The following graph shows a load-extension graph from a test on a polymer specimen 15mm in diameter and 45mm long. The rightmost datapoint represents the reading at the point of fracture. Load (KN) 20- 15 5- 0 0.4 Extension (mm) | | | | | | | | | | | Determine the following physical properties. Show all your working and express all your answers in appropriate Sl units where appropriate: 0.2 0.6 a) 0.2% offset yield strength b) The tensile strength c) The modulus of elasticity d) The % elongation 0.8 e) The engineering stress at fracture f) Estimate the modulus of resilience 1.0

The following graph shows a load-extension graph from a test on a polymer specimen 15mm in diameter and 45mm long. The rightmost datapoint represents the reading at the point of fracture. Load (KN) 20- 15 5- 0 0.4 Extension (mm) | | | | | | | | | | | Determine the following physical properties. Show all your working and express all your answers in appropriate Sl units where appropriate: 0.2 0.6 a) 0.2% offset yield strength b) The tensile strength c) The modulus of elasticity d) The % elongation 0.8 e) The engineering stress at fracture f) Estimate the modulus of resilience 1.0

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

The slope of the curve increases The slope of the curve decreases Vertical axis (y) 10 None of these 8 The slope of the curve will not change (5.0,5.0) 5.0 Horizontal axis (x) elasticity of curve: 0.25 This question is about the slope of a curve. The interactive above may be helpful in answering the following question. For a curve with an elasticity of 2.1, and a position of 2, as the x of point increases..... position of curve: 5 x of point: 5 Slope = 0.3
Name: Surname: Student ID: ME 216 MECHANICS OF MATERIALS QUIZ #2 6) QUESTION σ (MPa) 25.03.2019 490 420 350 280 210 140 70 (mm/mm) 0 0.002 0.004 0.006 0.008 0.010 A tension test was performed on a 7075-T6 aluminum tensile test specimen. The results are given in the figure above. According to this data, find: a. Young modulus of the material, b. Yield stress of the material (be careful, there is no any sharp yield point), c. Modulus of the resilience, d. If the material is stressed up to 420 MPa stress, what is the permanent elongation remained in the material if its length is 90 mm. e. For a 20 mm diameter bar made up of from the same material, what is the maximum load that can be applied without plastic deformation. ANSWER E=A ٤ E = 70.106 σ = 70 MPa at ε=0,0009 = 77,776 Pa 0.0009
Stress vs. Strain 517 465 414 362 310 155 103 51.7 5.00 10.0 15.0 20.0 30.0 35.0 40.0 45.0 50.0 Strain (%) Stress (MPa)
b) A sedan car weights 2100 kg with 45:55 distributions front to rear. The car has wheelbase of 1.9 m. With aid of following table showing the tyre cornering stiffness values for different loads: Load Cornering Stiffness (kg/deg) 300 131 400 149 500 174 600 203 i) Calculate the cornering coefficient for the front and rear tyres. ii) Calculate the understeering gradient. iii) Determine whether it is understeer or over steer. iv) Calculate critical or characteristic speed.
Hi please help with this question thank you I just can't seem to work it out even with the given formulas. Thanks! also, I have tried these but they seem to be incorrect, or maybe parts of the calculation may be incorrect since when I submit these all together and I can't really work out where I am going wrong: sigma x = 25.43 sigma y = 13.07 Txy = 49.124
The stress-strain diagram of a reinforcement steel having a cross-sectional diameter of 12 mmdiameter and 100 mm gage length is determined after its tensile strength test as follows. Based on the stressstrain diagram determine the followings properties of the material (Poisson’s ratio of the material is 0.32) : a) Modulus of elasticityb) Yield strengthc) Toughnessd) Resiliencee) Shear modulusf) Bulk modulusg) Ductility as described bypercent change in length
The following are the given information of the test specimen material: Specimen Material Specimen Width (mm) Specimen Thickness (mm) Specimen Initial Length (mm) Specimen Final Length (mm) aluminium 1.5 13.5 42.4 47.5 Grip Separation 50 mm Total Elongation of the specimen is given by Elongation = i-io/io x 100 l : Final length, in m l0 : Initial length, in m
Problem 1: (50 points) In Avengers 2, Captain America's improved shield is made from unobtainium, a new material that will soon be available in a store near you. Unobtainium has the normal stress- strain diagram shown. The proportional limit, the elastic limit and the yield point are identical in this material. o [MPa] 2001 175 150 125 100 Fig.1 Normal stress-strain 75 50 25 0 0. 0.05 0.1 0.15 0.2 0.25 Unobtainium has a Poisson's ratio of 0.3. a. [6pts] Identify the yield point oy, the ultimate stress ou and the fracture stress of. Include units. c. [25pts] A bar of unobtainium has a length of 1.5 m, a width of 100mm and a height of 50 mm, as shown. The cross-sectional area is 50mm 100mm = 5 x 10-³m². The bar is subjected to an axial force of 375 kN. Find the normal stress in a cross-section of the bar. b. [5pts] How do you find Young's modulus from this graph? Find its value with units. whesta fins dus al hoss P = 375 KN 1.5 m 0.3 50 mm 100 mm & [mm/mm] Is the material within its…
STATICS AND STENGITH OF MATERIALS Question: Load (kip) Elongation (in.) 0 0 A tension test was performed on a steel specimen having an original diameter of 0.503 in. and gage length of 2.00 in. The data is listed in the table beside. 1.50 0.0005 4.60 0.0015 8.00 0.0025 11.00 0.0035 11.80 0.0050 1. Plot the stress-strain diagram (better to use a software like MS Excel) and determine approximately 11.80 0.0080 12.00 0.0200 16.60 0.0400 a. the modulus of elasticity, 20.00 0.1000 21.50 0.2800 b. the yield stress, 19.50 0.4000 18.50 0.4600 c. the ultimate stress, d. the fracture stress. Use a scale of 1 in. = 20 ksi and 1 in. = 0.05 in.>in. 2. Redraw the elastic region (on the same graph), using the same stress scale but a strain scale of 1 in. = 0.001 in.>in.
The fatigue data for a ductile cast iron is given in the table below. Stress Amplitude [MPa] 248 236 224 213 201 193 193 193 Cycles to Failure 1 × 10³ 3 x 105 1 x 106 3 x 10° 1 x 107 a) Create a S-N curve from the data. b) What is the fatigue limit for this alloy? c) Determine the fatigue life at 230 Mpa and 175Mpa d) Determine the fatigue strength at 2x105 and 6x106 cycles 3 x 107 1 × 108 3 x 108
Question: a student has chosen the particular weight of 1.373 kg for gear and optimised is by getting the following plot. Taking into consideration these points. A) comment on the plot? B) why high contact stiffness ratio is important and high mesh as well? Contact stiffness 23.0 21.0 19.0 17.0 15.0 23 57 $2189 Normal module [mm] 4.600 82 79 72 2.862 67 61 62 42931 2325 53120 30 32 2627 54 18 21 ql0991 63 1.125 13.0 1.000 1.100 1.200 1.300 Total contact ratio 1.400 1.500 1.600 1.700
When an object is placed underwater, the pressure on the object is subjected to uniform stress on all sides. a) define volume strain, the bulk modulus, compressibility, and the relationship between the bulk modulus and compressibility. b) A 100 litre volume of liquid is subjected to a 2000 psi increase. If the compressibility of the liquid is 5 x 10-5 per atm, find the bulk strain and absolute decrease in the volume of the liquid c) explain the concepts of shear stress and shear strain? could these concepts be applied to the fluid problem above?