solucionario-capitulo-7

Steel Structures by S. Vinnakota

Chapter 7

page 7-1

CHAPTER 7 P7.1.

Determine the net area for the following tension members: (a) PL ½ x8 with holes for ¾ -in.-dia. bolts, as shown in Fig. 7.6.1a. (b) W12x120 with STD drilled holes for f -in.-dia. bolts, as shown in Fig. 7.6.1b. (c) PL ½ x8 with holes for ¾ -in.-dia. bolts with a gage, g = 2¾ in., pitch, p = 3 in., and a staggered pitch, s = 1½ in., as shown in Fig. 7.6.2a. (d) L4x3½ x ½ with holes for ¾ -in.-dia. bolts with pitch, p = 2½ in., and a staggered pitch, s = 1¼ in., as shown in Fig. 7.6.3a. (e) C8x18.75 with holes for ¾ -in.-dia. bolts with pitch, p = 2 ½ in., staggered pitch, s = 1¼ in., gage, g a = 1½ in., and gage, g b = 2½ in., as shown in Fig. 7.6.4a. Solution P7.1.a See Fig. 7.6.1a of the text book. Member: PL ½×8; Diameter of bolts, d = ¾ in. Assume standard punched holes. So, width of bolt holes, d e = ¾ + c = f in. Number of holes, n = 3 Gross area, A g = ½ × 8 = 4.00 in.2 From Eq. 7.6.3, net area, A n = 4.00 - 3 × f× ½ = 2.69 in.2

(Ans.)

See Fig 7.6.1b of the text book. Member: W12x120 From LRFDM table 1-1: A = 35.3 in.2; tf = 1.11 in.; tw = 0.710 in. Number of holes in the flanges = 2 + 2 = 4 Number of holes in the web = 3 Diameter of bolts, d = f in. Standard drilled holes. So, width of bolt holes, d e = f + 1/16 = 15/16 in. Gross area, Ag = 35.3 in.2 From Eq. 7.6.3, net area, An = 35.3 - 4 × (15/16) × 1.11 - 3 × (15/16) × 0.710 = 29.1 in.2

(Ans.)

P7.1.b

P7.1.c See Fig. 7.6.2a of the text book. Member: PL ½×8; Diameter of bolts, d = ¾ in. Assume standard punched holes. So, width of bolt holes, d e = ¾ + c = f in. Staggered pitch, s = 1 ½ in.; Gage, g = 2 ¾ in. Net areas for the two possible failure paths are calculated as follows, using Eqs. 7.6.3 and 7.6.4: path a-b-c-d: A n1 = 4.00 - 1× f × ½ = 3.56 in.2 path a-b-d-e:

A n2 = 4.00 - 3 × f × ½ + 2 ×

× ½ = 2.89 in.2

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 7

page 7-2

Net area, A n = min [ A n1, A n2] = min [3.56; 2.89] = 2.89 in.2

(Ans.)

P7.1.d See Fig. 7.6.3a of the text book. Member: L4×3½× ½ From LRFDM Table 1-7: A = 3.50 in.2; t = ½ in. Diameter of bolts, d = ¾ in. Staggered pitch, s = 1 ¼ in. Assume standard punched holes. So, width of bolt holes, d e = ¾ + c = f in. From Table 6.2.1: g1(4 in.) = 2½ in.; g 1(3.5 in.) = 2 in. From Fig. 7.6.3b, g ab = g a + g b - t = 2½ + 2 - ½ = 4.00 in. Net areas for the two possible failure paths are calculated as follows, using Eqs. 7.6.3 and 7.6.4: path a-b-c: A n1 = 3.50 - 1× f × ½ = 3.06 in.2 path a-b-d-e:

A n2 = 3.50 - 2 × f × ½ +

× ½ = 2.67 in.2

Net area, A n = min [ A n1, A n2] = min [3.06; 2.67] = 2.67 in.2

(Ans.)

P7.1.e See Fig. 7.6.4a of the text book. Member: C8x18.75 From LRFDM table 1-1: A = 5.51 in.2; tf = 0.390 in.; t w = 0.487 in.; g = 1½ in. Gages: ga = 1½ in. in the flange; g b = 2½ in. in the web. To determine the net area, two failure paths are identified: path 1: See Fig. 7.6.4b. From Eq. 7.6.8, g ab = g a + g b - ½ (tf + tw) = 1.50 + 2.50 - 0.5 × (0.390 + 0.487) = 3.56 in. t = ½ (tf + tw) = 0.5 × (0.390 + 0.487) = 0.439 in. A n1 = 5.51 - 2 × f × 0.487 - 2 × f × 0.390 + = 4.07 in.2

² controls

path 2 : Net area, A n = min [ A n1, A n2] = min [4.66; 4.07] = 4.07 in.2

(Ans.)



7.2.

Chapter 7

page 7-3

Determine the reduction coefficient, U, using Eq. 7.7.2 for the following tension members: (a) A L4x4x ½ connected to a gusset by two bolts with a pitch, p = 3 in. (Fig. 7.7.4b). (b) A W12x65 connected by two lines of bolts in each flange to gussets using a pitch, p = 3 in. (Fig. 7.7.4c). There are three bolts in each line. (c) A W12x35 connected by two lines of bolts in each flange to gussets using a pitch, p = 3 in. (Fig. 7.7.4e). There are three bolts in each line. (d) A WT7x45 connected by two lines of bolts in the flange to a gusset using a pitch, p = 3 in. (Fig. 7.7.4d). There are three bolts in each line. (e) A PL ½ x6 connected to a gusset by two 9 -in.-long longitudinal welds, as shown in Fig. 7.7.5a. (f) A L6x4x ½ with its long leg connected to a gusset by a 6 -in.-long transverse weld only (Fig. 7.7.5b). (g) A L6x4x ½ with its long leg connected to a gusset by a transverse weld and two longitudinal welds, having L w1 = 8 in. and L w2 = 4 in. (Fig. 7.7.5c).

Solution P7.2.a See Fig. 7.7.4b of the text book. Member L4x4x½. From Table 1-7 of the LRFDM, Pitch, p = 3 in.; number of bolts per line = 2 Length of connection, L con = 3.00 in. Connection eccentricity, From Eq. 7.7.2, reduction coefficient, (Ans.)

P7.2.b See Fig. 7.7.4c of the text book. Member: W12x65 with flanges connected to gussets. There are three bolts in each line. Pitch, p = 3 in. So, length of connection, L con = 3 × 3.0 = 9.00 in. From Fig. 7.7.2c of the text book, connection eccentricity, From Eq. 7.7.2, reduction coefficient, (Ans.)

P7.2.c See Fig. 7.7.4e of the text book. Member: W12x35 with flanges connected to gussets. There are three bolts in each line. Pitch, p = 3 in. So, length of connection, L con = 3 × 3.0 = 9.00 in.



Chapter 7

page 7-4

From Fig. 7.7.2c of the text book, connection eccentricity, From Eq. 7.7.2, reduction coefficient, (Ans.)

P7.2.d See Fig. 7.7.4d of the text book. Member: WT7x45 with its flange connected to a gusset. There are three bolts in each line. Pitch, p = 3 in. So, length of connection, L con = 3 × 3.0 = 9.00 in. Connection eccentricity, From Eq. 7.7.2, reduction coefficient, (Ans.)

P7.2.e See Fig 7.7.5a of the text book. Member: PL ½x6 6 width of plate, W pl = 6.00 in. Connection by two longitudinal welds only. Length of longitudinal weld, L lw = 9 in. L lw /W pl = 1.50 > 1.0 O.K. Also, from Eq. 7.7.3, reduction coefficient U = 0.87

(Ans.)

See Fig. 7.7.5b of the text book. Member: L6x4x ½ with only its long connected to gusset. Connection provided by a transverse weld only. So, from Eq. 7.7.4, A n = A ce = 6.00 × ½ = 3.00 in.2; U = 1.00

(Ans.)

P7.2. f

P7.2. g See Fig. 7.7.5c of the text book. Member: L6x4x ½ with only its long connected to gusset. From Table 1-7 of the LRFDM, Connection provided by a transverse weld and two unequal length longitudinal welds. L lw 1 = 8 in.; L lw 2 = 4 in. Length of connection, L con = max (8.00; 4.00) = 8.00 in. Connection eccentricity, From Eq. 7.7.2, reduction coefficient, (Ans.)



P7.3.

Chapter 7

page 7-5

A 14x ½ in. A242 steel plate is used as a tension member. The end connection to the gusset plate is by six f -in.-dia. bolts, arranged with staggered pitches of 2 in. and 1 ½ in., and gage distances of 2 ½ in. as shown in Fig. P7.3. Determine the design tensile strength of the member. Neglect block shear failure. See Figure P7.3 of the text book. Solution From LRFDM Table 2-2, ½ -in.-thick A242 steel plates are available in Grade 50 only, for which Fy = 50 ksi; F u = 70 ksi; Gross area, A g = ½ × 14.0 = 7.00 in.2 T d1 = N Fy A g = 0.90 × 50 × 7.00 = 315 kips Diameter of bolts, d = f in. Assuming standard punched holes, d e = f + c = 1.0 in. T d2 = N Fu U A n where U = 1.0 for a plate. Net areas for several failure paths are calculated as follows: path a-b-c: A n1 = 7.00 - 1× 1.0 × ½ = 6.50 in.2 path d-e-f-g: = 7.00 - 2× 1.0 × ½ = 6.00 in.2 6 A n2 = 6.00 × 6/5 = 7.20 in.2 path d-e-b-f-g: A n3 = 7.00 - 3× 1.0 × ½ +

× ½ × 2 = 5.73 in.2

path h-i-j-k-l:

= 7.00 - 3× 1.0 × ½ = 5.5 in.2 6

path h-i-e-f-k-l:

= 7.00 - 4× 1.0 × ½ +

A n4 = 5.5×6/3 = 11.0 in.2

× ½ × 2 = 5.40 in.2

6 A n5 = 5.40 × 6/5 = 6.48 in.2 path h-i-e-b-f-k-l: A n6 = 7.00 - 5× 1.0 × ½ +

× ½ × 2 = 5.13 in.2

² controls

T d2 = 0.75 × 70 × 1.00 × 5.13 = 269 kips

T d = min [315.0, 269] = 269 kips The design strength of the member therefore equals 269kips, and is controlled by the limit state of tension rupture. (Ans.)

Comment: If we assume A490-X type bolts, then, from Table 6.7.1, the design shear strength of a A490-X type bolt in single shear is, B dv = 33.8 kips C dv = 33.8 × 6 = 203 kips 7 Controls Also, the design bearing strength for interior and end bolts can be calculated using Tables 6.8.1 and 2 as follows: C db = [3(53.8) + 3(91.3)]× ½ × (70 ksi/58 ksi) = 263 kips T d = min [315.0, 269, 203 ] = 203 kips The design strength of the member therefore equals 203 kips, and is controlled by the limit state of bolt shear.



Chapter 7

page 7-6



P7.4.

