Steel Module 6b

Western Philippines University College of Engineering and Technology Civil Engineering Department STRUCTURAL STEEL DESIGN By Dr. Cesario A. Bacosa , Jr. Module 6b. 4 BEAMS AND OTHER FLEXURAL MEMBERS. Beams are members that support transverse loads. Among the many types of beams are joists, lintels, spandrels, stringers, and floor beams. Included in the items that need to be considered in beam design are moments, shears, torsion, crippling, buckling, lateral support, deflection, and perhaps fatigue. The selection of a beam section is usually done on the basis of flexure and then checked for other requirements. The bending stress in the beam can be computed using the flexure formula: Mc M M fb     Fb I I /c S where: fb = actual bending stress, MPa Fb = allowable bending stress, MPa M = bending moment at section under investigation, N-mm c = distance from neutral axis of bending to extreme fiber, mm I = moment of inertia about the neutral axis of bending, mm4 S = section modulus with respect to the axis of bending, mm3 NSCP Specifications: 506.1.1 Beams shall be distinguished from plate girders on the basis of the web slenderness ratio h/tw. When this value is greater than 2547 / Fy , allowable bending stress is given in Section 507. The allowable shear stresses and stiffener requirements are given in Section 506 unless tension field action is used, then the allowable shear stresses are given in Section 507. 506.1.2 This section applies to singly or doubly symmetric beams including hybrid beams and girders loaded in the plane of symmetry. It also applies to channels loaded in a plane passing through the shear center parallel to the web or restrained against twisting at load points and points of support. For members subject to combined flexural and axial force, see Section 508.2. 506.2 Allowable Stress: Strong axis bending of I-shaped members and Channels 506.2.1 Members with compact sections 506.2.1.1 For members with compact sections as defined in Section 502.6.1 (excluding hybrid beams and members with Fy > 448 MPa) symmetrical about, and loaded in, the plane of their minor axis the allowable stress is: Fb  0.66Fy (506-1) provided that the flanges are connected continuously to the web or webs and the laterally unsupported length of the compression flange Lb does not exceed the value Lc, as given by the smaller of: 200b f 137900 or (506-2) (d / Af ) Fy Fy 506.2.1.2 Members (including composite members and excluding hybrid members and members with Fy > 448 MPa) which meet the requirements for compact sections and are continuous over supports or rigidly framed to columns, may be proportioned for 9/10 of the negative moments produced by gravity loading when such moments are maximum at points of support, provided that, for such members, the maximum positive moment is increased by 1/10 of the average negative moments. This reduction shall not apply to moments produced by loading on cantilevers. If negative moment is resisted by a column rigidly framed to the beam or girder, the 1/10 reduction is permitted in proportioning the column for the combined axial and bending loading, provided that the stress fa due to any concurrent axial load on the member, does not exceed 0.15Fa.

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506.2.2 Members with Non-compact Sections 506.2.2.1 For members meeting the requirements of Section 506.2.1 except that their flanges are non-compact (excluding built-up members and members with Fy > 448 MPa), the allowable stress is:   b (506-3) Fb  Fy  0.79  0.000762 f Fy    2t f   506.2.2.2 For built-up members meeting the requirements of Section 506.2.1 except that their flanges are noncompact and their webs are compact or non-compact, (excluding hybrid girders and members with Fy > 448 MPa) the allowable stress is:  b Fb  Fy  0.79  0.000762 f  2t f  where:

Fy   kc 

kc 

(506-4)

4.05

 h / tw 

0.46

, if h / tw  70 , otherwise kc  1.0

506.2.2.3 For members with a non-compact section (Section 502.6) but not included above, and loaded through the shear center and braced laterally in the region of compression stress at intervals not exceeding: 200b f

Fy the allowable stress is: Fb  0.60Fy 506.2.3

(506-5)

Members with Compact or Non-compact Sections with Unbraced Length Greater than Lc

506.2.3.1 For flexural members with compact or non-compact sections as defined in Section 502.6.1, and with unbraced lengths greater than Lc as defined in Section 506.2.1, the allowable bending stress in tension is determined from Equation (506-5). 506.2.3.2 For such members with an axis of symmetry in, and loaded in the plane of their web, the allowable bending stress in compression is determined as the larger value from Equations (506-6) or (506-7) and (506-8), except that Equation (506-8) is applicable only to sections with compression flange that is solid and approximately rectangular in cross section and that has an area not less than the tension flange. Higher values of the allowable compressive stress are permitted if justified by a more precise analysis. Stresses shall not exceed those permitted by Section 507, if applicable. When 2 Fy ( L / rt )2  703270Cb L 3516330Cb : (506-6) Fb  Fy    0.60Fy    3 10550 x103 C  Fy rt Fy b   When L 3516330Cb :  rt Fy

