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Solutions Manual toStructural Loads2012 IBC and ASCE/SEI 7-10

This Solutions Manual was developed as a companion to the Structural Loads: 2012 IBC andASCE/SEI 7-10 textbook. To increase understanding of this material and for its mosteffective use, readers should study the Structural Loads textbook and reference thematerial as they read through this Solutions Manual.

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Solutions Manual to Structural Loads2012 IBC andASCE/SEI 7-10

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Preface

This manual contains solutions to the problems in Structural Loads - 2012 IBC and ASCE/SEI7-10 and is intended to aid in the understanding and application of the structural load provisionsin the 2012 IBC and ASCE/SEI 7-10 standard.

The solutions provided in this manual are organized by chapter number and they are numberedsequentially—the same as the problems in the book.

While every effort has been made to provide complete and accurate solutions, no guarantee ismade or implied concerning solution accuracy or the suitability of the loads calculated in thesolutions. This manual is intended to illustrate procedures for determining structural loads foreducational purposes only and not to provide solutions to specific designs.

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CHAPTER 2Load Combinations

2.1. Determine the strength design load combinations for a reinforced concrete beam on atypical floor of a multistory residential building using the nominal bending moments inTable 2.11. All bending moments are in foot-kips. Assume the live load on the floor is lessthan 100 psf.

Table 2.11 Design Data for Problem 2.1

SOLUTION

Table P2.1 Summary of Load Combinations Using Strength Design for Beam in Problem 2.1

2.2. Determine the strength design load combinations for a steel beam that is part of an ordinarymoment frame in an office building using the nominal bending moments and shear forcesin Table 2.12. All bending moments are in foot-kips and all shear forces are in kips.Assume the live load on the floor is less than 100 psf.


External Negative Positive Interior NegativeDead load, D –13.3 43.9 –53.2

Live load, L –12.9 42.5 –51.6

IBC Equation No. EquationLoad Combination

Exterior Negative

PositiveInterior

Negative16-1 1.4D –18.6 61.5 –74.5

16-2 1.2D + 1.6L –36.6 120.7 –146.416-3, 16-4, 16-5 1.2D + 0.5L –22.4 73.9 –89.6

16-6, 16-7 0.9D –12.0 39.5 –47.9

Bending Moment Shear ForceSupport Midspan Support

Dead load, D –57.6 41.1 11.8Live load, L –22.5 16.2 4.6

Wind, W ± 54.0 --- ± 4.8

1

2 Solutions Manual to Structural Loads

SOLUTION

Table P2.2 Summary of Load Combinations Using Strength Design for Beam in Problem 2.2

2.3. Given the information in Problem 2.2, determine the basic allowable stress design loadcombinations.

SOLUTION

Table P2.3 Summary of Load Combinations Using Basic Allowable Stress Design for Beam in Problem 2.3

IBC Equation No.

EquationLoad Combination


16-1 1.4D –80.6 57.5 16.516-2 1.2D + 1.6L –105.1 75.2 21.5

16-3

1.2D + 0.5L –80.4 57.4 16.5

1.2D + 0.5W –42.1 49.3 11.81.2D – 0.5W –96.1 49.3 16.6

16-41.2D + 1.0W + 0.5L –26.4 57.4 11.7

1.2D – 1.0W + 0.5L –134.4 57.4 21.316-5 1.2D + 0.5L –80.4 57.4 16.5

16-60.9D + 1.0W 2.2 37.0 5.8

0.9D – 1.0W –105.8 37.0 15.416-7 0.9D –51.8 37.0 10.6

IBC Equation No.



16-8, 16-10 D –57.6 41.1 11.816-9 D + L –80.1 57.3 16.4

16-11, 16-14 D + 0.75L –74.5 53.3 15.3

16-12D + 0.6W –25.2 41.1 8.9D – 0.6W –90.0 41.1 14.7

16-13D + 0.75(0.6W) + 0.75L –50.2 53.3 13.1

D – 0.75(0.6W) + 0.75L –98.8 53.3 17.4

16-150.6D + 0.6W –2.2 24.7 4.20.6D – 0.6W –67.0 24.7 10.0

16-16 0.6D –34.6 24.7 7.1

Chapter 2 3

2.4. Given the information in Problem 2.2, determine the alternative basic allowable stressdesign load combinations. Assume the wind loads have been determined using ASCE/SEI7.

SOLUTION

Table P2.4 Summary of Load Combinations Using Alternative Basic Allowable Stress Design for Beam in Problem 2.4

2.5. Determine the strength design load combinations for a reinforced concrete column that ispart of an intermediate moment frame in an office building using the nominal axial forces,bending moments and shear forces in Table 2.13. All axial forces are in kips, all bendingmoments are in foot-kips and all shear forces are in kips. Assume the live loads on thefloors are equal to 100 psf and SDS = 0.41g.


SOLUTION

Since the live loads on the floors are equal to 100 psf, f1 = 0.5.

Since an intermediate moment frame is used, the SDC is C or lower. Thus, ρ = 1.0.

The seismic load effect, E, is determined as follows:

For use in IBC Equation 16-5: E = Eh + Ev = ρQE + 0.2SDSD

= (1.0 × QE) + (0.2 × 0.41 × D) = QE + 0.08D

For use in IBC Equation 16-7: E = Eh – Ev = ρQE – 0.2SDSD

= (1.0 × QE) – (0.2 × 0.41 × D) = QE – 0.08D

IBC Equation No.



16-17, 16-21 D + L –80.1 57.3 16.4

16-18, 16-19D + L + 0.6ωW –38.0 57.3 12.7D + L – 0.6ωW –122.2 57.3 20.1

16-20D + L + 0.6ωW/2 –59.0 57.3 14.5

D + L – 0.6ωW/2 –101.2 57.3 18.3

16-22 0.9D –51.8 37.0 10.6

Axial ForceBending Moment

Shear Force

Dead load, D 167.9 21.3 2.3Live load, L 41.5 21.0 2.2Roof live load, Lr 14.9 --- ---

Wind, W ± 13.6 ± 121.0 ± 11.1Seismic, QE ± 36.4 ± 432.1 ± 42.2


Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.08D + 0.5L = 1.28D + QE + 0.5L

Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.08D = 0.82D + QE

Table P2.5 Summary of Load Combinations Using Strength Design for Column inProblem 2.5

2.6. Determine the strength design load combinations for a reinforced concrete shear wall in aparking garage using the nominal axial forces, bending moments and shear forces inTable 2.14. All axial forces are in kips, all bending moments are in foot-kips and all shearforces are in kips. Assume that ρ = 1.0 and SDS = 1.0g.


SOLUTION

Since the shear wall is in a parking garage, f1 = 1.0.

The seismic load effect, E, is determined as follows:

For use in IBC Equation 16-5: E = Eh + Ev = ρQE + 0.2SDSD

= (1.0 × QE) + (0.2 × 1.0 × D) = QE + 0.2D

For use in IBC Equation 16-7: E = Eh – Ev = ρQE – 0.2SDSD

= (1.0 × QE) – (0.2 × 1.0 × D) = QE – 0.2D

Substituting for E, IBC Equation 16-5 becomes: 1.2D + QE + 0.2D + 1.0L = 1.4D + QE + 1.0L.Similarly, IBC Equation 16-7 becomes: 0.9D + QE – 0.2D = 0.7D + QE

IBC Equation No.

Equation

Load Combination

Axial Force Bending Moment Shear Force

16-1 1.4D 235.1 29.8 3.2

16-2 1.2D + 1.6L + 0.5Lr 275.3 59.2 6.3

16-3

1.2D + 1.6 Lr + 0.5L 246.1 36.1 3.9

1.2D + 1.6 Lr + 0.5W 232.1 86.1 8.3

1.2D + 1.6 Lr – 0.5W 218.5 –34.9 –2.8

16-41.2D + 1.0W + 0.5L + 0.5Lr 243.3 157.1 15.0

1.2D – 1.0W + 0.5L + 0.5Lr 216.1 –84.9 –7.2

16-51.28D + QE + 0.5L 272.1 469.9 46.2

1.28D – QE + 0.5L 199.3 –394.3 –38.2

16-60.9D + 1.0W 164.7 140.2 13.2

0.9D – 1.0W 137.5 –101.8 –9.0

16-70.82D + QE 174.1 449.6 44.1

0.82D – QE 101.3 –414.6 –40.3

Axial Force Bending Moment Shear ForceDead load, D 645 0 0

Live load, L 149 0 0Seismic, QE 0 ± 4,280 ± 143

Chapter 2 5

Table P2.6 Summary of Load Combinations Using Strength Design for Shear Wall inProblem 2.6

2.7. Determine the strength design load combinations and the basic combinations for strengthdesign with overstrength factor for a simply supported steel collector beam in an assemblybuilding using the nominal axial forces, bending moments and shear forces in Table 2.15.All axial forces are in kips, all bending moments are in foot-kips and all shear forces are inkips. Assume SDS = 0.9g and Ω0 = 2.0.


SOLUTION

The governing load combination in IBC 1605.2 is Equation 16-2:

Negative bending moment:

1.2D + 1.6L = (1.2 × 80.6) + (1.6 × 42.1) = 164.1 ft-kips

Positive bending moment:

1.2D + 1.6L = (1.2 × 53.7) + (1.6 × 30.4) = 113.1 ft-kips

Shear force:

1.2D + 1.6L = (1.2 × 29.7) + (1.6 × 19.0) = 66.0 kips

The following basic combinations for strength design with overstrength factor are alsoapplicable (see IBC 1605.1 and 1605.2; ASCE/SEI 12.4.3.2 and 12.10.2.1):

• IBC Equation 16-5: (1.2 + 0.2SDS) D + Ω0QE + 1.0L

Axial force: Ω0QE = 2.0 × 241 = 482 kips tension or compression

IBC Equation No.

Equation

Load Combination

Axial Force Bending Moment Shear Force

16-1 1.4D 903.0 0 016-2 1.2D + 1.6L 1012.4 0 0

16-3, 16-4 1.2D + 1.0L 923.0 0 0

16-51.4D + QE + 1.0L 1052.0 4280.0 143.0

1.4D – QE + 1.0L 1052.0 –4280.0 –143.0

16-6 0.9D 580.5 0 0

16-70.7D + QE 451.5 4280.0 143.0

0.7D – QE 451.5 –4280.0 –143.0

Axial ForceBending Moment

Shear ForceNegative Positive

Dead load, D 0 80.6 53.7 29.7Live load, L 0 42.1 30.4 19.0

Seismic, QE ± 241 0 0 0


Negative bending moment:

(1.2 + 0.2SDS)D + 1.0L = (1.38 × 80.6) + (1.0 × 42.1) = 153.3 ft-kips

Positive bending moment:

(1.2 + 0.2SDS)D + 1.0L = (1.38 × 53.7) + (1.0 × 30.4) = 104.5 ft-kips

Shear force:

(1.2 + 0.2SDS)D + 1.0L = (1.38 × 29.7) + (1.0 × 19.0) = 60.0 kips

Note that the load factor on L must be equal to 1.0 because of the assembly occupancy.

• IBC Equation 16-7: (0.9 – 0.2SDS)D + Ω0QE

Axial force: Ω0QE = 2.0 × 241 = 482 kips tension or compression

Negative bending moment: (0.9 – 0.2SDS)D = 0.72 × 80.6 = 58.0 ft-kips

Positive bending moment: (0.9 – 0.2SDS)D = 0.72 × 53.7 = 38.7 ft-kips

Shear force: (0.9 – 0.2SDS)D = 0.72 × 29.7 = 21.4 kips

2.8. Given the information in Problem 2.7, determine the basic allowable stress design loadcombinations.

SOLUTION

The governing load combination in IBC 1605.3.1 is Equation 16-9:

Negative bending moment: D + L = 80.6 + 42.1 = 122.7 ft-kips

Positive bending moment: D + L = 53.7 + 30.4 = 84.1 ft-kips

Shear force: D + L = 29.7 + 19.0 = 48.7 kips

The following basic combinations for strength design with overstrength factor are alsoapplicable (see IBC 1605.1 and 1605.3.1; ASCE/SEI 12.4.3.2):

• IBC Equation 16-12: (1.0 + 0.14SDS)D + 0.7Ω0QE

Axial force: 0.7Ω0QE = 0.7 × 2.0 × 241 = 337.4 kips tension or compression

Negative bending moment: (1.0 + 0.14SDS)D = 1.13 × 80.6 = 91.1 ft-kips

Positive bending moment: (1.0 + 0.14SDS)D = 1.13 × 53.7 = 60.7 ft-kips

Shear force: (1.0 + 0.14SDS)D = 1.13 × 29.7 = 33.6 kips

• IBC Equation 16-14: (1.0 + 0.105SDS)D + 0.525Ω0QE + 0.75L


Negative bending moment: 1.1D + 0.75L = (1.1 × 80.6) + (0.75 × 42.1) = 120.2 ft-kips

Positive bending moment: 1.1D + 0.75L = (1.1 × 53.7) + (0.75 × 30.4) = 81.9 ft-kips

Shear force: 1.1D + 0.75L = (1.1 × 29.7) + (0.75 × 19.0) = 46.9 kips

• IBC Equation 16-16: (0.6 – 0.14SDS)D + 0.7Ω0QE


Chapter 2 7

Negative bending moment: (0.6 – 0.14SDS)D = 0.47 × 80.6 = 37.9 ft-kips

Positive bending moment: (0.6 – 0.14SDS)D = 0.47 × 53.7 = 25.2 ft-kipsShear force: (0.6 – 0.14SDS) D = 0.47 × 29.7 = 14.0 kips

2.9. Determine the strength design load combinations for a wood roof truss in a commercialbuilding with a curved roof using the nominal distributed loads in Table 2.16. All loads arein pounds per linear foot.


SOLUTION

Table P2.9 Summary of Load Combinations Using Strength Design for Roof Truss inProblem 2.9

LoadDead load, D 75Rain load, R 200Roof live load, Lr 100

Balanced snow load, S 125

IBC Equation No. Equation Load Combination16-1 1.4D 105.0

16-2, 16-4

1.2D + 0.5Lr 140.0

1.2D + 0.5S 152.5

1.2D + 0.5R 190.0

16-3

1.2D + 1.6Lr 250.0

1.2D + 1.6S 290.0

1.2D + 1.6R 410.0

16-5 1.2D + 0.2S 115.0

16-6, 16-7 0.9D 67.5


2.10.Given the information in Problem 2.9, determine the basic allowable stress design loadcombinations.

SOLUTION

Table P2.10 Summary of Load Combinations Using Basic Allowable Stress Design for Roof Truss in Problem 2.10

IBC Equation No. Equation Load Combination16-8, 16-9, 16-12 D 75.0

16-10

D + Lr 175.0

D + S 200.0D + R 275.0

16-11, 16-13

D + 0.75Lr 150.0

D + 0.75S 168.8

D + 0.75R 225.0

16-14 D + 0.75S 168.8

16-15, 16-16 0.6D 45.0

CHAPTER 3Dead, Live, Rain and Soil

Lateral Loads

3.1. Given the 10-story office building described in Example 3.1, determine reduced live loadsfor (a) column A3 and (b) column A6. Assume that the storage occupancy is on the eighthfloor and all other floors are typical floors. Use the basic uniform live load reductionprovisions of IBC 1607.10.1.

SOLUTION

Nominal Loads

• Roof: 20 psf in accordance with IBC Table 1607.1, since the roof is an ordinary flat roof thatis not occupiable.

• Eighth floor: storage load is given in the design criteria as 125 psf.

• Typical floor: 50 psf for office space in accordance with IBC Table 1607.1 and15 psf for moveable partitions in accordance with IBC 1607.5, since the live load does notexceed 80 psf. The partition load is not reducible; only the minimum loads in IBC Table1607.1 are permitted to be reduced (IBC 1607.10).

Part a: Determine reduced live load for column A3A summary of the reduced live loads is given in Table P3.1-a. Detailed calculations are givenbelow.

• Roof

The reduced roof live load, Lr, is determined by IBC Equation 16-26: Lr = LoR1R2 = 20R1R2

The tributary area, At, of column A3 = (28/2) × 22.5 = 315 square feet

Since 200 square feet < At < 600 square feet, R1 is determined by IBC Equation 16-28:

R1 = 1.2 – 0.001At = 1.2 – (0.001 × 315) = 0.89

Since F = 1/2 < 4, R2 = 1 (IBC Equation 16-30)

Thus, Lr = 20 × 0.89 × 1 = 17.8 psf

Axial load = 17.8 × 315/1,000 = 5.6 kips

9


Table P3.1-a Summary of Reduced Live Loads for Column A3

N = nonreducible live load, R = reducible live load* Roof live load reduced in accordance with IBC 1607.12.2** Live load > 100 psf is not permitted to be reduced (IBC 1607.10.1.2)

• Ninth floor

Reducible nominal live load = 50 psf

Since column A3 is an exterior column without a cantilever slab, the live load element fac-tor KLL = 4 (IBC Table 1607.10.1). Equivalently, KLL = influence area/tributary area = 28(25 + 20)/315 = 4.

Reduced live load, L, is determined by IBC Equation 16-23:

≥ 0.50Lo for members supporting one floor

With AT = 315 square feet and KLL = 4,

psf

Axial load = (L + 15)AT = 315 × (33.5 + 15)/1,000 = 15.3 kips

• Eighth floor

Since the eighth floor is storage with a live load of 125 psf, which exceeds 100 psf, the live load is not permitted to be reduced (IBC 1607.10.1.2).

Axial load = 125 × 315 / 1,000 = 39.4 kips

• Typical floors




≥ 0.40Lo for members supporting two or more floors

StoryLive Load

(psf) KLLAT

(sq ft)Reduction Multiplier

Reduced Live Load

(psf)

N + R(kips)

CumulativeN + R(kips)N R

10 --- 20 ---* --- 17.8 5.6 5.69 15 50 1,260 0.67 33.5 15.3 20.9

8 125 --- ---** --- --- 39.4 60.37 15 50 2,520 0.55 27.5 13.4 73.76 15 50 3,780 0.49 24.5 12.4 86.1

5 15 50 5,040 0.46 23.0 12.0 98.14 15 50 6,300 0.44 22.0 11.7 109.83 15 50 7,560 0.42 21.0 11.3 121.1

2 15 50 8,820 0.41 20.5 11.2 132.31 15 50 10,080 0.40 20.0 11.0 143.3

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

=

L 50 0.25 15

4 315×----------------------+⎝ ⎠

⎛ ⎞× 0.67 50× 33.5===

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

=

Chapter 3 11

The reduction multiplier is equal to 0.40 where KLL AT ≥ 10,000 square feet (see Figure 3.5).

Axial load = (L + 15)AT = 315 (L + 15)

Part b: Determine reduced live load for column A6A summary of the reduced live loads is given in Table P3.1-b. Detailed calculations are givenbelow.

• Roof

The reduced roof live load, Lr, is determined by IBC Equation 16-26:

Lr = LoR1R2 = 20R1R2

The tributary area, At, of column A6 = [(25/2) + 5] × (28/2) = 245 square feet


R1 = 1.2 – 0.001At = 1.2 – (0.001 × 245) = 0.96


Thus, Lr = 20 × 0.96 × 1 = 19.2 psf

Axial load = 19.2 × 245/1,000 = 4.7 kips

Table P3.1-b Summary of Reduced Live Loads for Column A6

N = nonreducible live load, R = reducible live load* Roof live load reduced in accordance with IBC 1607.12.2** Live load > 100 psf is not permitted to be reduced (IBC 1607.10.1.2)

• Ninth floor




StoryLive Load

(psf) (sq ft)

Reduction Multiplier

Reduced Live Load

(psf)

N + R(kips)


10 --- 20 ---* --- 19.2 4.7 4.79 15 50 833 0.77 38.5 13.1 17.8

8 125 --- ---** --- --- 30.6 48.47 15 50 1,666 0.62 31.0 11.3 59.76 15 50 2,499 0.55 27.5 10.4 70.1

5 15 50 3,332 0.51 25.5 9.9 80.04 15 50 4,165 0.48 24.0 9.6 89.63 15 50 4,998 0.46 23.0 9.3 98.9

2 15 50 5,831 0.45 22.5 9.2 108.11 15 50 6,664 0.43 21.5 8.9 117.0

TLLAK

KLLInfluence areaTributary area---------------------------------- 25 5+( ) 28×

245-------------------------------- 3.4= = =

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

=


With AT = 245 square feet and KLL = 3.4,

psf

Axial load = (L + 15)AT = 245 × (38.5 + 15)/1,000 = 13.1 kips

• Eighth floor

Since the eighth floor is storage with a live load of 125 psf, which exceeds 100 psf, the live load is not permitted to be reduced (IBC 1607.10.1.2).

Axial load = 125 × 245/1,000 = 30.6 kips

• Typical floors






Axial load = (L + 15)AT = 245(L + 15)

3.2. Given the information provided in Problem 3.1, determine the reduced live loads using thealternative uniform live load reduction provisions of IBC 1607.10.2. Assume a nominaldead-to-live load ratio of 2.

SOLUTION

Part a: Determine reduced live load for column A3A summary of the reduced live loads is given in Table P3.2-a. Detailed calculations are givenbelow.

L 50 0.25 15

3.4 245×---------------------------+⎝ ⎠

⎛ ⎞× 0.77 50× 38.5===

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

50 0.25 15

3.4AT

------------------+⎝ ⎠⎜ ⎟⎛ ⎞

×==

Chapter 3 13

Table P3.2-a Summary of Reduced Live Loads for Column A3

N = nonreducible live load, R = reducible live load* Roof live load reduced in accordance with IBC 1607.12.2** Live load > 100 psf is not permitted to be reduced [IBC 1607.10.2(1)]

• Roof



The tributary area, At, of column A3 = (28/2) × 22.5 = 315 square feet


R1 = 1.2 – 0.001At = 1.2 – (0.001 × 315) = 0.89


Thus, Lr = 20 × 0.89 × 1 = 17.8 psf

Axial load = 17.8 × 315/1,000 = 5.6 kips

• Ninth floor


Reduction factor, R, is determined by IBC Equation 16-24:

R = 0.08(A – 150) ≤ smaller of 60% (governs) or 23.1(1 + D/Lo) = 23.1(1 + 2) = 69.3%

R = 0.08 × (315 – 150) = 13%

Axial load = [Lo(1 – 0.01R) + 15]A = [50(1 – 0.13) + 15] × 315/1,000 = 18.4 kips

• Eighth floor

Since the eighth floor is storage with a live load of 125 psf, which exceeds 100 psf, the live load is not permitted to be reduced [IBC 1607.10.2(1)].

