I need handwritten solution if u skip any part then i swear, it's simple do all otherwise skip Figure Ql shows the stress strain curves of six (6) samples.1280A7010608.50403020E10F0.10.20.3StrainFigure Q1(i) The modulus of elasticity of sample DStress (10° psi)4.6.2.Stress (MPa) (ii) The strain at fracture C(iii) The tensile strength of sample E(iv) The yield strength of sample F(v) The fracture strength of sample B

Answered: Image /qna-images/answer/d1178f8a-70b2-4448-b961-3c47d774ae7b.jpg

Figure Q1 shows the stress strain curves of six (6) samples. SIx 12 80 70 10 60 В * 50 40 D 30 20 E 10 F 0.1 0.2 0.3 Strain Figure QI (i) The modulus of elasticity of sample D Stress (10 psi) 2. 4. A Stress (MPa)

Figure Q1 shows the stress strain curves of six (6) samples. SIx 12 80 70 10 60 В * 50 40 D 30 20 E 10 F 0.1 0.2 0.3 Strain Figure QI (i) The modulus of elasticity of sample D Stress (10 psi) 2. 4. A Stress (MPa)

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

From the shear stress - shear strain diagram shown below, all the following are true except: Shear Stress (MPa) 500 450 400 350 300 250 200 150 100 50 0. 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 Shear Strain (rad) Shear Stress-Strain Curve for Brass Select one: O a. Shear stress and Shear strain are 430 MPa and 1.72 rad at fracture. O b. Shear stress and Shear strain are 250 MPa and 0.05 rad at Yielding. Oc. The elastic zone ends at a shear stress of 400 MPa and shear strain of 1 rad Od. The ultimate Shear stress and Shear strain are 430 MPa and 1.72 rad.
Table B2: Stress-strain data for uniaxial compression test on Sample Normal stress Uniaxial strain (%) Lateral strain (%) (MPa) 0.0 0.0000 0.0000 5.0 0.0319 -0.0150 10.0 0.0720 -0.0275 15.0 0.1025 -0.0425 20.0 0.1450 -0.0600 25.0 0.1755 -0.0725 30.0 0.2150 -0.0875 35.0 0.2455 -0.1025 41.0 0.2815 -0.1125 46.0 0.3125 -0.1209 52.0 0.3515 -0.1285 58.0 0.4000 -0.1325 (f) Based on the plotted curve, calculate the average/tangent Young's modulus and Poisson's ratio for this sample, at 50 % UCS?
a. Plot the stress strain diagram b. Find the modulus of elasticity, ultimate stress and fracture stress and explain each term. Define modulus of toughness and find the same for the above specimen. Load (kN) Elongation (mm) 11.1 31.9 0.0175 0.0600 37.8 0.1020 40.9 0.1650 43.6 53.4 62.3 64.5 0.2490 1.0160 3.0480 6.3500
The following data are taken from a 20mm in diameter bar of length 250mm, the following results were recorded. Assume that the curve of the stress-strain diagram is linear from the origin to the first point. P (load in kN) 8(elongation in mm) 112 0.2 154 0.3 167 0.4 0.5 174 181 0.8 Determine the a. Stress at 187 kN load in MPa b. Strain at 167 kN load in mm/mm (expressed in scientific notation) c. Modulus of elasticity in MPa d. Modulus of resilience in N-mm/mm e. Modulus of toughness in N-mm/mm 234 oo OOO O0
Problem :- The following data was obtained tensile test on a specimen of 6 mm diameter and gauge length Lo=30 mm: load(KN) 0 40 80 95 90 135 150.2 145 130 120 elongation(mm) 0 0.022 0.13 0.25 0.55 2.01 3.88 4.12 5.22 6 Plot engineering and True (stress-strain) curves, then determine the (oy, ɛy), (ou, ɛu), (of, , ef), ductility, modulus of elasticity. Then select on the drawing ((strain hardening (n)) , (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region)?
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.
. Data was collected from the tensile test demonstration for Aluminum 6061-T6. Below are the resulting stress-strain diagrams based on the tensile test we performed. o (MPa) Al 6061-T6 Stress-Strain Curve o (MPa) 350 320 300 250 200 150 100 50 0 Estimate 0 0.05 0.1 0.15 ε (mm/mm) 300 270 250 220 200 150 140 100 a) The modulus of Elasticity, E b) The yield stress, d, using the 0.2% offset method c) The strain at failure, & failure d) The modulus of resilience (approximately!) 50 0 0 Al6061-T6 Stress-Strain Curve, Magnified N 0.0032 0.002 0.004 0.006 0.008 € (mm/mm) The right diagram shows the magnified elastic region of stress-strain diagram to enhance the details.
Problem 2: A specimen of steel having an initial diameter of 0.503 in was tested in tension using a gauge length of 2 inches. The following data were obtained for the tension. Plot the stress strain data and determine the modulus of elasticity, and the yield strength of the material. Elastic State Load P Ibf Elongation in 1 000 0.0004 2 000 0.0006 3 000 0.0010 4 000 0.0013 7 000 0.0023 8 400 0.0028 8 800 0.0036 9 200 0.0089
Given the stress strain curves of 4 samples containing different melanin compositions. The initial length of the models is 15 cm and their initial diameter is 10 mm. A. What is the modulus of elasticity of each of the curves? B. What will be the length of each material just before it breaks?
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. Figure σ (ksi) 0 33.2 45.5 49.4 51.5 53.4 € (in./in.) 0 0.0006 0.0010 0.0014 0.0018 0.0022 1 of 1 Part B Determine the modulus of elasticity. E = ksi Submit Part C Determine the modulus of resilience. Ur= in-lb in³ Submit Request Answer Provide Feedback Request Answer
Q1- The engineering tensile stress-strain data were obtained for AA6061-T6 aluminum alloy shown: note stress unit is (MPa). 314 294 266 226 287 325 329 337 334 a.(stress) 0 52 159 E.(strain) 0 1.07 1.70 2.23 3.07 4.80 7.13 9.13 11.33 12.97 14.40 16.10 Plot the stress-strain curves then determine the mechanical properties as followings, ay, ou, elastic zone, plastic zone, Young modulus, fracture stress and ductility if you know L - 30 mm, and Lf - 33.7 mm.
The following data was obtained tensile test on a specimen of 6 mm diameter and gauge length Lo=30 mm: 142 124.7 5.1 5.84 Using the graphic paper supplied, Plot engineering stress-strain curves, then determine the 0| 38 160.4 155 4.06 3.55 76.2 load(KN) elongation(mm) 00.02 92.7 107 149 0.12 0.25 0.50 |2.03 (gy sy), (gu, eu), (of, sf), modulus of elasticity, then detected in the draw ((strain hardening (n)), (Necking), (uniform plastic region and non-uniform plastic region) and (elastic & plastic) region))?