Loads, Building data and Material Properties Loading: NSCP 2010...

Loads, Building data and Material Properties

Loading: NSCP 2010

Minimum Design Loads

Table 1. Minimum Design Dead Loads

Component Unit Load, KN/m2

Ceillings

Plaster on Tile or Concrete 0.24

Acoustical Fiber Board 0.05

Suspended metal lath and gypsum plaster 0.48

Mechanical Duct Allowance 0.20

Covering, Roof and Wall

Asphalt Shingles 0.10

Insultaion (Urethane foam w/ Skin) 0.0009

Insultaion (Polystyrene foam) 0.0004

Water proofing membrane, Bituminous (smooth surface) 0.07

Cement tile Finished 0.77

Floor and Floor Finishes

Ceramic or Quarry Tile (20mm) 0.77

Marble and mortar on stone-concrete 1.58

Lightweight concrete, plain per mm 0.02

Linoleum or Asphaltic tile, 6mm 0.05

Subflooring, 19mm 0.14

Frame Partitions and Walls

Exterior stud walls with brick veneer 2.30

Windows, glass, frame and sash 0.38

Wood studs 2 x 4 in., ( 50 x 100 mm) plastered two sides 0.96

Wood studs 2 x 4 in., ( 5 x 10 mm) unplastered 0.19

Concrete Masonry units (Wall)

Masonry, Normal weight 21.20

Masonry Lightweight solid concrete 16.50

Table 2. Minimum Design Live Loads

Use or Occupancy Unit Load

Category Description KN/m2

Residential

Basic Floor Area 1.90

Bedrooms 2.00

Exterior Balconies 2.90

Decks 1.90

Storage 1.90

Restrooms ---- ---- 2.87

Stairs --- ---- 2.00

Table 3. Special Loads

Use or Occupancy Lateral load

Category Description KN/m2

Balcony Railings & guardrails

Exit Facilites 0.75

Other than exit facilities 0.30

components 1.20

Partitions & interior walls 0.25

Table 4. Minimum Roof Live Loads

Table 5. Minimum Densities for Design Loads

Material Density,KN/m3

Aluminum 26.70

Cement Board 7.10

Plywood 5.70

Laminated Red Wood (1/2") 28.00

Mortar Cement or lime (2") 20.40

Load Combinations: ASD

1.) D D = Dead Load

2.) D + L L = LiveLoad

3.) D + (Lr or S or R) Lr = Roof Live Load

4.) D + 0.75L + 0.75(Lr or S or R) R = Rain Load

5.) D + (W or 0.7E) W = Wind Load

6.) D + 0.75(W or 0.7E) + 0.75L + 0.75(Lr or S or R) E = Earthquake Load

Analysis:

Material Property:

Wood : Bayok

Grade = 63.00 % stress

Fb= 9.94 Mpa

Fc= 5.78 Mpa

Fv = 0.95 Mpa

Es = 3.94 Gpa

G = 0.44 Relative Density

Microsoft Excel 2010 and Graphical Rapid Analysis of Structures Program (GRASP) were used to anlyze basic and complex structures such as Purlins, Trusses, Beams and Girders, Columns, Trusses. (See Design Aids)

Design Aids

A. Analysis by The Coefficient Method

A.1 One way slabs & Continuous Beams (ACI CODE Section 408.4.3)

Analysis of Purlins: Pitch = h/L tan𝜃 =

ℎ

𝐿/2

Wind Load: Method 1: Ps = λKztIwPs9 Method 2: qz = 0.0000473KzKztKdIwV

2

Simplified ASCE: 𝜃 > 10o

Windward: Pn = P(1.3sin𝜃 − 0.5) Leeward: Pn = -0.6P (Suction) Duchemins Formula: Pn =

