CHAPTER G DESIGN OF MEMBERS FOR SHEAR

Transcription

1 G-1 CHAPTER G DESIGN OF MEMBERS FOR SHEAR INTRODUCTION This chapter covers ebs of singly or doubly symmetric members subject to shear in the plane of the eb, single angles, HSS sections, and shear in the eak direction of singly or doubly symmetric shapes. Most of the formulas from this chapter are illustrated by example. Tables for all standard ASTM A99 W-shapes and ASTM A36 channels are included in the Manual. In the tables, here applicable, and shear information is presented side-by-side for quick selection, design and verification. shear strengths have been increased slightly over those in the previous Specification for members not subject to shear buckling. strengths are essentially identical to those in the previous Specification. and ill produce identical designs for the case here the live load effect is approximately three times the dead load effect. G1. GENERAL PROVISIONS The design shear strength, φ v V n, and the alloable shear strength, V n /Ω v are determined as follos: V n = nominal shear strength based on shear yielding or shear buckling () (). Exception: For all current ASTM A6, W, S, and HP shapes except W44 30, W40 149, W36 135, W33 118, W30 90, W4 55, W16 6, and W1 14 for F y = 50 ksi: φ v = 1.00 () Ω v = 1.50 (). Section G does not utilize tension field action. Section G3 specifically addresses the use tension field action. Strong axis shear values are tabulated for W-shapes in Manual Tables 3- and 3-6, for S-shapes in Manual Table 3-7, for C-shapes in Manual Table 3-8 and for MC-shapes in Manual Table 3-9. Weak axis shear values for W- shapes, S-shapes, C-shapes and MC-shapes and shear values for angles, rectangular HSS and box members, and round HSS are not tabulated. G. MEMBERS WITH UNSTIFFENED OR STIFFENED WEBS As indicated in the User Note of this section, virtually all W, S and HP shapes are not subject to shear buckling and are also eligible for the more liberal safety and resistance factors, φ v = 1.00 () and Ω v = 1.50 (). This is presented in Example G.1 for a W-shape. A channel shear strength design is presented in Example G.. G3. TENSION FIELD ACTION A built-up girder ith a thin eb and vertical stiffeners is presented in Example G.8. G4. SINGLE ANGLES Rolled angles are typically made from ASTM A36 steel. All single angles listed in the Manual have a C v =1.0. A single angle example is illustrated in Example G.3.

2 G- G5. RECTANGULAR HSS AND BOX MEMBERS The shear height, h, is taken as the clear distance beteen the radii. If the corner radii are unknon, the outside radius is taken as 3 times the design thickness. An HSS example is provide in Example G.4. G6. ROUND HSS For all Round HSS and Pipes of ordinary length listed in the Manual, F cr can be taken as 0.6F y in Specification Equation G6-1. A round HSS example is illustrated in Example G.5. G7. WEAK AXIS SHEAR IS SINGLE AND DOUBLY SYMMETRIC SHAPES For a eak axis shear example see Example G.6 and Example G.7. G8. BEAMS AND GIRDERS WITH WEB OPENINGS For a beam and girder ith eb openings example see AISC Design Guide.

3 G-3 Example G.1a W-Shape in Strong-Axis Shear. Given: Verify the shear strength of a W4 6 ASTM A99 beam ith end shears of 48 kips from dead load and 145 kips from live load. Solution: Material Properties: W4 6 ASTM A99 F y = 50 ksi F u = 65 ksi Manual Table -3 Geometric Properties: W4 6 d = 3.7 in. t = in. Manual Table 1-1 Calculate the required shear strength V u = 1.(48.0 kips) + 1.6(145 kips) = 90 kips V a = 48.0 kips kips = 193 kips Take the available shear strength from Manual Table 3- φ v V n = 306 kips 306 kips > 90 kips o.k. V n /Ω v = 04 kips 04 kips > 193 kips o.k. Manual Table 3-

4 G-4 Example G.1b W-Shape in Strong-Axis Shear. The available shear strength, hich can be easily determined by the tabulated values of the Steel Construction Manual, can be verified by directly applying the provisions of the Specification. Except for very fe sections, hich are listed in the User Note, Specification Section G.1(a) is applicable to the I-shaped beams published in the Manual hen F y 50 ksi. C v = 1.0 Eqn. G- Calculate A A = dt = 3.7 in. (0.430 in.) = 10. in. Calculate V n V n = 0.6F y A C v = 0.6(50 ksi)(10. in. )(1.0) = 306 kips Eqn. G-1 Calculate the available shear strength φ v = 1.00 φ v V n = 1.00(306 kips) = 306 kips Ω v = 1.50 V n /Ω v = 306 kips / 1.50 = 04 kips Section G.1a

