CARMEL B. SABADO CE-164 PROF.Allan E. MilanoBSCE-5 Timber Excel Program*note:the boxes in yellow should be inputed by the designer,while blue ones are computed by the program.=)

DESIGN OF PRATT TRUSS

DATA:Type of wood: pahutanBending and Tension(Fb) 13.80 MPa

Shear(Fv) 1.34 MPa

Compression(Fc) 8.14 MPa

Modulus of Elasticity(E) 9100.00 MPa

Relative Density(G) 0.55Specific Gravity 5.40

LOADINGS:Wind Pressure 0.96 kPa

Minimum Roof Live Load 0.80 kPa

GI roofing 0.15 kPa

Residential Live Load 2.00 kPa

SPACING:Purlins 0.40 mTruss 2.75 mFloor Joist 0.40 m

DESIGN OF PURLINS

DATA: TRIAL DIMENSION:Span 12.00 m 150 x 200 mm

Height 2.00 m I= 1.00E+08

9.46

LOADINGS:Live load 0.32 Kn/mRoofing 0.06 Kn/mPurlin weight 0.16 Kn/m

0.54 Kn/m

Wn2 Wnt

kN/m3 WDL+LL

mm4

Theta, q;

WDL+LL

Load Combinations:Condition 1: DL + LL

= 0.53 kN/m governs!!Condition 2: DL + LL + WL

= 0.23 kN/m LOAD COMBINATION:Windward:

Pn = 1.3(sinq - 0.5)P -0.27

Leeward:

MOMENTS: Pn = -0.5P -0.48

9.6211285 Kn-m(WW) -0.11

0.178169 Kn-m (LW) -0.192

0.53

0.09

SHEAR:0.42

3.2070428 Kn 0.09

0.089084523 Kn

CHECK FOR BENDING:

= = 9.86 Kn-m < 13.80it is safe!=)

CHECK FOR SHEAR;

= = 0.16 Kn < 1.34it is safe!=)

CHECK FOR DEFLECTION:

= 0.44 mm

= 7.64 mm

it is safe!=) Therefore use

DESIGN OF TRUSS

WDL+LL

WDL+LL+WL

Mn = Mx = 1/8(WnLx2)

Wn1 = Pn(Spacing) Mt = My = 1/12(WnLy2) Wn1 = Pn(Spacing)

Wn2 = WDL+LL(cosq) Wnt = WDL+LL(sinq)

WN = Wn1 + Wn2

Vx = (1/2)WnLx Wt = Wnt

Vy = (1/2)WnLy

To be safe, Fb > Fact

To be safe, Fv > Fvact

****To be safe, Yall > Yact

Yact = (5/384)(WLn4/EI)Yallow = L/360

6Mx

bh2+6My

b2h

3Vx2bh

+3Vy2bh

TRIAL DIMENSION:75 x 200 mm

I= 5.00E+07 1.512

LOAD CARRIED BY THE TRUSS: 9.194q

Loadings:GI roofing = 5.0182791 Kn 9.069

Wt. of Purlins = 0.44517 KnMin. Roof LL = 26.764155 Kntotal = 32.227604 Kn

15.84

Weight of truss:Overall Length of Truss = 67.99 m q 16.05849

Weight of Truss = 5.5031106 KnTOTAL 37.73071 Kn

Windwardwind load = -9.19449308 Kn/mfx = 0.30231307 Knfy = 1.81387842 Kn

Leewardwind load = -16.0584931 Kn/mfx = 0.528 Knfy = 3.168 Kn

Load carried by the ceiling:Ceiling Load = 0.012 Kn/m

18.865 KN

Forces Due to DL + LL9.433 Kn 9.43268 Kn

mm4

M

9.433 Kn

9.433 Kn

9.433 Kn

ceiling load

3.0 3.0 3.0 m 3 m 3.000 m 3.0000

24 m

1.3541 KN

Forces Due to Wind Load 0.83 KN1.814 Kn 3.168 Kn

0.3023 KN 0.528

1.814 KN

0.3023 KN

1.814 Kn0.3023 KN

1.814 Kn0.3023 KN

A B C D FE

N

M

L

K

J

A B C D FE

N

M

L

K

J

3.0 3.0 3.0 m 3 m 3.000 m 3.0000

24 m

NOTE: the reactions and axial bar forces are solved using the GRASP SOFTWARE. The calculations and the drawings are not shown here.instead, a summary in table form is provided.

