Beam Design - Awning

Size: px

Start display at page:

Download "Beam Design - Awning"

Everett Peters
5 years ago
Views:

1 Beam Design - Awning 1. Beam Data Load Type: Uniform Dist. Load Support: Simple Beam Beam Type: Sawn Lumber Species: Douglas Fir-Larch Grade: DF No.2 Size: 4 x 12 Design Span (L): ft. Clear Span: ft. Total Span: ft. Bearing (l b ): 6 in. Quantity (N): 2 3. Design Options Lateral Support: unbraced Defl. Limits: Load Duration: 1.15 Exposure: wet Temperature: T <= 100 F Orientation: Vertical Incised Lumber: No Rep. Members: No 2. Design Loads Live Load: 100 plf Dead Load: 75 plf Selfweight: lbs Dist. Selfweight: plf Total Weight: lbs 4. Design Assumptions and Notes Code Standard: IBC 2015, NDS 2015 Bending Stress: Parallel to Grain Notes: 5. Adjustment Factors Factor Description F b F t F v F c F c E/E min C D Load Duration Factor C M Wet Service Factor 1 b c C t Temperature Factor C L Beam Stability Factor C F Size Factor C fu Flat Use Factor 1.1 d C i Incising Factor C r Repetitive Member Factor a) Adjustment factors per AWC NDS 2015 and NDS 2015 Supplement. b) When (F b)(c F) 1,150 psi, C M = 1.0. c) When (F c)(c F) 750 psi, C M = 1.0. d) Only applies when sawn lumber or glulam beams are loaded in bending about the y-y axis. ph. (425)

2 6. Beam Calculations Determine reference design values, sectional properties and self weight of beam: A = b x d,, b = Breadth of rectangular beam in bending (in.) d = Depth of rectangular beam in bending (in.) A = Cross sectional area of beam (in. 2 ) S x = Section modulus about the X-X axis (in. 3 ) S y = Section modulus about the Y-Y axis (in. 3 ) I x = Moment of inertia about the X-X axis (in. 4 ) I y = Moment of inertia about the Y-Y axis (in. 4 ) b = in. d = in. A = x = in. 2 S x = (3.500)(11.250) 2 /6 = in. 3 S y = (3.500) 2 (11.250)/6 = in. 3 I x = (3.500)(11.250) 3 /12 = in. 4 I y = (3.500) 3 (11.250)/12 = in. 4 Reference Design Values from Table 4A NDS Supplement (Reference Design Values for Visually Graded Dimension Lumber, 2" - 4" thick). Species & Grade F b F t F v F c F c E Emin G DF No The following formula shall be used to determine the density of wood (lbs/ft 3. (NDS Supplement Sec ) ρ w = Density of wood (lbs/ft 3 G = Specific gravity of wood (dimensionless) m.c. = Moisture content of wood (percentile) G = 0.5 m.c. = 28 % (Estimated moisture content at wet service conditions) ph. (425)

= 35.47 lbs/ft 3 Volume total = N[A x (L + l b )] = 2 x [39.38 x (258.00 + 6)] x (12 in./ft.) 3 = 12.

47 x 12.03 = 426.7 lbs Self Weight (W S ) = ρ w x Volume span = 35.47 x 11.76 = 417.0 lbs Distributed Self Weight (w s ) = = 19.

3 = lbs/ft 3 Volume total = N[A x (L + l b )] = 2 x [39.38 x ( )] x (12 in./ft.) 3 = ft 3 Volume span = N[A x L] = 2 x [39.38 x ] x (12 in./ft.) 3 = ft 3 Total Weight (W T ) = ρ w x Volume total = x = lbs Self Weight (W S ) = ρ w x Volume span = x = lbs Distributed Self Weight (w s ) = = plf Load, Shear and Moment Diagrams: ft. 100 plf 75 plf w s = plf ft ft lbs lbs ph. (425)

3.1) f b = M / S F b ' = F b (C D )(C M )(C t )(C L )(C F )(C i )(C r ) Beam is unbraced along its compression edge, lateral stability is

