CE456 HW 1 Solution

CE456 HW 1 Solution - 3.2-5 For a thickness oft = 3/8 in F...

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Unformatted text preview: ' 3.2-5 For a thickness oft = 3/8 in., F y = 50 ksi and F u = 70 ksi. First, compute the nominal strengths. For the gross section, Ag = 7.5(3/8) = 2. 813 in.2 P" = FyAg = 50(2.813) = 140. 7kips Net section: A. = 2.313 — (%) (1% + %)(2) = 1. 876 in.2 A. = An = 1.876 in.2 P" = FuAe = 70(1.876) = 131. 3kips a) The design strength based on yielding is ¢,P,, = 0.90(140.7) = 127 kips The design strength based on fracture is qb,P,, = 0.75(131.3) = 98.5 kips The design strength is the smaller value: ¢,P,, = 98. 5 kips Factored load: Combination 1: 1.4D = 1.4(25) = 35.0 kips Combination 2: 1.2D + 1.6L = 1.2(25) + 1.6(45) = 102 kips The second combination controls; Pu = 102 kips. Since Pu > ¢,P,., (102 kips > 98.5 kips), The member is unsatisfactory. 3.3—5 Gross section: Pn = F},Ag = 36(5.86) = 211.0 kips Net section: An = 5.86 — (—3—) (1 + 31—)(2): 4. 454 in.2 = _____1L= 1 (3+3+3) 0.8856 A; = AnU = 4.454(0.8856) = 3. 944 in.2 P" = FuAe = 58(3.944) = 228. 8 kips (a) The design strength based on yielding is ¢,P,, = 0.90(211.0) = 190 kips U=1— ©l><l The design strength based on fracture is (litPn = O.75(228. 8) = 172 kips The design strength is the smaller value: ¢,P,, = 172 kips Load combination 2 controls: Pu = 1.2D+ 1.6L =1.2(50)+1.6(100)= 220 kips Since Pu > ¢,P,,, (220 kips > 172 kips), The member is not adequate. 3.4—3 Gross section: Ag = 8(3/8) = 3.0 in.2, P” = FyAg = 36(3.0) = 108 kips - . - _ L i = i ' Net SCCtIOD. Hole dlameter— 2 + 8 8 1n. A" = Ag —th x (dor d’) = 3 — (3/8)(5/8) = 2. 766 in.2 or A: 3 - (3/8)(5/8) — (3/8)[5/8 — %] = 2. 954 in.2 or A" = 3 — (3/8)(5/8) - (3/8)|:5/8 — %] x 2 = 3. 141 in.2 or A" = [3 — (3/8)(5/8)(2)] x g— = 3. 03s in.2 or A” = (3 — (3/8)(5/8) —(3/8)[5/8— %](z)) x Use A? = An = 2.766 in.2 = 3.460 in.2 mlm n = FuAe = 58(2.766) = 160. 4kips I (a) Gross section: ¢,P,, = 0.90(108) = 97. 2 kips Net section: ¢,P,, = O.75(160.4) = 120 kips ¢,P,, = 97.2 kips 3.5-3 Tension member: Theshearareasare Agv=—Zé—(3 5+1. 5)x2= 4. 375 1n __:l_ _ _3_ _ = - 2 Anv—16[3.5+1.5 15 5(4+8 )]x2 3.227m. 3.0 (0. 5+0. 5)(% +%)] = 0.9297111.2 For this type of connection, Ubs = 1.0, and from AISC Equation J4-5, Rn = 0. 6F11Anv + UbsFuAnt = O.6(58)(3.227) + 1.0(58)(0.9297) = 166. 2 kips The tension area is Am = fil: with an upper limit of O.6FyAgv + UbsFuAm = O.6(36)(4.375)+1.0(58)(0.9297) = 148. 4 kips The nominal block shear strength of the tension member is therefore 148.4 kips. Gusset Plate: Ag. = %(3.5+2.5)x2 = 4.51112 A". = %[3.5 +2.5 —1.5(7/8)]x 2 = 3. 516 111.2 A,“ = —§—[3.0 — (0.5 + O.5)(7/8)] = 0.7969 in.2 From AISC Equation J4-5, Rn = 0. 6FuAnv + UbsFuAm = O.6(58)(3. 516) + 1.0(58)(O.7969) = 168. 6 kips with an upper limit of 0.6FyAgv + UbsFuAm = O.6(36)(4.5)+1.0(58)(0.7969) = 143. 4 kips The nominal block shear strength of the gusset plate is therefore 143.4 kips The gusset plate controls, and the nominal block shear strength of the connection is 143.4 kips (a) The design strength is ¢Rn = O.75(143.4) = 108 kips qun = 108 kips ...
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