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Tributary lateral load: The question is on a fictitious 1965 building in Albany, NY. It was designed using the 1964 State Building Construction Code, in which Allowable Stress Design was used.
This flat-roofed building has three stories: a 15 ft. tall ground story and two 12.5 ft. upper stories. A typical floor plan is shown in Figure 1a. The floors can be assumed to be rigid diaphragms, which is very important when determining force distribution. An elevation of an interior frame in the narrower, N-S direction is shown in Figure 1. The Wind Loads from the 1964 code are as follows:
At height above grade (feet) Exterior Walls (psf) >500 to 600 34 >400 to 500 33 >300 to 400 32 >200 to 300 30 >100 to 200 28 >60 to 100 24 >40 to 60 21 >25 to 40 18 >15 to 25 15 0 to 15 12
Use exterior windward forces only. That is, assume that leeward and sidewall pressures = 0 psf, all interior pressures = 0 psf, and all roof pressures = 0 psf. Within each story, half of the wind pressure is resisted by the upper floor and half of the wind pressure is resisted by the lower floor (which is the ground for the ground story). What is the North-South diaphragm wind shear (in pounds per lineal foot, plf) that acts at the top of the ground story at elevation = 15'-0"? Calculate the service load.
Step s for Solving Problem for Finding Maximum Power: 1. Use formula stress (σ) = force (Maximum Tension)/Area Where; Area = b.t that is, T max = σ.b.t 2. Unit mass (
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