**A F M r c i c s y r c = C e + C'e' r r C* = Q r r C = <|>A F - C r s y r C where e' = the level arm and is computed using a = n Q r . , t i U.OJq) DI C X C If not adequate, add studs un t i l factored moment resistance i s at least equal to the design bending moment. 107. Effect ive width b, Concrete compression a rea -v i PN.A S tee l compression a r e a , A c r Stee l tension a rea M l /-C.3. of concrete compress ion 0.850 cfa fCG of s t ee l compress ion C r = 0 . 8 5 M c a b , : Q r AsFy-Qr 4>FV t./2 e- L r 2 • T r = c r . c ; • C.G. of steel in tension UJ (a) Plastic Neutral Axis in Steel Flange Effect ive width b, a : Q r / ( O . B S ^ b , ) f C & o f C o n C r e t e c o m P r e s s i o n C.G. of steel compress ion t o l Concrete compress ion area I f t c i f 0.854>cfc PN.A . —y-Stee l compression a r e a , A c r Stee l tension area d, 1^1 *Fy * F V -C;=0.854> c fcab. :Q-r - r maximum axial tension force in the bottom chord r where T is the lesser of (a) T = .9 (area)(F ) r r y (b) T r = .85(.9) (area) F u For Grade 300W ->• F = 450 MPa u For Grade 350W, 380W ->• F = 480 MPa u Note a 5% reduction is applied to the factored tensile resistance to take into account the possibility of local stresses developed by web-chord connection. (4) For composite truss only: During the f i r s t iteration, an analysis for occupany loading i s performed for each t r i a l section by assum-ing the top steel chord is the same as the t r i a l bottom steel chord. 119. (5) For composite truss only: The effective width of concrete top flange is calculated using rules set forward in S16.1 clause 17.3.2. Ensure that the plastic neutral axis (P.N.A.) l i e s within the depth of the cover slab because S16.1 clause 17.4.2 limits the design of composite truss to 100% shear connection with only clause 17.4.3(a) being applicable (see Fig. 6.14). T:——:T Port Elevation 0.85 »„f o f concrete compression a 2_ C; = 0.85»ct;QO, C.G of steel in tension Cross -Section Figure 6.14 Force equilibrium of composite truss. Note: The top steel flange is neglected in the evaluation of force equilbrium. (6) Check the live load, superimposed dead load and deck/slab deflec-tion. For non-composite truss, the moment of inertia i s calculated using only the steel top and bottom chord and is reduced 15% for web deflection due to strain. For composite 120. truss, the moment of inertia i s computed using the transformed concrete top chord plus the steel top and bottom chord. The composite moment of inertia i s reduced 15% for creep and 15% for profiled deck. (7) Repeat Steps 1-6 un t i l up to five sections have been selected for the bottom chord so that later the least cost solution can be found. Select top chord: (Subjected to combined axial and bending forces) (1) Trusses are assumed to be laterally braced by bridging at the one-third points. The program determines the location of the c r i t i c a l top chord panel based on the maximum factored compressive force for each of the 3 load combinations. For composite construction and under occupancy loading, the steel top chord i s assumed to carry no compressive force. (2) A t r i a l section is selected based on the following criterias: (a) section type specified in the input data for truss. (b) If composite condition, the flange thickness must be able to support at least a 1/2" stud. (c) If composite condition, the width of HSS top chord must be atleast 3" wide to assist in shear stud placement. (d) If HSS chords, pick same width for top and bottom chord. (3) Check the adequacy of the t r i a l section under occupancy, concrete placement and deck placement for strength and sta b i l i t y in accord-ance with S16.1 clause 13.8.1 which requires that: Strength check: C^ /C + ttc /M < 1.0 f r f r x x e where C = .9 (area) (F ) 121. = factored compressive force = factored l o c a l moment at panel point x e M = f a c t o r e d moment r e s i s t a n c e based on c l a s s r x d e s i g n a t i o n and unsupported length, and -is ca l c u l a t e d i n accordance with S16.1 clause 13.5 and 13.6. S t a b i l i t y check: ZjQ + to M /M (1-C C/C ) < 1.0 f r x f r f ex x x m where = factored mid panel l o c a l moment xm = factored a x i a l compressive forces = the f a c t o r e d compressive r e s i s t a n c e based on s e c t i o n type and on the l a r g e r e f f e c t i v e s l e n d e r n e s s r a t i o , t h a t i s KL / r or KL / r x x y y (S16.1 clause 13.3.1, 13.3.2). C = e l a s t i c buckling strength ex lu^ = equivalent moment fac t o r equal to 1 M = f a c t o r e d moment r e s i s t a n c e based on c l a s s r x d e s i g n a t i o n and unsupported length, and i s calc u l a t e d i n accordance with S16.1 clause 13.5 and 13.6. Unsupported lengths and L^: During deck placement = panel width, = distance between bridging which Is set at 1/3 span. During concrete placement L = panel width, L =0 x y During occupancy L x = panel width, = 0. (4) Check a l l other top chord panels for strength and s t a b i l i t y f o r the 3 load combination. 