Chapter 05 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Beams By Dr. Jawad Talib Al-Nasrawi University of Karbala Department of Civil Engineering 71
Introduction Flexural Members/Beams: are defined as members acted upon primarily by transverse loading, often gravity dead and live load effects. Thus, flexural members in a structure may also be referred to as: Girders: usually the most important beams, which are frequently at wide spacing. Joists: usually less important beams which are closely spaced, frequently with truss type webs. Purlins: roof beams spanning between trusses. Stringers: longitudinal bridge beams spanning between floor beams. Girts: Horizontal wall beams serving principally to resist bending due to wind on the side of an industrial building, frequently supporting corrugated siding. Lintels: members supporting a wall over window or door openings. Figure(5-1): Common Beam Members 72
Sections Used as Beams Among the steel shapes that are used as beam include: W Shapes, which normally prove to be the most economical beam sections, and they have largely replaced channels and S Sections for beam usage. Channels are sometimes used for beams subjected to light loads, such as purlins, and in places where clearances available require narrow flanges. Another common type of beam section is the open web joist or bar joist. This type of section, which commonly used to support floor and roof slabs, is actually a light shop fabricated parallel chord truss. It is particularly economical for long spans and light loads. Figure(5-2): W-Section as a Beam Bending Stresses The basic design checks for beams includes checking:- Bending, Shear, and Deflection The loading conditions and beam configuration will dictate which of the preceding design parameters controls the size of the beam. When a beam is subjected to bending loads, the bending stress in the extreme fiber is define as:- f b = Mc = M (5-1) I S and the yield moment is defined as:- where f b =Maximum bending stress, M y =Yield moment, F y =Yield stress, M=Bending moment, c =Distance from the neutral axis to the extreme fiber, I =Moment of inertia, and M y = F y S (5-2) 73
S =Section modulus. The above formulation is based on the elastic behavior of the beam. Figure(5-3): Stress Distribution for Bending Members Plastic Moment A plastic hinge occurs when the entire cross section of the beam is at its yield point, not just the extreme fiber. The moment at which a plastic hinge is developed in a beam is called the plastic moment and is defined as:- M p = F y Z (5-3) where M p =plastic moment, and Z =plastic section modulus. Example (5-1): For the built-up shape shown in Figure, determine: (a) the elastic section modulus S and the yield moment M y. (b) the plastic section modulus Z and the plastic moment M z. (c) The shape factor. Bending is about the x-axis, and the steel is A572 Grade 50. 74
Solution: (a) Because of the symmetry, the elastic Neutral axis ( x-axis) is locate at mid depth of the cross section ( the location of the centroid). The moment of inertia of the cross section can be found by using the parallel axis theorem, and the results of the calculations are summarized in Table (5-1). Table (5-1): Calculations of the Elastic Section Modulus Component I (in 4 ) A (in 2 ) d (in) I + Ad 2 Flange 0.6667 8 6.5 338.7 Flange 0.6667 8 6.5 338.7 Web 72 - - 72 Sum 749.4 The elastic section modulus is: S = I c = 749.4 = 107 in 3 7 and the yield moment is: M y = F y S = 50(107) = 5350 in kips = 446 ft kips (b) Because this shape is symmetrical about the x-axis, this axis divides the cross section into equal areas and is therefore the plastic neutral axis. The centroid of the top half-area can be found by the principle of moments. Taking moments about the x-axis ( the neutral axis of the entire cross section) and tabulating the computations in Table (5-2), we get: Table(5-2) Ay (in 3 ) Y ( in) A(in 2 ) Component 52 9 61 6.5 3 8 3 11 Flange Web Sum y = Ay A = 61 = 5.545 in 11 a = 2y = 2(5.545) = 11.09 in and that the plastic section modulus is: The plastic moment is: Z = A a = 11(11.09) = 122 in3 2 75
M p = F y Z = 50(122) = 6100 in kips = 508 ft kips (c) The shape factor of a member cross section can be define as the ratio of the plastic moment to the yield moment. S. F = M p = 508 M y 446 = 1.14 The shape factor equals to 1.5 for rectangular cross sections and varies from about 1.1 to 1.2 for standard rolled beam sections. Example(5-2): Determine M y, M n, and Z for the steel tee beam shown below. Also, calculate the shape factor and the nominal load (w n )that can be placed on the beam for a 12-ft simple span. F y = 50 ksi. Figure(5-6): Beam Details Solution: Elastic Calculations: I = 8(1.5)3 12 A = 8 1 1 + 6(2) = 24 in2 2 12(0.75) + 12(4.5) y = = 2.625 in 24 + 8(1.5)(1.875) 2 + 2(6)3 12 + 2(6)(1.875)2 = 122.6 in 4 S = I c = 122.6 = 25.1 in3 4.875 M y = F y S = 50(25.1) = 104.6 ft kips 12 Plastic calculations (plastic neutral axis is at base of flange): Z = 12(0.75) + 12(3) = 45 in 3 M n = M p = F y Z = 50(45) 12 = 187.5 ft kips 76
