Structural analysis *answer question 3 Q1) Determine the horizontal and vertical components of reaction at the pins A, B, and C.10 kN/m4 mAQ2) Determine the horizontal and vertical components of reaction at pins A, B, and Cof the two-member frame.300 lbB-24-24-ft6kNB4 ft45°2 m2m2 mB2 mㄱ2 mCQ3) Determine the components of reaction at the roller support A and pin support C. Joint B is fixed connected.3 ft2 kN

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Answered: Q3) Determine the components of…

Q3) Determine the components of reaction at the roller support A and pin support C. Joint B is fixed connected. 2 m- 6 kN 2 m B 2 m 2 m 2 kN

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

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Consider the graph below for 3 test samples A, B and C of the same metal composition which have been cold-worked, but to different extents. If you had to sort the samples by the degree of cold-working they have undergone, how would you rank them? Stress (MPa) 600 500 400 300 200 100 Select one: A B 0.05 0.1 a. C
A tensile test generates the stress-strain diagram shown here. 600 Stress (MPa) 500 400 300 200 100 T 1 T 0 0.00 Stress (MPa) 0.04 500 400 300 200 100 0.000 0.08 Young's modulus numeric value equal to Young's modulus value obtain from 0.002 0.004 Yield strength numeric value equal to Yield strength value obtain from Tensile strength numeric value equal to Tensile strength value obtain from Strain Strain Assume a tensile specimen was originally 10 cm long with 2 cm x 2 cm square shape cross section. 0.12 1 0.006 0.16 0.20 The change in sample length while a stress of 480 MPa is applied to a tensile specimen [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ]
Consider the graph below for 3 test samples A, B and C of the same metal composition which have been cold-worked, but to different extents. If you had to sort the samples by the degree of cold-working they have undergone, how would you rank them? Stress (MPa) 600 500 400 300 200 100 05 0 Select one: O a. A
Question 1 You are working on a design team at a small orthopaedic firm. You have been asked to select a cobalt- chrome-molybdenum (CoCr) material that will not experience plastic deformation under a specific mechanical test, as follows... A tensile stress is applied along the long axis of a solid cylindrical rod that has a diameter of 10 mm. An applied load of some magnitude F produces a 7x10-³ mm change in diameter (see figure below, original shape is blue, elongated shape is unshaded). Q1E: Of those two materials (F75 CoCr alloy (as cast) and F90 CoCr alloy (hot forged)), which materials would you select to assure that the deformation is entirely elastic (No yield!)?
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answer e and f only please..
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Part D-Permanent elongation of the specimen if the specimen is loaded until its nominal stress value reaches 65 ksi, determine the increase in gauge longth after it is unloaded. The strain at 65 ksi is 06 in. /in. Express your answer to four significant figures and include the appropriate units. ▸ View Available Hint(s) Permanent elongation 6 Submit Previous Answers Request Answer % ?
10. Measure the diameter of impression using the profile projector and compute for the hardness number using the formula: Where: 2P P – Applied Load D - Diameter of steel ball (8.0 mm) BHN «D(D- /D² – d² d - Ave. diameter of indentation | Raw data: Diameter Specimen (carbon steel) d1 d2 Air cooled 5.13 5.22 Oil quenched Water quenched 4.12 4.36 3.74 3.81 Note: All dimensions are in mm.
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Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…