TIMBER TRUSS DESIGN PROCEDURE

Structural Timber Design Course – IEM Dec 2003

by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 1

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

7 m

Slope 22.5º

Spacing of truss 600 mm c/c

SG 5, Std Grade, Dry Timber



Load Determination

Dead Load - Long Term

On rafter : 0.7 kN/m2 on slope

On plan : 0.7 / cos 22.5º = 0.76 kN/m2

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 )



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

2.646 4.6 3.0

4.9

4.4

1.7

0.8

0.7 0.7 2.646

0.749

0.798

0.798

0.798

0.749

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



MEDIUM TERM 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

4.25 7.4 5.0

8.0

7.0

2.4

1.5

0.7 0.7 4.25

1.145

1.59

1.59

1.59

1.145

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

1.21

1.21

0.86 0.86

0.51

0.51+ 0.9

1.21



SHORT TERM

Grade Stresses (SG 5) .

σ m,g = 9.5 N/mm2

σ t,g = 5.7 N/mm2

σ c,g = 8.5 N/mm2

Emean = 9100 N/mm2

Emin = 6300 N/mm2

4.83 8.9 5.7 8.2

8.0

7.0 7.9

8.9 3.6

1.5 1.6 2.4

0.7 + 0.9 = 1.6

0.7 4.53

1.145

1.59

1.59

1.59

1.145

Normally (critical) only check

for :

Medium term

( DL + IL )

Short term

( DL + IL + PL)



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 mm2

RAFTER DESIGN Consider medium term load

Check for combine bending and axial force.

Rafter analysis :

Heel apex

0.75L

L = 1.9 m L = 1.9 m

0.125 wL2

0.0703 wL2

3.5

3.8 apex

22.5o

w kN /m



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 1.92 = 0.23 kNm

Applied bending stress,

σ m,a = M

Z

= 0.23 x 106 = 4.19 N/mm 54900

Under medium term , axial compressive force = 8.0 kN

Applied compressive stress,

σ c,a = P

A = 8000 = 2.35 N/mm2

3400

Effective length = 3 x 1.75 = 1.42 m

4 cos 22.5 o

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



Slenderress ratio , λ = Le = 1420 = 50.7 Ίxx 28

σ c// = 8.5 x 1.25 ( medium term )

= 10.625 N/mm2

E min = 6300 = 592.94 σ c// 10.625

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 = 13.0 N/mm2

σe = 2 E = 2 (6300) = 24.19 λ2 (50.7)2

Combine Compression and Bending ( Clause 12.6 )

σ m,a + σ c,a < 1

σ m, adm 1 - 1.5 σ c,a K8 σ c, adm σe

3.55 + 2.35 = 0.598 < 1

13 1 - 1.5 X 2.35 X 0.682 7.97 24.19

Therefore it is satisfactory



Consider portion over node point.

M = 0.125 wL2 = 0.125 x 0.906 x 1.752 = 0.347 kNm

Applied bending stress,

σ m,a = M

Z

= 0.347 x 106 = 6.32 N/mm 54900

Axial Compressive force ( Average lower and upper chord )

8 + 7 = 7.5 kN 2

Applied compressive stress,

σ c,a = P

A = 7500 = 2.21 N/mm2

3400



At node point , λ < 5.0 , rafter is designed as short column.

σ c, adm = 8.5 x 1.25 x 1.1 = 11.69 N/mm2

Combine Stress calculation for short column

σ m,a + σ c,a σ m, adm σ c, adm

= 6.32 + 2.21

13.0 11.69

= 0.68 < 0.9

The upper chord need not be checked because axial compressive force is

7kN < 8 kN for lower chord.

Whole rafter is satisfactory



DESIGN OF CEILING TIE

Ceiling tie – combined bending and tension.

Under long term – Loads 0.25 + 0.25

The BMD for UDL :

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

L= 2.33 L L

+ + +

0.1WL2 ,

W / unit length

0.08wL2 0.025w L2



σ m, a = M / Z

= 0.13 x 10 6

54900

= 2.39 N / mm2

Axial tensile force ( long term stress ) = 4.6 kN

σ t, a = 4600

3400

= 1.355 N / mm2

σ m, adm = 9.5 x 1 x 1.1 = 10.45 N /mm2

σ t, adm = 5.7 x 1 x 1.1 = 6.27 N / mm2

Combination :

= 2.39 + 1.355

10.45 6.27

= 0.45 < 1.0

*Satisfactory

At support , M = 0.1wL2

= 0.1 x 0.3 x 2.33 2

= 0.163 kNm

σ m,a + σ t,a < 1 σ m, adm σ t, adm



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 + 1.12 = 0.46 < 1

10.45 6.27

* Satisfactory

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

M 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

P

0.075 PL

0.175PL

2.33 2.33 2.33



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 + 2.62 = 0.75 < 1

15.68 9.41

Satisfactory



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 + 2.15 = 0.60 < 1

15.68 9.41

* Satisfactory

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TIMBER TRUSS DESIGN PROCEDURE