11 Timber Structures 11.1 INTRODUCTION Timber structures are used quite widely where timber is available cheaply. Tim- ber structures are less durable than steel of concrete structures. A well seasoned timber having higher strength to weight ratio and without any defects like knot, shake, crack, warping etc., is suitable for structural works. Table 11.1 gives the permissible limits for defects in structural timer as per IS: 883-1970. Table 11.1 Permissible defects in structural timber Select Grade I (Standard) Grade II grade Slope of grain 8 <1: 15 (Ordinary) Width | Narrow face and | Central half — | Depthof Slope of | 1/4 ofthe wide | of the width | checks | Slope of of grain | face | face close to top | _ of the wide and | grain a<1:20 and bottom edges face shakes | <1: 12 (mm) (mm) (mm) (mm) |_ ‘1 Knots, | 75 19 19 25° | Knots, checks [499 5 3 35 | checks, and and shakes | 150 38 38 50 | shakes should ae “] should rotbe | 200_| 44 oy es rire more than | 250 | 50 57 81 more NE Oe aj ian | the size | 300 34 ot 100 | 1.5 times | for: 350 37 | 81 |. 115 ".| thesize: | grade I t —| for | 400 63 | 87 131 | grade | | |262 Design of Stee! Structures Table 11.2 Permissible stresses for grade | timber Permissible stresses Location Group A Group B_— Group C (N/mm?) (1) Bending oj, and — Inside 18.2 12.3 8.4 Tension o, along grains ~ Outside 15.2 10.2 7.0 ~ Wet 12.0 8.1 6.0 (2) Shear All, horizontal# — 1.2 0.9 06 ~All, along grains 1.7 13 09 (3) Compression parallel — Inside 12.0 7.0 64 to grains, 0, ~ Outside 10.6 6.3 5.6 ~ Wet 88 58 46 (4) Compression — Inside 6.0 22 2.2 perpendicular to — Outside 46 18 i grains, O, - Wet 3.8 15 14 # The values of horizontal shear to be used only for beams. In all other cases, shears along grains to be used. 11.2. AVERAGE PERMISSIBLE STRESS AND MODIFICATION FACTORS Is: 883-1970 specifies that for ordinary design purpose the average permissible stress can be used for group A, B and C timbers instead of actual permissible stresses for that particular timber. Table 11.2 gives the average permissible stresses for grade I timber. For select grade and ordinary grade II timber these values are mutliplied by 1.16 and 0.84 respectively. IS: 883-1970 gives a num- ber of modification factors for timber design. The values in Table 11.2 are multiplie by modification factor k, (Table 11.3) if the slope of grains is more than | in 15 and by k, (Table 11.4) if the duration of load is not continuous. Indian timbers are classified into group A (E > 12.6 x 10? N/mm’), group B (98 x 10° < E< 12.6 x 10° N/mm’) and group (5.6 x 10° < E< 9.8.x 10° N/mm?) depending upon their modulus of elasticity as given in Table 11.5 for some of the Indian Timbers as per IS: 883-1970. Table 11.3 Modification factor k; for slope of grains Slope of grains ky Beams and Ties Post and Columns 1in 10 0.80 0.74 lin 12 0.90 0.82 lin 14 0.98 0.87 1 in 15 and less 1.00 1.00Timber Structures 263, Table 11.4 Modification factor k, for duration of load Duration of loading Modification factor ky Continuous 1.00 Two months 1.15 Seven days 1.25 Wind and Earthquake 1.33, Instantaneous or Impact 2.00 11.3 DESIGN OF TIMBER BEAMS Bending The usual bending Eq. 6.2, M = 6, x Z, is used for rectangular beams upto 30 cm deep. For deeper beams, Z is reduced by a form factor k. 2 ky = 0.81 x (32s) (1a) D> +550 where, D = depth of the beam in em. For I-beam and box girder (Fig. 11.1) form factor k, is multiplied with Z. D? +894 ky = 0.8 + 0.87 x | - 2 o yx (Bee ) (11.2) where, =p} (6— 8p, + 3p) (I- a) +.) (11.3) P = ratio of the thickness of the compression flange to the depth of the beam (d,/d) q, = ratio of the total thickness of web or webs to the overall width of beam (t)/t,) Shear The shear stress 7 can be calculated by Eq. 6.6. The maximum shear Str€SS Taq, Should be permissible values given in Table 11.2 or Table 11.5. For rectangular beams, the maximum shear stress is given by Eq. 11.4 15V. Toa = 5D (11.4a) a. = ie ie — 2H bY F i] |. 