Preliminary Area of Steel 1

[ronjon322]

I'm reviewing a set of calcs for a precast vault in which the designer is using the shortcut formula to calculate the area of steel required in the walls. However instead of using the typical As=Mu/4d with factored loads he is using As=Mu/1.76d with unfactored loads. Where is the 1.76 coming from? Has anyone seen this before and can shed some light on it for me? I'm assuming it's from an older ASD or even WSD method.

Appreciate ya

Measure twice, cut once.

 

[hokie66]

I am not familiar with those estimating tools, but vaults typically need more reinforcement than required structurally, for security reasons.

 

[Ingenuity]

ronjon322:

Not sure about 1.76. Seems very close to old masonry flexural design: A_s ≅ M_s/1.8d ...where M_s is a service-level (allowable/working) moment.


hokie66:

Maybe the vault is a burial vault...all DEAD load ...no chance of the 'occupant' escaping!

 

[JStephen]

I assumed the vault was a utility vault (fairly common), not a bank vault or burial vault.
Sometimes, it's worthwhile just to ask the person that wrote the equation, tho.

 

[PEinc]

For ASD, As = (M x 12)/(fs x j x d) = M / (1.76d)
M = ft-kips (with unfactored service loads)
fs = 24 ksi (allowable tension)
j ~ 0.88 for f'c ~ 3250 to 4000 psi
d = dimension from compressive face of concrete to center of tension steel.

Reference: Foundation Analysis and Design, Joseph E. Bowles, 1968, McGraw-Hill, page 160.

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