Question

General Information (Problems 2–4) The envelope of a four-story steel building (risk category II) is shown below. Resistance to lateral loads (wind and earthquake) in the direction shown is provided by a steel braced frame on the interior of the building. The gravity load system, floor and roof diaphragms, and the lateral load system for the orthogonal direction are omitted from the figure. GRAVITY BF Roof 15 ft - - - - - - 15 ft WIND 3 15 ft 15 ft EARTHQUAKE 108 ft 36 ft The uniform dead (D) and live (L) loads for the building are as follows: • Floors – qd = 90 psf, qu = 80 psf • Roof - 90 = 40 psf, qu = 25 psf

Answer 1

the value for the questions are as follows:

height of the building = 60 ft

wind speed = 140 mph

Kzt = 1

Kd = 0.85

exposure catagory = C

gust factor G = 0.85

external coefficient pressure Cp = 0.8 ( windward wall )

external coefficient pressure Cp = 0.25 ( leeward wall )

design wind pressure = p = q G Cp −qi (G Cpi) ( from ASCE 7-10)

since internal pressures need not to be evaluated, G Cpi = 0 ( given in question)

design wind pressure = p = q G Cp

q=0.00256 Kz Kzt Kd V2    ( from ASCE 7-10)

Kz = 1.13 (for exposure catogory C at height 60 ft from ASCE 7-10)

q = 0.00256 x 1.13 x 1 x 0.85 x 1402

q = 48.19 lb/ft2

design wind pressure for external coefficient pressure Cp = 0.8 ( windward wall ) ,

pw = 48.19 x 0.85x 0.8 = 32.76 lb/ft2

design wind pressure for external coefficient pressure Cp = 0.25 ( leeward wall ) ,

pl = 48.19 x 0.85x 0.25 = 10.24 lb/ft2

therefore,

magnitude of axial force P due to wind load on windward wall = pw x A

where A is the area of windward wall exposed to wind

Pw = 32.76 x 36 x 60

Pw = 70761.6 lb

Pw = 70.76 kip

magnitude of axial force P due to wind load on leeward wall = pl x A

where A is the area of windward wall exposed to wind

Pl = 10.24 x 36 x 60

Pl = 22118.4 lb

Pl = 22.12 kip

axial load P due to wind load are 70.76 kip and 22.12 kip on windward and leeward wall respectively