7 ~ 1a d | w | : c J /—______,| DS eal a I-beam Box-beam264 Design of Steel Structures ——————— pasn 9g 01 sures Buoye sways soseo somo |e UL # ‘Aquo sureaq 10} pasn aq 01 ssvays [eIUOZLOY Jo SenTEA BL + Seco Oe ry Oe o0) SL kl so 8008 86 '20Gl 96 O19 POOH 9 PL ter ci oF OS v9 60 got 09 OL 8 86. SLS mu ¢ Oley cL Lt so OF zs 80 90- "OS 95. 99 go SI'S wey > 61 ¥@ ey vl ov 60 08 TOL oPtl «616069 Bur ¢ St i OT fey 08 . "yo OL? "30 90 «49 gL 88 vs SIS ssosdkQ 7 i ie | oa. oc | OL SL. & 09 £0. “ Ob Se iol sé SHS mepood 1D Ste Ae OF Vo RE sgl o1 v6 OIL OFl 96 = OF'9 weL 9 9% 47 TE It; 99 08 co. el 60 = 86 PE0 SH EOL 008 Teselig. ¢ So er et Se BOUL owl FT ol wie Wel; Sol SIL AGL yneped > ey 1 6S «| BAIN OR 5 OL cele wl sol «Sel «9ST «= SIT Os'g — smdéjvong ¢ te ive Pe 99 “OR 83" <5 fa 88 Cle veh" gat “Ono oFON 7 Te {OS S90: OR Sor UL Ge $1 PUR; PSI? [BE F SOL . SRL nara 1 Se. Se Sr St pee S01 et 60! ZI OF: gor. Le soe res 7% Prt OS ue). oe ORD eee Ce st OC .. VSI OST. Ol - Sh’. SHEET V u0N1D90]_ UoNDI0] 01090] uo}»90] W9}20} UONDI0] —_UONvIO] UONDI0]1]v UIBAB J] 1D__UO'IDO] _wo!N90] UO!}I0] MUNA mM >PIEIMO APISMI, Y2M_—_2PISM_ApIsM|, SuoIY -UOMOH,, YM 2PM *PISUI quaquoa ae ¥ Y LY augif auasnxe YN D. 15V Tm = (11.4e) bxD, For beams notched at compression face when e < D, 1sV : far ore ep Nal (11.44) | D- ( = 2. D where, V = shear force at the section, b = width of the beam D = depth of the beam D, = depth of the beam at notch D3 = depth of the notch e = length of the notch measured along the span from inner edge of support to the farthest edge of the notch. Maximum shear force V is calculated as follows: For concentrated load W at a distance x from support, _ 1OW(L = x(x DY one 11.5) 9L[2 + (x / D)*) ! For uniformly distributed load w per unit length, i vew($-p) (11.6) 2 where, L = span of the beam. All concentrated loads in the vicinity of the support may be reduced by the reduction factors as follows. Distance of load from nearest support: $1.5D 2D 25D <3D Reduction factor for concentrated load: 0.6 0.4 0.2 Nil The maximum shear stress T,,, Should not exceed the permissible values given in Table 11.2 or 11.5. Bearing Permissible bearing stress parallel to the grain is taken as permissible compressive stress parallel to grain given in Table 11.2. But the permissible bearing stress perpendicular to the grain depends upon the length of bearing and its distance from the end of the member. IS 883-1970 specifies that the permissible compressive stress perpendicular to the grain may be multiplied by the modification factor kz given in Table 11.6 for bearing less than 150 mm in length and located 75 mm or more from the end of a member. For bearing under a washer or a small plate, the same coefficient k; may be taken for a bearing of length equal to the diameter of the washer or the width of the small plate.266 Design of Steel Structures Table 11.6 Modification factor k, for bearing stress Length of bearing(mm) 15 25 40. 