TIMBER TRUSS DESIGN PROCEDURE

Structural Timber Design Course – IEM Dec 2003

TIMBER TRUSS DESIGN PROCEDURE

1. Determine the dead and live loads acting on the truss 2. Compute the stresses 3. Determine the required sizes 4. Design the joints

Example : Standard Truss

7m

Slope 22.5º Spacing of truss 600 mm c/c SG 5, Std Grade, Dry Timber

by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn

1


Load Determination

 Dead Load - Long Term On rafter

: 0.7 kN/m2 on slope

On plan

: 0.7 / cos 22.5º = 0.76 kN/m 2

On ceiling tie : 0.25 kN/m2

 Live Load - BS6399 On rafter - 0.75 kN/m2 on plan ( medium term ) On ceiling tie :

1. 0.25 kN/m2 ( consider as long term ) 2. 0.9 kN point load ( short term )

# Assume wind load on rafter as less severe than live load in the design of the members.

 Wind Load ( very short term ) Taking design wind speed , V = 33 m/s For conservative approach , Cpi = 0.2 and Cpe = 0.9

CP3 Chap. V

- Rafter Wind Load = 0.613 x 10-3 x 332 x 0.9 = 0.6 kN/m2 ( -ve

)

- Ceiling Tie Wind Load = 0.613 x 10-3 x 332 x 0.2 = 0.134 kN/m2 ( -ve


)

2


Stress Computation 3 conditions of loading are required to calculate the member stresses : 1. Long Term ( only long term loads ) 2. Medium Term ( long term + medium term loads ) 3. Short Term ( all loads )

LONG TERM LOADING

On rafter

= 0.76 kN/m2 x 0.6m x 7m 4 bays = 0.798 kN

On ceiling = ( 0.25 + 0.25 ) x 0.6 x 7 3 bays = 0.7 kN LONG TERM

0.798 ( 0.76 + 0.75 ) x 0.6 x 7 / 4 bays

4.4 0.798

0.798 1.7 0.749

4.9

4.6 2.646

0.749

0.8

3.0 0.7


0.7

2.646

3


MEDIUM TERM

1.59 ( 0.25 + 0.25 ) x 0.6 x 7 / 3 bays

7.0 1.59

1.59 2.4 1.145

8.0

1.145

1.5

7.4

5.0

4.25

0.7

0.7

4.25

VERY SHORT TERM Rafter

= ( 0.76 + 0.75 – 0.36 ) x 0.6 x 0.7 = 1.21 kN 4

Ceiling Joist = ( 0.25 + 0.25 – 0.134 ) x 0.6 x 0.7 = 0.51 kN 3 1.21 1.21 1.21 0.86

0.86 0.51+ 0.9 0.51


4


SHORT TERM 1.59

7.0

7.9 1.59

1.59 3.6 1.145

2.4

8.0

8.9

5.7

4.83

1.145

8.2

0.7 + 0.9 = 1.6

Grade Stresses (SG 5) .

8.9

1.6

1.5

0.7

4.53

Normally (critical) only check for :

σ m,g = 9.5 N/mm2 σ t,g

= 5.7 N/mm2

σ c,g

2

 Medium term ( DL + IL )

= 8.5 N/mm

Emean = 9100 N/mm

 Short term 2

Emin = 6300 N/mm

( DL + IL + PL)

2


5


Example : Assume member size 38 x 100 Finished Size 35 x 97 From table of Properties : Zxx = 54900 mm3 ίxx = 28 mm ίyy = 10.1 mm A

= 3400 mm 2

RAFTER DESIGN Consider medium term load Check for combine bending and axial force.

apex

3.8

Rafter analysis :

22.5o 3.5 w kN /m

Heel

apex L = 1.9 m

L = 1.9 m

0.125 wL2

0.75L by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn

0.0703 wL2 6


Consider lower portion of rafter :

w = ( 0.76 + 0.75 ) x 0.6 = 0.906 kN/m L = 1.9 m M = 0.0703 x 0.906 x 1.9 2 = 0.23 kNm

Applied bending stress,

σ m,a

=

M Z

= 0.23 x 106 54900

= 4.19 N/mm

Under medium term , axial compressive force

= 8.0 kN

Applied compressive stress,

σ c,a

=

P A

= 8000 3400

= 2.35 N/mm2

Effective length = 3 x 1.75 4 cos 22.5 o

= 1.42 m

Rafter is fully restrained by tiling battens in the less stiff direction.