5 G-5 Example G.a C-Shape in Strong-Axis Shear. Given: Verify the shear strength of a C channel ith end shears of 17.5 kips from dead load and 5.5 kips from live load. Solution: Material Properties: C ASTM A36 F y = 36 ksi F u = 58 ksi Manual Table -3 Geometric Properties: C d = 15.0 in. t = in. Manual Table 1-5 Calculate the required strength V u = 1.(17.5 kips) + 1.6(5.5 kips) = 105 kips V a = 17.5 kips kips = 70.0 kips Take the available shear strength from Manual Table 3-8 φ v V n =117 kips 117 kips > 105 kips o.k. V n /Ω v = 77.6 kips 77.6 kips > 70.0 kips o.k. Manual Table 3-8

6 G-6 Example G.b C-Shape in Strong-Axis Shear. The available shear strength, hich can be easily determined by the tabulated values of the Steel Construction Manual, can be verified by directly applying the provisions of the Specification. C v is 1.0 for all rolled channels hen F y 36 ksi, and Specification Equation G-1 is applicable. C v = 1.0 Eqn. G- Calculate A A = 15.0 in.(0.400 in.) = 6.00 in. Calculate V n V n = 0.6F y A C v = 0.6(36 ksi)(6.00 in. )(1.0) = 130 kips Eqn. G-1 Calculate the available shear strength The values of φ v = 1.00 () and Ω v = 1.50 () do not apply to channels. The general values () and () must be used. φ v V n = 0.90(130 kips) = 117 kips V n /Ω v =130 kips / 1.67 = 77.6 kips

7 G-7 Example G.3 Angle in Shear. Given: Verify the shear strength of a (LLV) ASTM A36 angle ith end shears of 3.5 kips from dead load and 10.5 kips from live load. Solution: Material Properties: L5 3 4 angle ASTM A36 F y = 36 ksi F u = 58 ksi Manual Table -3 Geometric Properties: L5 3 4 angle d = 5.00 in. t = 0.50 in. Manual Table 1-7 Calculate the required shear strength V u = 1.(3.50 kips) + 1.6(10.5 kips) = 1.0 kips V a = 3.50 kips kips = 14.0 kips Note: There are no tables for angles in shear, but the shear strength can be calculated as follos: For angles in shear, use Specification Equation G-1 ith C v = 1.0. Section G4 Calculate A A = dt = (5.00 in.)(0.50 in.) = 1.5 in. Calculate V n V n = 0.6F y A C v = 0.6(36 ksi)(1.5 in. )(1.0) = 7.0 kips Eqn. G-1 Calculate the available shear strength φ v V n = 0.90(7.0 kips) = 4.3 kips 4.3 kips > 1.0 kips o.k. V n /Ω v = 7.0 kips / 1.67 = 16. kips 16. kips > 14.0 kips o.k.

8 G-8 Example G.4 Rectangular HSS in Shear. Given: Verify the shear strength of a HSS6 4 a ASTM A500 grade B member ith end shears of 11 kips from dead load and 33 kips from live load. The beam is oriented ith the shear parallel to the 6 in. dimension. Solution: Material Properties: HSS6 4 a ASTM A500B F y = 46 ksi F u = 58 ksi Manual Table -3 Geometric Properties: HSS6 4 a h = 6.00 in. = 4.00 in. t = in. Manual Table 1-11 Calculate the required shear strength V u = 1.(11.0 kips) + 1.6(33.0 kips) = 66.0 kips V a = 11.0 kips kips = 44.0 kips User note: There are no Manual Tables for shear in HSS shapes, but the shear strength can be calculated as follos: Calculate the nominal strength Section G5 For rectangular HSS in shear, use Section G.1 ith A = ht and k v = 5. If the exact radius is unknon, the radius is taken as 3 times the design thickness. h = d (3t ) = 6.00 in. (3)(0.349 in.) = 3.91 in. h/t = 3.91 in. / in. = kv E Fy =1.10 5(9,000 ksi 46 ksi) = < 61.8 Therefore C v = 1.0 Eqn. G-3 Note: most standard HSS sections listed in the manual have C v = 1.0 at F y 46 ksi.calculate A A = ht = (3.91 in.)(0.349 in.) =.73 in. Calculate V n V n = 0.6F y A C v = 0.6(46 ksi)(.73 in. )(1.0) = 75. kips Calculate the available shear strength Eqn. G-1 φ v V n = 0.90(75. kips) = 67.7 kips 67.7 kips > 66.0 kips o.k. V n /Ω v = 75. kips / 1.67 = 45.0 kips 45.0 kips > 44.0 kips o.k.

