Chapter 04 Structural Steel Design According to the AISC Manual 13 th Edition Analysis and Design of Compression Members By Dr. Jawad Talib Al-Nasrawi University of Karbala Department of Civil Engineering 40
Analysis of Compression Members There are three general modes by which axially loaded columns can fail. These are:- 1) Flexural Buckling ( also called Euler Buckling): is the primary type of buckling discussed in this chapter. Members are subject to flexure, or bending, when they become unstable. 2) Local Buckling: occurs when some part or parts of the cross section of a column are so thin that they buckle locally in compression before the other modes of buckling can occur. The susceptibility of a column to local buckling is measured by the width thickness ratios of the parts of its cross section. 3) Flexural Torsional Buckling: may occur in columns that have certain cross sectional configurations. These columns fail by twisting ( torsion) or by a combination of torsional and flexural buckling. Column Critical Buckling Load: There are two types of columns: 1) Short Column: the failure mode is crushing compression as shown in Figure (a). 2) Long Column: the failure mode is buckling at the mid-span of the member as shown in Figure (b). This is called a slender, or long, column, intermediate column fail by a combination of buckling and compression. 41
Euler Formula For a pure compression member, the axial load at which the column begins to bow outward is called the Euler Critical Buckling Load. The Euler critical buckling load for a column with pinned ends is: P e = π2 EI (4-1) L2 Where P e = Elastic critical buckling load, lb., E = Modulus of elasticity, 29000 ksi, I =Moment of inertia, in 4, and L= Length of the column brace points, in Knowing that I = Ar 2 and that the compression stress on any member is f c = P A, we can express the Euler critical buckling load in terms of stress as: F e = π2 E (4-2) L r 2 Where, F e, Euler elastic critical buckling stress, psi, A, Cross sectional area, in 2, r, Radius of gyration, The Euler equations above assumed that the ends of the column are pinned. For other end conditions, an adjustment or effective length factor, K, is applied to the column length. The effective length of a column is defined as KL, where K is usually determined by one of two methods: 1) AISC Table C C2.2, this table is especially useful for preliminary design when the size of the beams, girders, and columns are still unknown. 42
2) Alignment charts ( AISC Table C C2.3 and C C2.4 ), they provide more accurate values for the effective length factor than AISC Table C C2.2, but the process of obtaining these values is more tedious than the first method, and the alignment charts can only be used if the initial sizes of the columns and girders are known. This method will be discussed later. When the column end conditions are other than pinned, equations (4-1) and (4-2) are modified as follows: P e = π2 EI (4-3) (KL) 2 F e = π2 E (KL r) 2 (4-4) The term KL/r is called the slenderness ratio, and the AISC Specification recommends limiting the column slenderness ratio such that: KL r 20000 Example(4-1): a) A W10x22 is used as a 15-ft long pin-connected column. Using the Euler expression, determine the column s critical or buckling load. Assume that the steel has a proportional limit of 36 ksi. b) Repeat part (a) if the length is changed to 8-ft. Solution: a) Using a 15-ft long W10x22 ( A = 6.49 in 2, r x = 4.27 in, r y = 1.33 in ) Mkkn. r = r y = 1.33 kkn L r = 15(12) 1.33 = 135.34 Elastic or buckling stress is: F e = π2 299 1100 3 = 115. 663 ksi < the proportional limit or 36 ksi (1135.34) 2 OK column is in elastic range Elastic or buckling load=15.63(6.49)=101.4 kips b) Using an 8-ft long W10x22, L r = 8(12) 1.33 = 72.18 Elastic or buckling stress is: F e = π2 299 1100 3 = 54. 994 ksi > 366 ksi (662.118) 2 column is in inelastic range and Euler equation is not applicable. 43
