PART 6 -ANALYSIS AND DESIGN OF PURLINS (W/ CALCULATION OF C&C WIND LOADS FOR LOW-RISE BUILDINGS PDF

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NEW ERA UNIVERSITY DEPARTMENT OF CIVIL ENGINEERING PART 6 – ANALYSIS AND DESIGN OF PURLINS (W/ CALCULATION OF C&C WIND LOADS FOR LOW-RISE BUILDINGS) Using the building data from Part 3, design the required size of purlins to be placed on top of trusses spaced at 3.5m. h' = 15 tan 10 =2.64m 1...

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PART 6 -ANALYSIS AND DESIGN OF PURLINS (W/ CALCULATION OF C&C WIND LOADS FOR LOWRISE BUILDINGS Winfred Liwanag II

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NEW ERA UNIVERSITY

DEPARTMENT OF CIVIL ENGINEERING

PART 6 – ANALYSIS AND DESIGN OF PURLINS (W/ CALCULATION OF C&C WIND LOADS FOR LOW-RISE BUILDINGS) Using the building data from Part 3, design the required size of purlins to be placed on top of trusses spaced at 3.5m.

h' = 15 tan 10 =2.64m

10º

where:

h'

E = eave height = 7.5m R = ridge height = E+h’ = 10.14m Ht = mean roof height = E+(h’/2) = 8.82m

7.5m

15m

46m 15m

Building Perspective effective area for truss

3.5m

effective area for purlins

s Wall Truss Purlins Sag Rod

Roof Plan Consider loadings below: Superimposed DL (roof sheet, insulation and connections): Roof Live Load (maintenance and repairs): Wind Load: CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

0.90 KPa 0.60 KPa to be computed 1

NEW ERA UNIVERSITY

I. II. III.

DEPARTMENT OF CIVIL ENGINEERING

Provide purlin spacing. Determine wind load pressure acting on the purlins. Assuming no sag rod in place, Choose safe channel (LC) section with Demand Capacity Ratio (DCR) ≤ 0.80. Check strength and deflection requirements. Use the ASEP Steel Sections as a reference for LC shape section geometry and properties. Refer to Section 203.4 of NSCP 2015 (page 2-11) for the load combinations using Allowable Stress/Strength Design and Alternate Load Design. Assume allowable deflection as L/180. Assume that all loads will pass through the centroid of the channel section. Use ASTM A36 steel grade (Fy = 248 MPa, E = 200GPa)

Commentaries: Purlins is a horizontal structural member in a roof which supports the loads from the roof sheathing and are supported by rafters, trusses or building walls. As a rule of thumb, purlins are frequently proportioned for a ratio of depth to length of 1/32. Gravity, live and wind loads should be calculated by the designer and must be check against strength and deflection requirements of the code. Sag rods are usually furnished to transmit the gravity load of purlins. Additionally, sag rods are used to control deflection and stiffen purlins. Typical sag rods are 10-12mm in diameter. To be effective, the force in the sag rods must be carried and balanced by a corresponding force on the opposite side of the ridge (see Figure 1). Different steel sections might be used as a purlin (C, Z or LC sections) but the most common and locally available in the Philippine setting is the LC channel section as shown in the Figure 2.

Purlin Profile bf

c

d

tf

d = depth bf = flange width c = lipped tf = flange thickness

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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Figure 1. Typical Tie Rod Connector Detail

Figure 2. Typical Purlins and Sag Rod Detail

Y WIND

Y’

DEAD & LIVE

WN WT

X  X’

Figure 3. Forces on purlins

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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DEPARTMENT OF CIVIL ENGINEERING

