BEAMS. By.Ir.Sugeng P Budio,MSc 1

Transcription

1 BEAMS B.Ir.Sugeng P Budio,MSc 1

2 INTRODUCTION Beams are structural members that support transverse loads and are therefore subjected primaril to flexure, or bending. If a substantial amount of axial load is also present, the member is referred to as a beam-column. This figure (1) shows both a hotrolled shape and a built-up shape along with the dimensions to be used for the width-thickness ratios. If h t 970 w F the member is to be treated as a beam, regardless of whether it is a rolled shape or built-up. B.Ir.Sugeng P Budio,MSc

3 Figure (1) B.Ir.Sugeng P Budio,MSc 3

4 If h t 970 w F the member is considered to be a plate girder All of the standart hot-rolled shapes found in the Manual are in the first categor (beams). Most built-up shapes will be classified as plate girders, but some will be beams b the AISC definition. For beams, the basic relationship between load effects and strength can be written as M u where : M u controlling combination of factored load moments Ø b resistance factro for beams 0.90 M n nominal moment strength b The design strength Ø b M n is sometimes called the design moment. M n

5 BENDING STRESS AND THE PLASTIC MOMENT B.Ir.Sugeng P Budio,MSc 5

6 Figure () B.Ir.Sugeng P Budio,MSc 6

7 From elementar mechanics of materials, the stress at given point can be found from the flexure formula : f b M I x (1) Where M is the bending moment at the cross section under consideration, is the perpendicular distance from the neutral plane to the point of interest, and I x is the moment of inertia of the area of the cross section with respect to the neutral axis. For maximum stress, Equation (1) takes the following form : Mc M f max I I / c where c is the perpendicular distance from the neutral axis to the extreme fiber, and S x is the elastic section modulus of the cross section. x x M S x ()

8 The stress f maxx must not exceed F, and the bending moment must not exceed M F S x where M is the bending moment that brings the beam to the point of ielding. In figure above, a simpl supporte beam with a concentrated load at midspan is shown at succesive stages of loading. Once ielding begins, the distribution of stress on the cross section will no longer be linier, and ielding will progress from the extreme fiber toward the neutral axis.

9 Figure (3) B.Ir.Sugeng P Budio,MSc 9

10 The plastic neutral axis divides the cross section into two equal areas. For shapes that are smmetrical about the axis of bending, the elastic and plastic neutral axis are the same. The plastic moment M p is the resisting couple formed b the two equal and opposite forces : A Mp F (Ac)a F (A t )a F a F Z where : A total cross-sectional area a distance between the centroids of the two half-areas Z (A/)a plastic section modulus Figure (5)

11 Compute the plastic moment M p for a W10 x 60 of A36 steell Solution From the Dimensions and properties tables : A 17.6 in A in The centroid of half of this are can be found in the tables for WTshapes,which are cut from W-shapes. the relevant shape here is the WT5x30, and the distance from the outside faceof the flange to the centroid is inches (figure above) B.Ir.Sugeng P Budio,MSc 11

12 a Z d (0.884) 10. (0.844) 8.45 A a 8.8(8.45) This compares favorabll with the value of 74.6 given in the Dimensions and properties tables (the difference resulut from roundoff in the tabular values) ANSWER M p F Z 36(74.38) 678 in-kips 3 ft-kips 3 in in

13 Classification Of Shapes AISC classifies cross-sectional shapes as compact, noncompact, or slender. depending on the values of the width-thickness ratios. For I- and H- shapes, the ratio for the projecting flange (an unstiffened element) is b f /t f, and the ratio for the web ( a stiffened element) is h/t w. It can be summarized in a general wa as follows : Let λ width : thickness ratio λ p upper limit for the compact categor λ r upper limit for noncompact categor B.Ir.Sugeng P Budio,MSc 13

14 Width Thickness parameters ( for hot-rolled I- and H-shapes Element λ λ p λ r Flange bf 65 F t f 141 F 10 Web h t w 640 F 970 F B.Ir.Sugeng P Budio,MSc 14

