{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 9 - 7.4-3 A = ”dz/4 = ”(s/4W4 = 0.4418 in.2 Nominal...

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.4-3 A], = ”dz/4 = ”(s/4W4 = 0.4418 in.2 Nominal shear capacity of one bolt in double shear is R» = mm, = 48(0.4418) x 2 = 42. 41 kips The nominal bearing strength of one bolt (gusset plate thickness controls) is 24th,, = 2.4(3/4)(3/8)(58) = 39. 15 kips < 42.41 kips for shear Bearing strength controls. (a) LRFD solution: $12,, = 0.75(39.15) = 29. 36 kips/bolt Pu = 1.2D+ 1.6L =1.2(7)+1.6(14)= 30. 8 kips Number of bolts required is zigojsg = 1. 05 Use 2 bolts [7-41 7.6-5 Determine the nominal strengths for all limit states. For shear, Ag, = 7rd2/4 = 7r(1.125)2/4 = 0.994 in.2 Rn = FmAb = 48(0.994) = 47. 71 kips/bolt For4 bolts, Rn = 4(47.71) = 190. 8 kips Slip-critical strength: From AISC Table J3-1, the minimum bolt tension is T], = 56 kips. From AISC Equation 13-4, Rn = ”DuhmeN, = 0.35(1.13)(1.0)(56)(1.0) = 22. 15 kips/bolt For4bolts, R,1 = 4(22. 15) = 88. 6 kips Bearing: h=d+—1—=IL+L=1.188in. 16 s 16 Edgebolts: LC =Le——£L = 2——1—-—;3§ = 1.406 in. R, = 1.2LctFu) = 1.2(1.406)(1/2)(58) = 48. 93 kips/bolt Upper limit = 2.4thu) = 2.4(1. 125)(l/2)(58) = 78.3 kips/bolt > 48.93 kips/bolt.'. use Rn = 48.93 kips/bolt Otherbolts: L5 = s—h = 3 — 1.188 = 1. 812in. R,I = 1.2LctFu) = 1.2(1.812)(1/2)(58) = 63. 06 kips/bolt [7-12] 24th“) = 78.3 kips/bolt > 63.06 kips/bolt use R" = 63.06 kips/bolt Total Rn = 2(48.93) +2(63.06) = 224.0 kips Tension on the gross section: Ag = 0.5(6.5) = 3. 25 in.2 Pn = FyAg = 36(3.25)=117.0kips Tension on the net section: A. =A,, =Ag—thh = 3.25—2(%)(1%+%) =2.0in.2 P. = FHA, = 53(2.0) = 116.0kips Check block shear (tension member controls). PL‘/2X6‘/z 1 3/4" T'— 3" _Y_ 1 3/ " T t=5/8" Ag. = 2x %(3+2) = 5.0m.2 Am=2x%[3+2—1.5(1~é—+%)] = 3.125in.2 A... = %(3—1%)= 0.875 in.2 For this type of block shear, U1” = 1.0. From AISC Equation J4-5, R" = 0.6FuAnv + UbsFuAn, = 0.6(58)(3. 125) + 1.0(58)(0.875) = 159. 5 kips with an upper limit of 0.6FyAgv + UmFuAm = 0.6(36)(5.0)+1.0(58)(0.875) = 158. 8 kips The nominal block shear strength is therefore 158.8 kips. (a) LRFD solution. Compute design strengths: For bolt shear, ¢Rn = 0.75(190. 8) = 143. lkips For bolt slip, (M,I = 1. 00(88. 6) = 88. 6 kips (Slip is being treated as a serviceability limit state, so (it =1.0.) For bearing, ¢Rn = 0.75(224. 0) = 168.0kips [7—13] For tension on the gross section, (15R,l = 0.90(117.0) = 105. 3 kips For tension on the net section, 45R" = 0.75(116.0) = 87.0 kips Forblock shear, ¢Rn = 0.75(158.8) = 119. 