Axial load = 125 × 315 / 1,000 = 39.4 kips

• Typical floors


Axial load = [Lo(1 – 0.01R) + 15] × 315

StoryLive Load

(psf) A(sq ft)

Reduction Factor, R (%)

Reduced Live Load

(psf)

N + R(kips)


10 --- 20 ---* --- 17.8 5.6 5.69 15 50 315 13 43.5 18.4 24.08 125 --- ---** --- --- 39.4 63.4

7 15 50 630 38 31.0 14.5 77.96 15 50 945 60 20.0 11.0 88.95 15 50 1,260 60 20.0 11.0 99.9

4 15 50 1,575 60 20.0 11.0 110.93 15 50 1,890 60 20.0 11.0 121.92 15 50 2,205 60 20.0 11.0 132.9

1 15 50 2,520 60 20.0 11.0 143.9


Part b: Determine reduced live load for column A6A summary of the reduced live loads is given in Table P3.2-b. Detailed calculations are givenbelow.

• Roof



The tributary area, At, of column A6 = [(25/2) + 5] × (28/2) = 245 square feet


R1 = 1.2 – 0.001At = 1.2 – (0.001 × 245) = 0.96


Thus, Lr = 20 × 0.96 × 1 = 19.2 psf

Axial load = 19.2 × 245 / 1,000 = 4.7 kips

Table P3.2-b Summary of Reduced Live Loads for Column A6

N = nonreducible live load, R = reducible live load* Roof live load reduced in accordance with IBC 1607.12.2** Live load > 100 psf is not permitted to be reduced [IBC 1607.10.2(1)]

• Ninth floor


Reduction factor, R, is determined by IBC Equation 16-24:


R = 0.08 × (245 – 150) = 8%

Axial load = [Lo(1 – 0.01R) + 15]A = [50(1 – 0.08) + 15] × 245/1,000 = 15.0 kips

• Eighth floor

Since the eighth floor is storage with a live load of 125 psf, which exceeds 100 psf, the live load is not permitted to be reduced [IBC 1607.10.2(1)].

Axial load = 125 × 245 / 1,000 = 30.6 kips

StoryLive Load

(psf) A(sq ft)

Reduction Factor, R (%)

Reduced Live Load (psf)

N + R(kips)


10 --- 20 ---* --- 19.2 4.7 4.79 15 50 245 8 46.0 15.0 19.78 125 --- ---** --- --- 30.6 50.3

7 15 50 490 27 36.5 12.6 62.96 15 50 735 47 26.5 10.2 73.15 15 50 980 60 20.0 8.6 81.7

4 15 50 1,225 60 20.0 8.6 90.33 15 50 1,470 60 20.0 8.6 98.92 15 50 1,715 60 20.0 8.6 107.5

1 15 50 1,960 60 20.0 8.6 116.1

Chapter 3 15

• Typical floors


Axial load = [Lo(1 – 0.01R) + 15] × 245

3.3. A steel transfer beam supports an interior column that supports a flat roof and four floorsof office space. The tributary area of the column at each level is 600 square feet. The beamalso supports a tributary floor area of 300 square feet of office space at the first elevatedlevel. Using the basic uniform live load reduction provisions of IBC 1607.10.1, determinethe reduced live loads on the beam.

SOLUTION

The transfer beam carries both a concentrated live load from the column and a distributed liveload from the first level.

Concentrated live load from supported column

A summary of the reduced live loads is given in Table P3.3. Detailed calculations are givenbelow.

Table P3.3 Summary of Reduced Live Loads on Transfer Beam from Supported Column

N = nonreducible live load, R = reducible live load* Roof live load reduced in accordance with IBC 1607.12.2

• Roof



The tributary area, At, of the supported column = 600 square feet

Since At ≥ 600 square feet, R1 = 0.6.

Since F < 4, R2 = 1 (IBC Equation 16-30)

Thus, Lr = 20 × 0.6 × 1 = 12.0 psf

Axial load = 12.0 × 600 / 1,000 = 7.2 kips

• Typical floors


Since the supported column is an interior column, the live load element factor KLL = 4 (IBC Table 1607.10.1).

StoryLive Load

(psf) (sq ft)



N + R(kips)


R --- 20 ---* --- 12.0 7.2 7.2

4 15 50 2,400 0.56 28.0 25.8 33.03 15 50 4,800 0.47 23.5 23.1 56.12 15 50 7,200 0.43 21.5 21.9 78.0

1 15 50 9,600 0.40 20.0 21.0 99.0

TLLAK






Axial load = (L + 15)AT = (L + 15) × 600

Distributed live load from the first level

Nonreducible live load = 15 psf

Reduced live load, L, for an interior beam:

psf

Total live load = 58 psf

3.4. Given the information provided in Problem 3.3, determine the reduced live loads on thetransfer beam using the alternative uniform live load reduction provisions ofIBC 1607.10.2. Assume a nominal dead-to-live load ratio of 0.75.

SOLUTION






• Roof



StoryLive Load

(psf) A (sq ft)

Reduction Factor R, (%)


N + R(kips)


R --- 20 ---* --- 12.0 7.2 7.24 15 50 600 36 32.0 28.2 35.4

3 15 50 1,200 40 30.0 27.0 62.42 15 50 1,800 40 30.0 27.0 89.41 15 50 2,400 40 30.0 27.0 116.4

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

50 0.25 15

4 AT×--------------------+

⎝ ⎠⎜ ⎟⎛ ⎞

×==

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

50 0.25 15

2 300×----------------------+⎝ ⎠

⎛ ⎞× 0.86 50× 43.0====

Chapter 3 17




Thus, Lr = 20 × 0.6 × 1 = 12.0 psf

Axial load = 12.0 × 600 / 1,000 = 7.2 kips

• Typical floors

R = 0.08(A – 150) ≤ smaller of 60% or 23.1(1 + D/Lo) = 23.1(1 + 0.75) = 40% (governs)

Axial load = [Lo(1 – 0.01R) + 15] × 600


Nonreducible live load = 15 psf

Reduced live load, L:

R = 0.08(A – 150) = 0.08 × (300 – 150) = 12%

L = 50 × (1 – 0.12) = 44 psf

Total live load = 59 psf

3.5. Given the information provided in Problem 3.3, determine the reduced live loads using thebasic uniform live load reduction provisions of IBC 1607.10.1 assuming that all of thefloors support patient rooms in a hospital.

SOLUTION






• Roof



StoryLive Load

(psf) (sq ft)



N + R(kips)


R --- 20 ---* --- 12.0 7.2 7.2

4 --- 40 2,400 0.56 22.4 13.4 20.63 --- 40 4,800 0.47 18.8 11.3 31.92 --- 40 7,200 0.43 17.2 10.3 42.2

1 --- 40 9,600 0.40 16.0 9.6 51.8

TLLAK





Thus, Lr = 20 × 0.6 × 1 = 12.0 psf

Axial load = 12.0 × 600 / 1,000 = 7.2 kips

• Typical floors


Since the supported column is an interior column, the live load element factor KLL = 4 (IBC Table 1607.10.1).





Axial load = LAT = L × 600


Reduced live load, L, for an interior beam:

psf

3.6. Determine the reduced live load at each floor level of an edge column in an 8-story parkinggarage with a tributary area of 1,080 square feet at each level using the basic uniform liveload reduction provisions of IBC 1607.10.1.

SOLUTION


L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

50 0.25 15

4 AT×--------------------+

⎝ ⎠⎜ ⎟⎛ ⎞

×==

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

40 0.25 15

2 300×----------------------+⎝ ⎠

⎛ ⎞× 0.86 40× 34.4====

Chapter 3 19

Table P3.6 Summary of Reduced Live Loads on Edge Column in Parking Garage.

N = nonreducible live load, R = reducible live load* Live load on member supporting one level in a parking garage cannot be reduced (IBC 1607.10.1.3)

Since the supported column is an edge column without a cantilever slab, the live load elementfactor KLL = 4 (IBC Table 1607.10.1).


≥ 0.80Lo for members supporting two or more floors (IBC 1607.10.1.3)

Axial load = LAT = L × 1,080

3.7. Determine the rain load, R, on a roof similar to the one depicted in Figure 3.10 given thefollowing design data:

• Tributary area of primary roof drain = 3,000 square feet

• Rainfall rate i = 3.75 inches/hour

• 6-inch-wide (b) channel scupper

• Vertical distance from primary roof drain to inlet of scupper (static head distance, ds) =3 inches

SOLUTION

To determine the rain load, R, the hydraulic head, dh, must be determined, based on the requiredflow rate.

Using Equation 3.7, determine the required flow rate, Q:

Q = 0.0104Ai = 0.0104 × 3,000 × 3.75 = 117.0 gallons/minute

Assume that the hydraulic head, dh, is determined by Equation 3.8, which is applicable forchannel scuppers:

Q = 2.9bdh1.5

where b = width of the scupper.

StoryLive Load

(psf) (sq ft)



N + R(kips)


8 --- 40 ---* --- 40.0 43.2 43.27 --- 40 8,640 0.80 32.0 34.6 77.86 --- 40 12,960 0.80 32.0 34.6 112.4

5 --- 40 17,280 0.80 32.0 34.6 147.04 --- 40 21,600 0.80 32.0 34.6 181.63 --- 40 25,920 0.80 32.0 34.6 216.2

2 --- 40 30,240 0.80 32.0 34.6 250.81 --- 40 34,560 0.80 32.0 34.6 285.4

TLL AK

L Lo 0.25 15

KLLAT

---------------------+⎝ ⎠⎜ ⎟⎛ ⎞

40 0.25 15

4 AT×--------------------+

⎝ ⎠⎜ ⎟⎛ ⎞

×==


For a required flow rate of 117.0 gallons/minute and b = 6 inches, dh = 3.6 inches.

Alternatively, by interpolating the values in ASCE/SEI Table C8-1 for a flow rate of 117.0gallons/minute, dh = 3.5 inches.

The rain load, R, is determined by IBC Equation 16-36:

R = 5.2(ds + dh) = 5.2(3 + 3.6) = 34.3 psf

3.8 Given the information provided in Problem 3.7, determine the rain load, R, assuming 4-inch diameter drains are used as the secondary drainage system.

SOLUTION

To determine the rain load, R, the hydraulic head, dh, must be determined, based on the requiredflow rate.

Using Equation 3.7, determine the required flow rate, Q, for each drain, which covers an area of3,000 square feet:

Q = 0.0104Ai = 0.0104 × 3,000 × 3.75 = 117.0 gallons/minute

Interpolating the values in ASCE/SEI Table C8-1 for a 4-inch diameter drain with a flow rate of117.0 gallons/minute yields a hydraulic head dh = 1.4 inches.

The rain load, R, is determined by IBC Equation 16-36:

R = 5.2(ds + dh) = 5.2(3 + 1.4) = 22.9 psf

CHAPTER 4Snow and Ice Loads

4.1. Given the one-story warehouse described in Example 4.1, determine design snow loads fora roof slope of (a) 3 on 12 and (b) 6 on 12. Use the design data given in the example.

SOLUTION

Part a: Roof slope of 3 on 12

1. Determine ground snow load, pg.

From Figure 7-1 or Figure 1608.2, the ground snow load is equal to 20 psf for St. Louis, MO.

2. Determine flat roof snow load, pf , by Equation 7.3-1.

Use Flowchart 4.1 to determine pf.

a. Determine exposure factor, Ce, from Table 7-2.

From the design data, the terrain category is C and the roof exposure is partially exposed. Therefore, Ce = 1.0 from Table 7-2.

b. Determine thermal factor, Ct, from Table 7-3.

From the design data, the structure is kept just above freezing during the winter, so Ct = 1.1 from Table 7-3.

c. Determine the importance factor, Is, from Table 1.5-2.

From IBC Table 1604.5, the Risk Category is II, based on the occupancy given in the design data. Thus, Is = 1.0 from Table 1.5-2.

Therefore,

pf = 0.7CeCtIspg = 0.7 × 1.0 × 1.1 × 1.0 × 20 = 15.4 psf

Check if the minimum snow load requirements are applicable:

A minimum roof snow load, pm, in accordance with 7.3.4 applies to hip and gable roofs with slopes less than 15 degrees. Since the roof slope in this problem is equal to 14.04 degrees, minimum roof snow loads must be considered.

Since pg = 20 psf, pm = Ispg = 20 psf

21


3. Determine sloped roof snow load, ps, by Equation 7.4-1.

Use Flowchart 4.2 to determine roof slope factor, Cs.

a. Determine thermal factor, Ct, from Table 7-3.

From item 2 above, thermal factor Ct = 1.1.

b. Determine if the roof is warm or cold.

Since Ct = 1.1, the roof is defined as a cold roof in accordance with 7.4.2.

c. Determine if the roof is unobstructed or not and if the roof is slippery or not.

There are no obstructions on the roof that inhibit the snow from sliding off the eaves. Also, the roof surface is a rubber membrane. According to 7.4, rubber membranes are considered to be slippery surfaces.

Since this roof is unobstructed and slippery, use the dashed line in Figure 7-2b to deter-mine Cs:

For a roof slope of 14.04 degrees,

Therefore, ps = Cs pf = 0.93 × 15.4 = 14.3 psf. This is the balanced snow load for this roof.

4. Consider partial loading.

Since all of the members are simply supported, partial loading is not considered (7.5).

5. Consider unbalanced snow loads.

Flowchart 4.4 is used to determine if unbalanced loads on this gable roof need to be consid-ered or not.

Unbalanced snow loads must be considered for this roof, since the slope is equal to 14.04 degrees (2.38 degrees < 14.04 degrees < 30.2 degrees).

Since W = 128 feet > 20 feet, the unbalanced load consists of the following (see Figure 7-5 and Figure 4.5):

• Windward side: 0.3ps = 0.3 × 14.3 = 4.3 psf

• Leeward side: ps = 14.3 psf along the entire leeward length plus a uniform pressure

of psf, which extends from the ridge a distance of

feet where

hd = drift height from Figure 7-9 with W = 128 feet substituted for lu

= 0.43(W)1/3(pg + 10)1/4 – 1.5 = 3.6 feet

γ = snow density (Eq. 7.7-1)

= 0.13pg + 14 = 16.6 pcf < 30 pcf

S = roof slope run for a rise of one = 4

6. Consider snow drifts on lower roofs and roof projections.

Not applicable.

Cs 114.04 10–( )

60------------------------------ 0.93=–=

hdγ S⁄ 3.6( 16.6 ) 4⁄× 29.9==

8hd S 3⁄ 8( 3.6 4 ) 3⁄×× 19.2==

Chapter 4 23

7. Consider sliding snow.

Not applicable.

8. Consider rain-on-snow loads.

In accordance with 7.10, a rain-on-snow surcharge of 5 psf is required for locations where the ground snow load, pg, is 20 psf or less (but not zero) with roof slopes less than W/50.

In this problem, pg = 20 psf and W/50 = 128/50 = 2.56 degrees, which is less than the roof slope of 14.04 degrees. Thus, an additional 5 psf need not be added to the balanced load of 14.3 psf.

9. Consider ponding instability.

Since the roof slope in this problem is greater than 1/4 inch/foot, progressive roof deflection and ponding instability from rain-on-snow or from snow meltwater need not be investigated (7.11 and 8.4).

10.Consider existing roofs.

Not applicable.

In this problem, the minimum roof snow load of 20 psf governs since it is greater than thebalanced snow load, which is equal to 14.3 psf. The balanced and unbalanced snow loads aredepicted in Figure P4.1a.

Part b: Roof slope of 6 on 121. Determine ground snow load, pg.





Figure P4-1a Balanced and Unbalanced Snow Loads for Warehouse Building in Problem 4.1, Part a




From the design data, the structure is kept just above freezing during the winter, so

Ct = 1.1 from Table 7-3.



Therefore, pf = 0.7CeCtIspg = 0.7 × 1.0 × 1.1 × 1.0 × 20 = 15.4 psf


A minimum roof snow load, pm, in accordance with 7.3.4 applies to hip and gable roofs with slopes less than 15 degrees. Since the roof slope in this problem is equal to 26.57 degrees, minimum roof snow loads need not be considered.









Since this roof is unobstructed and slippery, use the dashed line in Figure 7-2b to deter-mine Cs:

For a roof slope of 26.57 degrees, ,






Unbalanced snow loads must be considered for this roof, since the slope is equal to 26.57 degrees (2.38 degrees < 26.57 degrees < 30.2 degrees).



Cs 126.57 10–( )

60------------------------------ 0.72=–=

Chapter 4 25


psf, which extends from the ridge a distance of

feet where

hd = drift height from Figure 7-9 with W = 128 feet substituted for lu

= 0.43(W)1/3(pg + 10)1/4 – 1.5 = 3.6 feet


= 0.13pg + 14 = 16.6 pcf < 30 pcf



Not applicable.


Not applicable.



In this problem, pg = 20 psf and W/50 = 128/50 = 2.56 degrees, which is less than the roof slope of 26.57 degrees. Thus, an additional 5 psf need not be added to the balanced load of 14.3 psf.




Not applicable.

hdγ S⁄ 3.6( 16.6 ) 2⁄× 42.3==

8hd S 3⁄ 8( 3.6 2 ) 3⁄×× 13.6==


The balanced and unbalanced snow loads are depicted in Figure P4.1b.

4.2. Given the one-story warehouse building and adjoining office building in Example 4.3,determine the design snow loads on the roof of the office building. Assume that the heightof the office building at the eave is (a) 15 feet and (b) 18 feet instead of 10 feet. Use thedesign data given in the example.

SOLUTION

Part a: Office building height at eave = 15 feet1. Determine ground snow load, pg.







From the design data, the structure is heated with an unventilated roof, so Ct = 1.0 from Table 7-3.



Figure P4.1b Balanced and Unbalanced Snow Loads for Warehouse Building in Problem 4.1, Part b

Chapter 4 27










Since Ct = 1.0, the roof is defined as a warm roof in accordance with 7.4.1.


In accordance with the design data, the roof surface is asphalt shingles. According to 7.4, asphalt shingles are not considered to be slippery.

Also, since the roof is unventilated with an R-value less than 30 ft2⋅h⋅°F/Btu, it is possi-ble for an ice dam to form at the eave, which can prevent the snow from sliding off of the roof (7.4.5). This is considered to be an obstruction.

Thus, use the solid line in Figure 7-2a to determine Cs:

For a roof slope of 2.38 degrees, Cs = 1.0.

Therefore, ps = Cspf = 1.0 × 14.0 = 14.0 psf.

In accordance with 7.4.5, a uniformly distributed load of 2pf = 2 × 14.0 = 28.0 psf must be applied on the 5-foot overhanging portion of the roof to account for ice dams. Only the dead load is to be present when this uniformly distributed load is applied.


It is assumed that the roof purlins are connected to the wood trusses by metal hangers, which are essentially simple supports, so partial loads do not have to be considered for the roof purlins. Therefore, with a spacing of 5 feet, the uniform snow load on a purlin is equal to 20.0 × 5.0 = 100 plf (minimum snow load governs).

The roof trusses are continuous over the masonry wall; thus, partial loading must be consid-ered (7.5). The balanced snow load to be used in partial loading cases is that determined by Equation 7.4-1, which is equal to 14.0 psf. With a spacing of 25 feet, the balanced and par-tial loads on a typical roof truss are

Balanced load = 14.0 × 25.0 = 350 plf

Partial load = one-half of balanced load = 175 plf

The balanced and partial load cases that must be considered for the roof trusses in this prob-lem are the same as those in Figure 4.21, including the ice dam load on the overhang, which was determined in item 3 above. Note that the minimum snow load of 20 psf is not applica-ble in the partial load cases and in the ice dam load case.



Not applicable.


Use Flowchart 4.7 to determine the leeward and windward drifts that form on the lower (office) roof.

a. Determine ground snow load, pg.

From item 1 above, the ground snow load is equal to 20 psf for St. Louis, MO.

b. Determine snow density, γ, by Equation 7.7-1.

γ = 0.13pg + 14 = (0.13 × 20) + 14 = 16.6 pcf < 30 pcf

c. Determine sloped roof snow load, ps, from Flowchart 4.3.

From item 3 above, ps = 14.0 psf.

d. Determine clear height, hc.

In this problem, the clear height, hc, is from the top of the balanced snow to the top of the warehouse eave (see Figure 4.12). The height of the balanced snow ishb = ps /γ = 14.0/16.6 = 0.8 foot.

Thus, hc = (5 – 25 tan 2.38°) – 0.8 = 3.2 feet

e. Determine if drift loads are required or not.

Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem, hc / hb = 3.2/0.8 = 4.0 > 0.2, so drift loads must be considered.

f. Determine drift load.

Both leeward and windward drift heights, hd, must be determined by the provisions of 7.7.1. The larger of these two heights is used to determine the drift load.

• Leeward drift

A leeward drift occurs when snow from the warehouse roof is deposited by wind to the office roof.

For leeward drifts, the drift height, hd, is determined from Figure 7-9 using the length of the upper roof lu = lupper. In this problem, lu = 256 feet and the ground snow load pg = 20 psf. Using the equation in Figure 7-9:

hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(256)1/3(20 + 10)1/4 – 1.5 = 4.9 feet

• Windward drift

A windward drift occurs when snow from the office roof is deposited adjacent to the wall of the warehouse building.

For windward drifts, the drift height, hd, is 75 percent of that determined from Figure 7-9 using lu = llower:

hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(30)1/3(20 + 10)1/4 – 1.5 = 1.2 feet

Thus, the leeward drift controls and hd = 4.9 feet.

Chapter 4 29

Since hd = 4.9 feet > hc = 3.2 feet, the drift width w = 4hd2 / hc = 4 × 4.92 / 3.2 = 30 feet,

which is greater than 8hc = 25.6 feet. Thus, use a drift width of 25.6 feet. Also, the drift height, hd, is set equal to hc = 3.2 feet in accordance with 7.7.1.

The maximum surcharge drift load pd = hdγ = 3.2 × 16.6 = 53.1 psf.

The total load at the step is the balanced load on the office roof plus the drift surcharge = 14.0 + 53.1 = 67.1 psf, which is illustrated in Figure P4.2a.

The snow loads on the purlins and trusses are obtained by multiplying the loads depicted in Figure P4.2a by the respective tributary widths. As expected, the purlins closest to the ware-house have the largest loads.


The provisions of 7.9 are used to determine if a load due to snow sliding off of the ware-house roof on to the office roof must be considered.

Load caused by snow sliding must be considered since the warehouse roof is slippery with a slope greater than 1/4 on 12. This load is in addition to the balanced load acting on the lower roof.

The total sliding load per unit length of eave is equal to 0.4pf W where pf is the flat roof snow load of the warehouse and W is the horizontal distance from the eave to the ridge of the warehouse roof. From Example 4.1, pf = 15.4 psf and W = 128 feet. Thus, the sliding load = 0.4 × 15.4 × 128 = 789 plf. This load is to be uniformly distributed over a distance of 15 feet from the warehouse eave (see Figure 4.16). Thus, the sliding load is equal to 789/15 = 52.6 psf.

The total load over the 15-foot width is equal to the balanced snow load on the office roof plus the sliding load = 14.0 + 52.6 = 66.6 psf. The total depth of snow for the total load is equal to 66.6/16.6 = 4.0 feet, which is essentially equal to the distance from the warehouse eave to the top of the office roof at the interface, which is 5 – (25 tan 2.38°) = 3.96 feet.

Figure P4.2a Balanced and Drift Loads on Office Roof, Problem 4.2, Part a


Thus, sliding snow is not blocked and the full load can be developed over the 15-foot length. See Figure 4.23 for the load case including sliding snow.



In this problem, pg = 20 psf and W/50 = 25/50 = 0.5 degree, which is less than the roof slope of 2.38 degrees. Thus, rain-on-snow loads are not considered.




Not applicable.

Part b: Office building height at eave = 18 feet1. Determine ground snow load, pg.
















Chapter 4 31
















Partial load = one half of balanced load = 175 plf



Not applicable.







γ = 0.13pg + 14 = (0.13 × 20) + 14 = 16.6 pcf < 30 pcf





Thus, hc = (2 – 25 tan 2.38°) – 0.8 = 0.16 foot


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem,hc / hb = 0.16/0.8 = 0.2; thus, drift loads must be considered.



i. Leeward drift



hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(256)1/3(20 + 10)1/4 –1.5 = 4.9 feet

ii. Windward drift



hd = 0.75[0.43(lu)1/3(pg+10)1/4 –1.5] = 0.75[0.43(30)1/3(20+10)1/4 –1.5] = 1.2 feet


Since hd = 4.9 feet > hc = 0.16 foot, the drift width w = 4hd2 / hc = 4 × 4.92 / 0.16 = 600

feet, which is greater than 8hc = 1.3 feet. Thus, use a drift width of 1.3 feet. Also, the drift height, hd, is set equal to hc = 0.16 foot in accordance with 7.7.1.


Chapter 4 33

The total load at the step is the balanced load on the office roof plus the drift surcharge = 14.0 + 2.7 = 16.7 psf, which is illustrated in Figure P4.2b.

The snow loads on the purlins and trusses are obtained by multiplying the loads depicted in Figure P4.2b by the respective tributary widths. As expected, the purlins closest to the ware-house have the largest loads.



Load caused by snow sliding must be considered since the warehouse roof is slippery with a slope greater than 1/4 on 12. This load is in addition to the balanced load acting on the lower roof.

The total sliding load per unit length of eave is equal to 0.4pf W where pf is the flat roof snow load of the warehouse and W is the horizontal distance from the eave to the ridge of the warehouse roof. From Example 4.1, pf = 15.4 psf and W = 128 feet. Thus, the sliding load = 0.4 × 15.4 × 128 = 789 plf. This load is to be uniformly distributed over a distance of 15 feet from the warehouse eave (see Figure 4.16). Thus, the sliding load is equal to 789/15 = 52.6 psf.

The total load over the 15-foot width is equal to the balanced snow load on the office roof plus the sliding load = 14.0 + 52.6 = 66.6 psf. The total depth of snow for the total load is equal to 66.6/16.6 = 4.0 feet, which is greater than the distance from the warehouse eave to the top of the office roof at the interface, which is equal to 2 – (25 tan 2.38°) = 0.96 foot. Thus, sliding snow is blocked and a fraction of the sliding snow is forced to remain on the upper roof. Based on the 0.96-foot available storage space, the total load on the lower roof over a 15-foot width is 0.96 × 16.6 = 15.9 psf.

Figure P4.2b Balanced and Drift Loads on Office Roof, Problem 4.2, Part b


See Figure P4.2c for the load case including sliding snow.







Not applicable.

Figure P4.2c Sliding Snow Load on Office Roof, Problem 4.2,Part b

Chapter 4 35

4.3. Given the information provided in Problem 4.2, determine the design snow loads on theroof of the office building for parts (a) and (b). Assume that the warehouse and officebuilding are separated horizontally a distance of 2 feet.

SOLUTION

Part a: Office building height at eave = 15 feet1. Determine ground snow load, pg.


2. Determine flat roof snow load, pf, by Equation 7.3-1.

































Not applicable.






γ = 0.13pg + 14 = (0.13 × 20) + 14 = 16.6 pcf < 30 pcf





Chapter 4 37

Thus, hc = (5 – 25 tan 2.38°) – 0.8 = 3.2 feet


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem,hc / hb = 3.2/0.8 = 4.0 > 0.2, so drift loads must be considered.


The provisions of 7.7.2 are used to determine leeward and windward drift heights since the adjacent structures are separated.

i. Leeward drift


Check if leeward drifts form on the lower roof (see Figure 4.14):

• s = 2 feet < 20 feet• s = 2 feet < 6h = 6 × (5 – 25 tan 2.38°) = 23.8 feet

Thus, leeward drifts must be considered.

For leeward drifts on separated structures, the drift height, hd, is the smaller of that determined from Figure 7-9 using the length of the upper roof lu = lupper and (6h – s)/6. In this problem, lu = 256 feet and the ground snow load pg = 20 psf. Using the equation in Figure 7-9:

hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(256)1/3(20 + 10)1/4 –1.5 = 4.9 feet

Also, hd = (6h – s)/6 = (23.8 – 2)/6 = 3.6 feet (governs)

The surcharge load is 3.6 × 16.6 = 59.8 psf. The horizontal extent of the leeward drift surcharge is the smaller of 6hd = 21.8 feet and 6h – s = 21.8 feet.

ii. Windward drift


For windward drifts on separated structures, the drift height, hd, is 75 percent of that determined from Figure 7-9 using lu = llower:

hd = 0.75[0.43(lu)1/3(pg+10)1/4 – 1.5] = 0.75[0.43(30)1/3(20+10)1/4 –1.5] = 1.2 feet

The surcharge load at the face of the warehouse is equal to 1.2 × 16.6 = 19.9 psf (see Figure 4.15). The truncated magnitude of pd at the face of the office is [1 – (2.0/4.8)] × 19.9 = 11.6 psf, which acts over a length equal to 4hd – s = 2.8 feet.



Load caused by snow sliding must be considered since the warehouse roof is slippery with a slope greater than 1/4 on 12, h/s = 3.96/2 = 1.98 > 1 and s = 2 feet < 15 feet. This load is in addition to the balanced load acting on the lower roof.

The total sliding load per unit length of eave is equal to 0.4pf W(15 – s)/15 where pf is the flat roof snow load of the warehouse and W is the horizontal distance from the eave to the ridge of the warehouse roof. From Example 4.1, pf = 15.4 psf and W = 128 feet. Thus, the


sliding load = 0.4 × 15.4 × 128 × (15 – 2)/15 = 684 plf. This load is to be uniformly dis-tributed over a distance of 15 – s = 13 feet from the warehouse eave (see Figure 4.17). Thus, the sliding load is equal to 684/13 = 52.6 psf.

The total load over the 13-foot width is equal to the balanced snow load on the office roof plus the sliding load = 14.0 + 52.6 = 66.6 psf. The total depth of snow for the total load is equal to 66.6/16.6 = 4.0 feet.







Not applicable.

Part b: Office building height at eave = 18 feet1. Determine ground snow load, pg.



Use Flowchart 4.1 to determine, pf.











Chapter 4 39
















The roof trusses are continuous over the masonry wall; thus, partial loading must be con-sidered (7.5). The balanced snow load to be used in partial loading cases is that deter-mined by Equation 7.4-1, which is equal to 14.0 psf. With a spacing of 25 feet, the balanced and partial loads on a typical roof truss are





Not applicable.







γ = 0.13pg + 14 = (0.13 × 20) + 14 = 16.6 pcf < 30 pcf





Thus, hc = (2 – 25 tan 2.38°) – 0.8 = 0.16 foot


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem,hc / hb = 0.16/0.8 = 0.2; thus, drift loads must be considered.




i. Leeward drift


Check if leeward drifts form on the lower roof (see Figure 4.14):

• s = 2 feet < 20 feet• s = 2 feet < 6h = 6 × (2 – 25 tan 2.38°) = 5.8 feet

Thus, leeward drifts must be considered.

For leeward drifts on separated structures, the drift height, hd, is the smaller of that determined from Figure 7-9 using the length of the upper roof lu = lupper and (6h – s)/6. In this problem, lu = 256 feet and the ground snow load pg = 20 psf. Using the equation in Figure 7-9:

hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(256)1/3(20 + 10)1/4 – 1.5 = 4.9 feet

Also, hd = (6h – s)/6 = (5.8 – 2)/6 = 0.63 foot (governs)

The surcharge load is 0.63 × 16.6 = 10.5 psf. The horizontal extent of the leeward drift surcharge is the smaller of 6hd = 3.8 feet and 6h – s = 3.8 feet.

ii. Windward drift


Chapter 4 41

For windward drifts on separated structures, the drift height, hd, is 75 percent of that determined from Figure 7-9 using lu = llower:

hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(30)1/3(20 +10)1/4 – 1.5] = 1.2 feet

The surcharge load at the face of the warehouse is equal to 1.2 × 16.6 = 19.9 psf (see Figure 4.15). The truncated magnitude of pd at the face of the office is [1 – (2.0/4.8)] × 19.9 = 11.6 psf, which acts over a length equal to 4hd – s = 2.8 feet.



Load caused by snow sliding need not be considered since the warehouse roof is slippery with a slope greater than 1/4 on 12, h/s = 0.96/2 = 0.48 < 1 and s < 15 feet.







Not applicable.

4.4. Determine the design snow loads for the buildings in Figure 4.35 given the design data inTable 4.3. The width of both roofs is 75 feet, and both roofs have a 1/4 on 12 slope into thepage. All roof members are simply supported.

Figure 4.35 Buildings in Problem 4.4



SOLUTION

Heated, unventilated building

1. Determine ground snow load, pg.From the design data, the ground snow load is equal to 40 psf.




From the design data, the terrain category is C. Assume that the roof exposure is par-tially exposed due to the presence of the adjoining taller building. Therefore, Ce = 1.0 from Table 7-2.


From the design data, the structure is heated, so Ct = 1.0 from Table 7-3.





A minimum roof snow load, pm, in accordance with 7.3.4 applies to monoslope roofs with slopes less than 15 degrees. Since the roof slope in this problem is equal to 1.19 degrees, minimum roof snow loads must be considered.

Since pg = 40 psf > 20 psf, pm = 20Is = 20 psf







Ground snow load: 40 psf

Terrain category: COccupancy: OrdinaryThermal condition: As noted in Figure 4.35

Roof exposure condition: Fully exposedRoof surface: Rubber membrane

Chapter 4 43



Since this roof is unobstructed and slippery, use the dashed line in Figure 7-2a to deter-mine Cs:






Flowchart 4.4 is used to determine if unbalanced loads on this gable roof need to be con-sidered or not.

Unbalanced snow loads need not be considered because the roof slope is less than 2.38 degrees.


Use Flowchart 4.7 to determine the leeward and windward drifts that form on the lower roof.


From the design data, the ground snow load is equal to 40 psf.


γ = 0.13pg + 14 = (0.13 × 40) + 14 = 19.2 pcf < 30 pcf




In this problem, the height of the balanced snow is hb = ps /γ = 28.0/19.2 = 1.5 feet.

Thus, hc = 15 – 1.5 = 13.5 feet


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem,hc / hb = 13.5/1.5 = 9.0 > 0.2; thus, drift loads must be considered.



i. Leeward drift


hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(70)1/3(40 + 10)1/4 – 1.5 = 3.2 feet


ii. Windward drift


hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(85)1/3(40 +10)1/4 – 1.5] = 2.7 feet


Since hd = 3.2 feet < hc = 13.7 feet, the drift width w = 4hd = 4 × 3.2 = 12.8 feet, which is less than 8hc = 109.6 feet. Thus, use a drift width of 12.8 feet.


The total load at the step is the balanced load on the office roof plus the drift surcharge = 28.0 + 61.4 = 89.4 psf.


Not applicable.



In this problem, pg = 40 psf so an additional 5 psf need not be added to the balanced load of 28.0 psf.


Since the roof slope in this example is equal to 1/4 inch/foot, progressive roof deflection and ponding instability from rain-on-snow or from snow meltwater need not be investigated (7.11 and 8.4).


Not applicable.

Unheated storage building

1. Determine ground snow load, pg.From the design data, the ground snow load is equal to 40 psf.




From the design data, the terrain category is C and the roof exposure is fully exposed. Therefore, Ce = 0.9 from Table 7-2.


From the design data, the structure is unheated, so Ct = 1.2 from Table 7-3.



Chapter 4 45



A minimum roof snow load, pm, in accordance with 7.3.4 applies to monoslope roofs with slopes less than 15 degrees. Since the roof slope in this problem is equal to 1.19 degrees, minimum roof snow loads must be considered.

Since pg = 40 psf > 20 psf, pm = 20Is = 20 psf









Since this roof is unobstructed and slippery, use the dashed line in Figure 7-2c to deter-mine Cs:



All other snow loads are not applicable.


4.5. Determine the design snow loads for the building in Figure 4.36 using the design data inTable 4.4. All roof members are simply supported.


SOLUTION


From the design data, the ground snow load, pg, is equal to 30 psf.



Figure 4.36 Building in Problem 4.5


Terrain category: BOccupancy: OfficeThermal condition: Heated

Roof exposure condition: Partially exposedRoof surface: Bituminous

Chapter 4 47


From the design data, the terrain category is B and the roof exposure is partially exposed. Therefore, Ce = 1.0 from Table 7-2.


From the design data, the roof is heated, so Ct = 1.0 from Table 7-3.



Therefore, pf = 0.7CeCtIpg = 0.7 × 1.0 × 1.0 × 1.0 × 30 = 21.0 psf








There are no obstructions on the roof that inhibit the snow from sliding off the eaves. Also, the roof surface is a rubber membrane. According to 7.4, bituminous membranes are considered to be slippery surfaces.

Since no information was provided regarding the R-value for the roof, use the solid line in Figure 7-2a to determine Cs:



4. Consider drift loading at parapet walls.

According to 7.8, drift loads at parapet walls and other roof projections are determined using the provisions of 7.7.1.

Windward drifts occur at parapet walls, and Flowchart 4.7 is used to determine the wind-ward drift load.

a. Determine snow density, γ, by Equation 7.7-1.

γ = 0.13pg + 14 = (0.13 × 30) + 14 = 17.9 pcf < 30 pcf

b. Determine clear height, hc.

The clear height, hc, is from the top of the balanced snow to the top of the parapet wall. The height of the balanced snow is equal to hb = ps/ γ = 21.0/17.9 = 1.2 feet.

Thus, hc = 5.0 – 1.2 = 3.8 feet.

c. Determine if drift loads are required or not.

Drift loads are not required where hc /hb < 0.2 (7.7.1). In this problem, hc /hb = 3.8/1.2 = 3.2 > 0.2, so drift loads must be considered.


d. Determine drift load.

Windward drift height, hd, must be determined by the provisions of 7.7.1 using three-quarters of the drift height, hd, from Figure 7-9 with lu equal to the length of the roof upwind of the parapet wall (7.8).

The equation in Figure 7-9 yields the following for the drift height, hd, based on a ground snow load pg = 30 psf and an upwind fetch lu = 100 feet:

hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(100)1/3(30 + 10)1/4 – 1.5] = 2.6 feet

Since hd = 2.6 feet < hc = 3.8 feet, the drift width w = 4hd = 4 × 2.6 = 10.4 feet.

The maximum surcharge drift load pd = hd γ = 2.6 × 17.9 = 46.5 psf.

The total load at the face of the parapet wall is the balanced load plus the drift surcharge = 21.0 + 46.5 = 67.5 psf.

Since the ground snow load, pg, exceeds 20 psf, the minimum snow load is 20Is = 20 × 1.0 = 20 psf (7.3), which is less than ps = 21.0 psf. Also, a rain-on-snow surcharge load is not required in accordance with 7.10.

4.6. Determine the design snow loads for the building in Figure 4.37 using the design data inTable 4.5. All roof members are simply supported.


SOLUTION

Snow loads on upper roof


From Table 1608.2 or Table 7-1, the ground snow load is equal to 60 psf for Fairbanks, Alaska.


Location: Fairbanks, Alaska

Terrain category: C

Occupancy: ResidentialThermal condition: Ventilated roof with R > 25

Roof exposure condition: Fully exposedRoof surface: Asphalt

Chapter 4 49






From the design data, the structure has a ventilated roof with R > 25, so Ct = 1.1 from Table 7-3.













There are no obstructions on the roof that inhibit the snow from sliding off the eaves. Also, the roof surface has asphalt shingles. According to 7.4, asphalt shingles are not considered to be slippery.

Since this roof is unobstructed and not slippery, use the solid line in Figure 7-2b to deter-mine Cs:








Unbalanced snow loads must be considered for this roof, since the slope is equal to 18.4 degrees.




of psf, which extends from the ridge a distance of

feet where

hd = drift length from Figure 7-9 with W = 30 feet substituted for lu

= 0.43(W)1/3(pg + 10)1/4 – 1.5 = 2.4 feet


= 0.13pg + 14 = 21.8 pcf < 30 pcf



Not applicable.


Not applicable.







Not applicable.

Snow loads on lower roof

1. Determine ground snow load, pg.From Table 1608.2 or Table 7-1, the ground snow load is equal to 60 psf for Fairbanks, Alaska.



hdγ S⁄ 2.4( 21.8 ) 3⁄× 30.2==

8hd S 3⁄ 8( 2.4 3 ) 3⁄×× 11.1==

Chapter 4 51


From the design data, the terrain category is C. Assume the roof exposure is partially exposed due to the presence of the taller portion of the building. Therefore, Ce = 1.0 from Table 7-2.







A minimum roof snow load, pm, in accordance with 7.3.4 applies to hip and gable roofs with slopes less than 15 degrees. Since the roof in this example is essentially flat, mini-mum roof snow loads must be considered.

Since pg = 60 psf, pm = 20Is = 20 psf










For a flat roof, Cs = 1.0.





Not applicable.





From item 1 above, the ground snow load is equal to 60 psf for Fairbanks, Alaska.


γ = 0.13pg + 14 = (0.13 × 60) + 14 = 21.8 pcf < 30 pcf




In this problem, the clear height, hc, is from the top of the balanced snow to the top of the eave of the upper building (see Figures 4.12 and 4.37). The height of the balanced snow is hb = ps /γ = 46.2/21.8 = 2.1 feet.

Thus, hc = 10 – 2.1 = 7.9 feet


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem,hc / hb = 7.9/2.1 = 3.8 > 0.2, so drift loads must be considered.



i. Leeward drift

A leeward drift occurs when snow from the upper roof is deposited by wind to the lower roof.


hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(60)1/3(60 + 10)1/4 – 1.5 = 3.4 feet

ii. Windward drift

A windward drift occurs when snow from the lower roof is deposited adjacent to the wall of the taller portion of the building.


hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(35)1/3(60 + 10)1/4 – 1.5] = 1.9

feet





Chapter 4 53


The provisions of 7.9 are used to determine if a load due to snow sliding off of the upper roof on to the lower roof must be considered.

Load caused by snow sliding must be considered, since the upper roof is not slippery with a slope greater than 2 on 12. This load is in addition to the balanced load acting on the lower roof.

The total sliding load per unit length of eave is equal to 0.4pf W where pf is the flat roof snow load of the upper roof and W is the horizontal distance from the eave to the ridge of the upper roof. It was determined above that pf = 41.6 psf and W = 30 feet. Thus, the sliding load = 0.4 × 41.6 × 30 = 499 plf. This load is to be uniformly distributed over a distance of 15 feet from the eave of the upper roof (see Figure 4.16). Thus, the sliding load is equal to 499/15 = 33.3 psf.

The total load over the 15-foot width is equal to the balanced snow load on the lower roof plus the sliding load = 46.2 + 33.3 = 79.5 psf. The total depth of snow for the total load is equal to 79.5/21.8 = 3.7 feet, which is less than the distance from the eave of the upper roof to the top of the lower roof at the interface, which is 10 feet. Thus, sliding snow is not blocked and the full load can be developed over the 15-foot length.



In this problem, pg = 60 psf so, rain-on-snow loads need not be considered.




Not applicable.


4.7. Determine the design snow loads for the building in Figure 4.38 using the design data inTable 4.6. The roof profile is an arc of a circle that has a radius of 75 feet. All roofmembers are simply supported.


SOLUTION

1. Determine ground snow load, pg.From the design data, the ground snow load, pg, is equal to 100 psf.




From the design data, the terrain category is B and the roof exposure is fully exposed. Therefore, Ce = 0.9 from Table 7-2.


From the design data, the structure is intermittently heated. Thus, Ct = 1.0 from Table 7-3.



