Larson-Miller parameter =12. 10° (R-h)Rupture lifetime = 1E -4hStress = 20 MPALarson-Miller parameter =2.2. 103 (K-h)103 T(20 + log I.)(°R - h)2530354045501000800Stress = 6.9 MPA1000Stress = 1: 103 psi600100650°C400730°C200925°C815°C1008010010604010-410-310-210-11 1010210310410510Rupture lifetime (h)20103 T(20 + log I:)(K - h)Stress (MPa)Stress (10³ psi)Stress (MPa) (a) Using Animated Figure, compute the rupture lifetime for an S-590 alloy that is exposed to a tensile stress of 100 MPa at 925°C.(b) Determine the same value from the Larson-Miller plot of Animated Figure, which is for this same S-590 alloy.(a) ihours(b) ihours

(a) Using Animated Figure, compute the rupture lifetime for an S-590 alloy that is exposed to a tensile stress of 100 MPa at 925°C. (b) Determine the same value from the Larson-Miller plot of Animated Figure, which is for this same S-590 alloy.

(a) Using Animated Figure, compute the rupture lifetime for an S-590 alloy that is exposed to a tensile stress of 100 MPa at 925°C. (b) Determine the same value from the Larson-Miller plot of Animated Figure, which is for this same S-590 alloy.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: (a) A material can be made stronger by applying plastic deformation-explain with the help of strain…

A: Strain hardening mechanism : When a metal is stressed past its yield point, it is called strain…

Q: Example: the low cycle fatigue of a certain steel is given by life cycle equation-2: (of/E)=0.005…

A: The fatigue strength of steel /Modulus of elasticity(E), The fatigue Ductility coefficient is, The…

Q: 14) Indicate ine correct answer: a) Copper and its alloys do not fail due to cycling loading if the…

A: Aluminum and copper based alloys are fail even for a small amplitude stresses due to cyclic loading.…

Q: During a high cycle fatigue test, a metallic specimen is subjected to cyclic loading with a mean…

A: Given, To find, σmin/σmax

Q: A three-point bending test was performed on an aluminum oxide specimen having a circular cross…

A: test 1aluminum oxide specimen 1circular cross section of radius 5.0 mm specimen fractured at a load…

Q: An elasto-plastic material with strain hardening behavior has a stress-strain relationship composed…

A: Hi there!! since you have asked multiple questions, we will solve the first question for you. If you…

Q: For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results…

A: Given KIC=80MPamσ=510MPaa=6mm Consider the formula KIC=Yσπa80=Y×5103.14×6Y=0.036

Q: 8.30 Ifa component fabricated from an S-590 alloy (Figure 8.30) is to be exposed to a tensile stress…

A: The given tensile stress is 300 Mpa Temperature- 6500 C Estimate its rupture strength using the…

Q: A setup for single-edge-notch-bending (SENB) tests is shown below. Support rollers provide…

A: Given Data: Thickness of the specimen is t=10 mm. Height of the specimen is W=10 mm. Modulus of…

Q: A ceramic specimen having a circular cross section of radius 2.7mm is subjected to a load of 900N.…

Q: A 1.5 cm diameter rod pull test was carried out. The measurements of the test, whose initial length…

A: The true stress is calculated as: True stress=F/Afinal σ1=16240×4π×12.0282=143…

Q: a) As an engineer, choose a desired fracture mode in metals with one (1) reason. b) Sketch the…

A: "Since you have posted a question with multiple sub-parts, we will solve the first three subparts…

Q: 8.11 SizN4 has a surface energy equal to 30 J/m? and an atomic spacing ao = 0.2 nm. Calculate the…

Q: A spherical steel pressure vessel (diameter500 mm, thickness 10 mm) is coated with brittle…

A: Step 1: Stress= E.Ɛ P = 2.05 x 10^5 x 150 X 10-6= 30 GPa Hoop stress = pd/2t = 30x0.5/2 x0.01 =750…

Q: During a high cycle fatigue test, a metallic specimen is subjected to cyclic loading with a mean…

