ASCE 7 wind tall buildings
© L. Prieto-Portar - 2008 EGN EGN - - 5439 The Design of Tall Buildings 5439 The Design of Tall Buildings Lecture 10 Lecture 10 ASCE 7 ASCE 7 - - 02 Solved Problem #3: 02 Solved Problem #3: Analytical Method 2 for Tall Buildings Analytical Method 2 for Tall Buildings The building for this third example is a 14-story office building in Tampa. It is 100 feet wide and 200 feet long and 160 feet tall, including a 3 foot parapet. The structure is a reinforced concrete rigid frame, with each floor and the roof providing a diaphragm reaction to lateral loads. Each floor is 11 feet high, and is clad with two glazing panels 5’ –6” high and 5 feet wide (aluminum mullions are spaced 5 feet on-center). The glazing is resistant to wind-borne debris impact. For simplicity, the office building is assumed to be in an Exposure B. It is not an essential facility, nor will it be occupied by 300 people in a single area at one time. Therefore, it is a Category II (see Table 1-1 on the next slide). [...]... and #16 References 1 American Society of Civil Engineers, Publication ASCE 7- 02, “Minimum Design Loads for Buildings and Other Structures”, Washington DC, 2002; 2 W C Bracken PE, Wind Load Design”, Florida Engineering Society, Tallahassee, 20 07; 3 K.C Mehta, J.M Delahey, “Guide to the Use of the Wind Load Provisions of ASCE 7- 02” ASCE Press, Washington DC, 2003 Arizona cactus flower ...Cp for wind normal to the 100 ft wall The second value of Cp = -0.18 is a small value and is not used in this example The roof Cp with wind normal to the 100 ft face, h/L = 1 57/ 200 ~ 0.8 Notice that interpolation is needed in Figure 6-6 The external pressure at the roof at a distance from 0 to h/2 = 1 57 ft / 2 = 79 ft from the edge with the wind normal to the 200 ft face The... roof at a distance from 79 ft to 100 ft from the edge with the wind normal to the 200 ft face The external design pressure p is, p = qz GC p = ( 35.1 psf )( 0.83)( −0 .70 ) = −20.4 psf Comparison of pressures between the two orthogonal faces How do the pressures change if some of the windows are breached? Internal pressure coefficients GCpi p = qz GC p − qi ( GC pi ) If the windows are breached by... Positive internal pressure = (30.0 psf)(+0.55) = 16.5 psf Wind Force on the Roof Parapet The pressure on the parapet with MWFRS pp is, p p = q p GC pn = ( 35.4 psf )(1.8 ) for the windward parapet p p = q p GC pn = ( 35.4 psf )( −1.1) for the leeward parapet Since the parapet is 3 feet high, the force on the parapet in the MWFRS case is as follows, Fwindward = q p GC pn hwall = ( 35.4 psf )(1.8 )( 3 feet . The Design of Tall Buildings 5439 The Design of Tall Buildings Lecture 10 Lecture 10 ASCE 7 ASCE 7 - - 02 Solved Problem #3: 02 Solved Problem #3: Analytical Method 2 for Tall Buildings Analytical. Publication ASCE 7- 02, “Minimum Design Loads for Buildings and Other Structures”, Washington DC, 2002; 2. W. C. Bracken PE, Wind Load Design”, Florida Engineering Society, Tallahassee, 20 07; 3 Equation 6-4, ( ) ( ) ( ) ( ) ( )( )( ) ( )( ) ( ) min 1 1 .7 1 1 .7 3.4 0.25 0.83 0.925 0.925 0.83 1 1 .7 1 1 .7 3.4 0.25 3.4 ( 6.5.8.1) 0.6 1 57 94.2 ( 6.5.8.1) 30 ( 6 2) 0.30 ( 6 2) 33 Q z v z Q v z g