Metal Structures Lecture XIII Steel trusses

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1 Metal Structures Lecture XIII Steel trusses

2 Contents Definition #t / 3 Geometry and cross-sections #t / 7 Types of truss structures #t / 15 Calculations #t / 29 Example #t / 57 Results of calculations - verification #t / 79 Death weight of truss #t / 90 Examination issues #t / 92

3 Definition Truss theory (idealization): Straight bars only; Hinge joints; Forces in joints only; Schwedler-Żurawski formula for straight bar: d M(x) / dx = Q(x) d Q(x) / dx = q(x) Photo: Author Des #2 / 2 Forces in joints only no loads along bar (q(x) = 0): q(x) = 0 Q(x) = const = C M(x) = C x + A Hinges: M(0) = 0 A = 0 ; M(L) = 0 C = 0 There are axial forces only M(x) = 0 ; Q(x) = 0

4 Truss real: Bars with imperfections; No ideal hinge joints; Gravity along bars; Photo: Author It s rather frame

5 Effort (calculation as a truss) Effort (calculation as a frame) Time of calculations (truss) << Time of calculations (frame) Because of these reasons, we calculate real truss as ideal truss.

6 Chord Web members = brace members = cross bars Chord Photo: waldenstructures.com There is common name for every not-chord bars (web member or brace member or cross bar). In Polish language exist two words: słupki (vertical bars) and krzyżulce (inclined bars). Common mistake is to design different type of cross-section only because of different name or direction - for web members important are only axial forces and lengths, not names.

7 There are many different shapes of trusses. Geometry and cross-sections Photo: steelconstruction.info Photo: tridenttruss.com Photo: e-plytawarstwowa.pl

8 Photo: domgaz.com.pl Photo: konar.eu Photo: i435.photobucket.com Photo: community.fansshare.net

9 Specific subtype are trusses with paralell chords. For these trusses, forces in top and bottom chord are similar. N Ed, top / N Ed, bottom 0,90 1,10 Photo: waldenstructures.com Photo: gillmet.com.pl

10 Initial drawings - geometry Des #2 / 9 Photo: Author h = L (1/10 ~ 1/15) H = L (1/5 ~ 1/10) a 5 o 30 o b 60 o or b 90 o max (h 1 ; h 2 ) 3,40 m max (L 1 ; L 2 ) 18,00 m max (L 3 ; L 4 ) 12,00 m

11 Cross-sections of truss members PN B EN 1993 Elements Each types of cross-sections are accepted Joints No additional requirements Additional requirements many types of cross-sections are not accepted

12 Chords Modern types of cross-sections (EN) Old types of cross-sections (PN-B) Web members Photo: Author

hermetically close the ends by welding to prevent

Such type of corrosion can develop without

13 Photo: architectureau.com Whenever we use hollow sections, we must hermetically close the ends by welding to prevent corrosion inside HS. Such type of corrosion can develop without apparent external symptoms and can cause damage of structure without warning signs. C D D D Photo: Błędy wykonawcze podczas realizacji konstrukcji stalowych, Litwin M, Górecki M, Budownictwo i Architektura 4 (2009) 63-72

14 Number of different cross-sections, designed in truss: 2-5: Chords Top and bottom - the same I-beam or Two different I-beams, first for top, second for bottom Top and bottom - the same RHS or Two different RHS, first for top, second for bottom Top and bottom - the same CHS or Two different CHS, first for top, second for bottom Brace members 1-3 different RHS 1-3 different CHS 1-3 different RHS 1-3 different CHS 1-3 different CHS

15 Truss purlins Types of truss structures Classical trusses Multi-chords trusses Truss-grates Truss-frames Photo: Author Spatial trusses Laced columns Span [m]

16 Truss purlins Photo: structural-steelbuilding.com Photo: CoBouw Polska Sp. z o. o.

17 Truss purlin #7 / 37 "Normal" truss - forces are applied in nodes; there are axial forces in chords and cross-bars only. Photo: construdare.com Truss purlin - continuous load from roofing; there are axial forces in chords and crossbars; in addition top chord is bending. Photo: Author

18 Wind Dead weight Snow Imposed load (D + S + I) cos a + W I-beam purlin: bi-axial bending. (D + S + I) sin a a #7 / 38 Truss purlin: horizontal force W sin a has very small value and can be neglected (acts on roofing, not purlins). All loads act in plane of truss. There is need wedge to install truss purlin in vertical position. Wind Dead weight Snow Imposed load D + S + I + W cos a W sin a a Photo: Author

19 Classical trusses Photo: steelconstruction.info Photo: domgaz.com.pl 2-chords flat trusses, used as a single roof girders. Photo: waldenstructures.com

20 Multi-chords trusses Photo: multimetalgb.ca Photo: steelconstruction.info

21 3 or 4 trusses connected each other in triangle or square cross-section. Often used for temporary structures (bandstand) or for masts and towers. Photo: conference-truss-hire.co.uk Photo: rktruss.com Photo: stretchtents.com.au Photo: eioba.pl

22 Truss-grates Photo: qdjinfei.en.made-in-china.com Photo: cdn8.muratorplus.smcloud.net

23 Complex of trusses of the same height, perpendicular each other. Photo: Author

24 Frame-truss "Classical" portal frame (two columns and roof girder), but made of trusses, not I-beams. Photo: bbsc303.arch.school.nz Photo: wikipedia

25 Spatial trusses Photo: urwishengineers.com Photo: miripiri.co.in

26 2 or 3 layers of bars + cross-bars between layers. Photo: civiltech.ir Photo: shreeengineering.in/

27 Photo: urwishengineers.com Layers can be flat, cylidrical or spherical. More information about this type of structures will be presented on II nd step of studies. Photo: wikipedia Photo: cnxzlf.com

28 Laced columns Photo: zksgrzelak.eu Structure looks like truss, but its behaviour and calculation model is completely different than for truss. For laced columns, battened columns and closely spaced build-up compression members we use specific algorithm of calculations #t / 43

29 Calculations N Ed / N b,rd 1,0 N b,rd = χ A f y / γ M0 N Ed / N t,rd 1,0 N t,rd = A f y / γ M1 Photo: Author χ - lecture #5: Flexural buckling Torsional buckling Flexural-torsional buckling

30 Critical length for compressed chord of truss is depend on plane of analysis. For vertical plane, critical length is equal distance between cross-bars. For horizontal plane, critical length is equal distance between bracings: Photo: Author

31 Critical length for compressed chord of truss is depend on plane of analysis and part of structure. For web members, critical length = distance between nodes. Flexural buckling of chords: Des #2 / 34 Photo: Author Top chords in compression; buckling in plane: critical length = distance between nodes

32 Flexural buckling of chords: Des #2 / 35 Photo: Author Top chords in compression; buckling out of plane: critical length = distance between horizontal bracings

33 Flexural buckling of chords: Des #2 / 36 Photo: Author Bottom chords in compression; buckling in plane: critical length = distance between nodes

34 Flexural buckling of chords: Des #2 / 37 Photo: Author Bottom chords in compression; buckling out of plane: critical length = distance between vertical bracings

35 Bracings are very important for behaviour of structures, including for trusses. For calculations, there are used special algorithms and models. More information about bracings will be presented on lecture #15. Photo: Author

36 Photo: Author #5 / 42 The result of calculations is buckling factor c. It is calculated in different way for different cross-sections. Flexural Buckling Flexural, torsional, flexural-torsional (hot rolled I) (welded I) c = c y = c z (only if l cr, y = l cr, z ) c = min( c y ; c z ) c = min( c y ; c z ; c T ; c z, T )

37 Photo: Author The same stiffness on each direction (J y = J z ): Flexural buckling (y); Flexural buckling (z) if l cr, y l cr, z (for example distance between bracings is not the same as between web members).

