Download - 32193180 Timber Truss Design
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KurtLouis
A.
Montemor
2010
TimberRoofin
gDesign
A detailed design of Timber Roofing using Philippine wood grades, in
accordance with the National Structural Code of the Philippines (NSCP).
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Timber Roofing Design
Dimensions of Members (b x h):
Purlin: 2 x 3Top Chord: 21 x 4Bottom Chord:21 x 4Web Member: 11 x 4
Purlin Design:
Influence Strip Width = 0.559m
0.559 m
h
b
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Dead Load: (along inclination of the roof)Roofing: Using Gage 24 G.I. Corrugated Sheet: 0.075 kPa Actual
Insulation: 10mm urethane foam w/ skin: 0.009 kPa Code SpecifiedTotal: 0.084 kPa
, =2
5 0.084 0.559 = 0.042 /, = 15 0.084 0.559 = 0.021 /
0.021 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
0.042 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
0.084 kPa
1
2
5
0.042 kN/m
0.021 kN/m
1
2
5
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Live Load: (along horizontal)Roof Live Load: 0.750 kPa Code Specified
, = 25 0.750 0.5 = 0.3354 /
,
=
1
5 0.750
0.5
= 0.1677
/
0.1677 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
0.3354 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
1
2
5
0.3354 kN/m
0.1677 kN/m
1
2
5
0.750 kPa
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Wind Load: (normal to roof inclination, suction nature)Suction Wind Load: 1.3513 kPa Code Specified
, = 1.3513 0.559 = 0.7554/
Load Orientation Diagram:*The loads are simultaneously acting, and they are oriented as follows:
W = Wind load (suction, normal to roof)
Dr = Roof Dead Load (vertical)Lr = Roof Live Load (vertical)
12
W
5
Dr , Lr
0.7554 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
1
2
5
1.3513 kPa
1
2
5
0.7554 kN/m
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Structural Analysis of Forces (Using STRAN):
*note:1. The purlin is assumed to be weightless.
2. The original output was edited for clarity. There have been inclusions of units and drawings
for reference.3. There are five (5) Load Combinations prescribed in the NSCP for the Allowable StressDesign, namely:
(1) Dead Load Only [D](2) Dead plus Live Load [D + L + Lr]
(3) Dead plus Wind or Earthquake Load [D + W or E/1.4](4) Dead plus Earthquake Load [0.9D E/1.4]
(5) Dead plus Live and Wind or Earthquake Load [D + 0.75(L + Lr+ W or E/1.4)]
However, since the roof elements do not form part of the earthquake-resisting system,Load Combination 4 is not considered anymore. Thus, only load combinations (1), (2) and (3)
are considered. Intuitively, Load Combination (5) tends to produce smaller values and is thus notaccounted for to save on calculation time.
Dead Load Only [D] Normal Component
==========================S T R A N -- Version 4.0
==========================Copyright 1998 by Russell C. Hibbeler
The output from this program is to be used for academic and educational purposes only, and is
not for professional use.
