CanMyWelderBendThisBarehanded.jpg

I am modeling a jig for a welding table.

The table is targeted to be 3/8" stainless plate with 5/8" holes in it, a 1.5" space, then a second identical plate.

I have some 2" thick blocks with 5/8" posts that pass through their corresponding holes on the top and bottom plate of the table.

I am trying to calculate the actual hole and post sizes I would really need to withstand my welder pushing against the blocks.

The attached image has a mockup of the scenario. I cannot find a bending/shear force calculation that matches my scenario online.

Any hints to point me in the best direction would be very helpful.

thank-you

I am modeling a jig for a welding table.

The table is targeted to be 3/8" stainless plate with 5/8" holes in it, a 1.5" space, then a second identical plate.

I have some 2" thick blocks with 5/8" posts that pass through their corresponding holes on the top and bottom plate of the table.

I am trying to calculate the actual hole and post sizes I would really need to withstand my welder pushing against the blocks.

The attached image has a mockup of the scenario. I cannot find a bending/shear force calculation that matches my scenario online.

Any hints to point me in the best direction would be very helpful.

thank-you

I am trying to calculate all the forces (in lbf) that the connecting rod adds to the engine which influences vibration.

I am dividing the rod in between the two ends into four sections and calculating for the middle point of each as if that point had all that sections weight.

I am calculating for every 15 degrees of crank rotation.

conrodforces.gif

Here’s the formulas I have but I’m not sure about the in-line force formula:

centrifugal force in lbf = (kg x (meters/sec velocity)^2 / arc radius in meters) x .225

in-line force in lbf = (kg x (meters/sec beginning velocity^2) / 2) x .74

I am smudging a bit by ignoring the curvature of movement for calculating the in-line force but I don’t care if it’s 10% off.

I think “inertia force” is the same as what I’m calling in-line force because to stop a moving object you have to exert a force equal to its inertia. Correct me if I’m wrong.

I am dividing the rod in between the two ends into four sections and calculating for the middle point of each as if that point had all that sections weight.

I am calculating for every 15 degrees of crank rotation.

conrodforces.gif

Here’s the formulas I have but I’m not sure about the in-line force formula:

centrifugal force in lbf = (kg x (meters/sec velocity)^2 / arc radius in meters) x .225

in-line force in lbf = (kg x (meters/sec beginning velocity^2) / 2) x .74

I am smudging a bit by ignoring the curvature of movement for calculating the in-line force but I don’t care if it’s 10% off.

I think “inertia force” is the same as what I’m calling in-line force because to stop a moving object you have to exert a force equal to its inertia. Correct me if I’m wrong.