
#1
Nov2713, 08:09 AM

P: 2

Hi, I found this forum while looking for a formula online and am hoping someone here can help me.
I'm having chairs built from sheet steel for a public art display and need to decide the thickness of sheet steel to use. The job shop is suggesting 1/4" steel (either A36 or stainless), but before I pay to have prototypes built I want to compare the estimated performance of 1/4" to 3/8" plate to see if I should use 3/8" instead. The chairs will be made from a single piece of laser cut steel, bent using a press brake into a form similar to that shown in the attached images, and secured to the floor. I want to calculate two figures: First, how much drop would there be at the back of the seat (due to the bend at the front of the seat) for a given maximum weight of say, 600 pounds, for either 1/4" or 3/8" steel. Second, for a given tolerance for the seat to drop at the back (such as a 1/4" drop between unweighted and weighted), what would be the maximum allowable weight? For these purposes I'm content to assume that all of the weight is placed about 80% of the distance from the seat front to seat back and, unless someone here suggests otherwise, for my current purposes I'm content to focus solely on the strain on the bend at the seat front and to ignore the strain on the bent support underneath the seat. The front bend will be the weak link. Finally, is there a formula to estimate the longevity of a steel bend? For example, I imagine that repeatedly placing 500 pounds on the seat would strain the front bend, and that the cumulative strain would eventually result in the back drop increasing to 1/2", and eventually failing altogether. Is there a way to estimate if the number of repeated 500 pound weightings before an additional drop of 1/4" would be closer to 10k, 100k or 1000k? All feedback and assistance is greatly appreciated. Thanks! Matt 



#2
Nov2713, 09:49 PM

Sci Advisor
HW Helper
P: 2,110

coldsquid: What is the exact width of the chair at the front bend?




#3
Nov2813, 01:51 AM

Sci Advisor
HW Helper
P: 2,110

coldsquid: I assumed the chair, at the seat front bend, is 460 mm wide.
(1) To answer your first question, with a 2670 N load applied at 488 mm from the seat front, for the 6.35 mm thick chair, the back of the seat deflects downward 103.23 mm. Or for the 9.525 mm thick chair, the back of the seat deflects downward 30.586 mm. However, the above results are invalid, because each chair is overstressed if you apply this load. (2) Therefore, let's move on to your second question. Using the 6.35 mm thick chair, for the back of the seat to deflect downward 6.35 mm, you would need to apply [(2670 N)/(103.23 mm)]*(6.35 mm) = 164.2 N, at 488 mm from the seat front. Or using the 9.525 mm thick chair, you would need to apply [(2670 N)/(30.586 mm)]*(6.35 mm)] = 554.3 N. And these item 2 results are valid, because each chair is not overstressed here in item 2. (3) I currently would say, you need to use a dynamic amplification factor of 1.50. Therefore, you must multiply a person's weight by 1.50, to obtain the applied load. Therefore, for the 6.35 mm thick chair, if you do not exceed a dynamic applied load of 1.50(700 N) = 1050 N, in which case the chair will deflect downward 40.60 mm, then I think the 6.35 mm thick chair might withstand something like 1e6 stress cycles (?). Likewise, for the 9.525 mm thick chair, if the person's weight does not exceed 1580 N, which means a dynamic applied load of 1.50(1580 N) = 2370 N, in which case the chair will deflect downward 27.15 mm, then I think the 9.525 mm thick chair might withstand something like 1e6 stress cycles (?). PS: Unlike what you imagined in your second to last paragraph, the chair elastic deflection generally will not increase with age; instead, it will remain constant with age. Then, at the end of its life span (say 1e6 stress cycles?), the chair will seemingly suddenly fracture, and collapse. 



#4
Nov3013, 06:06 AM

P: 2

Calculating requirements for a steel chair
Wow, thanks for the comprehensive answer, nvn. What unit is "N" in your formula? Kilos?




#5
Nov3013, 09:39 AM

Sci Advisor
HW Helper
P: 2,110

coldsquid: In standard units, the unit of force is newton (N). The unit name is lowercase newton; the unit symbol is uppercase N.



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