Chapter 7

page 7-7

The web of a MC13x35 steel tension member is connected to a gusset plate by four rows of ¾ -in.-dia. bolts using a gage of 3 in. and a pitch of 3 in. as shown in Fig. P7.4. Assume A588 steel and determine the design tensile strength of the member. Neglect block shear failure. See Figure P7.4 of the text book. Solution MC13×35: A = 10.3 in.2; tw = 0.447 in.; From LRFDM Table 2-1, for A588 steel, Fy = 50 ksi;

F u = 70 ksi

Limit state of yielding in gross section gives (Eq. 7.4.5), T d1 = N Fy A g = 0.90 × 50 × 10.3 = 464 kips Bolt diameter, d = ¾ in. Assuming standard punched holes, d e = ¾ + c = f in. Net area: path a-b-c-d: A n1 = 10.3 - 2× f×0.447 = 9.52 in.2 path e-f-g-h-j-k: = 10.3 - 4 × f × 0.447 = 8.74 in.2

6 An2 = 8.74 × 14/12 = 10.2 in.2 path a-b-g-h-c-d: A n3 = 10.3 - 4 × f × 0.447 + 2 × = 9.41 in.2

× 0.447

² controls

Length of connection, L con = 3 × 3.0 = 9.00 in. Connection eccentricity, From Eq. 7.7.2, reduction coefficient,

Limit state of fracture in net section gives (Eqs. 7.4.6 and 7.7.1), T d2 = N Fu U A n = 0.75×70×0.891×9.41 = 440 kips From Eq. 7.4.4, T d = min [T d1, T d2] = min [464, 440] = 440 kips Design tensile strength of the member is T d = 440 kips, and is controlled by the limit state of fracture in net section. (Ans.)



P7.5.

Chapter 7

page 7-8

A L4x4x ½ of A572 Grade 50 steel is used as a tension member. What is the design tensile strength of the member: (a) If only a transverse weld is used to connect the member to the gusset, as shown in Fig. P7.5? (b) If, in addition, two 6-in.-long longitudinal welds are provided? See Figure P7.5 of the text book. Solution A572 Gr 50 steel: Fy = 50 ksi, Fu = 65 ksi From LRFDM Table 1-7, for a L4×4×½: A = 3.75 in.2; Limit state of yielding in the gross section gives (Eq 7.4.5): T d1 = N Fy A g = 0.90×50×3.75 = 169 kips a.

Transverse weld only Since the tension load is transmitted only by the transverse weld Area of the directly connected elements, A ce = 4×½ = 2.00 in.2 Also, U = 1.0 From Eq. 7.7.4, A e = U A ce = 2.00 in.2 Limit state of yielding in the gross section gives (Eq. 7.4.6): T d2 = N Fu A e = 0.75×65×2.00 = 97.5 kips From Eq. 7.4.4, T d = min [T d1, T d2] = min [169, 97.5] = 97.5 kips The design strength of the member equals 97.5 kips, and is controlled by the limit state of fracture on the effective net area of the section. (Ans0)

b.

Transverse weld and longitudinal welds Length of longitudinal weld, L lw = 6.00 in. Length of connection, L con = L lw = 6.00 in. Connection eccentricity, From Eq. 7.7.2, reduction coefficient,

Limit state of fracture in net section gives (Eqs. 7.6 and 7.7.1), T d2 = N Fu U A n = 0.75 × 65 × 0.803 × 3.75 = 146 kips From Eq. 7.4.4, T d = min [T d1, T d2] = min [169, 146] = 146 kips Design tensile strength of the member is T d = 146 kips, and is controlled by the limit state of fracture in net section. (Ans.)



Chapter 7

page 7-9



P7.6.

Chapter 7

page 7-10

The web of a C12x25 steel tension member is connected to a gusset by three rows of f -in.-dia. A490-X bolts using a gage of 3 in. and a staggered pitch of 1½ in. as shown in Fig. P7.6. Assume A588 steel and determine the design tensile strength of the member. Neglect block shear rupture. See Figure P7.6 of the text book. Solution From LRFDM Table 2-1, for A588 Gr 50 steel: F y = 50 ksi; F u = 70 ksi From LRFDM Table 1-5, for a C12×25: A = 7.34 in.2; tw = 0.387 in.; Diameter of bolts, d = f in.

= 0.674 in.

Limit state of yielding in the gross section gives (Eq 7.4.5): T d1 = N Fy A g = 0.90 × 50 × 7.34 = 330 kips Length of connection, L con = 5 × 1.5 = 7.50 in. Connection eccentricity, Reduction coefficient,

Assume STD punched holes. d e = f + c = 1 in. To determine the net area, 3 possible failure paths are identified. We have, for: path a-b-c : A n1 = 7.34 - 1 × 1.00 × 0.387 = 6.95 in.2 path d-e-f-g :

= 7.34 - 2 × 1.0 × 0.387 = 6.57 in.2 6 A n2 = 6.57×

path d-e-b-f-g : A n3

= 7.34 - 3 × 1 × 0.387 + 2 ×

= 7.40 in.2

× 0.387

= 6.32 in.2 ² controls Limit state of fracture in net section gives (Eqs. 7.4.6 and 7.7.1), T d2 = N Fu U A n = 0.75 × 70 × 0.9 × 6.32 = 299 kips Bolts: Nine f -in.-dia. A490-X type bolts; single shear; bearing on 0.387-in.-thick web. There are six interior bolts with pitch, p = 3 in., two end bolts with L e = 3.5 in. and two end bolts with L e = 2 in. From Table 6.7.1, B dv = 33.8 kips 6 C dv = 9 × 33.8 = 304 kips Using the values given in Tables 6.8.1 and 6.8.2 for Fu = 65 ksi, the design bearing strength of the connectors is obtained as C db = [6×102 + 1×102 + 2×89.6] (70 ksi /65 ksi) (0.387) = 372 kips Strength of connectors, T d4 = min [304, 372] = 304 kips

From Eq. 7.4.3, T d = min [330, 299, 304] = 299 kips The design tensile strength of the member is 299 kips and is controlled by the limit state of fracture in the net section. (Ans.)



P7.7.

