Knowing that a 0.02in gap exists when the temperature is 75°F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -11ksi, (b) the corresponding exact length (2 decimal places in 10-3 inches) of the aluminum bar. 0.02 in. 14 in. 18 in. Bronze Aluminum A = 2.4 in² E = 15 × 106 psi a = 12 × 10-6/°F A = 2.8 in² E = 10.6 × 10° psi a = 12.9 × 10-6/°F

Knowing that a 0.02in gap exists when the temperature is 75°F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to -11ksi, (b) the corresponding exact length (2 decimal places in 10-3 inches) of the aluminum bar. 0.02 in. 14 in. 18 in. Bronze Aluminum A = 2.4 in² E = 15 × 106 psi a = 12 × 10-6/°F A = 2.8 in² E = 10.6 × 10° psi a = 12.9 × 10-6/°F

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.3P: A tensile test was performed on a metal specimen having a circular cross section with a diameter 0....

See similar textbooks

Similar questions

The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.
A tensile test was performed on a metal specimen having a circular cross section with a diameter 0. 510 inch. For each increment of load applied, the strain was directly determined by means of a strain gage attached to the specimen. The results are, shown in Table: 1.5.1. a. Prepare a table of stress and strain. b. Plot these data to obtain a stress-strain curve. Do not connect the data points; draw a best-fit straight line through them. c. Determine the modulus of elasticity as the slope of the best-fit line.
The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen with a circular cross section. The diameter was 3 8 inch and the gage length (The length over which the elongation is measured) was 2 inches. a. Use the data in Table 1.5.2 to produce a table of stress and strain values. b. Plot the stress-strain data and draw a best-fit curve. c. Compute the, modulus of elasticity from the initial slope of the curve. d. Estimate the yield stress.
A tensile test was performed on a metal specimen having a circular cross section with a diameter of 1 2 inch. The gage length (the length over which the elongation is measured) is 2 inches. For a load 13.5 kips, the elongation was 4.6610 3 inches. If the load is assumed to be within the linear elastic rang: of the material, determine the modulus of elasticity.
This figure shows typical stress-strain behaviour for a brass. The inset figure shows the behaviour at low strains. 500 70 Tosle strength 450 MPa 166,000 ps 60 400 300 MP 40 30 200 Yeld steg 250 MPa (,0a 200 20아 100- 20 100 10 0.20 030 040 If the cylindrical test sample is initially 14cm long with an initial radius of 2mm, estimate the true stress in MPa at the ultimate tensile stress. (Enter your answer to the nearest whole number) an es
2.58 Knowing that a 0.02-in. gap exists when the temperature is 75°F, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to 11 ksi, (b) the corresponding exact length of the aluminum bar. 0.02 in. 14 in. 18 in. ਕਰ Bronze A = 2.4 in² E-15 x 105 psi a = 12 x 10-6/"F Aluminum A = 2.8 in² E = 10.6 x 105 psi a = 12.9 × 10-5/"F
In a standard tensile test, an aluminum rod of 20-mm diameter is subjected to a tension force of P- 30 kN. Knowing that E- 70 GPa, and the change in diameter of the rod- 0.00955 mm, determine the poissons' ratio v. 20n diateter gage length-150 0.33 O 0.6 0.29 O 0.35
A specimen of steel 20 mm diameter with a gauge length of 200 mm is tested to destruction. It has an extension of 0.25 mm under a load of 80 kN and the load at elastic limit is 102 kN. The maximum load is 130 kN. The total extension at fracture is 56 mm and diameter at neck is 15 mm. Find . The stress at elastic limit. • Young's modulus. . Percentage elongation. Percentage reduction in area. Ultimate tensile stress. .
At what temperature will the aluminum and steel segments shown have stresses of equal magnitude after the 70-kip force is applied? Aluminum Steel A = 2 in.2 E = 10x 106 psi A = 3 in.2 E = 29 × 106 psi 70 kips 15 in. 10 in.-
Part A An aluminum 2014-T4 alloy bar 2.50 inches wide and 0.750 inches in thickness has a 1/4 inch radius hole centered on its width, as shown in the sketch. With an axial load, P = 8650 lbs applied, the maximum axial stress at the edges of the hole was experimentally determined to be 7670 psi What is the stress concentration factor, K, of the hole? 1.33 0.75 1.66 1.50 P 2.50 in. -Radius 1/4 in. P= 8650 lbs
The modulus of elasticity of mild steel is equal to .. .? .... .. a. 2 x 10ʻN/m? a. 2 x 10'N/m 2 b. 2 x 10° N/m 4 c. 2 x 10 N/mm 5 d. 2 x 10 N/mm Check
Sample: Malleable Steel (AISI 4145) Original Diameter: 6.14mm Gauge Length: 55 mm Final Length: 68.12 mm Final Diameter: 3.54mm Load (kN) Deformation (mm) Stress (MPa) 12.95 Strain (%) 0.3838 0.54 0.7841 0.95 26.48 1.3899 1.63 1.9485 2.07 3.3090 3.10 4.5821 4.22 5.9359 5.03 7.0340 5.51 8.2413 6.22 10.9446 7.22 13.1951 8.18 12.8228 8.77 12.2583 9.11 12.5915 9.61 13.2536 10.33 13.6636 10.71 13.9772 11.35 14.5433 12.63 15.1155 13.88 15.4970 15.10 15.6484 17.09 15.4031 17.79 14.7655 18.73 13.6721 19.28 10.4617 19.88