Terrain category: B

Occupancy: AssemblyThermal condition: Intermittently heatedRoof exposure condition: Fully exposed

Roof surface: Slippery

Chapter 4 55


From IBC Table 1604.5, the Risk Category is III for this assembly building. Thus, Is = 1.1 from Table 1.5-2.



A minimum roof snow load, pm, in accordance with 7.3 applies to curved roofs where the vertical angle from the eaves to the crown is less than 10 degrees. Since that slope in this problem is equal to 20.9 degrees, minimum roof snow loads need not be considered.

3. Determine sloped roof snow load, ps, from Eq. 7.4-1.







There are no obstructions on the roof that inhibit the snow from sliding off the eaves. Also, from the design data, the roof surface is defined to be slippery.

Use the solid line in Figure 7-2a to determine Cs since no information has been provided regarding R for the roof.

For the tangent slope of 41.8 degrees at the eave, the roof slope factor, Cs, is determined by the equation in C7.4 for warm roofs with Ct = 1.0:

Therefore, ps = Cs pf = 0.71 × 69.3 = 49.2 psf, which is the balanced snow load at the eaves (see Case 2 in Figure 7-3).

Away from the eaves, the roof slope factor, Cs, is equal to 1.0 where the tangent roof slope is less than or equal to approximately 30 degrees (see solid line in Figure 7-2a). Since the roof profile is an arc of a circle, this occurs at the following distance from the eaves at both ends of the roof:

feet

Therefore, in the 100 – (2 × 12.5) = 75 foot center portion of the roof, ps = Cs pf = 1.0 × 69.3 = 69.3 psf.

Cs 1.0 slope 30°–( )40°

--------------------------------– 1.0 41.8° 30°–( )40°

--------------------------------– 0.71= = =

50 752

tan230×

1 tan230+

-------------------------------– 12.5=


The balanced snow load is depicted in Figure P4.7, which is based on Case 2 in Figure 7-3 for a slope at the eaves equal to 30 through 70 degrees.


Flowchart 4.5 is used to determine if unbalanced loads on this curved roof need to be con-sidered or not.

Since the slope of the straight line from the eaves to the crown is greater than 10 degrees and less than 60 degrees, unbalanced snow loads must be considered (7.6.2).

Unbalanced loads for this roof are given in Case 2 of Figure 7-3 (see Figure 4.7). No snow loads are applied on the windward side.

On the leeward side, the snow load is equal to 0.5pf = 0.5 × 69.3 = 34.7 psf at the crown. The snow load is 2pf Cs|30 /Ce = 2 × 69.3 × 1.0/0.9 = 154.0 psf at the 30-degree point and is 2pf

Cs|eave /Ce = 2 × 69.3 × 0.71/0.9 = 109.3 psf at the eaves. The unbalanced snow loads are shown in Figure P4.7.

Figure P4.7 Balanced and Unbalanced Snow Loads, Problem 4.7

Chapter 4 57

4.8. Determine the design snow loads for the building in Problem 4.7. Assume that another roofabuts it within 3 feet of its eaves.

SOLUTION

The only difference occurs in the calculation of the unbalanced snow load at the location wherethe other roof abuts the curved roof.

From Figure 4.7, the snow load at the edge of the curved roof is equal to 2pf Cs|30 /Ce =2 × 69.3 × 1.0/0.9 = 154.0 psf. This snow load is constant from the 30-degree point to the edgeof the roof.

4.9. Determine the design snow loads for the building in Problem 4.7. Assume the following:R = 150 feet instead of 75 feet, and the distance to the eave = 143 feet instead of 50 feet.

SOLUTION

Given the information on the geometry of the roof in the problem statement, determine thefollowing geometric properties of the roof:

• Tangent roof slope at eave:

degrees

• Slope of straight line from eave to crown:

degrees

• Location from eave where tangent slope of roof = 70 degrees:

feet

• Slope of straight line from 70-degree point to crown:

degrees

• Location from eave where tangent slope of roof = 30 degrees:

feet

1. Determine ground snow load, pg.From the design data, the ground snow load, pg, is equal to 100 psf.



θeave arctan143

1502

1432

–

---------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

72.4==

θ arctan150 150

2143

2––

143------------------------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

36.2==

143 1502

tan270×

1 tan270+

----------------------------------– 2=

θ arctan150 150

2141

2––

141------------------------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

35.0==

143 1502

tan2× 30

1 tan230+

----------------------------------– 68=



From the design data, the terrain category is B and the roof exposure is fully exposed. Therefore, Ce = 0.9 from Table 7-2.


From the design data, the structure is intermittently heated. Thus, Ct = 1.0 from Table 7-3.


From IBC Table 1604.5, the Risk Category is III for this assembly building. Thus, Is = 1.1 from Table 1.5-2.



A minimum roof snow load, pm, in accordance with 7.3 applies to curved roofs where the vertical angle from the eaves to the crown is less than 10 degrees. Since that slope in this problem is equal to 36.2 degrees, minimum roof snow loads need not be considered.

3. Determine sloped roof snow load, ps, from Eq. 7.4-1.

No snow load is required for a distance of 2 feet from the eave at either end of the roof (that is, in the areas of the roof where the tangent slope is greater than 70 degrees; see 7.6.2).







There are no obstructions on the roof that inhibit the snow from sliding off the eaves. Also, from the design data, the roof surface is defined to be slippery.

Use the solid line in Figure 7-2a to determine Cs since no information has been provided regarding R for the roof.

For the tangent slope of 30 degrees, which is located 68 feet from the eave, the roof slope factor Cs = 1.0.

Therefore, ps = Cs pf = 1.0 × 69.3 = 69.3 psf, which is the balanced snow load over the 286 – (2 × 68) = 150 feet central portion of the roof (see Case 3 in Figure 7-3).


Flowchart 4.5 is used to determine if unbalanced loads on this curved roof need to be con-sidered or not.

Since the slope of the straight line from the 70-degree point to the crown is equal to 35 degrees, which is greater than 10 degrees and less than 60 degrees, unbalanced snow loads must be considered (7.6.2).

Unbalanced loads for this roof are given in Case 3 of Figure 7-3 (see Figure 4.8). No snow loads are applied on the windward side.

Chapter 4 59

On the leeward side, the snow load is equal to 0.5pf = 0.5 × 69.3 = 34.7 psf at the crown. The snow load is 2pf Cs|30 /Ce = 2 × 69.3 × 1.0/0.9 = 154.0 psf at the 30-degree point and decreases linearly to zero at the 70-degree point.

4.10. Determine the design snow loads on the roof of the building and the canopy in Figure4.39. Use the same design data as in Problem 4.6.

SOLUTION

Snow loads on roof


From Table 1608.2 or Table 7-1, the ground snow load is equal to 60 psf for Fairbanks, Alaska.





























Unbalanced snow loads need not be considered for this roof, since the slope is equal to 33.7 degrees.


Not applicable.


Not applicable.



Chapter 4 61





Not applicable.

Snow loads on canopy

1. Determine ground snow load, pg.From Table 1608.2 or Table 7-1, the ground snow load is equal to 60 psf for Fairbanks, Alaska.




From the design data, the terrain category is C. Assume the exposure of the canopy is partially exposed due to the presence of the building. Therefore, Ce = 1.0 from Table 7-2.


Since the canopy is an open air structure, Ct = 1.2 from Table 7-3.



Therefore, pf = 0.7CeCtIs pg = 0.7 × 1.0 × 1.2 × 1.0 × 60 = 50.4 psf


A minimum roof snow load, pm, in accordance with 7.3.4 applies to hip and gable roofs with slopes less than 15 degrees. Since the canopy in this problem is essentially flat, minimum roof snow loads must be considered.

Since pg = 60 psf, pm = 20Is = 20 psf








c. Determine if the canopy is unobstructed or not and if the roof is slippery or not.

There are no obstructions on the canopy that inhibit the snow from sliding off. Also, assume that the top surface of the canopy has asphalt shingles. According to 7.4, asphalt shingles are not considered to be slippery.

Since this canopy is unobstructed and not slippery, use the solid line in Figure 7-2b to deter-mine Cs:

For a flat canopy, Cs = 1.0.



Partial loading is not considered (7.5).


Not applicable.


Use Flowchart 4.7 to determine the leeward and windward drifts that form on the canopy.


From item 1 above, the ground snow load is equal to 60 psf for Fairbanks, Alaska.


γ = 0.13pg + 14 = (0.13 × 60) + 14 = 21.8 pcf < 30 pcf




In this problem, the clear height, hc, is from the top of the balanced snow to the top of the eave of the building. The height of the balanced snow is hb = ps /γ = 50.4/21.8 = 2.3 feet.

Thus, hc = 10 – 2.3 = 7.7 feet


Drift loads are not required where hc / hb < 0.2 (7.7.1). In this problem, hc / hb = 7.7/2.3 = 3.4 > 0.2, so drift loads must be considered.



i. Leeward drift

A leeward drift occurs when snow from the roof is deposited by wind to the canopy.


hd = 0.43(lu)1/3(pg + 10)1/4 – 1.5 = 0.43(40)1/3(60 + 10)1/4 – 1.5 = 2.8 feet

Chapter 4 63

ii. Windward drift

A windward drift occurs when snow from the canopy is deposited adjacent to the wall of the building.


hd = 0.75[0.43(lu)1/3(pg + 10)1/4 – 1.5] = 0.75[0.43(12)1/3(60 + 10)1/4 – 1.5] = 1.0 foot






The provisions of 7.9 are used to determine if a load due to snow sliding off of the roof on to the canopy must be considered.

Load caused by snow sliding must be considered, since the roof is not slippery with a slope greater than 2 on 12. This load is in addition to the balanced load acting on the canopy.

The total sliding load per unit length of eave is equal to (12/15) × 0.4pf W where pf is the flat roof snow load of the roof, W is the horizontal distance from the eave to the ridge of the roof, and the ratio (12/15) accounts for the permitted reduction since the length of the can-opy is less than 15 feet. With pf = 50.4 psf and W = 20 feet, the sliding snow load = (12/15) × 0.4 × 50.4 × 20 = 323 plf. This load is to be uniformly distributed over a distance of 12 feet from the eave of the roof. Thus, the sliding load is equal to 323/12 = 26.9 psf.

The total load over the 12-foot width is equal to the balanced snow load on the canopy plus the sliding load = 50.4 + 26.9 = 77.3 psf. The total depth of snow for the total load is equal to 77.3/21.8 = 3.6 feet, which is less than the distance from the eave of the roof to the top of the canopy, which is 10 feet. Thus, sliding snow is not blocked and the full load can be developed over the 12-foot length.



In this example, pg = 60 psf so rain-on-snow loads need not be considered.


Since the roof slope in this example is greater than 1/4 inch/foot, progressive roof deflection and ponding instability from rain-on-snow or from snow meltwater need not be investigated (7.11 and 8.4).


Not applicable.


4.11. Determine the weight of ice that can form on the support structure in Example 4.9.Assume the structural members are (a) HSS6.000 × 0.500 and (b) HSS6 × 4 × 1/2. Use thesame design data in Example 4.9.

SOLUTION

The general procedure in 10.8 is used to determine the ice weight, Di.

1. Determine nominal ice thickness, t, and the concurrent wind speed, Vc.From Figure 10-2, the nominal ice thickness, t, is equal to 1.0 inch and the concurrent wind speed, Vc, is equal to 40 mph for the structure located in Indianapolis, IN.

2. Determine the topographic factor, Kzt.

Since the building is located on a relatively flat site, Kzt = 1.0 in accordance with 10.4.5.

3. Determine the importance factor, Ii.

Since the commercial building is classified under Risk Category II, Ii = 1.0 from Table 1.5-2.

4. Determine height factor, fz.

The height factor is determined by Equation 10.4-4:

for 0 < z ≤ 900 feet

5. Determine the design ice thickness, td.

The design ice thickness is determined by Equation 10.4-5:

td = 2.0tIi fz(Kzt)0.35 = 2.0 × 1.0 × 1.0 × 1.08 × (1.0)0.35 = 2.2 inches

6. Determine the weight of ice, Di.

The weight of ice for a structural shape is determined by multiplying the area of ice deter-mined by Equation 10.4-1 and the density of ice, which is taken as 56 pcf (10.4.1).

Part a: HSS6.000 × 0.500

The diameter, Dc, for this round tube is equal to its diameter, which is 6 inches.

The area of ice is determined by Equation 10.4-1:

Ai = πtd(Dc + td) = π × 2.2 × (6.0 + 2.2) = 56.7 square inches

Thus, the weight of ice is the following:

Di = (56.7/144) × 56 = 22.1 plf

Part b: HSS6 × 4 × 1/2

Referring to Figure 10-1, the diameter of the cylinder circumscribing the structural steel tube inthis case is conservatively taken as the hypotenuse of the right triangle formed by the legs of thetube:

inches (the actual Dc = 6.5 inches)Dc 6( )24( )2

+ 7.2= =

fzz

33------⎝ ⎠⎛ ⎞ 0.10 70

33------⎝ ⎠⎛ ⎞ 0.10

1.08= = =

Chapter 4 65


Ai = πtd(Dc + td) = π × 2.2 × (7.2 + 2.2) = 65.0 square inches


Di = (65.0/144) × 56 = 25.3 plf

4.12.Determine the wind-on-ice load of a round chimney that is 60 feet tall and has an outsidediameter of 6 feet. Assume the chimney is located in Milwaukee, WI, on a relatively flatsite in Exposure C.

SOLUTION

The general procedure in 10.8 is used to determine the ice weight, Di.

1. Determine nominal ice thickness, t, and the concurrent wind speed, Vc.From Figure 10-2, the nominal ice thickness, t, is conservatively equal to 0.75 inch and the concurrent wind speed, Vc, is equal to 40 mph for the structure located in Milwaukee, WI.

2. Determine the topographic factor, Kzt.

Since the chimney is located on a relatively flat site, Kzt = 1.0 in accordance with 10.4.5.

3. Determine the importance factor, Ii.

Assume the chimney is classified under Risk Category II; thus, Ii = 1.0 from Table 1.5-2.

4. Determine height factor, fz.

The height factor is determined by Equation 10.1-4:

for 0 < z ≤ 900 feet

5. Determine the design ice thickness, td.

The design ice thickness is determined by Equation 10.4-5:

td = 2.0tIi fz(Kzt)0.35 = 2.0 × 0.75 × 1.0 × 1.06 × (1.0)0.35 = 1.6 inches

6. Determine the weight of ice, Di.

The weight of ice for a structural shape is determined by multiplying the area of ice deter-mined by Equation 10.4-1 and the density of ice, which is taken as 56 pcf (10.4.1).

The diameter, Dc, for this round chimney is equal to its diameter, which is 6 feet.


Ai = πtd(Dc + td) = π × 1.6 × (72 + 1.6) = 370.0 square inches


Di = (370.0/144) × 56 = 143.9 plf

fzz

33------⎝ ⎠⎛ ⎞ 0.10 60

33------⎝ ⎠⎛ ⎞ 0.10

1.06= = =


7. Determine velocity pressure, qz, at the centroid of the projected area of the chimney normal to the wind for the concurrent wind speed, Vc.

Equation 29.3-1 is used to determine qz:

qz = 0.00256KzKztKdVc2

The velocity pressure exposure coefficient, Kz, is determined from Table 29.3-1. At a height of 30 feet and Exposure C, Kz = 0.98.

For a round chimney, the wind directionality factor Kd = 0.95 (see Table 26.6-1).

Thus, qz = 0.00256 × 0.98 × 1.0 × 0.95 × 402 = 3.8 psf

8. Determine wind force coefficient, Cf.

The net force coefficient, Cf, for a round chimney is determined in accordance with Figure

29.5-1 corresponding to for all ice thicknesses (10.5.1).

For h/D = 60/6 = 10, Cf = 0.9 from linear interpolation.

9. Determine the gust factor, G, in accordance with 26.9.

Assuming the structure is rigid, G = 0.85.

10.Determine the wind-on-ice load, Wi, in accordance with 29.4.1.

The wind-on-ice load, Wi, is determined by Equation 29.5-1:

Wi = F = qzGCf Af = 3.8 × 0.85 × 0.9 × (60 × 6) = 1,050 lb

D qz 2.5≤

CHAPTER 5Wind Loads

5.1. Given the one-story warehouse described in Example 5.1, determine design wind pressureson the MWFRS in both directions using Part 1 of Chapter 27 assuming the roof slope isequal to (a) 1 inch/foot and (b) 3 inch/foot.

SOLUTION

Part a: Roof slope of 1 inch/footStep 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3 so

that this procedure can be used to determine the wind pressures on the MWFRS.

The building is regularly-shaped, that is, it does not have any unusual geometric irregu-larities in spatial form. Also, the building does not have response characteristics that make it subject to across-wind loading or other similar effects, and it is not sited at a location where channeling effects or buffeting in the wake of upwind obstructions need to be considered.

Thus, Part 1 of Chapter 27 may be used to determine the design wind pressures on the MWFRS.

Step 2: Determine the enclosure classification of the building.

The enclosure classification of the building depends on the number and size of openings in the precast walls.

Assume that there are two Type A precast panels on each of the east and west walls and two Type B precast panels on each of the north and south walls (see Figure 5.24). The openings in these walls are door openings. All other precast panels do not have door openings, and assume that there are no openings in the roof.

By definition, an open building is a building having each wall at least 80 percent open. It is evident that this building is not open since the area of openings on each wall is less than 80 percent of the gross area of the wall (see Figure 5.9). Therefore, it is enclosed or partially enclosed.

A building is defined as enclosed when it does not comply with the requirements for an open or partially enclosed building (26.2).

Calculate the area of each wall and the area of the openings in each wall:

• East and West walls

Ag = 20 × 148 = 2,960 square feet

Ao = 2[(12 × 14) + (3 × 7)] = 378 square feet

67


• North and South walls

Ag = (20 × 256) + [0.5 × 256 × (128 × tan 4.76°)] = 6,484 square feet

Ao = 2 × [(12 × 14) + (12 × 14)] = 672 square feet

• Roof

Ag = 2 × [148 × (128/cos 4.76°)] = 38,019 square feet

Ao = 0

• Totals

Ag = (2 × 2,960) + (2 × 6,484) + 38,019 = 56,907 square feet

Ao = 2 × (378 + 672) = 2,100 square feet

Check the conditions for a partially enclosed building:


Ao/Aoi = 378/(2,100 – 378) = 0.22 < 1.1 (condition is not met)

Ao= 378 square feet

> 4 square feet (governs) or 0.01Ag = 0.01 × 2,960 = 29.6 square feet (condition is met)

Aoi/Agi = (2,100 – 378)/(56,907 – 2,960) = 0.03 < 0.20 (condition is met)



Ao= 672 square feet



Since all of the conditions for each set of walls are not met and since the building is not open, the building is defined as enclosed.

Step 3: Use Flowchart 5.3 to determine the design wind pressures, p, on the MWFRS.

Since the roof angle is less than 10 degrees, the height at the eave is used as h (26.3). Thus, the wind pressures for this problem are exactly the same as those determined in Example 5.1.

Part b: Roof slope of 3 inch/footStep 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3 so


As shown in Part a of this problem, Part 1 of Chapter 27 may be used to determine the design wind pressures on the MWFRS.



Chapter 5 69


Ag = 20 × 148 = 2,960 square feet

Ao = 2[(12 × 14) + (3 × 7)] = 378 square feet


Ag = (20 × 256) + [0.5 × 256 × (128 × tan 14.04°)] = 9,216 square feet

Ao = 2 × [(12 × 14) + (12 × 14)] = 672 square feet

• Roof

Ag = 2 × [148 × (128/cos 14.04°)] = 39,054 square feet

Ao = 0

• Totals

Ag = (2 × 2,960) + (2 × 9,216) + 39,054 = 63,406 square feet

Ao = 2 × (378 + 672) = 2,100 square feet




Ao = 378 square feet





Ao = 672 square feet



Since all of the conditions for each set of walls are not met and since the building is not open,the building is defined as enclosed.


Since the roof angle is greater than 10 degrees, the mean roof height, h, is used in pres-sure calculations (26.3): h = [20 + (128 × tan 14.04°)] + 20/2 = 36 feet.

1. Determine whether the building is rigid or flexible in accordance with 26.9.2.

In Example 5.1, the building was determined to be rigid in both directions.

2. Determine the gust-effect factor, G, using Flowchart 5.1.

According to 26.9.4, gust-effect factor, G, for rigid buildings may be taken as 0.85 or can be calculated by Equation 26.9-6. For simplicity, use G = 0.85.


3. Determine velocity pressure qz for windward walls along the height of the building and qh for leeward walls, side walls and roof using Flowchart 5.2.

a. Determine the risk category of the building using IBC Table 1604.5.

Due to the nature of its occupancy, this warehouse building falls under Risk Category II.

b. Determine basic wind speed, V, for the applicable risk category from IBC Figure 1609A, B or C, or ASCE/SEI Figure 26.5-1A, B or C.

For Risk Category II, use Figure 1609A or Figure 26.5-1A. From either of these figures, V = 115 mph for St. Louis, MO.

c. Determine wind directionality factor, Kd, from Table 26.6-1.

For the MWFRS of a building structure, Kd = 0.85.

d. Determine exposure category.

Exposure C applies (see 26.7.3).

e. Determine topographic factor, Kzt.

Topographic factor Kzt = 1.0 (26.8.2).

f. Determine velocity pressure exposure coefficients Kz and Kh fromTable 27.3-1.

Values of Kz and Kh for Exposure C are summarized in Table P5.1a.

Table P5.1a Velocity Pressure Exposure Coefficient, Kz

g. Determine velocity pressure, qz and qh, by Equation 27.3-1.

qz = 0.00256KzKztKdV2 = 0.00256 × Kz × 1.0 × 0.85 × 1152 = 28.78Kz psf

A summary of the velocity pressures is given in Table P5.1b.

