Solution to Homework 5: Problem : Given and Required: Solution: a) Edge distance given =.5 > mimum =. (AISC Table J3.4) Spacg given = 3.0 or 2.5. > mimum = 2.67d b = 2.0. Spacg given = 3.0 or 2.5. > preferred = 3d b = 2.25. Therefore, all conditions are satisfied. b) Given lecture, F y = 50 ksi, F u = 70 ksi
For the bearg strength associated with the plate: t := 0.5 Fu := 70 ksi db := 0.75 6 At Edges: Lce :=.5 2 Lce =.094 φr 0.75 (.2 φrn = 34.453 47.25 Therefore φrn = 34.453 At Other Holes: Lce := 3 Lce = 2. 6 φr 0.75 (.2 φrn = 6.906 47.25 Therefore φr φrnmax φrn = 47.25 Bearg_Strength := 2 34.453+ 4 47.25 Bearg_Strength = 257.906
For the bearg strength associated with the gusset plate: If we assume that Le for gusset = Le for plate, Then: 3 t := Fu := 70 ksi db := 0.75 6 At Edges: Lce :=.5 2 Lce =.094 φr 0.75 (.2 φrn = 25.4 φrnm := φrnm = 35.437 Therefore φrn = 25.4 At Other Holes: Lce := 3 Lce = 2. 6 φr 0.75 (.2 φrn = 5.6 35.437 Therefore φr φrnmax φrn = 35.437 Bearg_Strength := 2 25.4 + 4 35.437 Bearg_Strength = 93.42 This is the acceptable answer: However, if we assume that Le for gusset is long as it appears to be from drawg, Then 35.437 Therefore φrn = 35.437 Bearg_Strength := 6 35.437 Bearg_Strength = 22.622 In both cases, the gusset plate controls
Problem 2 Given and Required: Solution: Step I: Determe Factored Load: D:= 40.4D = 56 L := 00.2D +.6L = 20 controls Step II: Check the bolts for shear capacity: 7 db := Ab := π db2 4 Design Shear Strength bolt: Shear Strength of Connection: Ab = 0.60 2 52 Vb = 26.475 F 4 Vb = 2.64 ksi φ := 0.75
Step III: Check bearg at strength of bolt holes angles: t := 0.5 Le := 2 s := 3 Fu := 5 Use Table 7-3 Use Table 7-2 ksi (A36 steel) Lce := Le 6 2 Lce =.53 φr 0.75 (.2 φrn = 39.966 Lcs := s 6 φrn = 39.966 79.9 t = 39.95 45.675 Lcs = 2.063 φr 0.75 (.2 Lcs t Fu) φr φrnmax φrn = 45.675 φrn = 53.3 9.3 t = 45.65 Bearg_Strength := 2 39.966+ 45.675 45.675 Bearg_Strength = 445.332 Step IV: Check bearg at strength of bolt holes gusset: 3 t := Fu := 5 ksi (A36 steel) Le := 2 Use Table 7-3 s := 3 Use Table 7-2 Lce := Le 6 2 Lce =.53 φr 0.75 (.2 φrn = 29.974 Lcs := s 6 φrn = 29.974 79.9 t = 29.963 Lcs = 2.063 34.256 φr 0.75 (.2 Lcs t Fu) φr φrnmax 9.3 t = 34.237 φrn = 34.256 φrn = 40.373 Bearg_Strength := 29.974+ 4 34.256 34.256 Bearg_Strength = 66.99 This is less than the factored load of 20, not acceptable
Problem 3 Given and Required: A WT7x9 of A572 Grade 50 steel is used as a tension member. It is connected to a 3/ gusset plate, also of A572 steel, with 7/ ch diameter bolts. The connection is through the flange of the tee. The connection must resist a service live load of 90 and a service dead load of 45. Assume, the nomal bearg strength will be 2.4d b tf u and answer the followg questions. a) How many A307 bolts are required? b) How many A325 bolts are required? c) How many A490 bolts are required? d) If relative costs of A307, A325, and A490 bolts are ratio of.0:.7:2.6, which type of bolt is most economical for connection? Solution: The thickness of the gusset plate is lower than the thickness of the flange. Therefore, bearg strength of the gusset will govern:
Step I: Determe Factored Load: D:= 45.4D = 63 L:= 90.2D +.6L = 9 controls Step III: Determe number of bolts based on bearg of gusset: 3 t := bolt: Fu := 65 φr ksi (A36 steel) 7 db := φrn = 3.39 9 φrn n = 5.5 6 bolts Step III: Determe number of bolts bases on shear capacity of connection: 7 db := Ab := π db2 4 a) F 24 ksi Design Shear Strength bolt: Ab = 0.60 2 Vb = 0.24 φ := 0.75 9 Vb n =.293 20 bolts b) F 4 ksi Design Shear Strength bolt: Vb = 2.64 c) F 60 ksi Design Shear Strength bolt: 9 Vb n = 9.47 Vb = 27.059 0 bolts 9 Vb n = 7.37 bolts d) For A307 bolts, Nomal cost: For A325 bolts, Nomal cost: For A325 bolts, Nomal cost: 20 = 20 0.7 = 7 2.6 = 20. units units units Therefore, A325 bolts are most economical for this connection: