Thanks, I've done some analysis, but not confident in my conclusion. We are mainly a contract fabrication shop and don't design our own products. This hasn't been tested in the shop. The mfg. manager suggested also to add some gusset plates at the mitred corners...
Thanks Desertfox. I don't like the lug connection either. More than likely the mitred corners is a groove weld and not a fillet weld. I am not familiar with using the strain energy methods to get deflection.
I'm going to try to clarify this again. I've uploaded a sketch of a lift device the shop wants to rate for use. It was mostly constructed of structrual tubing (round and square). 1.5 Sq with 0.125wall and 2"O.D. with 0.25" wall. The welds at B and C are 0.25 fillets welds. I don't the use of the...
A member of the shop had fabricated a lifting attachment from HSS tubing. If the part of the device are two sections mitered at 45 and welded together are these still considered two sections with the weld joint being a treated as a fixed support or one section in a FBD?
Customer drawing calls for minimum bend radius on 304 ASTM A666 stainless steel. ASTM specifications give a Bend Factor of 1 for Thicknesss>0.050 to <= 0.1874in. for Free Bending and Bend Factor of 3 for V-Block Bending. Is air bending regarded as "Free Bending"?
Compared the two formulae's again. If the bending stress were equal for both the simple supported and the fixed end, the simple supported beam would have half the load as the fixed end supported beam. But the deflection would be greater for the simple supported beam.
I've been reviewing tread ladder rung specifications and the companies are giving a safe allowable concentrated load calculated based on a simple beam calculation.
Wouldn't this over estimate the safe allowable load?
The flexible cable system has a pretension load requirement of 750 lbs. The load requirements are outlined by the manufacturer of the cable fall prevention equipment based on their own tests.
Our company is considering purchasing a flexible cable fall protection system. The manufacture requires the climbing structure be capable of supporting the loads imposed by the system. For calculation purposes the required bracket load may be assumed to be distributed evenly between the number...
I'm constructing a small rotating tool stand and planning to use two sets of tapered roller bearings (Timken L44643 2 bearings. The average loading would be less than 500 lbs. The nominal shaft size is 1.0 inch diameter, and the housing is 1.98 inch diameter. The housing would be stationary and...
Does anyone have an calculation to estimate the tonnage required to form a joggle or offset bend in sheet metal? Most texts will give an estimate between 3 - 6x the tonnage for air bending?
RPM of drive wheel = 0.283 rpm * 50.5" / 10" =~ 1.5 rpm
HP=(2.557)*(1.5)/(5252)=0.0007303 HP (1/1369 HP ?)
Not much horse power required at minumum, but like dimjim mentioned, whats the resistance of the two idlers? The 10" diameter idlers had bearings and little torque resistance.
I'm going to give a bit more information here:
Known quantities:
Cylinder weight: 1504 lbs.
Outside Diameter: 50.5 inches
Inside Diameter: 50.29 inches
Time to accelerate approximately 15 seconds (T)
Full speed rpm: 0.283 rpm (calculated from below)
- welding speed 45 ipm, perimeter of...
Thank you both for the information. I did pickup one of the available smart motion cheat sheets. Some of the information is unavailable for larger AC motor applications such as Load Inertia of the motor (at least in the catalogues I have in front of me).
I had recently fabricated a set of welding turning idlers for our weld shop using a set of 10 diameter wheels, 28 inches center to center. We are using it to turn 50" diameter tanks while welding. The tanks will be over 1500#, welding at a maximum of 45 ipm. I want to motorize the unit using a...
