1. A piston rod, made of AISI 3140 steel, OQT 1000 F, is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on: (a) ultimate strength, 1.62 in 1 5/8 in (b) yield strength. 1.23 in = 1 % in For AISI 3140 steel, OQT 1000 F Su = 152 5. ksi Nu = 8 Sy = 132 5. ksi Ny = 4

1. A piston rod, made of AISI 3140 steel, OQT 1000 F, is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on: (a) ultimate strength, 1.62 in 1 5/8 in (b) yield strength. 1.23 in = 1 % in For AISI 3140 steel, OQT 1000 F Su = 152 5. ksi Nu = 8 Sy = 132 5. ksi Ny = 4

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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A piston rod, made of AISI 3140 steel, OQT 1000 F , is subjected to a repeated, reversed load. The rod is for a 20-in. air compressor, where the maximum pressure is 125 psig. Compute the diameter of the rod using a design factor based on (a) ultimate strength, (b) yield strength. For AISI 3140, OQT 1000 F, Su=152.5 ksi, Sy=132.5 ksi, Nu=8, Ny= 4
The cylinder of a four stroke diesel engine has the followin, specifications : Cylinder bore :150 mm Maximum pressure : 3.5 MPa Cylinder material : GCI FG200 Factor of safety : 6 Poisson's ratio : 0.25 Determine the thickness of the cylinder wall. Also calculate the apparent and net circumferential stresses in the cylinder wall.
b) Determine the maximum gas pressure and the dimensions of cross -section of the connecting rod for a diesel engine with the following data: (i) Cylinder bore 120 mm and the Length of connecting rod = 360mm (ii) The critical buckling load = 407.1504 KN; (iii) The compressive yield stress =330 N/mm? ; Factor of safety = 6 %3D %3D
Given: Tie rods in a press machine are steel alloy AISI 5160 OQT 1000 - Diameter: D = 2.00 in. E = 30 x 106 psi - Length: L = 68.5 in. - Axial tensile force: F= 40,000 Ib. %3D Required: The elastic deformation of the tie rods is _in. 0.039 0.059 0.049 0.029
A helical spring made of oil hardened and tempered steel compression springs, made of circular wire, is subjected to an axial force, which varies from (2.5 KN) to (3.5 KN). Over this range of force, the deflection of the spring should be approximately (5 mm). The spring index can be taken as (5). The spring has square and ground ends. If ultimate tensile strength of (1050 N/mm2) and modulus of rigidity of (81370 N/mm2). The permissible shear stress for the spring wire should be taken as (50 %) of the ultimate tensile strength.Design the spring and calculate the following: 1. Wire diameter; 2. Mean coil diameter; 3. Number of active coils; 4. Total number of coils; 5. Solid length of the spring; 6. Free length of the spring: 7. pitch of coils (p); 8. Required spring rate; 9. Actual spring rate; and 10. Draw a neat sketch .of the spring showing various dimensions
6. Determine the size of a piston rod subjected to a total load of having cyclic fluctuations from 15 kN in compression to 25 kN in tension. The endurance limit is 360 MPa and yield strength is 400 MPa. Take impact factor= 1.25, factor of safety= 1.5, surface finish factor= 0.88 and stress concentration factor = 2.25. [Ans. 35.3 mm]
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…
cylindrical pressure vessel has inside diameter d_{o} = 500 mm and inside diameter di=490 mm.if it i made of carbon steel SAE 1040 OQT1300, find the maximum allowable internal pressure if the safety factor against yielding is not to be less than 4. Neglect the external pressure and assume that the longitudinal stresses are taken by the vessel itself. a) Solve using the MSST. b) Solve using the DET. See Appendix 3 for material properties.
Q12/ An electric motor drive an exhaust fan through an open belt drive. Following data are provided: Motor Pulley Fan Pulley Diameter (mm) 400 1600 Angle of contact (radians) 2.5 Coefficient of friction 0.3 Speed (rpm) Power transmitted (kW) 700 22.5 Calculate the width of 5 mm thick flat belt. Take permissible stress for the belt material as 2.3 MPa, and assuming the density of the belt is 1000 kg/m³.
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:
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
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