2𝑃 𝑠𝑖𝑛𝜃

1+𝑠𝑖𝑛2𝜃

(See Appendix A) Bending Stress: Mn =

𝑊𝑛𝐿𝑥2

8 ; fn =

6𝑀𝑛

𝑏𝑑2

Mt = 𝑊𝑡𝐿𝑦2

8 ; ft =

6𝑀𝑡

𝑏2𝑑

fb = fn + ft < Allowable bending stress (Fb) Shearing Stress: vn =

3𝑉𝑛

2𝑏𝑑

vt =

3𝑉𝑡

2𝑏𝑑

fv = 𝑣𝑛 2 + 𝑣𝑡 2 < Allowable shearing stress (Fv) Deflections: LiveLoad: yall =

𝐿

360 > yact

DeadLoad + LiveLoad: yall =

𝐿

240 > yact

Analysis of Trusses: Assumptions: Tenion Stress: ft =

𝑃

𝐴𝑛< F′t ; F't = CDCMCTCFCiFt

Compression Stress: fc =

𝑃

𝐴𝑔 < 𝐹′𝑐

Slenderness Factor Adjustments:

Where: Ps - Horizontal pressure (Windward & Leeward). λ - Adjustment factor for building height and exposure. Kz - Velocity exposure coefficient. Kzt - Topographic factor. Kd - Wind directionality factor. Iw - Importance factor. Ps9 - Simplified design wind pressure for exposure B @ h = 9, Iw = 1 (NSCP, fig.207-3). qz - Velocity pressure @ height z. V - Wind velocity (Kph) Pn - Wind pressure perpendicular to surface. Mn - Normal Moment. Mt - Tangential Moment. fn - Normal Stress. ft - Tangential Stress. v - Shearing stress. lu - Unsupported length. Ag - Gross Area. Ft - Allowable tensile stress. Fc - Allowable compression stress. Fb - Allowable bending stress. Fv - Allowable shearing strress. CD - Load duration Factor. CM - Wet service factor (CM=1,for dry safe factor). CT - Temperature factor (CT=1,for normal temp.). CF - Size factor(CF=1,for sawn lumber). Ci - insicing factor(Ci=0.8,for incised;Ci=1,for not incised). Cs - Slenderness factor. Ck - Support factor.

Analysis of Beams: Bending Stress: fb =

6𝑀

𝑏𝑑2 < F'b

Slenderness Factor Adjustments: (See Table 7 for value of le) Size Factor Adjusment: d > 300mm CF =

300

𝑑

1

9 ; d =depth of Beam in mm

fb = fb*CF Shearing Stress: 𝑓𝑣 =

3𝑉

2𝑏𝑑 < 𝐹𝑣

If notched beams: 𝑓𝑣 =

3𝑉

2𝑏𝑑′

𝑑

𝑑′ < 𝐹𝑣

Deflections:

LiveLoad: yall = 𝐿

360 > yact

DeadLoad + LiveLoad: yall = 𝐿

240 > yact

Analysis for Columns: Compression Stress:

fc = 𝑃

𝐴𝑔 < 𝐹′𝑐

Slenderness Factor Adjustments: (See Table 6 for value of k)

Table 6. Buckling Factors Ke:

Table 7. Effective Length of Beams

Klu = le

Table 8. Load duration factor

NSCP 2010: Wind Loads

Guidelines for Tile Toilet & Bathroom

Subfloor, Floor Joist, and Plumbing pipes

Waterproofing of wall and floor

Installation of Tiles

* You can use alternative material in waterproofing for economy.

Soure: www.google.com

1.) Install Plyboard (3/4"-1") and plumbing pipes.

2.) Lay down tar paper or plastic on the floor to preserved the wood floor from sucking to the moisture of the mortar berfore it dries. It is also important to nail down

3.) Mix mortar (Portland cement) and apply about 2" thick properly from the walls to the drain with fair amount of slope so that water flows out into the drain without making puddles. Let this dry for about 24 hours.

4.) Put rubber membrane (CPE) to make waterproof seal. Install cement board in wall properly with 1/4" clearance between the cement board and the membrane. Then repeat step 3.