1 2 2 . Steps 5-9 for composite truss only: (5) Determine the shear force required for 100% shear connection. In general, Q = 6A F where A = area of the steel bottom chord, r s y s (6) Find the maximum allowable stud diameter based on two criterions. F i r s t l y , S16.1 clause 17.3.5.5 specifies the diameter of a welded stud shall not exceed 2.5 times the thickness of the top steel flange. Secondly, the minimum stud projection above the top of the cellular steel deck is two stud diameter while maintaining a 25 mm concrete cover. (7) Check that the length of the stud i s >75 mm, i f not, the user i s warned to verify the stud capacity. (8) Determine the shear resistance, q^ of a single shear connector. The ultimate shear capacity is a function of components which make up the composite section, along with the orientation of the steel deck with respect to the steel top chord as discussed i n Chapter 1. (9) Determine the number of shear stud required for 100% shear connection. Only single shear stud per rib i s allowed due to the limited flange width of the steel top chord. (10) Repeat steps 2-9 until five sections are selected. The most economical top and bottom chord combinations w i l l be used. Web selection: (1) Select a t r i a l section based on the following requirements: a) section type specified in the input data for truss b) In truss where the chords are hollow structural section, the width of HSS web must be less than the width of both the top and bottom chord 1 2 3 . c) When HSS webs are specified, the width of a l l web members must be the same. d) Where chord are double angles, the gap between angles i s assumed to be 3/8" for the gusset plate when combined with single or double angle webs. e) Only equal leg angles (L's and 2L's) allowed for compression webs. If web is i n tension, check that the slenderness ratio does not exceed 300. Check the adequacy of the t r i a l section against the 3 load cases. The tensional resistance is governed by clause 16.5.7. T i s the lesser of (a) T = .9(area)(F ) r r y (b) T r = .85(.9)(area)(F u) If web is i n compression, check that the slenderness ratio does not exceed 200. For HSS web, check i t s s t a b i l i t y using S16.1 clause 13.8. .95 r where C^ = factored axial compressive force C^ = the factored compressive resistance based on HSS sec-tion and on the larger effective slenderness ratio, that i s KL /r or KL /r (clause 13.3.2). x x y y The 5% reserve capacity is for additional forces resulting from connection eccentricity. For angles L's or 2L's, check i t s strength and st a b i l i t y using CAM3-S136. 124. The general requirement i s : C f •x— < .95 (5% reserve for possible connection eccentricities) r In the above = factored axial compression force C = F (area) r a where: F^ is the average axial stress for compression member under concentric loading given in CAN3-S136, clause 12.6.1 as: a) F < F then F = F p o a p b) F > F then F = 2F - F 2/F p o a o o p where: F^ i s the lesser of the torsional-flexural buckling stress, F^and flexural buckling stress, F^ F = 4r [F +F - /(F +F ) z + 4BF F 1 (clause 12.6.3) st 2B L s t s t s t J F e = <()c(291,000)/Y2 (clause 12.6.4) F = .5QF o Q = local buckling factor defined i n Clause 4.9 F = basic design stress defined in clause 12.2 6.4.3 Trusses Output After the design of each truss, one page of detailed output data i s printed as shown in Fig. 6.15. Before control is transferred to another design module, a truss summary table is printed for a l l truss members designed to that point as shown in Fig. 6.16. A description of each f i e l d in the detailed truss output in Fig. 6.15 follows: TRUSS 2 B i SPAN (mi) - 9000 OUT-OUT STEEL DEPTH (mi) - 560 EFFECTIVE STEEL DEPTH (mi) - 496.5 STEEL MOMENT OF INERTIA (10A6mi4) - 158.32 EFFECTIVE COMPOSITE DEPTH itm) - 666.3 REDUCED COMPOSITE MOMENT OF INERTIA (10A6wi4) - 425.50 STUD DIAMETER USED (mi) - 16 TOTAL NUMBER OF STUDS - 18 for 190Z conn. CONC. UTILIZATION - .27 based on actual b«tto« chord force THEORETICAL PLACE SLAB COMPONENT SECTION LENGTH (m) OCCUP. DECK POUR TOP CHORD - HS76.2X76.2X6.35 9000 .36 .26 .74 .71 .55 .51 .55 .48 .29 .51 .51 .29 .48 .55 .Sl .55 1 TOM CHORD - HS76.2X50.8X6.35 R 9000 .94 .14 i MEMBERS -1 - 2 HS38.1X38.1X4.78 899 .98 . i i 2 - 3 HS50.8X50.8X3.8i 899 .90 .10 3 - 4 HS38.1X38.1X2.54 899 .98 .11 4 - 5 HS38.1X38.1X3.81 89? .86 .09 5 - 6 HS25.4X2S.4X2.54 899 .52 .06 6 - 7 HS25.4X25.4X3.18 899 .91 .10 7 - 8 HS25.4X2S.4X3.i8 899 .91 .10 8 - 9 HS25.4X25.4X2.54 89? .52 .06 9 - 10 HS38.iX38.iX3.31 899 .36 .09 10 - 11 HS38.1X38.1X2.54 89? .98 .11 11 - 12 HS50.8X50.8X3.8i 899 .90 .10 12 - 13 HS33.1X38.1X4.78 899 .98 .11 DECK/SLAB DEFLN (mi) - \ SUPER DEAD DEFLN (mi) -LIVE LOAD DEFLN (««) -GROSS MASS (kg) -GROSS UNIT MASS (kg/n) -DEAD LOAD (kN/n) - .272 20 UTILIZATION - .39 6 UTILIZATION - 8.00 6 UTILIZATION - .26 287 31.9 DECK/SLAB SPECIFICATIONS DECK DEPTH **