01d_Structural Steel_Engineering and Design

Core beyond web Core in web FIGURE 35 Interaction diagram for axial force and moment The interaction diagram is readily analyzed by applying the following relationships: dPldg =fyt\ dMpldg = -V2fytg\ ' dPldMp = -2Ig This result discloses that the change in slope along CB is very small, and the curvature of this arc is negligible Replace the true interaction diagram with a linear one Draw a vertical line AD = QA5Py, and then draw the straight line CD (Fig 35) Establish the equation of CD Thus, slope of CD = -Q.85Py/Mp; P = - 0.85/y^J/M^, or Mp = 1.18(1-PXPpM, The provisions of one section of the AISC Specification are based on the linear interaction diagram Ascertain whether the data are represented by a point on AD or CD; calculate M'p> accordingly Thus, Py = Afy= 13.24(36) = 476.6 kips (2.119.92 kN); PlPy = 84/476.6 = 0.176; therefore, apply the last equation given in step Thus, Mp = 55.0(36)712 = 165 ft-kips (223.7 kN-m); Mp = 1.18(1 - 0.176)(165) = 160.4 ft-kips (217.50 kN-m) This result differs from that in part a by 4.6 percent Load and Resistance Factor Method Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction, Inc., writing in Theory and Problems of Structural Steel Design, McGrawHill, states "In 1986 a new method of structural steel design was introduced in the United States with the publication of the Load and Resistance Factor Design Specification for Structural Steel Buildings Load and resistance factor design, or LRFD, has joined the old allowable stress design (ASD) method as a recognized means for the design of structural steel frameworks for buildings "Although ASD has enjoyed a long history of successful usage and is familiar to engineers and architects, the author and most experts prefer LRFD because it is a truer representation of the actual behavior of structural steel and unlike ASD, it can provide equivalent margins of safety for all structures under all loading conditions For these reasons its anticipated that LRFD will replace ASD as the standard method of structural steel design." The following selected procedures in this handbook cover structural steel design for buildings using the load and resistance factor design (LRFD) method drawn from the excellent Rokach book listed above And competent authorities on the LRFD method, listed below, are cited frequently in the Rokach book, and in this handbook, usually in abbreviated form: AISC: American Institute of Steel Construction, Inc., Chicago, IL AISC LRFD Specification' Load and Resistance Factor Design Specification for Structural Steel Buildings, published by AISC AISC LRFD Manual: Load and Resistance Factor Design Manual of Steel Construction, also published by AISC Equations in the following calculation procedures in this handbook are numbered as follows Those equations appearing in the AlSO LRFD Specification are accompanied by their AISC numbers in parentheses, thus (); other equations are numbered in brackets, thus [ ] It is recommended that the designer have copies of both the AlSO LRFD Specification and the AISC Manual on his or her desk when preparing any structural steel design using the LRFD method Both are available from the AISC at E Wacker Dr, Suite 3100, Chicago IL 60601 Abraham J Rokach writes, further, in his book cited above, "The ASD method is characterized by the use of one judgemental factor of safety A limiting stress (usually Fy) is divided by a factor of safety (FS, determined by the authors of the Specification) to arrive at an allowable stress Allowable stress = Fy/FS Actual stresses in a steel member are calculated by dividing forces or moments by the appropriate section property (e.g area or section modulus) The actual stresses are then compared with the allowable stresses to ascertain that Actual stress = allowable stress No distinction is made among the various kinds of loads Because of the greater variability and uncertainty of