Compressive and Tensile Stresses of an 'I' Beam

[123321]

Homework Statement

A symmetric I section beam is 60 mm deep with a second moment of area of 663x10-9 m^4 and a cross sectional area of 1600 mm^2. It is subject to a bending moment of 1.2 kN.m and an axial force of 25 kN (tension). Find the maximum tensile and compressive stresses.

Homework Equations

σ=My/I

The Attempt at a Solution

So everything is given except the 'y'. I am unsure how to work this out as I am only given the depth.

Also could someone explain the relevance of the given axial force?

 

[PhanthomJay]

Homework Statement

A symmetric I section beam is 60 mm deep with a second moment of area of 663x10-9 m^4 and a cross sectional area of 1600 mm^2. It is subject to a bending moment of 1.2 kN.m and an axial force of 25 kN (tension). Find the maximum tensile and compressive stresses.

Homework Equations

σ=My/I

The Attempt at a Solution

So everything is given except the 'y'. I am unsure how to work this out as I am only given the depth.

Also could someone explain the relevance of the given axial force?

The 'y' distance is measured from the centroid of the cross section. For what value of y is the bending stress a maximum in tension? In compression? What is the formula for axial stresses? The stresses from each are algebraically additive.

 

[nvn]

123321: y is one half of the I-beam depth. The tensile axial force is used to compute the uniform axial stress, which is part of the stress on the I-beam.

 

[123321]

Thanks for the quick replies guys!

So from you's have said, am I right to say that the y for compression is 30 mm?

How about the y for tension? Will it also be 30 mm because it is symmetrical?

I've never heard of axial stress, is it the same as torsion?

I'm so confused...

 

[AlephZero]

I've never heard of axial stress, is it the same as torsion?

Didn't your course deal with the axial stress in a rod? A beam can also act like a rod.

You need to find the stress from the axial load (uniform tension) and the bending stress (tension and compression) separately, then combine the two.

 

[123321]

Didn't your course deal with the axial stress in a rod? A beam can also act like a rod.

You need to find the stress from the axial load (uniform tension) and the bending stress (tension and compression) separately, then combine the two.

Nope, never heard of it.

How do I find the two bending stresses?

 

[123321]

Can someone help please? I've been working on this question all day and been getting nowhere...

 

[nvn]

123321: Hint 1: Try the equation listed in post 1. Hint 2: Look for a uniform axial stress equation in your textbook.

 

[123321]

Yeah, I'm still confused about the the two y's, for compression and tension. Can anyone explain?

 

[123321]

123321: Hint 1: Try the equation listed in post 1. Hint 2: Look for a uniform axial stress equation in your textbook.

I don't have a book and nothing's coming up on Google. :(

 

[nvn]

Hint 3: Regarding post 9, use plus and minus y, to obtain the two bending stresses.

 

[123321]

Hint 3: Regarding post 9, use plus and minus y, to obtain the two bending stresses.

Are they both 30 mm because it is a symmetrical beam? But ones -ve and the other +ve?

 

[nvn]

Yes.

 

[MarleyDH]

@123321,

A few tips:

1. Draw the section out, identify its Neutral Axis & Centroid. A neutral axis is the plane or axis through a section at which zero strain occurs. (At least in Euler Beam Theory, which you are dealing with at the moment.)

2. Grab a ruler or something and bend it. Note the behaviour. What happens at the top and what happens at the bottom? To make it easier for you to see it, place two dots about 2 cm apart from each other on the top and bottom. Apply the bending. What happens to the dots? This should tell you what stresses are in that portion of the "beam."

3. The symbol y in that equation (which is known as the bending equation) is a 'distance' variable. It starts at the neutral axis of the beam, where it is zero and 'travels' to where you want to investigate the stresses above or below the neutral axis.

 

[123321]

@123321,

A few tips:

1. Draw the section out, identify its Neutral Axis & Centroid. A neutral axis is the plane or axis through a section at which zero strain occurs. (At least in Euler Beam Theory, which you are dealing with at the moment.)