1 kips Tension on the net section controls. ¢R,. = 87.0 = P1, 1.2D+ 1.6L =1.2D+1.6(3D)= 87.0, Solution is: {D =14.5} P=D+L=14.5+3(14.5)=58.0kips P=58.0kips L BA . ~ " " \ 7.7-1 (a) LRFD Solution Pu =1.2D+1.6L = 1.2(40)+1‘6(80) = 176.0kips I Bolt shear (assume that the threads are in shear): Ab = Eda/4 = 7r(1.375)2/4 = 1.485 in.2 Shear capacity of one bolt is V ¢R,. = «my/1b = 0.75(48)(1.485) = 53. 46 kips/bolt [7-16] Slip-critical strength. Assuming Class A surfaces, ¢Rn = ¢pDuhflTbN5 = 1.00[O.35(1.13)(1.0)(85)(1.0)] = 33. 62 kips/bolt Number of bolts required = ‘37g7—gf = 5. 24 try 6 bolts. Minimum 5 acing= 2-2—d = 2.667 11 = 3. 667 in., 4 in. P 3 8 try Minimum edge distance from AISC Table 13.4 = 1% x 1% = 2. 41 in., try 2i in. 2 Try the following layout: 2 1/2" 7" Check bearing. To determine which component to check, compare the product of the thickness and the ultimate stress (the edge distances and spacings are the same for both components). For the gusset plate, z‘Fu = 0.5(58) = 29.0 kips/in. For the tension member, F u = 70 ksi for A242 steel, and thu = 0.448(70) = 31. 36 kips/in. > 29.0 kip/in cheek bearing on the gusset plate. h = 1—38— + 1—16 = 1.438 in. For the holes nearest the edge, L. = L. — g = 2.5 — % = 1. 781 in. (bR,l = ¢(1.2LctF,,) = 0.75(1.2)(1.781)(l/2)(58) = 46. 48 kips The upper limit is ¢(2.4th..) = 0.75(2.4)(1%)(1/2)(58) = 71.78 kips > 46.48 kips use (15R,l = 46.48 kips Forthe other bolts, Lc = s—h = 4 - 1.438 = 2. 562 in. (15R,l = ¢(l.2LEtFu) = 0.75(l.2)(2.562)(1/2)(58) = 66. 87 kips ¢(2.4thu) = 71.78 kips > 66.87 kips use ¢R,. = 66.87 kips [7-17] For the connection, the total bearing strength is 2(46.48) + 4(66. 87) = 360 kips > Pu = 175 kips (OK) \V Use 6 bolts as shown. 7.9-3 Pu = 1.2D+ 1.6L = 1.2(20)+1.6(60) = 120.0 kips Tu = %(120) = 72.0 kips, Vu = @120) = 96.0 kips Assume that the tension strength will control: [7-33] 1:1“: 1.3Fnz- Fm fl 5 Fur (15an =1.3(90)- %fi 5 9o =117—2.5f. s 90 m, = 0.75(117—2.5fv) s 075(90) = 37.75 —1.s75f. s 67.7 Letng'm: —%andfv = 29761, 72137 = 8775—1375(2—1167) 72 = 87,752.41, — 180.0 2 EA], = 2.872 in.2 2 The area of one bolt is Ab = ”(718 = 0.6013 in.2 Required N1, = Eff = 026807123 = 4.78 Try 6 bolts for symmetry. First, check the upper limit on Fin: =__96 =_9£_= ' f“ 2.45 6(0.6013) 26'“ 1‘51 FL, = 117 — 2. Sfi =117 — 2.5(26.61) = 50.48 ksi < 90 ksi (OK) Check shear: Vu/bolt = 96/6 = 16.0 kips/bolt ¢Rn = ¢anAb = 0.75(48)(0. 6013) = 21.6 kips/bolt> 16.0 kips/bolt (OK) Bearing strength (the WT flange thickness controls): ¢(2.4th,,) = 0.75(2.4)(7/8)(0.345)(65) = 35. 3 kips/bolt > 16.0 kips/bolt (OK) Use 6 bolts. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern