TIMBER TRUSS DESIGN PROCEDURE

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

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

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

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

σ

e =



2

E

=



2

(

6300) = 24.19

λ

2

(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

σ

e 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.

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