Summary of Bar Forces:

Top Chords Length DL + LL WL DL + LL + WLAJ 3.04 56.90 11.00 67.90JK 3.04 56.90 10.70 67.60KL 3.04 -57.30 -8.800 -66.10LM 3.04 -76.40 -11.70 -88.10MN 3.04 -76.40 -13.70 -90.10NO 3.04 -57.30 -10.30 -67.6OP 3.04 56.90 19.80 76.7

PI 3.04 56.90 19.30 76.20Bottom ChordS

AB 3.00 -56.20 -11.2 -44.30BC 3.00 56.60 11.90 71.10CD 3.00 75.40 14.50 88.40DE 3.00 70.70 13.00 83.70EF 3.00 70.70 13.00 83.70FG 3.00 75.40 15.60 91.00GH 3.00 56.60 11.7 68.30

HI 3.00 -56.20 -18.50 -74.70Verticals

BJ 0.50 -9.30 -1.900 -11.200KC 1.00 9.40 1.300 10.700LD 1.50 -3.10 -1.0 -4.10ME 2.00 0.002 0.000 0.002NF 1.50 -3.10 -1.800 -4.900OG 1.00 9.40 2.000 11.400HP 0.50 -9.30 -3.100 -12.400

DiagonalsBK 3.16 -118.80 -19.90 -138.70CL 3.35 -21.10 -2.900 -24.00DM 3.61 5.60 1.800 7.40MF 3.61 5.60 3.200 8.80NG 3.35 -21.10 -4.400 -25.50

OH 3.16 -118.80 -31.80 -150.60

Design of Truss MembersStresses Length

Top Chord -88.110 3.041

Bottom chord 91.000 / -74.700 3.000

Vertical 11.400 / -12.400 0.500

Diagonal 8.800 / -150.600 3.160

DESIGN OF Top Chord

TRIAL DIMENSION:150 x 200 mm

I= 1.00E+08

P= -88.110 KnL= 3041.38 mm

L/d = 20.275875

= 21.4413806 since L/d<K and L/d>11 it is short column

To be safe:Fc >= fc

Fc = 6.07

fc = P/A = 2.94 < 6.07 it is safe!=)

Therefore use 150 x 200 mm for BOTTOM CHORD

DESIGN OF Bottom Chord

TRIAL DIMENSION:

mm4

K=( π2 )( E6 fc ). 5

Fc= π2E

36( Ld )2

100 x 200 mm

I= 6.67E+07

P= 91.000 KnL= 3000.00 mm

L/d = 30

= 21.4413806 since L/d>K and L/d>11 it is long column

To be safe:Fc >= fc

Fc = 8.14

fc = P/A = 4.55 < 8 it is safe!=)

Therefore use 100 x 200 mm for BOTTOM CHORD

Check for Stress Reversals: To be safe:

>=

= 13.80 MPa

= 6.23 < 13.80 it is safe!=)

Since Fb > Ft, Use 100 x 200 mm for VERTICALS

DESIGN OF Verticals

TRIAL DIMENSION:75 x 200 mm

I= 5.00E+07

mm4

Fb ft

Fb

mm4

K=( π2 )( E6 fc ). 5

Fc= π2E

36( Ld )2

f t=P

(3 /5 ) Ag

P= 11.400 / -12.400 KnL= 2000.00 mm

L/d = 26.666667

= 21.4413806 since L/d>K and L/d>11 it is long column

To be safe:Fc >= fc

Fc = 3.51

fc = P/A = 0.76 < 3.51 it is safe!=)

Therefore use 75 x 200 mm for BOTTOM CHORD

Check for Stress Reversals: To be safe:

>=

= 13.80 MPa

= 1.38 < 13.80 it is safe!=)

