Structural Timber Design Course – IEM Dec 2003TIMBER 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 Wind Load ( very short term ) Taking design wind speed . 0.Structural Timber Design Course – IEM Dec 2003 Load Determination Dead Load .Long Term On rafter On plan : 0. V = 33 m/s For conservative approach .7 kN/m2 on slope : 0.BS6399 On rafter .75 kN/m2 on plan ( medium term ) On ceiling tie : 1.613 x 10-3 x 332 x 0.9 .7 / cos 22.134 kN/m2 ( -ve ) by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 2 .25 kN/m2 Live Load .2 = 0.Ceiling Tie Wind Load = 0. 0.5º = 0.2 and Cpe = 0.0.76 kN/m 2 On ceiling tie : 0.9 kN point load ( short term ) # Assume wind load on rafter as less severe than live load in the design of the members.9 = 0.Rafter Wind Load = 0.6 kN/m2 ( -ve ) CP3 Chap. Cpi = 0.25 kN/m2 ( consider as long term ) 2.613 x 10-3 x 332 x 0. V . 6 2.76 kN/m2 x 0.75 ) x 0.749 0.6 x 7 3 bays = 0.798 1.76 + 0.6 x 7 / 4 bays 4.798 ( 0.7 kN LONG TERM 0.4 0.7 0. Short Term ( all loads ) LONG TERM LOADING = 0.749 4.9 0.798 kN On rafter On ceiling = ( 0.6m x 7m 4 bays = 0.646 by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 3 .8 0.7 3.646 0.0 0.798 4. Medium Term ( long term + medium term loads ) 3.Structural Timber Design Course – IEM Dec 2003 Stress Computation 3 conditions of loading are required to calculate the member stresses : 1.25 + 0. Long Term ( only long term loads ) 2.7 2.25 ) x 0. 25 + 0.134 ) x 0.75 – 0.145 8.0 1.25 – 0.Structural Timber Design Course – IEM Dec 2003 MEDIUM TERM 1.5 1.25 VERY SHORT TERM Rafter = ( 0.21 kN 4 Ceiling Joist = ( 0.0 1.76 + 0.59 7.51+ 0.6 x 0.7 = 1.25 ) x 0.51 by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 4 0.145 1.0 0.7 5.7 = 0.6 x 7 / 3 bays 7.36 ) x 0.59 ( 0.59 2.9 0.4 4.21 1.21 0.86 .21 1.25 + 0.51 kN 3 1.86 0.4 1.25 0.6 x 0.7 4. 4 7. Normally (critical) only check for : σ m.Structural Timber Design Course – IEM Dec 2003 SHORT TERM 1.6 1.7 + 0.6 5.g σ c.53 Grade Stresses (SG 5) .9 1.0 1.145 8.g = 9.5 2.9 4.5 N/mm 2 Medium term ( DL + IL ) Short term Emean = 9100 N/mm 2 2 ( DL + IL + PL) Emin = 6300 N/mm by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 5 .9 1.5 N/mm2 σ t.59 3.145 8.7 0.59 8.7 N/mm2 = 8.59 7.9 = 1.0 1.2 4.g = 5.83 0.6 1.7 8. 5o 3.5 w kN /m Heel L = 1.8 apex Rafter analysis : 22.75L by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 0.125 wL2 0.9 m L = 1.1 mm A = 3400 mm 2 RAFTER DESIGN Consider medium term load Check for combine bending and axial force.Structural Timber Design Course – IEM Dec 2003 Example : Assume member size 38 x 100 Finished Size 35 x 97 From table of Properties : Zxx = 54900 mm3 ίxx = 28 mm ίyy = 10.9 m apex 0.0703 wL2 6 . 3. 0 kN Applied compressive stress. by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 7 .75 4 cos 22.19 N/mm Under medium term .906 kN/m L = 1.75 ) x 0.Structural Timber Design Course – IEM Dec 2003 Consider lower portion of rafter : w = ( 0.a = M Z = 0.a = P A = 8000 3400 = 2.76 + 0.23 kNm Applied bending stress.906 x 1. σ m.0703 x 0. axial compressive force = 8.35 N/mm2 Effective length = 3 x 1.23 x 106 54900 = 4.5 o = 1.9 m M = 0.42 m Rafter is fully restrained by tiling battens in the less stiff direction. σ c.9 2 = 0.6 = 0. a σ m.94 From table 10 ( MS 544 ) K8 = 0.682 σ c.97 = 0.a σ c.97 N/mm2 = 9.19 2 E = 2 (6300) (50. adm < 1 3.1 x 0.19 + 2.5 x 1.625 = 592. adm σe = 8.682 = 7.25 x 1.5 x 1.25 ( medium term ) = 10.625 N/mm2 E min σ c// = 6300 10. λ = Le Ίxx = 1420 28 = 50. adm 1 .5 x 1.5 X 2.7)2 λ 2 Combine Compression and Bending ( Clause 12.35 X 0.5 σ c.25 x 1.35 7.598 < 1 Therefore it is satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 8 .Structural Timber Design Course – IEM Dec 2003 Slenderress ratio .682 24.7 σ c// = 8.1.0 N/mm2 = 24.55 13 1 .a K8 σe + σ c.1 = = 13. adm σ m.6 ) σ m.1. 