MANILA: Room 206, JPD Building, CM Recto Avenue, Manila CEBU: 4/F J. Martinez Bldg., Osmeña Blvd., Cebu City Telephone Number: (02) 516 7559 (Manila) E-Mail: buksmarquez1 @yahoo.com (032) 254-9967 (Cebu) Review Module – Compression Members (ASD/LRFD) SLENDERNESS RATIO, SR ππ = πΎπ π EULER’S CRITICAL BUCKLING LOAD π 2 πΈπΌ πππ = (πΎπ)2 EULER’S CRITICAL BUCKLING STRESS π 2πΈ πΉππ = (ππ )2 END FIXITY FACTOR / EFFECTIVE LENGTH FACTOR a. Compute the effective slenderness ratio with respect to x – axis. b. Compute the effective slenderness ratio with respect to y – axis. c. Compute the Euler’s critical buckling load. d. What is the minimum length of column without exceeding the proportional limit? NSCP 2001 (Allowable Stress Design - ASD) SECTION 505 – COLUMNS AND OTHER COMPRESSION MEMBERS 502.8.1 For members whose design is based on compressive force, the slenderness ratio πΎπ/π preferably should NOT exceed 200. If this limit is exceeded, the allowable stress shall NOT exceed the value obtained from Equation (505-2). SITUATION 1. A build-up section with unbraced length of 6 m, used as a compression member, has the following properties: Assume that one end of the column is fixed and the other pinned (k = 0.80). Assumed both ends are pinned and proportional limit is 240 MPa. E=200GPa. I-Beam Properties: A = 22000 mm2 d = 730 mm bf = 310 mm Ix = 295 x 106 mm4 Iy = 30.6 x 106 mm4 505.3 ALLOWABLE STRESS 505.3.1 On the gross section of axially loaded compression members whose cross-sections meet the provisions of Table 502-1, when πΎπ/π, the largest effective slenderness ratio of any unbraced segment is less than πΆπ , the allowable stress is: [1 − πΉπ = 5 3 3 + β 8 2πΆπ2 πΎπ/π πΆπ ] πΉπ¦ (πππ − π) 1 (πΎπ/π)3 − β 8 πΆπ3 Where: πΆπ = √ Properties of each C-Section: A = 7350 mm2 d = 380 mm tw = 15 mm x = 24 mm Ix = 65 x 106 mm4 Iy = 5.08 x 106 mm4 a. Determine the effective slenderness ratio with respect to the strong axis. b. Determine the effective slenderness ratio with respect to the weak axis. c. Determine the Euler’s critical buckling load. d. Determine the minimum length of a column having this section so as not to exceed the proportional limit. (πΎπ/π)2 2π 2 πΈ πΉπ¦ (πππ − ππ) 505.3.2 On the gross section of axially loaded compression members, when πΎπ/π exceed πΆπ the allowable stress is: πΉπ = 12π 2 πΈ 23(πΎπ/π)2 (πππ − π) NSCP 2010/2015 (ASD & LRFD) SECTION 505 – DESIGN OF MEMBERS FOR COMPRESSION Design Compressive Strength Allowable Compressive Strength ππ ππ ππ Ωπ SITUATION 2. Built up column 12 m long consists of W 350 x 110 with two plates welded ππ = 0.9 (πΏπ πΉπ·) Ωπ = 1.67 (π΄ππ·) to form a box section. With respect to x – axis column is fixed, y – axis column is braced at mid height. Use the theoretical k values 505.3 COMPRESSIVE STRENGTH OF FLEXURAL BUCKLING OF Properties of W 350 x 110 MEMBERS WITHOUT SLENDER ELEMENTS 2 AThis = 12,340mm tf = 15 mmby 100000780997442 Poportional from limit =CourseHero.com 310 MPa study source was downloaded on 04-01-2022 19:57:32 GMT -05:00 The nominal compressive strength, Pn, shall be determined based on the Ix = 315 x 106 mm4 bf =250 m Iy = 49 x 106 mm4 limit state of flexural buckling. tw = 12 mm https://www.coursehero.com/file/66852488/Review-Module-39-Nov2020-Steel-Compression-Memberspdf/ MANILA: Room 206, JPD Building, CM Recto Avenue, Manila CEBU: 4/F J. Martinez Bldg., Osmeña Blvd., Cebu City Telephone Number: (02) 516 7559 (Manila) E-Mail: buksmarquez1 @yahoo.com (032) 254-9967 (Cebu) ππ = πΉππ π΄π (πππ. π. π) The Flexural buckling stress, Fcr, is determined as follows: 1. When kL E ο£ 4.71 or (Fe ο³ 0.44 Fy ) r Fy πΉπ¦ πΉππ = [0.658 πΉπ ] πΉπ¦ 2. When (πππ. π − π) kL E οΎ 4.71 or (Fe ο£ 0.44 Fy ) r Fy πΉππ = 0.877πΉπ (πππ. π − π) Ix= 695.11 x 106 mm4 Iy= 274.71 x 106 mm4 Sx=3719.18 x 103 mm3 Sy= 1396.18 x 103 mm3 rx=160.53 mm ry= 100.84 mm (Use the recommended k values) 1. Compute the critical slenderness ratio. Using NSCP 2001 2. Compute the allowable axial stress. 