Table P5.1b Velocity Pressure, qz

4. Determine pressure coefficients, Cp, for the walls and roof from Figure 27.4-1.

For wind in the E-W direction:

Windward wall: Cp = 0.8 for use with qz

Height above ground level, z (feet) Kz

36 1.0230 0.9825 0.94

20 0.9015 0.85

Height above ground level, z (feet) Kzqz

(psf)36 1.02 29.430 0.98 28.2

25 0.94 27.120 0.90 25.915 0.85 24.5

Chapter 5 71

Leeward wall (L/B = 256/148 = 1.7): Cp = –0.35 (from linear interpolation) for use

with qh

Side wall: Cp = –0.7 for use with qh

Roof (normal to ridge with θ = 14.04° > 10° and h/L = 36/256 = 0.14 < 0.25):

Windward: Cp = –0.54, –0.04 (from linear interpolation) for use with qh

Leeward: Cp = –0.46 (from linear interpolation) for use with qh

Note: The smaller uplift pressures on the windward portion of the roof due toCp = –0.04 may govern the design when combined with roof live load or snow loads. This pressure is not shown in this example but in general must be considered.

For wind in the N-S direction:


Leeward wall (L/B = 148/256 = 0.6): Cp = –0.5 for use with qh


Roof (parallel to ridge with h/L = 36/148 = 0.24 < 0.5):

Cp = –0.9, –0.18 from windward edge to h = 36 feet for use with qh

Cp = –0.5, –0.18 from 36 feet to 2h = 72 feet for use with qh

Cp = –0.3, –0.18 from 72 feet to 148 feet for use with qh

5. Determine qi for the walls and roof using Flowchart 5.2.

In accordance with 27.4.1, qi = qh = 29.4 psf for windward walls, side walls, leeward walls and roofs of enclosed buildings.

6. Determine internal pressure coefficients, GCpi, from Table 26.11-1.

For an enclosed building, GCpi = +0.18, –0.18.

7. Determine design wind pressures pz and ph by Equation 27.4-1.

Windward walls:

pz = qzGCp – qh(GCpi)

= (qz × 0.85 × 0.8) – 29.4(±0.18)

= (0.68qz 5.3) psf (external ± internal pressure)

Leeward wall, side walls and roof:

ph = qhGCp – qh(GCpi)

= (29.4 × 0.85 × Cp) – 29.4(±0.18)

= (25.0Cp 5.3) psf (external ± internal pressure)

A summary of the maximum design wind pressures in the E-W and N-S directions is given in Tables P5.1c and P5.1d, respectively.


Table P5.1c Design Wind Pressures, p, in E-W Direction

Table P5.1d Design Wind Pressures, p, in N-S Direction

* from windward edge to 36 feet† from 36 to 72 feet‡ from 72 to 148 feet

The minimum design wind loading prescribed in 27.1.5 must be considered as a load case as well (see Figure C27.4-1 and Figure 5.10).

5.2. Given the one-story warehouse described in Example 5.1, determine design wind pressureson (1) a solid precast wall panel and (2) an open-web joist purlin using Part 1 of Chapter 30assuming the roof slope is equal to (a) 1 inch/foot and (b) 3 inch/foot.

SOLUTION

Part 1: Determine design wind pressures on a solid precast wall panelPart a: Roof slope of 1 inch/foot

Step 1: Check if the building meets all of the conditions and limitations of 30.1.2 and 30.1.3and the additional conditions of 30.4 so that this procedure can be used to determine thewind pressures on the C&C.

LocationHeight above ground level,

z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)

Net pressure, p (psf)

+(GCpi) –(GCpi)

Windward wall

36 29.4 20.0 ±5.3 14.7 25.330 28.2 19.2 ±5.3 13.9 24.525 27.1 18.4 ±5.3 13.1 23.7

20 25.9 17.6 ±5.3 12.3 22.915 24.5 16.7 ±5.3 11.4 22.0

Leeward wall All 29.4 –8.8 ±5.3 –14.1 –3.5

Side walls All 29.4 –17.5 ±5.3 –22.8 –12.2

Roof36 (Windward) 29.4 –13.5 ±5.3 –18.8 –8.236 (Leeward) 29.4 –11.5 ±5.3 –16.8 –6.2


z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)


+(GCpi) –(GCpi)

Windward wall

36 29.4 20.0 ±5.3 14.7 25.330 28.2 19.2 ±5.3 13.9 24.5

25 27.1 18.4 ±5.3 13.1 23.720 25.9 17.6 ±5.3 12.3 22.915 24.5 16.7 ±5.3 11.4 22.0

Leeward wall All 29.4 –12.5 ±5.3 –17.8 –7.2Side walls All 29.4 –17.5 ±5.3 –22.8 –12.2

Roof

36 29.4 –22.5* ±5.3 –27.8 –17.2

36 29.4 –12.5† ±5.3 –17.8 –7.236 29.4 –7.5‡ ±5.3 –12.8 –2.2

Chapter 5 73


Also, the mean roof height h = 20 feet < 60 feet and the building is also a low-rise build-ing as defined in 26.2.

Thus, Part 1 of Chapter 30 may be used to determine the design wind pressures on the C&C.

Step 2: Use Flowchart 5.8 to determine the net design wind pressures, p, on the C&C.

Since the roof angle is less than 10 degrees, the height at the eave is used as h (26.3). Thus, the wind pressures on the solid precast wall panels for this part of the problem are exactly the same as those determined in Example 5.14.

Part b: Roof slope of 3 inch/foot


Part 1 of Chapter 30 may be used to determine the design wind pressures on the C&C.


1. Determine velocity pressure, qh, using Flowchart 5.2 with Kh determined by Table 30.3-1.

Since the values of Kh in Tables 27.3-1 and 30.3-1 are identical for Exposure C, the value of qh calculated in Part b of Problem 5.1 is also applicable in this problem; thus, qh = 29.4 psf.

2. Determine external pressure coefficients (GCp) from Figure 30.4-1 for Zones 4 and 5.

Pressure coefficients for Zones 4 and 5 can be determined from Figure 30.4-1 based on the effective wind area.

The effective wind area is defined as the span length multiplied by an effective width that need not be less than one-third the span length: 20 × (20/3) = 133.3 square feet (note: the smallest span length corresponding to the east and west walls is used, since this results in larger pressures).

The pressure coefficients from the figure are summarized in Table P5.2a.

Table P5.2a External Pressure Coefficients (GCp) for Precast Walls

The width of the end zone a = least of 0.1 (least horizontal dimension) = 0.1 × 148 = 14.8 feet or 0.4h = 0.4 × 36 = 14.4 feet (governs), which is greater than 0.04 (least hori-zontal dimension) = 0.04 × 148 = 5.9 feet or 3 feet (see Note 6 in Figure 30.4-1).

Zone(GCp)

Positive Negative4 0.80 –0.90

5 0.80 –1.00


3. Determine internal pressure coefficients (GCpi) from Table 26.11-1.

For an enclosed building, (GCpi) = +0.18, –0.18.

4. Determine design wind pressure, p, by Equation 30.4-1 on Zones 4 and 5.

p = qh[(GCp) – (GCpi)] = 29.4[(GCp) – (±0.18)]

Calculation of design wind pressures is illustrated for Zone 4:

For positive (GCp): p = 29.4[0.80 – (–0.18)] = 28.8 psf

For negative (GCp): p = 29.4[–0.90 – (+0.18)] = –31.8 psf

The maximum design wind pressures for positive and negative internal pressures are summarized in Table P5.2b. These pressures act perpendicular to the face of the precast walls.

Table P5.2b Design Wind Pressures, p, on Precast Walls

In Zones 4 and 5, the computed positive and negative pressures are greater than the minimum values prescribed in 30.2.2 of +16 psf and –16 psf, respectively.

Part 2: Determine design wind pressures on an open-web joist purlinPart a: Roof slope of 1 inch/foot

Since the roof angle is less than 10 degrees, the height at the eave is used as h (26.3). Thus, thewind pressures on the open-web joist purlins for this part of the problem are exactly the same asthose determined in Example 5.14.

Part b: Roof slope of 3 inch/foot


As shown previously, Part 1 of Chapter 30 may be used to determine the design wind pressures on the C&C.


1. Determine velocity pressure, qh, using Flowchart 5.2 with Kh determined by Table 30.3-1.

From Part 1 of this problem, qh = 29.4 psf.

Zone Design Pressure, p (psf)

4 0.80 28.8

–0.90 –31.8

5 0.80 28.8–1.00 –34.7

)( pGC

Chapter 5 75

2. Determine external pressure coefficients (GCp) from Figure 30.4-2B for Zones 1, 2 and 3.

Pressure coefficients for Zones 1, 2 and 3 can be determined from Figure 30.4-2B for gable roofs with a roof slope between 7 degrees and 27 degrees based on the effective wind area.

Effective wind area = larger of 37 × 8 = 296 square feet or 37 × (37/3) =456.3 square feet (governs).

The pressure coefficients from the figure are summarized in Table P5.2c.

Table P5.2c External Pressure Coefficients (GCp) for Open-web Joist Purlins


For an enclosed building, (GCpi) = +0.18, –0.18.

4. Determine design wind pressure, p, by Equation 30.4-1 on Zones 1, 2 and 3.

p = qh[(GCp) – (GCpi)] = 29.4[(GCp) – (±0.18)]

The maximum design wind pressures for positive and negative internal pressures are summarized in Table P5.2d.

Table P5.2d Design Wind Pressures, p, on Open-web Joist Purlins

The pressures in Table P5.2d are applied normal to the open-web joist purlins and act over the tributary area of each purlin, which is equal to 37 × 8 = 296 square feet. If the tributary area were greater than 700 square feet, the purlins could have been designed using the provisions for MWFRSs (30.2.3).

The positive pressures on Zones 1, 2 and 3 must be increased to the minimum value of 16 psf per 30.2.2.

5.3. The plan of a two-story building is shown in Figure 5.51. Determine the design windpressures on the MWFRS in the E-W direction by (a) Part 1 of Chapter 27 and (b) Part 2 ofChapter 27 using the provided design data in Table 5.86. The first floor is located 20 feet

Surface(GCp)

Positive Negative1 0.30 –0.802 0.30 –1.20

3 0.30 –2.00

Zone (GCp) Design Pressure, p (psf)

1 0.30 14.1–0.80 –28.8

2 0.30 14.1

–1.20 –40.6

3 0.30 14.1–2.00 –64.1


above ground and the roof, which has an angle less than 5 degrees from the horizontal, islocated 32 feet above ground. Use the floor heights as the reference heights and assume thebuilding is rigid.


SOLUTION

Part 1: Design Wind Pressures, Part 1 of Chapter 27Step 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3 so





The enclosure classification of the building is given in the design data as enclosed.



The building is defined as rigid in the design data.

Figure 5.51Plan of Two-story Building in Problem 5.3

Wind velocity, V: 115 mphExposure category: B

Enclosure classification: EnclosedOccupancy: ResidentialTopography: No hills or escarpments

Chapter 5 77





Due to the nature of its occupancy, this building falls under Risk Category II.


The wind velocity is given in the design data as 115 mph.


For the MWFRS of a building structure, Kd = 0.85


In the design data, the exposure category is given as B.


As noted in the design data, the building is not situated on a hill, ridge or escarp-ment. Thus, topographic factor Kzt = 1.0 (26.8.2).


Values of Kz and Kh for Exposure B are summarized in Table P5.3a.











32 0.7120 0.62

Height above ground level, z (feet) Kz qz (psf)

32 0.71 20.420 0.62 17.8



Roof (normal to ridge with θ < 10 degrees and h/L = 32/100 = 0.32 < 0.5):




Note: The smaller uplift pressures on the roof due to Cp = –0.18 may govern the design when combined with roof live load or snow loads. This pressure is not shown in this example, but in general must be considered.






Windward walls:


= (qz × 0.85 × 0.8) – 20.4(±0.18)




= (20.4 × 0.85 × Cp) – 20.4(±0.18)


A summary of the maximum design wind pressures in the E-W direction is given in Table P5.3c.

Chapter 5 79




Part 2: Design Wind Pressures, Part 2 of Chapter 27Step 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3

and the additional conditions of 27.5.2 and 27.5.4 so that this procedure can be used todetermine the wind pressures on the MWFRS.


Check the conditions for a Class 1 Building (27.5.2):

1. The building is enclosed. Assume that the building is a simple diaphragm building as defined in 26.2, that is, the windward and leeward wind loads are transmitted through a diaphragm to the MWFRS and there are no structural separations in the MWFRS.

2. Mean roof height h = 32 feet < 60 feet

3. In the E-W direction, L/B = 100/50 = 2.0, which is greater than 0.2 and less than 5.0

4. Topographic effect factor Kzt = 1.0

Therefore, the building is a Class 1 Building.

Check the condition for diaphragm flexibility (27.5.4):

Part 2 of Chapter 27 is to be applied to buildings with either rigid or flexible dia-phragms. Assume that a rigid diaphragm is provided in this problem.



z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)


+(GCpi) –(GCpi)

Windward wall

32 20.4 13.9 ±3.7 10.2 17.6

20 17.8 12.1 ±3.7 8.4 15.8

Leeward wall All 20.4 –5.2 ±3.7 –8.9 –1.5

Side walls All 20.4 –12.1 ±3.7 –15.8 –8.4

Roof

20 20.4 –15.6* ±3.7 –19.3 –11.9

20 20.4 –8.7† ±3.7 –12.4 –5.0

20 20.4 –5.2‡ ±3.7 –8.9 –1.5



1. Determine the risk category of the building using IBC Table 1604.5.


2. Determine basic wind speed, V, for the applicable risk category from IBC Figure 1609A, B or C, or ASCE/SEI Figure 26.5-1A, B or C.

From the design data, V = 115 mph.

3. Determine exposure category.


4. Determine topographic factor, Kzt.


5. Determine the net pressures in the E-W direction on the walls at the top ph and at the base po using Table 27.6-1.

• Along-wind net wall pressures

The along-wind net wall pressures are obtained by reading the values from Table 27.6-1 for Exposure B, a wind velocity of 115 mph, a mean roof height of 32 feet and L/B = 100/50 = 2.0:

At the top of the wall: ph = 18.9 psf (by linear interpolation)

At the bottom of the wall: po = 16.9 psf (by linear interpolation)

These pressures are applied to the projected area of the building walls in the direction of the wind (see Figure 27.6-1 and Figure 5.14).

Note 4 in Table 27.6-1 gives information on how to distribute the along-wind net wall pressures between windward and leeward wall faces based on the ratio L/B.

• Side wall external pressures

According to Note 2 in Table 27.6-1, the uniform side wall external pressures, which are applied to the projected area of the building walls in the direction nor-mal to the direction of the wind, are equal to 0.64ph = 12.1 psf for2.0 ≤ L/B ≤ 5.0.

It is important to note that these pressures do not include the effects from internal pressure.

• Roof pressures

Roof pressures are obtained by reading the values from Table 27.6-2 for Expo-sure C, a wind velocity of 115 mph, a flat roof and a mean roof height of 32 feet, and then modifying these pressures for Exposure B using the exposure adjust-ment factors in the table. Note that “NA” is given in Table 27.6-2 for Zones 1 and 2 for a roof with a slope less than 2:12. Thus, these Zones are not applicable on this flat roof; however Zones 3 through 5 are applicable.

Zone 3: p3 = –30.4 psf (from linear interpolation) applied normal to the roof area from the windward edge to 0.5h = 16 feet from the windward edge (see Figure

Chapter 5 81

27.6-1 and Table 27.6-2). Adjusted pressure p3 = 0.716 × (–30.4) = –21.8 psf (adjustment factor obtained from linear interpolation).

Zone 4: p4 = –27.0 psf applied normal to the roof area from 0.5h = 16 feet from the windward edge to h = 32 feet from the windward edge (see Figure 27.6-1 and Table 27.6-2). Adjusted pressure p4 = 0.716 × (–27.0) = –19.4 psf.

Zone 5: p5 = –22.2 psf applied normal to the remaining roof area (see Figure 27.6-1 and Table 27.6-2). Adjusted pressure p5 = 0.716 × (–22.2) = –15.9 psf.


5.4. Given the plan dimensions in Figure 5.51 and the design data in Table 5.87, determine thedesign wind pressures on the MWFRS in the E-W direction by (a) Part 1 of Chapter 27 and(b) Part 2 of Chapter 27 assuming that the building has five stories, each with a storyheight of 15 feet. The roof angle is less than 5 degrees from the horizontal. Use the floorheights as the reference heights and assume the building is rigid.

SOLUTION

Part 1: Design Wind Pressures, Part 1 of Chapter 27Step 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3 so








The building is defined as rigid in the design data.
















Values of Kz and Kh for Exposure B are summarized in Table P5.4a.












75 0.9160 0.8545 0.79

30 0.7015 0.57


75 0.91 26.260 0.85 24.5

45 0.79 22.730 0.70 20.215 0.57 16.4

Chapter 5 83





Note: The smaller uplift pressures on the roof due to Cp = –0.18 may govern the design when combined with roof live load or snow loads. This pressure is not shown in this example but in general must be considered.






Windward walls:


= (qz × 0.85 × 0.8) – 26.2(±0.18)




= (26.2 × 0.85 × Cp) – 26.2(±0.18)


A summary of the maximum design wind pressures in the E-W direction is given in Table P5.4c.





Part 2: Design Wind Pressures, Part 2 of Chapter 27Step 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3

and the additional conditions of 27.5.2 and 27.5.4 so that this procedure can be used todetermine the wind pressures on the MWFRS.


Check the conditions for a Class 2 Building (27.5.2):

1. The building is enclosed. Assume that the building is a simple diaphragm building as defined in 26.2, that is, the windward and leeward wind loads are transmitted through a diaphragm to the MWFRS and there are no structural separations in the MWFRS.

2. Mean roof height 60 feet < h = 75 feet < 160 feet

3. In the E-W direction, L/B = 100/50 = 2.0, which is greater than 0.5 and is equal to the upper limit of 2.0

4. Since the building is rigid, n1 ≥ 1.0 = 75/h = 75/75 = 1.0

5. Topographic effect factor Kzt = 1.0

Therefore, the building is a Class 2 Building.


z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)


+(GCpi) –(GCpi)

Windwardwall

75 26.2 17.8 ±4.7 13.1 22.5

60 24.5 16.7 ±4.7 12.0 21.4

45 22.7 15.4 ±4.7 10.7 20.1

30 20.2 13.7 ±4.7 9.0 18.4

15 16.4 11.2 ±4.7 6.5 15.9

Leeward wall All 26.2 –6.7 ±4.7 –11.4 –2.0

Side walls All 26.2 –15.6 ±4.7 –20.3 –10.9

Roof

20 26.2 – 20.1* ±4.7 –24.8 –15.4

20 26.2 –11.2† ±4.7 –15.9 –6.5

20 26.2 –6.7‡ ±4.7 –11.4 –2.0

Chapter 5 85

Check the condition for diaphragm flexibility (27.5.4):

Part 2 of Chapter 27 is to be applied to buildings with either rigid or flexible dia-phragms. Assume that a rigid diaphragm is provided in this problem.






From the design data, V = 115 mph.





5. Determine the net pressures in the E-W direction on the walls at the top (ph) and at the base (po) using Table 27.6-1.

• Along-wind net wall pressures

The along-wind net wall pressures are obtained by reading the values from Table 27.6-1 for Exposure B: a wind velocity of 115 mph, a mean roof height of 75 feet and L/B = 100/50 = 2.0:

At the top of the wall: ph = 26.2 psf (by linear interpolation)

At the bottom of the wall: po = 19.2 psf (by linear interpolation)

These pressures are applied to the projected area of the building walls in the direction of the wind (see Figure 27.6-1 and Figure 5.14).

Note 4 in Table 27.6-1 gives information on how to distribute the along-wind net wall pressures between windward and leeward wall faces based on the ratio L/B.

• Side wall external pressures

According to Note 2 in Table 27.6-1, the uniform side wall external pressures, which are applied to the projected area of the building walls in the direction nor-mal to the direction of the wind, are equal to 0.64ph = 16.8 psf for2.0 ≤ L/B ≤ 5.0.

It is important to note that these pressures do not include the effects from internal pressure.

• Roof pressures

Roof pressures are obtained by reading the values from Table 27.6-2 for Expo-sure C, a wind velocity of 115 mph, a flat roof and a mean roof height of 75 feet,


and then modifying these pressures for Exposure B using the exposure adjust-ment factors in the table. Note that “NA” is given in Table 27.6-2 for Zones 1 and 2 for a roof with a slope less than 2:12. Thus, these zones are not applicable on this flat roof; however Zones 3 through 5 are applicable.

Zone 3: p3 = –36.3 psf (from linear interpolation) applied normal to the roof area from the windward edge to 0.5h = 37.5 feet from the windward edge (see Figure 27.6-1 and Table 27.6-2). Adjusted pressure p3 = 0.764 × (–36.3) = –27.7 psf (adjustment factor obtained from linear interpolation).

Zone 4: p4 = –32.4 psf applied normal to the roof area from 0.5h = 16 feet from the windward edge to h = 32 feet from the windward edge (see Figure 27.6-1 and Table 27.6-2). Adjusted pressure p4 = 0.764 × (–32.4) = –24.8 psf.

Zone 5: p5 = –26.6 psf applied normal to the remaining roof area (see Figure 27.6-1 and Table 27.6-2). Adjusted pressure p5 = 0.764 × (–26.6) = –20.3 psf.


5.5. Given the data in Problem 5.3, determine the design wind pressures on the MWFRS in theE-W direction by (a) Part 1 of Chapter 28 and (b) Part 2 of Chapter 28. Assume continuousroof and floor diaphragms.

SOLUTION

Part a: Design Wind Pressures, Part 1 of Chapter 28Step 1: Check if the building meets all of the conditions and limitations of 28.1.2 and 28.1.3

and the additional condition that the building is a low-rise building so that this proce-dure can be used to determine the wind pressures on the MWFRS.


Two conditions must be checked to determine if a building is a low-rise building (26.2):

1. Mean roof height = 32 feet < 60 feet

2. Mean roof height = 32 feet < least horizontal dimension = 50 feet

Therefore, this building is a low-rise building.


Chapter 5 87

Step 2: Use Flowchart 5.5 to determine the net design wind pressures, p, on the MWFRS.

1. Determine velocity pressure, qh, using Flowchart 5.2.










The building is not situated on a hill, ridge or escarpment. Thus, topographic fac-tor Kzt = 1.0 (26.8.2).

f. Determine velocity pressure exposure coefficients, Kh, from Table 28.3-1.

