Calculate Factor of Safety for Steel Hanger w/ 550 MPa Strength

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Homework Help Overview

The discussion revolves around calculating the factor of safety for a steel hanger with a specified ultimate strength of 550 MPa, which is supporting a load of 64000 N. The cross-sectional area of the hanger is given as 7.4 cm², leading to questions about how to properly calculate the stress and factor of safety.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between load and area to find working stress, questioning whether any mathematical adjustments are needed for the cross-sectional area provided.
  • There are attempts to calculate stress using the formula stress = load/area, with some participants expressing confusion over unit conversions from cm² to m².
  • Several participants share their calculations and seek validation of their results, indicating a collaborative effort to understand the problem.

Discussion Status

The discussion is ongoing, with participants actively sharing their calculations and questioning unit conversions. Some have identified mistakes in their calculations and are seeking clarification on the correct approach to find the factor of safety.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There is a focus on ensuring correct unit usage and understanding the relationship between stress and the factor of safety.

walleye3
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Homework Statement



What is the factor of safety of a steel hanger having an ultimate strength of 550.00 MPa and supporting a load of 64000.000N. The steel hanger in question has a cross sectional area of 7.400cm squared

Homework Equations





The Attempt at a Solution


i understand the equation is ulitamate stress divided by allowable stress but the cross sectional area gives me problems. please show me a the steps i should take to find the factor of safety




well i went load/ area > 64000N/7.4cm squared = 8648.65 Pa


550 MPa = 550000kPa/8.648kPa (ultimate stress/allowable stress) =63594.11
63.59 as my answer is that correct ? if not please show me where i went wrong
 
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Given the load and the area, how do you find the working stress i.e. stress due to the load?
 
Load divided by area ...right ? , when it gives me cross sectional area do I need to do anything mathematically with it or do I plug that number in ?
 
walleye3 said:
Load divided by area ...right ? , when it gives me cross sectional area do I need to do anything mathematically with it or do I plug that number in ?

Well you have the formula

stress = load/area

And you are given both. You want to get the value of stress.
 
well i went load/ area > 64000N/7.4cm squared = 8648.65 Pa


550 MPa = 550000kPa/8.648kPa (ultimate stress/allowable stress) =63594.11
63.59 as my answer is that correct ? if not please show me where i went wrong
 
Check the units. cm2 ?
 
oh , i need metres squared
 
Load/Area= 64000N/0.074m= 864864.8649 Pa > 864.864 kPa
F/S = Ultimate stress/ Allowable Stress = 550 MPa/ 864.864 kPa > 550000 kPa/ 864.864kPa = 635.938


is (635.938 ) my answer ? (factor of safety) ??
 
walleye3 said:
Load/Area= 64000N/0.074m= 864864.8649 Pa > 864.864 kPa
F/S = Ultimate stress/ Allowable Stress = 550 MPa/ 864.864 kPa > 550000 kPa/ 864.864kPa = 635.938


is (635.938 ) my answer ? (factor of safety) ??



1 cm2 = 0.0001 m2
 
  • #10
1.0 cm2= ?? m2
 
  • #11
1 cm = 0.01 m ...
 
  • #12
okay thank you... i see my mistake
 
  • #13
64000 N/ 0.00074m2 = 86486486.49 Pa > 86486.48649 kPa
F/S Ultimate stress/Allowable Stress = 550 MPa/86486.48649 > 550000kPa /86486.48649= 6.359
 
  • #14
Allowable / Actual
 

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