Find Axial Force P: Stress Concentration Homework

[gv3]

Homework Statement

The stress distribution along the section of a bar is shown below. From this distribution, find the approximate axial force P in kip.

Homework Equations

δavg=P/A
A= (.6+.6)(.5)
δavg=30ksi[/B]

The Attempt at a Solution

P=(30ksi)(.6in²)=18kip

It says this answer is wrong but i don't see where i am going wrong here? .8in is the diameter of the whole so it is not considered when finding the area.[/B]

 

[Chestermiller]

Homework Statement

The stress distribution along the section of a bar is shown below. From this distribution, find the approximate axial force P in kip.
View attachment 203331

Homework Equations

δavg=P/A
A= (.6+.6)(.5)
δavg=30ksi[/B]

The Attempt at a Solution

P=(30ksi)(.6in²)=18kip

It says this answer is wrong but i don't see where i am going wrong here? .8in is the diameter of the whole so it is not considered when finding the area.[/B]

What answer is considered the correct answer?

 

[gv3]

What answer is considered the correct answer?

Not sure. I've still been unable to get the answer. I don't see any other way to approach this. My professor approached it the same way when he did a similar problem.

 

[Nidum]

This might be a silly . The average stress is actually a bit less than 30 ksi so your answer for force P is a little high . As near as can be judged from such a small diagram the average stress is actually about 28.7 ksi giving a value of force P of about 17.2 kip .

So maybe 17 kip is the 'correct' answer ?

 

[gv3]

This might be a silly . The average stress is actually a bit less than 30 ksi so your answer for force P is a little high . As near as can be judged from such a small diagram the average stress is actually about 28.7 ksi giving a value of force P of about 17.2 kip .

So maybe 17 kip is the 'correct' answer ?

The next question asked for the max stress/average stress and used 30ksi as he average stress which gave me the right answer. I am hesitant to try because i only have 2 attempts left.

 

[Mapes]

Have you been given an empirical equation such as the one in Table 6-1-2 here? In this case, by plugging in the numbers, the stress concentration factor would be 2.23, and the nominal stress would therefore be 16.12. But this is just an empirical fit, of course.