Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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- In a tensile test for an aluminum alloy, the sample used is 2 inches long and 0.5 inches in diameter. The proportional portion of the tension stress-strain diagram for an aluminum alloy is shown below. Determine the modulus of elasticity for this material: ____x106 lb/in2. Pay attention to units and calculate your answer to 1 decimal place for the unit specified above.arrow_forwardQ4/ A. Explain why ceramic materials do not response to the plastic deformation processes? Also clarify why flexural test is applied to ceramics instead of tensile test? B. Estimate whether a cylindrical aluminum oxide sample will fail or not when it subject to flexural test to achieve flexural strength of 390 MPa knowing that the sample radius is 2.5 mm, separation distance is 30 mm, and the applied load is 620 N respectively.arrow_forwardA three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm (0.20 in.); the specimen fractured at a load of 3000 N (675 lbf) when the distance between the support points was 40 mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15 mm (0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 40 mm (1.6 in.)?arrow_forward
- A circular nylon [E = 8 GPa; ν = 0.35; α = 29 ×10-6/°C] disk with a circular hole in its center is shown. At an initial temperature of Ti = 22°C, the outer diameter of the disk is D = 210 mm, the diameter of the central hole is d = 55 mm, and the thickness of the disk is t = 24 mm. At a final temperature of Tf = 46°C, determine:(a) the change in diameter Δd of the central hole (assume an increase Δd as positive, otherwise as negative).(b) the change in thickness Δt of the disk (assume an increase Δt as positive, otherwise as negative).(c) the normal strain in the circumferential direction around the outside of the disk.arrow_forwardBrinnell hardness testing can be used for both Ferris and nonferrous metals. The test for ferrous metals requires a 500 kg load for 30 seconds and the test for nonferrous metals requires a 3000 kg load for 15 seconds.arrow_forwardA student wants to measure the strain at a particular point K on a loaded specimen, so she affixes a 45° strain rosette at point K and obtains the following results: €A = 390 μrad EB = 275 μrad Ec = 325 prad After packing up and going home to analyse her results, she realises that the strain rosette was misaligned by 0 degrees CW (where a negative corresponds to CCW) from the x-y axes-ie. Gauge C is not parallel with the x-axis and is instead parallel with a different x' axis that is rotated by 0 degrees-where: 0 = 16 deg Instead of replacing the rosette and retaking measurements, the student realises she can use what she learned in ENGG2400 to find the strains she's interested in. b 45° 45° a) Draw a diagram of the situation, showing both the x-y axes and the strain rosette. b) Calculate Exx, Eyy and Yxy. c) Calculate the maximum in-plane shear strain and the associated average normal strain. I Xarrow_forward
- A ceramic sample with a rectangular section can support a load of up to 290 N under a common strength test for ceramics. Its flexural strength is then 78.3 MPa and the distance between support points is 45 mm. What would be the dimensions of this sample if its width were twice its height?arrow_forwardThe representative curves of the applied stress versus the number of cycles for different metals are shown in Figure Q2. Analyze both metals in terms of fatigue properties. 500 Stress amplitude (MN/m²) 400 300 200 100 0 1045 steel 10³ 104 2014-T6 aluminium 10⁹ 106 107 Number of cycles, N₂ Figure Q2: Representative curves of stress amplitude vs number of cycles 108 10⁹arrow_forwardA cylindrical specimen of a metal alloy 47.5 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 397 MPa causes the specimen to plastically elongate to a length of 54.3 mm. If it is known that the strain-hardening exponent for this alloy is 0.2, calculate the true stress (in MPa) necessary to plastically elongate a specimen of this same material from a length of 47.5 mm to a length of 55.8 mm. i MPaarrow_forward
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