# Roof Design

• January 2021
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Roof Design

Basis of Design Type of Occupancy: Residential Building Type of Truss: Howe Truss Project Location: Zone 2 Span of truss: ππ1 = 4.887 π ππ2 = 5.23 m

Height of Truss (rise): π»1 = 0.8 π π»2 = 1.3 π

Bay Distance: πΏ1 = 3.35 π πΏ2 = 3.777 π πΏ3 = 5.073 π

The Truss Truss 1

4

@

1.222

Purlin to purlin spacing,

0.8

h y 4.887

Ζ m

by ratio and proportion 0.8 = 4.887 y =

1.222 m

y 1.222 0.2

m

by Pythagorean theorem

c= h= for the angle, Ζ tan Ζ = Ζ=

1.23801 m

0.8 4.887 9.29684 degrees

Tributary Load Area πΏ1 πΏ2 ππΏπ΄ = ( + ) π₯ ππ1 2 2 3.35 π 3.777 π ππΏπ΄ = ( + ) π₯ 4.887 π 2 2 ππΏπ΄ = 18.46 π2 Load Considerations 1. Gravity Load (GL) πΊπΏ = ππΏπΏ + ππΆ + ππππ’πππππ  2. Loads ο·

Roof Live Load (RLL) UBC in Pa ππΏπ΄ = 18.46 π2 π = 9.297Β°

Horizontal Projection

RLL = 750 Pa Equivalent RLL = RLL x horizontal spacing Equivalent RLL = 750 Pa x 1.222 m Equivalent RLL = 916.3125

ο·

π π

Roof Cover (RC) From NSCP 2015

deck metal gage 20

Metal gage 20 = 0.12 kPa = 120 kPa RC = 120 Pa x inclined spacing RC = 120 Pa x 1.238 m RC = 148.5614

ο·

π π

Self-Weight of Purlins π = _________

ππ

From ASEP steel manual

π

ππ = π (9.81

π ) π 2

3. Wind Load (WL) ππΏ = π π₯ π Where: c = pressure coefficient q = wind pressure from UBC

values of βqβ

π»πππβπ‘ ππππ = π»1 + π»π π‘ππ‘ππ π»πππβπ‘ ππππ = 0.8 π + 15.5 π π»πππβπ‘ ππππ = 16.3 m π΄πππππ₯ππππ‘πππ¦ = 53.464 ft q = 30 psf q = 30 x 47.88 = 1436.4 Pa ο·

Windward Side:

π = 1.3 π₯ sin π β 0.5

ππΏ = [ 1.3 π₯ sin (9.297Β°) β 0.5 ] ( 1436.4 ππ ) [ ππΏ = β515.676

1.222 π ] cos(9.297Β°)

π π

Because of the negative value, there is no need to consider its critical design ο·

Leeward Side: π = 0.5 ( π π’ππ‘πππ ) ππΏ = 0.5 ( 1436.4 ππ ) [ ππΏ = 889.14

Truss 2

π π

1.222 π ] cos(9.297Β°)

5

@

1.046

Purlin to purlin spacing,

1.3

h y 5.23

Ζ m

by ratio and proportion 1.3 = 5.23 y =

1.046 m

y 1.046 0.26

m

by Pythagorean theorem

c= h= for the angle, Ζ tan Ζ = Ζ=

1.07783 m

1.3 5.23 13.9589 degrees

Tributary Load Area ππΏπ΄ = πΏ3 π₯ ππ2 ππΏπ΄ = 5.073 π π₯ 5.23 π ππΏπ΄ = 26.53 π2 Load Considerations 4. Gravity Load (GL) πΊπΏ = ππΏπΏ + ππΆ + ππππ’πππππ  5. Loads ο·

Roof Live Load (RLL) UBC in Pa ππΏπ΄ = 26.53 π2 π = 13.9589Β° RLL = 750 Pa

Horizontal Projection

Equivalent RLL = RLL x horizontal spacing Equivalent RLL = 750 Pa x 1.046 m Equivalent RLL = 784.5

ο·

π π

Roof Cover (RC) From NSCP 2015

deck metal gage 20

Metal gage 20 = 0.12 kPa = 120 kPa RC = 120 Pa x inclined spacing RC = 120 Pa x 1.0778 m RC = 129.34

ο·

π π

Self-Weight of Purlins π = _________

ππ

From ASEP steel manual

π

ππ = π (9.81

π ) π 2

6. Wind Load (WL) ππΏ = π π₯ π Where: c = pressure coefficient q = wind pressure from UBC

values of βqβ

π»πππβπ‘ ππππ = π»1 + π»π π‘ππ‘ππ π»πππβπ‘ ππππ = 1.3 π + 15.5 π π»πππβπ‘ ππππ = 16.8 m π΄πππππ₯ππππ‘πππ¦ = 55.104 ft q = 30 psf q = 30 x 47.88 = 1436.4 Pa ο·