3. Where: Fe = Elastic Critical buckling stress determined according to Eq. 505.3-4, Section 505.4, or the provisions of Section 503.2, as applicable, MPa. πΉπ = π 2πΈ (πΎπ/π)2 (πππ. π − π) SITUATION. A compression member is 3 meters long and pinned at both ends. Effective length factor k = 1.0. The section is made up of two- 150mm x 100mm x 10mm angle of unequal legs with long legs back to back and separated by a gusset plate 10 mm thick. Use A-36 steel, Fy= 248 MPa and E= 200,000 MPa. Properties of one angle 150mm x 100mm x 10mm A = 2400mm2 tf = 16 mm Ix = 5.576 x 106 mm4 x = 23.75 mm Iy = 2.026 x 106 mm4 y = 48.75 mm 1. Properties of W 14 x 142 A=26967.69 mm2 d=374.65 mm bf=393.70 mm tf=27.00 mm tw= 17.27 mm Determine the minimum radius of gyration. Compute the capacity of the column section. Using NSCP 2010 4. Compute the allowable axial stress. SITUATION. A W14x90 steel column is used to carry an axial dead load of 600 kN and live load of 1900 kN. The column is 9m long and is pinned at the top and bottom in both axes. Additional support has been added with lateral and torsional bracing about the y-axis and midpoint respectively. Determine the adequacy of the column section given Fy=345 MPa. Properties of W14x90 A=17097 mm2 Ix=415.8x106 mm4 Iy=150.7x106 mm4 Using: 1. NSCP 2015 LRFD NSCP 2001 ASD NSCP 2001 (Allowable Stress Design - ASD) Using NSCP 2001, 2. Determine the safe axial load of the compression member. 3. If the length is increased to 6 meters, determine the safe axial load of the compression member. COMBINED AXIAL COMPRESSION AND BENDING (Sec 508.2) 508.1.2 This Section pertains to doubly and singly symmetrical members only. See Section 505 for the determination of πΉπ and Section 506 for the determination of πΉππ₯ and πΉππ¦ . Using NSCP 2010, 4. Determine the allowable axial capacity of the compression member with Ωπ = 1.67. 5. Determine the design axial capacity of the compression member with ππ = 0.9. 6. If the length is increased to 6 meters, Determine the allowable axial capacity of the compression member with Ωπ = 1.67. 508.2 AXIAL COMPRESSION AND BENDING 508.2.1 Members subjected to both axial compression and bending stresses shall be proportioned to satisfy the following requirements: When ππ /πΉπ > 0.15 πΆππ¦ πππ¦ ππ πΆππ₯ πππ₯ + + ≤ 1.0 (πππ − π) π π π πΉπ (1 − ) πΉ (1 − π ) πΉ SITUATION. A W 14 x 142 is used as a column having length of 9 m long. It is hinged at the upper end and fixed at the lower end but there is a lateral bracing perpendicular to the minor axis of the W section at the 5.4 m above the bottom support. It is assumed to be pinned connected at the bracing point. Using A-36 steel Fy=248 MPa and the NSCP specifications. Es= 200 GPa. ′ πΉππ₯ ππ₯ ππ πππ₯ πππ¦ + + ≤ 1.0 0.60πΉπ¦ πΉππ₯ πΉππ¦ ′ πΉππ¦ ππ¦ (πππ − π) When ππ /πΉπ ≤ 0.15, Equation (508-3) is permitted in lieu of Equations (508-1) and (508-2): ππ πππ₯ πππ¦ + + ≤ 1.0 (πππ − π) πΉπ πΉππ₯ πΉππ¦ 508.2.2 In Equations (508-1), (508-2) and (508-3), the subscripts x and y, combined with subscripts b, m and e, indicates the axis of bending about which a particular stress or design properly applies, and This study source was downloaded by 100000780997442 from CourseHero.com on 04-01-2022 19:57:32 GMT -05:00 https://www.coursehero.com/file/66852488/Review-Module-39-Nov2020-Steel-Compression-Memberspdf/ MANILA: Room 206, JPD Building, CM Recto Avenue, Manila CEBU: 4/F J. Martinez Bldg., Osmeña Blvd., Cebu City Telephone Number: (02) 516 7559 (Manila) E-Mail: buksmarquez1 @yahoo.com (032) 254-9967 (Cebu) πΉπ = Axial compressive stress that would be permitted if axial force alone existed, MPa πΉπ = Compressive bending stress that would be permitted if bending moment alone existed, MPa 12π 2 πΈ πΉπ′ = 23(πΎππ /ππ )2 πΉπ′ = Euler stress divided by a factor of safety, MPa. As in the case of Fa, Fb, and 0.60Fy, F’e may be increased 1/3 in accordance with Section 501.5.2. ππ = The actual unbraced length in the plane of bending ππ = The corresponding radius of gyration πΎ = Effective length factor in the plane of bending ππ = Computed axial stress, MPa ππ = Computed compressive bending stress at the point under consideration, MPa. COLUMN CURVATURE COEFFICIENT, πͺπ The coefficient applied to the x or y axis of bending term in the interaction equation (508-1) and is dependent upon column curvature caused by the applied moments. This value shall be taken as follows: NSCP 2010/2015 (ASD & LRFD) SECTION 508 - DESIGN OF MEMBERS FOR COMBINED FORCES AND TORSION This Section addresses members subject to axial force and flexure about one or both axes, with or without torsion, and to members subject to torsion only. The interaction of flexure and compression in doubly symmetric members and singly symmetric members for which πΌ 0.1 ≤ π¦π ≤ 0.9 , πΌπ¦ that are constrained to bend about a geometric axis (x and/or y) shall be limited by Eqns. 508.1-1a and 508.1-1b, where Iyc is the moment of inertia about the y-axis referred to the compression flange. mm4. 