Shape Factor = M p or Z M n S = 45 25.1 = 1.79 M n = w nl 2 w 8 n = 8(187.5) = 10.4 k/ft 12 The values of the plastic section moduli for the standard steel beam sections are tabulated in AISC Manual Table 3-2. Classification of Shapes AISC classified cross-sectional shapes as: Compact, Noncompact, and Slender Depending on the values of the width-thickness ratios of the individual elements that form the shape. There are also two type of elements that are defined in the AISC Specification: Stiffened elements, and Unstiffened elements. For I-shape, the ratio for the projection flange ( an unstiffened element) is b f, 2t f and the ratio for the web ( a stiffened element) is h. t w The classification of shapes is found in Section B4 of the specification, Local Buckling, in Table B4-1. it can be summarized as follows: λ=width-thickness ratio λ p =upper limit for compact category λ r =upper limit for noncompact category If λ λ p and the flange is continuously connected to the web, the shape is compact; If λ p < λ λ r, the shape is noncompact; and If λ > λ r, the shape is slender. 77
Example (5-3): Determine the classification of a W18x35 and a W21x48 for F y = 50 ksi. Check both the flange and the web. Solution: From Part 1 of the AISC Manual, get: W18x35 W21x48 b f b f = 7.06 = 9.47 2t f 2t f h h = 53.5 = 53.6 t w t w Flange: Web: λ pf = 0.38 E F y = 0.38 29000 50 = 9.15 > 7.06 the W18x35 flange is compact λ rf = 1.0 E F y = 1.0 29000 50 = 24.0 > 9.47 > 9.15 the W21x48 flange is noncompact λ pw = 3.76 E F y = 3.76 29000 50 = 90.5 > 53.6 the W16x35 and W21x48 webs are compact 78
Bending Strength of Compact Shapes The basic design strength equation for beams in bending is: M u b M n M a M n Ω b ( For LRFD) ( For ASD) where M u = Ultimate Moment, b = 0.9 M a = Allowable Moment, Ω b = 1.67 M n = Nominal bending strength, b M n = Ultimate design bending strength, and M n Ω b = Allowable design bending strength. The nominal bending strength, M n, is a function of the following: 1) Lateral torsional buckling ( LTB), 2) Flange local buckling (FLB), and 3) Web local buckling (WLB). Flange local buckling and web local buckling are localized failure modes and are only of concern with shapes that have noncompact webs or flanges. Lateral torsional buckling occurs when the distance between lateral brace points is large enough that the beam fails by lateral, outward movement in combination with a twisting action (Δ and θ ), as shown in Figure ( 5-7). Figure (5-7) Figure ( 5-8) shows that beams have three distinct ranges, or zones, of behavior, depending on their lateral bracing situation. 79
Figure (5-8) The compact shapes defined as those whose webs are continuously connected to the flanges and that satisfy the following width-thickness ratio requirements for the flange and web: b f 0.38 E h and 3.76 E 2t f F y t w F y If the beam is compact and has continuous lateral support, or if the unbraced length is very short (L b L p ), the nominal moment strength, M n, is the full plastic moment capacity of the shape, M p. M n = M p = F y Z x ( AISC Equ. F2-1) Example (5-4): The beam shown in Figure(5-9) is a W16x31 of A992 steel. Its supports a reinforced concrete floor slab that provides continuous lateral support of the compression flange. The service dead load is 450 lb./ft. This load is superimposed on the beam; it does not include the weight of the beam itself. The service live load is 550 lb./ft. Does this beam have adequate moment strength? Figure(5-9) 80
Solution: Check the compactness: b f 2t f = 6.28 ( From Part 1 of the Manual) h 0.38 E F y = 0.38 29000 50 the flange is compact = 9.15 > 6.28 < 3.76 E ( the web is compact for all shapes in the Manual for F t w F y 65 ksi) y a W16x31 is compact The nominal flexural strength is: M n = M p = F y Z x = 50(54) = 2700 in. kips = 225 ft. kips Compute the max. bending moment: The total service dead load, including the weight of the beam, is: w D = 450 + 31 = 481 lb/ft M max = wl2 M 8 D = 0.481(30)2 = 54.11 ft. kips 8 M L = 0.550(30)2 = 61.88 ft. kips 8 M u = 1.2M D + 1.6M L = 1.2(54.11) + 1.6(61.88) = 164 ft. kips Or w u = 1.2(0.481) + 1.6(0.55) = 1.457 kips/ft M u = wl2 8 = 1.457(30)2 = 164 ft. kips 8 The design strength is: b M n = 0.9(225) = 203 ft. kips > 164 ft. kips the W16x31 is satisfactory The AISC Specification defines the unbraced length at which inelastic lateral-torsional buckling occurs as: L p = 1.76r y E ( AISC Equ. F2-5) F y L p is also the max. unbraced length at which the nominal bending strength equals the plastic moment capacity. The unbraced length at which elastic lateral-torsional buckling occurs is: L r = 1.95r ts E Jc 1 + 1 + 6.76 0.7F y S x h o 0.7F y S x h o E Jc 2 ( AISC Equ. F2-6) 81