Windward Side: π = 1.3 π₯ sin π β 0.5

ππΏ = [ 1.3 π₯ sin (13.9589Β°) β 0.5 ] ( 1436.4 ππ ) [ ππΏ = β288.59

1.046 π ] cos(13.9589Β°)

π π

Because of the negative value, there is no need to consider its critical design ο·

Leeward Side: π = 0.5 ( π π’ππ‘πππ ) ππΏ = 0.5 ( 1436.4 ππ ) [ ππΏ = 774.097

1.046 π ] cos(13.9589Β°)

π π

Design of Purlins Trial 2 C Section: C5x9 Properties: π = 13.39

ππ π

ππ₯ = 58.34 π₯ 103 ππ3 ππ¦ = 7.37 π₯ 103 ππ3 ππ₯ = 72.00 π β π ππ¦ = 17.00 π β π Loads: π )] π 2 π π π π = 1.6 (916.31 ) (1.238 π) + 1.2 [148.56 (1.238 π) + 13.39 (9.81 2 )] π π π  π π = 1628.258 π π = 1.6 (ππΏπΏ)(π) + 1.2 [ππΆ(π) + π (9.81

ππ₯ = ππ sin π ππ₯ = 1628.258

π sin (9.297Β°) π

ππ₯ = 404.7296

π π

ππ¦ = ππ cos π π cos (9.297Β°) π π ππ¦ = 1628.2584 π ππ¦ = 1628.258

Solution: Considering major axis

L πππ₯ =

π€πΏ2 8

πππ₯ =

1628.2584(5.073)2 8

πππ₯ = 5237.97 π β π Considering minor axis

L πππ¦

π€πΏ2 = 32

πππ¦

404.7296(5.073)2 = 32

πππ¦ = 325.495 π β π ππππ₯ = πΉπ¦ ππ₯ ππππ₯ = 12240 π β π πππ₯ = π·ππππ₯ πππ₯ = 0.9 β 12240 π β π πππ₯ = 11016 π β π ππππ¦ = min[(πΉπ¦ ππ¦ ); (1.6 πΉπ¦ ππ¦ )] ππππ¦ = min[(170 β 17); (1.6 β 170 β 7.37)] ππππ¦ = min[(2890 π β π); (2004.64 π β π)] ππππ¦ = 2004.64 π β π πππ¦ = π·ππππ¦ πππ¦ = 0.9 β 2004.64 π β π πππ¦ = 1804.176 π β π

Checking: πππ₯ πππ¦ + β€1 πππ₯ πππ¦ 5237.97 π β π 325.495 π β π + β€1 11016 π β π 1804.176 π β π 0.475 + 0.1804 = 0.6559 *Adequate

Trial 3 C Section: C4x7.25 Properties: π = 10.79

ππ π

ππ₯ = 37.53 π₯ 103 ππ3 ππ¦ = 5.62 π₯ 103 ππ3

ππ₯ = 46.00 π β π ππ¦ = 13.00 π β π Loads: π )] π 2 π π π π = 1.6 (916.31 ) (1.238 π) + 1.2 [148.56 (1.238 π) + 10.79 (9.81 2 )] π π π  π π = 1647.198 π π = 1.6 (ππΏπΏ)(π) + 1.2 [ππΆ(π) + π (9.81

ππ₯ = ππ sin π π sin (9.297Β°) π π ππ₯ = 397.3464 π ππ₯ = 1647.198

ππ¦ = ππ cos π π cos (9.297Β°) π π ππ¦ = 1589.5550 π ππ¦ = 1647.198

Solution: Considering major axis

L πππ₯ =

π€πΏ2 8

πππ₯ =

1589.5550(5.073)2 8

πππ₯ = 5142.4174 π β π Considering minor axis

L πππ¦ =

π€πΏ2 32

πππ¦ =

397.3464(5.073)2 32

πππ¦ = 319.557 π β π ππππ₯ = πΉπ¦ ππ₯ ππππ₯ = 7820 π β π πππ₯ = π·ππππ₯ πππ₯ = 0.9 β 7820 π β π πππ₯ = 7038 π β π ππππ¦ = min[(πΉπ¦ ππ¦ ); (1.6 πΉπ¦ ππ¦ )] ππππ¦ = min[(170 β 13); (1.6 β 170 β 5.62)] ππππ¦ = min[(2210 π β π); (1528.64 π β π)] ππππ¦ = 1528.64 π β π πππ¦ = π·ππππ¦ πππ¦ = 0.9 β 1528.64 β π πππ¦ = 1375.776 π β π