A: To find the value of R-ratio

Q: A thin-wall tube, made of a unidirectional graphite/epoxy lamina with a fiber direction 0 to its…

Q: At an axial load of 24 kN, a 50-mm-wide by 20-mm-thick polyimide polymer bar elongates 3.1 mm while…

A: Given, Axial load, P = 24 kN or 24000 N Cross section size = 50mm x 20mm Length = 210 mm

Q: Find out the working stress if a steel column carries a load of 1.6 MN with a factor of safety of 5.…

A: Given, Load = 1.6 MN Factor of safety FOS = 5 Ultimate stress = 472 MN/m2 Find, Working stress

Q: Determine the following: a. Young's modulus within the linear portion b. Tangent modulus at a stress…

A: a).Young's Modulus : At the proportional limit, stress is directly proportional to strain stress α…

Q: Determine the Basquin's law equation for the alloy. Use the following fatigue data, (about a…

Q: A chemicals company is looking to optimise the materials selection for their synthesis vessel with…

A: Yield before break is selected so it is related to tension

Q: Illustrate a typical stress-strain curve for brittle and ductile materials.

A: Stress strain curve is a characteristic property of material and does not depend upon shape, size of…

Q: An engineer is laying out a design for a ceramic component whose volume is 30.3 mm. In the processes…

A: Weibull modulus = 9.5Characteristic strength = 628Volume of reference sample =5.3 mm3If the maximum…

Q: 90 alloy component as shown in following figure that is subjected to a stress of 200 MPa (29,000…

A: Given data stress= 200 MPa rupture lifetime be 509 h

Q: A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel…

Q: a.) Plot the S-N curve for this alloy. b.) Does this material have an endurance limit? Explain. c.)…

A: As per company policy, we are supposed to answer only the first three subparts. Kindly repost to get…

Q: A structural component in the shape of a flat plate 20.8 mm thick is to be made from a metal with…

A: Given, Thickness of the flat plate = 20.8 mm Yield strength =σy=533 MPa Design stress=…

Q: 2. Using the two theoretical predictions of fracture strength, o = (equation 14.4) YE (equation…

Q: Example: the low cycle fatigue of a certain steel is given by life cycle equation-2: (of/E)-0.005…

A: Solution: Given Data: Fatigue strength coefficientSfModulus of ElasticityE=0.005Fatigue strngth…

Q: . Select the lightest L4 x 4 section that will safely support the service tensile loads Pp = 45 k…

Q: Given your understanding of what initiates and controls failure in materials, which of the following…

A: The given question asks which of the given following options increases the failure strength of a…

Q: At an axial load of 25 kN, a 50-mm-wide by 15-mm-thick polyimide polymer bar elongates 3.1 mm while…

A: Given that :-Required that :-(a) Modulus of elasticity (E) =?(b) Poisson ratio (v) =?(c) The change…

Q: A cylindrical metal specimen having an original diameter of 10.63 mm and gauge length of 50.1 mm is…

Q: Example: the low cycle fatigue of a certain steel is given by life cycle equation-2: (of/E)=0.005…

A: Given data:- The fatigue strength of steel σf/Modulus of elasticity(E), σfE=0.005 The fatigue…

Q: In unconfined compression test, the deformation dial gauge is used for stress calculation * O Ture O…

A: Introduction : The Unconfined Compression Test is a laboratory test that determines a rock…

Q: 14) Indi a) Copper and its alloys do not fail due to cycling loading if the applied stress amplitude…

A: Alloys of copper and aluminium fail under cyclic loading even for smaller stresses. Therefore the…

Q: During the tensile test a sample of metal having an initial diameter of 2.00 in., is reduced to a…

Q: The maximum measured Young’s modulus for a carbon nanotube is 950 GPa. How much stress would have…

A: Let’s consider an identical section for steel and carbon:

Q: A very large, steel plate of yield stress 200 MPa has a crack at the centre of length 18 mm,…

A: Solution:-The given values are as follows,Yield stress σys=200MPaγ=0.3Centre…

Q: A component made of Aluminium 6061-T651 has an edge crack with length equal to 15 mm, as shown The…