Problem Title: Dead Load Only [D] Normal Component
Structure Type: BEAM
1 2 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
Y
X
A B C D
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************************ NODE COORDINATES *
***********************Node No. 1:
X-Coordinate = 0
Node No. 2:X-Coordinate = 5/3 m
Node No. 3:
X-Coordinate = 10/3 m
Node No. 4:X-Coordinate = 5 m
************************
* MEMBER PROPERTIES *************************
Member No. 1:Near Node = 1
Far Node = 2Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m4Near Moment Released = No
Far Moment Released = No
Member No. 2:Near Node = 2
Far Node = 3Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
Member No. 3:Near Node = 3
Far Node = 4Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
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************************** STRUCTURE SUPPORTS *
*************************Support at node no. 1:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 2:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 3:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 4:
Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
*************************************** APPLIED LOADINGS ON STRUCTURE *
**************************************Distributed load on member no. 1:
Algebraic value of load closest to near node in member y' direction = -0.042 kN/mDistance from near node = 0Algebraic value of load farthest from near node in member y' direction = -0.042 kN/m
Distance from near node = 5/3 m
Distributed load on member no. 2:Algebraic value of load closest to near node in member y' direction = -0.042 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = -0.042 kN/m
Distance from near node = 5/3 m
0.042 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
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Distributed load on member no. 3:Algebraic value of load closest to near node in member y' direction = -0.042 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = -0.042 kN/m
Distance from near node = 5/3 m
******************************************************** MEMBER END DISPLACEMENTS IN GLOBAL DIRECTION *
*******************************************************Member No. 1:
Near end Y-displacement = 0Near end Rot-displacement = -2.57378679044378E-04 rad
Far end Y-displacement = 0Far end Rot-displacement = 8.57928930147941E-05 rad
Member No. 2:
Near end Y-displacement = 0Near end Rot-displacement = 8.57928930147941E-05 rad
Far end Y-displacement = 0Far end Rot-displacement = -8.57928930147942E-05 rad
Member No. 3:
Near end Y-displacement = 0Near end Rot-displacement = -8.57928930147942E-05 rad
Far end Y-displacement = 0Far end Rot-displacement = 2.57378679044378E-04 rad
************************************
* STRUCTURE SUPPORT REACTIONS *
************************************Support at node no. 1:Y-direction support reaction = 0.028 kN
Support at node no. 2:
Y-direction support reaction = 0.077 kN
0.042 kN/m
12 3
41 2 3
0.028 kN 0.077 kN 0.028 kN0.077 kN
A 7/600 kN-m7 600 kN-m7/600 kN-m 7/600 kN-mB C D
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Support at node no. 3:Y-direction support reaction = 0.077 kN
Support at node no. 4:
Y-direction support reaction = 0.028 kN
****************************************************************** MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM ******************************************************************
Member No. 1:Near end shear force = 0.028 kN
Near end moment = 0Far end shear force = 0.042 kN
Far end moment = -7/600 kN-m
Member No. 2:Near end shear force = 0.035 kN
Near end moment = 7/600 kN-mFar end shear force = 0.035 kN
Far end moment = -7/600 kN-m
Member No. 3:Near end shear force = 0.042 kN
Near end moment = 7/600 kN-mFar end shear force = 0.028 kN
Far end moment = 0
***************************************************************** STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *
****************************************************************Member No. 1:
Sum of near node Y-forces = -8.67361737988404E-19 kNSum of near node moments = 0
Sum of far node Y-forces = -5.20417042793042E-18 kNSum of far node moments = 1.73472347597681E-18 kN-m
Member No. 2:
Sum of near node Y-forces = -5.20417042793042E-18 kNSum of near node moments = 1.73472347597681E-18 kN-m
Sum of far node Y-forces = 3.46944695195361E-18 kNSum of far node moments = 1.73472347597681E-18 kN-m
Member No. 3:
Sum of near node Y-forces = 3.46944695195361E-18 kNSum of near node moments = 1.73472347597681E-18 kN-m
Sum of far node Y-forces = -8.67361737988404E-19 kNSum of far node moments = -1.73472347597681E-18 kN-m
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Dead Load Only [D] Tangential Component
==========================
S T R A N -- Version 4.0==========================
Copyright 1998 by Russell C. Hibbeler
The output from this program is to be used for academic and educational purposes only, and isnot for professional use.