5.) Seal or tape the cement board joints in wall with special adhesive netting. Then put one layer of cement board compound and let it dry for about 24 hours. Install the tiles using available adhesive cement or mortar properly from walls to the drain.

Factored Load used:

Wu = 1.2DL + 1.6LL

Ultimate Shear & Moments:

A. Analysis by The Coefficient Method

A.1 One way slabs & Continuous Beams (ACI CODE Section 408.4.3)

A.2 Two way slabs (Design of Concrete Structures,12 ed., Arthur H. Nilson)

Where:

Ca, Cb = Tabulated Moment Coefficients

wu = uniform factored load

La = Length of clear span in Short direction

Lb = Length of clear span in Long direction

Ma = Ca wu

Mb = Cb wu Lb2

Va = Ca wu

Vb = Cb wu

Positive Moment: End Spans Discontinuous end unrestrained wuln

2/11 Discontinuous end InteGral with Support wuln

2/14 Interior Spans Negative Moment: at exterior face of first interior support Two spans wuln

2/9 More than two spans wuln

2/10 at other faces of interior supports wuln

2/11 at face of all supports for slabs with spans not exceeding 3 meters; and stiffness to beam stiffness exceeds eight at each end of the span wuln

2/12 at interior face of exterior supports for members built integrally with supports: where support is a spandrel beam wuln

2/24 where support is a column wuln

2/16 Shear: at face of first interior supports 1.15wuln/2 at face of all other supports wuln/2 *ln - clear span

Truss @ Painitan Section,Palao Market,Iligan City Date Prepared:

A. Design of Purlins Checked By:

Rating:

Material Property: Model: Truss - T1

Wood = Bayok


Fb= 9.94 Mpa

Fc= 5.78 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

Assume Section: Spacing (s1) = 1.20 m

b = 75 mm Spacing (s2) = 1.12 m

d = 200 mm height (h) = 3.00 m

Length (L/2) = 8.00 m (consider half-span)

20.56 degress

Service Loads:

Dead loads: KN/m3

KN/m2

KN/m

Weight of Purlin 4.32 0.06

Asphalt Shingles 0.10 0.11

DLtotal 0.18

Live Loads: KN/m3

KN/m2

KN/m

Roof Slope : 20.56 degrees 0.75 0.84

LLtotal 0.84

Wind Loads: Method 2

Zone: 2 Wind Pressure :

V = 200 Kph qz = 0.0000473kzkztkdV2Iw = 1.892 Kpa

Kz = 1.00

Kzt = 1.00 Load Windward Leeward

Kd = 1.00 Pn -0.082 0.049 KN/m2

Iw = 1.00 Wn -0.099 0.059 KN/m

Loading:

Load Combinations Normal (Wn) Tangential (Wt) Condition

1. D 0.166 0.062 ---

2. D + Lr 0.955 0.358 governs

3. D + 0.75Lr + 0.75 W 0.684 0.210 ---

Sectiion:

Along X

Wn = 0.95 KN/m

L = 4.00 m

Lx = 4.00 m

Along Y

Wt = 0.36 KN/m

L = 4.00 m

Ly = 4.00 m

Purlins - Roof Framing Plan

s2

L

h

𝜃 = tan−1 ℎ

𝐿 =

W Dr , Lr

𝑊𝑝 = ɣ𝑏ℎ

*𝑃𝑛 = 𝑃(1.3𝑠𝑖𝑛𝛳 − 0.5) , Windward

*𝑃𝑛 = −0.6𝑃 , Leeward

𝑊𝑚𝑎𝑡′𝑙 = 𝑢𝑛𝑖𝑡 load x s2

𝑊𝑙 = 𝑢𝑛𝑖𝑡 load x s2

Design Loads:

Mn = WnLx2/8 = 1.367 KN-m

Mt = WnLy2/8 = 0.420 KN-m

Check for Bending Stress:

fn = 6Mn/bd2

= 2.734 Mpa

ft = 6Mt/b2d = 2.242 Mpa

fb = fn + ft = 4.976 Mpa

fb < Fb, Safe!

