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- An aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.An aluminum alloy [E = 69 GPa; v = 0.33; a = 23.0 x 10-6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 215 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 3.9 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by AT = 53°C, the longitudinal normal strain in the plate is found to be 2320 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L Answer: (a) P = i (b) Δd = = i d KN mmFor a point on a steel specimen, the principal stresses are known to be 01 = 360 MPa and O2 = 60 MPa. Calculate the minimum yield stress of the material according to the Tresca criterion. Give your answer in MPa to 3 significant figures.
- An aluminum alloy [E = 74 GPa; v = 0.33; a = 23.0 x 10-6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 265 mm, a cross-sectional area of A = 5300 mm², and a length of L= 4.2 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by AT = 57°C, the longitudinal normal strain in the plate is found to be 2920 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L P Answer: (a) P = i (b) Δd = i kN mmA steel 0.6 inch×1.2 inch steel 90 m long is subjected to a 45 KN tensile load along its lenght.If poison's ratio is 0.3 Find: A. The deformation along its lenght. B. The deformation along its thickness. C. The defirmation along uts width. D. The lateral strain.A steel component is subjected to alternate cyclical loading. The steel follows Basquin's law for high cycle fatigue, o, x N = C, (where the stress amplitude is in MPa). Ignore the geometric detail and assume that Marin's modifying factors are all equal to 1. You are given the minimum stress ain = -213 MPa, the maximum stress omax = 213 MPa. The material data are Tensile strength oUTS = 539 MPa, Basquin's constant c, = 875 MPa, Basquin's exponent a = 0.085. a) Calculate the stress ratio R, the stress amplitude o, in MPa and the mean stress am in MPa. The answers are acceptable with a tolerance of 0.01 for R and of 1 MPa the stresses. R: MPa MPа b) Calculate the corresponding life, in 10° cycles, (tolerance of 0.1 106 cycles) N :
- A steel with E = 29 000 ksi with a rectangular cross-section is bent over a rigid mandrel with R = 15 in as shown in the figure. If the maximum flexural stress in the bar is not to exceed the yield strength of 36 ksi, determine the allowable thickness h of the bar.The lowest stress at which permanent .deformation can be measured1.4-7 The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13 mm and a gage length of 50 mm (see figure for Prob. 1.4-3). At fracture, the elongation between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm. Plot the conventional stress-strain curve for the steefor the steel and determine the proportional limit, modulus of elastics of elastic- ity (i.e., the slope of the initial part of the stress-strain,tress-strain curve), yield stress at 0.1% offset, ultimate stress, percent, elongation in 50 mm, and percent reduction in area. 'ess, percent area. TENSILE-TEST DATA FOR PROB. 1.4-7 Elongation (mm) 0.005 0.015 0.048 Load (kN) 5 10 30 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracture
- At a point in a strained material, tensile stress of 100 MPa and compressive stress of 60 MPa are found to be principal stresses Maximum shear stress at that point is: O 60 MPa O 20 MPa O 40 MPa O 80 MPaThe stresses in a flat steel plate in a condition of plane stress are: o, = 10000 N/mm2 o = 6000 N/mm? = 8000 N/mm2 Find the magnitude and orientation of the principal stresses in the plane of the plate.1. Calculate the strain at the centroid of the tension steel in single layer if the effective depth is 250 mm and the depth of neutral axis is 100 mm. answer: 0.0045 2. Calculate the strain at extreme layer of steel if fy=415 MPa and the strength reduction factor is 0.80. answer: 0.0038