Problem Title: Dead Load Only [D] Tangential ComponentStructure Type: BEAM
********************* NODE COORDINATES *
********************Node No. 1:
X-Coordinate = 0Node No. 2:
X-Coordinate = 5/3 mNode No. 3:
X-Coordinate = 10/3 mNode No. 4:
X-Coordinate = 5 m
************************* MEMBER PROPERTIES *
************************Member No. 1:
Near Node = 1Far Node = 2
Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = NoFar Moment Released = No
Member No. 2:
Near Node = 2
12 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
Y
X
A B C D
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Far Node = 3Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
Member No. 3:Near Node = 3
Far Node = 4Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
************************** STRUCTURE SUPPORTS *
*************************Support at node no. 1:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 2:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 3:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
Support at node no. 4:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
0.021 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
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************************************** APPLIED LOADINGS ON STRUCTURE *
*************************************Distributed load on member no. 1:
Algebraic value of load closest to near node in member y' direction = -0.021 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = -0.021 kN/mDistance from near node = 5/3 m
Distributed load on member no. 2:
Algebraic value of load closest to near node in member y' direction = -0.021 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.021 kN/mDistance from near node = 5/3 m
Distributed load on member no. 3:
Algebraic value of load closest to near node in member y' direction = -0.021 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.021 kN/mDistance from near node = 5/3 m
*******************************************************
* MEMBER END DISPLACEMENTS IN GLOBAL DIRECTION ********************************************************
Member No. 1:Near end Y-displacement = 0
Near end Rot-displacement = -1.28689339522189E-04 radFar end Y-displacement = 0
Far end Rot-displacement = 4.2896446507397E-05 rad
Member No. 2:Near end Y-displacement = 0
Near end Rot-displacement = 4.2896446507397E-05 radFar end Y-displacement = 0
Far end Rot-displacement = -4.28964465073971E-05 rad
Member No. 3:Near end Y-displacement = 0
Near end Rot-displacement = -4.28964465073971E-05 radFar end Y-displacement = 0
Far end Rot-displacement = 1.28689339522189E-04 rad
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************************************ STRUCTURE SUPPORT REACTIONS *
***********************************Support at node no. 1:
Y-direction support reaction = 0.014 kN
Support at node no. 2:Y-direction support reaction = 0.0385 kNSupport at node no. 3:
Y-direction support reaction = 0.0385 kNSupport at node no. 4:
Y-direction support reaction = 0.014 kN
****************************************************************** MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM *
*****************************************************************Member No. 1:
Near end shear force = 0.014 kNNear end moment = 0
Far end shear force = 0.021 kNFar end moment = -7/1200 kN-m
Member No. 2:Near end shear force = 0.0175 kN
Near end moment = 7/1200 kN-mFar end shear force = 0.0175 kN
Far end moment = -7/1200 kN-m
Member No. 3:
Near end shear force = 0.021 kNNear end moment = 7/1200 kN-mFar end shear force = 0.014 kN
Far end moment = 0
0.021 kN/m
12 3
41 2 3
0.014 kN 0.0385 kN 0.014 kN0.0385 kN
A 7/1200 kN-m7 1200 kN-m7/1200 kN-m 7/1200 kN-mB C D
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***************************************************************** STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *
****************************************************************Member No. 1:
Sum of near node Y-forces = -4.33680868994202E-19 kN
Sum of near node moments = 0Sum of far node Y-forces = -2.60208521396521E-18 kNSum of far node moments = 8.67361737988404E-19 kN-m
Member No. 2:
Sum of near node Y-forces = -2.60208521396521E-18 kNSum of near node moments = 8.67361737988404E-19 kN-m
Sum of far node Y-forces = 1.73472347597681E-18 kNSum of far node moments = 8.67361737988404E-19 kN-m
Member No. 3:Sum of near node Y-forces = 1.73472347597681E-18 kN
Sum of near node moments = 8.67361737988404E-19 kN-mSum of far node Y-forces = -4.33680868994202E-19 kN
Sum of far node moments = -8.67361737988404E-19 kN-m
Dead plus Live Load [D+L+Lr] Normal Component
==========================
S T R A N -- Version 4.0==========================
Copyright 1998 by Russell C. HibbelerThe output from this program is to be used for academic and educational purposes only, and is
not for professional use.
Problem Title: Dead plus Live Load [D+L+Lr] Normal ComponentStructure Type: BEAM
12 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
Y
X
A B C D
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************************ NODE COORDINATES *
***********************Node No. 1:
X-Coordinate = 0
Node No. 2:X-Coordinate = 5/3 m
Node No. 3:X-Coordinate = 10/3 m
Node No. 4:X-Coordinate = 5 m
************************
* MEMBER PROPERTIES *************************
Member No. 1:Near Node = 1
Far Node = 2Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
Member No. 2:Near Node = 2
Far Node = 3
Youngs Modulus = 10100000 kPa
Moment of Inertia = 0.00000187 m4Near Moment Released = No
Far Moment Released = No
Member No. 3:Near Node = 3
Far Node = 4Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
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************************** STRUCTURE SUPPORTS *
*************************Support at node no. 1:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 2:Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 3:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
0.042 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead Load Normal Component
1
2 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
0.3354 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Live Load Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
+
=
0.3774 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead plus Live Load Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
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Support at node no. 4:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
*************************************
* APPLIED LOADINGS ON STRUCTURE **************************************Distributed load on member no. 1:
Algebraic value of load closest to near node in member y' direction = -0.3774 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.3774 kN/mDistance from near node = 5/3 m
Distributed load on member no. 2:
Algebraic value of load closest to near node in member y' direction = -0.3774 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.3774 kN/mDistance from near node = 5/3 m
Distributed load on member no. 3:
Algebraic value of load closest to near node in member y' direction = -0.3774 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.3774 kN/mDistance from near node = 5/3 m
*******************************************************
* MEMBER END DISPLACEMENTS IN GLOBAL DIRECTION ********************************************************
Member No. 1:Near end Y-displacement = 0
Near end Rot-displacement = -2.31273127312734E-03 radFar end Y-displacement = 0
Far end Rot-displacement = 7.70910424375795E-04 rad
Member No. 2:Near end Y-displacement = 0
Near end Rot-displacement = 7.70910424375795E-04 radFar end Y-displacement = 0
Far end Rot-displacement = -7.70910424375796E-04 rad
Member No. 3:Near end Y-displacement = 0
Near end Rot-displacement = -7.70910424375796E-04 radFar end Y-displacement = 0
Far end Rot-displacement = 2.31273127312734E-03 rad
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***********************************
* STRUCTURE SUPPORT REACTIONS ************************************
Support at node no. 1:Y-direction support reaction = 0.2516 kN
Support at node no. 2:Y-direction support reaction = 0.6919 kN
Support at node no. 3:Y-direction support reaction = 0.6919 kN
Support at node no. 4:
Y-direction support reaction = 0.2516 kN
****************************************************************** MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM *
*****************************************************************Member No. 1:
Near end shear force = 0.2516 kNNear end moment = 0
Far end shear force = 0.3774 kNFar end moment = -629/6000 kN-m
Member No. 2:
Near end shear force = 0.3145 kNNear end moment = 629/6000 kN-m
Far end shear force = 0.3145 kNFar end moment = -629/6000 kN-m
Member No. 3:
Near end shear force = 0.3774 kNNear end moment = 629/6000 kN-m
Far end shear force = 0.2516 kNFar end moment = 0
0.3774 kN/m
12 3
41 2 3
0.2516 kN
A 629/6000 kN-m B C D
0.2516 kN0.6919 kN 0.6919 kN
629/6000 kN-m 629/6000 kN-m 629/6000 kN-m
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***************************************************************** STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *
****************************************************************Member No. 1:
Sum of near node Y-forces = -1.38777878078145E-17 kN
Sum of near node moments = 0Sum of far node Y-forces = 1.38777878078145E-17 kNSum of far node moments = 0
Member No. 2:
Sum of near node Y-forces = 1.38777878078145E-17 kNSum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Member No. 3:
Sum of near node Y-forces = 0Sum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Dead plus Live Load [D+L+Lr] Tangential Component
==========================
S T R A N -- Version 4.0==========================
Copyright 1998 by Russell C. HibbelerThe output from this program is to be used for academic and educational purposes only, and is
not for professional use.