Check for Shearing Stress:

Vn = WnLx/2 = 1.91 KN

Vt = WtLx/2 = 0.72 KN

vn = 1.5Vn/bd = 0.19 Mpa

vt = 1.5Vt/bd = 0.07 Mpa

0.20 Mpa

fv < Fv, Safe!

Check for Deflection:

yall = L / 240 = 16.67 mm

yact = 5WnL4/384EI = 16.16 mm

yact < yall, Safe!

Use:

Purlins: 75 x 200 mm (Bayok lumber)

* Adapt the size of member to other purlins for aesthetic design.

𝑓𝑣 = 𝑣𝑛2 + 𝑣𝑡

2 =


B. Design of Truss Checked By:

Rating:

Material Property: Model: Truss - T1

Wood = Bayok


Fb= 9.94 Mpa

Fc= 5.78 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

Assume Section:

For wood :

Top Chord : 2 pccs

b = 25 mm

d = 200 mm Spacing (s) = 1.20 m

Bottom Chord : 2 pcs height (h) = 3.00 m

b = 25 mm Length (L/2) = 8.00 m

d = 200 mm 20.56 degress

Web : 1 pcs

b = 25 mm

d = 150 mm

For Steel :

ws = 77.3 KN3

∅s = 16 mm

Fy = 248 Mpa

Service Loads:

Dead loads: KN/m3

KN/m2

KN/m

Weight of Truss 4.32 0.1025

(Assumed)

DLtotal 0.10

Live Loads: KN/m3

KN/m2

KN/m

Roof Slope : 20.56 degrees ---- ---- 0.00

LLtotal 0.00

Design Loads:

w = DL = 0.10 KN/m

Loads from Purlins (Rp) = 1.5Vncosθ = 2.68 KN

T1 - Roof Framing Plan

𝜃 = tan−1 ℎ

𝐿 =

L

h

RP D

𝑊𝑇 = ɣ𝑏ℎ

A

B

C

D

E

F

J H

I

An

aly

sis:

U

sin

g G

rap

hic

al R

apid

An

alysi

s of

Str

uct

ure

s P

rogra

m (

GR

AS

P)

Rea

ctio

n @

Su

pp

ort

s

Mem

ber

Axia

l F

orc

e :

Design Loads:

Description Member L (mm) Area (mm2) Forces (KN)

Top Chord AB 2670.00 10000 35.4

Bottom Chord JH 2000.00 10000 6.1

Diagonal Web FH 3010.00 3750 7.5

Vertical Web FH 3000.00 201.06 8.6

* Choose Maximum Axial Load (GRASP)

Tensile Stress:

Compressive Stress:


KcE = 0.300 visually graded

KcE = 0.418 machine stress graded sawn lumber

Design for Truss Member:

Top Chord:

Compressive Stress: Tensile Stress:

k= 1.00 Hinge support CD = 1.25

klu/d = 6.68 Cs ≤ 50, Ok! CM = 1.00

CT= 1.00

17.52 CF= 1.00

Ci= 1.00

case 1: klu/d ≤ 11 F't = 12.43 Mpa

ft = 1.77 Mpa

F'c = 5.78 Mpa ft < Ft, Safe!

fc = 1.77 Mpa

fc < F'c, Safe!

Bottom Chord:



klu/d = 40.00 Cs ≤ 50, Ok! CM = 1.00

CT= 1.00

17.52 CF= 1.00

Ci= 1.00

case 3: klu/d ≥ K F't = 12.43 Mpa

ft = 0.31 Mpa


fc = 0.31 Mpa

fc < F'c, Safe!

Diagonal Web:



klu/d = 20.07 Cs ≤ 50, Ok! CM = 1.00

CT= 1.00

17.52 CF= 1.00

Ci= 1.00

case 3: klu/d ≥ K F't = 12.43 Mpa

ft = 1.00 Mpa


fc = 2.00 Mpa

fc < F'c, Safe!

𝑓𝑡 =𝑃

0.6𝐴𝑛 ≤ F′t ; F't = CDCMCTCFCiFt ; Ft = Fb

𝑓𝑐 =𝑃

𝐴𝑔 ≤ F'c

case 1: 𝑘𝑙𝑢

𝑑 ≤ 11 ; F'c = Fc

case 2: 11 ≤ 𝑘𝑙𝑢

𝑑 ≤ 𝐾 ; K = 0.671

𝐸

𝐹𝑐 ; F'c = Fc 1 −

1

3

𝑘𝑙𝑢

𝑑

𝐾

4


𝑑 ≥ 𝐾 ; F'c =

𝐾𝑐𝐸 𝐸

𝑘𝑙𝑢

𝑑

2

K = 0.671𝐸

𝐹𝑐 =

K = 0.671𝐸

𝐹𝑐 =

K = 0.671𝐸

𝐹𝑐 =

K = 0.671𝐸

𝐹𝑐 =

K = 0.671𝐸

𝐹𝑐 =

Diagonal Web: Steel

Ft = 0.60Fy

Ft = 148.8 Mpa

ft = F/As

ft = 42.77 Mpa

ft < Ft, Safe!

Use:

Top Chord: 2 - 25 x 200 mm (Bayok lumber)

Bottom Chord: 2 - 25 x 200 mm (Bayok lumber)

Diagonal Web: 1 - 25 x 150 mm (Bayok lumber)

Vertical Web: 16 mm∅ Plain Steel bars

16 mm∅ Plain Steel bars

2 - 25 x 200 mm (Bayok lumber)



Note: Adapt all sizes of member to other types of truss for aesthetic design.

A

B

C


C. Design of Truss Joints Checked By:

Rating:

Material Property: Model: Anchor Bolts Connections

Wood : Bayok


Fb= 9.94 Mpa

P= 5.78 Mpa

Q = 1.03 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

Steel :

Ft = 148.80 Mpa

∅s = 16 mm

T = 8.6 KN

θ = 20.56 degrees

Check if Steel is adequate :

ft = T/As

ft = 42.77 Mpa

ft < Ft, Safe!

Compressive Stress @ section AB ( r ) :

Figure:

r = 1.62 Mpa

Size of Washer :

Required Net Area : Figure:

An = T/r

An = 5322.34 mm2

Diameter of hole :

∅hole = ∅bolt + 2mm

∅hole = 18 mm

Gross Area :

Ag = An + Ahole

Ag = 5576.80 mm2

X2 = 5576.80 mm

2

X = 74.68 mm

say : 80 mm

Joint A - Truss T1

A B

Design

𝑟 = 𝑃𝑠𝑖𝑛2𝜃 +Q𝑐𝑜𝑠2𝜃 , Jacoby's Formula

X Dn

X

T

r r

A B

Thickness of Washer :

Dn = 1.5∅ + 3 Figure:

Dn = 27 mm

T1 = T2 = T/2

T1 = T2 = 4.3 KN

x1 = Dn/4

x1 = 7 mm

x2 = ∅hole

x2 = 18 mm

M = T2 (x2) - T1 (x1)

M = 48.375 KN-m

ft = 6M/bd2

, Ft =ft

ft = 6M/bt2

t = 5.61 mm b = X - ∅hole

say : 6 mm b = 62 mm

Use:

Steel :

Fbs = 124.00 Mpa

∅s = 22 mm

Washer :

t = 6 mm

27 mm 80 mm

80 mm

T

A B

T1

T2

x1

x2

∅hole A B

∅hole


D. Design of Truss Joints Checked By:

Rating:

Material Property: Model: Splicing Connections @ Bottom Chord

Wood : Bayok


Fb = 9.94 Mpa

P = 5.78 Mpa

Q = 1.03 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

F = 6.10 KN

Bolt :

∅b = 12 mm

Group = I

Pb = 8.38 KNQb = 4.70 KN

Figure : Main member 50 200 mm

3.05 KN

6 KN

3.05 KN

Side Plate 50 200 mm

No. of Bolts :

n = F/Pb

n = 0.73 pcs

Say : 12 pcs

No. of rows : 2

Check for Tension :

ft = F/An ,Fb = Ft Ag = bh (main member)

ft = 0.71 Mpa Ag = 10000 mm2

ft < Ft, Safe! Ahole = ∅bolt + 2mm

Ahole = 14 mm2

An = Ag - ∑Ahole

An = 8600 mm2

Check for Bolt Shear :

Pv = nbolt x Pb

Pv = 100.56 KN

Pv > F, Safe!