Chapter 7

page 7-11

A C12x25 channel shape is used as a tension member. At each end, the web of the channel is connected by six ¾ -in.-dia. bolts ( in two rows, using a pitch of 4 in. and a gage of 6 in.) to a thick gusset plate. In addition there are two holes in each flange (located on the normal gage line) to connect bracing members in the perpendicular plane. The holes in the flanges are staggered 2 in. with respect to the holes in the web, as shown in Fig. P7.7. Determine the design strength of the member. What is the maximum recommended length of the member as per LRFDS? See Figure P7.7 of the text book. Solution A36 steel: Bolts: C12 × 25:

Fy = 36 ksi; d = ¾ in.; A = 7.34 in.2; tf = 0.501 in.;

Fu = 58 ksi d e = ¾ + c = f in. r x = 4.43 in.; r y = 0.779 in. tw = 0.387 in.;

Limit state of yielding in the gross section gives (Eq 7.4.5): T d1 = N Fy A g = 0.90 × 36 × 7.34 = 238 kips Length of connection, L con = 4 × 2.0 = 8.00 in. Connection eccentricity, Reduction coefficient,

To determine the net area, two failure paths are identified : path 1-1: See Fig. 7.6.4b. From Eq. 7.6.8, g ab = g a + g b - ½ (tf + tw) = 1.75 + 3.0 - 0.5 × (0.501 + 0.387) = 4.31 in. t = ½ (tf + tw) = 0.5 × (0.501 + 0.387) = 0.444 A n1 = 7.34 - 2 × f × 0.387 - 2 × f × 0.501 + = 5.99 in.2

² controls

path 2-2: Limit state of fracture in net section gives (Eqs. 7.4.6 and 7.7.1), T d2 = N Fu U A n = 0.75×58×0.9×5.99 = 235 kips From Eq. 7.4.4, T d = min [T d1, T d2] = min[238; 235] = 235 kips The design strength of the member is 235 kips, and corresponds to the limit state of fracture in the net section. (Ans.) From Eq. 7.9.1, length of the member, L # 300 rmin = 300×0.779 = 234 in. = 19.5 ft The maximum recommended length of the member = 19.5 ft

(Ans.)



P7.8.

Chapter 7

page 7-12

Determine the design tensile strength of a pair of 6×4× ½ in. angles with their long legs connected to a e in. gusset plate (Fig. P7.8). Use A36 steel and f in.-dia. bolts. The bolts are arranged on standard gages. The tensile force T is transmitted to the gusset plate by the six bolts on lines A and B. Assume open holes in the outstanding legs (for connection to braces in the plane perpendicular to the plane of the paper). See Fig. P7.8 of the text book. Solution Member: 2L6×4×½ LLBB From LRFDM Table 1-7 for a L 6×4× ½: A L = 4.72 in.2; t = ½ in.; Diameter of bolts, d = f in. Staggered pitch, s = 2 in. Material: A36 steel Gross area, Ag = 2 × 4.72 = 9.44 in.2 Limit state of yielding in the gross section gives (Eq 7.4.5): T d1 = N Fy A g = 0.90 × 36 × 9.44 = 306 kips Assume standard punched holes. So, width of bolt holes, d e = f + c = 1.00 in. From Table 6.2.1: g2(6 in.) = 2¼ in.; g 3(6 in.) = 2½ in. Using Eqs. 7.6.3 and 7.6.4, net areas for the two possible failure paths are as follows,: path a-b-c: A n1 = 9.44 - 1× 1.00 × ½ × 2 = 8.44 in.2 path a-b-d-e-f: A n2 = 9.44 - 3 × 1.00 × ½ × 2 +

× ½× 2

= 6.84 in.2

Net area, A n = min [ A n1, A n2] = min [8.44; 6.84] = 6.84 in.2 As the member force is transmitted by bolts in the long leg only, reduction factor U < 1.0. For the connection with the long leg of the angle bolted to the gusset, connection eccentricity, Length of connection, L con = 5 × 2.0 = 10.0 in. From Eq. 7.7.2, reduction coefficient,

Limit state of fracture in net section gives (Eqs. 7.4.6 and 7.7.1), T d2 = N Fu U A n = 0.75×58×0.900×6.84 = 268 kips The bolts connecting the angles to the gusset are in double shear. From Table 6.7.1, the design shear strength of a in.-dia. A325X bolt in double shear is obtained as 54.1 kips. Design strength of the connectors corresponding to the limit state of bolt shear is, C dv = N B dv = 6 × 54.1 = 325 kips From Table 6.8.1, the design bearing strength of a f -in.-dia. interior bolt (STD holes, p = 4 in., t = 1 in., Fu = 58 ksi), is read as 91.3 kips. Similarly, from Table 6.8.2, the design bearing strength of a f -in.-dia. end bolt (STD holes, L e = 2 in., t = 1 in., Fu = 58 ksi), is read as 79.9 kips. So, the design strength of the connectors corresponding to the limit sstate of bearing on e -in.-thick gusset plate is obtained as: C db = [5×91.3 + 1×79.9] (e) = 335 kips PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 7

page 7-13

Strength of connectors, T d4 = min [325, 335] = 325 kips From Eq. 7.4.3, T d = min [306, 268, 325] = 268 kips The design tensile strength of the member is 268 kips and is controlled by the limit state of fracture in the net section. (Ans.)



P7.9.

Chapter 7

page 7-14

The end connection for a W14x82 tension member has two lines of f -in.-dia. bolts in each flange (Fig. P7.9). There are 3 bolts in each line arranged with a pitch, p = 4 in.; end distance, L e = 2 in.; and side distance, L s = 2 in. Determine the block shear rupture strength. See Figure P7.9 of the text book. Solution Connection is to the flanges of a W14×82: t f = 0.855 in. from LRFDM Table 1-1. A 992 steel: Fy = 50 ksi; F u = 65 ksi Diameter of bolts, d = f in. Pitch, p = 3 in.; end distance, L e = 2 in.; side distance, L s = 2 in. Assume standard punched hoes 6 d e = f + c = 1.0 in. There are four rectangular failure blocks (two in 10 -in.-long (= 2p + L e). A gt = 4 × 2.00 × 0.855 = A nt = 6.84 - 4 × ½ ×1.0 × 0.855 = A gv = 4 × 10.0 × 0.855 = A nv = 34.2 - 4 × 2½ × 1.0 × 0.855 = Tension rupture component, N Fu A nt Shear rupture component, N (0.6F u) A nv Tension yield component, N Fy A gt Shear yield component, N (0.6F y) A gv

= = = =

each flange). Each block is 2 -in.-wide (= L s), and 6.84 in.2 5.13 in.2 34.2 in.2 25.7 in.2

0.75×65×5.13 0.75×0.6×65×25.7 0.75×50×6.84 0.75×0.6×50×34.2

= = = =

250 kips 752 kips 7 Controls 257 kips; 770 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu)A nv + min [N Fy A gt; N Fu A nt] = 752 + min [257; 250] = 1002 kips The block shear rupture strength is 1002 kips and is controlled by fracture on both the shear and tension faces of the four rectangular blocks. (Ans.)



P7.10.

Chapter 7

page 7-15

Determine the block shear rupture strength of the 2L4x3½x3/8 LLBB tension member shown in Fig. P7.10. The angles are connected to a gusset by three f -in.-dia. bolts, using a pitch, p = 3 in; end distance, L e= 1½ in.; and side distance, L s = 1½ in. Assume A 572 Grade 60 steel. See Figure P7.10 of the text book. Solution 2L4×3½× d: t = d in. A572 Grade 60 steel : Fy = 60 ksi; Fu = 75 ksi Bolt diameter, d = f in. Pitch, p = 3 in.; end distance, L e = 1½ in.; side distance, L s = 1½ in. Assume standard punched holes. 6 d e = f + c = 1.0 in. There are two rectangular failure blocks (one in each angle). Each block is 1½ -in.-wide (= L s), and 7½ -in.-long (= 2p + L e). A gt = 2×1.50× d = 1.13 in.2 A nt = 1.13 - 2 × ½ × 1.0 × d = 0.750 in.2 A gv = 2×7.50× d = 5.63 in.2 A nv = 5.63 - 2× 2½ × 1.0 × d = 3.75 in.2 Tension rupture component, N FuA nt = 0.75 × 75 × 0.750 = 42.2 kips; Shear rupture component, N (0.6F u)A nv = 0.75 × 0.6 × 75 × 3.75 = 127 kips 7 controls Tension yield component, N FyA gt = 0.75 × 60 × 1.13 = 50.9 kips; Shear yield component, N (0.6F y)A gv = 0.75 × 0.6 × 60 × 5.63 = 152 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 127 + min [50.9; 42.2] = 169 kips The block shear rupture strength is 169 kips and is controlled by fracture on both the shear and tension faces of the two rectangular blocks. (Ans.)



P7.11.

Chapter 7

page 7-16

A ½ x6 in. plate is welded to a d --in.-thick gusset plate of A36 steel by ¼ in. fillet welds using E70 electrodes. Two 6 -in.-long longitudinal welds are provided in addition to a 6 -in.-long transverse weld. Determine the design tensile strength of the member. Include block shear strength of the gusset plate. See Figure P7.11 of the text book. Solution Member: PL ½×6 6 A g = ½ × 6 = 3.00 in.2 Gusset plate: d -in.-thick A36 steel: Fy = 36 ksi; Fu = 58 ksi Limit state of yielding in the gross section gives (Eq 7.4.5): T d1 = N Fy A g = 0.90 × 36 × 3.00 = 97.2 kips As the end connection is by welds using both transverse and longitudinal welds, A n = A g and U = 1.0. Limit state of rupture in the net section gives (Eq 7.4.6): T d2 = N Fu U A n = 0.75 × 58 × 1.0 × 3.00 = 131 kips Block shear rupture of the gusset plate Block shear failure consists of a single, rectangular block of the gusset plate. The block is 6-in.wide, 6½ -in.-long and d -in.-thick. We have: A gt = 6.0 × d = 2.25 in.2 = A nt A gv = 6.5 × d × 2 = 4.88 in.2 = A nv Tension rupture component, N Fu A nt = 0.75 × 58 × 2.25 = 97.9 kips; Shear rupture component, N (0.6F u) A nv = 0.75 × 0.6 × 58 × 4.88 = 127 kips 7 Controls Tension yielding component, N Fy A gt = 0.75 × 36 × 2.25 = 60.8 kips; Shear yielding component, N (0.6F y) A gv = 0.75 × 0.6 × 36 × 4.88 = 79.1 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 127 + min [60.8; 97.9] = 188 kips Strength of weld SMAW process using E70 electrodes. Weld size, w = ¼ in. From Table 6.19.1, design shear strength of unit length weld, W d = 5.57 kli N R lw = W d L lw = 5.57 × 6.0 × 2 = 66.8 kips N R tw = W d (1.0 + 0.5 sin 1.52) L tw = 5.57 (1.5) 6.0 = 50.1 kips T d4 = N R lw + N R tw = 66.8 + 50.1 = 117 kips

From Eq. 4.7.3, T d = min [T d1, T d2, T d3, T d4] = min [97.2, 131, 188, 117] = 97.2 kips The design strength of the member is 97.2 kips and is controlled by the limit state of yielding in the gross section. (Ans.)



P7.12.

