fyffe555

09-11-2004, 09:01 PM

Hi all, need some really basic physics help here.

I'm trying to do some really simple calculations on beam deflection for a new machine I'm doodling, maybe using igus plastic bearings on drill rod, and want to calculate the deflections based on rod diameter to determine best size/price.

All my text books are in storage, my maths and differential calculus is rusty, college was a long time ago and I can't find what I want on google so I thought I'd ask here. I want to calculate the deflection of a beam with fixed ends and a load a given but not equidistant distance from one end. Anyone have or point me to the equation for that?

For example, and it's been a while but I know that for a beam with fixed ends and centre loaded the deflection can be calculated as;

Deflection = (F*L^3)/(192*E*I)

Where;

F = Load

L = Length of beam

E = Youngs modulus for the material

I = Area moment of inertia

So for example a tool steel rod of 36" length and 1" dia with a 100lb load would give; Def = (100*36^3)/( 192*28,000,000*0.049) = 0.0177 inches

I seem to remember that for an intermediate load on a beam with fixed ends the equation included some content for the distance from one end but I can't figure it out and its bugging me.

For what it's worth;

A fixed end cantilever with load at end; Def=(F x L^3)/(3*E*I)

A simple supported beam with centre load Def=(F*L^3)/(48*e*I)

It's worth noting the value of firmly locating the ends of unsupported rod for linear bearings. For the above example with Fixed, immovable end locations for the rod will give a deflection of 0.0177 inches. A simple, non fixed end location will allow the same beam with the same load to deflect 0.071 inches. This would apply to a rotating beam such as a lead screw and shows why as ridged a location as possible for the ends helps reduce deflection.

So, any ideas on an intermediate load solution? thanks..

Andrew

I'm trying to do some really simple calculations on beam deflection for a new machine I'm doodling, maybe using igus plastic bearings on drill rod, and want to calculate the deflections based on rod diameter to determine best size/price.

All my text books are in storage, my maths and differential calculus is rusty, college was a long time ago and I can't find what I want on google so I thought I'd ask here. I want to calculate the deflection of a beam with fixed ends and a load a given but not equidistant distance from one end. Anyone have or point me to the equation for that?

For example, and it's been a while but I know that for a beam with fixed ends and centre loaded the deflection can be calculated as;

Deflection = (F*L^3)/(192*E*I)

Where;

F = Load

L = Length of beam

E = Youngs modulus for the material

I = Area moment of inertia

So for example a tool steel rod of 36" length and 1" dia with a 100lb load would give; Def = (100*36^3)/( 192*28,000,000*0.049) = 0.0177 inches

I seem to remember that for an intermediate load on a beam with fixed ends the equation included some content for the distance from one end but I can't figure it out and its bugging me.

For what it's worth;

A fixed end cantilever with load at end; Def=(F x L^3)/(3*E*I)

A simple supported beam with centre load Def=(F*L^3)/(48*e*I)

It's worth noting the value of firmly locating the ends of unsupported rod for linear bearings. For the above example with Fixed, immovable end locations for the rod will give a deflection of 0.0177 inches. A simple, non fixed end location will allow the same beam with the same load to deflect 0.071 inches. This would apply to a rotating beam such as a lead screw and shows why as ridged a location as possible for the ends helps reduce deflection.

So, any ideas on an intermediate load solution? thanks..

Andrew