Check for Bearing of bolt to main member :

fb = F/(Ap*n) Ap = ∅bolt x t , t = bmain member

fb = 0.85 Mpa Ap = 600 mm2

fb < P, Safe!

Joint B - Truss T1

Design

F F

x

x

b h

b h

Use :

∅bolt = 12 mm

nbolts = 12 pcs

12-12mm ∅ Bolts

F F


D. Design of Truss Joints Checked By:

Rating:

Material Property: Model: Splicing Connections @ Top Chord

Wood : Bayok


Fb = 9.94 Mpa

P = 5.78 Mpa

Q = 1.03 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

F = 35.40 KN

Bolt :

∅b = 16 mm

Group = I

Pb = 10.80 KNQb = 5.23 KN


17.7 KN

35 KN

17.7 KN


No. of Bolts :

n = F/Pb

n = 3.28 pcs

Say : 8 pcs

No. of rows : 2

Check for Tension :

ft = F/An ,Fb = Ft Ag = bh (main member)

ft = 4.32 Mpa Ag = 10000 mm2

ft < Ft, Safe! Ahole = ∅bolt + 2mm

Ahole = 18 mm2

An = Ag - ∑Ahole

An = 8200 mm2


Pv = nbolt x Pb

Pv = 86.4 KN

Pv > F, Safe!



fb = 5.53 Mpa Ap = 800 mm2

fb < P, Safe!

Joint C - Truss T1

Design

F F

x

x

b h

b h

Use :

∅bolt = 16 mm

nbolts = 8 pcs

8-16mm ∅ Bolts

F F


E. Design of Truss Joints Checked By:

Rating:

Material Property: Model: Notching Connections

Wood : Bayok


Fb = 9.94 Mpa

P = 5.78 Mpa

Q = 1.03 Mpa

Fv = 0.95 Mpa

Ew = 3.94 Gpa

Gw = 0.44

F = 7.5 KN

θ = 50 degrees

Web member 50 150 mm

8 KN

50 mm

Bot. Chord 50 200 mm

Check if dap is adequate :

Compressive Stress perpendicular to AB : Actual comp. Stress @ AB:

AC = h/sinϴ F1 = Fsinβ

AC = 150/sin(50) F1 = 4.26 KN

AC = 195.81 mm

r = AC/2 fAB = F1/A1

r = 97.91 mm fAB = 0.45 Mpa

α = 0.5asin(dap/r) Check with allowable comp. stress (r ):

α = 0.5asin(50/97.91)

α = 15.36 degrees φ = 90 - α

β = 34.64 degrees φ = 74.64 degrees

AB = cosα(AC)

AB = 188.82 mm

A1 = AB*b r = 1.093 Mpa

A1 = 9441.06 mm2 r > fab, Safe!

Joint B - Truss T1

F

Design

x

b h

x

b' h

𝑟 =𝑃𝑄

𝑃𝑠𝑖𝑛2𝜑 + Q𝑐𝑜𝑠2φ

𝜃

dap = 𝑟

𝑟

Compressive Stress perpendicular to BC :

BC = sinα(AC)

BC = 51.851 mm

A2 = BC*b'

A2 = 2592.55 mm2

Actual comp. Stress @ AB: Check with allowable comp. stress (r ):

F2 = Fcosβ α = 15.36 degrees

F2 = 6.17 KN

fBC= F2/A2

fBC = 2.38 Mpa r = 4.368 Mpa

r > fab, Safe!

