extra bracing for autocross
#21
damn, not done yet, haha... I cant weld the whole length in the back because its not touching the floor pan all the time, even when its in the front sub frame it raises it up a bit. Near the bulkhead area there is about an inch of space between the floor pan and the beam so I'm going to slice the carpet so it goes around the braces and inbetween the beam and the floorpan and then back on top under the seat so it looks nice. These theoretically should be ~8x stronger in the then normal flsfc's too.
#23
Also, how did you know where to start cutting into the floor at the front of the car?
Keep the pics coming.
#24
The moment of inertia is greater then 2x2 beam is 2 2 inch high hiehgts and 2 2 inch bases, with two 2x1 beams you have the extra heights of 2 inches in the center. So the equation for the inertia used in the bending beam equation is 1/12 b h^3 so with 4 2 inch heights verses 2 2 inch heights (per side) the numbers increase quiet a bit.
as for normal flsfc they are normally 1 inch high by 2 inches wide, so 1^3 verses 2 ^3 is about 8 times the resistance force of bending. I haven't taken twisting into account however but that would be higher to as well as the forces are greatest on the outside of the centroid (two inner heights are not as important rather the shape of the 2x2 verses the 1x2 lower sfc)
this is the equation I used:
I'm terrible at explaining engineering sometimes sorry if its confusing.
(I'm a mechanical engineer)
as for normal flsfc they are normally 1 inch high by 2 inches wide, so 1^3 verses 2 ^3 is about 8 times the resistance force of bending. I haven't taken twisting into account however but that would be higher to as well as the forces are greatest on the outside of the centroid (two inner heights are not as important rather the shape of the 2x2 verses the 1x2 lower sfc)
this is the equation I used:
I'm terrible at explaining engineering sometimes sorry if its confusing.
(I'm a mechanical engineer)
Last edited by stangalator; 11-04-2010 at 01:47 PM.
#25
Nope, that made sense...I'm an ME as well and was just wondering how you came to get those numbers.
But I wonder how placing the bars where you have versus under the car plays a role. It would be neat to have a test done with strain gauges on critical points of the car, one equipped with connectors in the typical location under the car with say MM SFC and placing the same size bars inside the car. To put some hard numbers to the theory and then extrapolate that into a precent increse of strength or percent decrese in twist.
Obviously test them on the same course at the same speeds and see what flexes less.
But I wonder how placing the bars where you have versus under the car plays a role. It would be neat to have a test done with strain gauges on critical points of the car, one equipped with connectors in the typical location under the car with say MM SFC and placing the same size bars inside the car. To put some hard numbers to the theory and then extrapolate that into a precent increse of strength or percent decrese in twist.
Obviously test them on the same course at the same speeds and see what flexes less.
#26
i just finished tonight, looks pretty good! ill get some pictures tomorrow since my camera died on me. I did the one jack stand test and its basically flat.
I was thinking about it and with both bars you have a much better resistance to bending then before since you move the centroid up. The upper and lower have a little bit of space between them so it kind of acts like a short I beam.
Its raining here so i couldnt really get on it around corners but it feels much stiffer going over a speed bump unevenly. Ill test it tomorrow when its dry and get some pictures too.
I was thinking about it and with both bars you have a much better resistance to bending then before since you move the centroid up. The upper and lower have a little bit of space between them so it kind of acts like a short I beam.
Its raining here so i couldnt really get on it around corners but it feels much stiffer going over a speed bump unevenly. Ill test it tomorrow when its dry and get some pictures too.
#27
The moment of inertia is greater then 2x2 beam is 2 2 inch high hiehgts and 2 2 inch bases, with two 2x1 beams you have the extra heights of 2 inches in the center. So the equation for the inertia used in the bending beam equation is 1/12 b h^3 so with 4 2 inch heights verses 2 2 inch heights (per side) the numbers increase quiet a bit.
as for normal flsfc they are normally 1 inch high by 2 inches wide, so 1^3 verses 2 ^3 is about 8 times the resistance force of bending. I haven't taken twisting into account however but that would be higher to as well as the forces are greatest on the outside of the centroid (two inner heights are not as important rather the shape of the 2x2 verses the 1x2 lower sfc)
this is the equation I used:
I'm terrible at explaining engineering sometimes sorry if its confusing.
(I'm a mechanical engineer)
as for normal flsfc they are normally 1 inch high by 2 inches wide, so 1^3 verses 2 ^3 is about 8 times the resistance force of bending. I haven't taken twisting into account however but that would be higher to as well as the forces are greatest on the outside of the centroid (two inner heights are not as important rather the shape of the 2x2 verses the 1x2 lower sfc)
this is the equation I used:
I'm terrible at explaining engineering sometimes sorry if its confusing.
(I'm a mechanical engineer)
here's a real good explanation of this for anyone who is interested
Let us look at two boards to intuitively determine which will deflect more and why. If two boards with actual dimensions of 2 inches by 8 inches were laid side by side - one on the two inch side and the other on the eight inch side, the board which is supported on its 2" edge is considerably stiffer than that supported along its 8" edge. Both boards have the same cross-sectional area, but the area is distributed differently about the horizontal centroidal axis.
effect of orientation upon the moment of inertia of a board
Calculus is ordinarily used to find the moment of inertia of an irregular section. However, a simple formula has been derived for a rectangular section which will be the most important section for this course.
Ixx = (1/12 ) (b)(h3)
In which the value b is always taken to be the side parallel to the reference axis and h the height of the section. This is very important to note! If the wrong value is assumed for the value of b, the calculations will be totally wrong.
Moment of Inertia
The importance of the distribution of the area around its centroidal axis becomes clear when comparing the values of the moment of inertia of a number of typical beam configurations. All of the members shown below are 2" x 6"; in cross section, equal in length and equally loaded.
beam configurations and moments of inertia
BUILT-UP SECTIONS
It is often advantageous to combine a number of smaller members in order to create a beam or column of greater strength. The moment of inertia of such a built-up section is found by adding the moments of inertia of the component parts. This can be done< B> if and only if the moments of inertia of each component area are taken about a common axis and if and only if the resulting section acts as one unit.
UNDER NO OTHER CONDITION CAN THEY BE ADDED!
Last edited by MineralGrey; 11-05-2010 at 04:36 AM.
#29
In simple terms, yes lol. You are doubling the ability of a 2"x1" beam to resist bending. Which becomes greater (or less deflection{distance the beam bent from rest}, however you want to spin it) than a single 2"x2" beam.
Like he said before it's due to the added 2" high walls in the center of the now created 2"x2" beam. A 2"x1" beam has a resistance to bending of MORE than half a 2"x2" beam. For the simple fact that if it was half, putting two 2"x1" beams together would net you a 2"x2" and this experiment would be uselss and he should have gone with a 2"x2" from the start.
Plus he has the strength of the welds that are helping.
Like he said before it's due to the added 2" high walls in the center of the now created 2"x2" beam. A 2"x1" beam has a resistance to bending of MORE than half a 2"x2" beam. For the simple fact that if it was half, putting two 2"x1" beams together would net you a 2"x2" and this experiment would be uselss and he should have gone with a 2"x2" from the start.
Plus he has the strength of the welds that are helping.