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,
- Show how this maybe is an open, partially enclosed, or enclosed building.
- Determine the internal pressure coefficients.
- Determine the external pressure coefficients for the design of main girders and shear walls.
- 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 = across-wind 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 | 0.82 | 1.32 | 0.12 | 1.68 | -1.38 | 0.18 |
| 12 | 0.76 | 1.22 | 0.05 | 1.61 | -1.38 | 0.18 |
| 8 | 0.67 | 1.08 | -0.05 | 1.52 | -1.38 | 0.18 |
| 4 | 0.57 | 0.92 | -0.16 | 1.41 | -1.38 | 0.18 |
| 0 | 0.57 | 0.92 | -0.16 | 1.41 | -1.38 | 0.18 |
Net
along wind pressures pz (kPa)
|
Afz (sqm)
|
Net
along wind loads Fz (kN)
|
Base
bending moment contribution My,z (kNm)
|
|||
with +Gcpi
|
with -Gcpi
|
with +Gcpi
|
with -Gcpi
|
with +Gcpi
|
with -Gcpi
|
|
1.56
|
1.57
|
20
|
31
|
31
|
620
|
620
|
1.5
|
1.5
|
40
|
60
|
60
|
960
|
960
|
1.43
|
1.43
|
40
|
57
|
57
|
684
|
684
|
1.33
|
1.34
|
40
|
53
|
54
|
424
|
432
|
1.22
|
1.23
|
40
|
49
|
49
|
196
|
196
|
1.22
|
1.23
|
20
|
24
|
25
|
0
|
0
|
Vx (kN) =
|
274 | 276 | 2884 | 2892 | ||
My (kNm) =
|
||||||
| Af,roof (sqm) | Roof loads 1, p (kPa) | Roof loads 2, p (kPa) | Vz = Roof loads 1 (kN) | Vz = Roof loads 2 (kN) | ||||
| with +Gcpi | with -Gcpi | with +Gcpi | with -Gcpi | with +Gcpi | with -Gcpi | with +Gcpi | with -Gcpi | |
| 100 | -2.04 | -0.47 | -1 | 0.56 | -204 | -47 | -100 | 56 |
| Vz (kN) = | -204 | 56 | ||||||
Spot any errors? Sound out in the comments. :)
28 comments:
Sir, diba yung roof pressure coefficient value must be multiplied by 0.80 to account for its area? So -0.18 should be -0.144? :)
The NSCP 2010 appears to say that the 0.8 reduction factor for area should be applied only to -1.3 and not to -0.18. The NSCP 2001 did not specify a -0.18 coefficient but it did say that the reduction factor applies only to the -1.3 coefficient.
Thanks, sir!
Hi, Ronjie. How would you evaluate the pressure coeff. for structures
with irregular polygon shapes, say a structure with a plan shape of the
letter "W" or "k". Most of the pressure coefficients in the code are
for uniform section such as Circle, rectangle, square, etc.
I
have my own approach on this (mostly based on engineering judgement),
but I would like to here it first on a wind load aficionado, such as
yourself.
John, thanks for your question. You are right, codes usually do not provide for such rather uncommonly used building shapes. You can of course use engineering judgment based on the coefficients for shapes given in the code, which are based on wind tunnel tests on those specific "regular" shapes. That said, the more accurate and most rational method is to use the wind tunnel approach (i.e. Method 3 in NSCP 2010, or ASCE7). Unless it is a shape that has been tested before and is published in literature, you might not be accounting for unexpected wind flow patterns that could increase wind pressures at specific locations, or you might not be realizing cost savings by using a more accurate approach. If you are dealing with tall buildings, it is easy to justify the relative low cost of doing a wind tunnel test.
Good Day Sir!
We're currently designing a building for a project and I hope it's okay to ask some concerns regarding the NSCP code.