Classification of Compression Sections Classification of Compression Sections for Local Buckling: Compression sections are classified as : 1) Non-slender element: is one where the width to thickness ratio (λλ) of its compression elements does not exceed (λ r ) from AISC Table B4.1a. 2) Slender element: is one where the width to thickness ratio (λλ) of its compression elements does exceed (λ r ) from AISC Table B4.1a. The limiting values for (λ r ) are given in AISC Table B4.1a. Note1: If the member is defined as a non-slender element compression member, we should refer to Section E3 of the AISC Specification. The nominal compressive strength is then determine based only on the limit state of flexural buckling. Note2: If the member is defined as a slender element compression member, we should refer to Section E7 of the AISC Specification. The nominal compressive strength shall be taken as the lowest value based on the limit state of flexural buckling, torsional buckling, and flexural torsional buckling. 44
Long, Short, and Intermediate Columns Columns are sometimes classed as being long, short, or intermediate. A brief discussion of each of these classifications is presented below. 1) Long Columns: the Euler formula predicts very well the strength of long columns where the axial buckling stress remains below the proportional limit. Such columns will buckle elastically. The slenderness ratio of long column is greater than 150.( KL r > 11500) 2) Short Columns: For very short columns, the failure stress will equal the yield stress and no buckling will occur. For a column to fall into this class, it would have to be so short as to have no practical application. Thus, no further reference is made to them here. The slenderness ratio of short column is smaller than 40. ( KL r < 400). 3) Intermediate Columns: Some of the fibers will reach the yield stress and some will not. The members will fail by both yielding and buckling, and their behavior is said to be inelastic. Most columns fall into this range. The slenderness ratio of intermediate column is between 40 and 150. ( 400 < KL r < 11500). AISC Requirements The basic requirements for compression members are covered in Chapter E of the AISC Specifications. The LRFD design strength and ASD allowable strength of a column may be determined as follows: P n = F cr A g (AISC Equation E3-1) LRFD Compression Strength ( c = 00. 99) ASD Compression Strength (Ω c = 11. 6666) c PP n = c FF cr AA g PP n = FF craa g Ω c Ω c The following expressions show how (F cr ), the flexural buckling stress of a column, may be determined for members without slender elements: a) If KL 4.71 E or F r F e 0.44F y, then y F cr = (00. 6658) F y Fe F y ( AISC Equation E3-2) 45
b) If KL > 4.71 E or F r F e < 0.44F y, then y F cr = 00. 86666F e ( AISC Equation E3-3) Where F e, elastic critical buckling stress determined according to AISC Equation E3-4. F e = π2 E ( KL r 2 AISC Equation E3-4) Note: AISC Table 4-22, provides values of critical stresses c F cr, and F cr Ω c practical (KL r) values. for 46
Analysis Procedure There are several methods available for the analysis of compression members and these are discussed below. The first step is to determine the effective length, KL, and the slenderness ratio, KL/r, for each axis of the column. Method 1: Use AISC Equations (3-2),(3-3), and (3-4), using the larger of K xl x and K y L y r y. Method 2: AISC available critical stress tables ( AISC Table 4-22). This gives the critical buckling stress, c F cr, as a function of KL r for various values of F y. For a given KL r, determine cf cr from the table using the larger of K xl x r x and K yl y r y. Knowing the critical buckling stress, the axial design capacity can be calculated from the equation, LRFD Compression Strength ( c = 00. 99) ASD Compression Strength (Ω c = 11. 6666) r x c PP n = c FF cr AA g PP n Ω c = FF craa g Ω c 47
Method 3: AISC available compression strength ( AISC Table 4-1 through Table 4-12). These tables give the design strength, c P n, of selected shapes for various effective length, KL, and for selected values of F y. Go to the appropriate table with KL, using the larger of K xl x r x ry and K y L y. Note 1: Ensure that the slenderness ratio for the member is not greater than 200, that is, KL r 20000. Note 2: Check that local buckling will not occur, and if local buckling limits are not satisfied, modify the critical buckling stress, c F cr using AISC Equations (E7-2 and (E7-3). a) If KL r 4.71 E QF y or F e 0.44QF y, then F cr = Q (0.658) QF y Fe F y ( AISC Equation E7-2) b) If KL r > 4.71 E QF y or F e < 0.44QF y, then F cr = 0.877F e ( AISC Equation E7-3) Where F e, elastic critical buckling stress, calculated using AISC Equations E3-4 and E4-4 for doubly symmetric members. AISC Equations E3-4 and E4-5 for singly symmetric members. AISC Equation E4-6 for unsymmetrical members, except for single angles where F e is calculated by E3-4. Q=1.0 for members with compact and non-compact sections, as defined in AISC Section B4, for uniformly compressed elements. = Q s Q a for members with slender element section, as define in AISC Section B4, for uniformly compressed elements. Note 3: Use column load tables ( i.e. the available sections are listed in these tables, but these are typically the most commonly used for building construction. 48
Note 4: AISC Equations (E3-2), (E3-3), and (E3-4) can be used in all cases for column shapes that have no slender elements. Column Analysis Using the AISC Equations Example (4-2): Calculate the design compressive strength of a W12x65 column, 20-ft long, and pinned at both ends. Use ASTM A572 steel. Solution: L x = L y = 20 ft K= 1.0 F y = 50 ksi A g = 19.1 in 2 Obtain the smallest radius of gyration, r, for W12x65 from AISC part 1. r x = 5.28 in, and r y = 3.02 in Use the smaller value, since KL is the same for both axes, KL r = (1)(20)(12) = 79.5 < 200 OK 3.02 Check the slenderness criteria for compression elements: b f = 12 in b = 12 = 6 in 2 t f = 0.605 in t w = 0.39 in h = d 2k des. = 12.1 2(1.2) = 9.7 in Note: k des., used for design, is smaller than k det., used for detailing. b t 0.56 E F y ; h t w 1.49 E F y ; Determine the flexural buckling stress, F cr : Since KL r 6 0.605 = 9.92 < 0.56 29000 = 13.48 OK 50 9.7 0.39 = 24.88 < 1.49 29000 = 35.88 OK 50 4.71 E F y = 4.71 29000 50 = 113.4 = 79.5 < 113.4, use AISC Equation E3-2 to determine F cr. F e = π2 E (KL r) 2 = π2 (29000) (79.5) 2 = 45.3 ksi 49
The design strength of the column is: F y F cr = 0.658F e F y = 0.65845.3 50 50 = 31.5 ksi LRFD, c = 00. 99 c PP n = c FF cr AA g = 0.9(31.5)(19.1) = 541 kkkkkkkk ASD, Ω c = 11. 6666 PP n = FF craa g = (31.5)(19.1) = 360.3 kkkkkkkk Ω c Ω c 1.67 Note 1: From AISC Table 4-22, F cr could be obtained directly by entering the table with KL/r=79.5 and F y = 500 ksi. A value of about F cr = 28. 4 ksi is obtained, which confirms the calculation above. Note 2: Alternatively, the design strength could be obtained directly AISC Table 4-1 (i.e. the column load tables). Go to the table with KL=20-ft and obtain P n = 5411 kips. Analysis of Columns Using AISC Tables Example (4-3): Determine the design compressive strength for a pin-ended HSS8 8 3 8 column of ASTM A500, Grade B steel with an unbraced length of 35-ft. Solution: Unbraced column length, L=35-ft Pin-ended column: K=1.0, KL=(1)(35)=35-ft ASTM A500 steel: F y = 466 ksi For an HSS8 8 3 8 from Part 1 of the AISC, we find: A g = 10.4 in 2, r x = r y = 3.1 in KL = (1)(35)(12) = 135.5 < 200 OK r 3.1 From AISC Table 4-22, with KL/r=135.5, and F y = 466 ksi obtained F cr = 112. 3 ksi; therefore, the design strength is: c P n = c F cr A g = 112. 3(1100. 4) = 1128 kips Alternatively, the design strength could be obtained directly from the AISC column load Table 4-4. Enter the table with KL=35-ft and obtain c P n = 1128 kips Alternate Check: ( b t = 1199. 99 from part 1 of the AISC ) b t 1.4 E ; 19.9 < 1.4 29000 = 35.1 OK F y 46 50
Determine the flexural buckling stress, F cr : 4.71 E = 4.71 29000 = 118.2 F y 46 Since KL = 135.5 > 118.2, use AISC Equation (E3-3) to determine F cr: F e = r π2 E = π2 (29000) (KL r) 2 (135.5) 2 = 15.6 ksi F cr = 0.877F e = 0.877(15.6) = 13.7 ksi c P n = c F cr A g = 00. 99(113. 66)(1100. 4) = 1128 kips Example (4-4): (a): Determine the LRFD design strength c P n and the ASD allowable design strength P n Ωc for the 50 ksi axially loaded W14x90 shown in the Figure. (b): Repeat part (a), using the column tables of Part 4 of the AISC. Solution: (a) From Part 1 and for W14x90, get: AA g = 26.5 kkn 2, r x = 6.14 kkn, r y = 3.7 kkn Determining effective lengths: K x L x = 0.8(32) = 25.6 ft K y L y = 1.0(10) = 10.0 ft K y L y = 0.8(12) = 9.6 ft Computing slenderness ratios: KL = (25.6)(12) = 50.03 r x 6.14 KL = (10)(12) = 32.43 r y 3.7 From AISC Table 4-22, with KL r = 50.03 and FF y = 50 ksi, get: c FF cr = 37.49 kkkkkk and F cr = 24.9 kkkkkk Ω c c PP n = c FF cr AA g = 37.49(26.5) = 993 kkkkkkkk Noting from part (a) solution that there are two different KL values ( KL x = 25.6 ft and KL y = 10 ft ). Equivalent KL y (from KL x )= KL x rx = 25.6 = 15.42 ft 1.66 ry > KL y = 10 ft ( r x get from the bottom of AISC Table 4-1 ) r y 51
