Parvinbhai 3. Given a Type 316 sheet annealed stainless steel hollow, rectangular column with 600 mmlength, 1 mm wall thickness, 1500 N concentric load, and a safety factor of 1.5, design alsuitable column if its boundary conditions are fixed-fixed and the ratio of side lengths of therectangle is 1.5:1. Use AISC recommended values and round dimensions to next largestwhole integers that maintain the ratio of side lengths.

The objective of the question is to design a suitable column given the load, material, and boundary…

Answered: 3. Given a Type 316 sheet annealed…

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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MACHINE DESIGN TOPIC A short compression member with Do = 2Di is tosupport a dead load of 25 tons. The material isto be 4130 steel, WQT 1100 F. Calculate theoutside and inside diameters in terms of incheson the basis of (a) yield strength, (b) ultimatestrength.*For 4130 steel, WQT 1100 FSu = 127 ksiSy = 114 ksi*Use the following Safety FactorsNu = 4Ny = 2
Uuestion 1: Calculate the axial load of spiral column with respect to first and second peaks. Materials used are C25 and S420 for longitudinal bars of 8- 024 and for spirals. Use the design strength of concrete and steel. 2. A, fywk 0, = D (s) 4. Ao P. = D (s) fa = f+4.0-0, GAP fa 4. f ynk (A min p, = 0.45. , min p, = 0.12. ywk D=550 mm H=600 mm Spiral : 010/85 mm 200 mm
Calculate the critical load of a steel circular column for pinned-pinned boundary conditions using Johnson formula. The length of column is 300 mm, diameter is 60 mm and slenderness ratio is 40. The modulus of elasticity of steel is 207 GPa and yield strength is 300 MPa. Use safety factor 2.5. Select one: O a. 756.6 kN b. 715.3 kN c. 798.4 kN
To calculate the maximum axial loads for a column based on yielding and buckling about both axes, and allowing for factors of safety. The critical load P for buckling of a column depends on the vertical distance between supports and how the column is connected to those supports. The length L in the T'EI formula Per =- is the distance between points of zero moment. For a pin- L2 supported column, this is the full height of the column. For a column that is not pinned at both ends, the points of zero moment are not at the supports. The equation for the critical load can be modified to use an effective length Le = KL, where K depends on the support conditions. For a column fixed at both ends, K = 0.5. For a column fixed at one end and pinned at the other, K = 0.7 (Figure 1). The equation for the critical load then becomes ’EI Pa (KL)²
Problem 4: 6.43. The simple expressions for Rockwell hardness num- bers in Table 6.9 involve indentation, 1, expressed in millimeters. A given steel with a BHN of 235 is also measured by a Rockwell hardness tester. Using a 16 -in.-diameter steel sphere and a load of 100 kg, the indentation t is found to be 0.062 mm. What is the Rockwell hardness number? Solution:
Q2 Design a long pole (diameters of hollow cross section, length) made of AISI 1020 CD steel capable of carrying a 600 kg electrical transformer. Then calculate the design safety factor.
The load on a helical spring is 1282 lb and the corresponding deflection is to be 3.4 inches. Rigidity modulus is 11 x 106 psi and the maximum intensity of safe torsional stress is 60,000 psi. If the wire diameter and the mean diameter are 0.5 in. and 3 in., respectively. Determine the number of active coils. Don't round your answer to preferred size. Round your answer to 4 significant figures
Condition #1: A structural support for a machine is subjected to a static compression load of 20 kN. The manufactured from a circular rod made from SAE 1040 Hot Rolled steel. Specify suitable diameter for the cross section of the rod based on the basic size. Steel data are available in Table A-10 from the textbook. support is A-10 Typical properties of carbon and alloy stcels. Ultimate strength, s Yield strength, s, Percent Material SAE no. Condition ksi MPa ksi MPa elongation 1020 Annealed Hot rolled 57 393 43 296 36 1020 65 448 48 331 441 36 1020 Cold drawn 75 517 64 20 1040 Annealed Hot rolled Cold drawn 75 517 51 352 30 BO40 90 621 60 82 414 1040 97 669 565 16 1040 WQT 700 WQT 900 WQT 1100 WQT 1300 127 118 876 93 641 19 1040 814 90 621 1040 107 738 80 552 24 1040 87 600 63 434 32 1080 Annealed OQT 700 OQT 900 OQT 1100 OQT 1300 89 614 1303 1234 1080 54 372 25 189 179 145 117 141 129 972 889 710 1080 12 1O80 13 1000 103 70 17 23 1080 807 483 1141 Annealed 87 600 772 1141 51 352 26…
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A 200 GPa young's modulus of a piston rod of diameter 20 mm and length 700 mm in a hydraulic cylinder is subjected to a compressive force of 10 kN due to the internal pressure. The end conditions of rod may be assumed as guided at the piston end and hinged at the other end.Calculate the factor of safety for the piston rod.

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