Problem Title: Dead plus Live Load [D+L+Lr] Tangential ComponentStructure Type: BEAM
12 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
Y
X
A B C D
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************************ NODE COORDINATES *
***********************Node No. 1:
X-Coordinate = 0
Node No. 2:X-Coordinate = 5/3 m
Node No. 3:
X-Coordinate = 10/3 m
Node No. 4:X-Coordinate = 5
************************
* MEMBER PROPERTIES *************************
Member No. 1:Near Node = 1
Far Node = 2Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m4Near Moment Released = No
Far Moment Released = No
Member No. 2:Near Node = 2
Far Node = 3Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
Member No. 3:Near Node = 3
Far Node = 4Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
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************************** STRUCTURE SUPPORTS **************************
Support at node no. 1:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
0.021 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead Load Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
+
0.1677 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Live Load Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
=
0.1887 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead plus Live Load Tangential Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
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Support at node no. 2:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
Support at node no. 3:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 4:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
************************************** APPLIED LOADINGS ON STRUCTURE *
*************************************Distributed load on member no. 1:
Algebraic value of load closest to near node in member y' direction = -0.1887 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.1887 kN/mDistance from near node = 5/3 m
Distributed load on member no. 2:
Algebraic value of load closest to near node in member y' direction = -0.1887 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.1887 kN/mDistance from near node = 5/3 m
Distributed load on member no. 3:
Algebraic value of load closest to near node in member y' direction = -0.1887 kN/mDistance from near node = 0
Algebraic value of load farthest from near node in member y' direction = -0.1887 kN/mDistance from near node = 5/3 m
*******************************************************
* MEMBER END DISPLACEMENTS IN GLOBAL DIRECTION ********************************************************
Member No. 1:Near end Y-displacement = 0
Near end Rot-displacement = -1.15636563656367E-03 radFar end Y-displacement = 0
Far end Rot-displacement = 3.85455212187897E-04 rad
Member No. 2:Near end Y-displacement = 0
Near end Rot-displacement = 3.85455212187897E-04 radFar end Y-displacement = 0
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Far end Rot-displacement = -3.85455212187898E-04 radMember No. 3:
Near end Y-displacement = 0Near end Rot-displacement = -3.85455212187898E-04 rad
Far end Y-displacement = 0
Far end Rot-displacement = 1.15636563656367E-03 rad
***********************************
* STRUCTURE SUPPORT REACTIONS ************************************
Support at node no. 1:Y-direction support reaction = 0.1258 kN
Support at node no. 2:Y-direction support reaction = 0.34595 kN
Support at node no. 3:Y-direction support reaction = 0.34595 kN
Support at node no. 4:
Y-direction support reaction = 0.1258 kN
*****************************************************************
* MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM ******************************************************************
Member No. 1:Near end shear force = 0.1258 kN
Near end moment = 0Far end shear force = 0.1887 kN
Far end moment = -629/12000 kN-m
Member No. 2:Near end shear force = 0.15725 kN
Near end moment = 629/12000 kN-mFar end shear force = 0.15725 kN
Far end moment = -629/12000 kN-m
Member No. 3:
Near end shear force = 0.1887 kN
0.1887 kN/m
12 3
41 2 3
0.1258 kN
A 629/12000 kN-m B C D
0.1258 kN0.34595 kN 0.34595 kN
629/12000 kN-m629/12000 kN-m 629/12000 kN-m
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Near end moment = 629/12000 kN-mFar end shear force = 0.1258 kN
Far end moment = 0
****************************************************************
* STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *****************************************************************Member No. 1:
Sum of near node Y-forces = -6.93889390390723E-18 kNSum of near node moments = 0
Sum of far node Y-forces = 6.93889390390723E-18 kNSum of far node moments = 0
Member No. 2:
Sum of near node Y-forces = 6.93889390390723E-18 kNSum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Member No. 3:
Sum of near node Y-forces = 0Sum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Dead plus Wind [D+W] Load Normal Component Only
==========================S T R A N -- Version 4.0
==========================Copyright 1998 by Russell C. Hibbeler
The output from this program is to be used for academic and educational purposes only, and isnot for professional use.