Now, from column tables (AISC Table 4-1) with KL y = 15.42 ft, find that c P n = 999911 kips Design Procedures Members not listed in AISC column tables: 1) Calculate the factored axial compression load or the required axial strength, P u, and assume a value for the critical buckling stress, c F cr, that is less than the yield stress, F y. 2) Determine the required gross area, A g req., which should be greater than or equal to P u. c F cr 3) Select a section from Part 1 of the AISC with A g > A g req.. Check that KL r 20000 for each axis. 4) For the section selected, compute the actual c F cr, using AISC Equations (E3-2) or (E3-3). 5) Compute the design strength, c P n = c F cr A g, of the selected shape. If c P n P u, the section selected is adequate; go to step 7. If c P n < P u, the column is inadequate; go to the step 6 6) Use c F cr, obtained in step 4, to repeat steps 1 through 5 until c P n P u. 7) Check local buckling ( see AISC Table B4-1 ). Members listed in AISC column tables: 1) Calculate PP u ( i.e. the factored load on the column). 2) Obtain the recommended effective length factor, K, from AISC Table C-C2.2 and calculate the effective length, KL, for each axis. 3) Enter the column load tables ( AISC Table 4-1 through Table 4-21) with a KL value that is the larger of KL y and KL x rx, and move horizontally until the lightest column ry section is found with a design strength c P n > P u. If is recommended to use the column load tables whenever possible because they are the easiest to use. Example (4-5): Using AISC Tables, select a W14 column of ASTM A572, Grade 50 steel, 14-ft long, pinned at both ends, and subjected to the following service loads: P D = 160 kips and P L = 330 kips 52
Solution: A572, Grade 50 steel: FF y = 50 kkkkkk Pinned at both ends, K=1.0 L=14-ft : KL=(1.0)(14)=14-ft The factored load is: PP u = 1.2PP D + 1.6PP L PP u = 1.2(160) + 1.6(330) = 720 kkkkkkkk From AISC Table 4-1, and for KL=14-ft, find the lightest W14 with c P n > P u. Obtained a W14x82 with c P n = 66664 kips > P u = 66200 kips. Example (4-6): Using AISC Column design Tables, select the lightest column for a factored compression load, P u = 11994 kips, and a column length, L=24-ft. Use ASTM A572, Grade 50 steel and assume that the column is pinned at both ends. Solution: K=1.0 ; KL=(1.0)(24)=24-ft For KL=24-ft, obtain the following c P n from the column load tables ( AISC Table 4-1): Selected Size c PP n, kips W8x?? W8x58 205 W10x?? W10x49 254 W12x?? W12x53 261 W14x?? W14x61 293 Always select the lightest column section if other considerations ( such as architectural restrictions on the max. column size) do not restrict the size. Therefore, use a W10x49 column Example (4-6): Select a W18 column of ASTM A36 steel, 26-ft long, and subjected to a factored axial load of 500 kips. Assume that the column is pinned at both ends. Solution: Since W18 shapes are not listed in the AISC column load tables, thus cannot use these tables to design this column. Procedure for column not listed in the column tables will be followed. PP u = 500 kkkkkkkk ; KL = (1.0)(26) = 26 ft Cycle 1: 1) Assume c FF cr = 20 kkkkkk < FF y 36 kkkkkk. 2) A g req. P u = 500 = 25 c F cr 20 in2. 53
