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Re: Beam design

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Santosh:

This is how I feel about it.

From basic equilibrium:

Case-1: Moment (Mu) + Axial Tension (Nu)
First assume that the axial force is tensile. Shift your axial force to the center of steel reabars. Now you have moved the tensile force, Nu, by (d - h/2) and have introduced a moment equal to Nu * (d-h/2), acting in the same direction as the external moment. To correct for the additional moment, subtract this moment from your actual design moment and you have a statically equivalent system. In this equivalent system, Nu is acting at the center of rebars and the applied moment has been reduced by Nu * (d-h/2). Design for the reduced moment and add additional reinforcement required for Nu.

Mus = Mu - Nu (d-h/2)
As1 = Reinforcement required for Mus
As2 = Nu / (Phi x fy)
As = As1 + As2
Provide As in the beam.

Case-2: Moment (Mu) + Axial Compression (Nu)
Now assume that the axial force is compressive. Again, shift your axial force to the center of steel reabars. Like before, you have moved the compressive force, Nu, by (d - h/2) and have introduced a moment equal to Nu * (d-h/2), but acting in the direction opposite to the external moment. To correct for the additional moment, add this moment to your actual design moment and you have a statically equivalent system. In this equivalent system, Nu is acting at the center of rebars and, this time, the applied moment has been increased by Nu * (d-h/2). Design for the increased moment and subtract additional reinforcement required for Nu.

Mus = Mu + Nu (d-h/2)
As1 = Reinforcement required for Mus
As2 = Nu / (Phi x fy)
As = As1 - As2
Provide As in the beam.

where
h = Total depth of the beam section.
d = Effective depth of the beam section.
fy = Yield stress of rebars
Phi = Capacity reduction factor of ACI 318.

Keep your units consistent. If you are working in the customary American units and if the moment is in Kip-feet, Nu should be in Kips and (d-h/2) should be in feet, so that you get Mus in Kip-feet. When computing As2, if the yield stress in KSI and Nu is in Kip, you would get your rebar area directly in square inches.

As to considering the beam with axial force as a column, it is correct that the basic equilibrium and compatibility equations would apply all the same. But the capacity reduction factor of ACI 318 would be different for a beam. Except for this, it is the same.

And, Santosh, I do not think that a young person with an engineering degree should be able to do it right away; one has to learn it. So do not take to heart the discouraging comment of a colleague. <Smile>

I hope it helps.

Rizwan Mirza
Rizwan Mirza, Consulting Engineers
Lahore, Pakistan

----- Original Message ----- From: "Santhosh Kumar Yedidi" <sant527(--nospam--at)gmail.com>
To: <seaint(--nospam--at)seaint.org>
Sent: Monday, January 09, 2006 2:41 PM
Subject: Beam design


How to do beam design with axial forces acting.

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