An Example Problem on Wind Load Calculation According to NSCP 2010

An example problem on wind load calculation according to NSCP 2010 ;) A 20-meter-high square-plan five-storey building with flat roof and 4m-high floors, located in Makati CBD, has sides of 10 meters length each, and a large open front door on the first floor that is 2m x 2m in dimension. Assuming that G = 0.85 and that torsion is negligible, 1. Show how this maybe is an open, partially enclosed, or enclosed building. 2. Determine the internal pressure coefficients. 3. Determine the external pressure coefficients for the design of main girders and shear walls. 4. Determine the base reactions due to along-wind loads acting on the front wall of the building. 1. The building satisfies all definitions of a partially enclosed building (NSCP 2010 Section 207.2). 2. The internal pressure coefficients for a partially enclosed building (GCpi) are +/- 0.55 (NSCP Figure 207-5). 3. The external pressure coefficients on MWFRS (from NSCP 2010 Figure 207-6) are as follows: - windward wall, Cp = 0.8 - leeward wall, Cp = -0.5 since L = 10m, B = 10m, and L/B = 1 - side walls, Cp = -0.7 - whole roof, Cp = -1.04 or -0.18 since h = 20m, h/L = 2, L <= h/2 = 10m, and Roof Area = 100 sq.m > 93 sq.m 4. The base reactions can be calculated after we calculate the design wind force at each level. However, taking x = along-wind direction, y = acrosswind direction, z = vertical direction, we already can deduce that Vy = 0, and Mx = 0. Additionally, Mz is given as zero. We only need to estimate Vx, Vz, and My. To calculate the design wind force at each level, we need to multiply net design wind pressures at each level with tributary areas. To get net design wind pressures, we calculate pressures on both windward and leeward faces. On each face, we need to calculate the net of external and internal pressures. To get external and internal pressures, we need first to calculate the velocity pressures at each level. To calculate by hand, it is easiest to do this in table form but with a computer, a spreadsheet makes it much easier: Assume: Exposure Terrain Category B Case 2, Iw = 1.0, Kd = 0.85, Kzt = 1.0, V = 200 kph Windward wall pz (kPa) Leeward wall pz (kPa) z (m) Kz qz (kPa) with +Gcpi with -Gcpi with +Gcpi with -Gcpi 20 0.88 1.42 0.18 1.75 -1.38 0.18 16 12 8 4 0 0.82 0.76 0.67 0.57 0.57 1.32 1.22 1.08 0.92 0.92 0.12 0.05 -0.05 -0.16 -0.16 1.68 1.61 1.52 1.41 1.41 -1.38 -1.38 -1.38 -1.38 -1.38 Net along wind pressures Net along wind loads Fz pz (kPa) (kN) Afz (sqm) with +Gcpi with -Gcpi with +Gcpi with -Gcpi 1.56 1.57 20 31 31 1.5 1.5 40 60 60 1.43 1.43 40 57 57 1.33 1.34 40 53 54 1.22 1.23 40 49 49 1.22 1.23 20 24 25 Vx (kN) = 274 276 My (kNm) = Af,roof (sqm) 100 0.18 0.18 0.18 0.18 0.18 Base bending moment contribution My,z (kNm) with +Gcpi with -Gcpi 620 620 960 960 684 684 424 432 196 196 0 0 2884 Roof loads 1, p Roof loads 2, p Vz = Roof Vz = Roof (kPa) (kPa) loads 1 (kN) loads 2 (kN) with with with with with with with with +Gcpi -Gcpi +Gcpi -Gcpi +Gcpi -Gcpi +Gcpi -Gcpi -2.04 -0.47 -1 0.56 -204 -47 -100 56 Vz (kN) = -204 56 2892 The first thing you need to know in calculating the wind load is the location of the structure or building. It will tell you the basic wind speed (V) that you’ll consider in your computation. Basic Wind Speed In table 207-1, the provinces or locations in the Philippines with its corresponding basic wind speed are shown. . Be mindful of certain popular spots in the country and their corresponding locations (province). the building is proposed at Boracay. grasslands. It depends mainly on the structure’s surrounding environment. Exposure