3) Select W18 section from part 1 of the AISC with AA g 25 kkn 2. Try W18x86 with (AA g = 25.3 kkn 2, r x = 7.77 kkn, r y = 2.63 kkn KL = (1.0)(26)(12) = 118.6 < 200 OK r min. 2.63 4) Go to AISC Table 4-22, with KL/r value from step 3 and obtain: c F cr = 115. 5 ksi. 5) c P n = c F cr A g = (15.5)(25.3) = 392 kips < PP u = 500 kkkkkkkk The selected column is not adequate. Therefore, proceed to cycle 2. Cycle 2: 1) Assume c FF cr = 15.5 kkkkkk ( from step 4 of the previous cycle). 2) A g req. P u = 500 = 32.2 c F cr 15.5 in2. 3) Try W18x119 with : AA g = 35.1 kkn 2 > AA g req. r x = 7.9 kkn, and r y = 2.69 kkn KL = (1.0)(26)(12) = 116 < 200 OK r min. 2.69 4) Go to AISC Table 4-22, with KL/r=116 and obtain c F cr = 1166 ksi. 5) The compression design strength is: c PP n = c FF cr AA g = (16)(35.1) = 561 kkkkkkkk > PP u = 500 kkkkkkkk OK Therefore, the W18x119 column is adequate. 6) Check the slenderness criteria for compression elements: b f = 11.3 kkn b = 11.3 = 5.65 kkn 2 t f = 1.06 kkn t w = 0.655 kkn h = d kkkk des. = 19 2(1.46) = 16.08 kkn b t 0.56 E ; 5.65 FF y 1.06 = 5.33 < 0.56 29000 = 15.9 OK 36 h 1.49 E ; 16.08 t w FF y 0.655 = 24.5 < 1.49 29000 = 42.2 OK 36 This implies that the column section is not slender; therefore, the design strength calculated above does not have to be reduced for slenderness effects. Example (4-7): Determine the design compressive strength of the following column: 54
W14x82 column, A572, Grade 50 steel, Unbraced length for strong ( x-x) axis bending=25-ft, Unbraced length for weak ( y-y) axis bending=12.5-ft, Solution: K x L x = (1.0)(25) = 25 ft K y L y = (1.0)(12.5) = 12.5 ft From Part 1 of the AISC, find the properties of W14x82: AA g = 24 kkn 2, r x = 6.05 kkn, r y = 2.48 kkn 1) Using AISC Table 4-22, KL x = (25)(12) = 49.6 < 200 OK r x 6.05 KL y = (12.5)(12) = 60.5 < 200 OK r y 2.48 Going into AISC Table 4-22, with KL/r=60.5, get: c F cr = 34. 4 ksi Column design strength, c PP n = c FF cr AA g = (34.4)(24) 2) Using the column load tables, KL x r = x r y 25 6.05 2.48 = 825 kips = 10.25 ft ( Note: r x r y is also listed at the bottom of the column load table ). KL y = 12.5 ft the larger value govern Solution( Continue): Going into the column load Table 4-1 for W14x82 (F y = 500 ksi) with KL=12.5 ft, get: The compression design strength, c P n, is 828 kips. Note that the column load tables also indicate whether or not the member is slender. For members that are slender, the column load tables account for this in the tabulated design strength; therefore, the local stability criteria does not need to be checked. Alignment Charts Alignment Charts or Nomo graphs: The alignment charts, or Nomo graphs, are an alternate method for determining the effective length factor, K. Two charts are presented in the AISC: Sidesway inhibited 55 KL c P n, kips 12 846 12.5 828 13 810
(i.e., buildings with braced frames or shear walls), AISC Table C-C2.3, and sidesway uninhibited (i.e., buildings with moment frames), AISC Table C-C2.4. 56
The following assumptions have been used in deriving these alignment charts, or Nomo graphs : 1) Behavior is purely elastic. 2) All members have a constant cross section. 3) All joints are rigid. 4) For columns in sidesway-inhibited frames (i.e. braced frames), rotations at opposite ends of the restraint beams or girders are equal in magnitude and opposite in direction, producing single-curvature bending. 5) For columns in sidesway-uninhibited frames, rotations at opposite ends of the restraining beams or girders are equal in magnitude and direction, producing double-or reverse curvature bending. 6) The stiffness parameters, L PP EI, of all columns are equal. 7) Joint restraint is distributed to the column above and below the joint in proportion to EI/L for the two columns. 8) All columns buckle simultaneously. 9) No significant axial compression force exists in the beams or girders. To use these charts, the relative stiffness, G, of the columns, compared to the girders ( or beams) meeting at the joint at both ends of each column, is calculated using the following equation: G = Total column kktkkffnekkkk at the jokknt Total gkkrder kktkkffnekkkk at the jokknt G = τ aei c L c τ gei g L g Where: E c, E g =Modulus of elasticity for columns and girders; I c =Moment of inertia of column in the plane of bending; I g =Moment of inertia of girder in the plane of bending; L c, L g =Unsupported or unbraced length of the columns and girders, respectively; τ a =Column stiffness modification factor for inelasticity from AISC Table 4-21. τ a =1.0 if the nine assumption above are satisfied; and τ g = Girder stiffness modification factor, Table below τ g =1.0 if the nine assumption above are satisfied. 57
