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RE: Torsion in Masonry Beam

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On a recent project, I had a 22 foot, single span, curved steel tube lintel. The tricky part was fastening the ends to take the torsion out. Fortunately I had a cross wall at each end that I used to hide a 4 foot length of tube (with holdown anchor bolts) in. These end members were then fully welded at right angles to the curved lintel to resist the torsion.

Just another idea......there are many ways to solve a problem.

Jim K.

-----Original Message-----
From: richard lewis [mailto:rlewistx(--nospam--at)juno.com]
Sent: Friday, April 16, 2004 11:13 AM
To: seaint(--nospam--at)seaint.org
Subject: Re: Torsion in Masonry Beam


A precast lintel wouldn't work here.  It needs to be cast in place.  The
CMU wall is on a radius.  I was trying to eliminate a trade by keeping it
masonry.  If I make the mason build to lintel height, wait for a concrete
crew to form and pour a lintel and then come back, it seems to me to be
added cost.  if I can get the CMU to take the torsion then the one trade
can just go wild on the wall until it's done.

Rich

On Thu, 15 Apr 2004 12:20:14 -0400 "Eric Ober" <eober(--nospam--at)holbertapple.com>
writes:
> Rather than 'pushing the envelope' of the masonry code, why not use 
> a
> precast lintel here, reinforced as appropriate, and figure out an
> appropriate bearing anchorage at the ends to resist the torsion?
> 
> Eric Ober
> 
> -----Original Message-----
> From: richard lewis [mailto:rlewistx(--nospam--at)juno.com] 
> Sent: Thursday, April 15, 2004 10:01 AM
> To: seaint(--nospam--at)seaint.org
> Subject: Torsion in Masonry Beam
> 
> I am designing a CMU lintel across an opening.  The opening is about 
> 11
> feet wide.  The wall is curved, putting the lintel into torsion 
> along
> with the typical bending and shear stress.  I am looking through the 
> ACI
> 530.1 code and they don't address torsion specifically.  I know 
> torsion
> is just shear stress so I was thinking of calculating the shear 
> stress
> from torsion and adding it to the beam shear and designing it that 
> way. 
> I pulled out my old "Beers and Johnston" strength text and looked 
> up
> torsion on a noncircular section.  The shear stress equation is
> stress=T/(C1*a*b^2).
> 
> My question is how to address this with the ACI code.  I am going to 
> put
> in stirrups for shear.  The above noted equation is for the maximum
> stress at the outer fiber.  The code equation is an average stress 
> over
> the shear area.  It states that Fv shall be less than 3*sqrt(f'm).  
> 
> 
> 1. Since the shear stress in torsion calculated above is for the 
> maximum
> stress, how should I apply it with the ACI maximum limit?  In order 
> to
> size the stirrups?
> 
> 2.  ACI states As=V*s/(Fs*d).  How should I combine the shear stress 
> and
> the beam shear to get a total 'V' for the shear stirrup equation?
> 
> Thanks for your insight!
> 
> Rich
> 
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