the supplied Normal table, apply them to get the two characteristic values, back out $\mu_R$ from the given relation, then treat the margin $M=R-Q$ as Normal (difference of independent Normals) to get $P_f=P(M<0)$.

  1. Standard-normal multipliers from the supplied table. $\Phi(1.28)=0.8997\approx0.90$ (so $z_{10\%}=1.28$ for the upper 10% tail of Q); interpolating the table between $z=1.64$ ($\Phi=0.9495$) and $z=1.65$ ($\Phi=0.9505$) for $\Phi=0.95$ gives $z_{5\%}=1.645$ (the lower 5% tail of R).
  2. (i) Characteristic values. $$Q_{10\%}=\mu_Q+z_{10\%}\sigma_Q=120+1.28(40)=\boxed{171.2\text{ kN}}$$ $$R_{5\%}=\gamma_T\,Q_{10\%}=1.20(171.2)=\boxed{205.4\text{ kN}}$$
  3. (ii) Mean of R. $R_{5\%}$ is the lower-tail value of R, $R_{5\%}=\mu_R-z_{5\%}\sigma_R$, so $$\mu_R=R_{5\%}+z_{5\%}\sigma_R=205.4+1.645(20)=205.4+32.9=\boxed{238.3\text{ kN}}$$
  4. (iii) Margin statistics. With $M=R-Q$ and $R,Q$ independent Normals: $$\mu_M=\mu_R-\mu_Q=238.3-120=\boxed{118.3\text{ kN}}$$ $$\sigma_M=\sqrt{\sigma_R^2+\sigma_Q^2}=\sqrt{20^2+40^2}=\sqrt{2000}=\boxed{44.7\text{ kN}}$$
  5. (iv) Probability of failure. Failure occurs when $M<0$ (resistance falls below load): $$z=\frac{\mu_M}{\sigma_M}=\frac{118.3}{44.7}=2.65,\qquad P_f=1-\Phi(2.65)$$ From the table, $\Phi(2.65)\approx 0.9960$ (interpolated between $\Phi(2.64)=0.99586$ and $\Phi(2.65)=0.99598$), so $$\boxed{P_f\approx 1-0.9959=0.41\%}$$
Load Q and resistance R — Normal densitiesload or strength [kN]Q (μ=120, σ=40 kN)R (μ=238.3, σ=20 kN)Q_10% = 171.2R_5% = 205.4
Figure 5.1 — the load (Q) and resistance (R) Normal densities and their characteristic (10%/5%) values, matching the exam's own sketch.
QuantityValue
Q10%171.2 kN
R5%205.4 kN
Mean of R238.3 kN
Mean / std. dev. of Margin118.3 kN / 44.7 kN
Probability of failure≈ 0.41%

(b) Partial factors of safety and LRFD

Part (a) used a single, lumped "total factor of safety" $\gamma_T$ relating one characteristic load value to one characteristic resistance value — conceptually simple, but it cannot distinguish how much uncertainty comes from the load side versus the resistance side, nor can it treat different load types (permanent, live, environmental) differently even though they have very different degrees of statistical scatter. Partial factors of safety resolve this by assigning a separate factor to each individual load type and to resistance, calibrated (typically via first-order reliability methods, FORM) so that the resulting design achieves a consistent, code-wide target reliability index $\beta$ (equivalently, a consistent target probability of failure) across many different combinations of load types and structural elements — rather than the single blanket factor implicitly assuming every failure mode and every load combination is equally uncertain.

LRFD (Load and Resistance Factor Design) is the design format built on exactly this idea: a limit-state design check of the form $$\phi R_n \ge \sum \gamma_i Q_i$$ where each factored load effect $\gamma_iQ_i$ (dead load, live load, environmental load, each with its own $\gamma_i>1$, reflecting how variable/how likely-to-be-exceeded that load type is) must not exceed the factored nominal resistance $\phi R_n$ (with resistance factor $\phi<1$, reflecting uncertainty in material strength, fabrication and the resistance model itself). Because the factors are calibrated probabilistically rather than picked by tradition, LRFD achieves much more uniform reliability across different structural members and load combinations than a single working-stress-design factor of safety ever could, which is exactly the motivation for classification-society and IACS rules moving toward partial-factor / LRFD-style formats for hull-girder and local-scantling design.