The one that would be designed here is the MWFRS part of the bldg. The building is a low rise type and the calculations will be done using the Method 1 based from NSCP 2010. It is said that Method 1 is applicable to both gable type and flat roof type buildings. The one indicated on Figure 2 of NSCP is a gable type building where there are loads A,B,C,D,E,F,G,and H applied.
a.) What will happen to the loads B and D for flat roof type of buildings? According to the table provided by NSCP there are negative horizontal pressures on these zones but since the eave height is equal to the height of the structure, i think B and D should be zero. |
b.) Will there be no pressure applied on the rear side of the building?
Adolfo, thanks for asking. I can't help but notice two things...
1. "WE... designing a building"? You have a license already? ;)
2. "NSCP code"... National Structural Code of the Philippines Code? :D
As to your two questions, reading the NSCP, I think it's quite clear what the answers are. As I have emphasized in class, just make sure to consider at least 4 wind directions. Good luck!
Sir, pare-pareho lang po ba talaga ung values ng Leeward wall pz sa kahit anong elevation?
Thank you po in advance. :)
NSCP Figure 207-6 answers your question. :)
Oh. Thank you so much Sir! :D
Hi ronjie, do you know where Table 207-11 is located in NSCP 2010?
Thanks, I think I do. How about you?
I have trouble finding it yesterday, it seems that the code has some typo graphical errors in sec. 207.7.1 which directs you to table 207-11. Table 207-11 does not exist but the intended content of table 207-11 is available in table 207-5.
Additional info. Eqn 207-38 & 207-39 is equal to 0.016/h and 0.23/h respectively. however Eqn 207-40 is equal to 0.007/n, in which h = ht of structure, and n = natural freq of structure. are Eqns 207-38 & 207-39 correct? or h should be replaced by n?
Yeah, thanks for pointing that out. I've started to use the latest NSCP only recently and I've also spotted a lot of typographical errors already. You are right, Table 207-11 should be Table 207-5, and note also that alpha-overbar and a small letter "L" in Times New Roman are mentioned in 207.7.1 but in Table 207-5, it's actually A-overbar and a small letter "L" in script font in there. There are also a number of instances where h or h <= 18 m or h >= 18 m appears to have been omitted.
As to your question on Equations 207-38, -39, and -40, note that 207-38 and -39 are formulas for one thing (Beta_s), while 207-40 is a formula for another thing (Beta_a). I think that is quite clear, so my answer to your question is: no, h should not be replaced by n.
Good day sir. do you have a sample problem like above to get the roof wind load for sawtooth type of buildings?
I don't have one, but I think the NSCP's figures for sawtooth roofed buildings is pretty straightforward. Thanks for commenting!
Good day sir! Im a bit confused. Can you show me how the wind pressure
Hi, can I show you how the wind pressure.....?
Good day sir! i'd like to ask if the qz/velocity pressure in the nscp acts perpendicular to the roof surface or perpendicular to the vertical projection of the roof..
You don't actually apply q_z on a roof surface, you apply q_h (which is q_z with z = h). The figures show or indicate that the NSCP wind pressures are either positive (pressing action) or negative (suction) -- which means they're always perpendicular to the surface.
thank you sir..so, how about if its a truss analysis? where should i start my computation sir? with reference to the Nscp..im at a lost in wind load computations.
For vertical truss structures, wind forces are always applied horizontally. For roof trusses, they are applied perpendicular to the surface receiving the wind pressures.
in reference with NSCP sir,wat is the equation to use for the wind load acting on a truss alone? And also In staad, Can i consider a single truss only considering loads(dead,live,wind) in tributary area?!
The answer to your first question is in the NSCP. For any structural analysis whether using STAAD or another software, of course you can consider a single truss considering only the tributary area but make sure to use the appropriate loads for that area.
What did you need this for? A homework or a project?
for a project sir. can i use fig. 207-1 in the computation of wind load and apply it to the truss considering only the tributary area?
for a project sir. im considering loads acting on a truss alone and not on d whole structure..in what part of the nscp will i refer to directly? if wimd velocity is 350kph. can i use ds formula (207-22):
p=Qh[(GCp) - (GCpi)]
to directly act normally to the roof?
or formula (207-15):
Qh=.0000473*Kz*Kzt*Kd*Iw*V^2?
Good day sir. I'd like to ask you about how to use the reduction factor to -1.3.
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