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- Compare the engineering and true secant elastic moduli for the natural rubber in Example Problem 6.2 at an engineering strain of 6.0. Assume that the deformation is all elastic.A 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.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 mm
- An aluminum alloy [E = 73 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 = 250 mm, a cross-sectional area of A = 6900 mm², and a length of L = 5.9 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 = 50°C, the longitudinal normal strain in the plate is found to be 2400 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Ad. L Answer: (a) P = i (b) Δd = KN mmAn 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.A strip of high-strength steel has a length of 30 cm and a cross section of 1 mm by 20 mm. The modulus of elasticity is 200 GPa and Poison’s ratio is 0.27. It is subjected to an axial load of 15000 N, and it is instrumented with two axial strain gauges with R= 120 Ω and a gauge factor of 2.10. The two fixed resistors are also 120 Ω, and the supply voltage is 2.5 V. The bridge is adjusted to zero voltage output before load is applied. Find the output of the bridge with load applied.
- An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0×10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 260 mm, a cross-sectional area of A = 14720 mm2, and a length of L = 5.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1680 µɛ. % D Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. L P Answer: Ad = i mmA certain steel has proportionality limit of 300 N/mm² in simple tension. Under a Three Dimensional Stress System, the principal stresses are 150 N/mm² (Tensile), 75 N/mm² (Tensile) and 30 N/mm² (Compressive), μ = 0.3. The factor of safety according to maximum shear stress theory would beThe total tensile load of 160kN of a steel is to be stress at the middle portion limited to 150 N/mm2, find the diameter of the middle portion and also the length of the middle portion if the total elongation of the rod is to be 0.2mm. consider the young modulus of a steel.
- 1.4-7 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 FractureAs shown, an aluminium alloy construction BCD with a circular cross section is fixed at end B and affected by a force of 150 N at the free end D. The diameter of the cross-section a-a is 20 mm. The yield strength of the material is 80 MPa: a) Determine the stresses at point A of the a-a cross-section. As indicated in the picture, draw the stress element in Cartesian coordinates and specify the stress values.(b) Calculate the factor of safety, n for Tresca, and the von Mises yield criterion to see if the structure would yield based on the stresses at point A.(c) In the major stress area, draw the yield loci of both criteria and indicate the operational stress state & why is the Rankine failure criterion inappropriate for aluminium alloys?An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown below. Axial loads are applied at the positions indicated. Find the maximum value of P in Newton that will not exceed a stress in steel of 150 MPa, in aluminum of 87 MPa, or in bronze of 97 MPa.