9 G-9 Example G.5 Round HSS in Shear. Given: Verify the shear strength of a round HSS ASTM A500 grade B member spanning 3 feet ith end shears of 30 kips from dead load and 90 kips from live load. Solution: Material Properties: HSS ASTM A500 Gr.B F y = 4 ksi F u = 58 ksi Manual Table -3 Geometric Properties: HSS D = 16.0 in. t = in. A g =17. in. Manual Table 1-13 Note: There are no Manual tables for Round HSS in shear, but the strength can be calculated as follos: Calculate the required shear strength V u = 1.(30.0 kips) + 1.6(90.0 kips) = 180 kips V a = 30.0 kips kips = 10 kips Calculate Fcr as the smallest of: 1.60E F cr = take L 5 v as half the span = 19 in. L 4 v D D t = 1.60(9,000ksi) 5 = 11 ksi 19 in in in in. Section G6 Eqn. G6-a Eqn. G6-b or F cr = or 0.78E ( ) 3 D/ t 0.78(9,000ksi) = 73 ksi 16.0 in in. = 3 F cr = 0.6F y = 0.6(4ksi) = 5.ksi controls Note: Equations G6-a and G6-b ill not normally control for the sections published in the Manual except hen high strength steel is used or the span is unusually long. Calculate V n Section G6 Fcr Ag V n = = ( 5. ksi)( 17. in. ) = 17 kips

10 G-10 Calculate the available shear strength Eqn. G6-1 φ v V n = 0.90(17 kips) = 195 kips 195 kips > 190 kips o.k. V n /Ω v = 17 kips / 1.67 = 130 kips 130 kips > 10 kips o.k.

11 G-11 Example G.6 Doubly-Symmetric Shape in Weak-Axis Shear. Given: Verify the strength of a W1 48 ASTM A99 beam ith end shears of 0 kips from dead load and 60 kips from live load in the eak direction. Solution: Material Properties: W1 48 ASTM A99 F y = 50 ksi F u = 65 ksi Manual Table -3 Geometric Properties: W1 48 b f = 8.14 in. t f = in. Manual Table 1-1 Calculate the required shear strength V u = 1.(0.0 kips) + 1.6(60.0 kips) = 10 kips V a = 0.0 kips kips = 80.0 kips For eak axis shear, use Equation G-1 and Section G.1(b) ith A = b f t f for each flange and k v = 1.. Section G7 Calculate A (multiply by for both shear resisting elements) A =b f t f = (8.14 in.)(0.430 in.) = 7.00 in Calculate C v Section G.1b b f /t f = 8.14 in. / in. = k E F v y = (9,000 ksi 50 ksi) = 9.0 > 18.9 therefore, C v = 1.0 Eqn. G-3 Note: For all ASTM A6 W, S, M, and HP shapes, hen F y < 50 ksi, C v = 1.0. Calculate V n V n = 0.6F y A C v = 0.6(50 ksi)(7.00 in. )(1.0) = 10 kips Calculate the available shear strength Eqn. G-1 φ v V n = 0.90(10 kips) = 189 kips 189 kips > 10 kips o.k. V n /Ω v = 10 kips / 1.67 = 16 kips 16 kips > 80.0 kips o.k.

12 G-1 Example G.7 Singly-Symmetric Shape in Weak-Axis Shear. Given: Verify the strength of a C9 0 ASTM A36 channel ith end shears of 5 kips from dead load and 15 kips from live load in the eak direction. Solution: Material Properties: C9 0 ASTM A36 F y = 36 ksi F u = 58 ksi Manual Table -3 Geometric Properties: C9 0 b f =.65 in. t f = in. Manual Table 1-5 Note: There are no Manual tables for this, but the strength can be calculated as follos: Calculate the required shear strength V u = 1.(5.00 kips) + 1.6(15.0 kips) = 30.0 kips V a = 5.00 kips kips = 0.0 kips For eak axis shear, use Equation G-1 and Section G.1(b) ith A = b f t f for each flange and k v = 1.. Section G7 Calculate A (multiply by for both shear resisting elements) A =b f t f = (.65 in.)(0.413in.) =.19 in. Section G7 Calculate C v b f /t f =.65 in. / in. = kv E F y = (9,000 ksi 36 ksi) = 34. > 6.4 Therefore, C v = 1.0 Section G.1b Eqn. G-3 Calculate V n V n = 0.6F y A C v = 0.6(36 ksi)(.19 in. )(1.0) = 47.3 kips Calculate the available shear strength Eqn. G-1 φ v V n = 0.90(47.3 kips) = 4.6 kips 4.5 kips > 30.0 kips o.k. V n /Ω v = 47.3 kips / 1.67 = 8.3 kips 8.3 kips > 0.0 kips o.k.

