DESIGN OF ROOF TRUSS Using 80 %stess grade molave (Vitex parviflora Juss) Bending and tension parallel to grain =24.0 MPa Modulus of elasticity in bending = 6.54 x MPa Compression parallel to grain = 15.4 MPa Compression perpendicular to grain =6.34 MPa Shear parallel to grain = 2.88 MPa A. PURLINS
tan = = 29.05 Weight of roofing = 80 Pa Velocity of wind = 200 kPa Relative density of molave = 0.64 Size of Purlins = 50 mm x 100 mm @ 0.80 m O.C Trusses @ 2.47 m apart 1. Wind load normal to roof: Pn = P = 0.0000473 P = 0.0000473 P = 1.892 kPa of vertical surface Pn = Pn = 1.49 kPa (normal to roof surface) 2. Maximum bending stress of purlins: X = 0.80 = 0.70 m
Wind load normal to the purlins: = Pn (spacing of purlins )(spacing of truss ) = 1.49(0.80)(2.47) =2.944 kN Weight of roofing: = 80(2.47)(0.70) = 138.32 N = 0.138 kN
3. TRUSSES 1. Total load carrying by the trusses excluding its weight P= P= P = 1544.53 MPa 2. Analysis of the truss members
x=
= 1.37 m
y=
= 1.144 m
= =
= 0.685 m
= y- = 1.144 – 0.685 = 0.459 m a. Forces acting normal to the top chord = P(spacing of trusses)( ) = 1544.53(2.47)(0.685) = 2613.27 N = P(spacing of trusses)(x) = 1544.53(2.47)(1.37) = 5226.535 N = P(spacing of trusses)( ) = 1544.53(2.47)(0.459) = 1751.08 N. b. Resultant and its location from 1. R= +3 + R = 2613.27 + 2(5226.535) + 1751.08 R = 20043.955 N R = (y) + (y + x) + (y+2x) + (y + 3x) 20043.955 ̅ = 5226.535(1.2 +1.2 +1.5 + 1.2 + 3) +2613.27(1.2 +.5) ̅ = 2.855 m x = 1.711 = 1.496 y =1.1711 = 0.831 c. Reactions: ∑ =0 (7.2) + 20043.955 = 12758.27 N ∑ =0 = 20043.955 = 9732.8 N ∑ =0 (7.2) + 20043.955 = 4764.08 N
(0.831) – 20043.955
(0.831) – 20043.955
(7.2 – 1.496) = 0
(1.496) = 0
CHECKING: ∑ =0 20043.955 = 12758.27 + 4764.08 17522.35 = 17522.35 OK! d. Analysing truss by the method of sections: For row truss, the maximum internal stress to be found at L/4 of the truss measured horizontally from point of maximum reaction:
Design for girts at second floor ceiling: (use the same kind of timber) Using 80% stress grade molave, allowable stress are as follows: Bending and tension parallel to grain = 24.0 MPa Modulus of elasticity in bending = 6.54x MPa Compression parallel to grain = 15.4 MPa Compression perpendicular to grain = 6.3MPa Shear parallel to grain = 2.88MPa
If shear controls: (assume b = 75 mm) = 2.88 =
d = 82.61 mm say 100 mm If bending controls: (assume b = 75 mm) = 24 = d = 99.57 mm say 100 mm Try 75 mm x 100 mm
Weight of girts = (0.075)(.10)(0.69)(9.81)(1000) = 50.77 N/m
Checking: By shear =
= 2.4 MPa < 2.88 MPa SAFE
By bending =
III.
= 23.97 MPa < 24.0 MPa SAFE
Design for columns Using 80% stress grade molave, allowable stress are as follows: Bending and tension parallel to grain = 24.0 MPa Modulus of elasticity in bending = 6.54x MPa Compression parallel to grain = 15.4 MPa Compression perpendicular to grain = 6.34 MPa Shear parallel to grain = 7.88 MPa Timber column size = 250 mm x 250 mm (use tpe of the same type of timber)
category of column: =
= 10.8
K = 0.671√
= 0.671√
= 13.83
k therefore the column is intermediate
For allowable stress of intermediate column Fc’ = Fc[1-
]
= 15.4 [1
= 13.49 MPa Allowable axial load P = Fc’A = 13.49( = 843,125 N > 13714.24 N SAFE
IV.
Design of second floor: Using 80% stress grade molave Floor load = 70 kPa Relative density of molave = 0.64 Spacing of floor joist = 0.40 m O.C Source: NSCP 2001 A. Design for floor joist 1. Size of joist by bending Live load WL = 7000 Pa (0.40) WL = 2800 N/m Relative stiffness: =
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