ANSWER TO QUESTION 5.3.1 5.3.2) Ultimate tensile stress. ANSWER TO QUESTION 5.3.2
Q: What is torsional stress?
A: When the twisting load is applied over the body or a shaft, then the shear stress generated is…
Q: The nature of the stresses
A: Given; A helical gear with pressure angle,α= 200 and β=300 helix angleAt the mid span of a shaft is…
Q: Why do the Ductile materials fail in shear?
A: Ductile materials are the materials which goes plastic deformation before the fracture. When any…
Q: Why is it important to find the absolute maximum shear stress in the material when it is subjected…
A: Shear stress: It is the force imposed per unit area in a body that cause deformation of a material…
Q: Part A Suppose that o, = 170 kPa and o, = 610 kPa. Determine the equivalent state of stress on an…
A:
Q: Do ductile materials fail in shear?
A: The type of materials that undergo considerable plastic deformation before fracture are known as…
Q: 1. What is the difference between direct and shear stress?
A: We are suppose to solve only one question. Please post other question as a separate question.
Q: What is the Allowable Stress Design?
A: In Allowable stress design, the maximum stress (σmax) that a member will endure under a given…
Q: Uniform and non uniform stress, gives proper solution explanation with diagrams.
A: Given data Uniform and non uniform stress, gives proper solution explanation with diagrams.
Q: Part 2 As the area under the stress-strain curve decreases, the toughness___________. stays the…
A: Toughness: The capacity to absorb the energy of the material before the failure is known as…
Q: The ability of a material to undergo large permanent deformation with the application of a tensile…
A: Elasticity: It is the ability of the material which allows to return to its original shape and size…
Q: What is the dynamic stress?
A: Stress: Stress is defined as the developed force in the specimen per unit area when that specimen is…
Q: In the case of plane stress, how can the plane-strain analysis be used?
A: Plane stress arises in case of thin components. Plane strain arises in case of very thick…
Q: Simple stress and strain pleasee help
A: Given Diameter = 150 mm Gauge length = 300 mm Scale for graph 25 mm = 50 mm 25 mm = 0.2 X 10-3 mm/mm…
Q: 3. Explain the concept and the relationship between state of stress at a point and transformation of…
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Q: Describe the Maximum Normal Stress Theory?
A: The Maximum Stress Theory is one of the failure theories postulated by Rankine. Hence this theory is…
Q: What is the tensile stress?
A: When there is loading over a member in the direction of its axis, then stress produced under such…
Q: What is yield stress?
A: Yield Stress: It is the amount of stress at the yield point of a material/component. The yield…
Q: In a tensile test on a ductile material Hook’s law is applicable up to _________ point Select one:…
A: Solution:
Q: Besides normal tensile stress, list three other types of material stresses? Are these typically all…
A: Stress: Stress is defined as the internal resistance per unit area of cross-section. There are…
Q: If we want to measure the stress of a flat non-uniform component, what should we pay attention to…
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Q: explain the state of plane stress and plane strain.
A: Given: stress and strain system To write: Plane stress and strain system
Q: - Differentiate between engineering stress and true stress.
A: Stress: It is the ratio of the load applied on the body per unit area
Q: What are the three equations that represent the general form of Hooke’s law for a Triaxial state of…
A: Just like 1D or 2D, Hooke's Law can also be applied to material undergoing three-dimensional stress…
Q: The ultimate tensile stress of mild steel compared toultimate compressive stress is O more O less O…
A: Concept: 1)Mild steel is a ductile material. 2) For ductile materials, ultimate strength in tensile…
Q: What is true stress and true strain?
A: To define: True stress True strain Stress: It is defined as the internal resistance developed at a…
Q: Briefly explain the state of plane stress and plane strain.
A: Plane stress is a two-dimensional condition of stress in which the entire amount of tension is…
Q: Why do we define engineering and true stresses for tension/compression loading but not for shear…
A: Engineering stress: It is the stress based on the initial or original cross sectional area (Ao)…
Q: differentiate allowable stress , ultimate stress and actual stress
A: To find : differentiate allowable stress , ultimate stress and actual stress
Q: 2) a) Draw a stress-strain curve for a low carbon steel and an aluminum after tensile test. Show…
A: The stress strain curve of low carbon steel and aluminium can be drawn as below, Here, x-axis and…
Q: With a suitable example explain the difference between True stress-strain and Engineering…
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Q: Q/ Does when the temperature increase, does ductility and deformation increase or decrease?
A: The impact behavior of plastic materials is strongly dependent upon the temperature. At high…
Q: Part A Suppose that o, = 50 kPa and o, = 740 kPa. Determine the equivalent state of stress on an…
A: Given Data⇒σx=50 KPaσy=740 Kpaζxy=400 Kpa
Q: A typical value for the yield stress of mild steel would be? Select one: O a. 200 GPa O b. 100 MPa O…
A: Option c is correct.
Q: When does the magnitude of stress increase?
A: When a body undergoes deformation under normal or shear forces/loads, then it undergoes…
Q: Between 1. Nominal (engineering) stress and true stress and 2. True strain and engineering strain…
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Q: Q5) material given the data from a tension test ? How would you determine the strain- hardening…
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Q: What is compressive principal stress?
A: In the three principal stresses , it is mentioned in the diagram as σ1, σ2 and σ3. σ1 is maximum…
Q: 6) Chapter 3: Problem 3.7: (USCS Units) The flow curve for a certain metal has a strain-hardening…
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Q: Define Tensile strength.
A: Given Data⇒ Tensile strength Problem statement⇒ Problem says Define Tensile strength.
Q: Write a short note on Strength, Malleability and Tensile Strain
A: STRENGTH: The strength of the material is it's bility to withstand an applied load without failure…
Q: What are the tensile forces?