13 G-13 Example G.8a Built-up Plate Girder ith Transverse Stiffeners Given: A built up ASTM A36 I-shaped girder spanning 56 ft. has a uniformly distributed dead load of 0.9 klf and a live load of.74 klf in the strong direction. The girder is 36 in. tall ith 1 in. 1½ in. flanges and a c in. eb. Determine if the member has sufficient available shear strength to support the end shear, ithout and ith tension field action. Use transverse stiffeners, as required. User note: This built-up girder as purposely selected ith a thin eb in order to illustrate the design of transverse stiffeners. A more conventionally proportioned plate girder ould have at least a ½ in. eb and slightly smaller flanges. Solution: Material Properties: Built-up girder ASTM A36 F y = 36 ksi F u = 58 ksi Manual Table -3 Geometric Properties: Built-up girder t = in. d = 36.0 in. b ft = b fc = 1.0 in. t f = 1.50 in. h = 33.0 in. Calculate the required shear strength at the support R u =[1.(0.9 klf) + 1.6(.74 klf)](8.0 ft) = 154 kips R a = (0.9 klf +.74 klf)(8.0 ft) = 10 kips Determine if stiffeners are required A = dt = (36.0 in.)(0.313 in.) = 11.3 in. h/t = 33.0 in. / in. = < 60 Therefore, k v = 5 Section G.1b 1.37 ke v / F y = (9,000 ksi) /(36 ksi) = > 86.9 therefore, use Specification Eqn. G-5 to calculate C v

14 G Ekv C v = ( h/ t ) F = 1.51(9, 000 ksi)(5) = (105) (36 ksi) y Eqn. G-5 Calculate V n V n = 0.6F y A C v = 0.6(36 ksi)(11.3 in. )(0.547) = 134 kips Eqn. G-1 Check the available shear strength ithout stiffeners φ v V n = 0.90(134 kips) = 10 kips 10 kips < 154 kips not o.k. Therefore, stiffeners are required. V n /Ω v = 134 kips / 1.67 = 80.0 kips 80.0 kips < 10 kips not o.k. Therefore, stiffeners are required. Manual Tables 3-16a and 3-16b can be used to select stiffener spacings needed to develop the required stress in the eb. Limits on the Use of Tension Field Action: Section G3.1 Consideration of tension field action is not permitted if any of the folloing are true: a) end panels in all members ith transverse stiffeners b) members hen a/h exceeds 3.0 or [60/(h/t )] c) A /(A fc +A ft ) >.5 d) h/b fc or h/b ft > 6.0 Select stiffener spacing for end panel Tension field action is not permitted for end panels, therefore use Table 3-16a. Use V u =φ v V n to determine the required stress in the eb by dividing by the eb area Use V a = V n /Ω v to determine the required stress in the eb by dividing by the eb area φvv A n V = u A = 154 kips 11.3 in. = 13.6 ksi Ω V v n A V = a A = 10 kips 11.3 in. = 9.03 ksi Use Table 3-16a from the Manual to select the required stiffener ratio a/h based on the h/t ratio φvvn of the girder and the required stress. Interpolate and follo an available stress curve, = A Vn 13.6 ksi for, = 9.03 ksi for, until it intersects the horizontal line for a h/t ΩvA value of 105. Project don from this intersection and take the maximum a/h value of 1.80 from the axis across the bottom. Since h = 33.0 in., stiffeners are required at (1.80)(33.0 in.) = 59.4 in. maximum. Therefore, use 59.0 in. Manual Table 3-16a Select stiffener spacing for the second panel Tension field action is alloed, but not required, since the second panel is not an end panel. Section G3.1

