For the stress-strain diagram, which region or point represents strain hardening? 80 (4) (5) 70- (3) (6) 60- (2) 50 - 40- (1) 30- 20 - 10 - 0- Strain O 4 O5 0 6 O 2 O 1 (1S) ssans
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- 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 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.If the gauge is made of a material having a yield stress equal 200 MPa, determine theminimum required diameter d. Using a factor of safety (F.S. = 2)
- Solve Sample Problem No. 1 using these load values: a. M D =175 kN-m,M L =1 90 kN-m b. Find the maximum design moment so that its tension controlled. (USE c = 3d / 8). Tension steel exactly experiences 0.005 strain.Strength of materials - ENGR 3321 The answer should be: 17.68 kN; 11.8 MN/m2 compressiveBar A was manufactured 2 mm less than bar B due to an error. The attachment of these bars to the rigid bar wouldcause a misfit of 2 mm. Calculate the initial stress in each assembly. Which of the two assembly configurationyou would recommend? Use modulus of elasticity of E = 80 GPa and diameter of the circular bars as 32 mm.
- An aluminum pipe must not stretch more than 0.05in when it issubjected to a tensile load. Knowing that E=10.1x10^6psi and that the maximum allowable normal stress is 14ksi, determine (a) the maximum allowable length of the pipe, and (b) the required area of the pipe if the tensile load is 127.5kips.If the gauge is made of a material having a yield stress equal to 200 Mpa, determine the minimum required diameter d. Using a Factor of safety (F.S=2)At room temperature (10 C)° a 0.25-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 130°C, determine the normal stress in the each rod * 0.25 250 mm 200 mm B Aluminum A 2000 mm² E = 75 GPa a = 23 × 10 6°C Stainless steel A = 800 mm² E = 190 GPa a = 17.3 X 10-6PC
- A simply supported beam having a span of 12 m carries a uniform dead load of 5.8 kN/m and a live load of 8.8.kN/m in addition to a concentrated dead load of 22 kN applied at its midspan. The beam has a width of 350 mm and an effective depth of 525 mm. Total depth of beam is 625 mm. it is reinforced with a tensile steel of 3040 mm2. a) Compute the cracking moment. b) compute the instantaneous deflection c) compute the immediate deflection due to sustained load d) Compute the total deflection after 5 years with a dependent factor of 2.0.The steel channel is used to reinforce the wood beam. Determine the maximum stress in thesteel and determine the maximum stress in the wood if the beam is subjected to a moment of M=1.2kN.m Est = 200GPa, Ew12GPa.If the gauge is made of a material having a yield stress equal 200 MPa, determine the minimumrequired diameter d. Using a factor of safety (F.S. = 2)