For a mean roof height of 32 feet and Exposure B, Kh = 0.71.

g. Determine velocity pressure, qh, by Equation 28.3-1.

qh = 0.00256KhKztKdV2 = 0.00256 × 0.71 × 1.0 × 0.85 × 1152 = 20.4 psf

2. Determine external pressure coefficients (GCpf) from Figure 28.4-1.

External pressure coefficients (GCpf) can be read directly from Figure 28.4-1 using a roof angle between 0 and 5 degrees in the E-W direction (Load Case A) and are summarized in Table P5.5a.

Table P5.5a External Pressure Coefficients (GCpf ) for MWFRS


For an enclosed building, (GCpi) = + 0.18, – 0.18.

Surface(GCpf)

Load Case A1 0.402 –0.693 –0.37

4 –0.291E 0.612E –1.07

3E –0.534E –0.43


4. Determine design wind pressures, p, by Equation 28.4-1.

p = qh[(GCpf) – (GCpi)] = 20.4[(GCpf) – (±0.18)]

Calculations of design wind pressures are illustrated for surface 1 for wind in the E-W direction (Load Case A):

For positive internal pressure: p = 20.4(0.40 – 0.18) = 4.5 psf

For negative internal pressure: p = 20.4[0.40 – (–0.18)] = 11.8 psf

A summary of the design wind pressures is given in Table P5.5b. Pressures are pro-vided for both positive and negative internal pressures. These pressures act normal to the surface.

Table P5.5b Design Wind Pressures, p, on MWFRS

The distance a = least of 0.1 (least horizontal dimension) = 0.1 × 50 = 5.0 feet (gov-erns) or 0.4h = 0.4 × 32 = 12.8 feet. This value of a is greater than 0.04 (least hori-zontal dimension) = 0.04 × 50 = 2.0 feet or 3 feet (see Note 9 in Figure 28.4-1).

According to Note 8 in Figure 28.4-1, when the roof pressure coefficients (GCpf) are negative in Zone 2 or 2E, they shall be applied in Zone 2/2E for a distance from the edge of the roof equal to 50 percent of the horizontal dimension of the building that is parallel to the direction of the MWFRS being designed or 2.5 times the eave height, he, at the windward wall, whichever is less. The remainder of Zone 2/2E extending to the ridge line must use the pressure coefficients (GCpf) for Zone 3/3E.


0.5 × 100 = 50 feet (governs)

2.5he = 2.5 × 32 = 80 feet

Therefore, in the E-W direction, Zone 2/2E applies over a distance of 50 feet from the edge of the windward roof, and Zone 3/3E applies over a distance of 50 feet in what is normally considered to be Zone 2/2E.

The minimum design loading of 28.4.4 must also be investigated (see Figure C27.4-1 and Figure 5.10).

Part b: Design Wind Pressures, Part 2 of Chapter 28Step 1: Check if the building meets all of the conditions and limitations of 28.1.2 and 28.1.3

and the additional conditions in 28.6.2 so that this procedure can be used to determinethe wind pressures on the MWFRS.

SurfaceDesign Pressure, p (psf)

(GCpf)(GCpi)

+0.18 –0.181 0.40 4.5 11.82 –0.69 –17.8 –10.4

3 –0.37 –11.2 –3.94 –0.29 –9.6 –2.2

1E 0.61 8.8 16.1

2E –1.07 –25.5 –18.23E –0.53 –14.5 –7.14E –0.43 –12.4 –5.1

Chapter 5 89


The following conditions are also relevant (see 28.6.2):

• The building is a simple diaphragm building as defined in 26.2, since windward andleeward wind loads are transmitted through the diaphragm to the MWFRS and thereare no separations in the MWFRS.

• The building is a low-rise building as defined in 26.2, since the mean roof height of32 feet is less than 60 feet and the mean roof height is less than the least horizontaldimension of 50 feet.

• The building is enclosed.

• It is given in the design data that the building is not a flexible building as defined in26.2.

• The building has a symmetrical cross-section in each direction, and the slope of theroof is less than 45 degrees.

• It can be shown that the building is exempted from the torsional load cases indicatedin Note 5 of Figure 28.4-1.


Step 2: Use Flowchart 5.6 to determine the net design wind pressures, p, on the MWFRS.









5. Determine design wind pressures, ps30, from Figure 28.6-1 for Zones A through H on the building.

Wind pressures, ps30, can be read directly from Figure 28.6-1 for V = 115 mph and a roof angle between 0 and 5 degrees. Since the roof is essentially flat, only Load Case


1 is considered (see Note 4 in Figure 28.6-1). These pressures, which are based on Exposure B, h = 30 feet and Kzt = 1.0 are given in Table P5.5c.

Table P5.5c Wind Pressures, ps30, on MWFRS

6. Determine net design wind pressures ps = λKzt ps30 by Equation 28.6-1 for Zones A through H.

From Figure 28.6-1, the adjustment factor for building height and exposure, λ, is equal to 1.02 (from linear interpolation) for a mean roof height of 32 feet and Expo-sure B. Thus,

ps = 1.02 × 1.0 × ps30 = 1.02 ps30

The horizontal pressures in Table P5.5d represent the combination of the windward and leeward net (sum of internal and external) pressures. Similarly, the vertical pres-sures represent the net (sum of internal and external) pressures.

Table P5.5d Wind Pressures, ps, on MWFRS

The net design pressures, ps, in this table are to be applied to the surfaces of the building in accordance with Cases A and B in Figure 28.6-1.

According to Note 7 in Figure 28.6-1, the total horizontal load must not be less than that determined by assuming ps = 0 in Zones B and D. Since the net pressures in Zones B and D in this problem act in the direction opposite to those in A and C, they decrease the horizontal load. Thus, the pressures in Zones B and D are set equal to 0 when analyzing the structure for wind in the E-W direction (Case A).

According to Note 2 in Figure 28.6-1, the load patterns for Case A and Case B are to be applied to each corner of the building; that is, each corner of the building must be considered a reference (windward) corner. Eight different load cases need to be examined (four in Case A and four in Case B).

The distance a = least of 0.1 (least horizontal dimension) = 0.1 × 50 = 5.0 feet (gov-erns) or 0.4h = 0.4 × 32 = 12.8 feet. This value of a is greater than 0.04 (least hori-zontal dimension) = 0.04 × 50 = 2.0 feet or 3 feet (see Note 9a in Figure 28.6-1).

The minimum design wind load case of 28.6.4 must also be considered: the load effects from the design wind pressures calculated above must not be less than the load effects assuming that ps = +16 psf in Zones A and C, ps = +8 psf in Zones B and D and ps = 0 psf in Zones E through H (see Figure C27.4-1 and Figure 5.10 for appli-cation of these pressures).

Horizontal pressures (psf) Vertical pressures (psf)A B C D E F G H

21.0 –10.9 13.9 –6.5 –25.2 –14.3 –17.5 –11.1

Horizontal pressures (psf) Vertical pressures (psf)A B C D E F G H

21.4 –11.1 14.2 –6.6 –25.7 –14.6 –17.9 –11.3

Chapter 5 91

5.6. A building with a mean roof height of 20 feet is located at the crest of a two-dimensionalridge that has the following properties: H = 100 feet, Lh = 40 feet and x = 0. Determine thetopographic factor, Kzt, assuming Exposure C and conditions 1 and 2 of 26.8.1 aresatisfied.

SOLUTION

Check if all five conditions of 26.8.1 are satisfied:

• It is given in the problem statement that the topography is such that conditions 1 and 2 aresatisfied.

• Condition 3 is satisfied since the building is located near the crest of the ridge.

• H/Lh = 100/40 = 2.5 > 0.2, so condition 4 is satisfied.

• H = 100 feet > 15 feet for Exposure C, so condition 5 is satisfied.

Since all five conditions of 26.8.1are satisfied, wind speed-up effects at the escarpment must beconsidered in the design, and Kzt must be determined by Equation 26.8-1:

Kzt = (1 + K1K2K3)2

where the multipliers K1, K2 and K3 are given in Figure 26.8-1 for Exposure C.

It was determined above that H/Lh = 2.5. According to Note 2 in Figure 26.8-1, where H/Lh >0.5, use H/Lh = 0.5 when evaluating K1 and substitute 2H for Lh when evaluating K2 and K3.

From Figure 26.8-1, K1/(H/Lh) = 1.45 for a 2-D ridge for Exposure C. Thus, K1 = 1.45 × 0.5 =0.73.

Since x/Lh = x/2H = 0/200 = 0, K2 = 1.00 for a 2-D ridge from Figure 26.8-1.

The multiplier K3 is determined at the height, z, corresponding to the top of the building. Thus,z/2H = 20/200 = 0.1. From Figure 26.8-1 for a 2-D ridge, K3 = 0.55.

Therefore, Kzt = [1 + (0.73 × 1.00 × 0.55)]2 = 1.96


5.7. A three-story police station is depicted in Figure 5.52. Determine the design windpressures on the walls and roof in both the N-S and E-W directions using Part 1 of Chapter27 and the design data in Table 5.87. The CMU walls around the perimeter of the buildinghave a nominal thickness of 8 inches.


SOLUTION

Step 1: Check if the building meets all of the conditions and limitations of 27.1.2 and 27.1.3 sothat this procedure can be used to determine the wind pressures on the MWFRS.




Wind velocity, V: 130 mphExposure category: C

Enclosure classification: EnclosedTopography: No hills or escarpments

Chapter 5 93





Since the building has CMU walls around its perimeter, assume it is rigid.





Due to the nature of its occupancy, this building falls under Risk Category IV.






In the design data, the exposure category is given as C.




Values of Kz and Kh for Exposure C are summarized in Table P5.7a.


g. Determine velocity pressure qz and qh by Equation 27.3-1.




35 1.0125 0.9415 0.85








Roof (parallel to ridge with θ = 0 degrees and h/L = 35/60 = 0.58):

Cp = –0.96, –0.18 (from linear interpolation) from windward edge to h/2 = 17.5 feet for use with qh

Cp = –0.87, –0.18 (from linear interpolation) from 17.5 feet to 60 feet for use with qh

For wind in the N-S direction:


Leeward wall (L/B = 160/60 = 2.67): Cp = –0.27 (from linear interpolation) for use with qh






Note: The smaller uplift pressures on the roof due to Cp = –0.18 may govern the design when combined with roof live load or snow loads. This pressure is not shown in this example but in general must be considered.


In accordance with 27.4.1, qi = qh = 37.1 psf for windward walls, side walls, lee-ward walls and roofs of enclosed buildings.


For an enclosed building, GCpi +0.18, –0.18.


Windward walls:



35 1.01 37.1

25 0.94 34.615 0.85 31.3

Chapter 5 95

= (qz × 0.85 × 0.8) – 37.1(±0.18)




= (37.1 × 0.85 × Cp) – 37.1(±0.18)


A summary of the maximum design wind pressures in the E-W and N-S directions is given in Tables P5.7c and P5.7d, respectively.


* from windward edge to 17.5 feet† from 17.5 to 60 feet

Table P5.7d Design Wind Pressures, p, in N-S Direction




z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)


+(GCpi) –(GCpi)

Windward wall

35 37.1 25.2 ± 6.7 18.5 31.9

25 34.6 23.5 ± 6.7 16.8 30.2

15 31.3 21.3 ± 6.7 14.6 28.0

Leeward wall All 37.1 –15.8 ± 6.7 –22.5 –9.1

Side walls All 37.1 –22.1 ± 6.7 –28.8 –15.4

Roof35 37.1 –30.2* ± 6.7 –36.9 –23.5

35 37.1 –27.4† ± 6.7 –34.1 –20.7


z (feet)q (psf)

External pressure

qGCp (psf)

Internal pressure

qh(GCpi) (psf)


+(GCpi) –(GCpi)

Windward wall

35 37.1 25.2 ± 6.7 18.5 31.9

25 34.6 23.5 ± 6.7 16.8 30.2

15 31.3 21.3 ± 6.7 14.6 28.0

Leeward wall All 37.1 –8.5 ± 6.7 –15.2 –1.8

Side walls All 37.1 –22.1 ± 6.7 –28.8 –15.4

Roof

35 37.1 –28.4* ± 6.7 –35.1 –21.7

35 37.1 –15.8† ± 6.7 –22.5 –9.1

35 37.1 –9.5 ± 6.7 –16.2 –2.8


5.8. Given the building in Problem 5.7, determine the design wind pressures on (a) a CMU walllocated in the first story, (b) an open-web joist and (c) a metal deck panel using Part 4 ofChapter 30.

SOLUTION

Part a: Determine design wind pressures on CMU wall in the first storyStep 1: Check if the building meets all of the conditions and limitations of 30.1.2 and 30.1.3

and the additional conditions of 30.7 so that this procedure can be used to determine thewind pressures on the C&C.


The following conditions of 30.7 must also be satisfied:

1. Mean roof height h = 35 feet < 160 feet

2. The building is enclosed

3. The building has a monoslope roof (slope = 3.6 degrees)

Thus, Part 4 of Chapter 30 may be used to determine the design wind pressures on the C&C.



Due to the nature of its occupancy, this building falls under Risk Category IV.




In the design data, the exposure category is given as C.



5. Determine design wind pressures, ptable, from Table 30.7-2 for Zones 4 and 5, which are the interior and end zones of walls, respectively, on the building.

The pressures, ptable, in Table P5.8a are obtained from Figure 30.7-2 for Exposure C with V = 130 mph, h = 35 feet and an effective wind area of 10 square feet.

Chapter 5 97

Table P5.8a Wind Pressures, ptable, on CMU Walls

6. Determine net design wind pressures, p, by Equation 30.7-1 for Zones 4 and 5.

p = ptable(EAF)(RF)Kzt

Since the building is located in Exposure C, the exposure adjustment factor (EAF) = 1.0.

The effective area reduction factor (RF) is determined from Table 30.7-2 based on the effective wind area, which in this case is equal to 15 × (15/3) = 75.0 square feet. Table P5.8b contains values of (RF) for an effective wind area of 100 square feet, which is conservative. The letters in parentheses after each value correspond to those given in the reduction factor graph and table that are a part of Table 30.7-2.

Table P5.8b Effective Area Reduction Factor (RF) for CMU Walls

The net design wind pressure is

p = ptable × 1.0 × (RF) × 1.0 = ptable × (RF)

A summary of the net design wind pressures on a CMU wall panel is given in Table P5.8c.

Table P5.8c Net Design Wind Pressures, p, on CMU Walls

Part b: Determine design wind pressures on an open-web joistStep 1: Check if the building meets all of the conditions and limitations of 30.1.2 and 30.1.3


As shown in Part a of this problem, Part 4 of Chapter 30 can be used to determine the design wind pressures on the C&C of this building.


1. The risk category, the basic wind speed, V, the exposure category, the exposure adjustment factor (EAF) and the topographic factor, Kzt, have all been determined in Part a of this problem and are used in calculating the wind pressures on the open-web joists.

2. Determine design wind pressures, ptable, from Table 30.7-2 for Zones 1, 2 and 3, which are located on the roof of the building.

Zone ptable (psf)4 44.0 –47.7

5 44.0 –73.8

ZoneSign Pressure

Plus Minus4 0.82 (D) 0.88 (C)5 0.82 (D) 0.76 (E)

Zone p (psf)4 36.1 –42.05 36.1 –56.1


The pressures, ptable, in Table P5.8d are obtained from Figure 30.7-2 for Exposure C with V = 130 mph, h = 35 feet and an effective wind area of 10 square feet.

Table P5.8d Wind Pressures, ptable, on Open-web Joists

3. Determine net design wind pressures, p, by Equation 30.7-1 for Zones 1, 2 and 3.



The effective area reduction factor (RF) is determined from Table 30.7-2 based on the effective wind area, which in this case is equal to the larger of the joist tributary area = 60 × 5 = 300 square feet or the span length multiplied by an effective width that need not be less than one-third the span length = 60 × (60/3) = 1,200.0 square feet (governs). Table P5.8e contains values of (RF) for an effective wind area of 1,000 square feet.

Table P5.8e Effective Area Reduction Factor (RF) for Open-web Joists



A summary of the net design wind pressures on an open-web joist purlin is given in Table P5.8f.

Table P5.8f Net Design Wind Pressures, p, on Open-web Joists

Part c: Determine design wind pressures on a metal deck panelStep 1: Check if the building meets all of the conditions and limitations of 30.1.2 and 30.1.3


As shown in Part a of this problem, Part 4 of Chapter 30 can be used to determine the design wind pressures on the C&C of this building.

Zone ptable (psf)1 21.7 –51.42 21.7 –66.33 21.7 –114.8

ZoneSign Pressure

Plus Minus1 0.80 (C) 0.70 (D)2 0.80 (C) 0.90 (B)

3 0.80 (C) 1.00 (A)

Zone p (psf)1 17.4 –36.02 17.4 –59.7

3 17.4 –114.8

Chapter 5 99


1. The risk category, the basic wind speed, V, the exposure category, the exposure adjustment factor (EAF) and the topographic factor, Kzt, have all been determined in Part a of this problem and are used in calculating the wind pressures on a metal deck panel.

2. Determine design wind pressures, ptable, from Table 30.7-2 for Zones 1, 2 and 3, which are located on the roof of the building.

The pressures, ptable, in Table P5.8g are obtained from Figure 30.7-2 for Exposure C with V = 130 mph, h = 35 feet and an effective wind area of 10 square feet.

Table P5.8g Wind Pressures, ptable, on Metal Deck Panel

3. Determine net design wind pressures, p, by Equation 30.7-1 for Zones 1, 2 and 3.



The effective area reduction factor (RF) is determined from Table 30.7-2 based on the effective wind area, which in this case is equal to the larger of the metal deck tributary area = 10 × 2 = 20 square feet or the span length multiplied by an effective width that need not be less than one-third the span length = 10 × (10/3) = 33.3 square feet (governs). Table P5.8h contains values of (RF) for an effective wind area of 30 square feet.

Table P5.8h Effective Area Reduction Factor (RF) for Metal Deck Panel



A summary of the net design wind pressures on the metal deck panel is given in Table P5.8i.

Table P5.8i Net Design Wind Pressures, p, on Metal Deck Panel

Zone ptable (psf)1 21.7 –51.4

2 21.7 –66.33 21.7 –114.8

ZoneSign Pressure

Plus Minus1 0.94 (C) 0.92 (D)

2 0.94 (C) 0.98 (B)3 0.94 (C) 1.00 (A)

Zone p (psf)1 20.4 –47.3

2 20.4 –65.03 20.4 –114.8


5.9. Given the building in Problem 5.7, assume that there is a roof overhang on the southelevation that is 5 feet in plan. Determine the design wind pressure on the roof overhangusing (a) Part 1 of Chapter 27 and (b) Part 4 of Chapter 30.

SOLUTION

Part a: Determine design wind pressures on overhang, Part 1 of Chapter 27It was shown in Problem 5.7 that Part 1 of Chapter 27 can be used to determine the design windpressures on this building.

Design wind pressures on the MWFRS of roof overhangs are determined in accordance with27.4.4. The positive external pressure on the bottom surface of a windward roof overhang isdetermined using Cp = 0.8; this is combined with the top surface pressures determined usingFigure 27.4-1 (see Figure 5.11).

The pressure at the top of the windward wall was determined in Problem 5.7 and is equal to 25.2psf (see Table P5.7c). Similarly, the pressure at the roof at the windward edge is equal to –30.2psf.

Part b: Determine design wind pressures on overhang, Part 4 of Chapter 30It was shown in Problem 5.8 that Part 4 of Chapter 30 can be used to determine the design windpressures on this building.

Design wind pressures on the C&C of roof overhangs are determined in accordance with30.7.1.3. The positive external pressure on the bottom surface of a windward roof overhang isdetermined as follows (see Figure 5.22):

ps = ptable|Zone 4 or 5(EAF)(RF)Kzt

The tabulated positive wall pressure is equal to 44.0 psf (see Table P5.8a). For Exposure C, EAF = 1.0. Also, assuming a one-foot width of overhang, the effective wind area = 5.0 squarefeet, so RF = 1.0 from Table 30.7-2. Therefore,

ps = 44.0 × 1.0 ×1.0 × 1.0 = 44.0 psf

The negative pressures on the roof overhang are as follows:

Zone 1 and 2: povh = ptable|Zone 1 or 2(EAF)(RF)Kzt

Zone 3: povh = 1.15ptable|Zone 3(EAF)(RF)Kzt

Using the negative pressures in Table P5.8d:

Zone 1: povh = –51.4 × 1.0 × 1.0 × 1.0 = –51.4 psf

Zone 2: povh = –66.3 × 1.0 × 1.0 × 1.0 = –66.3 psf

Zone 3: povh = 1.15 × (–114.8) × 1.0 × 1.0 × 1.0 = –132.0 psf

Chapter 5 101

5.10. Given the building in Problem 5.7, assume that there is one 3 foot-6 inch by 7 foot-0 inchopening and one 20 foot-0 inch by 8 foot-0 inch opening in each of the east and west wallsin the first story. Also in the first story are two 3 foot-6 inch by 7 foot-0 inch openings inthe south wall. There are no openings in the north wall or in the roof. Determine theenclosure classification of the building.

SOLUTION

Determine the enclosure classification of the building in accordance with 26.2.



Ag = (160 × 35) + (0.5 × 160 × 10) = 6,400 square feet

Ao = (7 × 3.5) + (8 × 20) = 185 square feet

• South wall

Ag = 45 × 60 = 2,700 square feet

Ao = 2 × (7 × 3.5) = 49 square feet

• North wall

Ag = 35 × 60 = 2,100 square feet

Ao = 0

• Roof

Ag = 60 × (160/cos 3.58°)] = 9,619 square feet

Ao = 0

• Totals

Ag = 6,400 + 2,700 + 2,100 + 9,619 = 20,819 square feet

Ao = (2 × 185) + 49 = 419 square feet



Ao/Aoi = 185/(419 – 185) = 0.8 < 1.1 (condition is not met)

Ao= 185 square feet

> 4 square feet (governs) or 0.01Ag = 0.01 × 6,400 = 64 square feet (condition is met)

Aoi/Agi = (419 – 185)/(20,819 – 6,400) = 0.02 < 0.20 (condition is met)

• South wall

Ao/Aoi = 49/(419 – 49) = 0.13 < 1.1 (condition is not met)

Ao= 49 square feet

> 4 square feet (governs) or 0.01Ag = 0.01 × 2,700 = 27 square feet (condition is met)

Aoi/Agi = (419 – 49)/(20,819 – 2,700) = 0.02 < 0.20 (condition is met)


Since all of the conditions of a partially enclosed building are not met, the building is defined asenclosed.

5.11. Given the building in Problem 5.7, assume that there is a mechanical unit located at thegeometric center of the roof. The unit has base dimensions of 15 by 15 feet, and its heightis 8 feet. Determine the wind forces on the mechanical unit.

SOLUTION

Use Flowchart 5.7 to determine the design wind force on the rooftop equipment.

1. Determine the velocity pressure, qh, evaluated at the mean roof height of the building.From Table 29.3-1, Kz = 1.01 (by linear interpolation) for Exposure C at a height of 35 feet above the ground level.

Velocity pressure, qh, is determined by Equation 29.3-1:

qh = 0.00256Kz Kzt KdV2 = 0.00256 × 1.01 × 1.0 × 0.90 × 1302 = 39.3 psf

where Kd = 0.90 for square-shaped rooftop equipment (see Table 26.6-1).

2. Determine force coefficients (GCr) for lateral forces and vertical uplift forces for buildings with h ≤ 60 feet.

• Lateral forces:

For both faces, Af = 8 × 15 = 120 square feet

0.1Bh = 0.1 × 160 × 35 = 560 square feet > Af

0.1Bh = 0.1 × 60 × 35 = 210 square feet > Af

Thus, use (GCr) = 1.9 in Equation 29.5-2.

• Vertical uplift forces:

Ar = 15 × 15 = 225 square feet

0.1BL = 0.1 × 60 × 160 = 960 square feet > Ar

Thus, use (GCr) = 1.5 in Equation 29.5-3.

3. Determine design wind forces by Equations 29.5-2 and 29.5-3.

Fh = qh(GCr)Af = 39.3 × 1.9 × (8 × 15) / 1,000 = 9.0 kips

Fv = qh(GCr)Ar = 39.3 × 1.5 × (15 × 15) / 1,000 = 13.3 kips

Chapter 5 103

5.12. A reinforced concrete chimney with an average outside diameter of 6 feet has a height 60feet above the ground surface, which is relatively flat. A preliminary analysis indicatesthat the approximate fundamental frequency of the chimney is 0.3 Hz. Assuming that theresponse characteristics are predominantly along-wind, determine the design wind forceson the chimney given a wind velocity of 115 mph, Exposure C and a damping ratio of 2percent.

SOLUTION

Use Flowchart 5.7 to determine the design wind force on the chimney.

1. Determine the velocity pressure, qz, evaluated at height z of the centroid of area, Af, of the chimney.The distance from the ground level to the centroid of the chimney = 60/2 = 30 feet

From Table 29.3-1, Kz = 0.98 for Exposure C at a height of 30 feet above the ground level.

Velocity pressure, qz, is determined by Equation 29.3-1:

qz = 0.00256Kz Kzt KdV 2 = 0.00256 × 0.98 × 1.0 × 0.95 × 1152 = 31.5 psf

where Kd = 0.95 for round chimneys (see Table 26.6-1).

2. Determine gust-effect factor, G.

Since the fundamental frequency is less than 1 Hz, the chimney is defined as flexible. Thus, the gust factor must be determined in accordance with 26.9.5.

a. Determine gQ and gv

gQ = gv = 3.4

b. Determine gR

Eq. 26.9-11

c. Determine Iz

z = 0.6h = 0.6 × 60 Table 26.9-1 for Exposure C

= 36 feet > zmin = 15 feet

Eq. 26.9-7 and Table 26.9-1 for Exposure C

gR 2 3 600n1,( )ln 0.577

2 3 600n1,( )ln---------------------------------------+=

2 3 600, 0.3×( )ln0.577

2 3 600, 0.3×( )ln---------------------------------------------- 3.9=+=

Iz c33z

------⎝ ⎠⎛ ⎞ 1 6⁄

=

0.203336------⎝ ⎠⎛ ⎞ 1 6⁄

0.20==


d. Determine Q

feet Eq. 26.9-9 and Table 26.9-1 for Exposure C

Eq. 26.9-8

e. Determine R

= 111.1 feet/sec Eq. 26.9-16 and Table 26.9-1 for Exposure C

Eq. 26.9-14

Eq. 26.9-13

Eq. 26.9-15a

Lz l z33------⎝ ⎠⎛ ⎞

∈=

5003633------⎝ ⎠⎛ ⎞ 1 5⁄

508.8==

Q 1

1 0.63B h+

Lz-------------⎝ ⎠⎛ ⎞ 0.63

+

-----------------------------------------------=

1

1 0.63+6 60+508.8---------------⎝ ⎠⎛ ⎞ 0.63

-------------------------------------------------- 0.92==

Vz bz

33------⎝ ⎠⎛ ⎞α 88

60------⎝ ⎠⎛ ⎞V=

0.653633------⎝ ⎠⎛ ⎞ 1 6.5⁄ 88

60------⎝ ⎠⎛ ⎞ 115×=

N1

n1Lz

Vz

-----------=

0.3 508.8×111.1

--------------------------- 1.37==

Rn

7.47N1

1 10.3N1+( )5 3⁄---------------------------------------=

7.47 1.37×

1 10.3 1.37 )×(+[ ]5 3⁄------------------------------------------------------- 0.11==

ηh

4.6n1h

Vz

----------------=

4.6 0.3× 60×111.1

--------------------------------- 0.75==

Rh1

ηh------

1

2ηh2

--------- 1 e2ηh–

–( )–=

10.75----------

1

2 0.752×

---------------------- 1 e2 0.75×–

–( ) 0.64=–=

Chapter 5 105

Eq. 26.9-15a

Eq. 26.9-15a

= 1.76 Eq. 26.9-12

f. Determine Gf

= 1.53 Eq. 26.9-10

3. Determine force coefficient, Cf, from Figure 29.5-1 for round chimney.

h/D = 60/6 = 10

From linear interpolation, Cf = 0.87.

4. Determine design wind force, F, by Equation 29.5-1.

F = qzGCfAf

= 31.5 × 1.53 × 0.87 × (6 × 60) / 1,000 = 15.1 kips

ηB

4.6n1B

Vz

-----------------=

4.6 0.3× 6×111.1

------------------------------ 0.08==

RB1

ηB------

1

2ηB2

---------- 1 e2ηB–

–( )–=

10.08----------

1

2 0.082×

---------------------- 1 e2 0.08×–

–( ) 0.95=–=

ηL

15.4n1L

Vz

--------------------=

15.4 0.3× 6×111.1

--------------------------------- 0.25==

RL1

ηL------

1

2ηL2

--------- 1 e2ηL–

–⎝ ⎠⎛ ⎞–=

10.25----------

1

2 0.252×

---------------------- 1 e2 0.25×–

–( ) 0.85=–=

R1β---RnRhRB 0.53 0.47RL+( )=

10.02---------- 0.11× 0.64× 0.95 0.53 0.47( 0.85 )×+[ ]×=

Gf 0.9251 1.7Iz gQ

2Q

2gR

2R

2++

1 1.7gvIz+-----------------------------------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

=

0.9251 1.7( 0.20 ) 3.4( 2

0.922 ) 3.9

2( 1.762 )×+××+

1 1.7( 3.4× 0.20 )×+------------------------------------------------------------------------------------------------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

=

D qz( )⁄ 6 31.5( )⁄ 1.1 2.5<==

CHAPTER 6Earthquake Loads

6.1. A typical floor plan of a five-story residential building is shown in Figure 6.46. Determinethe SDC of the structure for the cities identified in Table 6.25 assuming Site Class D.

Table 6.25 Locations for Problem 6.1

Figure 6.46 Typical Plan of Five-story Residential Building, Problem 6.1

City Latitude LongitudeBerkeley, CA 37.87 –122.28

Boston, MA 42.36 –71.06Chicago, IL 41.88 –87.63Denver, CO 39.74 –104.99

Houston, TX 29.76 –95.37New York, NY 40.72 –74.00

107


SOLUTION

The SDC is determined using Flowchart 6.4. A summary of the design accelerations and SDCare given in Table P6.1. Values of SS and S1 were obtained from the USGS website (https://geohazards.usgs.gov/secure/designmaps/us/) for each site.

Table P6.1 SDC for Locations in Problem 6.1

6.2. Given the information in Problem 6.1, determine V and Fx in both directions using theequivalent lateral force procedure for the building located in Berkeley, CA. The first storyheight is 12 feet and typical floor heights are 10 feet. Assume a 9-inch-thick reinforcedconcrete slab at all levels and that all concrete is normal weight (density = 150 pcf). Alsoassume a superimposed dead load of 30 psf on all floors and 10 psf on the roof. A glasscurtain wall system is used on all faces of the building, which weighs 8 psf. The SFRSconsists of moment-resisting frames in both directions.

SOLUTION

Use Flowchart 6.8 to determine the lateral seismic forces from the equivalent lateral forceprocedure. Since the building utilizes the same seismic-force-resisting system in both directions,V and Fx are the same in both directions.

1. The design accelerations and the SDC have been determined in Problem 6.1:SDS = 1.47, SD1 = 0.91, SDC = E

2. Determine the response modification coefficient, R, from Table 12.2-1.

Special reinforced concrete moment frames must be utilized in the building, since it is assigned to SDC E (system C5 in Table 12.2-1). In this case, R = 8. Note that there is no height limit.

3. Determine the importance factor, Ie, from Table 1.5-2.

For Risk Category II, Ie = 1.0.

4. Determine the period of the structure, T.

The approximate period of the structure, Ta, is determined in accordance with 12.8.2.1.

second

5. Determine long-period transition period, TL, from Figure 22-12.

City SS S1 SMS SM1 SDS SD1 SDCBerkeley, CA 2.20 0.91 2.20 1.36 1.47 0.91 E

Boston, MA 0.22 0.07 0.35 0.17 0.23 0.11 BChicago, IL 0.13 0.06 0.21 0.15 0.14 0.10 BDenver, CO 0.18 0.06 0.29 0.14 0.19 0.09 B

Houston, TX 0.07 0.04 0.16 0.09 0.08 0.06 ANew York, NY 0.28 0.07 0.44 0.17 0.29 0.12 B

Ta Cthnx

0.016 520.9× 0.56= ==

Chapter 6 109

For Berkeley, CA, TL = 8 seconds > Ta = 0.56 second.

6. Determine seismic response coefficient, CS.

The seismic response coefficient, CS, is determined by Equation 12.8-3:

The value of CS need not exceed that from Equation 12.8-2:

Also, CS must not be less than the larger of the following:

• 0.044SDSIe = 0.07 (governs)

• 0.01

• 0.5S1/(R/Ie) = (0.5 × 0.91)/(8/1.0) = 0.06

Thus, the value of CS from Equation 12.8-2 governs.

7. Determine effective seismic weight, W, in accordance with 12.7.2.

The member sizes and superimposed dead loads are given in Figure 6.46 and the design data.The effective weights at each floor level are given in Table P6.2. The total weight, W, is thesummation of the effective dead loads at each level.

Table P6.2 Seismic Forces and Story Shears

8. Determine seismic base shear, V.

Seismic base shear is determined by Equation 12.8-1:

V = CsW = 0.18 × 9,655 = 1,738 kips

9. Determine exponent related to structure period, k.

Since 0.5 second < T = 0.56 second < 2.5 seconds, k is determined as follows:

k = 0.75 + 0.5T = 1.03

10.Determine lateral seismic force, Fx, at each level x.

Fx is determined by Equations 12.8-11 and 12.8-12. A summary of the lateral forces, Fx, and the story shears, Vx, is given in Table P6.2.

LevelStory weight,

wx (kips)Height,hx (feet)

Lateral force,Fx (kips)

Story Shear,Vx (kips)

R 1,663 52 97,359 504 5044 1,993 42 93,638 485 9893 1,993 32 70,764 366 1,355

2 1,993 22 48,106 249 1,604

1 2,013 12 26,026 134 1,738

Σ 9,655 335,893 1,738

Cs

SD1

T R Ie⁄( )-------------------- 0.91

0.56 8 1.0⁄( )------------------------------ 0.20= = =

Cs

SDS

R Ie⁄------------ 1.47

8 1.0⁄-------------- 0.18= = =

wxhkx


6.3. Repeat Problem 6.2 for the building located in New York, NY.

SOLUTION

Use Flowchart 6.8 to determine the lateral seismic forces from the equivalent lateral forceprocedure. Since the building utilizes the same seismic-force-resisting system in both directions,V and Fx are the same in both directions.

1. The design accelerations and the SDC have been determined in Problem 6.1:SDS = 0.29, SD1 = 0.12, SDC = B


Ordinary reinforced concrete moment frames must be utilized in this the building, since it is assigned to SDC B (system C7 in Table 12.2-1). In this case, R = 3. Note that there is no height limit.




The approximate period of the structure, Ta, is determined in accordance with 12.8.2.1.

Ta = Cthnx = 0.016 × 520.9 = 0.56 second


For New York, NY, TL = 6 seconds > Ta = 0.56 second.




Also, CS must not be less than the larger of the following:

• 0.044SDSIe = 0.013 (governs)

• 0.01



Cs

SD1

T R Ie⁄( )-------------------- 0.12

0.56 3 1.0⁄( )------------------------------ 0.07= = =

Cs

SDS

R Ie⁄------------ 0.29

3 1.0⁄-------------- 0.10= = =

Chapter 6 111

The member sizes and superimposed dead loads are given in Figure 6.46 and the design data. The effective weights at each floor level are given in Table P6.3. The total weight, W, is the summation of the effective dead loads at each level.




V = CsW = 0.07 × 9,655 = 676 kips


Since 0.5 second < T = 0.56 second < 2.5 seconds, k is determined as follows:

k = 0.75 + 0.5T = 1.03



6.4. Given the information in Problems 6.1 and 6.2, determine the diaphragm forces for thebuilding located in New York, NY.

SOLUTION

Determine the diaphragm design seismic forces using Equation 12.10-1.

Diaphragm design force

where wi = weight tributary to level i and wpx = weight tributary to the diaphragm at level x.

Minimum Fpx = 0.2SDSIewpx = 0.2 × 0.29 × 1.0 × wpx = 0.0580wpx

Maximum Fpx = 0.4SDSIewpx = 0.1160wpx

LevelStory weight,

wx (kips)Height, hx (feet)

Lateral force, Fx (kips)

Story Shear, Vx (kips)

R 1,663 52 97,359 196 1964 1,993 42 93,638 189 3853 1,993 32 70,764 142 527

2 1,993 22 48,106 97 624

1 2,013 12 26,026 52 676

Σ 9,655 335,893 676

wxhkx

Fpx

Fi

i x=

n

∑

wi

i x=

n

∑

---------------wpx=


Since the exterior walls are glass, which weigh significantly less than the diaphragm weight ateach level, assume that the weight that is tributary to each diaphragm is identical to the weightof the structure at that level (i.e., wpx = wx).

A summary of the diaphragm forces is given in Table P6.4.

Table P6.4 Design Seismic Diaphragm Forces

*Maximum Fpx governs

6.5. A typical floor in a 20-story structural steel-framed office building is shown in Figure 6.47.Determine V and Fx in both directions using the equivalent lateral force procedure giventhe following design data:

• SS = 0.46, S1 = 0.20

• Site Class B

• Dead load per floor (includes superimposed dead load and cladding) =1,000 kips

• Dead load on roof (includes mechanical equipment and cladding) = 900 kips

• All story heights are 13 feet

• SFRS: steel concentrically braced frames as indicated in Figure 6.47

Levelwx

(kips)∑wx

(kips)Fx

(kips)∑Fx

(kips)∑Fx / ∑wx

wpx

(kips)Fpx

(kips)5 1,663 1,663 196 196 0.1160* 1,663 1934 1,993 3,656 189 385 0.1053 1,993 210

3 1,993 5,649 142 527 0.0933 1,993 1862 1,993 7,642 97 624 0.0817 1,993 1631 2,013 9,655 52 676 0.0700 2,013 141

Figure 6.47 Typical Plan of Twenty-story Office Building, Problem 6.5

Chapter 6 113

SOLUTION

Since the building utilizes the same seismic-force-resisting system in both directions, V and Fx

are the same in both directions.

Step 1: Determine the seismic ground motion values from Flowchart 6.3.

1. Determine the mapped accelerations SS and S1.The mapped accelerations are given in the design data as SS = 0.46 and S1 = 0.20.

2. Determine the site class of the soil.

The site class of the soil is given in the design data as Site Class B.

3. Determine soil-modified accelerations SMS and SM1.

Site coefficients Fa and Fv are determined from Tables 11.4-1 and 11.4-2, respectively. For Site Class B, Fa and Fv are equal to 1.0.

Thus,

SMS = 1.0 × 0.46 = 0.46

SM1 = 1.0 × 0.20 = 0.20

4. Determine design accelerations SDS and SD1.

From Equations 11.4-3 and 11.4-4:

Step 2: Determine the SDC from Flowchart 6.4.

1. Determine if the building can be assigned to SDC A in accordance with 11.4.1.Since SS = 0.46 > 0.15 and S1 = 0.20 > 0.04, the building cannot be automatically assigned to SDC A.

2. Determine the Risk Category from IBC Table 1604.5.

For an office occupancy, the Risk Category is II.

3. Since S1 < 0.75, the building is not assigned to SDC E or F.

4. Check if all four conditions of 11.6 are satisfied.

Check if the approximate period, Ta, is less than 0.8TS.

Use Equation 12.8-7 with approximate period parameters for “other structural systems”:

Ta = Cthnx = 0.02(260)0.75 = 1.30 seconds

where Ct and x are given in Table 12.8-2.

TS = SD1 / SDS = 0.133 / 0.307 = 0.43 second

1.30 seconds > 0.8 × 0.43 = 0.34 second

SDS23--- 0.46× 0.307= =

SD123--- 0.20× 0.133= =


Since this condition is not satisfied, the SDC cannot be determined by Table 11.6-1 alone (11.6).

5. Determine the SDC from Tables 11.6-1 and 11.6-2.

From Table 11.6-1, with 0.167 ≤ SDS < 0.33 and Risk Category II, the SDC is B.

From Table 11.6-2, with 0.133 ≤ SD1 < 0.20 and Risk Category II, the SDC is C.

Therefore, the SDC is C for this building.

Step 3: Use Flowchart 6.8 to determine the lateral seismic forces from the equivalent lateralforce procedure.

1. The design accelerations and the SDC have been determined in Step 1 of this problem.


Since the building is assigned to SDC C, this building frame system can utilize steel ordi-nary concentrically braced frames (system B3 in Table 12.2-1). In this case, R = 31/4. Note that there is no height limit.




It was determined in Step 2 of this problem that the approximate period of the structure, Ta, which is permitted to be used in the equivalent lateral force procedure, is equal to 1.30 sec-onds.


Since no location was provided in this problem, assume TL = 4 seconds (lowest value in Fig-ure 22-12) > Ta = 1.30 seconds.




Also, CS must not be less than the larger of 0.044SDSIe = 0.014 (governs) and 0.01 (Equation 12.8-5).



The effective weights at each floor level from the design data are given in Table P6.5. The total weight, W, is the summation of the effective dead loads at each level.

Cs

SD1

T R Ie⁄( )-------------------- 0.133

1.30 3.25 1.0⁄( )------------------------------------- 0.032= = =

Cs

SDS

R Ie⁄------------ 0.307

3.25 1.0⁄---------------------- 0.095= = =

Chapter 6 115




V = CsW = 0.032 × 19,900 = 637 kips


Since 0.5 second < T = 1.30 seconds < 2.5 seconds, k is determined as follows:

k = 0.75 + 0.5T = 1.40



LevelStory weight,

wx (kips)Height, hx (feet) wxhx

k Lateral force, Fx (kips)

Story Shear, Vx (kips)

R 900 260 2,163,741 66 6619 1,000 247 2,237,565 68 134

18 1,000 234 2,074,446 63 19717 1,000 221 1,914,914 58 25516 1,000 208 1,759,092 53 308

15 1,000 195 1,607,121 49 35714 1,000 182 1,459,150 44 40113 1,000 169 1,315,350 40 441

12 1,000 156 1,175,911 35 47611 1,000 143 1,041,047 32 50810 1,000 130 911,005 28 536

9 1,000 117 786,068 24 560

8 1,000 104 666,571 20 5807 1,000 91 552,915 17 5976 1,000 78 445,587 13 610

5 1,000 65 345,206 10 6204 1,000 52 252,583 8 6283 1,000 39 168,846 5 633

2 1,000 26 95,711 3 6361 1,000 13 36,268 1 637

Σ 19,900 21,009,098 637


6.6. The roof of a one-story commercial building is shown in Figure 6.48. Determine if thesimplified alternative method of 12.14 can be used to determine the seismic forces giventhe following design data:

• SS = 0.28, S1 = 0.10

• Site Class C

• All walls are 10-inches thick and are made of the same concrete

• Roof structure consists of open-web joists and metal deck

Determine the seismic base shear assuming the simplified alternative method can be used. For analysis purposes, assume the walls are fixed at the top and bottom.

SOLUTION

The simplified method is permitted to be used if the following 12 limitations are met:

1. The structure shall qualify for Risk Category I or II in accordance with Table 1.5-1.From Table 1.5-1, the Risk Category is II for a commercial building. O.K.

2. The Site Class shall not be E or F.

The Site Class is C in accordance with the design data. O.K.

Figure 6.48Roof Plan of One-story Commercial Building, Problem 6.6

Chapter 6 117

3. The structure shall not exceed three stories in height.

The structure is one story. O.K.

4. The SFRS shall be either a bearing wall system or building frame system in accordance with Table 12.14-1.

The SFRS is a bearing wall system. O.K.

5. The structure has at least two lines of lateral resistance in each of the two major axis direc-tions.

Concrete shear walls are provided along two lines in both directions. O.K.

6. At least one line of resistance shall be provided on each side of the center of mass in each direction.

The center of mass is approximately located at the geometric center of the building and walls are provided on each side of it in both directions. O.K.

7. For structures with flexible diaphragms, overhangs beyond the outside line of shear walls or braced frames shall satisfy: a ≤ d/5.

This diaphragm meets the conditions of a flexible diaphragm in accordance with 12.3.1.1. There are no overhangs, so this limitation is not applicable.

8. For buildings with diaphragms that are not flexible, the distance between the center of rigid-ity and the center of mass parallel to each major axis shall not exceed 15 percent of the greatest width of the diaphragm parallel to that axis.

Since the diaphragm is flexible, this limitation is not applicable.

9. Lines of resistance of the lateral force-resisting shall be oriented at angles of no more than 15 degrees from alignment with the major orthogonal axes of the building.

The shear walls in both directions are parallel to the major axes. O.K.

10.The simplified design procedure shall be used for each major orthogonal horizontal axis direction of the building.

The simplified design procedure is used in both directions. O.K.

11.System irregularities caused by in-plane or out-of-plane offsets of lateral-force-resisting elements shall not be permitted.

This building does not have any irregularities. O.K.

12.The lateral load resistance of any story shall not be less than 80 percent of the story above.

Since this is a one-story building, this limitation is not applicable. O.K.

Since all 12 limitations of 12.14.1.1 are satisfied, the simplified procedure may be used.

Determine the seismic base shear from Flowchart 6.9.

1. Determine SS, SDS and the SDC from Flowchart 6.4.SS = 0.28 from the design data

From Table 11.4-1, Fa = 1.2 for SS = 0.28. Thus, SDM = 1.2 × 0.28 = 0.34.


SDS = 2 × 0.34/3 = 0.23

According to 11.6, the SDC is permitted to be determined from Table 11.6-1 alone where the simplified design procedure is used.

For 0.167 ≤ SDS < 0.33 and Risk Category II, the SDC is B.

2. Determine the response modification factor, R, from Table 12.14-1.

For SDC B, a bearing wall system with ordinary reinforced concrete shear walls can be used (system A2). For this system, R = 4.

3. Determine the effective seismic weight, W, in accordance with 12.14.8.1.

Conservatively neglect any wall openings. Also assume that the weight of the roof structure including superimposed dead loads = 20 psf.

Weight of concrete walls tributary to roof diaphragm

kips

Weight of roof structure = 20 × 60 × 40 / 1,000 = 48 kips

W = 63 + 48 = 111 kips

4. Determine base shear, V, by the equation in 12.14.8.1.

kips

where F = 1 for a one-story building (see 12.14.8.1).

6.7. Given the information in Problem 6.6, determine if the simplified alternative method of12.14 can be used to determine the seismic forces when the roof deck consists of reinforcedconcrete framing. Determine the seismic base shear assuming the simplified alternativemethod can be used.

SOLUTION

The simplified method is permitted to be used if the following 12 limitations are met:

1. The structure shall qualify for Risk Category I or II in accordance with Table 1.5-1.From Table 1.5-1, the Risk Category is II for a commercial building. O.K.

2. The Site Class shall not be E or F.

The Site Class is C in accordance with the design data. O.K.

3. The structure shall not exceed three stories in height.

The structure is one story. O.K.

4. The SFRS shall be either a bearing wall system or building frame system in accordance with Table 12.14-1.

The SFRS is a bearing wall system. O.K.

1012------ 150× 10

2------× 40 3 20×( )+[ ] 1 000, 63=⁄×=

VFSDS

R-------------W 1 0.23× 111×

4----------------------------------- 6.4= = =

Chapter 6 119

5. The structure has at least two lines of lateral resistance in each of the two major axis direc-tions.

Concrete shear walls are provided along two lines in both directions. O.K.

6. At least one line of resistance shall be provided on each side of the center of mass in each direction.

The center of mass is approximately located at the geometric center of the building and walls are provided on each side of it in both directions. O.K.

7. For structures with flexible diaphragms, overhangs beyond the outside line of shear walls or braced frames shall satisfy: a ≤ d/5.

The diaphragm in this building is rigid, so this limitation is not applicable.

8. For buildings with diaphragms that are not flexible, the distance between the center of rigid-ity and the center of mass parallel to each major axis shall not exceed 15 percent of the greatest width of the diaphragm parallel to that axis.

Assume that the center of mass is at the geometric center of this building (note: the exact location of the center of mass should be computed where it is anticipated to be offset from the geometric center of the building). The 15-percent limitation is satisfied with respect to the north-south direction, since the center of rigidity and center of mass are on the same line due to the symmetrical layout of Walls C and D.

The center of rigidity must be located in the east-west direction, since Walls A and B are not identical. By inspection, the center of rigidity will be located closer to Wall A, since the stiffness of Wall A is greater than the stiffness of Wall B.

To locate the center of rigidity, the stiffnesses of Walls A and B must be determined. Assuming that the piers are fixed at the top and bottom ends, the stiffnesses (or rigidities) of the walls and piers can be determined by the following:

Total displacement of pier or wall i: δi = δfi + δvi

δfi = displacement due to bending =

δvi = displacement due to shear =

where hi = height of pier or wall

li = length of pier or wall

t = thickness of pier or wall

E = modulus of elasticity of pier or wall

Stiffness of pier or wall ki = 1/δi

In lieu of a more rigorous analysis, the stiffness of Wall A, which has an opening, is deter-mined as follows: first, the deflection of the wall is obtained as though it were a solid wall with no opening. Next, the deflection of a solid strip of wall that contains the opening is

hi

li

----⎝ ⎠⎜ ⎟⎛ ⎞ 3

Et-------------

3hi

li

----⎝ ⎠⎜ ⎟⎛ ⎞

Et-------------


subtracted from the total deflection. Finally, the deflection of each pier surrounded by the opening is added back.

Table P6.7 contains a summary of the stiffness calculations for Wall A. The pier designa-tions are provided in Figure P6.7.

Table P6.7 Stiffness Calculations for Wall A

Wall A stiffness = 1.009Et

Wall B stiffness =

The location of the center of rigidity in the east-west direction can be determined from the following equation:

where xi is the distance from a reference point to wall i.

Using the centerline of Wall A as the reference line:

feet

e1 = 30 – 15.2 = 14.8 feet

0.15 × 60 = 9.0 feet < 14.8 feet N.G.

Thus, the eighth limitation is not satisfied and the simplified procedure must not be used.

If this procedure were permitted to be used, the base shear would be computed the same as in Problem 6.6.

Figure P6.7Pier Designations for Stiffness Calculations, Problem 6.7

Pier/Wall hi (ft) li (ft) δfiEt = (hi/li)3 δviEt = 3(hi/li) δiEt ki/Et

1 + 2 + 3 + 4 10 40 0.016 0.750 0.766 ---1 + 2 + 4 7 40 –0.005 –0.525 –0.530 ---

1 7 10 0.343 2.100 --- 0.409

2 7 20 0.043 1.050 --- 0.9151 + 2 --- --- --- --- 0.755 1.324

0.991 → 1.009

Et

1020------⎝ ⎠⎛ ⎞ 3

31020------⎝ ⎠⎛ ⎞+

----------------------------------- 0.615Et=

xr

ki∑ xi

ki∑---------------=

xr0.615Et 40×

0.615Et 1.009Et+--------------------------------------------- 15.2= =

Chapter 6 121

6.8. A 15-kip cooling tower that is braced below its center of mass is supported on the roof ofan office building. Determine the seismic design force using SDS = 1.1.

SOLUTION

Flowchart 6.10 is used to determine the seismic design force.

1. Check if the weight of the cooling tower is greater than or equal to 25 percent of the effec-tive seismic weight of the office building.It is common that the weight of a cooling tower is much less than 25 percent of the effective seismic weight of the building that supports it. As such, the provisions of 13.6 may be used to determine the seismic design force on the cooling tower.

2. Determine SDS, SD1 and the SDC.

The design acceleration, SDS, is given in the design data and is equal to 1.1. The other infor-mation is not needed to determine the seismic design force.

3. Determine the component amplification factor, ap, from Table 13.6-1 for mechanical and electrical components.

For a cooling tower mounted on the roof of the office building that is braced below its cen-ter of mass, ap = 2.5.

4. Determine the component response modification factor, Rp, from Table 13.6-1 for mechani-cal and electrical components.

For a cooling tower mounted on the roof of the office building that is braced below its cen-ter of mass, Rp = 3.0.

5. Determine the component importance factor, Ip, in accordance with 13.1.3.

Since this cooling tower does not meet any of the conditions specified in 13.1.3 for Ip = 1.5, then Ip = 1.0.

6. Determine the horizontal seismic design force, Fp, applied at the cooling tower’s center of gravity by Equations 13.3-1, 13.3-2 and 13.3-3.

kips (governs)

0.3SDSIpWp = 5.0 kips

1.6SDSIpWp = 26.4 kips

0.3SDSIpWp Fp≤0.4apSDSWp

Rp

Ip------⎝ ⎠⎛ ⎞

------------------------------- 1 2zh---+⎝ ⎠

⎛ ⎞ 1.6SDSIpWp≤=

Fp0.4 2.5× 1.1× 15×

3.01.0-------⎝ ⎠⎛ ⎞

----------------------------------------------- 1 2+( ) 16.5==


6.9. A glass curtain wall panel spans 12 feet between floor levels in an office building thatutilizes special steel moment-resisting frames. It has been determined that the maximumdeflections at the top and bottom of one of the floors are δxAe = 1.2 inches and δyAe = 0.80inch. These deflections occur 48 feet and 36 feet above the base of the structure,respectively. Determine the seismic relative displacement.

SOLUTION

The requirements for seismic relative displacements that are given in 13.3.2 are applicable inthis case (see Figure 6.29, which pertains to relative displacements determined for one structureat multiple levels).

The seismic relative displacement, DpI, in this case is determined by Equations 13.3-5 and 13.3-6:

DpI = (δxA – δyA)Ie = (1.2 – 0.8) × 1.0 = 0.4 inch

where Ie = 1.0 for an office building.

According to 13.3.2.1, DpI need not exceed the following:

inches

where the allowable drift, ΔaA, is determined from Table 12.12-1 for a Risk Category IIstructure.

DpI

Ie hx hy–( )ΔaA

hsx------------------------------------

Ie hx hy–( ) 0.02hsx( )hsx

------------------------------------------------- 1.0 48( 36 ) 12 0.02××–× 2.9= = = =

CHAPTER 7Flood Loads

7.1. A 15-foot tall, vertical reinforced concrete wall is located in a V Zone that supports thebase of a Risk Category III structure. Determine the design flood loads on the wallassuming the following:

• Design stillwater elevation, ds = 6 ft-0 in.

• Base flood elevation (BFE) = 10 ft-0 in.

The local jurisdiction has adopted a freeboard of 2 feet above the BFE.

SOLUTION

The applicable flood loads are hydrodynamic, breaking wave, and impact.

Step 1: Determine water velocity, V.

Since the building is located in a V Zone, it is appropriate to use the upper bound average water velocity, which is given by ASCE/SEI Equation C5-2:

V = (gds)0.5 = (32.2 × 6.0)0.5 = 13.9 ft/sec

Step 2: Determine hydrodynamic load, Fdyn.

Since the water velocity exceeds 10 ft/sec, it is not permitted to use an equivalent hydro-static load to determine the hydrodynamic load (ASCE/SEI 5.4.3).

The hydrodynamic load, Fdyn, is determined by Equation 7.3:

Based on the recommendations of Section 8.5.9 of Reference 7.4, the drag coefficient, a, can be determined based on the ratio of the width of the object to the stillwater flood depth, ds, for objects that are not fully immersed. Assuming that the width of the wall is 60 feet, w/ds = 60/6 = 10. From Table 8-2 in Reference 7.4, a = 1.25 for 1 ≤ w/ds ≤ 12.

Assuming salt water, the hydrodynamic load is

kips

This load acts at 3 ft-0 in. below the stillwater surface of the water.

Fdyn12---aρV

2A=

Fdyn12--- 1.25× 64.0

32.2----------⎝ ⎠⎛ ⎞× 13.9

2× 60× 6 1 000,⁄× 87= =

123


Step 3: Determine breaking wave load, Ft.

The breaking wave load per unit length of a vertical wall, assuming that the space behind the wall is dry, is determined by ASCE/SEI Equation 5.4-6:

According to ASCE/SEI Table 5.4-1, the dynamic pressure coefficient, Cp, is equal to 3.2 for Risk Category III structures.

Therefore,

Ft = (1.1 × 3.2 × 64.0 × 62) / 1,000 + (2.4 × 64.0 × 62) / 1,000 = 8.1 + 5.5 = 13.6 kips/ft

This load acts at the stillwater elevation, which is 6 ft-0 in. above the ground surface.

Step 4: Determine impact load, F.

Both normal and special impact loads are determined.

1. Normal impact loads

ASCE/SEI Equation C5-3 is used to determine normal impact loads:

As noted in Section 7.4.6 of this publication, guidance on establishing the debris weight, W, is given in ASCE/SEI C5.4.5. It is assumed in this problem that W = 1,000 lb.

It is reasonable to assume that the velocity of the object, Vb, is equal to the velocity of the water, V. Thus, from Step 1 of this problem, Vb = 13.9 ft/sec.

The importance coefficient, CI, is obtained from Table C5-1. For a Risk Category III building, CI = 1.2.

The orientation coefficient CO = 0.8. This coefficient accounts for impacts that are oblique to the structure.

The depth coefficient, CD, is obtained from ASCE/SEI Table C5-2 or, equivalently, from Figure C5-1. For a V Zone, CD = 1.0.

The blockage coefficient, CB, is obtained from ASCE/SEI Table C5-3 or, equivalently, from Figure C5-2. Assuming that there is no upstream screening and that the flow path is wider than 30 feet, CB = 1.0.

The maximum response ratio for impulsive load, Rmax, is determined from ASCE/SEI Table C5-4. Using the recommended duration of the debris impact load, Δt, of 0.03 second (see ASCE/SEI C5.4.5) and assuming that the natural period of the building is 0.2 second, the ratio of the impact duration to the natural period of the building is 0.03/0.2 = 0.15. From Table C5-4, Rmax = 0.6 from linear interpolation.

Therefore,

kips

This load acts at the stillwater flood elevation.

Ft 1.1Cpγwds2

2.4γw+ ds2

=

FπWVbCICOCDCBRmax

2g Δt( )---------------------------------------------------------=

F π 1 000,× 13.9× 1.2× 0.8× 1.0× 1.0× 0.6×2 32.2× 0.03× 1 000,×

-------------------------------------------------------------------------------------------------------------- 13= =

Chapter 7 125

2. Special impact loads

ASCE/SEI Equation C5-4 is used to determine special impact loads:

Using a drag coefficient CD = 1.0 and assuming a projected area of debris accumulation A = 1.0 × 6.0 = 6.0 square feet, the impact force, F, per foot length on the wall is

kip/ft

This load acts at the stillwater flood level.

7.2. For the reinforced concrete wall in Problem 7.1, determine the design flood loads on thewall assuming that the wall is nonvertical where the longitudinal axis of the wall is 5degrees from vertical.

SOLUTION

The only difference in the determination of the flood loads occurs for the breaking wave load; inthis problem, the horizontal component of the breaking wave load is determined by Equation5.4-8:

Fnv = Ft sin2 α = 13.6 × sin2 (90 – 5) = 13.5 kips/ft

7.3. For the reinforced concrete wall in Problem 7.1, determine the design flood loads on thewall assuming that the wall is vertical and that the waves are obliquely incident to the wallwhere the horizontal angle between the direction of the waves and the vertical surface is 30degrees.

SOLUTION

The only difference in the determination of the flood loads occurs for the breaking wave load; inthis problem, the breaking wave load for a non-normally incident wave is determined byEquation 5.4-9:

Foi = Ft sin2 α = 13.6 × sin2 30 = 3.4 kips/ft

FCDρAV

2

2----------------------=

F1

64.032.2----------⎝ ⎠⎛ ⎞× 6.0× 13.9

2×

2 1 000,×---------------------------------------------------------- 1= =


7.4. A 125 by 200 foot commercial building is supported by 81 circular reinforced concretepiles that have a diameter of 12 inches each. The site has been categorized as Zone AE witha 100-year stillwater elevation of 10.5 feet NGVD. The lowest ground elevation at the siteis 6.0 feet NGVD and a geotechnical report has indicated that 8 inches of erosion isexpected to occur during the design flood. The local jurisdiction has not mandated anyadditional freeboard. Determine the design flood loads on an individual pile.

SOLUTION

The applicable flood loads are hydrodynamic, breaking wave, and impact.

From the design data, stillwater depth ds = 10.5 – [6.0 – (8/12)] = 5.17 feet

Step 1: Determine water velocity, V.

Since the building is located in an AE Zone, it is appropriate to use the upper bound average water velocity, which is given by ASCE/SEI Equation C5-2:

V = (gds)0.5 = (32.2 × 5.17)0.5 = 12.9 ft/sec

Step 2: Determine hydrodynamic load, Fdyn.

Since the water velocity exceeds 10 ft/sec, it is not permitted to use an equivalent hydro-static load to determine the hydrodynamic load (ASCE/SEI 5.4.3).

The hydrodynamic load, Fdyn, is determined by Equation 7.3:

Based on the recommendations in ASCE/SEI C5.4.3 and Section 8.5.9 of Reference 7.4, the drag coefficient, a, is taken as 1.2 for round piles.

Assuming salt water, the hydrodynamic load is

lb

This load acts at 2 ft-7 in. below the stillwater surface of the water.

Step 3: Determine breaking wave load, FD.

The breaking wave load is determined by ASCE/SEI Equation 5.4-4, which is applicable for vertical pilings and columns:

According to ASCE/SEI 5.4.4.1, the drag coefficient, CD, is equal to 1.75 for round piles.

The breaking wave height, Hb, is determined by ASCE/SEI Equation 5.4-2:

Hb = 0.78ds = 0.78 × 5.17 = 4.0 feet

Therefore, the breaking wave load on one of the piles is

lb

This load acts at the stillwater elevation, which is 5 ft-2 in. above the ground surface.

Fdyn12---aρV

2A=

Fdyn12--- 1.2× 64.0

32.2----------⎝ ⎠⎛ ⎞× 12.9

2× 5.17× 1212------⎝ ⎠⎛ ⎞× 1 026,= =

FD 0.5γwCDDHb2

=

FD 0.5 64.0× 1.75× 1212------⎝ ⎠⎛ ⎞× 4.0

2× 896= =

Chapter 7 127

Step 4: Determine impact load, F.

Both normal and special impact loads are determined.

1. Normal impact loads

ASCE/SEI Equation C5-3 is used to determine normal impact loads:

As noted in Section 7.4.6 of this publication, guidance on establishing the debris weight, W, is given in ASCE/SEI C5.4.5. It is assumed in this example that W = 2,000 lb.

It is reasonable to assume that the velocity of the object, Vb, is equal to the velocity of the water, V. Thus, from Step 1 of this problem, Vb = 12.9 ft/sec.

The importance coefficient, CI, is obtained from Table C5-1. For a Risk Category II building, CI = 1.0.

The orientation coefficient CO = 0.8. This coefficient accounts for impacts that are oblique to the structure.

The depth coefficient, CD, is obtained from ASCE/SEI Table C5-2 or, equivalently, from Figure C5-1. For an A Zone, CD = 1.0 for a stillwater depth greater than 5 feet.

The blockage coefficient, CB, is obtained from ASCE/SEI Table C5-3 or, equivalently, from Figure C5-2. Assuming that there is no upstream screening and that the flow path is wider than 30 feet, CB = 1.0.

The maximum response ratio for impulsive load, Rmax, is determined from ASCE/SEI Table C5-4. Using the recommended duration of the debris impact load, Δt, of 0.03 second (see ASCE/SEI C5.4.5) and assuming that the natural period of the build-ing is 0.3 second, the ratio of the impact duration to the natural period of the building is 0.03/0.3 = 0.10. From Table C5-4, Rmax = 0.4.

Therefore,

lb

This load acts at the stillwater flood elevation.

2. Special impact loads

ASCE/SEI Equation C5-4 is used to determine special impact loads:

Using a drag coefficient CD = 1.0 and assuming a projected area of debris accumulation A = 1.0 × 3.83 = 3.83 square feet, the impact force, F, on one pile is

lb

This load acts at the stillwater flood level.

FπWVbCICOCDCBRmax

2g Δt( )---------------------------------------------------------=

F π 2 000,× 12.9× 1.0× 0.8× 1.0× 1.0× 0.4×2 32.2× 0.03×

-------------------------------------------------------------------------------------------------------------- 13 425,= =

FCDρAV

2

2----------------------=

F1

64.032.2----------⎝ ⎠⎛ ⎞× 5.17× 12.9

2×

2------------------------------------------------------------- 855= =


Step 5: Determine flood load, Fa.

For the analysis of the entire (global) foundation with 81 piles, it is reasonable to assume that one of the piles will be subjected to the impact load, F, and the remaining piles will be subjected to the hydrodynamic load, Fdyn; thus, the total flood load is Fa = 13,425 + (80 × 1,026) = 95,505 lb.

For the case of an individual pile in the front row, Fa is equal to the larger of Fdyn and FD plus F: Fa = 1,026 + 13,425 = 14,451 lb.

7.5. For the design data given in Problem 7.4, determine the design floods on an individual pileassuming the building is designated an emergency shelter.

SOLUTION

The only difference occurs in the calculation of the normal impact load where CI = 1.3 for a RiskCategory IV structure:

lb

For the analysis of the entire (global) foundation with 81 piles, it is reasonable to assume thatone of the piles will be subjected to the impact load, F, and the remaining piles will be subjectedto the hydrodynamic load, Fdyn; thus, the total flood load is Fa = 17,452 + (80 × 1,026) = 99,532lb.

For the case of an individual pile in the front row, Fa is equal to the larger of Fdyn and FD plus F:Fa = 1,026 + 17,452 = 18,478 lb.

F π 2 000,× 12.9× 1.3× 0.8× 1.0× 1.0× 0.4×2 32.2× 0.03×

-------------------------------------------------------------------------------------------------------------- 17 452,= =

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