where, ( AISC Equ. F2-7) c = 1.0 ( for I-shapes) ( AISC Equ. F2-8a) c = h o 2 I y C w ( for Channel shapes) ( AISC Equ. F2-8b) F y = Yield strength E = Modulus of elasticity J = Torsional constant S x = Section modulus ( x-axis) I y = Moment of inertia ( y-axis) C w = Warping constant, and h o = Distance between flange centroids= d t f. For compact I-shapes and C-shapes when L p < L b < L r, the nominal flexural strength is: M n = C b M p M p 0.7F y S x L b L p M L r L p (AISC Equ. F2-2) p For compact I-shapes and C-shapes, when L b > L r, the nominal flexural strength is: M n = F cr S x M p ( AISC Equ. F2-3) where F cr = C bπ 2 E 1 + 0.078 Jc L b ( AISC Equ. F2-4) L b rts 2 S x h o C b =Moment gradient factor, factor to account for non-uniform bending within the unbraced length L b. 12.5M C b = max R 2.5M max +3M A +4M B +3M m 3.0 ( AISC Equ. F1-1) C where M max =Absolute value of the maximum moment in the unbraced segment, M A =Absolute value of the moment at the ¼ point of the unbraced segment, M B =Absolute value of the moment at the centerline of the unbraced segment, M C =Absolute value of the moment at the 3/4 point of the unbraced segment, R m =Section symmetry factor R m = 1.0, for doubly symmetric members ( I-shapes), R m = 1.0, for singly symmetric sections in single-curvature bending, R m = 0.5 + 2 I yc 2 I y r ts 2, for single symmetric shapes subjected to reverse curvature bending, and I yc =Moment of inertia of the compression flange about the y-axis. C b = 1.0, for cantilevers or overhangs where the free end is unbraced. 82
Some of C b values are given in Table 3-1 of the AISC. Example(5-5): Determine C b for the beam shown in Figure(5-10) parts (a) and (b). Assume the beam is a doubly symmetric member. Solution: Figure(5-10) 83
C b = 12.5M max 2.5M max + 3M A + 4M B + 3M C R m 3.0 12.5 1 C b = 8 2.5 1 8 + 3 3 32 + 4 1 8 + 3 3 = 1.14 32 12.5 1 C b = 12 2.5 1 12 + 3 3 96 + 4 1 24 + 3 3 = 2.38 96 Example (5-6): Determine the flexural strength of a W14x68 of A242 steel, Grade 50, subjected to: a) Continuous lateral support. b) An unbraced length of 20 ft with C b = 1.0 c) An unbraced length of 30 ft with C b = 1.0 Solution: Determine whether this shape is compact, noncompact, or slender: b f = 6.97 ( from Part 1 of the Manual) 2t f 0.38 E F y = 0.38 29000 50 = 9.15 > 6.97 the flange is compact The web is compact for all shapes in the Manual for F y 65 ksi a) Because the beam is compact and laterally supported (L b = 0 < L p ), the nominal flexural strength is: M n = M p = F y Z x = 50(115) = 5750 in. kips = 479.2 ft. kips 84
LRFD Solution: The design strength is: b M n = 0.9(479.2) = 431 ft. kips ASD Solution: The allowable moment strength is: M n = 479.2 = 288 ft. kips Ω b 1.67 b) L b = 20 ft and C b = 1.0. First, determine L p and L r : L p = 1.76r y E F y = 1.76(2.46) 29000 50 The following terms will be needed in the computation of L r : = I y C w S x = 121(5380) 103 = 104.3 in = 8.692 ft = 7.833 in 2 r ts = 7.833 = 2.799 in (r ts can also be found in the dimensions and properties tables in Part 1 of the Manual). h o = d t f = 14.0 0.72 = 13.28 in (h o can also be found in the dimensions and properties tables in Part 1 of the Manual). For a doubly-symmetric I-shape, c = 1.0 ( From AISC Equ. F2-6). L r = 1.95r ts E Jc 1 + 1 + 6.76 0.7F 2 y S x h o 0.7F y S x h o E Jc L r = 1.95(2.799) 29000 3.01(1.0) 0.7(50) 103(13.28) 1 + 1 + 6.76 0.7(50) 2 103(13.28) 29000 3.01(1.0) Since L p < L b < L r, then: L r = 351.3 in = 29.28 ft M n = C b M p M p 0.7F y S x L b L p L r L p M p 20 8.692 = 1.0 5750 (5750 0.7 50 103) 29.28 8.692 M n = 4572 in. kips = 381 ft. k < M p = 479.2 ft. k LRFD Solution: The design strength is: b M n = 0.9(381) = 343 ft. kips ASD Solution: The allowable moment strength is: M n = 381 = 229 ft. kips Ω b 1.67 85
c) L b = 30 ft and C b = 1. 0 Since L b > L r = 29.28, then: M n = F cr S x M p F cr = C bπ 2 E L 2 1 + 0.078 Jc L 2 b S x h o r ts b rts ( AISC Equ. F2-3) and LRFD Solution: ASD Solution: F cr = π2 (29000) 2 30 12 2.799 1 + 0.078 3.01(1) 2 12 30 103(13.28) 2.799 F cr = 33.9 ksi M n = 33.9(103) = 3492 in. k = 291 ft. k < M p = 479.2 ft. k b M n = 0.9(291) = 262 ft. kips M n = 291 = 175 ft. kips Ω b 1.67 Non-Compact Shapes The noncompact sections are those that have web-thickness ratios greater than λ p, but not greater than λ r. For the noncompact range, the width-thickness ratios of the flanges of W or other I-shaped rolled sections must not exceed λ r = 1.0 E F y, while those for the webs must not exceed λ r = 5.7 E F y. Other values are provided in AISC Table B4.1b for λ p and λ r for other shapes From AISC F3, Flange local buckling, if λ p < λ λ r, the flange is noncompact, buckling will be inelastic, and M n = M p M p 0.7F y S x λ λ p λ r λ p ( AISC Equ. F3-1) where λ = b f 2t f, λ p = 0.38 E F y, and λ r = 1.0 E F y Example(5-7): A simply supported beam with a span length of 45 ft is laterally supported at its ends and is subjected to the following service loads: Dead load=400 lb./ft ( including beam weight). Live load=1000 lb./ft. If F y = 50 ksi, is a W14x90 adequate? 86
