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- The figure below is a stress strain curve for an alloy of aluminum at various temperatures. a) What is the yield stress at room temperature? b) Assume a rod made of this material is 20 cm long and at room temperature. If the rod were put in tension, how long would it be when it breaks? c) If the rod has cross-sectional area 1 cm?, what force is required to break it (in tension)? Longitudinal RT 250F 350F 30 500F 0.002 0.004 0.006 0008 QO10 O012 Strain, in/in. A-93 Stress, ksiA.) It is the boundary condition for fixed support Choices: Both choices Y=0 Y’=0 Either of the choices B.) It is a type of simple stress acting parallel to an area Choices: Flexural stress Shear stress Axial stress Bearing stress C.) For concentric row of bolts in a coupling connection, the shear deformations in the bolts is proportional to the: Choices Modulus of rigidity Diameter of bolt circle Radial distance Number of bolts note: please help us in the word meaing sample problems thank youQuestion Three a) A wire 3m long and 2mm in diameter when stretched by weight of 10kg has its length increased by 0.28mm. Find the stress, strain and Young Modulus of the material of the wire. (g = 9.81m/s²) b) A steel column is 3.5m long and 0.4m diameter. It carries a load of 50MN. Given that the modulus of elasticity is 200GPa, calculate the compressive stress and strain, and determine how much the column is compressed.
- The bar of negligible weight is supported by two springs each having a stiffness 180 N/m The springs are originally unstretched, and the force is vertical as shown ( Figure 1) Figure C 30 N Im B 2m Part A Determine the angle the bar makes with the horizontal, when the 30-N force is applied to the bar Express your answer using three significant figures. VAX Submit Request Amer Return to Assignment vec Provide feedback ?Read the question carefully and give me both A&B right solutions otherwise leave it. A) Steel Alloy rod made of Stainless Steel 304 is subjected to 60 kip of Tensile force. Steel rod is having diameter of 1 in. and Length of 8 in. Determine change in its length and diameter. Consider material behaves elastically. For determining material properties use Mechanical property table. B) An aluminium specimen made up of 6061-T6 Material has a diameter of d = 20 mm and a gage length of L0= 300 mm. If a force of 135 kN elongates the gage length 0.9 mm, determine the modulus of elasticity. Also, determine contraction of diameter.And **The modulus of elasticity** (MPa) I need a clear step by step answer please :)
- ! How to compute the strain displacement matrix for each element?Suppose you purchase a 4 x 8 piece of plywood from Home Depot. You don't have a pickup truck (nor does your friend), but you do have a roof rack with cross bars on your car. So you will have to secure the plywood to the roof rack in order to get it home. Your friend (the one who does not have a pickup truck) suggests that you tie the plywood on sideways (so that the short edge is front to back) and drive at about 13 mph. Use standard sea-level conditions = 0:0023769 slugs/ft3 and = 3:7372 107 slugs/ft*s.(a) Estimate the drag on the plywood for the case suggested by your friend.(b) Estimate the drag on the plywood if you turn the sheet so that the long edge is front to back.(c) Explain why your friend suggested the sideways configurationConsider the brass alloy for which the stress-strain behavior is shown in figure below. A cylinder specimen of this material 10 mm in diameter and 120 mm long is pulled in tension with force of 15 kN. The Poisson's ratio is 0.35. Compute : a) The Tension stress of 15 kN force. b) The specimen elongation at force of 15 kN c) The reduction in specimen diameter at force of 15 kN d) The maximum Force where the specimen is broken e) The Force at Yield point. 500 70 Tensile strength 450 MPa (65,000 psi) 60 400 50 10 psi 300 MPa 40 40 200- 30 Yield strength 250 MPa (36,000 psi) 30 200 20 100 20 10 100 This figure is the detail figure from 10 "Stress - Strain OL 0. 0.005 diagram" with strain range from 0 - 0.005 0.10 0.20 0.30 0.40 Strain Stress (MPa) Isd 01) ssaS
- After the yield region in the stress-strain curve the strength up to ultimate value because: -Molecules orientation in the direction of applied stress -Molecules fails to withstand the applied stress -Molecules orthogonal to the axis of applied force The following figure shows typical engineering tensile stress vs. strain curves, where: * Stress Strain - Points A, C, E and F correspond to the tensile strength and elongation at yield - Points A, B, D and F correspond to the tensile strength and elongation at break Points A, C, E and F correspond to the tensile strength and elongation at break(Solid Mechanics) This finite element model is composed of 4 linear bar elements, each with a cross section of 10 mm² and material properties (E = 6 GPa, v=0.3). A weight of 200 N is applied and the nodes are named as shown below. (1) Compute the reaction forces. (2) What are the reaction forces when the weight is tripled to be 600 N? Find the solution without repeating the FEA. Explain why you can get the quick answer.please step by step solve this simple algebra. it s part of a stress and elasticity problem