Tributary lateral load, Mechanical Engineering

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:

555_Tributary lateral load.png

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.

Posted Date: 2/12/2013 6:48:02 AM | Location : United States







Related Discussions:- Tributary lateral load, Assignment Help, Ask Question on Tributary lateral load, Get Answer, Expert's Help, Tributary lateral load Discussions

Write discussion on Tributary lateral load
Your posts are moderated
Related Questions
Brea k Mean Effective Pressure (P bm ): Likewise, the brake mean effective pressure is given by   i p       = indicated power (kW) bp     = Break Powder (kW

The Ordinate Dimension Command We can use The Ordinate command to annotate co-ordinate points with X or Y values. This might be useful for setting-out on site plans.

What is Subsetting Subsetting : Vendors selected and implemented only portions of the whole of IGES, thus making exchange between two systems impossible without  prior agreemen

Relationship between two specific heat ? Sol:   dQ = dU + dW; for a perfect gas dQ at constant pressure dU at Constant volume; = mC v dT = mC v (T 2  - T 1 ) dW at

Illustrate the specification of PDDI (i)  The 3-D geometry of the product which is defined by a set of curves, surfaces and volumes. (ii) The topology which describes the se

(a) Illustrate governing Mechanism of Francis turbine. (b) Derive a mathematical expression for work done and efficiency of Francis runner. (c) Write differences between Kapl

Deadlock States Deadlock is a detrimental phenomenon in the shop floor; here part flow is inhibited because of unsuitable scheduling decisions made via the computer controller.

Find out the centroid of the shaded : Find out the centroid of the shaded area shown in Figure Solution Net area of shaded portion of Figure = The area of full

Submerged Arc Welding   Single pass filler welds upto (8.0 mm) maximum and groove welds made with a single pass or a single pass each side, may be made using an E7os-x elect

can we use electrical power chuck for job clamping in cnc lathe.what are the issues related