38 Photo: Author Hot-rolled Different stiffness on each direction (J y J z ): Flexural buckling (y); Flexural buckling (z); Additionally, is possible that l cr, y l cr, z

39 Photo: Author Welded Welded cross-section or less than two axes of symmetry: Flexural buckling (y) (axis u for L section); Flexural buckling (z) (axis v for L section); Torsional buckling; Flexural-torsional buckling; Additionally, is possible that l cr, y l cr, z l cr, T

40 Buckling length ratio μ for truss members: Element Direction Cross-section I H pipe closely-spaced build-up members Chord In plane 0,9 0,9 1,0 1,0 other Out of plane 1,0 0,9 1,0 1,0 Brace In plane 0,9 0,9; 1,0; 0,75 1,0; A 1,0; A Bolted connections Out of plane 1,0 1,0; 0,75 1,0; A 1,0; A Chords parallel and d truss bracing / d chord < 0,6 L sections: _ l eff, i = 0,5 + 0,7 l i _ l eff, n = 0,35 + 0,7 l n i = y, z n = u, v EN BB.1.1 Not recommended type of cross-sections because of requirements for joints

41 Photo: EN fig 1.1 Main axes for differen cross-sections. There is important, that axes are not horizontal and vertical for L-sections.

42 There were used special type of bars for old-type truss: closely spaced build-up compression members. Photo: Author Photo: img.drewno.pl Photo: EN fig 6.13 Special way of calculations; Resistance depends on distance between batten plates (a) and number of their planes.

43 We must analysed initial deformations for members with two or more chords (according to EN p.6.4). This means, these elements are always bent, even if axial force acts only. Photo: EN fig 6.7 M Ed II = e imperf N Ed More information will be presented on lecture #13 and # 19 #6 / 76 laced compression members, battened compression members, closely spaced build-up compression members Photo: EN fig 6.13

44 For these three types of members (laced compression members, battened compression members, closely spaced build-up compression members) we should use special way of calculations. Of course, we can put full geometry (chords and lacing system or batten plates) into computer programm, but each membes consist of many submembers n = 29 n = 41 Photo: Author Photo: Author

There is possible, that in structure we have more than 100 000

45 There is possible, that in structure we have more than sub-members; it means very long term of calculations. Photo: s9.flog.pl

46 Special way of calculations we no analysed sub-members, but one equivalent uniform element of effective geometry. Next, cross-sectional forces, calculated for this member, are recalculated to new values of cross-sectional forces applied to chord and lacing system or batten plates. Photo: EN fig 6.7 Photo: EN fig 6.13 laced compression members lecture #19; battened compression members lecture #19; closely spaced build-up compression members #t;

47 There are four possibilities for closely spaced build-up compression members: 2 or 1 batten planes: Small distance: a 70 i min a 15 i min Big distance: a > 70 i min a > 15 i min i min = i v (for one L section) Photo: Author EN tab. 6.9

48 For small distance: calculations as for uniform cross-sections, i.e.: J u1, J u1, J W, J T according to real cross-section; Flexural buckling (u1); Flexural buckling (v1); Torsional buckling; Flexural-torsional buckling; Additionally, is possible that l cr, u1 l cr, v1 l cr, T Photo: Author

49 No batten plates: calculation as for two independent L section Axial force = 0,5 N Ed Flexural buckling (u); Flexural buckling (v); Torsional buckling; Flexural-torsional buckling; Additionally, is possible that l cr, u l cr, v l cr, T Photo: Author

50 Photo: Author Batten plates exist, but on big distance:

51 Buckling about y - y axis (material axis; axid goes through chords): No imperfections; Global axial force N Ed only; Critical length = length of member; Moment of inertia for cross-section = 2 moment of inertia for one chord; Photo: Author

52 Buckling about z - z axis (immaterial axis; axis goes out of chords): No imperfections; Global axial force N Ed only; Critical length = length of member; Moment of inertia for cross-section = effective moment of inertia for cross-section; Photo: Author

