40.0030.00100.00-100.00P = 1000 NA=167d=140.00100.00--b 20.00200.00Weld Strength P = 273 N/mm^2 Electrod E60Safety factor S₁ = 3Force P = 1000 NUsing by SOLIDWORKS

Answered: Image /qna-images/answer/6bcee5cb-775a-462b-bd09-7784b67765a3.jpg

Answered: 40.00 30.00 100.00- 100.00 P = 1000 N…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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Material Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6AL-4V. All of the bars have identical cross section areas of 1 in^2. An axial stress of 20 ksi is placed on each of the bars. Which bar is the strongest? AISI 4130 Steel Aluminum 6061 T4 Ti-6AL-4V All of the bars have the same strength
Material Sy Sut E AISI 4130 100 ksi 125 ksi 29 x 10^3 ksi AISI 1025 36 ksi 55 ksi 29 x 10^3 ksi Aluminum 2014 T6 53 ksi 60 ksi 10.8 x 10^3 ksi Aluminum 6061 T4 16 ksi 30 ksi 9.9 x 10^3 ksi Ti-6AL-4V 126 ksi 134 ksi 16 x 10^3 ksi You have a bar of AISI 4130 steel, a bar of Aluminum 6061 T4, and a bar of Ti-6AL-4V. All of the bars have identical cross section areas of 1 in^2. An axial stress of 20 ksi is placed on each of the bars. Which bar will yield? AISI 4130 Steel Aluminum 6061 T4 Ti-6AL-4V All of the bars will yield
answer 35mm
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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la. In the design of a pressure vessel, the leak-before-break criterion is used considering the circumferential wall thickness and the plain strain fracture toughness. The design is limited by the yield strength gs and the factor of safety (SF). Derive expressions for the critical crack size. 1b. Calculate the design stress and stress intensity factor if the safety factor is 1.5. Data: gxS = 586 MPa and KIC = 33 MPa m1/2.
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Given is a thin-walled tube with the following data: Wall thickness- 2mmABC is a sector with a radius of 50 mmAOD & COD is a triangleTorque is 1000 N.mLength of the thin walled tube is 0.5 mUse G=24 GPa for aluminium Determine the following:a. Enclosed Areab. Perimeter of the thin walled tubec. Shear flowd. Shear stresse. Angle of twist
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The engineering stress-engineering strain tensile curve for a stainless steel type 304 alloy is shown. Calculate the constants K and n for the true stress and true strain curve before necking (or = K (&)"). Engineering Stress (MPa) 1000 900- 800- 700- 600- 500- 400- 300- 200- 100- of 0.0 02 0.4 Enginering Strain (mm/mm) 0.6 0.8 1.0
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40.00 30.00 100.00- 100.00 P = 1000 N A=167 d=140.00 100.00- -b 20.00 200.00 Weld Strength P = 273 N/mm^2 Electrod E60 Safety factor S₁ = 3 Force P = 1000 N Using by SOLIDWORKS