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please send handwritten solution for Q1 part a
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- 1. An element whose material has Sy = 280 MPa, Sut = 330 MPa, and Suc = 460 MPa is loaded as shown. Determine the safety factors of the following elements based on these criteria. (a) MSST (b) MDET (c) MNST = -40 MPa T Txy = 40 MPa σx = 30 MPaIn the plastic range of a material, if the load is removed the material will: a. reach failure and break O b. continue up to elastic limit O C. go back to its original shape o d. remain in its current shapeProblem 1 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Aluminum Steel A=800mm2 A=1000mm2 Bronze A=700mm? 4P 2P 500mm 600mm 700mm 1. What is the maximum value of P that will not exceed the axial stress bronze of 105 MPa? 2. What is the maximum value of P that will not exceed the axial stress in aluminum of 90 MPa? 3. If P=10KN, what is the axial force to be carried by the aluminum in KN? 4. If P is 5KN, what is the axial stress of steel?
- A bar of solid circular cross section is loaded in tension by forces P (see figure). P L The bar has a length L = 14.0 in. and diameter d = 0.65 in. The material is a magnesium alloy having a modulus of elasticity E = 6.4 x 106 psi. The allowable stress in tension is allow = 17,300 psi, and the elongation of the bar must not exceed 0.04 in. What is the allowable value of the forces P (in lb)?A tensile force of 5 kN produce an elastic reduction in diameter of 4 x 10-4 mm on a metal alloy specimen that has a diameter of 6 mm. Determine: The modulus of elasticity of the alloy (Unit: GPa) . The Poisson' s ratio of this alloyA material obeys hook's law up to а. Elastic limit O b. Plastic limit O c. Limit of proportionality O d. Yield point
- S Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness f. Which metal is more ductile? Why? 900 Metal A 600 Metal B 300 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Strain, m/m FIGURE P1.16 Stress, MPaA circular bar of magnesium alloy is 750mm long. The stress-strain diagram for the material is shown in the figure. The bar is loaded in tension to an elongation of 6.0mm, and then the load is removed. (a)What is the permanent set of the bar? (b)If the bar is reloaded, what is the proportional limit? (High: Use the concepts illustrated in figures below)Mechanics of Deformable Bodies The rigid bar of negligible weight is pinned at O and attached to two vertical rods. Assuming that the rods were initially stress-free, what is the largest load P that can be applied without exceeding stresses of 150 MPa in the steel rod and 70 MPa in the bronze rod? 21 -1.5 →1.5 Bronze Steel A = 900 mm2 E= 200 GPa L= 1.5 m A = 300 mm? E = 83 GPa L=2 m
- An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 780 mm. The cross-sectional dimensions of the beam are b₁ = 36 mm, d₁ = 104 mm, b₂ = 22 mm, d₂ = 22 mm, and a = 7 mm. For this material, the allowable tensile bending stress is 16 MPa, and the allowable compressive bending stress is 10 MPa. Determine the largest moment M that can be applied as shown to the beam. b2 ✓ a M d₁ B L N.m A Answer: M = d₂ b₁An aluminum alloy rod has a circular cross section with a diameter of 7mm. The rod is subjected to a tensile load of 5kN. Assume that the material is in the elastic region and E = 69 GPa. If Poisson's Ratio is 0.38, what will be the lateral strain? E=6/8 v= -€(lateral)/e(axial) V ominorbooomesAn extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 620 mm. The cross-sectional dimensions of the beam are b1 = 39 mm, d, = 82 mm, b = 23 mm, dɔ = 23 mm, and a = 8 mm. For this material, the allowable tensile bending stress is 20 MPa, and the allowable compressive bending stress is 12 MPa. Determine the largest moment M that can be applied as shown to the beam. b2 lak M di B L b1 Answer: M = i N.m