Since Fb > Ft, Use 75 x 200 mm for VERTICALS

DESIGN OF Diagonals

TRIAL DIMENSION:150 x 200 mm

I= 1.00E+08

P= 8.800 / -150.600 KnL= 3160.00 mm

L/d = 21.066667

Fb ft

Fb

mm4

Fc= π2E

36( Ld )2

f t=P

(3 /5 ) Ag

K=( π2 )( E6 fc ). 5

= 21.4413806 since L/d<K and L/d>11 it is short column

To be safe:Fc >= fc

Fc = 5.62E+00

fc = P/A = 0.29 < 5.62.E+00 it is safe!=)

Therefore use 150 x 200 mm for BOTTOM CHORD

Check for Stress Reversals: To be safe:

>=

= 13.80 MPa

= 8.37 < 13.80 it is safe!=)

Since Fb > Ft, Use 150 x 200 mm for VERTICALS

Fb ft

Fb

K=( π2 )( E6 fc ). 5

Fc= π2E

36( Ld )2

f t=P

(3 /5 ) Ag

12.17

2.00 m

12.00 m

-0.27 kN/m (+) Windward (-) Leeward

-0.48 kN/m

-0.11 kN/m

-0.192

0.53 kN/m

0.09 kN/m

0.42 kN/m

0.09 kN/m

13.80 Mpa

1.34 Mpa

150 X 200 mm thick purlins

16.05849

2.64

Kn

9.433 Kn

9.43 kn

9.43 kn

3.0000 m 3.0000 3.0000

Kn

0.528 KN3.17 Kn

0.528 KN

3.17 kn

0.528 KN

3.17 kn

0.528 KN

G H

I

P

O

G H I

P

O

3.0000 m 3.0000 3.0000

NOTE: the reactions and axial bar forces are solved using the GRASP SOFTWARE. The calculations and the drawings are not shown here.

it is short column

it is long column

it is long column

it is short column

CARMEL B. SABADO CE-164 PROF.Allan E. MilanoBSCE-5 PROJECT 2 OCT. 19, 2009

EXCEL SOLUTIONS

1.) Design the vertical member using steel bar and its square/rectangular washer if Fs=100 MPa for steel (assume hole diameter = bolt diameter ).

a.) Use steel bar Given: Fs 100 MPa

P = 11.4 kN

Solution :

Fs = P As = P = 11400 N = 114 mm2

As Fs 100 MPa

= = 12.0478 mm ø4 π

= 16

b.) Design WasherGiven: Fs = 100 MPa Fq = 2.07 MPa

Fb = 13.8 MPa Fv = 1.1 MPa

Fp = 8.14 MPa E = 9100 MPa

STEEL :

Fb = 100 MPa

Solution :

Assume hole diameter = steel diameter

16 mm ø P = 11.4 kN

0.00026 100 = 20.10619 kN

4

T = P + P = 11.4 + 20.1062 = 15.7531 kN

2 2

Design washer : Assume it to be square

Fq = T ; Aw = T = 15.7531

Aw Fq 2.07

Aw = 7610.19 ( Net Area )

Find b of washer :

Aw = Anet =

4

7610.19 = 256

As = π dr2 dr 120 mm2 ( 4 )

Therefore, use dr mm ø steel rod.

dh =

Ps = As Fs = π

mm2

b2 - π ( dh )2

b2 - π

4

b = 88.38 mm

75 mm2

Revise as rectangular washer:

Aw = Anet =

4

7610.19 = 256

4

L = 104.15 mm

SAY L = 110 mm

WASHER : 75 mm x 110 mm

Find thickness, t :

t = 6M ; Fb = Fs

M = T ( b - dn ) ; = 1.5(16) +3

8 dn = 27 mm

M = 15.7531 0.11 - 0.027

8

M = 163.438 N.mm

t = 6 163.438 ; t = 9.44 mm

0.11 100 SAY t = 10 mm

Therefore, use rectangular washer dimension:

110 x 75 x 10 mm

2.) Design the wooden splicing at member 6, using wooden side plates .