15 Then if λ λ p and the flange is countinuousl to the web, the shape is compact if λ p < λ λ r, the shape is noncompact ; and if λ > λ r, the shapes is slender The categor is based on the worst width-thickness ratio of the cross section. For example, if the web is compact and the flange is noncompact, the shape is classified as noncompact. Table has been extracted from AISC Table and contains width-thickness raios for hotrolled I- and H- shapes cross sections. B.Ir.Sugeng P Budio,MSc 15

16 Bending Srength Of Compact Shapes A beam can fail b reaching M p and becoming full plastic, or it can fail b buckling in one of the following was : 1. Lateral-torsional buckling (LTB), either elasticall or inelasticall. Flange local buckling (FLB), elasticall or inelasticall 3. Web local buckling (WLB), elasticall or inelasticall We begin with compact shapes, defined as those whose webs are continuousl connected to the flanges and that satisf the following width-thickness ratio requirement for the flange and the web : b t f f 65 F and h t w 640 F B.Ir.Sugeng P Budio,MSc 16

17 The first categor, laterall supported compact beams, is quite common and is the simple case. AISC gives the nominal strength as : M n M p where : M p F Z<1.5M The limit of 1.5 M for M p is to prevent excessive working load deformations and is satisfied when : F Z 1.5F S or Z S 1.5 B.Ir.Sugeng P Budio,MSc 17

18 The beam shown in Figure above is a W16 x 31 of A36 steel. It supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange. The service dead load, including the weight of the beam, is 500 lb/ft. The service live load os 550 lb/ft. Does this beam have adequate moment strength? Solution For simpl supported, uniforml loaded beam, the maximum bending moment occurs at midspan and is equal to M max 1 8 wl B.Ir.Sugeng P Budio,MSc 18

19 where w is the load in units of force per units length, and L is the span length. M M D L (30) wdl 56.5 ft kips (30) ft kips 8 Since the dead load is less than 8 times the live load, load combination controls : M u 1.M D + 1.6M L 1.(56.5)+1.6(61.88)166 ft-kips Alternativel, the loads can be factored at the outset : w M u u 1.w 1 8 w u D L + 1.6w L 1.480(30) 8 1.(0.500) + 1.6(0.550) ft kips kips/ft B.Ir.Sugeng P Budio,MSc 19

20 Check for compactness: b t f 65 F f h t 6.3 w a F W16x31 (for all is 6.3 the shapes) compact flange for A36 Since the beam is compact and laterall supported, is compact steel M n M p F Z x 36(54) 1944 in-kips 16 ft-kips B.Ir.Sugeng P Budio,MSc 0

21 Check for M p < 1.5M Z S x x b M n OK (16) 146ft kips 166 ft kips N.G Answer : Since the design moment is less than the factored load moment, the W16x31 is unsatisfactor B.Ir.Sugeng P Budio,MSc 1

22 The moment strength of compact shapes is a function of the unbraced length L b, defined as the distance between points of lateral support, or bracing. In this book, points of lateral support will be indicated b an x, as shown in Figure (10), The relationship between nominal strength M n and unbraced length is shown in next Figure (11). B.Ir.Sugeng P Budio,MSc

23 If the unbraced length is nograter than Lp, to be defined presentl, the beam is considered to have full lateral support, and Mn Mp. If Lb is greater than Lp but less than or equal to the parameter Lr, the strength is based on inelastic LTB. If Lb is graterthan Lr, the strength is based on elastic LTB. The equation for the theoretocal elastic latera-torsional buckling strength can be found in the Theor of Elastic stabilit and, with some notational changes, is as follows : B.Ir.Sugeng P Budio,MSc 3

24 M n π L b EI G πe + Lb I C w Figure (11) B.Ir.Sugeng P Budio,MSc 4

25 where : L b unbraced length (in.) G shear modulus 11,00 ksi for structural steel J torsional constant (in. 4 ) C w warping constant (in. 6) If the moment is greater than that corresponding to first ield, the strength is based on inelastic behaviour. The moment corresponding to first ield is taken as : M r F L S x where F L is the smaller of (F f F r ) or F w. In this expression, the flange ield stress is reduced b F r, the residual stress. For a nonhbrid member, F f F w F, and F L will alwas vbe equal to F F r. For example, AISC Equation above will be written as M r (F F r )S x B.Ir.Sugeng P Budio,MSc 5

