1. Explain the storage and loss modulus of viscoelastic materials in your own words.2. Show that phase lag is equal to when considering purely viscous materials.Hint:Use Equations 6.1 and 6.2 provided in the introduction along with the strain rate question =(n) is the viscosity and represents the measurement of resistance to deformation with respect to time).

“Since you have posted multiple questions, we will provide the solution only to the first question…

Answered: 1. Explain the storage and loss modulus…

Engineering

Mechanical Engineering

1. Explain the storage and loss modulus of viscoelastic materials in your own words.

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

1) Draw (using a normal graph paper) a conventional stress-strain diagram for ANY metallic material (e.g. steel, aluminium, copper, brass, iron, tungsten). The diagram should be as accurate as possible using a suitable scale (e.g. 1cm: 10 N). 2) Calculate the Modulus of Elasticity, Modulus of Toughness and Modulus of Resilience for the material from the stress-strain diagram. Show your calculations in detail on a separate A4 piece of paper.
1. For the stress-strain curve shown below, please estimate the properties indicated. (a) Fracture Strain Please do your work on a separate sheet of paper, and put your answers in the boxes on the right. Be sure to include the proper symbol and units. Stress Strain 70 60 50 Stress (ksi) 240 30 20 10 70 0 0.000 60 50 Stress (ksi) 40 20 10 KULL 0 0.000 0.010 0.050 0.100 Strain (in/in) Stress Strain 0.020 0.030 Strain (in/in) 0.040 0.150 0.050 (b) Ultimate Tensile Stress (c) Fracture Stress (d) Proportional Limit (e) Elastic Modulus (1) Yield Stress (g) Tensile Toughness (Modulus of Toughness) (h) Modulus of Resilience
A given model is used to study the deformation of a polymer at E2 = 100 MPa, n1*= 500 MPa.s and n2*= 400 MPa.s Please plot the stress and relaxation versus time response when: a) a strain of 1.5 applied to the model for 100 seconds b) the stress behavior for the next 100 seconds after the strain is removed.
A 11 in. inner diameter, 0.35 " wall thickness pipe is under a pressure of 2.5 ksi where strain gages installed along axial and circumferential directions register strains of 180 and 900 micro-strains (x10^-6), respectively. A) What is the Poisson's Ratio of this material and it's Elastic Modulus? B) In a uniaxial test, the pipe's material is observed to yield at a longitudinal strain of 0.1 in/in. Assuming a factor of safety of 2, the pipe can withstand impact energy of _______ lb - in per foot without suffering permanent deformation. C) If the pipe is depressurized and then subjected to a torque of 50 lblb - ft.ft., it will experience a shear strain of ________ rad.
A 11 in. inner diameter, 0.35 " wall thickness pipe is under a pressure of 2.5 ksi where strain gages installed along axial and circumferential directions register strains of 180 and 900 micro-strains (x10^-6), respectively. A) What is the Poisson's Ratio of this material and it's Elastic Modulus? B) In a uniaxial test, the pipe's material is observed to yield at a longitudinal strain of 0.1 in/in.in/in. Assuming a factor of safety of 2, the pipe can withstand impact energy of _______ lblb - in.in. per foot without suffering permanent deformation. C) If the pipe is depressurized and then subjected to a torque of 50 lblb - ft.ft., it will experience a shear strain of ________ rad.
(a) Two different materials designated A, and B, are tested in tension using test specimens having diameters of 0.505 cm and gage lengths of 2.0 cm (Figure 1). At failure, the distances between the gauge length marks are 2.13 cm (sample A) and 2.48 cm (sample B). Also, at the failure cross-sections, the diameters are found to be 0.484 cm (sample A) and 0.398 cm (sample B), respectively. Classify each material as brittle or ductile using your judgement.
(a) Two different materials designated A, and B, are tested in tension using test specimens having diameters of 0.505 cm and gage lengths of 2.0 cm (Figure 1). At failure, the distances between the gauge length marks are 2.13 cm (sample A) and 2.48 cm (sample B). Also, at the failure cross-sections, the diameters are found to be 0.484 cm (sample A) and 0.398 cm (sample B), respectively. i. Calculate the percent elongation and percent of area reduction in each specimen. a. Sample A b. Sample B
the creep data for two different samples. You have been asked to analyse the results, if you know that the original length of specimen is 17.4625 mm and the original width is 4.8 mm and the original thickness 1.6 mm and the cross sectional area is 7.68 mm: 1. Compare the lifetime for these samples.2. Determine the rapture lifetime for the samples below. 3. Calculate the creep rate for all of them
6. A ceramic part for a jet engine has a yield strength of 648 MPa and a plane strain fracture toughness of 8.49 MPam 4. To be sure that the part does not fail, we plan to ensure that the maximum applied stress is only one-third of the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.27 mm long. Assuming that the Y constant is 1.4, does our nondestructive test have the required sensitivity? Explain.
1) Fic 7.11 ypical engineering strain behavior to ture, potnt F. The sille strength 75 s licated at point M. The circular insets et the geometry of the deformed 75 pecimen at various ts along the curve. Please indicate in the stress-strain diagram given below the stress levels that can fail the material due to creep rather than instantaneously. (Please note that this is NOT the same question as the one in the final homework you were given)
Consider a cylindrical specimen of a steel alloy with 8.5 mm diameter and 80 mm long that is pulled in tension. Estimate the following mechanical properties using Fig. 1: a. Modulus of Elasticity and Resilience in MPa and psi b. Ultimate Tensile Strength in MPa and psi c. Fracture Strength in MPa and psi d. Ductility or % elongation at fracture in MPa and psi 2000 10³ psi MPa 300 2000 200 1000 100 0 0.000 0.005 0.010 0.015 Strain 0.020 0.040 0.060 Strain Fig. 1 Engineering Stress-Strain Curve Stress (MPa) 1000 0 0.000 Stress 0.080 300 200 100 0 Stress (10³ psi)
After the yield region in the stress-strain curve the strength up to ultimate value because: -Molecules orientation in the direction of applied stress -Molecules fails to withstand the applied stress -Molecules orthogonal to the axis of applied force The following figure shows typical engineering tensile stress vs. strain curves, where: * Stress Strain - Points A, C, E and F correspond to the tensile strength and elongation at yield - Points A, B, D and F correspond to the tensile strength and elongation at break Points A, C, E and F correspond to the tensile strength and elongation at break