JRCEng
Mechanical
- Nov 7, 2019
- 6
Hello All,
New Member here, and I have seen a few of the other post about scissor lift on gratings but am still unclear on how to go about determining the load capacity of grating given a contact patch of about 4.5" (tire width) by 4" (tire length). The scissor lift is a Genie 1930 model, and weights 2748lbs, I'm assuming a concentrated load of 687 (one tire, which is not conservative, because tipping could occur, but even so it fails).
The grating is McNICHOLS (3/16 by 1 1/2 with a c-c distance of 1.1875"). Its Galvanized steel with an allowable design stress of 18,000psi, and Young's Modulus of 29E6 psi. The largest span that I measured was 76".
I utilized the Metal Bar Grating Engineering Design Manual found here
I used example 2 with the only changing being K ( I used 3.78947 whereas they used 10.1), there are only that many bars supporting load for the tire width given above 4.5".
Mw = FSw = 4796.08 lb in
C = 4Mw/L = 252.42 lbs (much less than 687lbs)
Dc = 0.398607"
For my load I just divided 687/252.42 because load/displacement linear relationship, and then multiplied Dc by this value to get
Dactual = 1.08485" (which is absurd)
I'm mechanical, not structural so I'm not sure if this math checks out, it makes intuitive sense to me, but any advice would be greatly appreciated. Thank you!
New Member here, and I have seen a few of the other post about scissor lift on gratings but am still unclear on how to go about determining the load capacity of grating given a contact patch of about 4.5" (tire width) by 4" (tire length). The scissor lift is a Genie 1930 model, and weights 2748lbs, I'm assuming a concentrated load of 687 (one tire, which is not conservative, because tipping could occur, but even so it fails).
The grating is McNICHOLS (3/16 by 1 1/2 with a c-c distance of 1.1875"). Its Galvanized steel with an allowable design stress of 18,000psi, and Young's Modulus of 29E6 psi. The largest span that I measured was 76".
I utilized the Metal Bar Grating Engineering Design Manual found here
I used example 2 with the only changing being K ( I used 3.78947 whereas they used 10.1), there are only that many bars supporting load for the tire width given above 4.5".
Mw = FSw = 4796.08 lb in
C = 4Mw/L = 252.42 lbs (much less than 687lbs)
Dc = 0.398607"
For my load I just divided 687/252.42 because load/displacement linear relationship, and then multiplied Dc by this value to get
Dactual = 1.08485" (which is absurd)
I'm mechanical, not structural so I'm not sure if this math checks out, it makes intuitive sense to me, but any advice would be greatly appreciated. Thank you!