Girder Stiffness Modification Factors, τ g Girder Far End Condition Girder Stiffness Modification Factor, τ g Sidesway Uninhibited (i.e. Unbraced or Moment Frames) Far End is Fixed 2 3 Far End is Pinned 0.5 Sidesway Inhibited (i.e. braced Frames) For pinned column base, use G A = 1100 For fixed column base, use G A = 11. 00 Far End is Fixed 2.0 Far End is Pinned 1.5 Example(4-8): For the two-story moment frame shown down, the preliminary column and girder sizes have been determined as shown. Assume in-plane bending about the strong axes for the columns and girders, and assume columns supported by spread footings. The factored axial loads on columns BF and FJ are 590 kips and 140 kips, respectively, and F y = 500 ksi. 1. Determine the effective length factor, K, for columns BF and FJ using the alignment charts, assuming elastic behavior. 2. Determine the effective length factor, K, for columns BF and FJ using the alignment charts, assuming inelastic behavior. 58
Solution: The moments of inertia for the given column and girder sections are as shown: 1) Elastic Behavior Column BF: Joint B: pinned support, G A = 10 Joint F: τ a = 1.0 and τ g = 1.0 Member Section I x (in 4 ) Length (ft) I x L FJ W12x50 391 20 19.6 BF W12x72 597 15 39.8 IJ, EF W18x35 510 20 25.5 JK W18x40 612 30 20.4 FG W18x50 800 30 26.7 τ aei c L G B = c E[(1.0)(39.6) + (1.0)(19.6)] τ = gei g E[(1.0)(25.5) + (1.0)(26.7)] = 1.14 L g Entering the alignment chart For unbraced frames ( AISC Table C-C2.4), with G A = 1100 and G B = 11. 114, joining these two points with a straight line, get: K=1.93 as shown in Figure 59
Column FJ: τ a = 1.0 and τ g = 1.0 Joint F: Joint J: G A = τ aei c L c τ gei g L g L g = E[(1.0)(39.6) + (1.0)(19.6)] E[(1.0)(25.5) + (1.0)(26.7)] = 1.14 τ aei c L G B = c E[(1.0)(19.6)] τ = gei g E[(1.0)(25.5) + (1.0)(20.4)] = 0.43 Entering the alignment chart for unbraced frames ( AISC Table C-C2.4), with G A = 11. 114 and G B = 00. 43, joining these two points with a straight line, get: K=1.93 as shown in Figure above. 2) Inelastic Behavior Column Size A g, in 2 P u, kips P u Ag τ a BF W12x72 21.1 590 28 0.804 FJ W12x50 14.6 140 9.6 1.0 Column BF: Joint B: pinned support, G A = 10 Joint F: τ a = 0.804 and τ g = 1.0 τ aei c L G B = c τ = gei g L g E[(0.804)(39.6) + (1.0)(19.6)] E[(1.0)(25.5) + (1.0)(26.7)] = 0.99 Entering the alignment chart for unbraced frames ( AISC Table C-C2.4), with G A = 1100 and G B = 00. 9999, joining these two points with a straight line, get: K=1.9 as shown below. Column FJ: Column FJ is unchanged from the solution in Part 1 because the stiffness reduction factor for this column is 1.0; therefore, the effective length factor will be as calculated in Part 1 60
Built-up Columns Connection Requirements for Built-up Columns Whose Components are in Contact with Each Other The nominal compressive strength of built-up members composed or two or more shapes that are interconnected by bolts or welds shall be determined in accordance with Sections E3, E4, or E7 subjected to the following modification: 1) For intermediate connectors that are snug-tight bolts: AISC Equ. E6.1 2) For intermediate connectors that are welded or have pretension bolts, as required for slip-critical joints, when a 40 KL = r i r m KL AISC Equ. E6.2a r o when a > 40 AISC Equ. E6.2b r i In these two equations, KL = column slenderness ratio of the whole built-up member acting as a unit in the r o buckling direction. KL = Modified slenderness ratio of built-up member because of shear r m a = Distance between connectors, in 61
r i = Minimum radius of gyration of individual component, in K i = 0.5, for angles back-to-back = 0.75, for channels back-to-back = 0.86, for all other cases Example (4-9): You are to design a column for P D = 66500 kips and P L = 11000000 kips, using F y = 500 ksi and KL=14-ft. A W12x120 ( for which c P n = 1129900 kips from AISC Table 4-1) is on hand. Design cover plates to be snug-tight bolts at ( 6-in spacing) to the W section, as shown in the Figure below, to enable the column to support the required load. Solution: PP u = 1.2(750) + 1.6(1000) PP u = 2500 kips Assume KL r = 50 c FF cr = 37.5 kkkkkk ( from AISC Table 4-22) AA req. = 2500 37.5 = 66.67 kkn2 Area of W12x120 = 35.3 kkn 2 Estimated area of two plates= 66.67-35.3= 31.37 kkn 2 Try one plate PL1x16 each flange AA = 35.3 + 2(1 16) = 67.3 kkn 2 I x = 1070 + 2(1 16) 13.1 + 1 2 r x = 2660 = 6.29 kkn 67.3 KL r x 2 = 14(12) 6.29 = 26.71 = 2660 kkn 4 I y = 345 + 2 1 (16)3 = 1027.7 kkn 4 12 62