26 As shown in Figure, the boundar between elastic and inelastic behavior will be for an unbraced length of L r, which is the value of L b obtained from Equation when M n is set equal to M r. The following equation is obtained : L r X X 1 r ( F F ) π S x 4C I X w 1 r 1+ EGJA Sx GJ 1+ X ( F F ) r where As with columns, inelastic behavior of beams is more complicated than elastic behavior, and empirical formulas are often used. With one minor modification, the following equation is used b AISC : B.Ir.Sugeng P Budio,MSc 6

27 where M L n p M p 300 F r ( M M ) p r L L b r L L p p The nominal bending strength of compact beams is completel described b Equation subject to an upper limit of M p for inelastic beams, provided the applied moment is uniform over the unbraced length L b. Otherwise there is a moment gradient, and Equation must be modified b a factro C b. this factro is given in AISC as C b.5m max 1.5M max + 3M + 4M A B + 3M C B.Ir.Sugeng P Budio,MSc 7

28 where : M max absolute value of the maximum moment within the unbraced length (including the end points) M A absolute value of the moment at the quarter point of the unbraced length M B absolute value of the moment at the midpoint of the unbracedlength M C absolute value of the moment at the three-quarter point of the unbraced length. when the bending moment is uniform, the value of C b is 1.5M C b 1.0.5M + 3M + 4M + 3M B.Ir.Sugeng P Budio,MSc 8

29 Determine C b for a uniforml loaded, simpl supported beam with lateral support at its ends onl. Because of smmetr, the maximum moment is at midspan : M M max B 1 8 wl Also because of smmetr, the moment at the quarter point is equal to the moment at three-quarter point. From Figure ; B.Ir.Sugeng P Budio,MSc 9

30 1.5M C wl wl 8 wl 8 L 4 wl 4 L wl M M max b C A Answer : C b 1.14 B.Ir.Sugeng P Budio,MSc M 4M 3M.5M C C B A max b

31 Figure (1) B.Ir.Sugeng P Budio,MSc 31

32 Figure (13)next shows the value of C b for several common cases of loading and lateral support. For unbraced cantilever beams, AISC specifies a value of C b of 1.0. A value of 1.0 is alwas conservative, regardless of beam configuration or loading, but in some cases it ma be excessivel conservative. The complete specification of nominal moment strength for compact shapes can be summarized. B.Ir.Sugeng P Budio,MSc 3

33 Figure (13) B.Ir.Sugeng P Budio,MSc 33

34 ( ) 1.5 p cr n p b r p b r p p b n p n M M M M L L L L M M M C M M M M For L b L p For L p <L b L r For L b > L r 1 1 ) / ( 1 / b b x b w b b b cr r L X X r L X S C C I L E GJ EI L C M + + π π

35 Figure (14) B.Ir.Sugeng P Budio,MSc 35

36 The ons ants X 1 and X have been previousl defined and are tabulated in the Dimensions and properties Tables The effect of C b on the nominal strength is illustrated in Figure (14). Although the strength is directl proportional to C b, this graph clearl shows the importance of observing the upper limit of M p, regardless of which equation is used for M n B.Ir.Sugeng P Budio,MSc 36

37 Example Solution Determine the design strength Ø b M n for a W14x68 of A4 steel suject to the following conditions. a) Continous lateral support b) Unbraced length0ft;c b 1.0 c) Unbraced length0ft;c b 1.75 a) From Part 1 of the Manual, aw14 x 68 is seen to be in shape group and is therefore available with a ield stress F of 50 ksi Determine whether shape is compact,noncompact,or slender < this shape is compact and B.Ir.Sugeng P Budio,MSc 37

38 M n M p F Z 5750 in kips 479. ft kips x 50(115) Check for M p 1.5M Z S x S x Answer : n M n 0.90(479.)431 ft-kips b) L b 0 ft, and C b 1.0, Compute L p and L r : (.46) Lp in F 50 ft B.Ir.Sugeng P Budio,MSc 38