It is evident as that the purlins are subjected to biaxial or unsymmetric bending. Dead and live loads should be resolve in its equivalent normal and tangential forces in line with the principal axes X and Y (see Figure 3). On the other hand, wind loads are directly normal with the Xaxis. It is assumed that the purlins are simply supported at the trusses where the maximum bending moment about each axis is WL2/8. If sag rods are used, they will provide lateral support with respect to X-axis (strong axis) bending and will act as transverse supports for Y-axis (weak axis) bending, requiring that the purlin be treated as a continuous beam. Among the notable changes in NSCP 2010, the wind speed map and the importance factor for wind loads have been replaced with three wind contour maps based on difference and newly defined return periods. The new wind maps are now specified at the strength design level (Load Resistance Factored Design) rather than the service design level (Allowable Stress Design). A load factor of 1.6 for strength design has already been integrated into the design wind speeds specified in the NSCP 2015 maps, so the wind load factor for the strength design load combinations has actually gone from 1.6 in NSCP 2010 to 1.0 in NSCP 2015, and the wind load factor for the allowable stress design load combinations has gone from 1.0 to 0.6 in correspondence. Summarizing Section 203.4, the load combination to be included are as follows: For Allowable Stress Design: (Note: No increase in allowable stresses) i. DL + RLL ii. DL + 0.6WL1 iii. DL + 0.6WL2 For Alternate Load Design: (Note: 1/3 increase in allowable stresses shall be permitted) iv. DL + 0.75 [ RLL + 0.6(WL1)] v. DL + 0.75 [ RLL + 0.6(WL2)] vi. 0.6DL + 0.6WL1 vii. 0.6DL + 0.6WL2 where: DL RLL WL1 WL2

= = = =

Dead Load Roof Live Load Wind Load (Suction) Wind Load (Pressure)

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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DEPARTMENT OF CIVIL ENGINEERING

Solution: I.

Provide purlin spacing. Purlins are usually spaced at 600-800mm in line with the slope of the roof (see Figure 2). For the sake of this discussion, we will use S = 600mm

II.

Determine wind load acting on purlins. As defined in the code... COMPONENTS AND CLADDING (C&C): Components receive wind loads directly or from cladding and transfer the load to the MWFRS. Cladding receives wind loads directly. Examples of components include fasteners, purlins, girts, studs, roof decking, and roof trusses. Therefore, calculations made in Part 3 & 4 is not applicable because it focuses on MWFRS. Section 207E of NSCP 2015 (page 2-130) will be used. Furthermore, since the mean roof height (h = 8.82m) is less than 18m, Part 2: Low-Rise Buildings (Simplified) Section 207E.5 (page 2-137) is applicable and design wind pressures for C&C will be:

where:

𝑝𝑛𝑒𝑡 = 𝜆𝐾𝑧𝑡 𝑃𝑛𝑒𝑡9

λ = adjustment factor for building height and exposure from Figure 207E.5-1 Kzt = topographic factor Pnet9 = net design wind pressure from Figure 207E.5-1

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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a) Basic Wind Speed, V Since occupancy category is IV, refer to Figure 207A.5-1A (NSCP 2015 page 2-38)

Therefore, design wind speed, V = 240kph

b) Exposure Category Referring to Section 207A.7.2 (NSCP 2015 page 2-42) Open and flat terrain is equivalent to Exposure C.

c) Topographic Factor, Kzt Referring to Section 207A.8 (NSCP 2015 page 2-46) For flat terrain, Kzt = 1.0

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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d) Define Roof Zones Using Figure 207E.5-1 (page 2-169), the roof angle θ = 10º will fall under Gable Roof (7º < θ ≤ 45º) having three different zones (interior, end and corner zones).

Additionally, the width of end zone surface “a” shall be equal to 10 percent of least horizontal dimension or 0.4h, whichever is smaller, but not less than either 4% of leas horizontal dimension or 0.9m. where: h = mean roof height, in meters, except that eave height shall be used for θ ≤ 10°. Therefore: Width “a” Smaller of a = 2(0.1)(30m) = 6m Or (0.4)(7.5) = 3m [governs] But not less than (0.04)(30) = 1.2m Or 0.9m a = 3m CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

7

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DEPARTMENT OF CIVIL ENGINEERING

e) Net Design Wind Pressure, Pnet9 Using Figure 207E.5-1 (page 2-170), the basic wind speed, V = 240kph will fall into 250kph column (closest). Effective wind area of purlins = 0.6m (3.5m) = 2.1m2 for pressure

for suction

By Interpolation: Zone 1 2 3

Pressure (KPa) 1.084 1.084 1.084

Suction (KPa) -1.847 -3.026 -4.553

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

8

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DEPARTMENT OF CIVIL ENGINEERING

f) Adjustment Factor, λ From Figure 207E.5-1 (page 2-171)

By Interpolation:

λ = 1.39

g) Design Wind Pressure, Pnet

Zone 1 2 3

𝑝𝑛𝑒𝑡 = 𝜆𝐾𝑧𝑡 𝑃𝑛𝑒𝑡9

Pressure (KPa) 1.39(1.0)(1.084) = 1.507 1.39(1.0)(1.084) = 1.507 1.39(1.0)(1.084) = 1.507

Suction (KPa) 1.39(1.0)(-1.847) = -2.567 1.39(1.0)(-3.026) = -4.206 1.39(1.0)(-4.553) = -6.329

Zone 1 Zone 2 Zone 3

a a a = 3m a

a

For simplicity, we will use the maximum wind pressure at zone 3 to design all the purlins. The designer may use different pressure at each zone to make the design more economical.

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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III.

DEPARTMENT OF CIVIL ENGINEERING

Assuming no sag rod in place, Choose safe channel (LC) section with Demand Capacity Ratio (DCR) ≤ 0.80. Check strength and deflection requirements.

a) Initial LC section, L = 3.5m Depth/Length = 1/32 = L/32 = 3500/32 Therefore: Initial depth = 109.375mm say 150mm Referring to ASEP sections. Try LC150x65x20x4.5 for initial section. Section Properties: Weight, W Area, A Ix Sx Iy Sy d bf tf c

= = = = = = = = = =

9.98kg/m = 97.87 N/m 1,272mm2 4,206x103 mm4 56.1x103 mm3 643x103 mm4 14.5x103 mm3 150 mm 65 mm 4.5 mm 20 mm

Y’

Y

10º W

WN

WT

X 10º X’

b) Resolving equivalent forces in reference to axis X and Y. Superimposed DL: Roof Live Load: Wind Load:

0.90 KPa 0.60 KPa 6.329 KPa (Suction) 1.507 KPa (Pressure)

Normal Load Per Linear Meter: Due to DL: Tributary width = Purlin spacing = 600mm Vertical DL, WDL = Selfweight + Superimposed DL WDL = 97.87 N/m + 0.90 KPa (0.6m) (1000N/1kN) WDL = 637.87 N/m Normal Component, WDL(N) = WDL cos α = 637.87 N/m cos10º WDL(N) = 628.18 N/m Due to RLL: WRLL = 0.60 KPa (0.6m) (1000N/1kN) WRLL = 360 N/m Normal Component, WRLL(N) = WRLL cos α = 360 N/m cos10º WRLL(N) = 354.53 N/m CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

10

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DEPARTMENT OF CIVIL ENGINEERING

Due to WL: WWL1 = 6.329 KPa (0.6m) (1000N/1kN) = 3,797.4 N/m WWL2 = 1.507 KPa (0.6m) (1000N/1kN) = 904.2 N/m

(Suction) (Pressure)

Tangential Load Per Linear Meter: Due to DL: WDL = 637.87 N/m Tangential Component, WDL(T) = WDL sin α = 637.87 N/m sin 10º WDL(T) = 110.76 N/m Due to RLL: WRLL = 360 N/m Tangential Component, WRLL(T) = WRLL sin α = 360 N/m sin10º WRLL(T) = 62.52 N/m

Y’

Y

10º WN

W

WT

c) Combination of loads

203.4.1 Basic Load

Load Combination DL + RLL DL + 0.6WL1 DL + 0.6WL2

203.4.2 Alternate Basic Load

DL + 0.75 [ RLL + 0.6(WL1)]

DL + 0.75 [ RLL + 0.6(WL2)]

0.6DL + 0.6WL1

0.6DL + 0.6WL2

WN (N/m)

WT (N/m)

628.18 + 354.53 = 982.71 628.18 + 0.6(-3,797.4) = -1,650.26 628.18 + 0.6(904.2) = 1,170.7 628.18 + 0.75[354.53 + 0.6(-3,797.4)] = -814.75 628.18 + 0.75[354.53 + 0.6(904.2)] = 1,300.97 0.6(628.18) + 0.6(3,797.4) = -1,901.53 0.6(628.18) + 0.6(904.2) = 919.43

110.76 + 62.52 = 173.28 110.76

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

110.76 110.76 + 0.75[62.52] = 157.65 110.76 + 0.75[62.52] = 157.65 0.6(110.76) = 66.46 0.6(110.76) = 66.46

11

X 10º X’

NEW ERA UNIVERSITY

DEPARTMENT OF CIVIL ENGINEERING

d) Interaction Criteria “Biaxial bending” occurs when a beam is subjected to a loading condition that produces bending about both the major (strong) axis and the minor (weak) axis. To deal with combined loading, we look ahead to Section 508.2 Unsymmetric Members Subjected to Flexure of NSCP 2015 (page 5-78). The specification deals with combined loading primarily through the use of interaction formulas, which account for the relative importance of each load effect in relation to the strength corresponding to that effect. If there is bending about both the x and y axes, the interaction approach requires that the sum of ratios for the two effects be less than 1.0; that is, For ASD: |

𝑓𝑏𝑥

𝐹𝑏𝑥

where:

|

𝑓𝑏𝑥

𝐹𝑏𝑥 ∗1.33

+𝐹

+

𝑓𝑏𝑦

𝐹𝑏𝑦

𝑓𝑏𝑦

| ≤ 1.0 (for basic load combo)

| ≤ 1.0 (for alternate basic load combo)

𝑏𝑦 ∗1.33

fbx, fby = required flexural stress at the specific axis using ASD load combinations Fbx, Fby = allowable flexural stress By checking width-limiting thickness ratio to determine whether it is a compact or noncompact section. The equation below should be satisfied for a compact section, 𝑏𝑓 170 ≤ 2𝑡𝑓 √𝐹𝑦

170 65 ≤ 2(4.5) √248 7.22 ≤ 10.80

If the section is compact, the flexural capacity (Fbx if along x-axis and Fby for y-axis) of the purlin is equal to 0.66Fy, if it is non-compact section, its flexural or bending moment capacity is equal to 0.60Fy Fb = 0.66Fy Fb = 0.60Fy

(compact section) (non-compact section)

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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e) Determine bending forces. Let us first use the DL + RLL under basic load combination to define our algorithm. For bending along X-axis: 𝑊𝑁 𝐿2 𝑀𝑋 = (𝑠𝑖𝑚𝑝𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑) 8

982.72(3.5)2 𝑀𝑋 = = 1,504.79𝑁. 𝑚 8

𝑓𝑏𝑥

𝑓𝑏𝑥

𝑓𝑏𝑥

𝑀𝑋 = 𝑆𝑋 1,504.79 1000𝑚𝑚 ( ) = 1𝑚 56.1𝑋103 𝑁 = 26.82 = 26.82𝑀𝑃𝑎 𝑚𝑚2

Y’

Y

10º W

WN X 10º X’

WT

For bending along Y-axis: 𝑀𝑌 =

173.28(3.5)2 = 265.34𝑁. 𝑚 8 𝑀𝑌 = 𝑆𝑌 265.34 1000𝑚𝑚 ( ) = 1𝑚 14.5𝑋103

𝑀𝑌 = 𝑓𝑏𝑦

𝑓𝑏𝑦

𝑊𝑇 𝐿2 (𝑠𝑖𝑚𝑝𝑙𝑦 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑) 8

𝑓𝑏𝑦 = 18.30𝑀𝑃𝑎

It should be noted that for bending along X-axis (Mx), the addition of sagrod has no effect in bending moment which is always considered as a simply supported M = WL2/8. A different scenario for bending along Y-axis (My), if one sagrod is added at midspan the purlin will be treated as a continuous beam. The maximum moment @ midspan will be computed using three moment equation: 𝑀1 𝐿1 + 2𝑀2 (𝐿1 + 𝐿2 ) + 𝑀3 𝐿2 +