Problem Title: Dead plus Wind Load [D+L] Normal Component
Structure Type: BEAM (assuming a 50mmx75mm section)
12 3
4
1 2 3(0,0) (1.6667,0) (3.3333,0) (5,0)
Y
X
A B C D
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************************ NODE COORDINATES *
***********************Node No. 1:
X-Coordinate = 0
Node No. 2:X-Coordinate = 5/3 m
Node No. 3:
X-Coordinate = 10/3 m
Node No. 4:X-Coordinate = 5
************************
* MEMBER PROPERTIES *************************
Member No. 1:Near Node = 1
Far Node = 2Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m4Near Moment Released = No
Far Moment Released = No
Member No. 2:Near Node = 2
Far Node = 3Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
Member No. 3:Near Node = 3
Far Node = 4Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m
4
Near Moment Released = No
Far Moment Released = No
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************************** STRUCTURE SUPPORTS *
*************************Support at node no. 1:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
0.042 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead Load Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
+
0.7554 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Wind Load Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
=
0.7134 kN/m
1.6667 m 1.6667 m 1.6667 m
5 m
Dead plus Wind Load Normal Component
12 3
41 2 3
(0,0) (1.6667,0) (3.3333,0) (5,0)
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Support at node no. 2:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction
Support at node no. 3:
Restrained in Y-direction, Support displacement = 0Free in Rot-direction
Support at node no. 4:Restrained in Y-direction, Support displacement = 0
Free in Rot-direction*************************************
* APPLIED LOADINGS ON STRUCTURE **************************************
Distributed load on member no. 1:Algebraic value of load closest to near node in member y' direction = 0.7134 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = 0.7134 kN/m
Distance from near node = 5/3 m
Distributed load on member no. 2:Algebraic value of load closest to near node in member y' direction = 0.7134 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = 0.7134 kN/m
Distance from near node = 5/3 m
Distributed load on member no. 3:Algebraic value of load closest to near node in member y' direction = 0.7134 kN/m
Distance from near node = 0Algebraic value of load farthest from near node in member y' direction = 0.7134 kN/m
Distance from near node = 5/3 m
******************************************************** MEMBER END DISPLACEMENTS IN GLOBAL DIRECTION *
*******************************************************Member No. 1:
Near end Y-displacement = 0Near end Rot-displacement = 4.37176070548236E-03 rad
Far end Y-displacement = 0Far end Rot-displacement = -1.45725356849415E-03 rad
Member No. 2:
Near end Y-displacement = 0Near end Rot-displacement = -1.45725356849415E-03 rad
Far end Y-displacement = 0Far end Rot-displacement = 1.45725356849415E-03 rad
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Member No. 3:Near end Y-displacement = 0
Near end Rot-displacement = 1.45725356849415E-03 radFar end Y-displacement = 0
Far end Rot-displacement = -4.37176070548236E-03 rad
***********************************
* STRUCTURE SUPPORT REACTIONS ************************************Support at node no. 1:
Y-direction support reaction = -0.4756 kN
Support at node no. 2:Y-direction support reaction = -1.3079 kN
Support at node no. 3:
Y-direction support reaction = -1.3079 kN
Support at node no. 4:Y-direction support reaction = -0.4756 kN
