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IRISSAK [1]
3 years ago
9

A 100-mm-diameter cast Iron shaft is acted upon by a 10 kN.m bending moment, a 8 kN.m torque, and a 150 kN axial force simultane

ously. The ultimate tensile strength of the shaft material is 210 MPa, and the ultimate compressive strength of the shaft material is 750 MPa, determine the safety factor against failure for the above types of loading using a proper theory of failure.
Engineering
1 answer:
konstantin123 [22]3 years ago
4 0

Answer:

Are you smart You seem smart

Explanation:

im ur dad

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A solid cylindrical workpiece made of 304 stainless steel is 150 mm in diameter and 100 mm is high. It is reduced in height by 5
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Answer:

45.3 MN

Explanation:

The forging force at the end of the stroke is given by

F = Y.π.r².[1 + (2μr/3h)]

The final height, h is given as h = 100/2

h = 50 mm

Next, we find the final radius by applying the volume constancy law

volumes before deformation = volumes after deformation

π * 75² * 2 * 100 = π * r² * 2 * 50

75² * 2 = r²

r² = 11250

r = √11250

r = 106 mm

E = In(100/50)

E = 0.69

From the graph flow, we find that Y = 1000 MPa, and thus, we apply the formula

F = Y.π.r².[1 + (2μr/3h)]

F = 1000 * 3.142 * 0.106² * [1 + (2 * 0.2 * 0.106/ 3 * 0.05)]

F = 35.3 * [1 + 0.2826]

F = 35.3 * 1.2826

F = 45.3 MN

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