2. Grab a ruler or something and bend it. Note the behaviour. What happens at the top and what happens at the bottom? To make it easier for you to see it, place two dots about 2 cm apart from each other on the top and bottom. Apply the bending. What happens to the dots? This should tell you what stresses are in that portion of the "beam."

3. The symbol y in that equation (which is known as the bending equation) is a 'distance' variable. It starts at the neutral axis of the beam, where it is zero and 'travels' to where you want to investigate the stresses above or below the neutral axis.

That rings a bell.

Thanks

 

[123321]

So I've found the max tensile and compressive stresses to be ±54.3 MPa

Any clues as to which one is +ve and which one is -ve?

 

[MarleyDH]

So I've found the max tensile and compressive stresses to be ±54.3 MPa

Any clues as to which one is +ve and which one is -ve?

That is purely your choice, just remember which one you made positive and which one you made negative. However, by convention tensile stresses are usually taken as +ve.

Just remember stress is taken as a vector (though this is not 100% correct) and therefore it has a magnitude and direction. (σ = F/A, the bold symbols are vectors.)

 

[123321]

That is purely your choice, just remember which one you made positive and which one you made negative. However, by convention tensile stresses are usually taken as +ve.

Just remember stress is taken as a vector (though this is not 100% correct) and therefore it has a magnitude and direction. (σ = F/A, the bold symbols are vectors.)

OK, thanks

Do you know how I should deal with the axial force? Someone previously mentioned that I should deal with the compressive/tensile stress and axial force separately and then combine at the end. I'm confused by this suggestion.

 

[MarleyDH]

Oh I know how to deal with it, but you need to figure that one out. ;-)

Here are a couple of questions that you should ask yourself:

Q: What types of forces do you get? (Hint: There are 3, but 2 have the word "force" attached to them where the other doesn't.)

Q: What type of stresses do these forces produce? (Hint: There are only 2 types of stresses)

Q: What are the units of stress? (Hint: What is MPa [Megapascals] in SI units? This one should answer your question on how to deal with the axial force.)

Someone previously mentioned that I should deal with the compressive/tensile stress and axial force separately and then combine at the end.

This is correct. Stresses are vectors and therefore vector algebra applies.

 

[123321]

Oh I know how to deal with it, but you need to figure that one out. ;-)

Here are a couple of questions that you should ask yourself:

Q: What types of forces do you get? (Hint: There are 3, but 2 have the word "force" attached to them where the other doesn't.)

Q: What type of stresses do these forces produce? (Hint: There are only 2 types of stresses)

Q: What are the units of stress? (Hint: What is MPa [Megapascals] in SI units? This one should answer your question on how to deal with the axial force.)

1. Normal force, shear force and torsion

2. Normal and shear stress

3. MPa = MN/m^2 so do I divide axial force by the cross-sectional? Then how do I combine it with the tensile/compressive stresses?

 

[MarleyDH]

1. Normal force, shear force and torsion

2. Normal and shear stress

3. MPa = MN/m^2 so do I divide axial force by the cross-sectional? Then how do I combine it with the tensile/compressive stresses?

You're getting there. ;-)

1. Axial Force, Shear Force and Moment. (Torsion is a moment about the axis of a member)

2. Shear Stress yes, but normal not so much. The word is Direct Stresses.

3. Yep, but typically its represented as N/mm^2. Yes you do divide the axial force by the cross-sectional area, but you have to note what the axial force is doing. For example, is it putting the member into tension or compression?

Another point to think about with regards to the axial force is:

You're dividing the force by the ENTIRE cross-sectional area, by doing this you are implicitly assuming that the stress is uniform over the cross-section.

Whereas with the bending formula, the "y" kept changing depending upon where you wanted to investigate the stress.

If you were to draw the stress distribution for the stresses due to bending how would it look? (Hint: Plot it on a cartesian plane, with the y-axis representing the distance from the neutral axis and the x-axis representing the stress.)

From what you've said: Stress = Force/Area. Does the value of the stress depend on a distance variable? No. Therefore its uniform. If its uniform, its unchanging and that means it can only take on a rectangular shape.

So then you simply add the two diagrams together and you see what get.

 

[pongo38]

Does the equation stress=N/A ring a bell? Where could it apply here?