53 Buckling about y 1 - y 1 axis (own axis of chord): Bow imperfections; Axial force N ch, Ed in one chord; Bending moment from imperfections M ch, Ed in one chord; Shear force from imperfections V ch, Ed in one chord; Critical length = distance of battens; Moment of inertia for cross-section = moment of inertia for one chord about axis of one chord; Photo: Author

54 N ch, Ed = N Ed / M II Ed z s A ch / (2 J eff ) M II Ed = N Ed e 0 / [1 - (N Ed / N cr ) - (N Ed / S V )] e 0 = L / 500 N cr = p 2 E J eff, / (m L) 2 Numbers of modules between battens or lacings 3 Length of modules should be equal Recommended is odd number of modules EN

55 EN (6.73), (6.74) S V = min { 24 X / [1 + 4 J ch, v z s / (n J b a )] ; 2 p X } J eff = 2 z s2 A ch + 2 m eff J ch n - number of battens planes X = E J ch,v / a 2 z s - distance between centre of gravity for total cross-section and centre of gravity for one chord X ch - geometical characteristic of one chord J b - moment of inertia for crosssection of batten l ~ l = m L / i 0 i 0 = [J / ( 2 A ch ) ] m eff l / 75 1,0 J = 2 z s2 A ch + 2 J ch EN tab. 6.8

56 V Ed = p M Ed II / (n L) h 0 = 2 z s For chord: V ch, Ed = V Ed / 2 M ch, Ed = a V Ed / 4 For batten: V b, Ed = V Ed a / (2 h 0 ) M b, Ed = a V Ed / 2 Photo: EN fig 6.11

57 Example S 235 L 60x40x5 A ch = A (L 60x40x5) = 4,79 cm 2 J u1 (L 60x40x5) = 19,75 cm 4 ; i u1 = 2,03 cm J v1 (L 60x40x5) = 3,50 cm 4 ; i v1 = 0,85 cm J y1 (L 60x40x5) = 17,2 cm 4 ; i y = 1,89 cm J z1 (L 60x40x5) = 6,11 cm 4 ; i z = 1,13 cm J u (2L 60x40x5) = 93,29 cm 4 ; i u = 3,14 cm J v (2L 60x40x5) = 24,44 cm 4 ; i v = 1,60 cm Photo: Author No batten plates #t / Batten plates in one plane #t / Batten plates in two planes #t / L = 4,000 m d = 8 mm a = 400 mm N Ed = 20,000 kn L 60x40x5 I st class of cross-section

58 No batten plates: calculation for single L-section chord (according to example, #5 / 45): N Ed, 1 = N Ed / 2 = 10,000 kn N Rd = A f y = 112,565 kn m u = m v = m T = 1,00 L 0u = L 0v = L 0T = 4,000 m N cr, u = p 2 EJ u / (m u L 0u ) 2 = 25,584 kn N cr, v = p 2 EJ v / (m v L 0v ) 2 = 4,534 kn J W, J T approximation according to #5 / 41 J W = 0,727 cm 6 J T = 0,375 cm 4

59 z s = 2,73 cm i 0 = (i u2 + i v2 ) = 2,20 cm i s = (i 02 + z s2 ) = 3,50 cm N cr, T = [p 2 EJ w / (m T l 0T ) 2 + GJ T ] / i s2 = 248,036 kn m = min[ (m v / m T ) ; (m T / m v )] = 1,000 x = 1 - (m z s2 / i s2 ) = 0,392 N cr, zt = {N cr, v + N cr, T - [(N cr, v + N cr, T ) 2-4 N cr, v N cr, T x] } / (2 x) = 4,484 kn

60 A f y = N Rd = 112,565 kn l u = (A f y / N cr, u ) = 2,098 l v = (A f y / N cr, u ) = 4,983 l T = (A f y / N cr, T ) = 0,674 l vt = (A f y / N cr, vt ) = 5,010 L 60x40x5 tab. 6.1, 6.2, a u = a v = a T = a vt = 0,49