Fb 

1172100Cb

 L / rt 

For any value of L/rt : 82740Cb Fb   0.60Fy  Ld / Af 

2

 0.60Fy

(506-7)

(506-8)

where: L = distance between lateral supports. For cantilevers braced against twist only at the support, L may conservatively be taken as the actual length. rt = radius of gyration of a section comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plane of the web, mm Af = area of compression flange, mm2 2

M  M  Cb  1.75  1.05  1   0.3  1   2.3 M  2  M2 

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Cb is the bending coefficient, M1 is the smaller and M2 the larger end moment of the unbraced length, taken about the strong axis of the member, and where M1/M2 is positive when M1 and M2 have the same sign (reverse curvature bending) and negative when they are of opposite sign (single curvature bending). When the bending moment at any point within an unbraced length is larger than that at both ends of this length, Cb = 1.0. When computing Fbx to be used in Equation (508-1), Cb may be computed by the equation given above for frames subject to joint translation, and it shall be taken as unity for frames braced against joint translation. Cb may conservatively be taken as unity for cantilever beams. 506.2.3.3 For channel bent about their major axis, the allowable compressive stress is determined from Equation (506-8). 506.2.3.4 For hybrid plate girders, Fy for Equations (506-6) and (506-7) is the yield stress of the compression flange. Equation (506-8) shall not apply to hybrid girders. Section 506.2.3 does not apply to tee sections if the stem is in compression anywhere along the unbraced length. 506.3 Allowable Stress: Weak axis bending of I-shaped members, solid bars and rectangular plates 506.3.1 Lateral bracing is not required for members loaded through the shear center about their weak axis nor for members of equal strength about both axes. 506.3.1.1 Members with Compact Sections 506.3.1.1.1

Fb  0.75Fy

For doubly symmetrical I and H-shape members with compact flanges (Section 502.6) continuously connected to the web and bent about their weak axes (except members with Fy > 448 MPa); solid round and square bars; and solid rectangular sections bent about their weaker axes, the allowable stress is: (506-9)

506.3.1.2 Members with Non-compact Sections 506.3.1.2.1

Fb  0.60Fy

For members not meeting the requirements for compact sections of Section 502.6 and not covered in Section 506.4, bent about their minor axis, the allowable stress is: (506-10)

506.3.1.2.2

Doubly symmetrical I- and H-shape members bent about their weak axes (except members with Fy > 448 MPa) with non-compact flanges (Section 502.6) continuously connected to the web may be designed on the basis of the allowable stress of:   b Fb  Fy 1.075  0.0019 f Fy  (506-11)   2t f  

506.4

Allowable stress: Bending of box members, rectangular tubes and circular tubes

506.4.1 Members with compact sections 506.4.1.1 For member bent about their strong or weak axes, members with compact section as defined in Section 502.6 and flanges continuously connected to the webs, the allowable stress is: Fb  0.66Fy (506-12) 506.4.1.2 To be classified as a compact section, a box-shaped member shall have, in addition to the requirements in Section 502.6, a depth not greater than 6 times the width, a flange thickness not greater than 2 times the web thickness and a laterally unsupported length Lb less than or equal to:  b   M  b Lc  13445  8275 1   8275   , mm (506-13)  Fy  M F 2  y   

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506.4.2 Members with non-compact sections 506.4.2.1 For box-type and tubular flexural members that meet the non-compact section requirements of Section 502.6, the allowable stress is: Fb  0.66Fy (506-14) 506.4.2.2 Lateral bracing is not required for a box section whose depth is less than 6 times its width. Lateral support requirements for box sections of larger depth-to-width ratios must be determined by special analysis. NSCP Table 5. Limiting Slenderness Ratio of Elements as a Function of Fy Specification Section and Ratios Section 506.1.1

Fy (MPa) 248

290

317

345

414

448

162

150

143

137

125

120

170 / Fy

10.8

10.0

9.6

9.2

8.4

8.1

500 / Fy

31.7

29.3

28.0

26.9

24.5

23.6

1680 / Fy

106.7

98.7

94.4

90.5

82.6

79.4

675 / Fy

42.8

39.6

37.9

36.3

33.2

31.9

703270Cb / Fy

53 Cb

49 Cb

47 Cb

45 Cb

41 Cb

40 Cb

3516330Cb / Fy

119 Cb

110 Cb

105 Cb

101 Cb

92 Cb

89 Cb

200 / Fy

12.7

11.7

11.2

10.7

9.8

9.4

250 / Fy

15.8

14.7

14.0

13.4

12.3

11.8

333/ Fy

21.2

19.6

18.7

18.0

16.4

15.8

625 / Fy

39.7

36.7

35.1

33.7

30.7

29.5

832 / Fy

52.8

48.9

46.7

44.8

40.9

39.3

634 / Fy

42.2

39.0

37.3

35.8

32.7

31.4

91.7

78.6

71.7

66.0

55.0

50.8

361

310

283

260

217

200

322

282

261

243

207

192

333

309

295

283

258

248

2547 / Fy

Table 502-1

Section 506.1.2

Table 502-1

Table 502-1

Appendix 502.5.2b 22750/ Fy

89630/ Fy Section 507.1

96530 / Fy ( Fy  16.5)

5250 / Fy

Problem 401. Find the total uniform load that can be carried by a W310x97 on a simple span of 6 m with a) Fy = 248 MPa and b) Fy = 345 MPa. Assume the section has full lateral support for its compression flange.

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Solution: bf

a) using Fy = 248 MPa 1) Verify if section is a beam or plate girder [NSCP 506.1.1]. h 277.2   28.0  162 (Table 5) tw 9.9

tf d

Therefore, the section is classified as a beam. 2) Check compactness From Problem 301 the section is compact.

tw

3) Allowable bending stress [NSCP 506.2.1.1] Fb  0.66Fy  0.66(248)  163.68 MPa

Properties of W310x97:

4) Moment capacity M x  Sx Fb  (1440 x103 )(163.68)106  235.70 kN-m

d = 308 mm bf = 305 mm

5) Load capacity

1 M  wL2 : 8 w  52.40 kN/m b)

1 235.70  w(6) 2 8

tf = 15.4 mm

using Fy = 345 MPa

1) Verify compactness From Example 3.1 the section is non-compact. 2) Allowable bending stress [NSCP 506.2.2.1]   b Fb  Fy  0.79  0.000762 f Fy    2t f  





Fb  345 0.79  0.000762(9.90) 345  224.30 MPa 3) Moment capacity M x  Sx Fb  (1440 x103 )(224.30)106  322.92 kN-m 4) Load capacity 1 1 M  wL2 : 322.92  w(6) 2 8 8 w  71.80 kN/m Problem 402. Determine the allowable bending moment for a W360x101 beam for unbraced simple spans of 1) 3.00 m, 2) 8.00 m. Use A36 steel (Fy = 248 MPa) and the NSCP specifications. Solution:

bf

1) Lb  3.00 m 1.1) Verify if section is a beam or plate girder [NSCP 506.1.1] h 320.4   30.5  162 (Table 5) tw 10.5

tf d

Therefore, the section is classified as a beam. 1.2) Check adequacy of lateral support [NSCP 506.2.1.1] 200b f 200(255)   3238 mm Fy 248

tw

Properties of W360x101: d = 357 mm bf = 255 mm tf = 18.3 mm 5

137900 137900   7268 mm (d / Af ) Fy  357  248    18.3(255)  since (Lb  3.00)  ( Lc  3.238) , laterally supported

1.3) Check compactness: Flange compactness: Table 502-1 item (1) and Table 5 bf 255   6.97  10.8 , flange is compact. 2t f 2(18.3) Check web compactness: Table 502-1 item (10) and Table 5 d 357   34.0  106.7 , web is compact. t w 10.5 Therefore the section is a compact section. 1.4) Allowable bending stress [NSCP 506.2.1.1]

Fb  0.66Fy  0.66(248)  163.68 MPa 1.5) Moment capacity M x  Sx Fb  (1690 x103 )(163.68)106  276.62 kN-m 2) Lb  8.00 m 2.1) Check adequacy of lateral support [NSCP 506.2.1.1] since (Lb  8.00)  ( Lc  3.238) , laterally unsupported 2.2) Allowable bending stress [NSCP 506.2.3.2] 1 1 At  Af  Aw  255(18.3)  (320.4)(10.5)  5227 mm2 6 6 1 1 I f  I y   50.6 x106   25.2 x106 mm4 2 2 rt 

If At



25.3x106  69.57 mm 5227

L 8000   115.0 rt 69.57

simply supported, Cb = 1.0

Referring to Table 5,

L  119 Cb  119 1.0  119.0    115.0  r  t  L  53 Cb  53 1.0  53.0    115.0   rt 

2 248(115.0)2  Use (506-6): Fb     248  0.60(248) 3  3 10550 x10 (1.0)  Fb  88.23 MPa  148.80 MPa

82740Cb 82740(1.0)   135.12 MPa < 148.8 MPa ( Ld / Af ) 8000(357) 18.3(255) use the larger value from (506-6) and (506-8), thus Fb  135.12 MPa Check also (506.8): Fb 

