Roof Design

Roof Design

Basis of Design Type of Occupancy: Residential Building Type of Truss: Howe Truss Project Location: Zone 2 Span of truss: 𝑆𝑇1 = 4.887 𝑚 𝑆𝑇2 = 5.23 m

Height of Truss (rise): 𝐻1 = 0.8 𝑚 𝐻2 = 1.3 𝑚

Bay Distance: 𝐿1 = 3.35 𝑚 𝐿2 = 3.777 𝑚 𝐿3 = 5.073 𝑚

The Truss Truss 1

4

@

1.222

Purlin to purlin spacing,

0.8

h y 4.887

Ɵ 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, Ɵ tan Ɵ = Ɵ=

1.23801 m

0.8 4.887 9.29684 degrees

Tributary Load Area 𝐿1 𝐿2 𝑇𝐿𝐴 = ( + ) 𝑥 𝑆𝑇1 2 2 3.35 𝑚 3.777 𝑚 𝑇𝐿𝐴 = ( + ) 𝑥 4.887 𝑚 2 2 𝑇𝐿𝐴 = 18.46 𝑚2 Load Considerations 1. Gravity Load (GL) 𝐺𝐿 = 𝑅𝐿𝐿 + 𝑅𝐶 + 𝑆𝑊𝑝𝑢𝑟𝑙𝑖𝑛𝑠 2. Loads 

Roof Live Load (RLL) UBC in Pa 𝑇𝐿𝐴 = 18.46 𝑚2 𝜃 = 9.297°

Horizontal Projection

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



𝑁 𝑚

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



𝑁 𝑚

Self-Weight of Purlins 𝑊 = _________

𝑘𝑔

From ASEP steel manual

𝑚

𝑆𝑊 = 𝑊 (9.81

𝑚 ) 𝑠2

3. Wind Load (WL) 𝑊𝐿 = 𝑐 𝑥 𝑞 Where: c = pressure coefficient q = wind pressure from UBC

values of “q”

𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 𝐻1 + 𝐻𝑐 𝑡𝑜𝑡𝑎𝑙 𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 0.8 𝑚 + 15.5 𝑚 𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 16.3 m 𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦 = 53.464 ft q = 30 psf q = 30 x 47.88 = 1436.4 Pa 

Windward Side:

𝑐 = 1.3 𝑥 sin 𝜃 − 0.5

assuming wind load is pressure

𝑊𝐿 = [ 1.3 𝑥 sin (9.297°) − 0.5 ] ( 1436.4 𝑃𝑎 ) [ 𝑊𝐿 = −515.676

1.222 𝑚 ] cos(9.297°)

𝑁 𝑚

Because of the negative value, there is no need to consider its critical design 

Leeward Side: 𝑐 = 0.5 ( 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 ) 𝑊𝐿 = 0.5 ( 1436.4 𝑃𝑎 ) [ 𝑊𝐿 = 889.14

Truss 2

𝑁 𝑚

1.222 𝑚 ] cos(9.297°)

5

@

1.046

Purlin to purlin spacing,

1.3

h y 5.23

Ɵ 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, Ɵ tan Ɵ = Ɵ=

1.07783 m

1.3 5.23 13.9589 degrees

Tributary Load Area 𝑇𝐿𝐴 = 𝐿3 𝑥 𝑆𝑇2 𝑇𝐿𝐴 = 5.073 𝑚 𝑥 5.23 𝑚 𝑇𝐿𝐴 = 26.53 𝑚2 Load Considerations 4. Gravity Load (GL) 𝐺𝐿 = 𝑅𝐿𝐿 + 𝑅𝐶 + 𝑆𝑊𝑝𝑢𝑟𝑙𝑖𝑛𝑠 5. Loads 

Roof Live Load (RLL) UBC in Pa 𝑇𝐿𝐴 = 26.53 𝑚2 𝜃 = 13.9589° RLL = 750 Pa

Horizontal Projection

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



𝑁 𝑚

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



𝑁 𝑚

Self-Weight of Purlins 𝑊 = _________

𝑘𝑔

From ASEP steel manual

𝑚

𝑆𝑊 = 𝑊 (9.81

𝑚 ) 𝑠2

6. Wind Load (WL) 𝑊𝐿 = 𝑐 𝑥 𝑞 Where: c = pressure coefficient q = wind pressure from UBC

values of “q”

𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 𝐻1 + 𝐻𝑐 𝑡𝑜𝑡𝑎𝑙 𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 1.3 𝑚 + 15.5 𝑚 𝐻𝑒𝑖𝑔ℎ𝑡 𝑍𝑜𝑛𝑒 = 16.8 m 𝐴𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑙𝑦 = 55.104 ft q = 30 psf q = 30 x 47.88 = 1436.4 Pa 

Windward Side: 𝑐 = 1.3 𝑥 sin 𝜃 − 0.5

assuming wind load is pressure

𝑊𝐿 = [ 1.3 𝑥 sin (13.9589°) − 0.5 ] ( 1436.4 𝑃𝑎 ) [ 𝑊𝐿 = −288.59

1.046 𝑚 ] cos(13.9589°)

𝑁 𝑚

Because of the negative value, there is no need to consider its critical design 

Leeward Side: 𝑐 = 0.5 ( 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 ) 𝑊𝐿 = 0.5 ( 1436.4 𝑃𝑎 ) [ 𝑊𝐿 = 774.097

1.046 𝑚 ] cos(13.9589°)

𝑁 𝑚

Design of Purlins Trial 2 C Section: C5x9 Properties: 𝑊 = 13.39

𝑘𝑔 𝑚

𝑆𝑥 = 58.34 𝑥 103 𝑚𝑚3 𝑆𝑦 = 7.37 𝑥 103 𝑚𝑚3 𝑍𝑥 = 72.00 𝑁 − 𝑚 𝑍𝑦 = 17.00 𝑁 − 𝑚 Loads: 𝑚 )] 𝑠2 𝑁 𝑁 𝑚 𝑃 = 1.6 (916.31 ) (1.238 𝑚) + 1.2 [148.56 (1.238 𝑚) + 13.39 (9.81 2 )] 𝑚 𝑚 𝑠 𝑁 𝑃 = 1628.258 𝑚 𝑃 = 1.6 (𝑅𝐿𝐿)(𝑐) + 1.2 [𝑅𝐶(𝑐) + 𝑊 (9.81

𝑊𝑥 = 𝑊𝑇 sin 𝜃 𝑊𝑥 = 1628.258

𝑁 sin (9.297°) 𝑚

𝑊𝑥 = 404.7296

𝑁 𝑚

𝑊𝑦 = 𝑊𝑇 cos 𝜃 𝑁 cos (9.297°) 𝑚 𝑁 𝑊𝑦 = 1628.2584 𝑚 𝑊𝑦 = 1628.258

Solution: Considering major axis

L 𝑀𝑟𝑥 =

𝑤𝐿2 8

𝑀𝑟𝑥 =

1628.2584(5.073)2 8

𝑀𝑟𝑥 = 5237.97 𝑁 − 𝑚 Considering minor axis

L 𝑀𝑟𝑦

𝑤𝐿2 = 32

𝑀𝑟𝑦

404.7296(5.073)2 = 32

𝑀𝑟𝑦 = 325.495 𝑁 − 𝑚 𝑀𝑛𝑐𝑥 = 𝐹𝑦 𝑍𝑥 𝑀𝑛𝑐𝑥 = 12240 𝑁 − 𝑚 𝑀𝑐𝑥 = 𝛷𝑀𝑛𝑐𝑥 𝑀𝑐𝑥 = 0.9 ∗ 12240 𝑁 − 𝑚 𝑀𝑐𝑥 = 11016 𝑁 − 𝑚 𝑀𝑛𝑐𝑦 = min[(𝐹𝑦 𝑍𝑦 ); (1.6 𝐹𝑦 𝑆𝑦 )] 𝑀𝑛𝑐𝑦 = min[(170 ∗ 17); (1.6 ∗ 170 ∗ 7.37)] 𝑀𝑛𝑐𝑦 = min[(2890 𝑁 − 𝑚); (2004.64 𝑁 − 𝑚)] 𝑀𝑛𝑐𝑦 = 2004.64 𝑁 − 𝑚 𝑀𝑐𝑦 = 𝛷𝑀𝑛𝑐𝑦 𝑀𝑐𝑦 = 0.9 ∗ 2004.64 𝑁 − 𝑚 𝑀𝑐𝑦 = 1804.176 𝑁 − 𝑚

