# Design of an Industrial Truss

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1 Design of an Industrial Truss Roofing U 2 U 3 Ridge U 4 Sagrod 24 U 1 U 5 L 0 L 1 L 2 L 3 L 4 L 5 L 6 = 120 Elevation of the Truss Top Cord Bracing Sagrod Purlin at top, Bottom Cord Bracing at bottom Design Truss = 60 Column Building Plan Total Span of the Truss = = 120, Total Height of the Truss = 24, Spacing = 30 Pitch Angle = tan 1 (24/60) = Dead Loads: Roofing = 2 psf, Purlins = 1.5 psf, Sagrods + Bracings = 1 psf Basic Wind Speed V = 120 mph Material Properties: f c = 3 ksi, f y = 40 ksi 1. Design of Purlins and Sagrods Purlin length = Spacing of truss = 30

2 Dead Load: GI Sheet roofing = 2.0 psf, Self weight of Purlins = 1.5 psf Total Load = 3.5 psf Purlin Spacing S p = Length of Top Cord U 1 U 2 = { (24/3) 2 } = UDL on purlins, w DL = 3.5 S p = lb/ft w DLx = w DL sin = 28 lb/ft, w DLy = w DL cos = 70 lb/ft Wind Load: Basic wind speed, V = 120 mph Basic wind pressure, q = V 2 = = psf Wind pressure for windward surface, p = 0.7q, for 0 20 p = ( )q, for p = ( )q, for p = 0.9q, for (1.1(a)~(d)) Wind pressure for leeward surface, p = 0.7q, for any value of....(1.2) Here, pitch angle = tan 1 (24/60) = ; i.e., Wind pressure for windward surface, p WW = ( ) = psf Wind pressure for leeward surface, p LW = = psf UDL on windward surface, w WW = S p = lb/ft UDL on leeward surface, w LW = S p = lb/ft y w WW = lb/ft y w LW = lb/ft y w DLx = 28 lb/ft w DL x x x w DLy = 70 lb/ft Dead Load Windward Wind Load Leeward Wind Load

3 Load Combination and Biaxial Bending: It is clear from the preceding analyses that leeward side is more critical for wind loading. Therefore, the combination of dead load and leeward wind load provides the governing design condition for purlins. Combined UDL in y-direction, w y = lb/ft w y = lb/ft y Combined UDL in x-direction, w x = 28 lb/ft Span of the purlin is S t = 30 It is simply supported in y-direction w x = 28 lb/ft M xx = w y S 2 /8 = k = k, at midspan x Due to sagrod, it is 2-span continuous in x-direction M yy = w x S 2 /32 = 0.79 k = 9.45 k, also at midspan Design of Purlins: The absolute maximum tensile/compressive due to biaxial bending is zz = M xx /S xx + M yy /S yy = /S xx /S yy.(1.3) where S xx and S yy are the section moduli about the x- and y-axes. Allowable bending stress f b = 0.66 f y = 26.4 ksi Table 1.1: Calculation for Optimum Purlin Section Section S xx (in 3 ) S yy (in 3 ) zz (ksi) f b (ksi) Comments C Use much larger section C Use slightly larger section C OK The section C is chosen as the optimum purlin section. Self weight of purlin = 33.9 lb/ft, i.e., 33.9/21.54 = 1.57 psf assumed 1.5 psf, OK. Design of Sagrods: The maximum axial force in the sagrods = (5/8) w DLx S = (5/8) = 525 lb = k The minimum size of the sagrods = 3/8 Allowing 1/16 reduction for bolt threads, the rod area = (3/8 1/16) 2 /4 = in 2 Maximum stress = 0.525/0.077 = 6.84 ksi Allowable f s = 20 ksi The sagrod is chosen to have 3/8 -diameter, the weight of sagrods and bracings should be within the assumed limit of 1.0 psf.