39 From the torsion Properties Tables, J 3.0 in. 4 C w 5380 in. 6 Although X 1 and X are tabulated in the Dimensions and Properties Tables in Manual, the will be computed here for illustration. B.Ir.Sugeng P Budio,MSc 39

40 X X L 1 r π S x π C 4 I 4 ( F r w EGJA 9,000 (11,00 )(3.01)(0 ) ksi S x GJ X 1 F r.46 (301 ) (50 10 ) ,00 x3.0 ) ( ksi ) X 1 + ( F F r ) (50 10 ) 316.8in 6.40 B.Ir.Sugeng P Budio,MSc 40 ft

41 Since L p < L b < L r, the strength is based on inelastic LTB. M M r n ( F F (50 10)(103) ft kips C b M p r ) S x ( M p M p r ) ( ) ft kips M L L b r L L Answer : Ø b M n 0.90(39.4) 353 ft-kips p p B.Ir.Sugeng P Budio,MSc 41

42 c) L b 0 ft, and C b The design strength for C b 1.75 is 1.75 times the design strength for C b 1.0 M n 1.75(39.4) ft-kips > M p 479. ft-kips The nominal strength cannot exceed M p, therefore, use a nominal strength of M n 479. ft-kips : Answer : Ø M n 0.90(479.) 431 ft-kips B.Ir.Sugeng P Budio,MSc 4

43 BENDING STRENGTH OF NONCOMPACT SHAPES B.Ir.Sugeng P Budio,MSc 43

44 The strength corresponding to both limit states must be computed and the smaller value will control. From AISC, λ p 65 F Where : λ r F 141 F r M r ( F F r ) S x Fr residual stress 10 ksi for rolled shapes These terms have been specialized for nonhbrid beams B.Ir.Sugeng P Budio,MSc 44

45 A simpl supported beam with a span length of 40 feet is laterall supported at its end and is subjected to the following service loads. Dead load 400 lb/ft (including the weight of the beam) Live load 1000/ft If ASTM A57 Grade 50 steel is used, is a W14 x 90 adequate? B.Ir.Sugeng P Budio,MSc 45

46 Factored load moment : w 1. w w 1.(0.400) + 1.6(1.000).080 kips / M u u 1 8 w L u D L.080(40) ft kips ft B.Ir.Sugeng P Budio,MSc 46

47 Determine whether shape is compact, noncompact,or slender λ λ p b f t f 65 F λ r F 141 F r Since λ p < λ < λ r, this is a noncompact shape Check the capacit based on the limit state of flange local buckling. B.Ir.Sugeng P Budio,MSc 47

48 M M M p r n F Z ( F M p 50(157) x 1 (50 10)(143) Fr ) Sx 1 λ λ p ( M p M r ) λr λp 654. ft kips ft kips ( ) ft kips The design strength based on FLB is therefore Ø b M n 0.90(640.5) 576 ft-kips B.Ir.Sugeng P Budio,MSc 48

49 Check the capacit based on the limit state of lateraltorsional buckling.from the Load Factor Design Selection Tables, L p 15, L r 38.4 ft L b 40 ft > L r, failure is b elastic LTB From part 1 of manual I 4 36 in. J 4.06 in. 4 C w 16,000 in. 6 For a uniforml loaded, simpl supported beam with lateral support at the ends, C b 1.14 B.Ir.Sugeng P Budio,MSc 49

50 AISC Equation : M n π πe Cb EI GJ + I C w M p L b L b π x π 9, ,000 (36 )(11,00 )(4.06) + 40(1) x (541) 618.0in kips ft kips (36 )(16,000 ) M p 654. ft-kips > ft-kips (OK) B.Ir.Sugeng P Budio,MSc 50

51 Since < 640.5, LTB controls : Ø b M n 0.90(515.0) 464 ft-kips > M u 416 ft-kips (OK) Answer : Since M u < Ø b M n, the beam has adequate moment strength B.Ir.Sugeng P Budio,MSc 51