15 G-15 Calculate the required shear strength at the start of the second panel, 59 in. from end V u [ ] 59.0 in. 1.(0.9klf)+1.6(.74klf) = 154k 1 in./ft = 17 kips 59.0 in. Va = 10 k - (0.9klf +.74klf) 1 in./ft = 84.0 kips Check the available shear strength ithout stiffeners From previous calculation φ v V n = 0.90(134 kips) = 10 kips 10 kips < 17 kips not o.k. Therefore additional stiffeners are required. Use V u =φ v V n to determine the required stress in the eb by dividing by the eb area From previous calculation V n /Ω v = 134 kips / 1.67 = 80.0 kips 80.0 kips < 84.0 kips not o.k. Therefore additional stiffeners are required. Use V a = V n /Ω v to determine the required stress in the eb by dividing by the eb area φvv A n V = u A = 17 kips 11.3in. = 11. ksi V v n Ω A V = a A = 84.0 kips 11.3 in. = 7.43 ksi Use Table 3-16b from the Manual to select the required stiffener a/h ratio based on the h/t ratio of the girder and the required stress. Interpolate and follo an available stress curve, φvvn Vn = 11. ksi for, = 7.43 ksi for, until it intersects the horizontal line for A ΩvA a h/t value of 105. Because the available stress does not intersect the h/t value of 105, the maximum value of 3.0 for a/h may be used. Since h = 33.0 in., an additional stiffener is required at (3.0)(33.0 in.) = 99.0 in. maximum from the previous one. Manual Table 3-16b Select stiffener spacing for the second panel Tension field action is alloed, but not required, since the next panel is not an end panel. Section G3.1 Calculate the required shear strength at the start of the third pane, 158 in. from end V u = 154 kips in. - [1.(0.90 klf) + 1.6(.74 klf)] 1 in./ft = 81.7 kips V a = 10 kips in. - (0.90 klf +.74 klf) 1 in./ft = 53.8 kips

16 G-16 Check the available shear strength ithout stiffeners From previous calculation φ v V n = 0.90(134 kips) = 10 kips 10 kips > 81.7 kips o.k. Therefore additional stiffeners are not required. From previous calculation V n /Ω v = 134 kips / 1.67 = 80.0 kips 80.0 kips > 53.8 kips o.k. Therefore additional stiffeners are not required. The four Available Shear Stress tables, Manual Tables 3-16a, 3-16b, 3-17a and 3-17b, are useful because they permit a direct solution for the required stiffener spacing. Alternatively, you can select a stiffener spacing and check the resulting strength, although this process is likely to be iterative. In the proof belo, the stiffener spacings that ere selected from the charts in the example above are used. Manual Table 3-16

17 G-17 Example G.8b Built-up Plate Girder ith Transverse Stiffeners The stiffener spacings from Example G.8a, hich ere easily determined from the tabulated values of the Steel Construction Manual, are verified belo by directly applying the provisions of the Specification. Verify the shear strength of end panel 5 5 k v = 5 + = 5 + = 6.54 a/ h 1.80 ( ) ( ) Section G.1b h/t = 33.0 in. / in. = 105 Check a/h limits 59 in. a/h = = in. 60 a/h = 1.80 < = ( h/ t ) Therefore, use k v = = 6.13 Tension field action is not alloed since the panel is an end panel. Since h/ t > 1.37 v y ( )( ) ( ) 1.51Ekv ( )( ) h t F = , 000 ksi 6.54 = 0.71 (105) 36 ksi C v = ( / ) y ke/ F = ,000 ksi / 36 ksi = 99.5, V n = 0.6F y A C v = 0.6(36ksi)(11.3 in. )(0.71) = 176 kips Eqn. G-5 Eqn. G-1 Calculate the available shear strength φ v V n = 0.90(176 kips) =159 kips 159 kips > 154 kips o.k. V n /Ω v = (176 kips) / 1.67 = 106 kips 106 kips > 10 kips o.k. Verify the shear strength of the second panel a/h for the second panel as k v = 5 + = 5 + = 5.56 a/ h 3.0 ( ) ( ) Check a/h limits 99 in. a/h = = in. 60 a/h = 3.00 < = ( h/ t ) Therefore use k v = = 6.13 Section G.1

18 G-18 Since h t > 1.37 ( )( ) ( ) / ke/ F = , 000 ksi / 36 ksi = 91.7, v 1.51Ekv C v = ( h/ t ) F = 1.51(9, 000 ksi)(5.56) = (105) (36 ksi) y y Since h t > 1.10 ke/ F = (9,000ksi) /(36ksi) = 73.6, use Eqn. G3- / v 1 C v Vn = 0.6FyA Cv ( a/ h) = 0.6(36 ksi)(11.3 in. ) (3.0) + V n = 176 kips y Eqn. G-5 Section G3. Eqn. G3- Calculate the available shear strength φ v V n = 0.90(176 kips) =158 kips 156 kips > 17 kips o.k. V n /Ω v = (176 kips) / 1.67 = 105 kips 105 kips > 84.0 kips o.k.