Solution: Determine whether the shape is compact, noncompact, or slender: λ = b f 2t f = 10.2, λ p = 0.38 E F y = 9.15, and λ r = 1.0 E F y = 24.1 Since λ p < λ λ r, this shape is noncompact. Check the capacity based on the limit state of flange local buckling: M p = F y Z x = 50(157) = 7850 in. kips M n = M p M p 0.7F y S x λ λ p λ r λ p 10.2 9.15 M n = 7850 (7850 0.7 50 143) = 7650 in. kips 24.1 9.15 = 637.5 ft. kips Check the capacity based on the limit state of lateral-torsional buckling: From the Z x table: L p = 15.2 ft and L r = 42.6 ft L b = 45 ft > L r failure is by elastic LTB From Part 1 of the Manual: I y = 362 in 4, r ts = 4.11 in, h o = 13.3 in, J = 4.06 in 4, C w = 16000 in 6 For a uniformly loaded, simply supported beam with lateral support at the ends: C b = 1.14 ( AISC Table 3-1) For a doubly-symmetric I-shape, c=1.0 ( AISC Equ.F2-4) F cr = C bπ 2 E L 2 1 + 0.078 Jc L 2 b S x h o r ts b rts F cr = 1.14π2 (29000) 2 1 + 0.078 4.06(1) 2 12 45 45 12 4.11 143(13.3) 4.11 F cr = 37.2 ksi From AISC Equ. F2-3: M n = F cr S x = 37.2(143) = 5320 in. k < M p = 7850 in. k This is smaller than the nominal strength based on flange local buckling, so lateraltorsional buckling controls. LRFD Solution: The design strength is: b M n = 0.9(5320) = 4788 in. kips = 399 ft. kips The factored load and moment are: w u = 1.2w D + 1.6w L = 1.2(0.4) + 1.6(1) = 2.08 k/ft 87
M u = w ul 2 8 = 2.08(45)2 8 = 527 ft. k > 399ft. k N. G Summary of Moment Strength This summary is for compact and noncompact shapes (noncompact flanges) only ( no slender shapes): 1) Determine whether the shape is compact. 2) If the shape is compact, check for lateral-torsional buckling as follows: Using L p = 1.76r y E F y, If L b L p, there is no LTB, and M n = M p If L p < L b L r, there is inelastic LTB, and M n = C b M p M p 0.7F y S x L b L p L r L p M p If L b > L r, there is inelastic LTB, and M n = F cr S x M p Where F cr = C bπ 2 E L b 2 S x h o r ts 1 + 0.078 Jc L b rts 2 3) If the shape is noncompact because of the flange, the nominal strength will be the smaller of the strength corresponding to flange local buckling and lateral-torsional buckling. a) Flange local buckling: If λ λ p, there is no FTB, If λ p < λ λ r, the flange is noncompact, and M n = M p M p 0.7F y S x λ λ p λ r λ p b) Lateral-torsional buckling: Using L p = 1.76r y E F y, If L b L p, there is no LTB, If L p < L b L r, there is inelastic LTB, and M n = C b M p M p 0.7F y S x L b L p L r L p M p If L b > L r, there is elastic LTB, and M n = F cr S x M p 88
Where F cr = C bπ 2 E 1 + 0.078 Jc L b rts 2 Design for Shear Generally, shear is not a problem in steel beams, because the webs of rolled shapes are capable of resisting rather large shearing forces. Perhaps it is well, however, to list the most common situation where shear might be excessive: 1) Should large concentrated loads be placed near beam supports. 2) Probably the most common shear problem occurs where two members ( as a beam and column) are rigidly connected together so that their webs lie in a common plane. 3) Where beams are notched or coped, as shown in Figure(5-11). L b 2 S x h o r ts Figure(5-11) 4) Theoretically, very heavily loaded short beams can have excessive shears, but practically, this does not occur too often unless it is like case 1. 5) Shear may very well be a problem even for ordinary loading when very thin webs are used, as in plate girders or in light-gage cold-formed steel members. From mechanics of materials, the general formula for shear stress in a beam is: f v = VQ Ib where f v = Shear stress at the point under consideration, V = Vertical shear at a point along the beam under consideration, I = Moment of inertia about the neutral axis, and b = Thickness of the section at the point under consideration The AISC specification allows the design for shear to be based on an approximate or average shear stress distribution as shown in Figure(5-12b), where the shear stress is concentrated only in the vertical section of the beam. In the AISC specification, the shear yield stress is taken as 60% of the yield stress, F y. The nominal shear strength of unstiffened or stiffened webs is specified as: V n = 0. 6F y A w C v ( AISC Equ. G2-1) 89
Figure(5-12): Shear in a beam where A w = Area of the web dt w d = Overall depth of the beam C v =Web shear coefficient The value of C v depends on whether the limit state is web yielding, web inelastic buckling, or web elastic buckling. For the special case of hot-rolled I-shapes with: h 2.24 E t w F y The limit state is shear yielding, and C v = 1.0, v = 1.0, Ω v = 1.5 Most W shapes with (F y 50 ksi) fall into this category. Except for the following shapes for (F y = 50 ksi): W12x14, W16x26, W24x55, W30x90, W33x118, W36x135, W40x149, and W44x230 For all other doubly and singly symmetric shapes, except for round HSS v = 0.90, Ω v = 1.67 and C v is determine as follows: For h 1.1 k ve, there is no web instability, and t w F y C v = 1.0 ( AISC Equ. G2-3) For 1.1 k ve F y < h 1.37 k ve, inelastic web buckling can occur, and t w F y C v = 1.1 k ve Fy h tw For h 1.37 k ve, there limit state is elastic web buckling, and t w F y ( AISC Equ. G2-4) 90