6𝐴1 𝑎̅1 6𝐴2 𝑎̅2 + =0 𝐿1 𝐿2

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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DEPARTMENT OF CIVIL ENGINEERING

where: 1

2

𝑀1 = 𝑀3 = 0 𝐿1 = 𝐿2 = 𝐿/2

3

3 6𝐴1 𝑎̅1 𝑊𝐿3 𝑊(𝐿1 )3 𝑊(𝐿⁄2) = = = 𝐿1 4 4 4 3 𝑊𝐿 = 32 3 6𝐴2 𝑎̅2 𝑊𝐿3 𝑊(𝐿2 )3 𝑊(𝐿⁄2) = = = 𝐿2 4 4 4 3 𝑊𝐿 = 32

-WL^2/32

Thus, 0 + 2𝑀2 (𝐿) + 0 +

𝑊𝐿3 𝑊𝐿3 + =0 32 32

𝑀2 = 𝑀𝑌 =

𝑊𝐿2 𝑊𝑇 𝐿2 𝑜𝑟 32 32

The same applies for sagrod in third points, using three moment equation the bending moment at Y-axis will be: 𝑊𝑇 𝐿2 𝑀𝑌 = 90 Applying the interaction criteria for DL+RLL load combination: |

𝑓𝑏𝑥 𝑓𝑏𝑦 + | ≤ 1.0 𝐹𝑏𝑥 𝐹𝑏𝑦

18.30 26.82 + | ≤ 1.0 | 0.66(248) 0.66(248) |0.164 + 0.119| ≤ 1.0 |0.28| ≤ 1.0

The DCR = 0.28 is safe yet uneconomical. Smaller or lighter section should be adopted.

CE 525SE Special Topics in Structural Engineering (Wind Engineering) Prepared by: Engr. Winfred Liwanag II, M. Eng

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The complete results for trial section LC150x65x20x4.5 will be as shown in the table below. Bending Moments Location

Normal

Load Combinations

Actual Stress

Tangential

Normal

Allowable

Tangential

Stress Ratio Normal

Tangential

Combined

Fbx

Fby

fbx / Fbx

fby / Fby

DCR

Mx

My

fbx

fby

1507.16

265.35

26.88

18.28

163.94

163.94

0.164

0.112

0.28

2. D + 0.6WL1

-2524.57

42.41

45.02

2.92

163.94

163.94

0.275

0.018

0.29

3. D + 0.6WL2

1795.02

42.41

32.01

2.92

163.94

163.94

0.195

0.018

0.21

4. D + 0.75(RLL+0.6WL1)

-1245.20

60.36

22.20

4.16

218.05

218.05

0.102

0.019

0.12

4. D + 0.75(RLL+0.6WL2)

1994.50

60.36

35.57

4.16

218.05

218.05

0.163

0.019

0.18

6. 0.6D + 0.6WL1

-2910.29

25.44

51.90

1.75

218.05

218.05

0.238

0.008

0.25

7. 0.6D + 0.6WL2

1409.31

25.44

25.13

1.75

218.05

218.05

0.115

0.008

0.12

@ Midspan 1. DL + RLL

Note: Might not be exact with your hand calcs. This result came from excel spreadsheet.

f) Deflection check.

Y’

Y

10º

Along X-axis: 1𝑚 4 5𝑊𝑁 𝐿4 5 ∗ [982.72 ∗ (1000𝑚𝑚 )](3,500) = ∆𝑋 = 384(200,000)(4,206𝑋103 ) 384𝐸𝐼𝑋

∆𝑋 = 2.28𝑚𝑚

W

WN

WT

Along Y-axis:

1𝑚 5 ∗ [173.28 ∗ ( )](3,500)4 5𝑊𝑇 𝐿4 1000𝑚𝑚 ∆𝑌 = = 384𝐸𝐼𝑌 384(200,000)(643𝑋103 )

∆𝑌 = 2.63𝑚𝑚