D Flat. For example. Exposure Category The next step is to take its exposure category. or other terrain with numerous closely spaced obstruction having the size of single-family dwellings or larger. unobstructed areas and water surfaces. If you don’t know what province Boracay is located at. This category includes smooth mud flats and salt flats. wooden area. and all water surfaces in regions with records of extreme typhoons. Exposure C Open terrain with scattered obstruction having heights generally less than 9 meters. This category includes flat open country. Exposure B Urban and suburban area.Test/Exam Tip! The professor may only give you the location of the structure. then it will be a big problem for you as a designer. Importance Factor . Every structure has its importance factor (Iw) depending on its occupancy category. Refer to tables 207-3 (Iw) and 103-1 (Occupancy Category). . . .Wind Directionality Factor. K d This is solely based on what structural type/part you are designing. Main Wind Force Resisting Systems are structural elements that support large area exposed to the wind. Components and Cladding are the structural elements that support small areas exposed to the wind. the “Main Wind Force Resisting System (MWFRS)” and “Components and Claddings” are being considered since we are dealing mostly with buildings and warehouses. . Meanwhile.Test/Exam Tip! More often that not. K /K z h The velocity pressure exposure coefficients (Kz or Kh) goes larger as the height above the ground increases.Exposure Coefficient. consider two cases (Case 1 and Case 2). . When your structure lands in exposure B. All possible or applicable values of “height above ground level” should be taken into consideration. Test/Exam Tip! . 1.2 and 207. read 207. (used most of the time) For rigid buildings. L = Horizontal dimension of building measured normal/perpendicular to wind direction B = Horizontal dimension of building measured parallel to wind direction H = Mean Roof Height (Height from the ground to the middle part of the roof) .5. use G = 0.7. Topographic Factor. Pressure Coefficient The pressure coefficient (Cp) are based on the enclosure category of the structure and location on a structure for which a pressure is to be computed. NOTE: For more detailed computation of Kzt.5. Kzt is usually equals to 1. Click here if you don’t know how to interpolate. G For stiff buildings and structures.Interpolate the desired value if the height falls between two known values.85.1. K zt For simplicity purposes.8. read 207. Gust Factor.5.7. Leeward. use Cp=0.8 *Interpolate the desired coefficient if angle is between two known values. use eave height (height of the structure excluding the roof) For roof slopes greater than 80 degrees. Windward. Side Wall .*If angle is less than or equals to 10 degrees. Sign Conventions Positive – The pressure is towards the structure. Positive and Negative Internal Pressures . GCpi *Values of GCpi shall be used with either qz or qh (You’ll learn the use of this in the next part of this tutorial). Negative – The pressure is pointing away from the structure (suction). *Two values must always be considered: positive and negative. you must consider these two cases: .In computing the wind load of a structure. Wind Normal/Perpendicular to Ridge 2.3 x 10-6 Kz Kzt Kd V2 Iw – – – formula 2 Eave height = height of wall (excluding the roof) Mean roof height (h) = height of wall + (height of roof/2) Consider the following wind directions: 1. We’ll be talking about the working formulas and how to apply it. P = Wind pressure (this is what we want to get at the end) P = qGCp – q(GCpi) – – – formula 1 q = 