Percent correction :

%cor. = fBC/r

%cor. = 54.491

corrected dap = 27.25 mm

Try: dap = 50 mm

Use :

Web member 50 x 150 mm

8 KN

50 mm

Bot. Chord 50 x 200 mm

𝑟 =𝑃𝑄

𝑃𝑠𝑖𝑛2𝜑 + Q𝑐𝑜𝑠2φ

𝜃

dap =

Service Loads:

Dead loads: KN/m3

KN/m2

KN/m

Weight of Truss 4.32 0.0000

Ceiling 1.0000 (Assumed)

DLtotal 1.00

Live Loads: KN/m3

KN/m2

KN/m

Roof Slope : degrees ---- ---- 0.00

LLtotal 0.00

Design Loads:

w = DL = 1.00 KN/m

0.83 KN

Analysis: Using Graphical Rapid Analysis of Structures Program (GRASP)

Reaction @ Supports:

Member Axial Forces:

Design Loads:

Description Member L (mm) Area (mm2)Inertia (mm4)Forces (KN)

Top Chord AB 2150.00 0 0 9.3

Bottom Chord AC 1400.00 0 0 2.7

Web FI 2350.00 0 0 4.7

* Choose Maximum Axial Load (GRASP)

Tensile Stress:

𝑊𝑇 = ɣ𝑏ℎ

Loads from Purlins (Rp) = 1.5Vncosθ =

Compressive Stress:


KcE = 0.300 visually graded

KcE = 0.418 machine stress graded sawn lumber

Design for Truss Member:

Top Chord:


k= 1.00 Hinge support CD = 1.25 (Refer Table 8.)

klu/d = #DIV/0! #DIV/0! CM = 1.00

CT= 1.00

17.52 CF= 1.00 (See Design Aids)

Ci= 1.00

#DIV/0! F't = 12.43 Mpa

ft = #DIV/0! Mpa

F'c = #DIV/0! Mpa #DIV/0!

fc = #DIV/0! Mpa

#DIV/0!

Bottom Chord:



klu/d = #DIV/0! #DIV/0! CM = 1.00

CT= 1.00


Ci= 1.00

#DIV/0! F't = 12.43 Mpa

ft = #DIV/0! Mpa


fc = #DIV/0! Mpa

#DIV/0!

𝑓𝑡 =𝑃

0.6𝐴𝑛 ≤ F′t ; F't = CDCMCTCFCiFt ; Ft = Fb

𝑓𝑐 =𝑃

𝐴𝑔 ≤ F'c


𝑑 ≤ 11 ; F'c = Fc

case 2: 11 ≤ 𝑘𝑙𝑢

𝑑 ≤ 𝐾 ; K = 0.671

𝐸

𝐹𝑐 ; F'c = Fc 1 −

1

3

𝑘𝑙𝑢

𝑑

𝐾

4


𝑑 ≥ 𝐾 ; F'c =

𝐾𝑐𝐸 𝐸

𝑘𝑙𝑢

𝑑

2

K = 0.671𝐸

𝐹𝑐 =

K = 0.671𝐸

𝐹𝑐 = K = 0.671

𝐸

𝐹𝑐 =

Web:



klu/d = #DIV/0! #DIV/0! CM = 1.00

CT= 1.00


Ci= 1.00

#DIV/0! F't = 12.43 Mpa

ft = #DIV/0! Mpa


fc = #DIV/0! Mpa

#DIV/0!

Use:

Top Chord: - x mm (Bayok lumber)

Bottom Chord: - x mm (Bayok lumber)

Web: - x mm (Bayok lumber)

* Adapt all sizes of member to other types of truss for aesthetic design.