7


Slenderress ratio , λ =

σ c//

Le Ίxx

= 1420 28

= 50.7

= 8.5 x 1.25 ( medium term ) = 10.625 N/mm2

E min σ c//

= 6300 10.625

= 592.94

From table 10 ( MS 544 )

 K8

= 0.682

σ c, adm

= 8.5 x 1.25 x 1.1 x 0.682 = 7.97 N/mm2

σ m, adm

= 9.5 x 1.25 x 1.1

σe

=

2 E

=

2

= 13.0 N/mm2

2 (6300)

= 24.19

(50.7)2

λ

Combine Compression and Bending ( Clause 12.6 )

σ m,a σ m, adm

+

1 - 1.5 σ c,a K8 σe

3.55 13

σ c,a

< 1

σ c, adm

+

1 - 1.5 X 2.35 X 0.682

2.35

= 0.598

< 1

7.97

24.19

Therefore it is satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn

8


Consider portion over node point. M = 0.125 wL2 = 0.125 x 0.906 x 1.75 2 = 0.347 kNm

Applied bending stress,

σ m,a

=

M Z

= 0.347 x 106 54900

= 6.32 N/mm

Axial Compressive force ( Average lower and upper chord )

8 + 7 = 7.5 kN 2

Applied compressive stress,

σ c,a

=

P A

= 7500 3400

= 2.21 N/mm2


9


At node point , λ < 5.0 , rafter is designed as short column. = 8.5 x 1.25 x 1.1 = 11.69 N/mm2

σ c, adm

Combine Stress calculation for short column

σ m,a

+

σ m, adm

= 6.32

+

13.0

=

0.68

σ c,a σ c, adm

2.21 11.69

< 0.9

The upper chord need not be checked because axial compressive force is 7kN < 8 kN for lower chord.

 Whole rafter is satisfactory


10


DESIGN OF CEILING TIE

Ceiling tie

– combined bending and tension.

Under long term – Loads 0.25 + 0.25

The BMD for UDL : W / unit length

L= 2.33

L

L

0.1WL2 ,

+

0.08wL2

+

+

0.025w L2

Check Outer Bay

W

= ( 0.25 + 0.25 ) x 0.6 = 0.3 kN/m

L

= 7/3 = 2.33

M

= 0.08wL 2 = 0.08 x 0.3 x ( 2.33 ) 2 = 0.13 kNm


11


σ m, a

= M/Z = 0.13 x 10 6 54900 = 2.39 N / mm2

Axial tensile force ( long term stress )

σ t, a

= 4.6 kN

= 4600 3400 = 1.355 N / mm2 = 9.5 x 1 x 1.1 = 10.45 N /mm 2

σ m, adm σ t, adm

= 5.7 x 1 x 1.1 = 6.27 N / mm2

Combination

:

= 2.39

+ 1.355

10.45

6.27

σ m,a + σ t,a σ m, adm σ t, adm

<1

= 0.45 < 1.0

*Satisfactory

At support ,

M

= 0.1wL2 = 0.1 x 0.3 x 2.33 2 = 0.163 kNm


12


Axial tensile force

= 4.6 + 3.0

= 3.8 kN

2

σ m, a

= 0.163 x 10 6 54900 = 2.97 N / mm2

σ t, m

= 3800

= 1.12 N / mm 2

3400

Combination :

2.97

+

10.45

1.12

= 0.46 < 1

6.27

* Satisfactory

Under short term - Loads = point load 0.9 kN + UDL

P

0.075 PL 0.175PL 2.33 M

2.33

2.33

at center of ceiling tie due to UDL , M = 0.025 wL 2 = 0.025 x 0.3 x 2.33 2 = 0.041 kNm


13


M due to point load 0.9 kN ,

M = 0.175 PL = 0.175 x 0.9 x 2.33 = 0.367 kNm

*∑M

= 0.408 kNm

σ m, a

= 0.408 x 10 6 = 7.43 N / mm 2 54900

Axial tensile force , ( max ) = 8.9 kN

σ t, a

= 8900 = 2.62 N / mm 2 3400

Permissible stresses :

σ m, adm

= 9.5 x 1.5 x 1.1 = 15.68 N / mm2

σ t, adm

= 5.7 x 1.5 x 1.1 = 9.41 N / mm2

Combination : 7.43 15.68

+ 2.62

= 0.75 < 1

9.41

 Satisfactory


14


At support , M for UDL , M = 0.1wL2 = 0.1 x 0.3 x 2.332 = 0.163 kNm

M for point load , M = 0.075 PL = 0.075 x 0.9 x 2.33 = 0.157 kNm

*∑ M

= 0.321 kNm

σ m, a

= 0.321 x 106 = 5.85 N / mm2 54900

Axial tensile force =

8.9 + 5.7

= 7.3 kN

2 σ t, a

= 7300 = 2.15 N / mm2 3400

Combination , 5.85 15.68

+

2.15

= 0.60

< 1

9.41

* Satisfactory


15

TIMBER TRUSS DESIGN PROCEDURE

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