32193180 Timber Truss Design

7/29/2019 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


3/68

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


4/68

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


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


5/68

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


6/68

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


7/68

************************ 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


Far Node = 3Youngs Modulus = 10100000 kPaMoment of Inertia = 0.00000187 m

4

Near Moment Released = No




4




8/68

************************** 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 = 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


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)


9/68

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


******************************************************** 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:



************************************

* STRUCTURE SUPPORT REACTIONS *

************************************Support at node no. 1:Y-direction support reaction = 0.028 kN


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


10/68

Support at node no. 3:Y-direction support reaction = 0.077 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


Near end moment = 7/600 kN-mFar end shear force = 0.035 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


11/68

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


12/68


4





4








Support at node no. 3:Restrained in Y-direction, Support displacement = 0




0.021 kN/m

1.6667 m 1.6667 m 1.6667 m

5 m


12 3

41 2 3

(0,0) (1.6667,0) (3.3333,0) (5,0)


13/68

************************************** 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




*******************************************************

* 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


Near end Rot-displacement = 4.2896446507397E-05 radFar end Y-displacement = 0

Far end Rot-displacement = -4.28964465073971E-05 rad





14/68

************************************ STRUCTURE SUPPORT REACTIONS *

***********************************Support at node no. 1:


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:


****************************************************************** 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. 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


15/68


****************************************************************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


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


16/68


***********************Node No. 1:

X-Coordinate = 0



Node No. 4:X-Coordinate = 5 m

************************




4




Far Node = 3

Youngs Modulus = 10100000 kPa

Moment of Inertia = 0.00000187 m4Near Moment Released = No




4




17/68




Support at node no. 2: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)


18/68



*************************************

* APPLIED LOADINGS ON STRUCTURE **************************************Distributed load on member no. 1:









*******************************************************












19/68

***********************************

* STRUCTURE SUPPORT REACTIONS ************************************







*****************************************************************Member No. 1:

Near end shear force = 0.2516 kNNear end moment = 0


Member No. 2:

Near end shear force = 0.3145 kNNear 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


20/68


****************************************************************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


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


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


21/68


***********************Node No. 1:

X-Coordinate = 0


Node No. 3:


Node No. 4:X-Coordinate = 5

************************







4





4




22/68

************************** STRUCTURE SUPPORTS **************************



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)


23/68







************************************** APPLIED LOADINGS ON STRUCTURE *

*************************************Distributed load on member no. 1:









*******************************************************








24/68

Far end Rot-displacement = -3.85455212187898E-04 radMember No. 3:


Far end Y-displacement = 0


***********************************

* STRUCTURE SUPPORT REACTIONS ************************************






*****************************************************************

* MEMBER END FORCES IN LOCAL MEMBER COORDINATE SYSTEM ******************************************************************


Near end moment = 0Far end shear force = 0.1887 kN





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


25/68


Far end moment = 0

****************************************************************

* STATICS CHECK AT NEAR NODE & FAR NODE OF EACH MEMBER *****************************************************************Member No. 1:


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


Member No. 3:



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


26/68


***********************Node No. 1:

X-Coordinate = 0


Node No. 3:


Node No. 4:X-Coordinate = 5

************************







4





4




27/68




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)


28/68






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








******************************************************** 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:




29/68




***********************************

* 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


Y-direction support reaction = -1.3079 kN

Support at node no. 4:Y-direction support reaction = -0.4756 kN


*****************************************************************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


30/68

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


****************************************************************Member No. 1:



Member No. 2:



Member No. 3:


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.


31/68

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


32/68

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


33/68

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


34/68

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


35/68

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 /


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


37/68


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

Documents

32193180 Timber Truss Design