jacktbg
Mechanical
- Jun 14, 2017
- 34
I'm trying to follow this to get a finger to the wind on whether I can apply a few thousand pound pointload (on a steel plate) onto my 7" reinforced slab...
All online calculators I can find have a minimum dowel size of 3/4". The actual dowel size and spacing is 5/8" and 16" spacing...
Where I'm coming up short is the bearing stress on dowels. When you calculate Beta, you use the equation B = (Kc*db/(4*Eb*ib))^(1/4)
My numbers are B=(1.5E^6 psi * .3125 in / (4*2.9E^6 psi * .0155 in^4))^(1/4)
Every example I can find has that work out as unitless, but how can it be perfectly unitless with the in^4 in the denominator??
When I then take what I get (and assume I just calculate the number and calll it unitless) and put it into the fd(actual) calculation I get
fd actual = kc*Pc*(2+Bz)/(4B^3*Eb*Ib)
fd actual = 1.5e^6 psi * 731.78 lbs * (2+.1204*.125 in)/(4*.1204^3*2.9e^6 psi * .0155 in^4)
My number ends up astronomical and the units are nonsensical. Can anyone help me figure out where I'm going wrong???
Thanks a million!
All online calculators I can find have a minimum dowel size of 3/4". The actual dowel size and spacing is 5/8" and 16" spacing...
Where I'm coming up short is the bearing stress on dowels. When you calculate Beta, you use the equation B = (Kc*db/(4*Eb*ib))^(1/4)
My numbers are B=(1.5E^6 psi * .3125 in / (4*2.9E^6 psi * .0155 in^4))^(1/4)
Every example I can find has that work out as unitless, but how can it be perfectly unitless with the in^4 in the denominator??
When I then take what I get (and assume I just calculate the number and calll it unitless) and put it into the fd(actual) calculation I get
fd actual = kc*Pc*(2+Bz)/(4B^3*Eb*Ib)
fd actual = 1.5e^6 psi * 731.78 lbs * (2+.1204*.125 in)/(4*.1204^3*2.9e^6 psi * .0155 in^4)
My number ends up astronomical and the units are nonsensical. Can anyone help me figure out where I'm going wrong???
Thanks a million!