47. Wind Parallel to Ridge Wind Normal/Perpendicular To Ridge . Take note that you MUST read part 1 and part 2 of this tutorial in order to fully understand this topic.Let’s now go straight to the computation process. Compute q for every applicable height (stated in table 207-4) including the mean roof height (h).Windward Wall P = qzGCp – qh(GCpi) P = qzGCp – qh(-GCpi) 1. the eave height of a warehouse is 6. In this case. q6.5 .q6. you’ll need compute for four values of q: q0-4.5 meters and its mean roof height is 13. .5. For example.5.5 and q13. 8. The value of Cp for windward wall is 0.18).2. 4.e.85 unless otherwise stated. 3. G is usually equals to 0.  Positive GCpi – Positive Internal Pressure . Compute two values of P considering the positive and negative values of GCpi (i. +0.18 and -0. 18). 3. qh.e.7. 2. .85 unless otherwise stated. G is usually equals to 0.  Positive GCpi – Positive Internal Pressure  Negative GCpi – Negative Internal Pressure Side Wall Same procedure with the computation for leeward wall but equate Cp to -0. +0.18 and -0. Negative GCpi – Negative Internal Pressure Leeward Wall P = qhGCp – qh(GCpi) P = qhGCp – qh(-GCpi) 1. Compute only 1 value of q considering the mean roof height. 4. The value of Cp for leeward wall will depend on the value of your L/B. Compute two values of P considering the positive and negative values of GCpi (i. For enclosed buildings P = qhGCp – qh(GCpi) P = qhGCp – qh(-GCpi) For partially enclosed buildings Positive Internal Pressure: P = qhGCp – qz(GCpi) .Roof Use the table above for the values of Cp for both windward and leeward. qz = level of the highest opening in the building that could affect the positive internal pressure. Calculate for the wind pressures at all applicable horizontal distances from the windward edge. Compute for qh. Negative Internal Pressure: P = qhGCp – qh(-GCpi) Wind Parallel To Ridge P = qhGCp – qh(GCpi) P = qhGCp – qh(-GCpi) 1. . 2. Wind pressures at the windward and leeward walls will be dealt the same way as above.1. The building is on flat terrain. Given: The enclosed office building shown in Figure 7. It means all the forces will be suction or pointing away from the structure. Figure 7.4. The building is located in a region with a wind speed (3-sec gust) of 120 mph. The exposure is Exposure C.1. As you have noticed. all the values are negative.3.1.1 Building Definition . 4.4. Wanted: The wind pressures applied to the surfaces and the net forces applied to the building. but first we will define a few parameters.1 ft . Solution: To solve this problem. we need to independently look at two different wind directions. Determine critical elevations:  Mean Roof Height: h = 2*11' + (3/12)*25'/2 = 25. 01(max(h.18 (ASCE 7-05 Figure 6-5)  qh = 29.7 psf  qhGCpi = + 5. qhGCpi:  GCpi = + 0.125 16. Buildings)  I = 1. G Kz 0.5 ft  Mean 1st Floor Height: h = 11'/2 = 5.85 (ASCE 7-05 Table 6-4.5 Compute the Internal Pressures.5 5.866 0.1 26.849 qz (psf) 29.946 0.7 27. Category II building)  Kz = varies with elevation = 2.00256 Kz Kzt Kd V2 I (ASCE 7-05 Equation 6-15)  Kzt = 1 (Flat Terrain)  Kd = . (ASCE 7-05 Tables 6-2 and 6-3) Roof 2nd flr 1st flr h (ft) 25.5).6 . Mean 2nd Floor Height: h = 11' + 11'/2 = 16.0 (ASCE 7-05 Table 6-1.15)/900)(2/9.5 ft Compute the Velocity Pressures.34 psf Determine the Gust Factor. qz = . 