52 Shear Strength The shear strength of a beam must be sufficient to satisf the re V u Ø v V n V u maximum shear based on the controlling combination of factored Ø v resistance factor for shear 0.90 V n nominal shear strength Consider the simple beam of Figure(15). At a distance x from the left end and at the neutral axis of the cross section, the state of stress is as shown in part d. since this element is located at the neutral axis, it is not subjected to flexural stress. From elementar mechanics of materials, the shearing stress is given b B.Ir.Sugeng P Budio,MSc 5

53 B.Ir.Sugeng P Budio,MSc 53

54 f v VQ It f v vertical and horizontal shearing stress at the point of interest V vertical shear force at the section under consideration Q first moment, about the neutral axis, of the area of the cross section between the point of interest and the top or bottom of the cross section I moment of inertia about the neutral axis t width of the cross section at the point of interest B.Ir.Sugeng P Budio,MSc 54

55 Taking the shear ield stress as 60% of the tensile ield stress, the stress in the web at failure can be written as V n fv Aw where A w area of the web. The nominal strength corresponding to this limit state is F V n 0.6 F A w For h/t w 418/ F, there is no web instabilit, and V n 0.6F A w For 418/ F < h/t w 53/ F, inelastic web buckling can occur, V n 0.6F A w 418/ h / t w F B.Ir.Sugeng P Budio,MSc 55

56 For 53/ F < h/t w 60, the limit state is elastic web buckling : V n 13,000A ( h / t ) w w A w area of the web dt w d overall depth of the beam B.Ir.Sugeng P Budio,MSc 56

57 B.Ir.Sugeng P Budio,MSc 57

58 Check the beam of Example for shear B.Ir.Sugeng P Budio,MSc 58

59 From example, w u.080 kips/ft, and L 40 ft. A W14x90 with F 50 ksi,is used. For a simpl supported, uniforml loaded beam, the maximum shear occurs at the support and is equal to the reaction : V wu L.080(40) u kips From the Dimensions and properties Tables in Part1 of the Manual, the web width-thickness ratio of a W14x90 is : 418 F h t w B.Ir.Sugeng P Budio,MSc 59

60 Since h/t w is less than 418/ F, the strength is governed b shear ielding of the web V n 0.60 F A w 0.6 F (dt w ) 0.6(50)(14.0)(0.440) kips Ø v V n 0.90(185.1) 167 kips > 41.6 kips (OK) Answer : Since the shear design strength is greater than the factored load shear, the beam is satisfactor B.Ir.Sugeng P Budio,MSc 60

61 Block Shear Figure (18) B.Ir.Sugeng P Budio,MSc 61

62 AISC Block Shear Rupture Strength,gives two equation for the block shear design strength : R R n n [ 0.6F ] Agv + Fu Ant [ 0.6F A + F A ] u nv gt Ø 0.75 A gv gross area in shear (in Figure(18),length AB times the web thickness) A nv net area in shear A gt gross area in tension (in Figure(18),length BC times the web thickness) A nt net area in tension The gorvening equation is the one which has the larger fracture term B.Ir.Sugeng P Budio,MSc 6

63 Determine the maximum factored load reaction, based on block shear, that can be resisted b the beam shown in Figure (19) B.Ir.Sugeng P Budio,MSc 63

64 Effective hole diameter ¾ + ⅛ ⅞ in Gross and net tension areas : A gv (+3+3+3)t w 11(0.300) in. A nv ( x ⅞)(0.300).381 in. AISC Equation ØR n Ø[0.6F A gv + F u A nt ] 0.75[0.6(36)(3.300)+58(0.438)] 0.75[ ]64.1 kips B.Ir.Sugeng P Budio,MSc 64

65 AISC Equation ØR n Ø[0.6 F u A nv + F A gt ] 0.75[0.6(58)(.381)+36(0.375)] 0.75[ ] 7.3 kips The fracture term in AISC Equation is the larger one (that is, 8.86>14.14) so this equation governs Answer : Maximum factored load reaction design strength for block shear 7.3 kips B.Ir.Sugeng P Budio,MSc 65

66 Deflection For the common case of a simpl supported, uniforml loaded beam such as that in Figure, the maximum vertical deflection is given b wl 4 EI The following values are tpical maximum allowable total (service dead load plus service live load) deflections. B.Ir.Sugeng P Budio,MSc 66