2.3) Moment capacity M x  Sx Fb  (1690 x103 )(135.12)106  228.35 kN-m

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Problem 403. A W360x45 section with Fy = 345 MPa is loaded as shown. Determine the adequacy of the section using NSCP specifications if lateral support is provided only at points 1, 2 and 3. Neglect weight of beam. Solution: 1) Determine the shear and moment: 1.1) Set up the 3-moment equations

bf

 6 Aa   6 Ab  M 0 L0  2M1  L0  L1   M 2 L1      0  L 0  L 1

 6 Aa   6 Ab  M1L1  2M 2  L1  L2   M 3 L2      0  L 1  L 2 1.2) Evaluate the three moment factors  6 Aa   6 Ab  80(3) 2 6  22   1080 kN-m2      6  L 1  L 1  6 Ab  30(6)  1620    4  L 2 1.3) Apply the moment equations in 1.1 2M1  0  6  M 2 (6)  1080  0

W360x45:

tf

d = 352 mm d bf = 171 mm tf = 9.8 mm tw 80 kN 30 kN/m

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38.57 kN-m 1

3m 3m L1 = 6 m

29.28 kN

M 2  102.86 kN-m

72.86 kN

107.1 4

29.28

2) For unbraced length L1 = 6.00 m 2.1) Verify if section is a beam or plate girder [NSCP 506.1.1] h 332.4   48.2  137 (Table 5) tw 6.9

3

L2 = 6 m

157.86 kN

M1 (6)  2M 2  6  6  1080  1620  0 M1  38.57 kN-m

2

-50.72

-72.86 88.44

49.27

Therefore, the section is classified as a beam. 2.2) Check compactness: Flange compactness: Table 502-1 item (1) and Table 5 bf 171   8.72  9.2 , flange is compact. 2t f 2(9.8)

-38.57 -102.86

Check web compactness: Table 502-1 item (10) and Table 5 d 352   51.0  90.5 , web is compact. t w 6.9 Therefore the section is a compact section. 2.3) Check adequacy of lateral support [NSCP 506.2.1.1] 200b f 200(171)   1841 mm Fy 345 137900 137900   1903 mm (d / Af ) Fy  352    345  9.8(171)  since (Lb  6.00)  (Lc  1.841) , laterally unsupported

2.4) Allowable bending stress [NSCP 506.2.3.2] 1 1 At  Af  Aw  171(9.8)  (332.4)(6.9)  2058 mm2 6 6 1 1 I f  I y  8.18 x106   4.09 x106 mm4 2 2

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rt 

If At



4.09 x106  44.58 mm 2058

L 6000   134.6 : rt 44.58

M 1 38.57   0.375 (reverse curvature bending) M 2 102.86

Cb  1.75  1.05  0.375   0.3  0.375   2.186  2.3 2

L  101 Cb  101 2.186  149.3    134.6   rt 

L  45 Cb  45 2.186  66.5    134.6  r  t 

2 345(134.59)2  use (506-6): Fb     345  0.60(345) 3  3 10550 x10 (2.186)  Fb  136.51 MPa  207 MPa 82740Cb 82740(2.186)   143.44 MPa < 207 MPa 6000(352) ( Ld / Af ) 9.8(171) use the larger value from (506-6) and (506-8), thus Fb  143.44 MPa Check also (506-8): Fb 

2.5) Moment capacity M x  Sx Fb  (691x103 )(143.44)106  99.12 kN-m

M neg

2.6) Adjust design moment [NSCP 506.2.1.2] 9  (102.86)  92.57 kN-m < 99.12 kN-m OK 10 1  38.57  102.86  M pos  49.27     56.34 kN-m < 99.12 kN-m OK 10  2 

3) For unbraced length L2 = 6.00 m 3.1) Allowable bending stress [NSCP 506.2.3.2] M1 0  0 Cb  1.75  2.3 M 2 102.86

L  101 Cb  101 1.75  133.6    134.6   rt  1172100Cb 1172100(1.75) Use (506-7): Fb    0.60(345) 2 (134.59) 2  L / rt  Fb  113.23 MPa  207 MPa

82740Cb 82740(2.186)   143.44 MPa < 207 MPa 6000(352) ( Ld / Af ) 9.8(171) use the larger value from (506-7) and (506-8), thus Fb  143.44 MPa Check also (506-8): Fb 

2.5) Moment capacity M x  Sx Fb  (691x103 )(143.44)106  99.12 kN-m 2.7) Adjust design moment [NSCP 506.2.1.2] 9 M neg  (102.86)  92.57 kN-m < 99.12 kN-m OK 10 1  102.86  0     93.58 kN-m < 99.12 kN-m OK 10  2  Therefore the section is satisfactory. M pos  88.44 

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