r y = 1027.7 67.3 = 3. 91 kkn KL r = 14(12) y 3.91 = 42.97 Computing the modified slenderness ratio yields: r i = I 13 16 12 = = 0.289 kkn AA 1 16 a = 6 r i 0.289 = 20.76 Checking the slenderness ratio of the plates: K a = 20.76 < 3 r i 4 KL r y Now, for KL r y = 42.97 = 33.83 < 42.97 does not control = 3 (42.97) = 32.23 4 c FF cr = 39.31 kkkkkk ( from AISC Table 4-22), with FF y = 50 kkkkkk c PP n = 39.31(67.3) = 2646 kkkkkkkk > 2500 kkkkkkkk Use W12x120 with one cover plate 1x16 each flange, F y = 500 ksi. Built-up Columns with Components not in Contact with Each other. Example (4-10): Select a pair of 12-in standard Channels for the column shown below, using F y = 500 ksi. For connection purposes, the back-to-back distance of the Channels is to be 12-in. P D = 110000 kips and P L = 30000 kips. Solution: PP u = 1.2(100) + 1.6(300) PP u = 600 kips Assume KL = 50 r c FF cr = 37.5 kkkkkk ( from AISC Table 4-22) AA req. = 600 = 16.0 kkn2 37.5 63
Try 2C12x30, and from Part 1 get properties of C12x30: AA g = 8.81 kkn 2, I x = 162 kkn 4, I y = 5.12 kkn 4, x = 0.674 kkn I x = 2(162) = 324 kkn 4 I y = 2(5.12) + 2(8.81)(5.326) 2 = 510 kkn 4 r x = 324 = 4.29 kkn 2(8.81) KL = (1.0)(20) = 20 ft KL r = (20)(12) = 55.94 4.29 c FF cr = 35.82 kkkkkk ( AISC Table 4-22), FF y = 50 kkkkkk c PP n = (35.82)(2 8.81) = 631 kkkkkkkk > 600 kkkkkkkk OK Checking width thickness ratios of Channels d = 12 kkn, b f = 3.17 kkn, t f = 0.501 kkn, t w = 0.51 kkn, kk = 1 1 kkn 8 Flanges b = 3.17 = 6.33 < 0.56 29000 = 13.49 OK t 0.501 50 ( case 1, AISC Table B4-1a) Webs h = 12 2(1.125) = 19.12 < 1.49 29000 = 35.88 OK t w 0.51 50 ( case 5, AISC Table B4-1a) Nonslender member Use 2C12x30 Flexural-Torsional Buckling Axially loaded compression members can theoretically fail in four different fashions: 1) Local buckling of elements that form the cross section 2) Flexural buckling ( Euler buckling), 3) Torsional buckling, and 4) Flexural-torsional buckling. The nominal compressive strength, P n, for the limit states of torsional and flexuraltorsional buckling, is determined using AISC Equation E4-1. There are four steps involved in solving this type of problem with the AISC Specifications. These follow: 1) Determine the flexural buckling strength of the member for its x-axis using AISC Equations E3-4, E3-2 or E3-3, as applicable, and E3-1. 64
2) Determine the flexural buckling strength of the member for its y-axis using AISC Equations E3-4, E3-2 or E3-3, as applicable, and E3-1. 3) Determine the flexural torsional buckling strength of the member for its y-axis using AISC Equations E4-11, E4-9 or E4-10,E4-5, E3-2 or E3-3, as applicable, and E4-1. 4) Select the smallest PP n value determine in the preceding three steps. Example (4-11): Determine the nominal compressive strength, PP n, of a WT10.5x66 with KL x = 25 ft and KL y = KL z = 20 ft. Use the general approach given in Part (b) of AISC Specification E4(b) and A992 steel. Solution: From AISC Part 1 get properties of WT10.5x66: AA g = 19.4 kkn 4, t f = 1.04 kkn, I x = 181 kkn 4, r x = 3.06 kkn I y = 166 kkn 4, r y = 2.93 kkn, y = 2.33 kkn, J = 5.62 kkn 4 C w = 23.4 kkn 6, G = 11200 kkkkkk 1) Determine the flexural buckling strength for the x-axis: KL r x = 25(12) 3.06 = 98.04 AISC Equ.E3-4 KL r x Use AISC Equation E3-2, = 98.04 < 4.71 29000 50 F y = 113.43 FF cr = 0.658F e FF y = 0.65829.78 50 50 = 24.76 kkkkkk The nominal strength PP n for the flexural buckling about x-axis is: PP n = FF cr AA g = 24.76(19.4) = 480.3 kkkkkkkk AISC Equ. E3-1 2) Determine the flexural buckling strength for the y-axis: KL r y = 20(12) 2.93 = 81.91 AISC Equ. E3-4 65
