An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 31 MPa√//m. It has been determined that fracture results at a stress of 227 MPa when the maximum (or critical) internal crack length is 2.47 mm. a) Determine the value of Yo√ra for this same component and alloy at a stress level of 340 MPa when the maximum internal crack length is 1.24 mm. i M. MPa √m b) Under these circumstances will the component fail? No Yes Unable to Determine

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 31 MPa√//m. It has been determined that fracture results at a stress of 227 MPa when the maximum (or critical) internal crack length is 2.47 mm. a) Determine the value of Yo√ra for this same component and alloy at a stress level of 340 MPa when the maximum internal crack length is 1.24 mm. i M. MPa √m b) Under these circumstances will the component fail? No Yes Unable to Determine

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Related questions

Q: A 250mm plate of a 4340 steel is tested for quality control purposes. It fails at a stress of 550MPa…

A: given; ⇒for maximum toughness of plate (K1)=?⇒plate fails at stress (σ)=550MPa⇒internal flaw…

Q: A hot rolled steel tension test sample has a diameter of 6 mm and the original gage lenght of 30 mm.…

A: Original diameter of the specimen, Do=6mm Original Length of the specimen, Lo=30mm Fracture…

Q: Please match different stresses to the stress value corresponding to the different points on the…

Q: Determine the tensile yield strength (0.2% offset) and the maximum strength of a metal alloy having…

A: Yield Strength : Yield strength is the maximum stress that can be applied before it begins to…

Q: A machine component is made of Aluminium alloy having an ultimate tensile stress Ou = 350MPa, a…

A: Given data:Ultimate tensile stress = Fracture toughness = Geometry factor = Cyclic tensile stress =…

Q: Q1/An Tensile Test applied on carbon steel sample, the results of this test represented by…

A: The area of the specimen is The formula for stress and strain is

Q: Q1. A typical graph of strain (e) versus time for a metal under creep conditions is shown below. င်…

A: DISCLAIMER FOR MULTIPLE Since you have asked multiple question, we will solve the first question for…

Q: Given: Metal Specimen diameter x-section (circular)= 0.55 in Gage Length= 2.030 in Fracture…

Q: I just can't find The percent elongation at fracture.

A: a)From the stress-strain curve,Under the proportionality limit the stress is directly proportional…

Q: A certain force F is applied at D. OABC bar is made of a brittle material with 210 MPa ultimate…

Q: hree cases of plane stress at yield or fracture are shown in the figures. Examine each case and…

A: Maximum Normal stress theory: This criterion of failure is accredited to British engineer (1850), W…

Q: A structural beam is built from steel with a yield strength of 345 MPa. The structure has a minimum…

A: Consider Y as the geometric parameter, σy as the yield strength, KIC as the fracture toughness and a…

Q: Problem 1 The fatigue data for a brass alloy are given as follows : Stress amp, 170 148 130 114 92…

A: (a,b) The above plot was made in MS Excel by following the steps outlined in the question. (c) The…

Q: 6. The following engineering stress-strain data were obtained for 0.2% C plain carbon steel. (a)…

Q: The figure below shows the tensile engineering stress-strain behavior for a steel alloy. (a) What is…

A: Given -Engineering stress strain curve for steel alloyTo find -Modulus of elasticity of steel…

Concept explainers

Thermal Properties of Materials

In mechanical engineering, the thermal property is represented as the physical quantity that is relevant to the application of heat. The thermal properties are evaluated by variations in any specific material due to low heat or high heat. The quantities that influence the thermal property of the materials are mass, density, temperature, and many others.

Question

### Aircraft Component Fracture Analysis

An aircraft component is fabricated from an aluminum alloy with a plane strain fracture toughness of 31 MPa√m. Fracture occurs at a stress of 227 MPa when the maximum (critical) internal crack length is 2.47 mm.

#### Problem Statement

**a) Calculation:**
Determine the value of \( Y \sigma \sqrt{\pi a} \) for the same component and alloy under a stress level of 340 MPa when the maximum internal crack length is 1.24 mm.

*Input Box:*
- Enter the calculated value in MPa√m.

**b) Failure Assessment:**
Evaluate whether the component will fail under these conditions.

*Options:*
- No (selected)
- Yes
- Unable to Determine

This exercise involves understanding the critical stress intensity factor and applying fracture mechanics concepts to assess material failure under specified conditions.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Types of Properties of Engineering Materials

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

640 480 160 0.002 0,004 0.006 0.008 0.010 Strain (mm/mm) A tensile test specimen of Aluminum alloy having a gauge length of 280 mm was tested to fracture. Stress and strain data obtained during the test are shown in the attached Figure. **Answers are approximate - do your best to read the graph. Determine: (a) the modulus of elasticity in GPa (4sigs). Enter this value into D2L answer box. (b) the proportional limit in MPa (3sigs). (c) the ultimate strength in MPa (3sigs). (d) Ductility of material based on percent elongation. (3sigs) Your Answer: Stress (MPa)
The answer is one of the options below please solve carefully and circle the correct option Please write clear .
1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of Resilience
Determine the percentage of ductility of a metal alloy having the following tensile stress-strain diagram. 600 500E 400 500 300 400E 300 200 200 100 100 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain EN Stress (MPa) Stress (MPa)
Review Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. No elements selected a(ksi) 50 + 40 - 30 Figure 1 of 1 20 10 + o (ksi) € (in./in.) ex 10(in./in.) 0.5 1.0 1.5 2.0 2.5 33.2 0.0006 45.5 0.0010 Press (ENTER) to select this element. Press ESC) to return to the main menu. Press CTRL+Q) to quit the application. 49.4 0.0014 51.5 0.0018 53.4 0.0022
1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020
5. The following data were collected from a standard 0.505-in.-diameter test specimen of a copper alloy (initial length lo= 2.0 in.). After fracture, the total length was 3.014 in. and the diameter was 0.374 in. Load (Ib) Al (in.) 00000 3,000 6,000 0.00167 0.00333 7,500 0.00417 9,000 10,500 0.0090 0.040 12,000 0.26 12,400 11,400 0.50 (maximum load) 1.02 (fracture) a) Plot the data as engineering stress versus engineering strain. b) Compute the modulus of elasticity. c) Determine the yield strength at a strain offset of 0.002. d) Determine the tensile strength of this alloy. e) What is the approximate ductility, in percent elongation? f) Compute the modulus of resilience. g) Compute from the data and plot true stress versus true strain diagram.
Figure 1 shows the tensile testing results for different materials. All specimens have an initial diameter of 12 mm and an initial gauge length of 50 mm. 300 250 Low carbon steel Network polymer 200 Crystalline polymer 150 Amorphous polymer 100 50 5 10 15 20 25 30 Strain (%) Figure 1: Stress-strain curve b. Determine the following parameters for each material: • the tensile strength the 0.2% offset yield strength the modulus of elasticity • the ductility Stress (MPa) LO
The following data was obtained as a result of tensile testing of a standard 0.505 inch diameter test specimen of magnesium. After fracture, the gage length is 2.245 inch and the diameter is 0.466 inch. a). Calculate the engineering stress and strain values to fill in the blank boxes and plot the data. Load(lb) Gage Length (in) Stress (kpsi) Strain 0 2 1000 2.00154 2000 2.00308 3000 2.00462 4000 2.00615 5000 2.00769 5500 2.014 6000 2.05 6200 (max) 2.13 6000 (fracture) 2.255 b). Calculate the modulus of elasticity c). If another identical sample of the same material is pulled only to 6000 pounds and is unloaded from there, determine the gage length of the sample after unloading.
Please explain each step briefly.
Review Data taken from a stress-strain test for a ceramic are No elements selected o(ksi) given in the table. The curve is linear between the origin and the first point. 50 40 30 Figure 1 of 1 20 10 σ (ksi) e (in./in.) ex 10-*(in./in.) 0.5 1.0 1.5 2.0 25 33.2 0.0006 45.5 0.0010 Press (ENTER to select this element. Press (ESc to return to the main menu. Press (CTRL+Q) to quit the application. 49.4 0.0014 51.5 0.0018 53.4 0.0022
ASAP