5 Wind Load: As calculated in the design of purlins and sagrods, Wind pressure for windward surface = psf, and for leeward surface = psf Concentrated load (suction) on windward surface, P WW = S p S t /1000 = kips Concentrated load (suction) on leeward surface, P LW = S p S t /1000 = kips These are halved at end joints; i.e., P WW(end) = 6.84 kips, and P LW(end) = 8.34 kips 6.84 k 8.34 k k Wind k k k 6.84 k 8.34 k Concentrated wind ( ) loads at truss joints 3.34 Wind forces and support reactions (kips) due to wind ( ) loads Wind forces and support reactions (kips) due to wind ( ) loads

7 Design Concept of Truss s: The design of truss members is carried out using the following equations. s under tension: t = F t /A all(t) = 0.5 f y (3.1(a)~3.1(b)) The acceptable design condition is t s under compression: c = F c /A all(t) Slenderness Ratio, = L e /r min, and c = (2E/f y) If c, all(c) = f y [1 0.5 ( / c ) 2 ]/[5/3 + 3/8 ( / c ) 1/8( / c ) 3 ] If c, all(c) = 0.52 ( 2 E/ 2 ) (3.2(a)~3.2(d)) The acceptable design condition is c Here F t = Tensile force, F c = Compressive force, E = Modulus of elasticity A = Cross-sectional area, L e = Effective length of member, r min = Minimum radius of gyration t = Tensile stress, c = Compressive stress, all = Allowable stress all(c) For the material properties used for design; i.e., E = ksi, f y = 40 ksi all(t) = 0.5 f y = 20 ksi, c = (2E/f y) = , = / c = / If 1, all(c) = 40 [ ]/[5/3 + 3 /8 3 /8] 2 If 1, all(c) = / = 10.4/ 2 (3.3(a)~3.3(d)) Rather than designing for all the members individually, only one section will be chosen for all the bottom cord members, one section for all the top cord members and one section for all the other (vertical and diagonal; i.e., web ) members. Since the design truss has a long span, design sections will be chosen from double angle sections rather than single angle sections, which are chosen for smaller trusses.

8 Design of Bottom Cord s: For the bottom cord members (L 0 L 1 ~ L 5 L 6 ), the maximum tensile force = kips and the maximum compressive force = kips. Since all the bottom cord members are of equal length (i.e., if effective length factor k is taken = 1, then L e = 20 ft = 240 inch), the maximum forces are taken as the design forces. Although the design compressive force is much larger and is likely to be governing condition in this case, the sections are designed for both tension and compression for completeness of design. Section A (in 2 ) Table 3.2: Bottom Cord Design Chart t (ksi) all(t) (ksi) c (ksi) r min (in) all(c) (ksi) Comments 2½ 2½ 1/ Use much larger section 4 4 3/ Use larger section / Use smaller section 5 5 5/ OK Design of Top Cord s: For the top cord members (L 0 U 1 ~ U 5 L 6 ), the maximum tensile force = kips and the maximum compressive force = kips. Since all the top cord members are of equal length (i.e., if k = 1, L e = ft = inch), the maximum forces are taken as the design forces. The sections are designed for both tension and compression. A Section (in 2 ) Table 3.3: Top Cord Design Chart t all( t) c r min all(c) (ksi) (ksi) (ksi) (in) (ksi) Comments 2½ 2½ 1/ Use much larger section 4 4 5/ Use larger section 4 4 3/ OK

9 Design of Vertical and Diagonal s: For the vertical and diagonal members (U 1 L 1 ~ U 5 L 5, U 1 L 2 ~ U 5 L 4 ), the maximum tensile force = kips and the maximum compressive force = kips. The maximum tension acts on U 2 L 3 and U 4 L 3 while the maximum compression acts on U 3 L 3, which is 24 ft long. Although U 2 L 3 and U 4 L 3 are slightly longer (i.e., ft long), the maximum compressive forces on them are much smaller (i.e., 7.86 kips). Therefore the effective length of the design members is taken as L e = 24 ft = 288 inch. The sections are designed for both tension and compression although it is likely to be governed by compression. A Section (in 2 ) Table 3.4: Web Design Chart t all( t) c r min all(c) (ksi) (ksi) (ksi) (in) (ksi) Comments 2½ 2½ 1/ Use much larger section 4 4 1/ Use larger section 4 4 5/ OK 5/ / Bottom Chord s Top Chord s 4 5/8 4 4 Web s