A: Tensile force is the force applied on a material which stretches the material. The tensile force…
Q: a component is made of 6061 T6 aluminum sheet LT. A axial tensile stress of 50 ksi is applied at…
A: The deformation of a material in response to a tensile, compressive, or torsional force is…
Q: What is Multiaxial Stress?
A: Multiaxial stress states due to the action of various external loads exist at the considered point…
Q: Describe the relationship between stress and strain according to Hook's law?
A: The relationship between stress and strain according to Hook's law
Q: Write the definitions for engineering stress, true stress,engineering strain, and true strain for…
A: Engineering stress is defined as the force ( applied load ) per unit original cross section of the…
Q: Describe the behavior of tested materials (mild steel and aluminium) as it responds to increasing…
A: following is the answer to the above problem-
Q: What is Normal Stress?
A: Normal Stress: When the stress is acting in the direction perpendicular to the body then it is said…
Q: Q15- A material with a yield stress of 70 MPa is subjected to three principal (normal) stresses of…
A:
Q: In stress and strain diagram for a perfectly ductile material, the yield strength, and ultimate…
A: In the above statement, it mentions about stress strain curve of perfect ductile material have well…
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- Consider the tensile stress-strain diagram below. If samples 1 and 2 are identical except that they were tested at different temperatures, which sample was tested at a lower temperature? Note: This question will be marked negatively so consider your answer carefully. A. Sample 1 B. Both were tested at the same temperature/ Alby is by die dieselfde temperatuur getoets C. Sample 2Force P and length change AL data are given in table below for the initial portion of a tension test on 7075-T651 Al alloy. The diameter before testing was 9.07 mm., and the gage length Linitial for t length change measurement was 50.8 mm. Determine the yield strength with offset limit, and %total elongation. P, kN AL mm 7.22 14.34 0.0839 0.1636 21.06 0.241 26.8 0.308 31.7 0.380 34.1 0.484 35.0 0.614 36.0 0.924 36.5 1.279 36.9 1.622 37.2 1.994Stress Strain Diagram The Data shown in the table have been obtained from a tensile test conducted on a high-strength steel. The test specimen had a diameter of 0.505 inch and a gage length of 2.00 inch. Using software. plot the Stress-Strain Diagram for this steel and determine its: A= TTdT(050s A %3D 1. Proportional Limit, 2. Modulus of Elasticity, 3. Yield Strength (SY) at 0.2% Offset, 4. Ultimate Strength (Su), 5. Percent Elongation in 2.00 inch, 6. Percent Reduction in Area, 7. Present the results (for Steps 1-6) in a highly organized table. e Altac ie sheet (as problelle 4 A = 0.2.002 BEOINNING of the effort Elongation (in) Elongation (In) Load Load #: #3 (Ib) (Ib) 1 0.0170 15 12,300 0.0004 1,500 16 12,200 0.0200 0.0010 3. 3,100 17 12,000 0.0275 0.0016 4,700 18 13,000 0.0335 5. 6,300 0.0022 19 15,000 0.0400 0.0026 6. 8,000 20 16,200 0.055 0.0032 9,500 21 17,500 0.0680 0.0035 8. 11,000 22 18,800 0.1080 0.0041 11,800 23 19,600 0.1515 0.0051 24 20,100 0.2010 10 12,300 0.0071 25…
- The table below shows the deformation data that resulted from applying a pure tensile load to a brass alloy rod with an initial length of 30 mm and a diameter of 10 mm. This rod was subjected to necking and beyond necking deformation. The data at fracture is shown in the last row. Applied Load (N) Length (mm) Diameter (mm) 75,100 33.8 68, 250 34.9 50,200 35.7 6.7 Solve for the TRUE STRESS/ES and TRUE STRAINS. 8.3 7.8Part B The stress-strain diagram for a steel alloy having an original diameter of 1.0 in. and a gage length of 7 in. is shown in the figure below. (Eigure 1) Determine the load on the specimen that causes yielding. Express your answer to three significant figures and include appropriate units. HA Py- Value Units Submit Request Answer Part C Figure t oft> Determine the ultimate load the specimen will support. Express your answer to three significant figures and include appropriate units. NO 70 60 50 P. - Value Units 40 30 20 Submit Bequest Answer 10 (in/in) o G04 0m 12 016 020 024 02 a aa omams Provide FeedbackA tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the ultimate strength. Stress (ksi) 60 50- 40- 30- 20 10- 0 0 0 Upper scale O 66 ksi 28 ksi O 72 ksi O 53 ksi O 62 ksi Lower scale. 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (in./in.)
- A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the proportional limit. Stress (ksi) 40 30- 20 10- 0- 0.0 0.0 Upper scale 0.020 0.002 22 ksi O 15 ksi E 19 ksi O 10 ksi 13 ksi Lower scale- 0.040 0.060 0.080 0.100 0.004 0.006 0.008 0.010 Strain (in./in.) 0.120 0.0121. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 -0.0019 13.092 0.0762 -0.0057 21.819 0.1270 -0.0095 30.547 0.1778 32.729 0.7620 34.911 3.0480 30.01 mm Ø F Rigid Plate Cylindrical Alloy - Steel casement Figure 1: Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.ASAP
- A tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 66 % - Percentage of Reduction in Area = 38 % - Final length after fracture = 34.6 mm - Final Diameter after fracture = 4.43 mm & - Ultimate stress = 364 MPa Determine the Initial length, Initial diameter and Maximum load. ii) Final Area (in mm2) = iii) Initial Area (in mm2) =The following data were obtained from the tensile test of Aluminum alloy. The initial diameter of testspecimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.