32 N/mm Axial Compressive force ( Average lower and upper chord ) 8 + 7 = 7.347 kNm Applied bending stress. σ m.906 x 1.21 N/mm2 by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 9 .125 x 0.Structural Timber Design Course – IEM Dec 2003 Consider portion over node point.125 wL2 = 0.a = M Z = 0. M = 0. σ c.5 kN 2 Applied compressive stress.75 2 = 0.a = P A = 7500 3400 = 2.347 x 106 54900 = 6. a σ m.0 .0 + 2. adm = 8. adm = 6.69 = 0.32 13.5 x 1. λ < 5.9 The upper chord need not be checked because axial compressive force is 7kN < 8 kN for lower chord. adm + σ c.1 = 11.Structural Timber Design Course – IEM Dec 2003 At node point . rafter is designed as short column.69 N/mm2 Combine Stress calculation for short column σ m.a σ c.68 < 0. σ c. Whole rafter is satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 10 .21 11.25 x 1. 33 0.025w L2 W L = ( 0.3 kN/m = 7/3 = 2.6 = 0.33 ) 2 = 0.25 + 0.25 ) x 0. Under long term – Loads 0.3 x ( 2.33 = 0.13 kNm M by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 11 .1WL2 . L L + + + 0.08wL 2 = 0.25 + 0.08 x 0.Structural Timber Design Course – IEM Dec 2003 DESIGN OF CEILING TIE Ceiling tie – combined bending and tension.25 The BMD for UDL : W / unit length L= 2.08wL2 Check Outer Bay 0. 45 N /mm 2 = 5. adm σ t.5 x 1 x 1. a = 4600 3400 = 1. adm σ t.27 N / mm2 σ m.0 *Satisfactory At support .45 + 1.27 = 0.39 N / mm2 Axial tensile force ( long term stress ) = 4.45 < 1.3 x 2. a = M/Z = 0.1 = 10.163 kNm by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 12 .Structural Timber Design Course – IEM Dec 2003 σ m. M = 0.a σ m.1wL2 = 0.33 2 = 0.6 kN σ t.13 x 10 6 54900 = 2.1 x 0.1 = 6. adm <1 σ m.355 N / mm2 = 9.39 10.a + σ t.7 x 1 x 1.355 6. adm Combination : = 2. 0 2 = 3.12 N / mm 2 σ t.041 kNm by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 13 .175PL 2.45 + 1.97 10.163 x 10 6 54900 = 2.27 = 0.33 2.33 2 = 0.025 x 0. M = 0.97 N / mm2 = 1.9 kN + UDL P 0.8 kN σ m.6 + 3.3 x 2.025 wL 2 = 0. m = 3800 3400 Combination : 2.33 at center of ceiling tie due to UDL .Loads = point load 0.12 6.075 PL 0.46 < 1 * Satisfactory Under short term . a = 0.Structural Timber Design Course – IEM Dec 2003 Axial tensile force = 4.33 M 2. a Permissible stresses : = 9.43 15.9 kN .9 kN = 8900 = 2.43 N / mm 2 54900 σ m.62 N / mm 2 3400 σ t.9 x 2.408 x 10 6 = 7.5 x 1. a Axial tensile force .41 = 0.68 + 2.7 x 1. ( max ) = 8.5 x 1.175 PL = 0.175 x 0.68 N / mm2 = 5.367 kNm *∑M = 0.1 = 15. adm Combination : 7. M = 0.1 = 9.75 < 1 Satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 14 .41 N / mm2 σ m. adm σ t.5 x 1.33 = 0.408 kNm = 0.Structural Timber Design Course – IEM Dec 2003 M due to point load 0.62 9. M = 0.163 kNm M for point load . M for UDL .3 kN σ t.60 < 1 * Satisfactory by David Yeoh of Kolej Universiti Teknologi Tun Hussein Onn 15 .157 kNm *∑ M = 0.321 kNm = 0.85 N / mm2 54900 σ m.332 = 0. a Axial tensile force = 8.85 15.7 2 = 7. 5.15 9.3 x 2. M = 0.9 x 2.9 + 5.Structural Timber Design Course – IEM Dec 2003 At support .68 + 2.1 x 0.1wL2 = 0.321 x 106 = 5. a = 7300 = 2.075 x 0.41 = 0.15 N / mm2 3400 Combination .075 PL = 0.33 = 0.