10 4. Design of Bracings and Connections The truss members designed in the previous section are supported against out-of-plane loads by several bracings, joined to each other by welded plate connections and connected to column or wall supports. This section discusses the design of these so-called non-structural members. Design of Bracings: The bracings connect joints of two successive trusses in order to provide structural support against out-of-plane loadings. The bracing system used for the truss illustrated here is shown below, consisting of three types of bracings; i.e., (i) Bottom cord bracings connect the corresponding bottom joints (e.g., L 0 with L 0 ) (ii) Top cord bracings connect the bottom joints (e.g., L 0 ) with top joints (e.g., U 2 ) diagonally (iii) Vertical bracings connect the bottom joints (e.g., L 3 ) with top joints (e.g., U 3 ) vertically Bottom Chord Tie Bottom Chord Bracing Top Chord Bracing Vertical Bracing Since the structural analysis of the bracing system is complicated, the design follows simplified guidelines, according to which the slenderness ratio (L e /r min ) 400 for bracings under tension and 300 for bracings under compression. In the absence of accurate calculations, the more conservative second criterion (i.e., r min L e /300) is chosen for design here. The design (using single equal angles) is best carried out in a tabular form as shown below. Table 4.1: Design Chart for Bracings Length, L (ft) Effective Length, L e (in) r min (in) Chosen Section BC Bracing 30 (0.7 30/2 12 =) L2½ 2½ 3/16 TC Bracing { ( ) =} ( /2 12 =) L4 4 1/4 V Bracing { ( ) =} ( /2 12 =) L3 3 3/16

11 Design of Connections: The joints provided here are actually gusset plates joining two or more members with welded connections. The design is based on the following material and structural properties Allowable shear stress f v = 0.3 f y = = 12 ksi The gusset plate should be designed to resist the maximum possible design load condition. In this case, however, the thickness of the plate is approximately estimated based on the maximum axial force kips. If allowable tensile stress = 0.5 f y = = 20 ksi, and plate thickness = 0.25, then the maximum required width of the plate = 64.63/( ) 13. This should not be too large, based on the weld-lengths calculated subsequently. Thickness of gusset plate is chosen tentatively as = 0.25 and thickness of weld t = 0.25 The length of weld, L w = P/(f v 0.707t) = P/( ) = P/2.12 Table 4.2: Weld Design Chart Type Bottom Cord Top Cord Vertical Diagonal y : L y Design Force Weld Weld Length Size (in) (kips) Length (in) Ratio (in) L 0 L 1, L 5 L 6 * , 8.50 L 1 L 2, L 4 L 5 * 5 5 5/8 1.48: , 8.50 L 2 L 3, L 3 L 4 * , 6.50 L 0 U 1, U 5 L , U 1 U 2, U 4 U /4 1.27: , 8.50 U 2 U 3, U 3 U , 6.50 U 1 L 1, U 5 L , 0.25 U 2 L 2, U 4 L , 1.00 U 3 L /8 1.23: , 3.25 U 1 L 2, U 5 L , 1.00 U 2 L 3, U 4 L , 1.25 [* Design forces assumed equal]

12 1/4 21 Gusset plate (0.25 thick) L 0 U 1 (4 4 3/4 ) 1/4 8.5 L 0 L 1 (5 5 5/8 ) 1/4 10 1/4 20 Joint L 0 1/4 14 U 2 U 3 (4 4 3/4 ) U 3 U 4 (4 4 3/4 ) 1/ / / U 3 L 3 (4 4 5/8 ) Joint U 3

14 The base plate is supported on a concrete pedestal and connected to the column by four reinforcements to resist the entire tensile and shear force. Allowable tensile stress = 0.5 f y = 20 ksi and allowable shear stress = 0.3 f y = 12 ksi Required area (based on tensile force) = 29.29/(4 20) = in 2 Required area (based on shear force) = 3.34/(4 12) = 0.07 in 2 Provide 4 #6 (i.e., 3/4 diameter) anchor bolts (Area = 0.44 in 2 each). Allowable tensile force per anchor = = 8.8 kips Allowable bond force per unit length = 35 f c = lb/in = 1.92 k/in Development length = 8.8/1.92 = 4.59 Provide anchorage of 6 for each bolt. The base plate will be connected to the gusset plate by the section similar to the bottom cord (i.e., a 5 5 5/8 double angle section), also with 3/4 diameter bolts to transfer the maximum support reaction (= kips) by shear. Required area = 29.29/12 = 2.44 in 2, i.e., provide 3-3/4 diameter bolts in double shear.

15 Hinge Support L 0 Gusset plate (0.25 thick) L 0 U 1 (4 4 3/4 ) L 0 L 1 (5 5 5/8 ) 3/4 Bolts Base Plate (0.5 thick) 6 Anchorage Concrete Pedestal Masonry Column Roller Support L 6 Gusset plate (0.25 thick) L 6 U 5 (4 4 3/4 ) L 5 L 6 (5 5 5/8 ) 6 Roller

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