67 Plastered construction : Unplastered floor construction : Unplastered roof construction : L 360 L 40 L 180 Where L is the span length B.Ir.Sugeng P Budio,MSc 67

68 Check the deflection of the beam shown in Figure(1). The maximum permissible total deflection is L/40 B.Ir.Sugeng P Budio,MSc 68

69 Maximum permissible deflection L 40 30(1) in Since it is more convenient to express the deflection in inches rather than feet, use units of inches in the following equation : Total service load lb/ft kips/ft B.Ir.Sugeng P Budio,MSc 69

70 Maximum total deflection : wl EI ( / 1,000 )( 30 x 1 (510 ) ) in in. (OK) Answer : the beam satisfied the deflection criterion B.Ir.Sugeng P Budio,MSc 70

71 Design Beam design entails the selection of a cross-sectional shape that will have enough strength and will meet the serviceabilit requirement. As far strength is concerned, flexure is almost alwas more critical than shear, so the usual practice is to design for flexure and the check shear. The design process can be outlined as follows : B.Ir.Sugeng P Budio,MSc 71

72 1. Compute the factored load moment M u. This will be required design strength Ø b M n. The weight of the beam is part of the dead load but is unknown at this point. A value ma be assumed, or the weight ma be ignored initiall and checked after a shape has been selected.. Select a shape that is satisfies this strength requirement. This can be done on one of two was : Assume a shape,compute the design strength,and compare it with the factored load moment. Revise if necessar. The trial shape can be easil selected in onl a limited number of situations (Example) Use the beam design charts in Part 3 of the manual. This is preferred method and will be explained following Example 3. Check the shear strength 4. Check the deflection. B.Ir.Sugeng P Budio,MSc 7

73 Select a standart hot-rolled shape of A36 steel for the beam shown in Figure (). The beam has continuous lateral support and must support a uniform service live load of 5 kips/ft. the maximum permissible live load deflection is L/360. B.Ir.Sugeng P Budio,MSc 73

74 Assume a beam weight of 100 lb/ft w u 1.w D + 1.6w L 1.(0.100) + 1.6(5.000) 8.10 kips/ft 1 M u w u L (30) ft kips required bm n Assume a compact shape. For a compact shape and continuous lateral support, M n M p Z x F 1.5M B.Ir.Sugeng P Budio,MSc 74

75 From Ø b M n M u Ø b F Z x M u Z Tr a W30x108 M u F 913.5(1) 0.90(36) x b 338.3in. This shape is compact as assumed (noncompact shapes are marked as such in the table), and Z x /S x 346/ this is less than 1.5,therefore M n M p as assumed. Since the weight is slightl larger than assumed, the required strength will be recomputed, although the W30x108 has more capacit than originall required, and it will almost certainl be adequate. B.Ir.Sugeng P Budio,MSc 75 3

76 w u 1.(0.108) + 1.6(5.000) 8.130kips ft M u 8.130(30) ft kips From the Load Factor Design selection Table, Ø b M p Ø b M n 934 ft-kips > ft-kips (OK) In lieu of basing the search on the required section modulus, the design strength Ø b M n could be used, since it directl proportional to Z x and is also tabulated. Check shear : V u wu L 8.130(30) 1 kips B.Ir.Sugeng P Budio,MSc 76

77 From the Tables of Uniform Load Constants, Ø v V n 316 kips > 1 kips OK Check deflection : The maximum permissible live load deflection is L/360 30(1)/360 1 in. OK wll EI x (5.000 /1)(30x1) 9,000(4470) in. 1 in. Answer : Use a W30 x 108 B.Ir.Sugeng P Budio,MSc 77

78 TUGAS KELAS STRUKTUR BAJA 1 Oktober 011 Tentukan dimensi profil baja mutu A 50 untuk balok seperti terlihat pada gambar di atas. Lateral support pada balok tersebut hana pada tumpuan saja dan harus memikul beban seperti pada gambar. Lendutan ang diijinkan adalah L/360. B.Ir.Sugeng P Budio,MSc 78