Why can maximum shear stress in a web be replaced by the average shear stress?

In summary, the average web shear stress (V/dt) is greater than the max shear stress (VQ/It) when using the web area only (V/Aw).
  • #1
mtnmama
3
0
I've been reading my strength of materials text about design of steel beams. I understand everything up to the point where the text tells me to use the average web shear stress (V/dt) instead of max shear stress (VQ/It) to determine if a given beam is within the allowable shear stress. I calculated both to determine how close they are and found the average greater than the max! I've checked several times and don't see an error. What am I missing?
 
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  • #2
mtnmama said:
I've been reading my strength of materials text about design of steel beams. I understand everything up to the point where the text tells me to use the average web shear stress (V/dt) instead of max shear stress (VQ/It) to determine if a given beam is within the allowable shear stress. I calculated both to determine how close they are and found the average greater than the max! I've checked several times and don't see an error. What am I missing?
Are you using a wide flange (I) beam? Max shear stress at the neutral axis (VQ/It) is approximately equal to the average shear stress when using the web area only (V/Aw).
 
Last edited:
  • #3
Yes, it is a wide flange. When I calculate an average and a max stress on the flange of a 8 ft section of W18 x 50 with a uniform load of 1.85 klb/in I get an average of 16.22 klb/sq in. The max I get is 6.56 klb/sq in. Is that considered close, or have I made an error somewhere?
 
  • #4
Please show your working.
 
  • #5
I found some errors in Q. What I get now is:

V = wL/2 = [1.85 klb/in * 8 ft *12 in/ft]/2 = 88.80 kips
I = 400 in^4 (table)
Q = sum of first moments of upper half of beam, ie the flange and half the web
= Q of flange + Q of 1/2 web
= A of flange x y to centroid + A of 1/2 web x y to centroid
= (7.495 in * .57) * 8.71 + (8.425*.355*4.2125)
= 37.21 + 12.60
= 49.81 in^3

Max stress = VQ/(It) = 88.80 * 49.81 / (800 * .355) = 15.57 ksi

Avg stress = V/A of web = 88.89/(.355 * 15.42) = 16.22 ksi

These two values are close enough to verify that the max stress can be approximated by avg stress. If you don't see any errors in my equations, I will continue to use the approximation with confidence.

Thank you.
 

1. Why is it important to understand the concept of maximum shear stress and average shear stress in a web?

The concept of maximum shear stress and average shear stress in a web is important because it helps in understanding the load-bearing capacity and structural integrity of the web. It also allows for the design and analysis of the web to ensure it can withstand the expected stresses and forces.

2. How does the maximum shear stress differ from the average shear stress in a web?

The maximum shear stress is the highest stress that occurs in a web, usually at its weakest point or where it experiences the most stress. On the other hand, the average shear stress is the average of all the shear stresses acting on the web, taking into account the distribution of stress across its entire surface.

3. Can the maximum shear stress be replaced by the average shear stress in all cases?

No, the maximum shear stress cannot be replaced by the average shear stress in all cases. In some cases, the maximum shear stress may be significantly higher than the average shear stress, indicating potential failure points in the web. It is important to consider both values in the design and analysis of a web.

4. What are the benefits of replacing the maximum shear stress with the average shear stress in a web?

Replacing the maximum shear stress with the average shear stress can provide a more conservative and accurate estimate of the web's load-bearing capacity. It also helps to distribute the stress more evenly across the entire web, reducing the likelihood of failure at specific points.

5. Are there any limitations to using the average shear stress in place of the maximum shear stress in a web?

Yes, there are limitations to using the average shear stress in place of the maximum shear stress in a web. The average shear stress may not accurately represent the stress distribution in complex or non-uniformly loaded webs. In these cases, it is important to consider the maximum shear stress to ensure the web can withstand the expected stresses.

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