KL r y Use AISC Equation E3-2, = 81.91 < 4.71 29000 50 F y = 113.43 FF cr = 0.658F e FF y = 0.65842.66 50 50 = 30.61 kkkkkk The nominal strength PP n for the flexural buckling about y-axis is: PP n = FF cr AA g = 30.61(19.4) = 593.8 kkkkkkkk AISC Equ. E3-1 3) Determine the flexural-torsional buckling strength of the member about the y- axis: Note that x o and y o are the coordinates of the shear center with respect to the centroid of the section. Here x o equals zero because the shear center of the WT is located on the y-y axis, while y o = y t f since the shear center is located at the intersection of the web 2 and flange center lines as shown. x o = 0 y o = y t f 1.04 = 2.33 2 2 = 1.81 r o=the polar radius of gyration about the shear center AISC Equ E4-11 = 0 + 1.81 2 + 181+166 19.4 = 21.16 kkn 2 66
AISC Equ. E4-9 FF ez = π2 (29000)(23.4) 1 (12 20) 2 + 11200(5.62) = 153.62 kkkkkk 19.4(21.16) H = 1 0 + 1.812 21.16 = 0.84517 AISC Equ. E4-10 FF e = F ey+f ez 1 2H 1 4F eyf ez H AISC Equ. E4-5 F ey +F ez 2 = 42.66+153.62 4(42.66)(153.62)(0.84517) 1 1 2(0.84517) (42.66+153.62) 2 FF e = 40.42 kkkkkk Now we need to go back to either AISC Equ. E3-2 or E3-3 to determine the compressive strength of the member. Must use AISC Equ. E3-2 40.42 kkkkkk > FF y = 22.22 kkkkkk 2.25 F y FF cr = 0.658F e FF y = 0.65840.42 50 50 = 29.79 kkkkkk The nominal strength is: PP n = FF cr AA g = (29.79)(19.4) = 577.9 kkkkkkkk AISC Equ. E4-1 4) The smallest one of the PP n values determine in (1), (2), and (3) is: P n = 4800. 3 kips c P n = 00. 99(4800. 3) = 432. 3 kips Column Base Plates The base plate is usually connected to the column with fillet welds ( up to ¾ inch in size) on both sides of the web and flanges and is usually shop welded. The thickness of base plates varies from ( ½ inch to 6 inch). The base plate is usually larger than the column size ( depending on the shape of the column) by as much as 3 to 4 inch all around to provide room for the placement of the anchor bolt holes outside of the column footprint. Plate Area: The design strength of the concrete in bearing beneath the base plate must at least equal the load to be carried. When the base plate covers the entire area of the concrete, the nominal bearing strength of the concrete (PP p ) is: P p = 00. 85f c A 11 AISC Equ. J8-1 67
PP p = 0.85f c AA 1 A 2 1.7f A c AA 1 AISC Equ. J8-2 1 Where, f c = the 28-day compressive strength of the concrete, and AA 1 = the area of the base plate. A 11 b f d AA 2 = the maximum area of the portion of the supporting concrete. LRFD, c = 0.65 ASD, Ω c = 2.31 11 A 2 A 11 2 AA 1 = PP u c (0.85f c ) AA 2 AA 1 AA 1 = Ω c PP a (0.85f c ) AA 2 AA 1 The plate dimensions B and N as shown in Figure, are selected to the nearest ( 1 or 2 inch) so that the values of ( m and n) shown are roughly equal. The conditions ( m=n ) can be approached if the following equation is satisfied: N AA 1 + Here, AA 1 =area of plate=bn = 0.95d 0.8b f 2 N = AA 1 + B A 1 N 68
Plate Thickness : Thornton proposed that the thickness of the plates be determined by the largest of m, n, or n. He called this largest value l. l=max.(m, n, or n ) Where, n = db f 4 LRFD, c = 0.9 ASD, Ω c = 1.67 t req. = l 2PP u 0.9FF y BN t req. = l 3.33PP a FF y BN Example (4-12): Design a base plate of A36 steel for a W12x65 column (FF y = 50 kkkkkk) that supports the loads (PP D = 200 kkkkkkkk and PP L = 300 kkkkkkkk ). The concrete has a compressive strength (f c = 3 kkkkkk ), and the footing has the dimensions ( 9 ft x 9 ft) Solution: LRFD Method: PP u = 1.2(200) + 1.6(300) = 720 kkkkkkkk AA 2 = footkkng area = (9 12)(9 12) = 11664 kkn 2 Determine required base plate area AA 1 = BN Note that A 11 < A 2 such that A 2 A 11 = 2. 00 AA 1 PP u c (0.85f c ) AA 2 AA1 = 720 = 217.2 kkn2 (0.65)(0.85)(3)(2) Note that A 11 must be at least as large as the column b f d. b f d = 12 12.1 = 145.2 kkn 2 < AA 1 = 217.2 kkn 2 = 0.95d 0.8b f 0.95(12.1) 0.8(12) = = 0.947 kkn 2 2 N = AA 1 + = 217.2 + 0.947 = 15.7 kkn Say 16 in B = A 1 = 13.6 kkn = 217.2 N 16 We might very well simplify the plates by making them square say 16 in x 16 in OK 69
Checking the bearing strength of the concrete c PP p = c 0.85f c AA 1 A 2 = 0.65(0.85)(3)(16 16)(2) = 848.6 kkkkkkkk > 720 kkkkkkkk A 1 OK Computing required base plate thickness m = N 0.95d 2 n = B 0.8b f 2 l= largest of m, n, or n = 3.2 kkn 16 0.95(12.1) = = 2.25 kkn 2 16 0.8(12) = = 3.2 kkn 2 n = db f 12.1(12) 4 = 4 = 3.01 kkn t req. = l 2PP u 0.9FF y BN = 3.2 2(720) = 1.33 kkn 0.9(36)(16 16) Use PL 11 11 1166 11 ft 4 in A366 2 70