A cylindrical steel bar 8 mm in diameter isloaded 1000 cycles per day with a loadof 15560 N. How long until fatigue failuretakes place? Note: stressforce/area500450400350Steel Allby300250200150Brass Alloy100501.00E-07Cycles to Failure, N1.00E+041.00E+051.00E-061.00E+08100E+091.00E+10, The bar should not fatigue.) The bar will fail in one cycle.) 100 days) 1000 daysStress Amplitude, MPa

Answered: Image /qna-images/answer/baafc49d-3bf0-442d-8708-07583b0dba6a.jpg

Answered: A cylindrical steel bar 8 mm in…

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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Calculate each of the following under 20 kN load on a structure given below where all pins have a diameter of 10 mm. a) What is maximum stress on bar BD? b) Check the Pin strength at Location D against shearing. (Use ?Y=300 MPa, n=2 - factor of safety for Pin). Is it secure/safe or not? c) Check the Pin and Bearing strength against shearing at Location C. Are they secure/safe or not? (Use ?Y=300 MPa, n=2 - FOS for pin/bearing and ?Allow=300 MPa for bearing).
B) A simply supported shaft is shown in the following Figure. A constant magnitude distributed unit load p is applied as the shaft rotates. The shaft is also subject to a steady torque of Tmax 2000-N -m., find the diameter of shaft required to obtain a safety factor of 2.5 in fatigue loading if the shaft is steel of Sut 814 MPa and Sy 703 MPa. The dimensions are in cm, the distributed force in N/cm, and the torque is in N-m. No stress concentrations are present, the shaft is machined, the required reliability is 90%, and the shaft operates at room temperature Figure: 1= 20-cm: b = 18-cm ; a = 16 cm Applied distributed load p:= 1000 N em ; 区区

A steel bar is desired to be formed until it has a final length of 1.8m. What length sould the bar in m be before the forming stress is removed? The yield strength is 427 MPa, the tensile strength is 572 MPa, and the youngs modulus 210 GPa. Round of to tenths
15.28 For the fixture shown in Figure P15-28, the inside surfaces of the eight holes are increased in temperature by 80 °F. Determine the von Mises stresses throughout the fixture. What is the largest von Mises stress in the fixture? Is there concern for fail- ure due to yielding of the material? Assume the material is aluminum alloy 6061-0 (annealed). Fix the inside surface of the upper hole. 60.60 0.30 R 0,60 Y tax Lx $0.50 Figure P15-28 0.25 R மத் 0.10 0.40 (All dimensions in inch units) 0.50 in. R chamfer typ. 500 R 150 0.625 1.500 625- 0.775 -2.50 -0.30 3.00 0.60 dia. -0.60 dia. -0.150 R (All holes 0.60 in. dia. (6) with 0.50 in. rad. chamfer)
A solid round bar with diameter of 2 in has a groove cut to a diameter of 1.8 in, with a radius of 0.1 in. The bar is not rotating. The bar is loaded with a repeated bending load that causes the bending moment at the groove to fluctuate between 0 and 25 000 lbf in. The bar is hot-rolled AISI 1095, but the groove has been machined. Determine the factor of safety for fatigue based on infinite life and the factor of safety for yielding. Plars & .
A ¼ inch diameter steel rod is to carry an axial load of 2000 lbs. If the yield strength is 38 ksi, will the rod yield?
A round bar of length /= 100 mm, diameter d 20 mm, is subjected to an axial force P- 500 N, a transverse force F5000 N, and a torque T-50 N.m. The material is steel with Sy 480 MPa. Using the distortion energy (Von Mises) yield criterion, calculate the safety factor at points A and B. 20 D. Select one: O a. NA = 7.897, Ng = 4.958 O b. NA = 0.624, Ng- 4.352 Oc. NA 4.768, Ng 2.346 O d. NA = 0.749, Ng =5.223
9:17 A Mild steel specimen of circular cross-section is subjected to tensile test with following data ultimate load of 235 kN, final area - 499 mm?, initial area - 672.8 mm?, final length - 387 mm, initial length - 299 mm. Find the Initial Diameter, % of Elongation in length, % of Reduction in area & Ultimate strength. 3 Marks)
A rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 14 mm, and the rod length was 440 mm. When the applied torque reached 225.3 N-m, a shear strain of 2105 microradians was measured in the specimen. What was the angle of twist in the specimen? Calculate the shear stress in the specimen. Answer: T= 418.16 Use Hooke's Law to calculate the shear modulus of the specimen. answer: MPa G = i MPa
1. A 0.015 meter diameter bar of steel is stress tested and the following stress-strain curve is produced Stress (psix 10 60 50 8 20 10 1 MPa 300 100 psi (x 10%) 50 200-30 OL 40- 20- 10 0.05 0.005 1 L 0.010 0.10 Strain a. What is the ultimate tensile strength (MPa)? d. What is the yield strength(MPa)? 0.15 c. What is the percent plastic elongation at the time of fracture? e. What is the Modulus of Elasticity (GPa)? 400 300 200 Stress (MPa) b. What is the greatest load that can be place on this bar before failure(N)? 100 10 0.20
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Example: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025

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