50 75 100 2150 Modification factor 1.67 14 1.25 12 113 - 11 1.0 When the direction of stress is at an angle to the direction of grain in any struc- tural member, then the permissible bearing stress in the member should be cal- culated by Eq. 11.7. On, XO, op X Fen (7) Oe = 6 .)sin’ 0+ 6, cos’ @ where, 6, = permissible bearing stress at an angle 6 to the grain, O, ermissible bearing stress parallel to the grain, 6,,, = permissible bearing stress perpendicular to the grain. Bearing length Flexural members except roof timbers which are supported directly on masonry or concrete should have a bearing length of not less than 75 mm. Members supported on corbels, off-sets and roof timbers on a wall should bear immediately on and be fixed to wall plate of at least 75 x 40 mm size. Stiffening All flexural members having a span exceeding 50 times its width or depth exceeding 3 times its width, shall be laterally restrained from twisting or buckling at a distance not exceeding 50 times its width. Deflection The deflection of all flexural members supporting brittle material like asbestos sheets, tiles etc. should not exceed 1/360 of the span. The deflec- tion in the case of other flexural members should not exceed 1/240 of the span and 1/180 of the freely hanging length in the case of cantilevers. For checking up the deflection in case of beams and joists, the loads taken should be twice the dead load plus 3/4 live load as specified in IS 883. 11.4 TIMBER COLUMNS Solid timber columns are the simplest to provide. Built-up columns. Box col- umns and Spaced columns are used for more effective use of timber. Columns are classified as short, intermediate and long columns as given in Table 11.7. ad, Solid columns Box columns Built-up column Fig. 11.2Timber Structures 267 Table 11.7. Design data for timber columns Permissible Compressive Stress Limit of slenderness ratio Type Solid rectangular columns L (a) Short pc op (b) Intermediate 11<=SKy o,j1-4{4 D 3 ted L 0.329E (©) Lon; =>K, . c. (Ld? Built up and box columns (a) Short Gy 4 (&) Intermediate ¢,,|1-4] —4 3 Ky ya? +a? ou. 0.320UE _ (z a? + a2 ) Spaced columns Le (a) Short a d,). 11.4.2. Box and Built-up Columns Box and Built-up columns are made by fastening the planks at interval < 6 x thickness of the planks. Hollow core of Box columns is blocked by solid piece of timber at ends and at least at one intermediate point. 11.4.3 Spaced Columns Spaced compression members are usually used in timber trusses. They consist of two or more members separated by spacer blocks as shown in Fig. 11.3. A single spacer block should be located within the middle 1/10th of the total length. If more than one spacer block are used, their spacing should not exceed half the distance between end block. jj} Spacer block Disc dowel End block fee eft Spaced column Fig, 11.3 Effective length, L, = L/.{2.5, when the connectors of the end blocks are at a distance 1 < Lo/20 = L//3, when the connectors are at a distance / < Lo/10 to Lp/20Timber Structures 269 The slenderness ratio of individual member of a spaced column should not be more than 80. 11.5 COMBINED BENDING AND AXIAL STRESS Structural members subjected to both bending stress and axial compression or tension stress should be designed to comply with the instraction formula as follow. (1.11a) O by cal bee: (11.11b) On Cu where Op. cat For, calr Fac, cal ANd Fy, cai Ate the calculated stresses in bending compression, bending tension, axial compression and axial tension respec- tively. Gj., Fy, Fge and O,, are the permissible stress in bending compression, bending tension, axial compression and axial tension respectively. Example 11.1 Find the form factor and load carrying capacity of the cross-sections of beams of teak, as shown in Fig. 11.4. 