1.51Ek v C v = ( AISC Equ. G2-5) h t 2 F w y where k v the web plate shear buckling coefficient, is specified in the AISC Specification G2.1b. For webs without transverse stiffeners and with h < 260: t w k v = 5 Except that k v = 1. 2 for the stem of T-shapes. For all steel shapes, C v = 1. 0, except for the following for F y = 50 ksi : M10x7.5, M10x8, M12,10, M12x10.8, M12x11.8, M12.5x11.6, and M12.5x12.4 Example (5-8): Check the beam in Example(5-7) for shear. Solution : From the dimensions and properties tables in Part 1 of the Manual, the web width-thickness ratio of a W14x90 is: h t w = 25.9 and the web area is: A w = dt w = 14(0.44) = 6.16 in 2 h 2.24 E F y = 2.24 29000 50 = 54 Since < 2.24 E the strength is governed by shear yielding of the web and C t w F v = y 1.0. ( as pointed out in the Specification User Note, this will be the case for most W shapes with F y 50 ksi ). The nominal shear strength is: V n = 0.6F y A w C v = 0.6(50)(6.16)(1.0) = 184.8 kips LRFD Solution: Determine the resistance factor v h Since < 2.24 E t w F v = 1.0 y and the design shear strength is: v V n = 1.0(184.8) = 185 From Example(5-7), w u = 2.08 kips/ft and L=45 ft. For a simply supported, uniformly loaded beam, the max. shear occurs at the support and is equal to the reaction: V u = w ul 2 = 2.08(45) = 46.8 kips < 185 kips O. K 2 Answer: The required shear strength is less than the available shear strength, so the beam is satisfactory. 91
The values of v V nx and V nx Ωv with F y = 50 ksi are given for W shapes in the Manual Table 3-2. A very useful Table(3-6) is provided in Part 3 of the AISC Manual for determining the max. uniform load each W shape can support for various spans. Beam Design Tables The design bending strength of W-shapes and C-shapes with respect to the unbraced length is given in AISC, Tables 3-10 and 3-11, respectively. These tables assume a moment gradient factor of C b =1.0, which is conservative for all cases, and yield strengths of F y =50 ksi for W-shapes and F y =36 ksi for C-shapes. For beams with C b greater than 1.0, multiply the moment capacity calculated using these tables by the C b value to obtain the actual design moment capacity of the beam for design moments that correspond to unbraced lengths greater than Lp. Note that C b M n must always be less than M p. AISC, Tables 3-2 through 3-5 can be used to select the most economical beam based on section properties. AISC, Table 3-2 lists the plastic section modulus, Z x, for a given series of shapes, with the most economical in one series at the top of the list in bold font. AISC, Table 3-6 provides a useful summary of the beam design parameters for W-shapes. The lower part of the table provides values for ØMp, ØMr, ØVn, Lp, and Lr for any given shape. The upper portion of the table provides the maximum possible load that a beam may support based on either shear or bending strength. AISC, Table 3-6 can also be used to determine the design bending strength for a given beam if the unbraced length is between Lp and Lr. When the unbraced length is within this range, the design bending strength is: b M n = b M p BF L b L p where BF is a constant found from AISC Table 3-6. Note that this equation is simpler version of (AISC Equ. F2-2) Deflection The deflections of steel beams are usually limited to certain maximum values. Among the several excellent reasons for deflection limitations are the following: 1) Excessive deflections may damage other materials attached to or supported by the beam in equation. 2) The appearance of structures is often damaged by excessive deflections. 3) Extreme deflections do not inspire confidence in the persons using a structure. 4) It may be necessary for several different beams supporting the same loads to deflect equal amounts. 92
The student should note that deflection limitations fall in the serviceability area. Therefore, deflections are determined for service loads, and thus the calculations are identical for both LRFD and ASD designs. For the common case of a simply supported, uniformly loaded beam, the max. vertical deflection is: = 5 wl 4 384 EI Deflection formulas for a variety of beams and loading conditions can be found in Part 3 of the Manual. Beam Design Procedure The design process can be outlined as follows:- 1) Determine the service and factored loads on the beam. Service loads are used for deflection calculations and factored loads are used for strength design. The weight of the beam would be unknown at this stage, but the self-weight can be initially estimated and is usually comparatively small enough not to affect the design. 