If in a tensile loading application, the factor of safety for one member is set as FS = 2.2 based on the material's tensile (failure) strength as 6 (fail) 614 MPa, then the load member's allowable stress in this application will be: MPa. Calculate your answer to 1 decimal place. (allow)

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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f in a tensile loading application, the factor of safety for one member is set as FS = 2.2 based on the material's tensile (failure) strength as    = 614 MPa, then the load member's allowable stress in this application will be:   =  _______  MPa. Calculate your answer to 1 decimal place. 

If in a tensile loading application, the factor of safety for one member is set as FS = 2.2 based on the material’s tensile (failure) strength as \( \sigma_{(fail)} = 614 \, \text{MPa} \), then the load member's allowable stress in this application will be: \( \sigma_{(allow)} = \, \_\_\_\_\_ \, \text{MPa} \). Calculate your answer to 1 decimal place.
Transcribed Image Text:If in a tensile loading application, the factor of safety for one member is set as FS = 2.2 based on the material’s tensile (failure) strength as \( \sigma_{(fail)} = 614 \, \text{MPa} \), then the load member's allowable stress in this application will be: \( \sigma_{(allow)} = \, \_\_\_\_\_ \, \text{MPa} \). Calculate your answer to 1 decimal place.
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Given:Factor of safety, FS=2.2Tensile failure strength, σfail=614 MPaFind:The load member's allowable stress, σallow

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could you give the equation and work for this to get the answer please

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