6.5.2 as well as for the side walls.8 for all elevations. so Cp = -0.1.1.2 N/S Building Section The pressure coefficients for the walls are found in ASCE 7-05 Figure 6-6 (pg 49) For the Windward wall (P1 & P2). For the Leeward wall (P5 & P6).50 for all elevations. In this case L/B = 50'/90' = 0. Cp is 0.1.4. .4.1) Wind in the N/S Direction: For this part of the problem we need to determine pressure coefficients for the locations shown in Figure 7.4. G = 0.8. For the sidewalls (not shown in Figure 7. the value of C p is -0.2). Figure 7.556.85 (ASCE 7-05. Cp is dependent on the ration of L/B.7 in all cases. These coefficients are then combined with the gust factor and velocity pressures to obtain the external pressures in each region. 7 29.7 29. For the Leeward side.7 29.7 -0.1 18.6 -17.5 -17.8 -0. qGCp.6 -4.7 q (psf) 26.For the roof.6 psf) and negative internal pressure .5 -0. From the Figure we get that the values of Cp for the Windward side of the roof is -0.0 degrees. Cp is -0. We can now compute the external pressures. We also need to know that h/L = 25.70 and -0.8 0.7 qGCp (psf) 18.5 -0.50. The following table shows the computation results: Windward Wall Windward Roof Leeward Roof Leeward Wall Side Walls Pressure Cp P1 P2 P3 P3 P4 P5 P6 P7 0.18 -0.5 psf) and positive internal pressure  Case III includes the maximum windward pressure (-17.6 -12.1 29.6 27.7 29. This is close to 15 degrees and probably not worth interpolating between the values given in ASCE 7-05 Figure 6-6.7 29.6 Combining with the internal pressures you get the following four load cases where:  Case I includes the maximum windward pressure (-17.18.1'/50' = 0.5 -12.6 psf) and positive internal pressure  Case II includes the minimum windward pressure (-4.6 -12. These values represent two different load cases.50. the slope angle is 14. for each surface.5 -0. 9 -17.3 -7.54 -16.8 12.9 -17.29 -41.19 -7.3 -12.3 -7. the sign is important.3 shows the location of each of the resulting forces.19 -17.86 -16.1 -23.4 23.55 -28.1.3 -12.19 -17. All forces are normal to their respective surfaces.3 -7.8 0. making the area calculation easier. Case IV includes the minimum windward pressure (-4.4.76 -17.90 -41.38 It is often useful to resolve each force into it's global components so that they can be easily added vectorially.9 -17. In this building all but the gable ends are rectangles.9 -17.62 12.0 -9.99 -53.62 12.4 23.76 -32.0 Case III (psf) 23. Figure 7.55 1.60 -17.38 Case IV (k) 23.3 Areas (ft2) 990 990 2319 2319 990 990 1413 Net Force Case I (k) 12. Note that we are computing actual surface areas (as opposed to projected areas) in each of the cases below.85 -7.1 13. Figure 7.4.5 psf) and negative internal pressure The net forces are found by multiplying the appropriate pressures by the areas over which they act.8 -12. Pressure Windward Wall Windward Roof Leeward Roof Leeward Wall Side Walls P1 P2 P3 P4 P5 P6 P7 Net Pressures: Case I Case II (psf) (psf) 12.3 -7.3 Case IV (psf) 23.8 -7.19 23.19 23.9 -17.1. Negative signs indicate a force that is outward from the surface and a positive sign is inward.76 -32.9 -23.9 -17.8 13.76 -17.19 -7.99 -22.3 Building Forces for N/S Wind .46 Case II (k) 12.85 -7.0 -23.3 -7.60 -17.46 Case III (k) 23. Also. (k) 0.38 17.00 0.46 32.38 N/S (k) 23.76 17.00 -17.46 N/S (k) 12.00 0.36 0.99 -12.00 51.00 vert.55 0.45 4.00 0.70 40.00 .00 0.00 vert.00 vert.00 27.09 7.00 0.92 4.62 12.00 0.00 0.00 0.55 -6. (k) 0.09 17.99 -5.00 -32.00 0.00 0.34 0.93 10.00 0.00 0.19 23.00 0.00 0.00 0.09 7.00 0.00 0.00 0.00 0.00 vert.69 16.19 0.62 12. (k) 0.00 0.38 N/S (k) 23.55 10.00 -32.00 0.19 0.22 40.00 0.00 -1.00 Case III E/W (k) 0.00 0.00 0.38 17.00 Case IV E/W (k) 0.00 0.00 0.00 0.00 0.76 0.34 0.00 0.36 0.00 0.46 N/S (k) 12.80 16.00 0.00 0.00 0.46 32.00 0.00 0.19 7.00 0.00 0.00 0.19 23.00 22.09 17.Force Windward Wall Windward Roof Leeward Roof Leeward Wall Side Walls F1 F2 F3 F4 F5 F6 F7a F7b Case I E/W (k) 0.00 0.00 -17.76 17. (k) 0.00 0.76 0.19 7.00 Case II E/W (k) 0. 