the live load and other loads in comparison with the dead load, a uniform reliability for all structures is not possible " Briefly, LRFD uses a different factor for each type of load and another factor for the strength or resistance Each factor is the result of a statistical study of the variability of the subject quantity Because the different factors reflect the degrees of uncertainty in the various loads and the resistance, a uniform reliability is possible." DETERMININGIFAGIVENBEAM IS COMPACT OR NON-COMPACT A designer plans to use a W6 x 15 and a W12 x 65 beam in (a) A6 steel (Fy = 36 ksi [248 MPa)], and (b) with Fy = 50 ksi [344.5 MPa]) and wishes to determine if the beams are compact or non-compact Calculation Procedure: For the W6 x 15 beam Analyze the W6 x 15 beam Refer to the AISC Manual table, namely "Limiting Width-Thickness Ratios for Beams" and its illustration "Definition of widths (b and h) and thickness", the flanges of a W shape are given by 65 A Xp= VFy where A^ = limiting width-thickness ratio for compact section Substituting for each of the two beams, we have { —P= = 10.8 if Fy = 36 ksi (248 MPa) 65 —7= =9.2 if Fy = 50 ksi (344.5 MPa) Compute the data for the web of a W shape Using the same equation as in Step 1, for the web of a W shape r 64p J Xp ~VFy-\ [ 640 —/= = 106.7 if Fyv = 36 ksi (248 MPa) V36 640 V50 =90 ' lf Fy =5°ksl (344'5 M?a) Determine if the beam is compact From the Properties Tables for W Shapes, in Part of the AISC LRFD Manual (Compact Section Criteria): for a W6 x 15 b bf flange = ^f = 11.5 f hc web —=21.6 *W Since flange (bit = 11.5) > (A7, 10.8), the W x 15 beam is noncompact in A36 steel Likewise, it is noncompact ifFy = 50 ksi (344.4 Mpa) For the WU x 65 beam Compute the properties of the beam shape From the AISC Manual "Properties Tables for W Shapes", for a W12 x 65 b bf flange & - = -^ = 9.9 t 2tf h web —c = 24.9 tw (a) In A3 steel flange A = 10.8 See W6 x 15 web A^ = 24.9 Since flange (bit = 9.9) < (Ap = 10.8), and web (hjtw = 24.9) < (\p = 106.7), a W12 x 65 beam is compact in A3 steel (b) However, if Fy = 50 ksi (344.5 MPa) flange A.P = 9.2 See W6 x 15 web A^ = 90.5 Because flange (bit = 9.9) > (\p = 9.2), a W12 x 65 beam is noncompact if Fy = 50 ksi (344.5 MPa) Related Calculations: The concept of compactness, states Abraham J Rokach, MSCE, AISC, relates to local buckling Cross-sections of structural members are classified as compact, noncompact, or slender-element sections A section is compact if the flanges are continuously connected to the web, and the width-thickness ratios of all its compression elements are equal to, or less than, A^ Structural steel members tih compact sections can develop their full strength without local instability In design, the limit state of local buckling need not be considered for compact members This procedure is the work of Abraham J Rokach, MSCE, AISC, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor DETERMINING COLUMN AXIAL SHORTENING WITH A SPECIFIED LOAD A WlO x 49 column, 10ft (3m) long, carries a service load of 250 kips (113.5 Mg) What axial shortening will occur in this column with this load? Calculation Procedure: Choose a suitable axial displacement equation for this column The LRFD equation for axial shortening of a loaded column is Pl Shortening, A = —K where A = axial shortening, in (cm); P = unfactored axial force in member, kips (kg); / = length of member, in (cm); E = modulus of elasticity of steel = 29,000 ksi (199.8 MPa); Ag = cross sectional area of member, sq in (sq cm) Compute the column axial shortening Substituting, Pl 250 kips x (IQ.O ft x 12 in/ft) Shortening, A = — = 29?000 kips/in2 x 14 in2 = 0.072 in (0.183 cm) Related Calculations: Use this equation to compute axial shortening of any steel column in LRFD work This procedure is the work of Abraham J Rokach, MSCE, American Institute of Steel Construction DETERMINING THE COMPRESSIVE STRENGTH OFA WELDED SECTION The structural section in Fig 36a is used as a 40-ft (12.2-m) column Its effective length factor Kx = Ky=LQ Determine the design compressive strength if the steel is A3 Calculation Procedure: Choose a design compressive strength The design compressive strength is given by: 4JPn = WrA8 The values of ^f cr can be obtained from the Table, "Design Stress for Compression Members of 36 ksi Specified Yield-Stress Steel, = 0.85" in the AISC Manual, ifKl/r is known With Kl = 1.0 x 40.0 ft x 12 in/ft = 480 in (1219 cm), then r= v? A = (18 in) - (17 in)2 = 35.0 in2 Ix = Iy = I= ^ in)2-2 °7in)4 - 1788 in* (225.8 sq cm) Find the Kl/r ratio for this section With the data we have, r= /1788 in4 r-r = 7.15 in V 35.0 in4 KI _ 480in T-TUhT 67 - t = I in (typical) FIGURE 36 Determine the design compressive strength of this section Using the suitable AISC Manual table, namely "Design Stress for Compression Members of 36 ksi Specified Yield-Stress Steel, c/>c = 0.85," and interpolating, for KlIr = 67.2, cFcr = 24.13 ksi (166.3 Mpa) the design compressive strength ^fn = 24.13kips/in2 x 35.0 in2 = 845 kips (3759 kN) Related Calculations This procedure is the work of Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor DETERMINING BEAM FLEXURAL DESIGN STRENGTH FOR MINOR- AND MAJOR-AXIS BENDING For a simply supported W24 x 76 beam, laterally braced only at the supports, determine the flexural design strength for (a) minor-axis bending and (Z?) major-axis bending Use the "Load Factor Design Selection Table for Beams" in Part of the AISC LRFD Manual Calculation Procedure: Determine if the beam is a compact section The W24 x 75 is a compact section This can be verified by noting that in the Properties Tables in Part of the AISC LRFD Manual both bf/2tfand hjtw for a W24 x 76 beam are less than the respective flange and web values of \p for Fy = 36 ksi (248 MPa) Find the flexural design strength for minor-axis bending For minor- (or y-) axis bending, Mny = Mpy ZyFy regardless of unbraced length (Eq [56]) The flexural design strength for minor-axis bending of a W24 x 75 is always equal to 234 /I tw ^F; C c = JMOO_ " (Mj^ Substituting, we obtain T, K n ^ , r^ 44,000 26,400£ 0^'"G^-^ MF A FIGURE 37 The case of no-stiffeners corresponds to k - Check the original assumptions for doubling the shear strength To double the shear strength, I fc = x = 10 Then in AISC Equation A-G3-4, *=5+cir10 This implies a/h = 1.0 or a= k; thus, the clear distance between transverse web stiffeners a = I1 = 55 in (142.2 cm) Checking the original assumption we obtain /A = J6"L = i28.9W234 / I (tw 0.44 in J \ VFy /BL1233) V 36 / O.k Design the stiffener, trying a pair of stiffener plates Stiffener design can be performed thusly Because tension field action is not utilized, the equation Ist > at^j must be satisfied, where '-(S?-"" y-^-2-05 Ist > 56 in x (0.44 in)3 x 0.5 - 2.34 in4 (97.4 cm4) Try a pair of stiffener plates, 2.5 in x 0.25 in (6.35 x 0.635 cm), as in Fig 38 The moment of inertia of the stiffener pair about the web centerline = St 0.25 in x (5.44 in)3 J2 = 33g ^4 > ^ ^4 ok (139.4 cm4 > 97.4 cm4) o.k Web Stiffeners FIGURE 38 Try a single stiffener plate Using the plate in Fig 39, which is 3.5 x 0.25 in (8.89 x 0.635 cm), the moment of inertia of the stiffener about the face of the web is 4= 0.25 in x (3.5 in)3 ^ = 3.57 in4 > 2.34 in4 o.k (148.6 cm4 > 97.4 cm4) o.k Related Calculations This procedure is the work of Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor DETERMINING THE DESIGN MOMENT AND SHEAR STRENGTH OFA BUILT-UP WIDE-FLANGE WELDED BEAM SECTION For the welded section in Fig.37a (selected from the table of Built-Up Wide-Flange Sections in Part of the AISC LRFD Manual), determine the design moment and shear strengths Bending is about the major axis; Cb = 1.0 The (upper) compression flange is continuously braced by the floor deck Steel is A36 Calculation Procedure: Check the beam compactness and flange local buckling Web Stiffener FIGURE 39 Working with the "Flexural Strength Parameters" table in the Appendix of the AISC LRFD Specification, the compactness of the beam (for a doubly symmetric I shape bending about its major axis) should first be checked: Flange A_*_£_JiJn 9.o t 2tf x in For the definition of b for a welded I shape, see the AISC LRFD Manual Flange L-W-&"* For the flange, A < Ap Therefore, the flange is compact, and Mn* = Mpx for the limit state of flange local buckling (FLB) Web Ap p = ^ = 7*L»L = 128.o tw V XiS in Web 640 640 970 970 A= ^TVsT106-7 A= v/TV36 = 161-7 Web For the web, (Ap = 106.7) < (A = 128.0) < (Ar = 161.7) The web is noncompact: Mn < M^ < Mpx; for the limit state of web local buckling (WLB); M^ is determined from AISC LRFD Manual Eq (A-F 1-3) Analyze the lateral bracing relating to the limit state of lateraltorsional buckling (LTB) For this continuously braced member Lb = O; Mn^ = Mpx for LTB Summarizing: Limit State LTB FLB WLB M^ Mpx Mpx MrxQ2 Use formula (Hl-Io), which, for Muy = O, reduces to _^ + ! J^sio $fn Wn* Determine the design flexural strength To determine 0.3D (the overall dimension) = 2.25 in > (0.3 x 6.625 in = 1.99 in) = 2.25 in (5.72 cm) The slenderness parameter A _ Kl [F^ ^ VT V = 10.0 ft x 12 in/ft / 51.4 ksi 2.25 in x TT V 35>744 ksi Because A c < 1.5 Fcr = (0.65V2c)Fmy = 0.658'0-64)2 x 51.4 ksi = 43.2 ksi (297.6 MMPa) The design compressive strength 4,Pn = Wf0, = 0.85 x 5.6 in2 x 43.2 kips/in2 = 205 kips > 200 kips required (J)JPn = 205 kips (911.8 kN) for this case is also tabulated on p 4-100 of the AISC LRFD Manual) The 6-in (15.2 cm) standard-weight concrete-filled pipe-column is satisfactory Related Calculations This procedure is the work of Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor DETERMINING DESIGN COMPRESSIVE STRENGTH OF COMPOSITE COLUMNS Determine the design compressive strength of a W8 x 40 (A36 steel) encased in a 16 x 16 in (40.6 x 40.6 cm) (fcr = 3.5 ksi) (24.1 MPa) normal-weight concrete column in Fig 46 Reinforcement is four No (Grade 60) bars longitudinally, and No ties at 10 in (25.4 cm) horizontally; Calculation Procedure: Check the minimum requirements for the column Checking minimum requirements (a) For a W8 x 40, A89 = 11.7 in2, total area = 16 in x 16 in = 256 in2 (1652 cm2) —: = 4.6% > 4% minimum 256 in o.k (b) Lateral tie spacing = 10 in (25.4 cm) < 2/3 x 16 in outer dimension = 10.7 in (27.2 cm) o.k Minimum clear cover = 1.5 in (3.8 cm) o.k Horizontal No.3 bars: Ar = 0.ll in2 per bar > 0.007 in2 x 10 in spacing = 0.07 in2 (0.45 cm2) o.k Vertical No.7 bars: Ar =0.60 in per bar > 0.007 in2 x 11.4 in spacing = 0.08 in2 (0.52 cm2) f o.k (c) 3.0 ksi < (fc =3.5 ksi) < 8.0 ksi for normal weight concrete o.k (d) Use Fy, = 55 ksi (378.9 MPa) for reinforcement in calculations, even though actual Fyr = 60 ksi (413.4 MPa) for Grade 60 bars FIGURE 46 Determine the modified yield stress and modulus of elasticity Determine Fmy and Em: Ar Ac Fmy = Fy + clFyr — +c2f^—A AS s where Ar = the cross-sectional area of four No longitudinal bars = x 0.6 in2 = in2 (15.5 cm2) A5 = cross-sectional area of W8 x 40 = 11.7 in2 (75.5 cm2) A0 = 16 in x 16 in - (11.7 in2 + 2.4 in2) = 242 in2 (1561 cm2) For concrete-encased shapes, C1 = 0.7 and C2 = 0.6 2 in + 0.6 x 3.5 ksi x ^B_ 242 in Fm, = 36 ksi + 0.7 x 55 ksi x-j^5_ = 87.3 ksi (601.5 MPa) Em = E + c£e^AS where C3 = 0.2 for concrete-encased shapes EC = w1-5 V^ = 1451-5 V3^5 = 3267 ksi (24,577 MPa) for 3.5-ksi normal-weight (145 lb/ft3) (2320 kg/cu m) concrete Em = 29,000 ksi + 0.2 x 3267 ksi x 242 in2/! 