Engineering

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of and a yield strength of 1400 MPa (205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm (0.16 in.). If the design stress is one-half the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of and a yield strength of 1400 MPa (205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm (0.16 in.). If the design stress is …

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of and a yield strength of 1400 MPa (205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm (0.16 in.). If the design stress is one-half the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection. Read More »

Steady-state creep rate data are given here for some alloy taken at 200�C (473 K):If it is known that the activation energy for creep is 140000 J/mol, compute the steady-state creep rate at a temperature of 240 �C (513 K) and a stress level of 47 MPa.

Steady-state creep rate data are given here for some alloy taken at 200�C (473 K): If it is known that the activation energy for creep is 140000 J/mol, compute the steady-state creep rate at a temperature of 240 �C (513 K) and a stress level of 47 MPa. EXPERT ANSWER

A cylindrical 4340 steel bar is subjected to reversed rotating-bending stress cycling, which yielded the test results presented in Animated Figure 8.21. If the maximum applied load is 5,000 N. compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.25 and that the distance between loadbearing points is 55.0 mm. mm Cycles to failure = 3.2E 3 Maximum stress = 0 Mpa 700 Ti-5Al-2.5Sn titanium alloy 4340 steel Maximum stress, S (MPa) 1045 steel Ductile cast iron 70Cu-30Zn brass 2014-T6 Al alloy EQ21A-T6 Mg alloy 101_10510610_100100* Cycles to failure. N

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A cylindrical 4340 steel bar is subjected to reversed rotating-bending stress cycling, which yielded the test results presented in Figure 8.20. If the maximum applied load is 5000 N, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.25 and that the distance between load-bearing points is 55.0 mm. Use the following relation between the stress () and the load (F)   16FL , (d0) is the cylinder  d 03 diameter, and (L) is the distance.

A cylindrical 4340 steel bar is subjected to reversed rotating-bending stress cycling, which yielded the test results presented in Figure 8.20. If the maximum applied load is 5000 N, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.25 and that the distance between …

A cylindrical 4340 steel bar is subjected to reversed rotating-bending stress cycling, which yielded the test results presented in Figure 8.20. If the maximum applied load is 5000 N, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.25 and that the distance between load-bearing points is 55.0 mm. Use the following relation between the stress () and the load (F)   16FL , (d0) is the cylinder  d 03 diameter, and (L) is the distance. Read More »

The spring mechanism is used as a shock absorber for a load applied to the spring is originally unstretched and the drawbar slides along the smooth guide posts CG: and EF. The ends of all springs are attached to their respective members. Determine the force in spring H when the P = 50 kN force is applied. Express your answer to three significant figures and include appropriate units. Determine the force in spring AG (and, by symmetry, BF) when the P = 50 kN force is applied. Express your answer to three significant figures and include appropriate units. What is the required diameter of the shank of bolts CG and EF if the allowable stress for the bolts is sigma_allow = 135 MPa? Express your answer to three significant figures and include appropriate units

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A rotating shaft of 25-mm diameter is simply supported by bearing reaction forces R and R. The shaft is loaded with a transverse load of 13 kN as shown in the figure. The shaft is made from AISI 1045 hot-rolled steel. The surface has been machined. Determine (a) the minimum static factor of safety based on yielding. (b) the endurance limit, adjusted as necessary with Marin factors. (c) the minimum fatigue factor of safety based on achieving infinite life. (d) If the fatigue factor of safety is less than 1 (hint: it should be for this problem), then estimate the life of the part in number of rotations. 200 mm -50 mm- 25 mm 13 kN, Problem 6-10 R Not to scale

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Technology – Production Name. Folsot .Family name.. ALMOZYO…..Note… Exercise N1: A firm that assembles computers and computer equipments is about to start production of two types of microcomputers. Each type will require assembly time, inspection time, and storage space. The amounts of each of these resources that can be devoted to the production of the microcomputers are limited. The manager of the firm would like to determine the quantity of each microcomputer to produce in order to maximize the profit generated by sales of these microcomputers. The manager has the following information: Type 1 Type 2 Profit per unit 3 Assembly time per unit 2 hours 5 hours Inspection time per unit 3 hours 8 hour Storage space per unit 1 cubic feet 2 cubic feet The company resources are as follow: Resource Amount available Assembly time 124 hours 28 hours Inspection time 42 cubic feet Storage space Complete the following table and elaborate the associated mathematical formulation Assembly time Inspection time Storage 7 Space Profit per unit per unit per unit Type 1 Type 2 Exercise N2: Solve this problem by the geometric method Maximise Z=60 xy +50x Subject to 4x+10 x 5100 2 xq+x2S22 3x +3 x 539 X;20: x 20 1/2 Points of the first line 4x +10 x2=100 Points of the first line 2 x + x2=22 Points of the first line 3x +3 X2=39 20+ 10+ (0:0) 10 A(11:0) 20 Intersection between 4x +10 X2=100 and 3x +3 X2=39 Intersection between 2 x + X2=22 and 3×2+3 x2=39 ZO= ZA= ZB ZC- ZD. The optimal solution is the point

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