#REF! KN

#REF! KN

#REF! KN


No. of Bolts :

n = F/Pb

n = #REF! pcs

K = 0.671𝐸

𝐹𝑐 = K = 0.671

𝐸

𝐹𝑐 =

Design

x

x

b d

b d

Say : 12 pcs

No. of rows : 3

Check for Tension :

ft = F/An ,Fb = Ft Ag = bd (main member)

ft = #REF! Mpa Ag = 25000 mm2

#REF! Ahole = ∅bolt + 2mm

Ahole = 24 mm

An = Ag - ∑Ahole

An = 17800 mm2


Pv = nbolt x Pb

Pv = 261.6 KN

#REF!



fb = #REF! Mpa Ap = 2200 mm

#REF!

Use :

∅bolt = 22 mm

nbolts = 12 pcs

Project Title : Long Span Truss Design Date Prepared : Sept - 29 - 2012

Section : Design of Notch Connection Date Submitted :

Subsection : Design of Notch of Connection A Rating :

Structure Data

Web Member Detail:

b = mm

h = mm

ϴ = ◦

Chord Member Detail:

bc = mm

hc = mm

NOTE : ϴ is the angle of inclination of the web

member with respect to the horizontal

Member Properties :

Fc|| = MPa

FcL = MPa

Load

P = kN

Assume depth of Dap :

hd = mm

Analysis

T = 30 KN

θ = 30 degrees

Bayok


ft = T/As

ft = 543.93 Mpa

ft > Fbs, Not Safe!


Figure:

r = 2.22 Mpa

Size of Washer :


An = T/r

An = 13528.75 mm2


Diamete of hole :

∅hole = 10 mm

Gross Area :

Ag = An + Ahole

Ag = 13613.37 mm2

X2 = 13613.37 mm

2

X = 116.68 mm

say : 120 mm


Dn = 1.5∅ + 3 Figure:

Dn = 15.57 mm

T1 = T2 = T/2

T1 = T2 = 15 KN

x1 = Dn/4

x1 = 4 mm

x2 = ∅hole

x2 = 10 mm

M = T2 (x2) - T1 (x1)

M = 97.3125 KN-m

fbs = 6M/bd2

, Fbs =fbs

fbs = 6M/bt2

t = 21.07 mm b = X - ∅hole

say : 12 mm b = 110 mm

Use:

Steel :

Fbs = 124.00 Mpa

∅s = 22 mm

Washer :

∅hol

15.57 mm 120 mm

120 mm


ft = T/As

ft = 80.37 Mpa

ft > Fbs, Not Safe!


Figure:

r = #REF! Mpa

Size of Washer :


An = T/r

An = #REF! mm2

Diamete of hole :

∅hole = 24 mm

Gross Area :


Ag = An + Ahole

Ag = #REF! mm2

X2 = #REF! mm

2

X = #REF! mm

say : 120 mm


Dn = 1.5∅ + 3 Figure:

Dn = 35.7 mm

T1 = T2 = T/2

T1 = T2 = 15 KN

x1 = Dn/4

x1 = 9 mm

x2 = ∅hole

x2 = 24 mm

M = T2 (x2) - T1 (x1)

M = 223.125 KN-m

fbs = 6M/bd2

, Fbs =fbs

fbs = 6M/bt2

t = 25.15 mm b = X - ∅hole

say : 12 mm b = 96 mm

Use:

Sept - 29 - 2012

Steel :

Fbs = 124.00 Mpa

∅s = 22 mm

Washer :

35.7 mm 120 mm

∅hol

Group = I

120 mm

Bolt :

∅b = 22

Pb = 21.80

Qb = 7.21

Material Property:

Wood : Bayok


Fb= 9.94 Mpa

Fc= 5.78 Mpa

Fv = 0.95 Mpa

Es = 3.94 Gpa

G = 0.44 Relative Density

Steel:

Fy = 248.00 Mpa

∅s = 16 mm

Bolts :

Group = I

∅b = 12 mm ∅b = 16 mm

Pb = 8.38 KN Pb = 10.80 KN

Qb = 4.70 KN Qb = 5.23 KN

1

Loads, Building data and Material Properties Loading: NSCP 2010...

Documents

Transcript of Loads, Building data and Material Properties Loading: NSCP 2010...