long span pratt
DESCRIPTION
an excel program that will help the user in designing a long-span timber pratt truss..=)TRANSCRIPT
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