Determining the state of stress at a point

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Discussion Overview

The discussion focuses on determining the state of stress at a specific point on a cantilever beam, particularly at point A on the cross section at section a−a. Participants explore various aspects of stress analysis, including axial, bending, and shear stresses, as well as the necessary calculations involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest using the formula σ = P/A for axial stress but note that this is not applicable for lateral loads, which create bending and shear stresses.
  • There is a discussion about calculating shear stress using the formula τ = VQ / IT, with participants sharing their values for V, Q, and I, but expressing uncertainty about their calculations.
  • One participant requests assistance in calculating Q, indicating a struggle with understanding how to derive it correctly.
  • Another participant emphasizes the importance of measuring distances correctly for calculating Q, specifically the distance from the centroid of the area to the centroid of the entire beam.
  • Participants discuss the calculation of the bending moment (M) and the neutral axis location, with some expressing confusion about the correct formulas and units to use.
  • There is a back-and-forth regarding the correct values for y, M, and I, with participants confirming their calculations and seeking validation from others.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate both shear and bending stresses, but there is no consensus on the specific values and methods for calculating Q, M, and I, leading to ongoing confusion and debate.

Contextual Notes

Some calculations are presented in millimeters, while others suggest converting to meters for consistency. There are unresolved issues regarding the correct application of formulas and the interpretation of distances related to centroids.

Who May Find This Useful

This discussion may be useful for students and practitioners involved in structural engineering or mechanics, particularly those working on stress analysis in beams.

Bluestribute
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Homework Statement


Determine the state of stress at point A on the cross section at section a−a of the cantilever beam. Take P = 350kN .

Homework Equations


Um . . . σ = P/A ?

The Attempt at a Solution


Yeah I'm not quite sure. I thought it was P/A, using 350000N as "P" and 300000mm^2 as A, but alas, it wasn't. Because this section of the course uses the P/A definition, I don't know if I have to use an equation with moment of Inertia and things like that. But I'm not sure, really at all, where to start for something simple like this. I need like a kick in the right direction to have a "duh" moment.
And more duh moments to follow . . . because this section is confusing the hell out of me.
 

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Bluestribute said:

Homework Statement


Determine the state of stress at point A on the cross section at section a−a of the cantilever beam. Take P = 350kN .

Homework Equations


Um . . . σ = P/A ?

The Attempt at a Solution


Yeah I'm not quite sure. I thought it was P/A, using 350000N as "P" and 300000mm^2 as A, but alas, it wasn't. Because this section of the course uses the P/A definition, I don't know if I have to use an equation with moment of Inertia and things like that. But I'm not sure, really at all, where to start for something simple like this. I need like a kick in the right direction to have a "duh" moment.
And more duh moments to follow . . . because this section is confusing the hell out of me.

σ = P/A is good only for loads applied axially.

Here, the load P is applied laterally and sets up two kinds of stress: Bending stress and shear stress are created by the load P at point A in section a-a.

If you want to tackle this problem, I would recommend first that you determine the reactions at the fixed end of the beam and then construct the shear force and bending moment diagrams for the indicated load.
 
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So VQ/IT?

I have my V at 350kN.
My Q is at 3000000 (100mm * 100mm * 300mm*) <- I suck at Q calcs
My I equalled 225000000 (1/12 * 100 * 300^3)
T = . . . 100mm?

Not quite sure I have the right numbers . . . But the answer these gave was wrong
 
Bluestribute said:
So VQ/IT?

The shear stress τ = VQ / I t

I have my V at 350kN.
My Q is at 3000000 (100mm * 100mm * 300mm*) <- I suck at Q calcs
My I equalled 225000000 (1/12 * 100 * 300^3)
T = . . . 100mm?

Not quite sure I have the right numbers . . . But the answer these gave was wrong

It's better to calculate the section properties using meters rather than millimeters. I would adjust T and I so that the magnitudes are expressed in meters and m4, respectively, and calculate Q in m3.

Q is the first moment of the cross sectional area which extends from the outermost fiber to the horizontal line drawn thru point A. This moment must be calculated about the centroid of the cross section. Q is a maximum at the centroid of the cross section, and Q = 0 at the outermost fiber.

After calculating the shear stress, then the bending moment and the corresponding bending stress must be calculated. It helps to find the reactions at the fixed end and construct the shear force and bending moment diagrams.
 