Checking: 𝑀𝑟𝑥 𝑀𝑟𝑦 + ≤1 𝑀𝑐𝑥 𝑀𝑐𝑦 5237.97 𝑁 − 𝑚 325.495 𝑁 − 𝑚 + ≤1 11016 𝑁 − 𝑚 1804.176 𝑁 − 𝑚 0.475 + 0.1804 = 0.6559 *Adequate

Trial 3 C Section: C4x7.25 Properties: 𝑊 = 10.79

𝑘𝑔 𝑚

𝑆𝑥 = 37.53 𝑥 103 𝑚𝑚3 𝑆𝑦 = 5.62 𝑥 103 𝑚𝑚3

𝑍𝑥 = 46.00 𝑁 − 𝑚 𝑍𝑦 = 13.00 𝑁 − 𝑚 Loads: 𝑚 )] 𝑠2 𝑁 𝑁 𝑚 𝑃 = 1.6 (916.31 ) (1.238 𝑚) + 1.2 [148.56 (1.238 𝑚) + 10.79 (9.81 2 )] 𝑚 𝑚 𝑠 𝑁 𝑃 = 1647.198 𝑚 𝑃 = 1.6 (𝑅𝐿𝐿)(𝑐) + 1.2 [𝑅𝐶(𝑐) + 𝑊 (9.81

𝑊𝑥 = 𝑊𝑇 sin 𝜃 𝑁 sin (9.297°) 𝑚 𝑁 𝑊𝑥 = 397.3464 𝑚 𝑊𝑥 = 1647.198

𝑊𝑦 = 𝑊𝑇 cos 𝜃 𝑁 cos (9.297°) 𝑚 𝑁 𝑊𝑦 = 1589.5550 𝑚 𝑊𝑦 = 1647.198

Solution: Considering major axis

L 𝑀𝑟𝑥 =

𝑤𝐿2 8

𝑀𝑟𝑥 =

1589.5550(5.073)2 8

𝑀𝑟𝑥 = 5142.4174 𝑁 − 𝑚 Considering minor axis

L 𝑀𝑟𝑦 =

𝑤𝐿2 32

𝑀𝑟𝑦 =

397.3464(5.073)2 32

𝑀𝑟𝑦 = 319.557 𝑁 − 𝑚 𝑀𝑛𝑐𝑥 = 𝐹𝑦 𝑍𝑥 𝑀𝑛𝑐𝑥 = 7820 𝑁 − 𝑚 𝑀𝑐𝑥 = 𝛷𝑀𝑛𝑐𝑥 𝑀𝑐𝑥 = 0.9 ∗ 7820 𝑁 − 𝑚 𝑀𝑐𝑥 = 7038 𝑁 − 𝑚 𝑀𝑛𝑐𝑦 = min[(𝐹𝑦 𝑍𝑦 ); (1.6 𝐹𝑦 𝑆𝑦 )] 𝑀𝑛𝑐𝑦 = min[(170 ∗ 13); (1.6 ∗ 170 ∗ 5.62)] 𝑀𝑛𝑐𝑦 = min[(2210 𝑁 − 𝑚); (1528.64 𝑁 − 𝑚)] 𝑀𝑛𝑐𝑦 = 1528.64 𝑁 − 𝑚 𝑀𝑐𝑦 = 𝛷𝑀𝑛𝑐𝑦 𝑀𝑐𝑦 = 0.9 ∗ 1528.64 − 𝑚 𝑀𝑐𝑦 = 1375.776 𝑁 − 𝑚

Checking: 𝑀𝑟𝑥 𝑀𝑟𝑦 + ≤1 𝑀𝑐𝑥 𝑀𝑐𝑦

5142.4174 𝑁 − 𝑚 319.557 𝑁 − 𝑚 + ≤1 7038 𝑁 − 𝑚 1375.776 𝑁 − 𝑚 0.7307 + 0.2323 = 0.963 *Adequate and economical