2) Determine the factored shear and moments on the beam. 3) Select a shape that satisfies strength and deflection criteria. One of the following methods can be used: a) For shapes listed in the AISC beam design tables, select the most economical beam to support the factored moment. Then check deflection and shear for the selected shape. b) Determine the required moment of inertia. Select the most economical shape based on the moment of inertia calculated, and check this shape for bending and shear. c) For shapes not listed in the AISC beam design tables, an initial size must be assumed. An estimate of the available bending strength can be made for an initial beam selection; then check shear and deflection. A more accurate method might be to follow the procedure in step b above. 4) Check the shear strength. 5) Check the deflection. Example(5-9): For the floor plan shown in Figure (5-13), design members B1 and G1 for bending, shear, and deflection. Compare deflections with L/240 for total loads and L/360 for live loads. The steel is ASTM A992, grade 50; assume that C b = 1.0 for bending. The dead load ( including the beam weight) is assumed to be (85 psf) and the live load is (150 psf). Assume that the floor deck provides full lateral stability to the top flange of B1 Ignore live load reduction. Use the design tables in the AISC where appropriate. 93
Figure (5-13): Floor Plan for Example (5-9) Solution: Since the dead load is more than half of the live load, the total load deflection of L/240 will control. Summary of loads ( see Figure(5-14). Figure(5-14): Loading for B1 and G1 p a = 85 + 150 = 235 psf ( total service load) p u = 1.2(85) + 1.6(150) = 342 psf ( total service load) Design of Beam B1: Tributary width= 6 8 = 6.67 w a = 6.67(0.235) = 1.57 kips/ft w u = 6.67(0.342) = 2.28 kips/ft V u = w ul 2 = 2.28(30) = 34.2 kips 2 M u = w ul 2 8 = 2.28(30)2 = 257 ft. kips 8 From AISC Table 3-10, for: M u = 257 ft. kips and L b = 0, try with W18x40, the most economical size for bending ( b M n = 294 ft. kips, and I = 612 in 4 ) 94
M = wl2 8 = 5wL4 384EI = 5(8M)L2 384EI 8M = wl2 = 40ML2 (1728) 384(29000)I = L(12) 240 = ML2 161.1I ML2 161.1I I req. = ML2 8.056 = wl2 64.44 = 1.57(30)2 = 658 in 4 64.44 The required moment of inertia is greater than the moment of inertia of the W18x40, which is 612 in 4 ; therefore, a new size needs to be selected. From AISC Table 1-1, get: W16 50, I = 659 in 4 W18 46, I = 712 in 4 W21 44, I = 843 in 4 Select W24 55, I = 1350 in 4 Try W21x44 is the lightest: I = 843 in 4 > I req. = 658in 4 O. K From AISC Table 3-10, find, b M n = 358 ft. kips > M u = 257 ft. kips O. K Checking shear, note that a W21x44 does not have a slender web; therefore, the design shear strength is determined from equation G2-1, with C v = 1.0 and v = 1.0: v V n = v 0.6F y A w C v = 1.0(0.6)(50)(0.35)(20.7)(1.0) = 217 kips > V u = 34.2 kips O.K Alternatively, the shear strength can be found from AISC Table 3-6: v V n = 217 kips same as above a W21x44 is selected for member B1 Design of Beam G1: Tributary width= (6 8 )(30) = 200ft 2 P s = 200(0.235) = 47 kips P u = 200(0.342) = 68. 4 kips V u = P u = 68.4 kips M u = P ul 3 = 68.4(20) = 456 ft kips 3 For M u = 456 ft kips and L b = 6.67 ft, and from AISC Table 3-10, select a W24x55 the W24x55 is the most economical size for bending, with b M n = 460 ft kips Checking deflection: = PL3 28EI = 47[20(12)] 3 28(29000)(1350) 95 = 0.593in < L 240 = 1in
Check Shear: The design shear strength is determine from Equation G2-1, with C v = 1.0 and v = 0.9: v V n = v 0.6F y A w C v = 0.9(0.6)(50)(0.395)(23.6)(1.0) = 251 kips > V u = 68.4 kips O.K Alternatively, the shear strength can be found from AISC Table 3-6: v V n = 251 kips same as above a W24x55 is selected for member G1 Webs and Flanges with Concentrated Loads If flange and web strengths do not satisfy the requirements of AISC Specification Section J.10, it will be necessary to use transverse stiffeners at the concentrated loads. These situations are discussed as follow: 1) Local Flange Bending The flange must be sufficiently rigid so that it will not deform and cause a zone of high stress concentrated in the weld in line with the web as shown in Figure (5-15). Figure (5-15) The nominal tensile load that may be applied through a plate welded to the flange of a W section is to be determined by the expression to follow, in which F yf is the specified minimum yield stress of the flange (ksi) and t f is the flange thickness (in): ( AISC Equ. J10-1) Ø=0.9 ( LRFD) and Ω=1.67 ( ASD) It is not necessary to check this formula if the length of loading across the beam flange is less than 0.15 times the width b f or if a pair of half-depth or deeper web stiffeners are provided. 2) Local Web Yielding Local web yielding is illustrated in Figure (5-16). The nominal strength of the web of a beam at the web toe of the fillet when a concentrated load or reaction is applied is to be determined by one of the following two expression, in which k is the distance from the outer edge of the flange to the web toe of the fillet, N is the length of bearing (in) of the force parallel to the plane of the web, F yw 96
