---
title: "Reliability from stress-strength interference"
subject: "Machine Design II"
chapter: 4
key: "semester_7/machine_design_ii:stress-strength-interference-reliability"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/stress-strength-interference-reliability
site: BCT Engineering
---

# Reliability from stress-strength interference

Machine Design II · Chapter 4: Risk and Reliability of Design

- Asked in: year not given, 3+5 marks
- Video: [Reliability prediction using Stress Strength Interference (Analytical Method)](https://www.youtube.com/watch?v=WeTHHIF7tI4) — Institute of Quality and Reliability
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/stress-strength-interference-reliability
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-4

## Question

(a) Explain the stress-strength interference method of probabilistic design. (b) The load-induced stress in a link is normally distributed with mean 250 MPa and standard deviation 25 MPa. The strength of the material is normally distributed with mean 350 MPa and standard deviation 30 MPa. Find the reliability of the link and the probability of failure. What mean strength is needed for a reliability of 0.999 if the standard deviations are unchanged?

## Short answer

Stress-strength interference models stress and strength distributions; failure occurs where they overlap. For μ_S=350, σ_S=30, μ_s=250, σ_s=25 MPa, margin M has μ_M=100, σ_M=39.05 MPa, z=2.56, giving reliability 0.995 and failure probability 0.0052. For R=0.999, μ_S ≈ 371 MPa.

## Answer

#### (a) Stress-strength interference
In probabilistic design, stress $s$ and strength $S$ are random variables with distributions. Failure occurs when stress exceeds strength. The region where the two distribution curves overlap is the **interference**; its area is related to the probability of failure.

```text
 f
 |   stress     strength
 |    /\          /\
 |   /  \        /  \
 |  /    \__/\__/    \
 +-------------------------> MPa
       overlap = interference
```

Reliability $R = P(S > s)$. A larger gap between the means or smaller scatter gives higher reliability. The factor of safety based on means alone does not show this.

#### (b) Numerical
For normal variables the margin $M = S - s$ is also normal:

$$
\mu_M = 350 - 250 = 100\ \text{MPa}
$$

$$
\sigma_M = \sqrt{30^2 + 25^2} = \sqrt{1525} = 39.05\ \text{MPa}
$$

Standard normal variate (safety index):

$$
z = \frac{\mu_M}{\sigma_M} = \frac{100}{39.05} = 2.56
$$

From the normal table $\Phi(2.56) = 0.9948$, so

$$
R = 0.9948,\qquad P_f = 1 - R = 0.0052\ (\text{about } 1 \text{ in } 190)
$$

**Mean strength for $R = 0.999$:** $z = 3.09$.

$$
\mu_S - 250 = 3.09 \times 39.05 = 120.7 \;\Rightarrow\; \mu_S = 370.7\ \text{MPa}
$$

**Answer:** $R = 0.995$, $P_f = 0.0052$; required mean strength about 371 MPa.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In stress-strength interference, what does the overlap of the two distributions represent?
   A. The region where strength exceeds stress
   B. The region related to the probability of failure, where stress can exceed strength
   C. The factor of safety
   D. The mean margin
   Answer: B. Failure occurs when stress exceeds strength, and the overlapping area is linked to the failure probability. (Key point: Interference idea)

2. For the margin M = S - s with strength SD 30 MPa and stress SD 25 MPa, what is the SD of M?
   A. 5 MPa
   B. 55 MPa
   C. 27.5 MPa
   D. 39.05 MPa, from sqrt(30^2 + 25^2)
   Answer: D. Variances of independent normal variables add, so sigma_M = sqrt(900 + 625) = 39.05 MPa. (Key point: Standard deviation of margin)

3. What is the reliability for mean margin 100 MPa and sigma_M = 39.05 MPa?
   A. 0.995, from z = 100/39.05 = 2.56
   B. 0.50
   C. 0.84
   D. 0.0052
   Answer: A. z = 2.56 and Phi(2.56) = 0.9948, so R = 0.995. The failure probability is 0.0052. (Key point: Reliability from z)

4. For R = 0.999 (z = 3.09), what mean strength is needed with stress mean 250 MPa?
   A. 350 MPa
   B. about 371 MPa
   C. about 330 MPa
   D. about 400 MPa
   Answer: B. mu_S - 250 = 3.09 x 39.05 = 120.7 MPa, so mu_S = 370.7 MPa. (Key point: Strength for R = 0.999)

## Related in Machine Design II

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- [Helical spring under fluctuating load](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-fatigue-soderberg-design) — Machine Design II, chapter 6
- [Exponential reliability and system reliability](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Areliability-exponential-series-parallel) — Machine Design II, chapter 4
- [Gearbox shaft design with spur gear](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Agearbox-shaft-design-asme) — Machine Design II, chapter 5
- [Spring materials and yield strength estimation](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Aspring-materials-yield-strength) — Machine Design II, chapter 6
- [Design of a helical compression spring](https://www.bctengineer.com/question.md?key=semester_7%2Fmachine_design_ii%3Acompression-spring-design-static) — Machine Design II, chapter 6

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