Checking: πππ₯ πππ¦ + β€1 πππ₯ πππ¦

5142.4174 π β π 319.557 π β π + β€1 7038 π β π 1375.776 π β π 0.7307 + 0.2323 = 0.963 *Adequate and economical

Design of Sagrods 5π€πΏ 8 5 β 397.3464 β 5.073 = 8

ππ ππ = ππ ππ

ππ ππ = 1259.84 π ππ πππππ₯ = πππ’πππππ  ππ ππ ππ πππππ₯ = 5 β 1259.84 π ππ πππππ₯ = 6299.18 π ππ πππππ₯ = π·π ππ 1259.84 π = 0.9 β 170 β π = 7.24 ππ Say π = 8 ππ

π 2 π 4

Design of Truss ο·

Considering Forces on Truss 1

P = Wtotal (bay distance) P= 6.33707 kN P/2 = 3.16854 kN Ceilings Suspended steel channel system = Gypsum Board, 1.4β thick = Total = CL = y1 = y2 = y3 =

100 50 150 613.929

Pa Pa Pa N

0.2 m 0.4 m 0.6 m

Solve forces using SW Truss R1 = 18.23 kN R2 = 13.91 kN Top Chord: Most Critical Force =

Bottom Chord:

67.677 kN

C

Most Critical Force =

64.426 kN

T

7.903 kN 24.811 kN

T C

Web Members: Most Critical tension = Most Critical compression =

Considering Forces on Truss 2

P = Wtotal (bay distance) P= 6.33707 kN P/2 = 3.16854 kN Ceilings Suspended steel channel system = Gypsum Board, 1.4β thick = Total = CL = y1 = y2 = y3 = y4 = Solve forces using SW Truss R1 = 21.88 kN R2 = 17.55 kN

0.26 0.52 0.78 1.04

100 50 150 613.929

Pa Pa Pa N

m m m m

Top Chord: Most Critical Force =

60.424 kN

C

Bottom Chord: Most Critical Force =

56.278 kN

T

11.599 kN 20.576 kN

T C

Web Members: Most Critical tension = Most Critical compression =

ο·

Design of Bottom Chord ππ’ = π·πΉπ¦ π΄π Pu = 64.426 kN Fy = 248 Mpa 64.426 kN = 0.9 (248 πππ) β π΄π π΄π = 288.647 ππ2

From aisc-shapes-database use: L51x51x3.2 for bottom chord Properties: π΄ = 317 ππ2 ππ¦ = 15.7 ππ ο·

Design of Top Chord

Trial 1 Use: L51x51x3.2

Properties: π΄ = 317 ππ2 ππ¦ = 15.7 ππ

0.1 (1238 ππ) 200 000 πππ β€ 4.71 β 15.7 248 πππ 78.854 < 133.755 πΉπ¦

πΉππ = (0.658 πΉπ ) πΉπ¦ πΉπ =

π2πΈ ππΏ 2 (π)

=

π 2 (200 000 πππ) (

πΉπ = 317.5 πππ

0.1 (1238 ππ) 2 ) 15.7 248 πππ

πΉππ = (0.658 317.5 πππ ) (248 πππ) πΉππ = 178.8 πππ π·π π = 0.9 (178.8 πππ) (317 ππ2 ) π·π π = 51.02 ππ

<

π π’ = 67.677 ππ

inadequate Trial 2 Use: L51x51x4.8 Properties: π΄ = 466 ππ2 ππ¦ = 15.5 ππ

0.1 (1238 ππ) 200 000 πππ β€ 4.71 β 15.5 248 πππ 79,872 < 133.755 πΉπ¦

πΉππ = (0.658 πΉπ ) πΉπ¦ πΉπ =

π2πΈ ππΏ 2 (π)

=

π 2 (200 000 πππ) (

0.1 (1238 ππ) 2 ) 15.5

πΉπ = 308.42 πππ

248 πππ

πΉππ = (0.658 308.42 πππ ) (248 πππ) πΉππ = 177.32 πππ π·π π = 0.9 (177.32 πππ) (466 ππ2 ) π·π π = 74.368 ππ

<

π π’ = 67.677 ππ

Use: L51x51x4.8

for top chord

ο·

Design of Web Members

Tension: 11.599 kN Ag =

T

51.9668 mm^2

Initial assumption : From aisc-shapes-database L51x51x3.2 Properties: A= ry =

317 mm^2 15.7 mm

Compression : 24.811 kN

78.854251 Fe = Fcr = Pn = Ξ¦ Pn =

use :

C

< 317.453 178.833 56.7 51.0209

133.755 Mpa Mpa kN kN adequate

L51x51x3.2

for web members

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