Can you calculate Q for me (showing exactly which numbers you used). No matter what numbers I use, in any type of problem (not just this one), I can NEVER get the Q value that's required. And I know it's required because they say, "Q = ". Still can't get it.
 
Bluestribute said:
Can you calculate Q for me (showing exactly which numbers you used). No matter what numbers I use, in any type of problem (not just this one), I can NEVER get the Q value that's required. And I know it's required because they say, "Q = ". Still can't get it.

You've got an area which is 100 mm wide x 100 mm deep from the outer fiber. Call this area Ashear.

You should be able to determine the location for the centroid of this 100 mm x 100 mm area.

You want to calculate the first moment of the 100 mm x 100 mm area, measuring from the centroid of the whole section to the centroid of Ashear. Call this distance ybar.

The moment Q = Ashear * ybar

Please show your calculations for Q when you reply. :smile:
 
A = 100x100 = 10,000 mm^2
Y, for a square, is centered, so 50mm

10,000x50 = 500,000

That's why I can't calculate Q . . .
 
Bluestribute said:
A = 100x100 = 10,000 mm^2
Y, for a square, is centered, so 50mm

10,000x50 = 500,000

That's why I can't calculate Q . . .
You're not measuring ybar correctly. ybar must be measured from the centroid of the 100 mm x 100 mm area to the centroid of the whole beam. That distance is not 50 mm.
 
50mm + 0.5m/2, so 300mm for Ybar?
 
  • #10
Bluestribute said:
50mm + 0.5m/2, so 300mm for Ybar?

Sigh, no.

The centroid of the whole beam is 150 mm from the outer fiber.
The centroid of the 100 mm x 100 mm area is 50 mm from the outer fiber.
What's the difference in these two measurements?
 
  • #11
So 100mmx100mm under A.
A's Ybar is at 50mm
The entire cross section Ybar is at 150mm
So Y is 100mm? And A is 100x100? So Q = 100x100x100?

And if Q is 100^3, would the thickness be 100 (the base there)?
 
  • #12
Bluestribute said:
So 100mmx100mm under A.
A's Ybar is at 50mm
The entire cross section Ybar is at 150mm
So Y is 100mm? And A is 100x100? So Q = 100x100x100?

And if Q is 100^3, would the thickness be 100 (the base there)?
Yes.
 
  • #13
Ok, so that would give me τ, but what about σ?
 
  • #14
Bluestribute said:
Ok, so that would give me τ, but what about σ?
Do you know the formula for calculating σ ?
 
  • #15
My/I right? What's y though?
 
  • #16
Bluestribute said:
My/I right? What's y though?
y is the distance from the neutral axis of the beam to the point at which σ is calculated.

In this case, y is the distance from the neutral axis of the beam to point A. The neutral axis is also located at the centroid of the beam.

M is the bending moment at section a-a, which has yet to be calculated. I = b h3 / 12 for this cross section.
 
  • #17
y: 150 (ybar) - 100 (location of A) = 50

M: (1/2)P - 0.5*350kN = 175 kN

I: I'm actually decent enough at these. (100)(300)^3(1/12)

Just confirming I'm using the correct numbers/signs
 
  • #18
Bluestribute said:
y: 150 (ybar) - 100 (location of A) = 50

M: (1/2)P - 0.5*350kN = 175 kN

I: I'm actually decent enough at these. (100)(300)^3(1/12)

Just confirming I'm using the correct numbers/signs
You should always show units with the quantities you calculate.

Your calculation of M is suspect. How did you arrive at this formula for M?

(Note: when writing formulas, it's always better to show M = whatever. What you have written for M is a bit of a head-scratcher.)
 
  • #19
I just wrote M as force*distance.

ybar in mm
 
  • #20
Bluestribute said:
I just wrote M as force*distance.

ybar in mm

Bending moment has units of force * distance; in this case kN-m. In order to calculate the correct value for σ, both y and I will have to be converted to m and m4, repectively.
 
  • #21
Well conversions aside is that the right formula? I'm going to put these in N-mm for the required units of MPa, but if I'm not calculating using the right formulas, my conversions don't matter.
 
  • #22
Got it . . . but it was negative.
I'll probably have another very similar problem soon with an I beam. Unless I can transfer these correctly to an I beam
 

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