Problem 4.24 30 Components subjected to large contact forces (e.g., brake disks) are often made of high- grade steel coated with a thin film of a very hard material, such as diamond. A common test to assess the mechanical properties (hardness, elastic moduli, etc.) of the coating is the nanoindentation test, which consists of making a controlled dent in the film using a nail-like object, called an indentor. During the test, the force applied to the indentor and the indentation depth are measured. The graph shows the curves interpolating the loading and unloading data for a diamond film on steel. Since the unloading curve does not go back to the origin of the plot, the film is permanently deformed during the indentation process. Determine the energy lost to permanent deformation if the indentation force is given by Unloading 25 20 Loading 15 10 20 40 60 80 100 120 during loading. 300-145x+ 18x² during unloading. where F is expressed in uN and the indentation depth x is given in nm. Indentation Depth (nm) Fi = Figure P4.24 Indentation Force (10-4 N)

Problem 4.24 30 Components subjected to large contact forces (e.g., brake disks) are often made of high- grade steel coated with a thin film of a very hard material, such as diamond. A common test to assess the mechanical properties (hardness, elastic moduli, etc.) of the coating is the nanoindentation test, which consists of making a controlled dent in the film using a nail-like object, called an indentor. During the test, the force applied to the indentor and the indentation depth are measured. The graph shows the curves interpolating the loading and unloading data for a diamond film on steel. Since the unloading curve does not go back to the origin of the plot, the film is permanently deformed during the indentation process. Determine the energy lost to permanent deformation if the indentation force is given by Unloading 25 20 Loading 15 10 20 40 60 80 100 120 during loading. 300-145x+ 18x² during unloading. where F is expressed in uN and the indentation depth x is given in nm. Indentation Depth (nm) Fi = Figure P4.24 Indentation Force (10-4 N)

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

Similar questions

QUESTION 15 A bimetallic bar is made of two equal length steel and brass bars of same width firmly bonded together. The cross-sectional area of the brass bar is twice that of the steel bar. The Young's modulus of steel is twice that of brass. The coefficient of linear expansion of brass is twice that of steel. When the compound bar is subjected to a temperature rise of 50°C, what is the ratio of stress generated in steel to stress generated in brass? Note: Please study lecture slides and tutorial problem on thermal stress - strain O a. 1:(-2) O b. 1 : (-1) C. 2: (-1) O d.4:(-1) QUESTION 16
Determine the indentation diameter and the surface area of indentation of a copper material subjected to a Brinell hardness test with a test force of 8.2 kN using a hardened steel ball indentor of 9 mm. Take the Brinell Hardness Number as 902. Solution: i) Indentation Diameter (in mm) = ii) Surface Area of Indentation (in mm^2) =
A tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 65 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 36.1 mm - Final Diameter after fracture = 4.25 mm & - Ultimate stress = 401 MPa SOLUTION: Initial Diameter (in mm) = Ultimate Load (in N) =
The tensile modulus of a 0°, 90°, and 45° graphite/epoxy ply is measured as follows to give E₁ = 26.25 Msi, E2 = 1.494 Msi, Ex = 2.427 Msi for the 45° ply, respectively. 1. What is the value Ex for a 30° ply? 2. Can you calculate the values of v12 and G12 from the previously measured three values of elastic moduli?
Q1. (i) Initial strain Please enter your answers here Q1. (i) Your Answers: Please enter your supplied question data Q1. (ii) E0 B (hr1¹Pa-¹) n Fracture 0.003 4.60E-34 Mechanics of Materials 2 Tutorial 6 Q1. (ii) E General guidance on the Tutorial Portfolio assignment was provided at the beginning of Tutorial Exercise 1 and is available separately on-line on the module website. B 3.2 A Q2. σ (Pa) σo (Pa) 780,000,000 This tutorial exercise assesses fundamental understanding of the creep and stress relaxation response of metals subjected to either constant load or constant strain. It is designed to reinforce learning of principles set out in your current stage 2 lecture notes in "Creep Failure and Stress Relaxation" and to build on knowledge previously presented in the stage 1 module "Solids & Structures". C Q3. t (hr) For further information you may refer to Chapter 21 (Creep and Viscoelasticity) of Benham, Crawford, and Armstrong (1996), 2nd Ed. Q1. A typical graph of strain (e) versus…
A 2-cm square cross-section steel part having a Brinell hardness of 200 and a length of 5 cm is to be ground by adhesive wear using a tool of electroplated chromium with a square head of 2 mm x 2 mm. If the normal force applied is 60 kN and the depth to be removed from the surface is 4 mm, determine the number of passes needed (parallel to the length direction) for poor lubrication. Select one: O a. N = 785 O b. N = 897 O C.N = 1046 O d. N = 1256
Sintered silicon nitride parts are expected to experience tensile stresses 50% of the ultimate tensile strength of the material. What will be the maximum allowable size of internal flaws for these parts not to fail in service? Given the following properties of the material: Specific surface energy: 0.3 J/m2 its Youngs Modulus: 304 GPa Ultimate tensile Strength: 570 MPa Select one: a. 1.43 µm b. 0.71 µm c. 15 µm d. 85 µm e. 9.5 µm
Q1/A structural part is 1-meter-long and subjected to a 50 KN load in which this part must be deformed elastically without experiencing any permanent deformation. If you know that part is made of steel, brass, aluminum, and Titanium alloys and the yield strengths and densities of these alloys are: 860 MPa, 7.9 g/cm³; 415 MPa, 8.5g/cm³; 310 MPa, 2.7 g/cm³; and 550 MPa, 4.5 g/cm³ respectively. Based on these criteria, rank the alloys from the heaviest to the lightest in weight.
1. Ductile materials are tougher than brittle ones II. Tensile tests are not used when the material is brittle III. Brittle materials reach much higher ultimate stresses in tension than in compression Which statement(s) given above is/are true?
Question 2 In designing prosthetic sockets, the latter will need to be experimentally tested for their structural integrity. Figure 2 shows one such design of a prosthetic socket which is made of carbon fibre composite. Strain gauges are installed to record the strains at various locations of the legs during walking and the readings are recorded using a telemetry system to detemine the critical stressed area. At a particular strain gauge location indicated in Figure 2, the readings recorded by one of the 45° strain gauge rosettes are: Ea = 2500 x 10*, es = 1500 x 10°, & = -950 x 10* Using Mohr's Cicle or otherwise, detemine: (a) the principal strains and the direction of the maximum principal strain relative to the gauge "a". (b) the corresponding principal stresses and sketch the results on a properly oriented element. You may assume that the prosthetic socket is made of polypropylene whose Young's modulus of 1.0 GPa and Poisson ratio of 0.3. Figure 2
Please help. Do not just guess. Prove your answer. I will give a good feedback. thank you :)
The following data were recorded during the tensile test of a test specimen with a diameter of 12.8 mm. The gage length is 50.8 mm. The given data are as follow; Given: Test specimen material: Aluminum Diameter of test specimen: Do = 12.8 mm Length of test specimen: Lo = 50.8 mm Force and Elongation data for test specimen Assumptions: 1. The given data is accurate and the material is isotropic 2. The direction of applied force is parallel to the length of the cylinder Requirement: To interpret and plot the stress Vs Strain curve for a given specimen DATA: LENGTH, mm STRESS, MPa LOAD, N 0 50.8 7 330 50.851 15 100 50.902 23 100 50.952 30 400 51.003 34 400 51.054 38 400 51.308 41 300 51.816 44 800 52.832 46 200 53.848 47 300 54.864 47 500 55.88 36 100 56.896 44 800 57.658 42 600 58.42 36 400 59.182 STRAIN