Shear Design Example - Reinforced Concrete Beams using ACI 318-19

This video presents a complete example problem for conducting the shear design of reinforced concrete beams according to ACI 318-19. First, the demands are computed using a shear envelope. Then the proper shear limits are checked to ensure the beam dimensions are allowable and that the proper stirrup spacing is selected. Finally stirrup spacing is computed along the length of the beam to satisfy the shear demands.
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Chapters:
0:00 Introduction
1:16 Shear Envelope
5:50 Concrete Capacity Vc
7:40 Shear Limits
9:15 Minimum Shear Steel
11:09 Capacity with Min Steel
12:20 Required Stirrup Spacing
14:10 Solution

[@mujtabashanan3548]

I am shifting from the British standard BS8110 and the European EC2 to the American standards ACI, your videos have been very helpful, I would like to express my gratitude for your efforts.

[@StructuresProfH]

Thank you! I'm glad you've found these useful.

[@alfredjrtovar1499]

This video should be shown to every prospective engineering student, guarentee people will go a different route.

[@FEAscent]

The way you explain difficult topics easily is remarkable. Thank you very much for your time.

[@StructuresProfH]

You are most welcome!

[@hanxiao3246]

This is one of the best video for RC shear presented by one of the best structural engineering professors!

[@StructuresProfH]

Thanks, Han!

[@aboudeh21]

yeah exactly

[@agiti]

grate & sufficient explanation how to be economical & safe design metedology ❤

[@MahmoudSAnagy]

very good and clear. i hope to use si units code in separate lectures series that will make it easier for some people like me work outside usa.

[@robertsamuelcolladodejesus5475]

Thanks for the explanation

[@diegovalarezo6806]

great video, thanks for sharing.

[@kisolojulius235]

Well done bro

[@armandojosuegallonunez3178]

Wow!! thank you bro

[@yaochian_]

If considering SDC D E F was required ,in this situation how to calculate Vu for demand

[@civilengineeringwitharas4776]

Professor I hope you start a series of tutorials on Steel design with AISC. It will be much much better if you use a software side by side. Thanks in advance

[@StructuresProfH]

I was thinking about doing something like that, though perhaps not for a few months at least. Do you have any software recommendations?

[@civilengineeringwitharas4776]

@@StructuresProfH I would recommend to use Tekla Tedds for that purpose.

[@tk-wl7xq]

@@StructuresProfH sap2000?

[@michaeladams2878]

Great video but what is this number 0.0156 that you don't define?

[@StructuresProfH]

That’s the value of rho_w (the reinforcement ratio) where the two V_c equations are equal. For rho_w less than 0.0156, the 2*root(fc’) equation will be greater, and for rho_w greater than 0.0156, the second equation will be greater. This isn’t in the code, and it’s not necessary to know, but it is nice to keep in mind if you are trying to maximize V_c.

[@michaeladams2878]

@@StructuresProfH Thanks, that was very informative video!

[@michaelsabbagh2673]

Hello prof. Shall the size reduction factor be considered for pile cap design?

[@StructuresProfH]

It's a bit confusing, but it looks to me that ACI-318-19 does require you to use the size factor for pile cap design. Section 13.2.6.2 states that the size factor can be ignored for shallow foundations (specifically, "one-way shallow foundations, two-way isolated footings, two-way combined footings, and mat foundations"). It says nothing about deep foundations, which have their own section 13.4. The pile cap section 13.4.6.3 just sends you to 22.5 for shear capacity, with no mention of ignoring the size effect factor. I'm not sure if this was intended, but it seems that's what we have. If it is any consolation, it gets even worse if you look into plain concrete. Inexplicably, plain concrete does NOT have a size effect factor, even though size effects would (per theory) be a bigger concern for plain concrete sections compared to reinforced concrete sections. Because of this, the plain concrete provisions in Section 14.5.5 can give you HIGHER strength than the provisions in 22.5 when you have lightly reinforced sections. For example, say I have d = 24", reinforcement ratio of 0.005, and no shear reinforcement. Section 22.5.5 gives me 1.05*sqrt(f'c)*bw*d for the capacity, whereas for plain concrete I would get 1.33*sqrt(f'c)*bw*h... and h > d so I get another minor benefit to plain concrete. Even accounting for the different strength reduction factors (phi = 0.75 for shear but only 0.6 for plain concrete), the plain concrete comes out on top, leading to the perverse incentive whereby I can remove reinforcement and gain strength per the code! Don't do that, by the way.

[@abod--lr6kx]

Thank you. What is the minimum spacing for stirrups ?

[@StructuresProfH]

Maybe someone else in the comments can jump in - I couldn't find a code provision specifically for stirrups (weird?) even though this is clearly defined for ties, spirals, etc., in Section 25.7 of ACI 318-19. Anyway, I would maintain a minimum clear spacing of at least 1 inch, which is consistent with the provisions for ties and spirals. You need to make sure the concrete can flow and fill around the reinforcement.

[@abod--lr6kx]

@@StructuresProfH Thank you professor, you are the best 👍

[@alejandroc.deltoro7440]

I looked into the ACI code but still don't understand what Nu is, and why in this problem equals to 0 🤔. Could anyone explain that?

[@StructuresProfH]

Nu is the axial force in the beam when subjected to the factored shear Vu. For the example in this video, the beam has no axial force, meaning Nu = 0. It's rather odd that ACI 318-19 uses two different symbols for axial load. Normally, Pu is the factored axial load, but for some select shear and torsion equations, the code uses Nu. It's a subtle distinction. Definitions for these are in Chapter 2 of ACI 318-19 as follows: Nu = factored axial force normal to cross section occurring simultaneously with Vu or Tu Vu = factored axial force

[@ibrahimelbaz6680]

Thank you Professor. In step 5 you multiply Vn = Vc + Vs by phi. Why? In Eq 22 . 5 . 1 . 1 in the code, there is no phi

[@StructuresProfH]

Chapter 22 in the code always gives the nominal strength. The strength reduction factor is called out in the design strength equations given in the respective member chapters. For example, Section 9.5.1.1 for beams states that phi*Vn must be greater than or equal to Vu, where the phi factors are defined in Chapter 21. Similar equations for the "Design Strength" are found in Section X.5 for other members, where X is the chapters 7 through 14. It's a bit repetitive and convoluted, but it is all there.

[@ibrahimelbaz6680]

@StructuresProfH Thank you professor 🙏

[@agiti]

it will be better if u put some metric lectures too for Africans like me

[@alfredjrtovar1499]

Im officallly changing my major, this shit is extremely boring!!!!! Math is fun, but this is just plain cruel