61 F u = [1 + a u (l u - 0,2) + l u2 ] / 2 = 3,165 F v = [1 + a v (l v - 0,2) + l v2 ] / 2 = 14,085 F T = [1 + a T (l T - 0,2) + l T2 ] / 2 = 0,843 F vt = [1 + a vt (l vt - 0,2) + l vt2 ] / 2 = 14,229 c u = min{1/[f u + (F u2 - l u2 )] ; 1,0} = 0,181 c v = min{1/[f v + (F v2 - l v2 )] ; 1,0} = 0,037 c T = min{1/[f T + (F T2 - l T2 )] ; 1,0} = 0,741 c vt = min{1/[f vt + (F vt2 - l vt2 )] ; 1,0} = 0,036 c = min(c u ; c z ; c T ; c vt ) = 0,036

62 A f y = 112,565 kn c A f y = 4,052 kn N Ed / A f y = 0,089 OK. N Ed / c A f y = 2,212 > 1,000 Wrong, buckling, destruction!

63 Batten plates in one plane; Accordin to #t / 47, limit between small distance and long distance is equal 15 i min = 15 i v = 128 mm; Distance betwen batten plates a = 400 mm a > 15 i min big distance There is even number of modules, but odd number is only recommended, not obligatory. A ch = (L 60x40x5) = 4,79 cm 2 J ch, u = J u1 (L 60x40x5) = 19,75 cm 4 J ch, v = J v1 (L 60x40x5) = 3,50 cm 4 J u = J u (2L 60x40x5) = 93,29 cm 4 J v = J v (2L 60x40x5) = 24,44 cm 4

64 e 0 = L / 500 = 8 mm z s = 2,73 cm m = 1 i 0 = [J v / ( 2 A ch ) ] = 1,60 cm l = m L / i 0 = 250 #t / 55 m eff = 0 J eff = 2 z s2 A ch + 2 m eff J ch, v = 71,40 cm 4 X = E J ch,v / a 2 = 45,938 kn J batten = / 12 = 83,33 cm 4

65 S V = min {24 X / [1 + 4 J ch, v z s / (n J b a )] ; 2 p X } = 288,637 kn N cr = p 2 E J eff / (m L) 2 = 92,491 kn M II Ed = N Ed e 0 / [1 - (N Ed / N cr ) - (N Ed / S V )] = 0,224 knm N ch, Ed = N Ed / M II Ed z s A ch / J eff = 18,203 kn V ch, Ed = p M II Ed / n L = 0,088 kn M ch, v, Ed = V ch, Ed a / 4 = 0,022 knm

66 Photo: Author Axis v goes through chords material axis Axis u goes out of chords immaterial axis

67 v - v flexural buckling only axial force; moment of inertia J v N Ed = 20 kn N Rd = A f y = 2 A ch f y = 225,130 kn L cr,u = L = 4,000 m i v = (J v / A) = 1,60 cm l u1 = (L cr,v / i v ) (1 / l 1 ) = 2,662 Buckling curve c a = 0,49 F v = 4,646 c v = 0,118 N Ed / (c v N Rd ) = 0,753 < 1,000 OK

68 u - u flexural buckling only axial force; moment of inertia = J eff N Ed = 20 kn N Rd = A f y = 2 A ch f y = 225,130 kn L cr,u = L = 4,000 m i u = (J eff / A) = 2,73 cm l = 1,560 Buckling curve c a = 0,49 F = 2,050 c = 0,296 N Ed / (c N Rd ) = 0,431 < 1,000 OK

69 v 1 - v 1 flexural buckling interaction axial force - bending moment interaction flexuraltorsional buckling - lateral buckling N ch, Ed = 18,203 kn M ch, v, Ed = 0,022 knm V ch, Ed = 0,088 kn L cr, v1 = a = 0,800 m L cr, LT = a = 0,800 m m z1 = m LT = 1,0 i v1 = (J v1 / A ch ) = 0,85 cm N ch, Rd = A ch f y = 112,565 kn M ch, Rd, v1 = 0,415 kn V ch, Rd A ch f y / 3 = 64,989 kn

