M3= lkg f2=1 N-s/m MI= Ikg Kl=1 N/m fl=1 N-s/m K2=1 N/m F(t M2= 1kg f3=1 N-s/m
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- Data:P = 2 kipd = 3.6 inchw = 305 millimeterst = 0.03 meterr = 5.08 centimeterh = 0.6667 feet a) Calculate the maximum static stress in the material (in MPa). b) Estimate the theoretical stress concentration factor at the fillet c) Estimate the theoretical stress concentration factor at the holeVerify if my answers are correct: P =1000 N Q =1300 N C =216000 N.mmThis is part of alarger super position problem and I was able to solve for a moment and a concentrated load but this distributed load is confusing me. E = 200 GPa I = 216*10^6 mm^4 w = 81 kN/m d = 1.5 m I ended up getting an answer near -12 mm of diflection but that is an incorrect according to the grader. What would the equation for the elastic curve of this problem be? What is its value at point B? thank you so much!
- Mechanical design The maximum load in a rotating shaft is subjected to bending moment of 1 kN.m and a torque of 1 kN.m. The member is made of steel the Ultimate Tensile Strength 440 MPa, Yield Tensile Strength 370 MPa. With a safety factor of 4, kt = km = 1.5 and keyway stress consternation is presented. According to the maximum shear stress theory, the shaft diameter will be: a. 160 mm b. 74 mm C. 145 mm d. 92.5 mm e. 60 mm f. 66 mmL=9m W=5kN/m 360 UB 50.7Pls answer thank you! Find the diameter of a steel line shaft to transmit 10 horsepower at 150 revolutions per minute with a torsional deflection not exceeding 0.08 degree per foot of length.
- A link of a valve gear has to be curved in one plane, for the sake of clearance. Estimate the maximum tensile and compressive stress in the link if the thrust is 2500 N. (Cam bridge) ANSWER: maximum tensile 38.0 MN/m2; maximum compressive 46.0 MNIm2. Please show the solution to get the answer.Help in filling table: Material Yield Strength (MPa) Elastic Modulus-E (GPa) Poisson’s ratio Ultimate Strength (MPa) Steel 0.305 Aluminum 69 0.33 Copper 123 0.33 Brass 88 0.333) A flange coupling is to connect two 60mm shafts. The hubs of the coupling are 120mm, 100mm thick, & the flange webs are 20 mm thick. Six M16 bolts fasten the flange from a 165 bolt circle. The keyway is 15mm shorter than the hub’s thickness & the key is 15mm x 15mm. The coupling transmits 45kW @ 450rpm. The yield point in shear for all materials is half of its yield point in tension or compression. Yield tensile strength is 0.75 of ultimate strength of the materials. The Sult is 448N/mm2 . Find: (a) the shear stress in key; (b) the bearing stress in key; (c) the shear stress in bolts; (d) bearing stress in flange.
- Repeat Problem 11.2-14 using L = 12 ft, ß = 0.25 kips/in., ßRl= 1.5ßL2, and ßR2= 2 ßR1.A pulley is keyed to a shaft midway between two anti-friction bearings. The bending moment at the pulley varies from – 170 N-m to + 510 N-m.The shaft is made of cold drawn steel having an ultimate strength of 540 MPa and yield strength of 400 MPa, Take σe = 0.5 σu. Determine the required diameter for an indefinite life based on Goodman equation. The actual stress concentration factor for the keyway is 1.6. The factor of safety is 1.5. Take size factor = 0.85 and surface finish factor = 0.88.The shaft and the flange of a marine engine are to be designed for flange coupling, in which the flange is forged on the end of the shaft. The following particulars are to be considered in the design: Power of the engine=2.4 MW Speed of the engine 80 r.p.m Permissible shear stress in bolts and shaft = 60 MPa Number of bolts used =6 Pitch circle diameter of bolts=1.6* Dimeter of shaft Find 1.diameter of shaft; 2.diameter of bolts; 3.thickness of flange; and 4.diameter of flange. (25 degrea