Shear design in a wood beam

[Buleeek]

Hello everyone,

I would like to get a clarification on shear design in a wood beam. I know that per NDS2015 it should be calculated "d" from the support and portion of the uniform load beyond that should be omitted. Does it mean that the distance is calculated from the interior face ? Attached is an example of simply supported beam with uniform load. Can someone explain to me what is the actual value of the force V taken to calculate the stress ?

Thank you,

 

[dik]

Yes, calculated d from the support... this is a heads up incase you have point loads within that distance... The shear stress is used to calculate horizontal shear (parallel to the grain for splitting) at that location. It is the shear divided by the rectangular area (for rectangular beams/joists) multiplied by a factor of 1.5. The 1.5 increases the average shear stress to the maximum for a parabolic stress distribution.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[Buleeek]

dik,

What is the shear force value then?

 

[CivilSigma]

Draw the shear force diagram along the entire span, and if your support is 6" from the edge of the beam, find the shear value there.
An online calculator can help you with that

 

[Buleeek]

It is still unclear to me whether "d" should be calculated from the edge of the BEAM or the interior edge of the support. I found two sources showing different approach and that's why I'm confused. NDS2015 3.4.3.1(a) states "...uniformly distributed loads within a distance from SUPPORTS (?) equal to the depth of the bending member, d, shall be permitted to be ignored".

Thanks,

 

[CivilSigma]

It's a distance d from the face of the support.

See the third figure here:

https://www.quora.com/Why-is-shear-in-beams-checked-at-distance-D-from-the-face-support

Codes allow this because if I remember correctly, the shear force in the beam within the distance "d" is carried by the support member below.

 

[msquared48]

(10/2 - 7.25/12)50 x 1.5 = V

Mike McCann, PE, SE (WA, HI)

 

[dik]

This is the shear force, caused by the load at the (face of the beam plus the depth of the beam).


Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

That means that any load occurring between the face of support and the critical section for shear need not be checked for shear, but would need to be checked for bearing.

BA

 

[XR250]

You could also need to check bending. Depending on the magnitude of the point load and other loads, it may be an issue.

 

[Buleeek]

Thank you all,

I'm getting ready for my PE exam and I want to be sure I got it right.

 

[dik]

BART... I was taught that the depth was a bit of a flag to make sure you had no point loads in that area... not just a carte blanche to ignore loads in the area.

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

 

[BAretired]

dik,

If that were the case, why not just say the critical section for shear is the face of support?



BA