18cm 30cm. pS cm 5 J 8 Some, | en omnes) 1 = ae (a) (b) § © 4 Fig. 11.4 Solution: Permissible stress for group B (Teak) timber from Table 11.2. , = 12.3 N/mm? = 1.23 kN/om? Rectangular beam: Form factor (Eq. 11.1) 2 D* +894 _ 0.81(36* +894) = 0.96 Ky =0.81x = = : D> +550 (36? +550)270 Design of Steel Structures 2 Moment of resistance = 0, ZK = 1.23 x Boe 0.96 = 4591 KN cm 2 T-beam: Form factor, K, = 0.8 + 0.8 y| 2, * 84 _ D? +550 where, Y= PP (6 -8p, + 3p?) A -at Sut 0 ei era) 2 Y= (3 (s-1+2\1-2)+3 = 0.23 8 oh 6)" 6 2 = 0.8 +08 x 0.23 Fes . | =0.83 or zi 40° +550 30 x 40° 30° = 20x40" _ 25% 30" 19375 x 10 em* 12 12 10.375 x 10* Z= aan = 5187.5 cm? 20 Moment of resistance = 1.23 X 5187.5 x 0.83 = 5296 KN. om Circular section 3 Moment of resistance = 0,2 = 1.23 x (2) = 3260.4 KN, cm Example 11.2 Design one sal timber joist of clear span 3 m at a spacing of 50 cm in a root. The bearing at each end is 10 em. The dead load of roof covering is 2 KN/m? and live load is 2.5 kN/m’. Solution: Loads: Live =2.5 x 0.5 x 1 = 1.25 KN/m Dead = 2 0.5 x 1 = 1.0 kN/m Self weight, assuming 10 x 20 cm beam = 8.65 x 0.1 x 0.2 = 0.173 KN/m Total u.d.L., w = 2.423 KN/m 2 Maximum bending moment = “= eee 2.91 kNm For Sal timber, permissible bending stress ©, = 1.68 kN/cm? (Table 11.5)Timber Structures 271 Zrequired = 221X100 _ 173.9 om? 1.68 Taking d/b =2 oo ae3 go 6 b = 6.38 cm, d = 12.76 em Using 7 x 13 em Sal beam Check for shear Permissible shear stress = 0.9 Nimm? (Table 11.5). Vertical shear force (Eq. 11.6) =173.2cm! Vew (4-2)=2.428 03) = 3.44 kN Maximum horizontal shear (Eq. 12.6) 15V_ 1.5% 344x108 bD 70x 130 = 0.57 N/mm? +0.9 N/mm? which is safe. Check for bearing: Bearing area. = 7X 10 = 70cm? End reaction = ah x 2,423 = 3.76 KN Permissible bearing stress = permissible compressive stress normal to grain = 4,5 N/mm? = 0.45 kN/em? 3:78 _ 9.35 cm? $70 cm? Bearing area required = 0.45 which is safe. 5 wit Check for deflection: Maximum deflection 54. = = ——— eck for deflection: Maximut Ser ET w =2X 1.173 + 0.75 x 1.25 = 3.28 kKN/m = 3.28 x 10 kN/em (increasing dead load by 2 and reducing live load by 3/4) E = 12.7 x 10° N/mm? = 1.27 kN/em? = 1281.6 cm* 7x13 12 -2 4 WS 5 x 3.28 x10 * x 310 oe max “384 x 1.27 x 10° x 1281.6 . 0 Permissible deflection = —— = 0.86 cm 360 Increasing Ts x x 1281.6 = 3615 cem*272__Design of Stool Structures 3 4 bd _ 8O 3615 om? je 2 or b = 8.58 cm and d= 18.16 cm Provide 9 x 18 cm Sal beam Example 11.3 Determine the load carrying capacity (including self weight) of the flitched beams shown in Fig. 11.5 (a) and (b) over an effective span of 6 m. Timber used is Andaman Padak. 1cm mild steel plate a \ ae lcm mild steel plate en Ee -Sa— 4 va 7 E IL 5 5 5 5 e 3 & a 1 em mild steel plate J 15cm (a) (b) © Fig. 11.5 “Solution: For Andaman Padak E, = 1.12 x 10° kN/em?, oj, = 1.72 kN/em? For mild steel E, = 2x 10' kN/em’, o, = 16.5 kN/cm™ Case (a) Steel and timber will stretch together ~. Strain in steel = Strain in timber 2) or, ab} aj (2). ~. Maximum stress at the extreme fiber of timber Sor 7 0924 kN/om? $1.72 kN/cm? x E, Oh, antes = Go Tha = 165% which is all right. Moment of resistance of the composite sectionTimber Structures 273 = (% Z)steei + (F% Zimnber 2 2 =16.5x2x 2 4 9.924 x BAS = 7029 kN.cm = 70.29 kN.m Equating moment of resistance and maximum bending moment; 2 — = 70.29 kN.m at = 15.62 kN/m 6x6 U.d.k the beam can carry, w Case (b) Transforming timber section into steel section as shown in Fig. 11.5(c). thickness of equivalent steel section 112x108 =15x 2x104 = 0.84 cm (since M/EI = /R = constant ) 15x 32% (15-0.84)30° 12 12 1 _16.5x9100 Moment of resistance = x = y 16 Equating moment of resistance and bending moment, = 9100 cm* = 9384.4 KN cm = 93.844 KN/m a = 93.844 kNm U.dl. the beam can carry, w = 22-844%8 _ 99.85 kvm 6x6 Example 17.4 Determine the safe axial load on a circular sal column of diameter 20 cm and length 3.5 m. Solution For sal, E=1.27 x 10° kN/em® Op = 1.06 kN/em? x Sectional area of column, A=] X20? = 314.16 cm? Size of equivalent square column, d= J314.16 = 17.7 cm L 11S J S Ke, it is an intermediate column274 Design of Steel Structures 4 Permissible load =4 0, yet 3\ Ked 119.77)" = 314.16 x 1.06 }1-+| "| | = 284.38 kN 3\ 243 Example 11.5 Design a 5 m long rectangular box columns built up by 5 cm thick Deodar planks to carry on axial load of 400 kN. Solution: Axial load = 400 kN Assuming permissible compressive stress along grain, 0, = 0.7 kN/em? Area required = 8 57t cot 07 Side of a square column = 2 4.5 = 33.5 em 5x4 Try 35 x 35 cm box column as shown in Fig. 11.6 FI dy Fig. 11.6 5000) a Oy =11.6 ya? +a3 3s? 425° 3 UE _ |0.6x0.95%10" _ 1g og cp 2 5x 0.78 L oe = < Ky; it is an intermediate column. fae +az +dz K. 4 iE L Permissible load =Ao,|1 {1 | 3\ Kyyd? +d}Timber Structures 275 4 = (30 x5 x 4)0.78 | 1 aes 3(18.99 = 446 KN > 400 KN which is safe. Example 11.6 Design a spaced column 3 m long of kail timber to carry 100 KN. Solution: Length of column = 3 m c Assuming end connectors are within 7 from ends, effective length, Taking thickness of planks = 5 cm = 37.95 For Kail, E =68x 10° kN/em? xp = 0.52 KN/emn? Ky) = 0.702 [258 = 0.702,[22* 6810" = 40.14 Sp 0.52 L. d Permissible compressive stress 11 $< Kp, it is an intermediate column 4 a = 0.38 kN/cm? 14 Area required = 100 263.16 om? 0.38 Total width required = — = 52.63 cm Provide 2 planks 5 x 26.5 cm or 3 planks 5 x 18 cm. One spacer block at the centre and disc dewel connectors are provided at the ends. 17.6 NAILED JOINTS IS 2366-1963 and SP: 33 (S&T)-1986 give specifications for the fabrication, finishing and maintenance of nail jointed timber construction for structural pur- pose. The nails used for timber joints should be plain headed nails specified in IS 273-1961 specification for Mild Steel Wire Nails. Nails may be driven di- rectly on soft woods without prebore. In hard woods, holes are drilled before nailing. Table 11.8 give the prebore suitable for various nail sizes.276 Design of Steel Structures Table 11.8 Prebore suitable, for various nail sizes Nail shank dia Size (dia.) of prebore in mm. suitable for (mm) Corres- Hardwood Corres- Soft wood __Corres- ponding ponding ponding (a) (bf — SWG (ay (bo) SWG @e ba swe 5.00 5.16 6 400 410 8 3.55 3.78 9 450 4:57 7 955 408 9 315 3.40 10 4.00 4.10 8 3.15 3.40 10 280 3.05 11 Boos 76 9 280" 1305 td 250. 2.76 12 315 340 10 250 276 12 224 241 13 230 305 i 224 241-13, 2.00 211 14 2.50 2.76 12 2.00 2.11 14 1.80 1.88 1S 2.24 2.41 13 1.80 1.88 15 1.60 1.65 16 2.00 2.11 14 1.60 1.65 16 1.40 1.47 7 * Rational metric values ~ Standard wire gauge number # Exact equivalent metric values for standard wire gauge number 11.7 ARRANGEMENT OF NAILS IN A JOINT Figure 11.7 through 11.9 show the arrangement of nails in lengthening and node joints as per SP:33 (S&T) 1986, clause 6.2.11.1. The spacing of the nails should be generally as follows Along grains € 5 d in compression + 10d in tension Perpendicular to € 5 d grains or load Edge distance ¢ 5d End distance € 10 d in compression + 12 din Tension where, d is diameter of nails, which should be between 1/6 to 1/11 of the least thickness of the member. The size of the members should be within the follow- ing range. Minimum thickness — 15 mm Maximum thickness (except spacer blocks) — 100 mm Maximum width — 8 x member thickness Combined thickness of gusset ¢ 1.5 x thickness of main member plate on cither side of joint in monochord type of construction Combined thickness of all spacer blocks and/or splice plates in split chord type construction € 1.5 x total thickness of all the main members at the jointTimber Structures 277 oie plate us Packing te Effective end distance 10d min. id, 10d _,1, 10d 1, 10d T \ . | | | feet +444 |b as tee+ +t++4 $\ feet 4¢4e | ft +444 P| _ a Effective edge distance 5d min. () Split chord type butt joint subject to compression Effective end distance — ee Oct ) 12d j 12d, 10d yLOd 10d | 12d, 12d min._ J tn Pt Seales 5d min. (ii) Split chord type butt joint subject to tension Fig. 11.7278 Design of Steel Structures a Unloaded (i) Compression member edge (ii) Tension member Fig. 11.8 11.8 STRENGTH OF NAILED JOINT The allowable load for a nailed joint in withdrawal or lateral resistance is equal to sum of the allowable load for the individual nails, provided that the centroid of the group of fixing nails lies on the axis of the members and the arrangement of nails is as per Sec 11.7. The permissible latcral strength of mild steel wire naiis are given in Table 11.9 for Indian species of timber, which will apply to nails that have their points cut flush with the faces, For nails clenched across the grain, the strength may be increased by 20%.. Timber Structures 279 Timber fish plate Bottom chord - Web members Fea ae Od 10d Spacing of nails at a node joint of a timber truss Fig. 11.9 Example 11.7 Design a timber truss joint as shown in Fig. 11.10. The members con- sist of Padauk timber bottom chord 8 x3 cm and slings 4 x 3 cm. Solution: From Table 11.9, for Node joint: Strength per nail (9 SWG) for Padauk = 1.4 kN Number of nail required for member OB = member OC = = = 1,75280 Design of Steel Structures Table 11.9 Permissible lateral strengths (in double shear) 9 SWG nails of size 8 to 10 cr driven in timber. S.No. Trade name Strength per Nail (kN) Permanent Temporary constructi¢ construction Lengthening or Lengthening joint Node joint Node joint 1 Kala-siris (H) 14 05 2.2 2 Sal (H) 1.0 0.5 19 3. Babul (H) 15 1 34 4 Kokko (H) 2.0 07 24 5. Eucalyptus (H) 17 1.0 3.0 6 Padauk (H) 19 14 23 7 Bijasal (H) 15 1.2 27 | 8 Teak (H) 14 08 13 9 Deodar* (S) 09 04 15 10 Cypress (H) 0.6 0s 18 11 Mango* (S) Ll 09 16 12 Kail* (S) 07 03 09 13 Chir* (S) BL 1.0 1.6 * Species require no pre-boring for nail penetration. H-=Hard wood $= Soft wood B € 1.25 kN 2.45 KN 0 A D 5.0kN 3.35 KN Fig. 11.10 50-335 _ va Minimum 2 nails for slings and 3 nails for bottom chord are provided as shown in F W111 member AD = 1.18 11.9 BOLTED CONSTRUETION i Fig. 11.12 through 11.14 show the general arrangement of bolts in structu joints of timber. The spacing of bolts, end distance and edge distances in bolt joints are as follow.Timber Structures 281 Gusset plate 2.5 em thick on both sides 4x3 cm slings Bt 7! 4d ° 4d Xe 4d | 4d End dist. ih End dist. ‘din sat de. admin Fig. 11.12 Lengthening Jointin compression Ht + aged ee Fig. 11.13 Lengthening joint in tension Spacing of bolts along row Parallel or Perpendicular to grains > 4d282. Design of Steel Structures 2 Unloaded edge r Pod oe min. Ad + Ls | 4d Je—rfente—af Teg’ "5g, Loaded edge min | Min 25d for t/d =2 5.0d for 1/d = 6 Loaded edge Y “| “7 | min. | 15d min. Ja} of —o} 15d min. 2.5d for t/a ‘Unloaded edge 5.0d for t/d = 6 Fig. 11.14 Spacing of botts at node joints Spacing between rows ofbolts Parallel to grains, A > (N — 4)d, or 2.5d, which- ever is greater. It is also governed by the net area at critical section which should be 80% of the total area in bearing under all bolts. Perpendicular to grans > 2.5 d for t/d = 2 20.5 d for tid = 6 where, d = diameter of the bolts ¢ = thickness of the main member N = total number of bolts in the joint, A = spacing as shown in Fig. 11.12 and 11.13.Timber Structures 283 End distance —_In Tension 2 5 d for hard wood. 2 7d for soft wood. In Compression 2 4d for all species. Edge distance Parallel to grain loading, B 2 1.5 d, or half the distance between row of bolts, whichever is greater. Perpendicular to grain loading, > 4 d where, B = spacing as shown in Fig. 11.12 and 11.13 For inclined member, the spacing shown for perpendicular and parallel to grain of wood