( both sides ) and 16 mm diameter boltsGiven : P = 91 kN at member 2 Group 2 Apitong

Member 6: 100 mm x 200 mm From Table : ( Double Shear )

L = 200 mm 4.359

d = 16 mm 3.069

100 mm

Since b > Lchord = 75 mm, use b =

bL - π ( dh )2

75 L - π

bFb

dn = 1.5 dh + 3

P// =

QL =

Solution :P/2

l P

P/2

a.) Find number of bolts :

= P/2

n = 91 2

4.359

n = 10.44 bolts = 12 bolts in 2 rows

b. ) Find thickness of wooden side plate :

100 mm

P

L

Check from Tension :

b.1 ) Fb = P/2

3/5 Ag

13.8 = 45.5

3/5 Ag

Ag = 5495.17 = L t ; L = 100

t = 5495.17

100

t = 54.95 mm

Say t = 55 mm

b.2 ) Fb = P/2

Anet

Anet = 45.5

13.8

Anet = 3297.10

Anet = ; = 2 ( cross-section )

; =

= 16 +

= 19 mm

3297.10 = 100 - 2 19 t

t = 53.18 mm

Say t = 55 mm

Use the larger value for thickness. Therefore, t = 55 mm

IF t = 55 mm

P// ( n )

mm2

mm2

( L - # holes [ øh ] ) t #holes

øh øb + 3

øh

øh

Anet = 100 - 2 19 55

Anet = 3410

= 45.5

3410

= 13.34 MPa < Fb = 13.8 MPa

DETAILS OF WOODEN SIDE PLATE :grain

edge s s From minimum requirements :

Use : g = 2.5 d if l/d = 2

s = 4 d g =

s = 4 16 l/d = 200 16 =

s = 64 mm o. c. g = 5 d

g = 5 16

g = 80 mm o. c.

Edge : e = 1.5 d ( // to grain )

e = 1.5 16

e = 24 mm ( edge )

End : e = 5 d ( tension ; hardwood )

e = 5 16

e = 80 mm ( end )

3.) Design the notching at joint A :Given : P = 67.9 kN ( C )

Fp = 8.14 MPa

Fq = 2.07 MPa

Solution :

θ = 6

θ = 9.46

mm2

fb

fb

s ≥ 4 d ( // to grain )

5 d if l/d ≥ 6

tan-1 2.4

θPy

P = 67.9 kN

Px

θ/2 = 4.73

Py = = 67.9 sin 4.73 = 5.60

Px = = 67.9 cos 4.73 = 67.67

For section BC :

= 4.73

= pq

= 8.14 2.07

8.14 4.73 + 2.07

= 7.98 MPa

Px = 7.98

= 67.67

75 7.98

Psin θ/2

Pcos θ/2

αBC

rBC

psin2αBC + qcos2αBC

sin2 cos2

rBC

Equate δBC = rBC

75 BCrequired

BCrequired

75

75

BC

AC

αBC

αAC

B A

C

= 113.05 mm

For section AC :

= 90 - 4.73 = 85.27

= pq

= 8.14 2.07

8.14 85.27 + 2.07

= 2.08 MPa

Py = 2.08

= 5.60

75 2.08

= 35.89 mm

Use the larger one :

Use BCrequired = 113.05 mm

Try from BC :

= 113.05 cos 4.73

= 112.67 mm

Try from AC :

= 35.89 cos 85.27

= 2.96 mm

Therefore, use = 70.48 mm

SAY = 113 mm

BCrequired

αAC 90 - θ/2 =

rAC

psin2αAC + qcos2αAC

sin2 cos2

rAC

Equate δAC = rAC

75 ACrequired

ACrequired

ACrequired

hrequired :

hrequired

hrequired

hrequired

hrequired

hrequired

hrequired

PROF.Allan E. Milano

OCT. 19, 2009

1.) Design the vertical member using steel bar and its square/rectangular washer

From Table : ( Double Shear )

kN/bolt

kN/bolt

mm

( cross-section )

3

OK!

end

g

P/2

end

12.5

0.5

3

kN

kN

4.73

85.27

DESIGN OF PEAK JOINT

Data needed for the design of peak joint

= -90.10 kN

= -24.000 kN

Top Chord member = 150mm x 200mm

Diagonal Chord member = 150mm x 200mm

Compression Parallel to Grain, p = 8.14 MPa

Compression Perpendicular to grain, q = 2.07 MPa

Sloving for diagonal stress, r:

= 21.8 ˚

= 40.23 ˚

= = (-90.1)(cos 21.8)

= = (-90.1)(sin 21.8)

= = (-24)(cos 30.96)

= = (-24)(sin 30.96)

P1

P2

θ1

θ2

P1x P cos θ1

P1y P sin θ1

P2x P cos θ2

P2y P sin θ2

3" x 8"

-90.1kN

-24.0kN

= AB / 200 AB

= BC /200 BC

= DE /200 DE

= EF / 200 EF

using Hankinson's formula, we have:

=

= 5.796 MPa

=

= 3.661 MPa

= = 2.135 MPa

At plane BC and EF:(actual stress)

Bearing Area:

= 141.29(200 - 28) = 24302.0598

= 121.03(200 - 28) = 20817.7143

A = 45119.77403

= = 101.98 kn

= = 1.130 Kpa

2.135 MPa ok!

= = 68.2

= = 49.77

sin θ1

cos θ1

sin θ2

cos θ2

rBC (8.14 x 2.07) / (8.14(sin2 21.8) + 2.07(cos2 21.8))

rEF p q / (p sin2θ + q cos2θ) = (8.14 x 2.07) / (8.14(sin2 30.96) + 2.07(cos2 30.96))

rBF rBC - rBF

Af

Af

mm2

Px P1x + P2x

factual Px / A

Since factual <

β1 90˚ - 11.31˚

β1 90˚ - 30.96˚

=

= 2.307 MPa

=

= 3.005 MPa

= = 5.312 MPa

At plane AB and DE:(actual stress)

Bearing Area:

= = 4884.24784

= = 4884.24784

A = 9768.49568

= = 48.961 kN

= = 5.012 kN

5.312 MPa ok!

DESIGN OF SIDE PLATE & NO. OF BOLTS

Dressed Dimension:

For bolt shear:

from NSCP table: P/bolt = 12.01 kN/bolt

using metal side plates:

= 1.25(12.01) = 15.0125

rAB p q / (p sin2θ + q cos2θ) = (8.14 x 2.07) / (8.14(sin2 78.69) + 2.07(cos2 78.69))

rDE p q / (p sin2θ + q cos2θ) = (8.14 x 2.07) / (8.14(sin2 78.69) + 2.07(cos2 78.69))

rAE rAB + rDE

A1 50*110 - (π/4)(282)

A2 50*110 - (π/4)(282)

mm2

Py P1y + P2y

factual Py / A

Since factual = 12.992 >

3 5/8" x 7 1/2"

Psp

PP

no. of bolts, n

n = = 97.249 / 15.0125

= 6.82325 say 7 bolts

Check for bending:

= =

= 1403.223

P = = (25.90)(1403.2)

= 20487.0558 kN

Since P > 105.0875 kN ok!

Thickness of Side Plate:

for each plate:

P / 2 = 105.088 / 2 = 52.5437 kN

σ = P / A 124

124 MPa =

t = 3.07058 mm say 3

P / Psp

Ab (3 5/8" x 7 1/2") - 2 (3/4" x 3 5/8")

mm2

Fb Ab

allowable σ =

52.57 x 103 / (138 x t)

7 1/2

3 5/8

= 83.657 kN

= 33.460 kN

= 18.323 kN

= 15.501 kN

75mm x 110mm 4mm washer

16 mm Φ-90.1kN

-24.0kN

3" x 8"

3" x 8" 3" x 8"

= 50.367 mm

= 141.29 mm

= 88.606 mm

= 121.03 mm

˚

˚

21.8) + 2.07(cos2 21.8))

(8.14 x 2.07) / (8.14(sin2 30.96) + 2.07(cos2 30.96))

mm2

mm2

kN/bolt

(8.14 x 2.07) / (8.14(sin2 78.69) + 2.07(cos2 78.69))

(8.14 x 2.07) / (8.14(sin2 78.69) + 2.07(cos2 78.69))

mm2

mm2

21.75

MPa

mm

in2