1.7 in2 = 42,513 ksi (292,915 MPa) The modified radius of gyration rm = >>(W8 x 4°) ^ 0.3 x 16 in (overall dimension) = 2.04 in > 4.80 in (12.2 cm) = 4.80 in (12.2 cm) The slenderness parameter H c JE^ Rn* V Em = 15.0 ft x 12 in/ft / 87.3 ksi 4.8OmX17 V42>513ksi The critical stress Fcr = (0.658*c2)F^ = 0.658^0-54)2 x 87.3 ksi = 77.2 ksi (531.9 MPa) Compute the design compressive strength The design compressive strength JJPn = Wf0, = 0.85 x 11.7 in2 x 77.2 kips/in2 (531.9 MPa) = 768 kips (5292 MPa) (^0Pn = 768 kips for this case is also tabulated on p 4-73 of the AISC LRFD Manual) The 768-kip design strength is considerably more than the 238-kip (1640 Mpa) design strength of a noncomposite W8 x 40 column under the same conditions Related Calculations This procedure is the work of Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor ANALYZING A CONCRETE SLAB FOR COMPOSITE ACTION A W18 x 40 interior beam is shown in Fig 47 Steel is A36, beam span is 30 ft O in (9.14 m), and beam spacing 10 ft O in (3.04 m) The beams are to act compositely with a 5-in (12.7-cm) normal-weight concrete slab;/,' = 5.0 ksi (41.3 kN) Determine: (a) The effective width of concrete slab for composite action; (b) Vh (the total horizontal shear force to be transferred) for full composite action; (c) The number of 0.75-in (1.9-cm) diameter shear studs required if Fu = 60 ksi (413.4 kN) Calculation Procedure: Find the effective width of concrete slab for composite action For an interior beam, the effective slab width on either side of the beam centerline is the minimum of -| - -^p = 3.75 ft = 45 in (114.3 cm) I -iMIU 5.00 ft (1.52m) The effective slab width is x 45 in = 90 in (228.6 cm) Determine the total horizontal shear force for full composite action In positive moment regions, Vh for full composite action is the smaller of 0.85/c'^c = 0.85 x ksi x (90 in x in) = 1913 kips (8509 kN) FIGURE 47 Afy = 11.8 in2 x 36 ksi = 425 kips (1890 kN) Vh = 425 kips (189OkN) Find the number of shear studs required The nominal strength of a single shear stud [from Eq (/5-7)] is Qn = Q.5AscVftEc32.7 in2 required for bearing on concrete) =AI A1 123 in2 " T™^ = 15-oin(38-lcm) By increasing the length of bearing of the beam to 15 in (38.1 cm), the bearing plate can be eliminated Related Calculations This procedure is the work of Abraham J Rokach, MSCE, Associate Director of Education, American Institute of Steel Construction SI values were prepared by the handbook editor P ART HANGERS, CONNECTORS, AND WIND-STRESS ANALYSIS In the following Calculation Procedures, structural steel members are designed in accordance with the Specification for the Design, Fabrication and Erection of Structural Steel for Buildings of the American Institute of Steel Construction In the absence of any statement to the contrary, it is to be understood that the structural-steel members are made of ASTM A36 steel, which has a yield-point stress of 36,000 lb/in2 (248.2 MPa) Reinforced-concrete members are designed in accordance with the specification Building Code Requirements for Reinforced Concrete of the American Concrete Institute DESIGN OFAN EYEBAR A hanger is to carry a load of 175 kips (778.4 kN) Design an eyebar of A440 steel Calculation Procedure: Record the yield-point stresses of the steel Refer to Fig for the notational system Let subscripts and refer to cross sections through the body of the bar and through the center of the pin hole, respectively Eyebars are generally flame-cut from plates of high-strength steel The design provisions of the AISC Specification reflect the results of extensive testing of such members A section of the Specification permits a tensile stress of 0.60/J, at and 0.45/J, at 2, where fy denotes the yield-point stress ... as the standard method of structural steel design. " The following selected procedures in this handbook cover structural steel design for buildings using the load and resistance factor design (LRFD)... successful usage and is familiar to engineers and architects, the author and most experts prefer LRFD because it is a truer representation of the actual behavior of structural steel and unlike ASD,...Structural Steel Buildings Load and resistance factor design, or LRFD, has joined the old allowable stress design (ASD) method as a recognized means for the design of structural steel frameworks