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Unformatted text preview: 7.43
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 [741 7.65 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 Slipcritical strength: From AISC Table J31, the minimum bolt tension is T], = 56 kips.
From AISC Equation 134, 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 [712] 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 J45, 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 . ~ " " \ 7.71 (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 [716] Slipcritical 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= 22—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 [717] 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. 7.93
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: [733] 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. ...
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 Fall '07
 Kargahi

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