Design of Sagrods 5𝑤𝐿 8 5 ∗ 397.3464 ∗ 5.073 = 8

𝑇𝑠𝑎𝑔 = 𝑇𝑠𝑎𝑔

𝑇𝑠𝑎𝑔 = 1259.84 𝑁 𝑇𝑠𝑎𝑔𝑚𝑎𝑥 = 𝑛𝑝𝑢𝑟𝑙𝑖𝑛𝑠 𝑇𝑠𝑎𝑔 𝑇𝑠𝑎𝑔𝑚𝑎𝑥 = 5 ∗ 1259.84 𝑁 𝑇𝑠𝑎𝑔𝑚𝑎𝑥 = 6299.18 𝑁 𝑇𝑠𝑎𝑔𝑚𝑎𝑥 = 𝛷𝑇 𝑃𝑛 1259.84 𝑁 = 0.9 ∗ 170 ∗ 𝑑 = 7.24 𝑚𝑚 Say 𝑑 = 8 𝑚𝑚

𝜋 2 𝑑 4

Design of Truss 

Loadings and Analysis of Truss

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, 1.4” 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, 1.4” 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 =



Design of Bottom Chord 𝑃𝑢 = 𝛷𝐹𝑦 𝐴𝑔 Pu = 64.426 kN Fy = 248 Mpa 64.426 kN = 0.9 (248 𝑀𝑝𝑎) ∗ 𝐴𝑔 𝐴𝑔 = 288.647 𝑚𝑚2

From aisc-shapes-database use: L51x51x3.2 for bottom chord Properties: 𝐴 = 317 𝑚𝑚2 𝑟𝑦 = 15.7 𝑚𝑚 

Design of Top Chord

Trial 1 Use: L51x51x3.2

Properties: 𝐴 = 317 𝑚𝑚2 𝑟𝑦 = 15.7 𝑚𝑚

0.1 (1238 𝑚𝑚) 200 000 𝑀𝑃𝑎 ≤ 4.71 √ 15.7 248 𝑀𝑃𝑎 78.854 < 133.755 𝐹𝑦

𝐹𝑐𝑟 = (0.658 𝐹𝑒 ) 𝐹𝑦 𝐹𝑒 =

𝜋2𝐸 𝑘𝐿 2 (𝑟)

=

𝜋 2 (200 000 𝑀𝑃𝑎) (

𝐹𝑒 = 317.5 𝑀𝑝𝑎

0.1 (1238 𝑚𝑚) 2 ) 15.7 248 𝑀𝑃𝑎

𝐹𝑐𝑟 = (0.658 317.5 𝑀𝑝𝑎 ) (248 𝑀𝑝𝑎) 𝐹𝑐𝑟 = 178.8 𝑀𝑝𝑎 𝛷𝑃 𝑛 = 0.9 (178.8 𝑀𝑃𝑎) (317 𝑚𝑚2 ) 𝛷𝑃 𝑛 = 51.02 𝑘𝑁

<

𝑃 𝑢 = 67.677 𝑘𝑁

inadequate Trial 2 Use: L51x51x4.8 Properties: 𝐴 = 466 𝑚𝑚2 𝑟𝑦 = 15.5 𝑚𝑚

0.1 (1238 𝑚𝑚) 200 000 𝑀𝑃𝑎 ≤ 4.71 √ 15.5 248 𝑀𝑃𝑎 79,872 < 133.755 𝐹𝑦

𝐹𝑐𝑟 = (0.658 𝐹𝑒 ) 𝐹𝑦 𝐹𝑒 =

𝜋2𝐸 𝑘𝐿 2 (𝑟)

=

𝜋 2 (200 000 𝑀𝑃𝑎) (

0.1 (1238 𝑚𝑚) 2 ) 15.5

𝐹𝑒 = 308.42 𝑀𝑝𝑎

248 𝑀𝑃𝑎

𝐹𝑐𝑟 = (0.658 308.42 𝑀𝑝𝑎 ) (248 𝑀𝑝𝑎) 𝐹𝑐𝑟 = 177.32 𝑀𝑝𝑎 𝛷𝑃 𝑛 = 0.9 (177.32 𝑀𝑃𝑎) (466 𝑚𝑚2 ) 𝛷𝑃 𝑛 = 74.368 𝑘𝑁

<

𝑃 𝑢 = 67.677 𝑘𝑁

adequate

Use: L51x51x4.8

for top chord



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 = Φ Pn =

use :

C

< 317.453 178.833 56.7 51.0209

133.755 Mpa Mpa kN kN adequate

L51x51x3.2

for web members