4.66 14. You will also notice that the internal pressure has no effect on the net horizontal force. Wind in the E/W Direction Figure 7.1. In this case we combined all the leeward wall segments into one because they all have the same pressures.66 62.54 Note that the maximum uplift and maximum horizontal force do not occur in the same load cases! Do not combined the two cases.4.4 E/W Building Section . Figure 7. The net force in the lateral direction is zero since the forces on the side walls will cancel each other.00 65.4 defines the pressures (with the exception of the lateral/side wall pressures) that need to be computed for wind loading from the E/W direction.00 58.1.00 58.Sum 0.03 0.58 0.06 0.29 44.29 92.00 65. design for each individually. 8 0.8 0.18 -0.7 29.The pressure coefficients are taken from ASCE 7-05 Figure 6-6.34 -0.6 -4.6 27.7 q (psf) 26.6 .7 qGCp (psf) 18.9 -0.3 -0.1 18.7 29.7 29.1 29.2 -22.7 29. Windward Wall Roof Leeward Wall Side Walls Pressure Cp P8 P9 P10 P11 P11 P12 P12 P13 P13 P14 P15 0.5 -0.6 -17. Note that the coefficient for the leeward wall is obtained by interpolation with an L/B ratio of 1.5 -8.7 -4.5 -12.8 -0.7 29.7 29.7 29.18 -0.8.6 -4.5 -7.5 20.7 29.18 -0. Note that some of the pressures are applied to differently oriented surfaces. Figure 7.4.1. in terms of direction relative the surface. based on combinations of maximum/minimum roof pressures and +internal pressures. are as follows: .5 for force applications.1.5 Building Forces for E/W Wind The net forces on each surface. we have chosen to label on as "a" and the other as "b". See Figure 7. Four cases are computed. When the same pressure is applied to a different surface.4. 3 -7.98 -11.98 0.36 0.8 13.50 Case III (k) 12.20 6.32 0.00 0.55 -1.00 vert.2 -2.9 -23.00 45.00 34.2 -12.50 0.00 83.2 -3.00 4.3 Area (ft2) 550 550 156 647 647 647 647 1026 1026 1256 1980 1980 Case I (k) 7.28 -2.00 0.88 13.00 0.82 0.00 -0.00 34.00 24. (k) 0.52 0.58 11.8 25.88 13.00 0.00 0.00 17.3 Case IV (psf) 23.88 4.01 7.21 2.00 0.39 -6.00 0.98 0.3 -12.21 2.00 10.88 13.36 -24.00 0.00 0.88 13.00 0.0 Case II (psf) 12.00 0.50 0.40 2.00 0.00 44.00 1.8 0.00 0.00 0.36 Case IV (k) 12.80 -0.1 14.82 9.06 0.13 -17.0 Case III (psf) 23.9 -17. (k) 0.00 -0.1 14.01 N/S (k) 0.9 -12.00 0.46 0.55 -1.00 0.00 0.21 0.00 0.01 7.21 -3.50 -0.82 0.00 4.83 12.58 17.4 23.00 0.00 45.0 -28.56 2.00 17.50 -0.8 25.00 24.00 0.00 0.00 0. (k) 0.00 vert.13 0.20 6.36 0.50 -0.00 34.36 -24.01 N/S (k) 0.40 -4.50 -45.21 2.14 0.36 -24.09 3.01 N/S (k) 0.00 0.9 -23.0 -23.8 0.60 -11.8 0.00 34.00 vert.9 -9.43 Case III E/W (k) 12.01 N/S (k) 0.52 0.22 -11.8 -9.00 0.0 -23.14 -1.9 -12.00 17.47 -45.00 0.3 -12.36 -24.52 0.55 -0.00 0.13 -10.81 3.80 0.56 4.12 -18.00 0.4 23.00 0.5 0.00 0.39 -6.20 0.55 2.60 .00 0.00 0.21 2.20 6.47 0.20 0.00 0.46 -2.20 9.31 Case IV E/W (k) 12.00 0.00 35.47 0.06 -24.23 -13.0 -17.00 0.01 7.26 11.00 0.83 0.9 -9.50 -45.00 0.8 13.98 0.39 -10.8 -3.00 0.06 0.00 0.00 vert.23 -17.60 -13.Pressure Windward Wall Roof Leeward Wall Side Walls P8 P9 P10 P11a P11b P12a P12b P13a P13b P14 P15a P15b Case I (psf) 12.72 -2.00 0.8 0.09 3.52 0.3 -7.3 -2.50 -45.13 -0.88 10.55 0.13 0.00 0.00 0.72 1.26 12.00 4.21 0.82 -4.9 -9.9 -9.70 -2.00 -3.22 -4.55 1.00 0.32 -18.09 3.34 Case II E/W (k) 7.06 -24.70 -4.00 2.00 0.28 -4.36 Restating the forces in terms of the global coordinate system we get: Pressure Windward Wall Roof Leeward Wall Side Walls Sum F8 F9 F10 F11a F11b F12a F12b F13a F13b F14 F15a F15b Case I E/W (k) 7.00 6.50 Case II (k) 7.00 0.8 -28. (k) 0.47 -45.9 -13.8 0.00 0.50 -45.12 -11.00 0.13 -0.39 -6.50 -0.32 -6.5 -17.00 0.01 7.9 -13.21 2.9 -9.09 3.2 -12.32 0.81 -2.3 -17.00 0.