with a stress of 52.89 and diameter of 4cm and thickness of 2mm (P) load of 2MPa how was the answer of 5.289MPa achieved? I am using this to answer another similar question to find the Load (P) i have got the information of stress = 18.70MPa , diameter of 1.3m and wall thickness of 29mm
Design Against Fluctuating Loads
Machine elements are subjected to varieties of loads, some components are subjected to static loads, while some machine components are subjected to fluctuating loads, whose load magnitude tends to fluctuate. The components of a machine, when rotating at a high speed, are subjected to a high degree of load, which fluctuates from a high value to a low value. For the machine elements under the action of static loads, static failure theories are applied to know the safe and hazardous working conditions and regions. However, most of the machine elements are subjected to variable or fluctuating stresses, due to the nature of load that fluctuates from high magnitude to low magnitude. Also, the nature of the loads is repetitive. For instance, shafts, bearings, cams and followers, and so on.
Design Against Fluctuating Load
Stress is defined as force per unit area. When there is localization of huge stresses in mechanical components, due to irregularities present in components and sudden changes in cross-section is known as stress concentration. For example, groves, keyways, screw threads, oil holes, splines etc. are irregularities.
with a stress of 52.89
and diameter of 4cm and thickness of 2mm
(P) load of 2MPa
how was the answer of 5.289MPa achieved?
I am using this to answer another similar question to find the Load (P) i have got the information of stress = 18.70MPa , diameter of 1.3m and wall thickness of 29mm
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