4 1.) Bending: Members subject to bending stresses shall be proportioned so that the actual bending stress or moment shall not exceed the adjusted bending design value: f b F b ' (NDS Sec ) f b = M / S F b ' = F b (C D )(C M )(C t )(C L )(C F )(C i )(C r ) Beam is unbraced along its compression edge, lateral stability is considered below: Slenderness Ratio for bending member R B: l u = Unbraced Length = ft. l u /d = = l e = 1.63l u + 3d = 1.63(258.0) + 3(11.25) = in. = ft. (NDS Table 3.3.3) R b = = R b = < 50 ph. (425)

5 Euler-based ASD critical buckling value for bending members: E miny ' = E miny (C M )(C t )(C i ) = (0.9)(1)(1) = psi F be = = psi F bx * = F bx (C D )(C M )(C t )(C F )(C i )(C r ) = (900)(1.15)(1)(1)(1.1)(1)(1) = psi Beam stability factor: C L = = F bx ' = (900)(1.15)(1)(1)(0.989)(1.1)(1)(1) = psi f b = = psi f b = psi < F bx ' = psi (CSI = 0.81) 2.) Shear: Members subject to shear stresses shall be proportioned so that the actual shear stress parallel to grain or shear force at any cross section of the bending member shall not exceed the adjusted shear design value: f v F v ' (NDS Sec ) f v = F v ' = F v (C D )(C M )(C t )(C i ) F vx ' = (180)(1.15)(0.97)(1)(1) = psi Shear Reduction: Uniformly distributed loads within a distance, d, from supports equal to the depth of the bending member shall be pemitted to be ignored. Concentrated loads within a distance equal to the depth of the bending member from supports shall be permitted to be multiplied by x/d where x is the distance from the beam support face to the load. See NDS 2015, Figure 3C. f v * = = psi f v * = psi < F vx ' = psi (CSI = 0.18) No Reduction in Shear (conservative): f v = = psi f v = psi < F vx ' = psi (CSI = 0.20) ph. (425)

6 3.) Deflection: Bending deflections calculated per standard method of engineering mechanics for live load and total load: LL Allowable: L/360 TL Allowable: L/240 E x ' = E x (C M )(C t )(C i ) = (0.9)(1)(1) = psi Δ LL = = 0.40 in. (L/d) LL = / 0.40 = 642 Δ LL = 0.40 in = L/642 < L/360 Δ TL = = 0.78 in. (L/d) TL = / 0.78 = 330 Δ TL = 0.78 in = L/330 < L/240 4.) Bearing: Members subject to bearing stresses perpendicular to the grain shall be proportioned so that the actual compressive stress perpendicular to grain shall be based on the net bearing area and shall not exceed the adjusted compression design value perpendicular to grain: f c F c ' (NDS Sec ) f c = F c ' = F c (C M )(C t )(C i ) F c x ' = (625)(0.67)(1)(1) = psi A b = b x l b = 3.5 x 6 = in 2 f c = = 50.9 psi f c = 50.9 psi < F c x ' = psi (CSI = 0.12) *Disclaimer: The calculations produced herein are for initial design and estimating purposes only. The calculations and drawings presented do not constitute a fully engineered design. All of the potential load cases required to fully design an actual structure may not be provided by this calculator. For the design of an actual structure, a registered and licensed professional should be consulted as per IRC 2012 Sec. R and designed according to the minimum requirements of ASCE The beam calculations provided by this online tool are for educational and illustrative purposes only. Medeek Design assumes no liability or loss for any designs presented and does not guarantee fitness for use. ph. (425)

Beam Design - FLOOR JOIST

Beam Design - FLOOR JOIST 1. Beam Data Load Type: Uniform Dist. Load Support: Simple Beam Beam Type: Sawn Lumber Species: Douglas Fir-Larch Grade: DF No.2 Size: 2 x 10 Design Span (L): 11.83 ft. Clear