is the specified minimum yield stress (ksi) of the web, and t w is the thickness of the web: Figure (5-16) If the force is a concentrated load or reaction that causes tension or compression and is applied at a distance greater than the member depth, d, from the end of the member, then: R n = (5k + N)F yw t w ( AISC Equ. J10-2) Ø=1.00 ( LRFD) and Ω=1.50 ( ASD) If the force is a concentrated load or reaction applied at distance d or less from the member end, then: R n = (2. 5k + N)F yw t w ( AISC Equ. J10-3) 3) Web Crippling When web crippling occurs, it is located in the part of the web adjacent to the loaded flange. Thus, it is thought that stiffening the web in this area for half its depth will prevent the problem. Web crippling is illustrated in Figure (5-17). Figure (5-17) The nominal web crippling strength of the web is to be determined by the appropriate equation of the two that follow ( in which d is the overall depth of the member). If the concentrated load is applied at a distance greater than or equal to d/2 from the end of the member, then: 97
( AISC Equ. J10-4) Ø=0.75 ( LRFD) and Ω=2.00 ( ASD) If the concentrated load is applied at a distance less than d/2 from the end of the member, then: For N 0. 2, d ( AISC Equ. J10-5a) For N d > 0. 2, Ø=0.75 ( LRFD) and Ω=2.00 ( ASD) ( AISC Equ. J10-5b) Ø=0.75 ( LRFD) and Ω=2.00 ( ASD) If one or two web stiffeners or one or two doubly plates are provided and extend for at least half of the web depth, web crippling will not have to be checked. 4) Sidesway Web Buckling Should compressive loads be applied to laterally braced compression flanges, the web will be put in compression and the tension flange may buckle, as shown in Figure (5-18). Figure (5-18) Should members not be restrained against relative movement by stiffeners or lateral bracing and be subject to concentrated compressive loads, their strength may be determined as follows: 98
h tw When the loaded flange is braced against rotation and L b 2. 3, bf Ø=0.85 ( LRFD) and Ω=1.76 ( ASD) ( AISC Equ. J10-6) h tw When the loaded flange is braced against rotation and L b > 2. 3, the limit state bf of web sidesway buckling does not apply. When the loaded flange is not braced against rotation and h tw L b bf 1. 7, ( AISC Equ. J10-7) Ø=0.85 ( LRFD) and Ω=1.76 ( ASD) When the loaded flange is not braced against rotation and h tw L b bf > 1. 7, the limit state of web sidesway buckling does not apply. It is not necessary to check Equations ( J10-6 and J10-7) if the webs are subjected to distributed load. Furthermore, these equations were developed for bearing connections and do not apply to moment connection In these expressions: C r = 960000 ksi when: M u < M y (LRFD) or 1. 5M a < M y ( ASD) at the location of the force C r = 480000 ksi when: M u M y (LRFD) or 1. 5M a M y ( ASD) at the location of the force 5) Compression Buckling of the Web The equation to follow is applicable to moment connections, but not to bearing ones. ( AISC Equ. J10-8) Ø=0.90 ( LRFD) and Ω=1.67 ( ASD) If the concentrated forces to be resisted are applied at a distance from the member end that is less than d/2, then the value of R n shall be reduced by 50 percent. 99
Example (5-10): A W21x44 has been selected for moment in the beam shown in Figure (5-19). Lateral bracing is provided for both flanges at beam ends and at concentrated loads. If the end bearing length is 3.5 inch and the concentrated load bearing lengths are 3 inch, check the beam for web yielding, web crippling, and sidesway web buckling. Solution: Using a W21x44( d = 20. 7 in, b f = 6. 5 in, t w = 0. 35 in t f = 0. 45 in, k = 0. 95 in ) LRFD Figure (5-19) R u = 1.2(1.044 k/ft) 15 + 1.6(35) 2 = 65.4 k Concentrated load P u = 1.6(35) = 56 k ASD R a = (1.044 k/ft) 15 + 35 2 = 42.83 k Concentrated load P a = 35 k Local Web Yielding: N =bearing length of reactions=3.5 in. N =3.0 in. for concentrated loads At end reactions ( AISC Equ. J10-3): R n = (2. 5k + N)F yw t w = (2. 5 0. 95 + 3. 5)(50)(0. 35) = 102. 8 kips LRFD = 1. 00 ASD Ω = 1. 5 R n = 1.0(102.8) = 102.8 k > 65.4 k O.K At concentrated loads ( AISC Equ. J10-2): R n = (5k + N)F yw t w 100 R n = 102.8 = 68.5 k > 42.83 k Ω 1.5 O.K
= (5 0. 95 + 3. 0)(50)(0. 35) = 135. 6 kips LRFD = 1. 00 R n = 1. 0(135. 6) = 135. 6 k > 56 k O.K ASD Ω = 1. 5 R n = 135.6 = 90. 4k > 35 k Ω 1.5 O.K Web Crippling: At end reactions ( AISC Equ. J10-5a) since N 0. 2: d N 3. 5 = = 0. 169 < 0. 2 d 20. 7 1.5 R n = 0. 4(0. 35) 2 3. 5 35 1 + 3 0. 20. 7 0. 45 29000(50)(0. 45) = 90. 3 kips 0. 35 LRFD = 0. 75 R n = 0. 75(90. 3) = 67. 7 k > 65. 4 k O.K At concentrated loads ( AISC Equ. J10-4): 1.5 R n = 0. 8(0. 35) 2 3. 0 35 1 + 3 0. 20. 7 0. 45 LRFD = 0. 75 29000(50)(0. 45) = 173. 7 kips 0. 35 R n = 0. 75(173. 7) = 130. 3k > 56 k O.K ASD Ω = 2. 0 R n = 90.3 = 45. 1k > 42. 83 k Ω 2.0 O.K ASD Ω = 2. 0 R n = 130.7 = 86. 8k > 35 k Ω 2.0 O.K Sidesway Web Buckling: The compression flange is restrained against rotation. 101
