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

[mtnmama]

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?

 

[PhanthomJay]

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/A_w).

 

[mtnmama]

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?

 

[pongo38]

Please show your working.

 

[mtnmama]

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.