may be used as a guide and the bolts are arranged at the joints with respect to loading direction. Staggering of bolts shall be avoided as far as possible in case of members loaded parallel to grains of wood. Staggering is preferable for loads acting perpendicular to grains of wood to avoid spilling due to weathering effects. Forcible driving of bolts is avoided by providing 1 mm oversizes prebore holes. The design of bolted joints will be governed by the area required for bearing on timber. The allowable bearing compressive pressure Gj, or 0), on timber under a bolt as per LS. Handbook SP: 33 (S&T) - 1986 is as follow. For parallel to grain loading, Oj, = Oy, ky, (11.13) For normal to grain loading, Oj, = Oy ki, ki2 dn.l4) where, ,,, Or G,,= permissible compressive stress on timber parallel or normal to grains. (Table 11.2) k,, and ky, = modification factors (Table 11.10 and 11.11) For inclined grain loading similar to Eq. 11.7, permissible bearing stress op, at an angle 6 to grains is given by Eq. 11.15. Fp X Fon Ope = (11.15) ©, Sin’ O+6,,.cos” 8 Table 11.10 Modification factor k,, Ld Parallel Perpendicular — L/d Parallel Perpendicular to grain __to gain to grain to grain 1.0 1.0 10 55 0.72 0.487 15 1.0 0.962 6.0 0.65 0.456 2.0 1.0 0.88 65 0.58 0.43 25 10 0.8 1.0 0.52 0.406 3.0 10 0.725 15 0.46 0.39 3.5 1.0 0.662 8.0 04 0.375 4.0 0.96 0.6 8.5 0.36 0.362 45 0.9 0.556 90 0.34 0.343 5.0 0.8 0.518 95 0.315 0.33 L = Length of bolt in main member.‘ug 284 Design of Steel Structures Table 11.11 Modification factor k,» (Diameter factor) Diameter of bolt (mm) 6 10 12 16 20 22 25 kip af 26 355 315 305 30 29 Example 11.8 Design a lengthening bolted joint for a timber truss compression mem- ber carrying 12 kN in an inside location. The thickness of the member is 3.5 cm. Timber used is teak. Solution: From Table 11.5, for teak 0, = 8.8 N/mm? Using 10 mm @ bolts, length of bolt in main member = 3.5 mm For Lid = 3.5/1 = 3.5, modification factor, k;, = 1.0 (Table 11.10) Permissible bearing stress parallel to grain, Oy, = O,, X ky, = 8.8 N/mm? Safe load per bolt = o,,, x diameter x length of bolt = 8.8 x 10 x 35 = 3080 N 12000 Number of bolts ired = ——— = 4 bolt lumber of bolts required = Soe is Fig. 11.15 shows the arrangement of the bolts. Example 11.9 Determine the safe load a 10 mm $M.S. bolt can carry in a node joint -10 mm 9 bolts 17.5 mm thick fish plate 35mm }=— 12kN 70 '40!40!40! 70 1 70 T40 '40 T40! 70 1” Fig. 11.15 of a roof truss consisting of 3.5 cm thick tintber members in an inside location. Solution: Safe compression stress perpendicular to grains in Teak from Table 11.5, 0.4 = 4 N/mm? For, Lid = 35/10 = 3.5, factor k,, = 0.662 (Table 11.10) From Table 11.11, diameter factor ky) = 3.6 Safe load on bolt normal to grain loading = (zy ky, kig) XL xd = 4x 0,662 x 3.6 x 10 x 35 = 3336.5 N Example 11.10 Design the bolted joint and the members for joint 1 of a bridge truss shown in Fig, 11.16,Timber Structures 285 15m Wii™ 3720 kN 20 kN [*--———__ 4@ 1.5m ¢/e ——————__+ Fig. 11.16 Solution: Using Deodar wood, the permissible stresses Fy from Table 11.5, for wet outside location, E=9.5 x 10° N/mm’, o, = 7N/mm? Op = 5.6 Nim’, Oi, = 1.7 Nimm?. 1 Aa Wonsidering free body diagram in Fig. 11.17 for joint 1. 