70 N cr, v = p 2 EJ v / (m v a) 2 = 36,079 kn N cr, T = [p 2 EJ w / (m T a) 2 + GJ T ] / i s2 = 249,881 kn m = min[ (m v / m T ) ; (m T / m v )] = 1,000 x = 1 - (m z s2 / i s2 ) = 0,392 N cr, vt = {N cr, v + N cr, T - [(N cr, v + N cr, T ) 2-4 N cr, v N cr, T x] } / (2 x) = 25,889 kn l vt = (A f y / N cr, vt ) = 2,085 a vt = 0,49 F vt = 3,135 c vt = 0,183 M cr = i s (N cr, z N cr, T ) = 2,815 knm l LT = (W y f y / M cr ) = 0,384 F LT = [1 + a LT (l LT -0,2) + l LT2 ] / 2 = 0,605 c LT = 0,932

71 Interaction between flexural-torsional buckling and lateral buckling lecture #18 Rough approximation (EN NA.20): N ch, Ed / (c N Rd ) + M ch, Ed / (c LT M Rd ) 0,8 N ch, Ed / (c N Rd ) + M ch, Ed / (c LT M Rd ) = 0, ,057 = 0,941 > 0,800 Wrong, buckling, destruction!

72 Interaction between V and M lecture #16 Battens resistance, welds between battens and chords lecture #19

73 Batten plates in two planes; Accordin to #t / 47, limit between small distance and long distance is equal 70 i min = 70 i v = 597 mm; Distance betwen batten plates a = 400 mm a < 70 i min small distance J b, u = J u1 (2L 60x40x5) = 93,29 cm 4 J b, v = J v1 (2L 60x40x5) = 24,44 cm 4 J w (2L 60x40x5) = 59,02 cm 6 (calculations according to Mechanics of Materials, thin-walled theory) J T 2 J T (L 60x40x5) = 0,750 cm 4

74 N Rd = 2 A f y = 225,130 kn m u = m v = m T = 1,00 L 0u = L 0v = L 0T = 4,000 m N cr, u = p 2 EJ u1 / (m u L 0u ) 2 = 120,847 kn N cr, v = p 2 EJ v1 / (m v L 0v ) 2 = 31,659 kn z s = 2,73 cm i 0 = (i u2 + i v2 ) = 2,20 cm i s = (i 02 + z s2 ) = 3,50 cm N cr, T = [p 2 EJ w / (m T l 0T ) 2 + GJ T ] / i s2 = 495,740 kn m = min[ (m v / m T ) ; (m T / m v )] = 1,000 x = 1 - (m z s2 / i s2 ) = 0,392 N cr, zt = {N cr, v + N cr, T - [(N cr, v + N cr, T ) 2-4 N cr, v N cr, T x] } / (2 x) = 30,448 kn

75 l u = (A f y / N cr, u ) = 1,365 l v = (A f y / N cr, u ) = 2,667 l T = (A f y / N cr, T ) = 0,674 l vt = (A f y / N cr, vt ) = 2,706 L 60x40x5 tab. 6.1, 6.2, a u = a v = a T = a vt = 0,49

76 F u = [1 + a u (l u - 0,2) + l u2 ] / 2 = 1,717 F v = [1 + a v (l v - 0,2) + l v2 ] / 2 = 4,661 F T = [1 + a T (l T - 0,2) + l T2 ] / 2 = 0,843 F vt = [1 + a vt (l vt - 0,2) + l vt2 ] / 2 = 4,776 c u = min{1/[f u + (F u2 - l u2 )] ; 1,0} = 0,362 c v = min{1/[f v + (F v2 - l v2 )] ; 1,0} = 0,118 c T = min{1/[f T + (F T2 - l T2 )] ; 1,0} = 0,741 c vt = min{1/[f vt + (F vt2 - l vt2 )] ; 1,0} = 0,114 c = min(c u ; c z ; c T ; c vt ) = 0,114

77 A f y = 225,130 kn c A f y = 25,844 kn N Ed / A f y = 0,089 OK. N Ed / c A f y = 0,774 < 1,000 OK.

78 Summary: No batten plates: effort 2,212 > 1,000 Battens in one plane (big distance): effort 0,941 > 0,800 1,176 > 1,000 Battens in two planes (small distance): effort 0,774 < 1,000 Conclusions: Batten plates in one planes is more effective than no batten plates, but the best are two planes of batten plates

79 Results of calculations - verification F i F ch-up F wm F ch-down Sx = 0 ; F wm / F ch-down 0 Photo: Author F ch-down F ch-up

80 q Photo: Author There is possible to estimate force in top and bottom chords; we use beam-analogy q = S F i / L M max = q L 2 / 8 or q L 2 / (8 cos 2 a) F ch-down F ch-up M max / H

81 Photo: Author Similar shape of axial forces in chord and bending moment for single-span beam; the same for axial forces in web members and shear force in single-span beam.

82 But this estimation is not true for symmetrical supports - there are additional axial forces in chords. Photo: Author

83 There is big difference between symmetrical supports and unsymmetrical supports (approximation) Photo: Author

84 N top : Photo: Author N c, Ed, sym 1,5 N c, Ed, unsym N bottom : N t, Ed, sym 0,5 N t, Ed, unsym

85 Photo: Author There is big horizontal reaction for symmetrical support. Because of this reaction, truss will behave as for unsymmetrical supports after short time of exploatation (deformations of columns, local destruction of masonry or concrete structure around anchor bolts). Calculation as for symmetrical supports means, that forces taken under consideration in design project are completely different than real forces in structure: for bottom chord real are much more bigger than theoretical. This means big probability destruction of bottom chord and collapse total structure. Better way is to model the truss with unsymmetrical supports - this is closer its real behaviour.

86 There is difference between two directions of web members Photo: Author

87 Deformations: elongation (tensile force) and abridgement (compressive force): Transmission of forces form chords to supports and zero force members: Photo: Author

88 Stifness of I-beam: J I = J top flange + J web + J bottom flange For symmetrical cross-section: J top flange = J bottom flange 2 (h 1, I / 2) 2 A flange Photo: Autor J I (h 1, I2 / 2) A flange + J web Stifness of truss (Konstrukcje metalowe, M. Łubiński, A Filipiak, W. Żółtowski, Arkady 2000): J truss 0,7 [A top chord A bottom chord / (A top chord + A bottom chord )] (h 1, truss / 2) 2 For symmetrical cross-section: A top chord = A bottom chord = A chord J truss 0,7 [A chord2 / (2 A chord )] (h 1, truss / 2) 2 = 0,7 (h 1, truss2 / 2) A chord

89 Stifness of I-beam: Stifness of truss: J I (h 1, I2 / 2) A flange + J web J I 0,7 (h 1, truss2 / 2) A chord For different types of I-beams, J web = 7% - 25% of J flange For J truss = J I, we need A chord >> A flange and h 1, truss >> h 1, I h 1, I = L / 20 - L / 25 ; h 1, truss = L / 10 - L / 15 h 1, truss > h 1, I A chord - CHS, RHS, I-beam ; A flange - plate A chord >> A flange

90 Death weight of truss I st proposal (PN B 02001): g T = [ 2 / a + 0,12 (g + q)] L / 100 g T, g (roofing + purlins), q (snow + wind) [kn/m 2 ], characteristic values a (distance between trusses), L [m]

91 II nd proposal: F ch-down F ch-up M max / H A ch-down A ch-up = A = M max / (H f y ) g ch-down g ch-up g web members = A d steel L g T = 3 d steel L M max / (H f y )

92 Examination issues Types of truss structures Algorithm of calculation for compressed closely spaced build-up compression members

93 Laced members - słup wielogałęziowy skratowany Battened members - słup wielogałęziowy z przewiązkami Closely spaced build-up members - pręt wielogałęziowy

94 Thank you for attention 2016 Tomasz Michałowski, PhD

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