Chapter 7

page 7-17

A MC 6x18 channel tension member of A242 corrosion resistant steel is welded to a d -in.-thick gusset plate of A36 steel using a 6-in.-long transverse weld and two 8-in.-long longitudinal welds as shown in Fig. P7.12. Assume E70 fillet welds of ¼ in leg size. Determine the design strength of the member and the gusset plate. Include block shear rupture strength of the member. See Figure P7.12 of the text book. Solution MC 6×18: A = 5.29 in.2; tw = 0.379 in. = d in.; d = 6.0 in.; tf = 0.475 in. A242 steel member: Fy = 50 ksi; F u = 70 ksi A36 steel gusset: Fy = 36 ksi; Fu = 58 ksi Fillet welds: SMAW process using E70 electrodes. FEXX = 70 ksi; weld size, w = ¼ in. Length of longitudinal welds, L lw = 8 in. Length of transverse weld, L tw = 6 in. Limit state of yielding in gross section gives (Eq. 7.4.5), T d1 = N Fy A g = 0.90 × 50 × 5.29 = 238 kips Net area, A n = A g = 5.29 in.2 Length of connection, L con = L lw = 8.00 in. Connection eccentricity, From Eq. 7.7.5, reduction coefficient,

Limit state of fracture in net section gives (Eqs. 7.4.6 and 7.7.1), T d2 = N Fu A e = 0.75 × 70 × 0.860 × 5.29 = 239 kips Block shear strength of the member Block shear failure of the member consists of the web section (rectangular element d × t w) at the front-end of the weld failing in tension and two flange sections (rectangular sections L lw × tf) failing in shear at the web-flange junction of the C-shape, leaving a rectangular block of the web (L lw × d) welded to the gusset. We have: A gt = 0.379 × 6.00 = 2.27 in.2 = A nt A gv = 0.475 × 8.00 × 2 = 7.60 in.2 = A nv Tension rupture component, N Fu A nt Shear rupture component, N (0.6F u) A nv Tension yielding component, N Fy A gt Shear yielding component, N (0.6F y) A gv

= = = =

0.75 × 70 × 2.27 0.75 × 0.6 × 70 × 7.60 0.75 × 50 × 2.27 0.75 × 0.6 × 50 × 7.60

= = = =

119 kips 239 kips 7 controls 85.1 kips 171 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs C = N (0.6Fu) A nv + min [N Fy A gt; N F u A nt] = 239 + min [85.1; 119] = 324 kips



Chapter 7

page 7-18

Block shear strength of gusset plate Block shear failure of the gusset consists of a single, rectangular block of the gusset plate. The block is 6-in.-wide,8½ -in.-long, and d -in.-thick. We have: A gt = A nt = d × 6.00 = 2.25 in.2 A gv = A nv = d × 2 × (8 + ½) = 6.38 in.2 Tension rupture component, N Fu A nt = Shear rupture component, N (0.6F u) A nv = Tension yielding component, N Fy A gt = Shear yielding component, N (0.6F y) A gv =

0.75 × 58 × 2.25 0.75 × 0.6 × 58 × 6.38 0.75 × 36 × 2.25 0.75 × 0.6 × 36 × 6.38

= = = =

97.9 kips 167 kips 7 Controls 60.8 kips 103 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs GP = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 167 + min [60.8; 97.9] = 228 kips Strength of weld From Table 6.19.1, design shear strength of a unit length ¼ -in. fillet welds using E70 electrodes is, W d = 5.57 kli. Strength of longitudinal welds, N R lw = W d L tw = 5.57 × 8.00 × 2 = 89.1 kips Strength of transverse weld, N R tw = W d (1.0 + 0.5 sin 1.52) L tw = 5.57 ×1.50 ×6.00 = 50.1 kips Design strength of the welds, T d4 = N R lw + N R tw = 89.1 + 50.1= 139 kips

T d = min [T d1, T d2, T bs C, T bs GP, T d4] = min [238, 239, 324, 228, 139] = 139 kips The design strength of the member is 139 kips, and is controlled by the limit state of weld shear strength. (Ans.)



P7.13.

Chapter 7

page 7-19

A 8x½ in. plate tension member is connected to a d -in.-thick gusset plate by five A325-N bolts of f in. diameter (Fig. P7.13). Assume A36 steel for the member and the gusset. Determine the tensile strength of the member. Include block shear rupture strength of the member and the gusset. See Figure P7.13 of the text book. Solution Plate 8×½ in. 6 A g = 8 × ½ = 4.00 in.2 Bolts: Type: A325-N Diameter, d = f in. 6 d e = 1 in. assuming standard punched holes Pitch, p = in.; gage, g = 2 ½ in. End distance, L e = 1¼ main plate; = 2 in. gusset plate Limit state of yielding in the gross section gives, T d1 = N F y Ag = 0.90 × 36 × 4.00 = 130 kips Two possible failure paths are identified. For, path a-b-c-d: A n1 = 4.00 - 2 × 1.0 × ½ path a-b-e-c-d: A n2 = 4.00 - 3 × 1.0 × ½ + 2 ×

= 3.00 in.2 ² controls × ½ = 3.40 in.2

T d2 = N Fu U A n = 0.75 × 58 × 1.0 × 3.00 = 131 kips Block shear strength of the member Consider a rectangular block having two 4½ -in.-long horizontal surfaces in shear, and one 5-in.long vertical surface in tension. A gt = 5.0 × 0.5 = 2.50 in.2 A nt = 2.50 - 2 × ½ × 1.0 × 0.5 = 2.00 in.2 A gv = 2 × 4.25 × 0.5 = 4.25 in.2 A nv = 4.25 - 2 × 1½ × 1.0 × 0.5 = 2.75 in.2 Tension rupture component, N Fu A nt = 0.75×58×2.00 = 87.0 kips 7 Controls Shear rupture component, N (0.6F u) A nv = 0.75×0.6×58×2.75 = 71.8 kips Tension yield component, N Fy A gt = 0.75×36×2.5 = 67.5 kips; Shear yield component, N (0.6F y) A gv = 0.75×0.6×36×4.25 = 68.9 kips