Q: A solid cylindrical shaft has a diameter of d = 20 mm and is made from steel with an ultimate…

A: Given data:- The given Solid cylindrical shaft has a diameter, d=20 mm The Ultimate Strength, Su=850…

Q: Brinnell hardness testing can be used for both Ferris and nonferrous metals. The test for ferrous…

A: Answer:- Brinell Hardness Test:- This test is most commonly used to test materials that have a…

Q: A Brinell hardness tester with a diameter of 10 mm and a load of 500 kg gives an impression with a…

A: Given data in question Diameter = 10 mm Load = 500 kg Indentation Diameter = 1.62 mm To find out…

Question

Larson-Miller parameter =
12. 10° (R-h)
Rupture lifetime = 1E -4h
Stress = 20 MPA
Larson-Miller parameter =
2.2. 103 (K-h)
103 T(20 + log I.)(°R - h)
25
30
35
40
45
50
1000
800
Stress = 6.9 MPA
1000
Stress = 1: 103 psi
600
100
650°C
400
730°C
200
925°C
815°C
100
80
100
10
60
40
10-4
10-3
10-2
10-1
1 10
102
103
104
105
10
Rupture lifetime (h)
20
103 T(20 + log I:)(K - h)
Stress (MPa)
Stress (10³ psi)
Stress (MPa)

(a) Using Animated Figure, compute the rupture lifetime for an S-590 alloy that is exposed to a tensile stress of 100 MPa at 925°C.
(b) Determine the same value from the Larson-Miller plot of Animated Figure, which is for this same S-590 alloy.
(a) i
hours
(b) i
hours

Expert Solution

This question has been solved!

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

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

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

Similar questions

6. Derive the resolved shear stress(RSS) equation for dislocation at an arbitrary plane at angle (e) with horizontal under nominal axial tensile stress(o) and plot the variation of Schmidt factor with
In a bending test for a brittle material, the distance between the supports is 300mm, the cross sectional area of the specimen is 1200 ??2and the thickness is 30mm. find the transverse rupture strength if the applied load was 1 ton
A copper specimen subjected to the Brinell Hardness Test using hardened steel ball indenter of diameter 13 mm and the indentation diameter 3.78 mm is measured using an optical magnifying lens with a ruler. Draw the Brinell Hardness Test setup neatly and determine the force applied on the specimen. Take Brinell Hardness Number for copper as 1037. Surface Area of Indentation (in mm2) = Applied Force (in N) =
A specimen of metal of original length 80mm and original diameter 21mm was subjected to tensile a test. The load and extension at the limit of proportionality were found to be 105,000N and 0.082mm. The yield point was at a load of 112,00ON. The maximum load before the material failed was 199,000N. The final length at failure was 91mm. Calculate: i. the stress at the limit of proportionality i. the yield stress iii. the ultimate tensile stress iv. the value of Young's Modulus V. the strain at failure
Nilo
Answer Q(i), (ii) & (iii) explaining detailly each step
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 450 mm (17.72 in.)? Assume a value of 0.22 for the strain-hardening exponent, n. i mm
Example: the low cycle fatigue of a certain steel is given by life cycle equation-2: (ay/E)=0.005 -0.07 b= -0.08 c= -0.7 a. What is the value of the transition fatigue life, in this case 2 N/when EE - Ep b. What is the total strain amplitude at the transition fatigue life?
A cylindrical specimen of a metal alloy 47.5 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 397 MPa causes the specimen to plastically elongate to a length of 54.3 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.5 mm to a length of 55.8 mm. i MPa
"Torsion Test on Mild Steel"What are the different failure modes of the specimens
Example 8.4. A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find (i) The stress at elastic limit. (ii) Young's modulus. (iii) Percentage elongation.
In a tensile test for an aluminum alloy, the sample is 2 inches long and 0.5 inches in diameter. The proportional portion of the tension stress-strain diagram for an aluminum alloy is shown below. It the diameter change of the sample was also monitored during the above test, and it was found that the lateral strain of the sample is 1/3 of its axial longitudinal strain, what is the Poisson' ratio of the material under test: ___. Calculate your answer to 2 decimal place.