30 kN 30 XL Fy=0, Fin = Fa = 42.426 KN Fig. 11.17 = 42426 N (compression) D F,=0,Fy3 = Fy, x cos 45° = 30000 N (tension) Force = 4285.7 mm? Memier 13: Area required = Sane Try 75 x75 mm section. If 16 mm dia. bolts are used in one row, Net area = (75 - 16) x 75 = 4425 mm? > 4285.7 mm? which is all right Member 12; Assuming solid column of 75 mm thickness, Slenderness ratio, L/d = 2121.32/75 = 28.28 = E 9500 _ k= 0.702 oe 0.702)-2F = 28.91 Since 11 < Lid < ke, it is an intermediate column, 4 o,,x{1-4x{—£ 3 ligxd 4 5.6 x [ -4>(33) = 3.89 N/mm? Permissible compressive stress w 3 (28.91 42426 : Area required = ——— = 10906.4 mm’ —286 Design of Steel Structures Provide 75 x 150 mm section, area = 11250 mm? Bolted connection: Using 16 mm dia. M.S. bolts, Lid for bolts = 75/16 = 4.69 For parallel to grain loading, k,; = 0.9 ~ (0.9 - 0.8) x 2 = 0.862 For normal to grain loding, k,, = 0.556 — (0.556 - 0.518) x “ = 0.54 For normal to grain loading, k, = 3.15 For parallel to grain loading, 0, = ,, ki; 5.6 X ky) 0.862 = 4.83 N/mm? For normal to grain loading, oj, = 0, ky, ky2 = 1.7 x 0.54 x 3.15 = 2.89 N/mm? For loading inclined at 45° to grains, permissible bearing stress Gs bp X Fon He Opp Sin? 8 + Fy, COO = 4.83 x 2.89 © 4.83 x sin? 45° + 2.89cos? 45° = 3.16 N/mm? Permissible load on bolt = 3.61 x 16 x 75 = 4332 N Bolts required for member 12 = 42426/4332 = 9.8 say 10 bolts Bolts required for member 13 = 30000/4332 = 6.9 say 7 bolts Provide the bolts as shown in Fig. 11.18. 35mm 2-40 mm thick fish plate on either sides Lj 7-16 mm ¢ bolts @ 65mm c/c 125mm Fig. 11.18m6 = wp | kK ——__—_ wz ——____| | mounosy smo om ware & sony siso3 om UL 2/9 Wd 9] ® STeAu's'prd Guru QZ sean ‘sped ouivs O21 ofS wo 9, 9/9 W99 ® sow | ; pd wo 7S ww QF] —> tua ve] gum 0z-ZI eles Be ee ee Ee PP 6 ww oz—z1 4 usu 9 x WU 000Z +) sp wozr=d(z) it yousg=d(1) ft aretd qa _ : sau |, * |e) wut 9x Ox OL WSET ; | spd gumoz | 6 4 ie + | +) p a ag ee =e PM JO Sais 0G UO: wu 01, | ee ee = = ; | anton ao LK. 2p 61 ® + H + |e | a ww Oz i 7 oe ee = r =F ; Fd y al i i | */*) wus gt x O€1 x OL YSE@ \ Resa aaa ae ese a seth eee eset ee tplse as eslees sehfea senses cease Touyns Suureag t id sou gam Jo sapts yi0q uo said axeyd san arp] imoyZnonp smos om UT a ‘yo ino eorrasoatyy, yo ind eons1oau, wut Z1 x 0S) 1x06 d yskz 3 WU Z1 ® N4%00p saa 's'p'd bunts QZ : 2p) wi 9OZ | @ gam Jo sapis syed s9ytiy ww 71 GE SIeUIDITE UO UNL 9x Spx OL VSEL ww Ot x SLx S7LVSEE (spud yi0q we parol) s9Uygns [eOTUDA, sous Supeag ”Timber Structures 287 This type of joint can be used for temporary timber truss bridge which is to carry light load. Joints of timber truss bridges for heavy loads require special attention as they tend to nectors Q1L1 Q11.2 Q3 Qua QULs Q116 QUL7 Qs Qu become loose due to vibrations, reversal of loads and weathering. Steel con- and seats are provided at joints for durability. PROBLEMS Design a timber beam of Deodar for a clear span of 5 m to carry u.dl. of 15 kN/m excluding its self weight, What will be the size of permissible defect in the beam. Assume suitable data if required. Design a fitched beam using 2.5 cm thick Eucalyptus planks and two mild steel plates of 0.5 cm thickness to carry a u.d.l. of 8000 N over a span of 5 m. Find the moment of resistance of chir timber beam of the cross-sections as follows. (i) Rectangular beam 20 x 40 cm. (ii) T beam of 20 cm wide flange, 40 cm deep overall, 5 cm thick, (iii) Square beam 20 x 20 cm, with load acting along a diagonal. Design a circular Deodar column 4 m long to carry an axial load of 250 KN. Design the column for Q.11.4. as a built up column using 5 cm thick planks of sal. Design a spaced compression members for a roof truss to carry 20 KN using chir planks 2.5 cm thick. The length of the compression member is 2.5 m. Design lengthening joint for a 150 x 150 mm teak wood member to cerry a compression of 25 KN. Resolve Q. 11.7 if the member has also to carry tension due to stress reversal. Calculate safe load for the joint shown in Fig. 11.14 if the thickness of middle member is 3.5 cm and the thickness of side members is half of the middle (horizontal) member. Diameter of bolts is 10 mm and timber used in teak.