N Fu A nt > N (0.6Fu) A nv, from Eq. 7.8.1, T bs M = N Fu A nt + min [N (0.6Fy) A gv; N (0.6Fu) A nv] =

87.0 + min [68.9; 71.8] = 156 kips

Block shear strength of the gusset plate Block shear failure of the gusset consists of a single, rectangular block having two 5-in.-long horizontal surfaces, and one 5-in.-long vertical surface. A gt = 5.0× d = 1.88 in.2; A nt = 1.88 - 2 × 1.0 × d = 1.13 in.2 2 A gv = 2×5.0× d = 3.75 in. ; A nv = 3.75 - 2 × 1.5×1.0× d = 2.63 in.2 Tension rupture component, N F u Ant = 0.75 × 58×1.13 = 49.2 kips Shear rupture component, N (0.6F u) Anv = 0.75 × 0.6 × 58 × 2.63 = 68.6 kips 7 controls Tension yielding component, N F y Agt = 0.75 × 36 × 1.88 = 50.8 kips Shear yielding component, N (0.6F y)A gv = 0.75 × 0.6 × 36 × 3.75 = 60.8 kips



Chapter 7

page 7-20

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs GP = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 68.6 + min [50.8; 49.2] = 118 kips Strength of bolts From Table 6.7.1, for a f -in.-dia. A325-N type bolt in single shear, B dv = 21.6 kips Shear strength of connectors, C dv = 5 × 21.6 = 108 kips 7 controls Using the bearing strength values given in Tables 6.8.1 and 6.8.2 for f -in.-dia. bolts bearing on F u = 58 ksi steel plates, we obtain for: Bolts bearing on ½ in. main plate, C db M = [3 × 40.8 + 2 × 91.3]× ½ = 153 kips Bolts bearing on d in. gusset plate, C db GP = [2 × 79.9 + 3 × 91.3] × d = 163 kips Design strength of connectors, C d = 108 kips

From Eq. 7.4.3, T d = min [T d1, T d2, T bs M, T bs GP, T d4] = min [130, 131, 156, 118, 108] = 108 kips The design strength of the member is 108 kips and itis controlled by the limit state of bolt shear. (Ans.)



Chapter 7

page 7-21

P7.14. a) Determine the design tensile strength of a C15x40 truss member of A572 Grade 42 steel shown in Fig.P7.14. The end connection to the gusset is by ten f -in.-dia. A490-X type bolts. They are arranged in three rows with a gage of 3 in., pitch of 2½ in., and an end distance of 1½ in. Include block shear rupture strength. b) Redesign the end-connection to increase the design strength of the member to the extent possible, if there are no architectural restrictions. See Fig. P7.14 of the text book. Solution A572 Gr 42 steel: Fy = 42 ksi, Fu = 60 ksi C15 × 40: A = 11.8 in.2; t w = 0.520 in.; t f = 0.650 in.;

T d1 = N F y A g = 0.90 × 42 × 11.8 = 446 kips To calculate the net area, two possible failure paths are identified. We have, for path a-b-c-d: A n1 = 11.8 - 2 × 1.0 × 0.52 = 10.8 in.2 path e-f-b-c-g-h: A n2 = 11.8 - 4 × 1.0 × 0.52 + 2 ×

× 0.52 = 10.3 in.2 ² controls

Length of connection, L con = 2 × 2.5 = 5.0 in. Connection eccentricity,

T d2 = N Fu U A n = 0.75 × 60 × 0.844 × 10.3 = 391 kips Block shear strength A gt = 9.0 × 0.52 = 4.68 in.2 A nt = 4.68 - (2 + ½ + ½ )× 0.52 × 1.0 + 2 ×

× 0.52 = 3.66 in.2

A gv = 2×(1.5 + 2.5) × 0.52 = 4.16 in.2 A nv = 4.16 - 2 × (1 + ½ ) × 0.52 = 2.60 in.2 Tension rupture component, N Fu A nt Shear rupture component, N (0.6F u) A nv Tension yielding component, N Fy A gt Shear yielding component, N (0.6F y) A gv

= = = =

0.75 × 60 × 3.66 0.75 × 0.6 × 60 × 2.60 0.75 × 42 × 4.68 0.75 × 0.6 × 42 × 4.16

= = = =

165 kips 7 Controls 70.2 kips 147 kips 78.6 kips

N F u Ant > N (0.6F u) Anv, from Eq. 7.8.1, T bs = NF uA nt+ min [N(0.6F y)A gv; N(0.6F u)A nv] = 165 + min [78.6; 70.2] = 235 kips Strength of connectors L e = 1.5 in.; p = 3 in. From Table 6.7.1, the design shear strength of a f -in.-dia. A490-X type bolt in single shear is, B dv = 33.8 kips C dv = 33.8 × 10 = 338 kips 7 controls PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 7

page 7-22

Also, the design bearing strength for interior and end bolts bearing on F u= 60 ksi steel plates can be determined using Tables 6.8.1 and 6.8.2 as follows: C db = [4 (53.8) + 6 (91.3)] × (60 ksi/58 ksi) × 0.520 = 410 kips Design strength of the connectors is C d = 338 kips

T d = min [T d1, T d2, T d3, T d4] = min [446, 391, 235, 338] = 235 kips Design strength of the member, as given, is 235 kips.