h t w lb b f = (20. 7 2 0. 95) 0. 35 5(12) 6. 5 = 5. 82 > 2. 3 Sidesway web buckling does not have to be checked The preceding calculations can be appreciably shortened if use is made of the Manual tables numbered (9-4) and entitled Beam Bearing Constants. In these tables, values are shown for R 1, R 2, R 3, R 1 Ω, R 2 Ω, R 3 Ω, and so on. The values given represent parts of the equations used for checking web yielding and web crippling and are defined on page 9-19 in the Manual. Instructions for use of the tables are provided on pages 9-19 and 9-20 of the Manual. Then, those expressions and the table values are used to check the previous calculations. LRFD ASD R n = R 1 + l b R 2 = 41. 6 + 3. 5(17. 5) = 102. 8 kips R n Ω = R 1 Ω + l b R 2 Ω = 27. 7 + 3. 5(11. 7) = 68. 6 kips Beam Bearing Plates The material used for a beam support can be concrete, brick, or some other material, but it usually will be concrete. This material must resist the bearing load applied by the steel plate. The basic design checks for beam bearing are web yielding and web crippling in the beam, plate bearing and plate bending in the plate, and bearing stress in the concrete or masonry. The nominal bearing strength specified in AISC J8 is the same as that given in the ACI Code 2005. 102
If the plate covers the full area of the support, the nominal strength is: P p = 0. 85f c A 1 ( AISC Equ. J8-1) If the plate does not covers the full area of the support, the nominal strength is: P p = 0. 85f c A 1 A 2 1. 7f A c A 1 1 f c = 28-day compressive strength of the concrete A 1 =bearing area A 2 =full area of the support, A 2 A 1 2. 0 For the design of such a plate, its required area A 1 is:- ( AISC Equ. J8-2) LRFD c = 0. 65 ASD Ω c = 2. 31 R u A 1 = A c 0.85f 1 = Ω cr a c 0.85f c Note that the strength reduction factor given in the AISC specification for bearing on concrete is 0.60. However, ACI 318 recommends a value of 0.65, which will be used here. Plate Thickness:- The required thickness of a 1- inch wide strip of plate can be determined as follows:- Z of a 1-inch wide piece of plate of (t) thickness=(1) t 2 t (2) = t2 4 4 The moments M u and M a are computed at a distance k from the web center line and equated, respectively, to b F y Z and F y Z/Ω c ; the resulting equations are then solved for the required plate thickness. LRFD b = 0. 9 ASD Ω b = 1. 67 103
b M p M u = R u A 1 n n 2 0.9F y t 2 4 R un 2 2BN t 2R un 2 0.9BNF y M p M Ω a = R a n n b A 1 2 F y t 2 4 1.67 R an 2 2BN t 3.34R an 2 BNF y or t 2.22R un 2 BNF y The design procedure for bearing plates can be summarized as follows: 1. Determine the location of the load relative to the beam depth. 2. Assume a value for the bearing plate length, N. 3. Check the beam for web yielding and web crippling for the assumed value of N; adjust the value of N as required. 4. Determine the bearing plate width, B, such that the bearing plate area, A 1 =BN, is sufficient to prevent crushing of the concrete or masonry support. 5. Determine the thickness,t p, of the beam bearing plate so that the plate has adequate strength in bending. Example (5-11): Design a bearing plate to distribute the reaction of a W21x68 with a span length of ( 15 ft 10 inches) center to center of supports. The total service load, including the beam weight, is ( 9 kips/ft), with equal parts dead and live load. The beam is to be supported on reinforced concrete walls with f c = 36 ksi for the plate. Solution by LRFD: w u = 1. 2w D + 1. 6w L = 1. 2(4. 5) + 1. 6(4. 5) = 12. 6 kips/ft And the reaction is:- R u = w ul 12. 6(15. 83) = = 99. 73 kips 2 2 Determine the length of bearing N required to prevent web yielding:- From AISC Equ. J10-3, the nominal strength for this limit state is:- R n = (2. 5k + N)F y t w For R n R u 1. 0[2. 5(1. 19) + N](50)(0. 43) 99. 73 Resulting in the requirement: N 1. 66 in 104
Use AISC Equ. J10-5 to determine the value of N required to prevent web crippling. Assuming N/d > 0. 2 and try the second form of the equation, J10-5b. For R n R u 29000(50)(0.685) 0.43 0. 75(0. 4)(0. 43) 2 1 + 4N 1.5 0. 43 0. 2 21. 1 0. 685 99. 73 This results in the requirement: N 3. 0 in Check the assumption: N d = 3 = 0. 14 < 0. 2 21. 1 For N/d 0. 2, use AISC Equ. J10-5a: For R n R u N. G 29000(50)(0.685) 0.43 0. 75(0. 4)(0. 43) 2 1 + 3 N 1.5 0. 43 21. 1 0. 685 99. 73 Resulting in the requirement: N 2. 59 in, and N 2. 59 = = 0. 12 < 0. 2 O. K d 21. 1 Try N=6 in. determine dimension B from a consideration of bearing strength. If we conservatively, assume that the full area of the support is used, the required plate area A 1 can be found as follows:- c P p R u From AISC Equ. J8-1, P p = 0. 85f c A 1. then, c (0. 85f c A 1 ) R u 0. 65(0. 85)(3. 5)A 1 99. 73 A 1 51. 57 in 2 The min. value of dimension B is:- B = A 1 51. 57 = = 8. 6 in > b N 6 f = 8. 27 (B b f ) Try B=10 in., then compute the required plate thickness:- 105
n = B 2k 2 = 10 2(1. 19) 2 = 3. 81 in t = 2.22R un 2 = 2.22(99.73)(3.81)2 = 1.22 in BNF y 10(6)(36) Use a PL 1 1 4 6 10 If we were to check to see if the flange thickness alone is sufficient, we would have n = b f 2 k = 8.27 2 1.19 = 2.95 in t = 2.22R un 2 = 2.22(99.73)(2.95)2 = 1.04 in > t BNF y 8.27(6)(36) f N. G 106