Long Span Pratt
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Transcript of Long Span Pratt
![Page 1: Long Span Pratt](https://reader033.fdocuments.us/reader033/viewer/2022061118/546a6159b4af9f9c748b4688/html5/thumbnails/1.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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= 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
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12.17
2.00 m
12.00 m
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-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
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16.05849
2.64
Kn
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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
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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.
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it is short column
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it is long column
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it is long column
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it is short column
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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
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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 =
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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
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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
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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
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= 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
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PROF.Allan E. Milano
OCT. 19, 2009
1.) Design the vertical member using steel bar and its square/rectangular washer
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From Table : ( Double Shear )
kN/bolt
kN/bolt
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mm
( cross-section )
3
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OK!
end
g
P/2
end
12.5
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0.5
3
kN
kN
4.73
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85.27
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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
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= 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˚
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=
= 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
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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
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= 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"
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= 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
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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
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21.75
MPa
mm
in2