****************************************************************** MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM *
*****************************************************************Member No. 1:
Near end shear force = -0.4756 kNNear end moment = 0
Far end shear force = -0.7134 kN
Far end moment = 1189/6000 kN-m
Member No. 2:
Near end shear force = -0.5945 kNNear end moment = -1189/6000 kN-m
Far end shear force = -0.5945 kNFar end moment = 1189/6000 kN-m
0.7134 kN/m
12 3
41 2 3
0.4756 kN
A1189/6000 kN-m
B C D
0.4756 kN1.3079 kN 1.3079 kN
1189/6000 kN-m1189/6000 kN-m
1189/6000 kN-m
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Member No. 3:Near end shear force = -0.7134 kN
Near end moment = -1189/6000 kNFar end shear force = -0.4756 kN
Far end moment = 0
***************************************************************** STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *
****************************************************************Member No. 1:
Sum of near node Y-forces = -1.38777878078145E-17 kNSum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Member No. 2:
Sum of near node Y-forces = 0Sum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 0
Member No. 3:
Sum of near node Y-forces = 0Sum of near node moments = 0
Sum of far node Y-forces = 0Sum of far node moments = 2.77555756156289E-17 kN
Determine Which Load Combination Governs:
*Calculate resultant reaction on support A (node 1) to know governing load combination:
Dead Load Only:
= ,2 + ,2 = 0.028 2 + 0.014 2 = 0.031 Dead+Live Load:
= ,2 + ,2 = 0.2516 2 + 0.1258 2 = 0.2813 Dead+Wind Load:
= ,2 + ,2 = 0.4756 2 + 0 2 = 0.4756 *By comparison, the Governing Load Combination for the Purlin Design is the Dead+WindLoad [D + W] Combination. However, it cannot be concluded from this section that [D+W] will
also govern in the truss design, since additional [D+L] load is coming from the ceiling, which isnot carried by the purlin but is carried on the bottom chord of the truss.
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Purlin Design:
Governing Load Combination: Dead plus Wind [D+W] Load Combination
*note that the dead load has both normal and tangential component, but the effect of Dead Load[D] Only Tangential Component is not included in the design of purlins, since its effect is to
decrease the design stresses, and adds complication to the calculations.
, = 11896000 = 0.1982 , = 1.3079 Properties of M15 Lumber: (values are valid for a 10-year design period)Fb = 15 MPa
Ft = 9 MPa (parallel to grain)Ft = 0.29 MPa (perpendicular to grain)
Fc = 12 MPa (parallel to grain)Fc = 4.3 MPa (perpendicular to grain)
Fv = 2.2 MPa (parallel to grain)E = 10.1 GPa (20
thpercentile)
0.7134 kN/m
12 3
41 2 3
0.4756 kN
A1189/6000 kN-m
B C D
0.4756 kN1.3079 kN 1.3079 kN
1189/6000 kN-m1189/6000 kN-m
1189/6000 kN-m
Dead plus Wind Load [D+W] Normal Component
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Therefore the stresses are:
Bending Stress:
, = 6
2
=60.1982 106
50
50
2
= 9.51 = 15 < (!)Compressive Stress (perpendicular to grain):
= = 1307.9025002 = 0.52 = 4.30 < (!)Shearing Stress (parallel to grain):
,=
3
2=
3
1307.90
25050= 0.78
= 2.20 < (!)Deflection:*Note that by inspection, the end members 1 and 3 will have the maximum deflection. Also, thetangent to the elastic curve at B rotates by a small amount of 0.001457 rad clockwise (with the
assumption that the section is 2x3, refer to structural analysis) thus it cannot be assumed to bea fixed support.
Normal Component:
For members 1 and 3,
Since it is possible to set-up one equation that expresses the moment for the entire length of thepurlin, the Double Integration Method will be used.
