Depends upon what kind of assembly it is. There are dozens of different load factors to select from, not including exit load factors.
For instance, a dance hall of that size might accommodate 370 people (factor of 7 square feet per person), but only if the main entrance also provides for emergency exit of 245 of those people (2/3), meaning it would need to have 2 doors that were both 3 feet wide. There are many other factors, depending upon type of assembly, what floor it's on, etc.
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That is 1000 x 2.6 x 1 = 2600 cubic feet. Sand weighs 100 pounds per cu ft, so it takes 100 x 2600 = 260,000 pounds, or 130 tons. Sand is deliverd by the cubic yard; that is about 100 yards of sand
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This question makes no sense. A 90 lb load per sq. ft. refers to an area. To calculate the stresses on beams supporting an area you have to know the spacing between adjacent beams as well as the span. p.s. This qn when fixed belongs in mechanical engineering.
That depends on the type of lighting . Fluorscent lighting for example , will give of little heat and fillament with generate far more .
It depends upon whether or not the wall is load bearing. If load bearing, the distance between standard studs is nominally 16 inches. If not load bearing, it can be 24 inches. There may be complications at the ends of the wall and doors and windows will also increase the number of 2x4's that are needed. A good rule of thumb is to figure one per foot. So in your case that would be 16.