508.1.1 Doubly and Singly Symmetric Members in Flexure and Compression 1. πΉππ a. For compression members in frames subject to joint translation (sidesway), πΆπ = 0.85 ππ ≥ 0.2 ππ ππ 8 πππ₯ πππ¦ + ( + ) ≤ 1.0 ππ 9 πππ₯ πππ¦ (πππ. π − ππ) ππ < 0.2 ππ b. For rotationally restrained compression members in frames braced against joint translation and NOT subject to transverse loading between their supports in the plane of bending. πΆπ = 0.6 − 0.4(π1 /π2 ) 1. πΉππ where π1 /π2 is the ratio of the smaller to larger moments at the ends of the portion of the member unbraced in the plane of bending under consideration. π1 /π2 is positive when the member is bent in reverse curvature, negative when bent in single curvature. Where: Pr = required axial compressive Strength, N Pc = available axial compressive Strength, N Mr = required flexural strength, N-mm Mc = available flexural strength, N-mm x = subscript relating symbol to strong axis bending y = subscript relating symbol to weak axis bendin π1 = (+) π2 ππππππ πππππππππ (REVERSE CURVATURE) π1 = (−) π2 ππππππ πππππππππ ππ πππ₯ πππ¦ +( + ) ≤ 1.0 2ππ πππ₯ πππ¦ (πππ. π − ππ) SITUATION 1. The figure below shows a framing plan of a warehouse. The columns are 7 meters long and are rigidly attached to the base and pin-connected at the top. Sidesway is uninhibited about the x-axis (K=1.20) but prevented about the y-axis through cross braces (K=1.0). Properties of the W-Section columns: A = 10000 mm2 Ix = 225 × 106 mm4 d = 350 mm Iy = 25 × 106 mm4 bf = 200 mm rT = 55 mm t f = 15 mm Fy = 248 MPa t w = 10 mm rx = 150 mm E = 200 GPa ry = 50 mm c. For compression members in frames braced against joint translation in the plane of loading and subjected to transverse loading between their supports, the value of πΆπ may be determined by an analysis. However, in lieu of such analysis, the following values are permitted: i. For members whose ends are restrained against rotation in the plane of bending, πΆπ = 0.85 The following loads for column (1) has been obtained from analysis, Axial Load, P = 70 kN Moment at the base, Mx = 30 kN β m, My = 0 Assume that the columns are compact with respect to bending. a. Determine the allowable axial compressive stress in column (1). b. Determine ratio-05:00 of actual to allowable axial and bending stresses in column This study source was downloaded by 100000780997442 from CourseHero.com on 04-01-2022 19:57:32the GMT (1). ii. For members whose ends are unrestrained against rotation in the plane of bending, πΆπ = 1.0 https://www.coursehero.com/file/66852488/Review-Module-39-Nov2020-Steel-Compression-Memberspdf/ MANILA: Room 206, JPD Building, CM Recto Avenue, Manila CEBU: 4/F J. Martinez Bldg., Osmeña Blvd., Cebu City Telephone Number: (02) 516 7559 (Manila) E-Mail: buksmarquez1 @yahoo.com (032) 254-9967 (Cebu) c. If My = 15 kN β m, determine the total ratio of actual to allowable axial and bending stresses in column (1). SITUATION 2. A W 250x58 column carries an eccentric load of 200 kN with an eccentricity from y axis = 75mm and eccentricity from x axis = 50mm. Fy = 400 MPa, Fbx = 0.60Fy, Fby = 0.75Fy. Assume k = 1.0 Cm = 0.85 Unsupported length = 6m A = 7420 mm2 Sx = 693 x 103 mm3 rx = 108 mm 6 4 3 3 Ix = 87.3 x 10 mm Sy = 186 x 10 mm ry = 50.4 mm a. Determine the allowable axial compressive stress. b. Determine the interaction value for both axial and bending. This study source was downloaded by 100000780997442 from CourseHero.com on 04-01-2022 19:57:32 GMT -05:00 https://www.coursehero.com/file/66852488/Review-Module-39-Nov2020-Steel-Compression-Memberspdf/ Powered by TCPDF (www.tcpdf.org)