Double Integration Method: = 0.7134 /2 0.4756 = 0.35672 0.4756
0.7134 kN/m
A B
5/3 m
RBy = 1.3079 kN
MB = 0.1982 kN-m
RAy = 0.4756 kN
C
x
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22 = 22 =
22 = 1 0.35672 0.4756Using indefinite integration,22 = 1 0.35672 0.4756 = = 1 0.11893 0.23782 + 1 = 0.11893 0.23782 + 1 1 = 1 0.11893 0.23782 + 1
=
=
1
1189
40000
4
1189
15000
3 +
1
+
2
= 1189400004 1189150003 + 1 + 2 2Boundary Conditions:
At = 0,= 0, therefore:0 = 1189
4000004 1189
1500003 + 10 + 2 =
At
=
5
3
,
= 0, therefore:
0 = 118940000534 11891500053
3
+ 1 53 + 0 = Therefore,
= 1 0.11893 0.23782 + 118914400= 1
1189
400004 1189
150003 + 1189
14400
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There is a point C in the elastic curve (see figure above) where the deflection is maximum, at this
point the rotation angle of the tangent to the curve is zero (0). Hence, = 00 =
1 0.11893 0.23782 + 118914400The roots are:1 = 0.5245,2 = 1.7811,3 = 0.7434Note that the roots 1 = 0.5245 and 2 = 1.7811 are extraneous roots. The first root isunreal because there is no such thing as negative distance. The second root is also unrealbecause the range of x is 0 5
3 only. Thus, 3 = 0.7434 is the correct root. Therefore,
=1
1189
400004
1189
150003
+
1189
14400= 1 118940000 0.74344 118915000 0.74343 + 118914400 0.7434=
0.0379 3 = 0.0379 3
10.1 1061.873 1064 = 0.00200 = 2 But = 240 = 1666 .67240 = 6.94 (note that it is also safe with the criteria L/360)Therefore, < !
0.7134 kN/m
A B
5/3 m
RBy = 1.3079 kN
MB = 0.1982 kN-m
RAy = 0.4756 kN
C
x
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Ceiling Joist Design:
Timber Roofing Parts:
Purlin Web or Cross BracingTruss Ceiling Joist System
A1:
Dead Load:
Ceiling: Marine Plywood: 0.036 kPa Code SpecifiedCeiling Joist:
Use M15 Lumber: [weightless] Assumed
Live Load:Mechanical Duct Allowance 0.20 kPa Code Specified
= 12 = 0.036 + 0.20 12 = 0.236 /
A1
B1
0.236 kN/m
1m
RA RB
A1 0.236 kN 0.236 kN
1m 1m 1m
B1
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Calculate Reactions:
= = 12 0.236 /1 = 0.118
=1
82 =1
8 0.236 /12= 0.0295 /
Therefore the stresses are:
Bending Stress:
= 6
2
=60.0295 106
50
50
2
= 1.42 < (!)Shearing Stress (parallel to grain):
= 32 = 3118 25050 = 0.0708 < (!)Calculate Deflection:
= 54384
=
50.236 14384
10100000
555 109
4
= 0.00055
= 0.55
= 360 = 1000360 = 2.78 < !B1:
*note that the load is coming from A1*use M15 Lumber
Calculate Loads:
= 2
1
= 0.236
Calculate Reactions:
= = 12 0.236 2 = 0.236 = 0.236 1.5 0.236 0.5= 0.236
0.236 kN/m
1m
RA RB
0.236 kN 0.236 kN
1m 1m 1m
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Therefore the stresses are:
Bending Stress:
= 6
2
=60.236 106
50
50
2
= 11.33 < (!)Shearing Stress (parallel to grain):
= 32 = 3236 25050 = 0.14 < (!)Calculate Deflection:
By inspection, the maximum deflection occurs at point C, thus by Moment-Area Method: = = = 12 10.236 23 + 0.50.236 1.25 = 0.226
3
Substituting the values of E and I,
= = 0.226 310100000 555 109 = 0.0403 = 40.35
=
360
=3000
360= 8.33
> , !Revise the design for higher moment of inertia:
Using 2x3 wood, I = 1.87*10-6 m4,
= = 0.226 310100000 1.87 1064 = 0.0112 = 11.96 > , !
0.236 kN-m/ EI
1m 1m 1m
0.236 kN 0.236 kN
C
A B
1.5m
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Revise the design for higher moment of inertia:Using 2x4 wood, I = 4.44*10-6 m4,
= = 0.226 310100000 4.44 1064 = 0.0050 = 5.04