Redesign of the member Design strength of connectors (T d4) is limited by the shear strength of bolts. By adding four more bolts, in a vertical row, the revised design strength of connectors is, C d = C dv = 33.8 × 14 = 473 kips To increase T d2 , increase the bolt pitch to 4 in. Using the revised values for A n (= 10.8 in.2) and U (= 0.90), we obtain: T d2 = N Fu U A n = 0.75 × 60 × 0.90 × 10.8 = 437 kips Block shear strength A gt = 9.0 × 0.52 = 4.68 in.2 A nt = 4.68 - (2 + ½ + ½ ) × 0.52 × 1.0 + 2 ×

× 0.52 = 4.55 in.2

A gv = 2×(12.0) × 0.52 = 12.5 in.2 A nv = 12.5 - 2 × (2 + ½ ) × 1.0 × 0.52 = 10.5 in.2 Tension rupture component, N Fu A nt = Shear rupture component, N (0.6F u) A nv = Tension yielding component, N Fy A gt = Shear yielding component, N (0.6F y) A gv =

0.75 × 60 × 4.55 0.75 × 0.6 × 60 × 10.5 0.75 × 42 × 4.68 0.75 × 0.6 × 42 × 10.4

= = = =

205 kips 284 kips 147 kips 197 kips

7 Controls

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv+ min [N (Fy) A gt; N Fu A nt] = 284 + min [147; 205] = 431 kips

Thus, the design strength of the member can be increased to T d = min [T d1 , T d2 , T d3 , T d4] = min [446, 437, 431, 473] = 431 kips Provide fourteen f -in.-dia. A490-X bolts, with a pitch of 4 in., end distance of 4 in. and a gage of 3-in. in 4 rows (3 + 4 + 4 + 3), increasing the design tensile strength of the member to 431 kips. (Ans.)



P7.15.

Chapter 7

page 7-23

The axial tensile forces in the vertical member of a pedestrian truss bridge, from the code specified service loads, have been calculated as: 40 kips from dead load, 50 kips live load and 30 kips from wind. Select a 1 in.-thick plate of A36 steel of suitable width. Assume a single line of bolts of ¾ in. diameter. Solution Service loads: D = 40 kips; L = 50 kips; W = 30 kips Factored load, T u = max{ 1.4D; 1.2D + 1.6L; 1.2D + 1.6W + 0.5L} = max {56.0; 128; 121} = 128 kips Member: 1 -in.-thick plate a.

Preliminary selection Limit state of yield in the gross section requires that,

Limit state of fracture in the net section requires that,

Diameter of bolt, d = ¾ in. Assume standard punched holes. 6 d e = f in. U = 1.0 for plate member with bolted end connection. Single line of bolts. So,

Ag $ 3.95 in.2 Select a 4 × 1 in. plate of A36 steel. b.

(Ans.)

Checks T d1 = N Fy A g = 0.90 × 36 × 4.0 × 1.0 = 130 > 128 kips T d2 = N Fu U A n = 0.75 × 58 × 1.0 × (4.0 - 1× f×1.0) = 136 kips > 128 kips The selection is O.K.



P7.16.

Chapter 7

page 7-24

The web member of a roof truss is subjected to tensile forces of 40 kips and 60 kips under service dead load and live load, respectively. Select the lightest WT5, if the end connection to gusset is by two lines of ¾ -in.dia. bolts in the flange. Assume that there will be more than three bolts in each line. Neglect block shear failure. Solution Service loads: D = 40 kips; L = 60 kips Factored load, T u = max [1.4D; 1.2D + 1.6L] = max[1.4×40; 1.2×40 + 1.6×60] = max[84.0; 144] = 144 kips WT shape: A992 steel 6 Fy = 50 ksi; Fu = 65 ksi a.

Preliminary selection Limit state of yield in the gross section requires that:

Limit state of fracture in the net area requires that:

Diameter of bolts, d = ¾ in. Assume standard punched holes. 6 d e = f in. Connection is to the flange of a WT. Assume tf . 0.5 in. It is given that there will be more than 3 bolts in each line of bolts. Assume U = 0.85.

7 controls

A g $ 4.35 in.2 So, tentatively select a WT5×16.5 (A g = 4.85 in.2; tf = 0.435 in.; b f = 4.87 in.; d = 7.96 in.). b.

Checks T d1 = N Fy A g = 0.90 × 50 × 4.85 = 218 kips > 136 kips

O.K.

As

T d2 = N Fu U A n = 0.75 × 65 × 0.85 × (4.85 - 2 × f × 0.435) = 169 kips > 136 kips So, select a WT 5×16.5 of A992 steel.

O.K. (Ans.)



P7.17.

Chapter 7

page 7-25

The factored tensile load on a truss bottom chord is 480 kips. Select the lightest W10 with its flanges connected to gusset plates at each end, with two lines of ¾ -in.-dia. bolts in each flange. The bolts are A490-N type. Neglect block shear failure. Solution Factored load, T u = 480 kips A992 steel: Fy = 50 ksi; F u = 65 ksi a.


Limit state of fracture in the net section requires that:

Diameter of bolts, d = ¾ in. Assume standard punched holes. 6 d e = f in. Entering LRFDM Table 1-1 with A req = 10.7 in.2, we observe that the W10s that provide a gross area slightly greater than 10.7 in.2, have b f . 8 in., and d . 10 in. resulting in b f /d . 0.80 > b. Also, tf . e in. Assume there will be more than 3 bolts in each line, so U = 0.90.

7 controls

Ag $ 13.1 in.2 So, tentatively select a W10×45 (A g = 13.3 in.2; tf = 0.620 in.; b f = 8.02 in.; d = 10.1 in.). b.

Checks T d1 = N Fy A g = 0.90 × 50 × 13.3 = 599 kips

> 480 kips

O.K.

From Table 6.7.1, the design shear strength of a ¾ -in.-dia. A490-N type bolt in single shear is B dv = 19.9 kips. So, the number of bolts required,

More than 6 bolts are needed in each line ( > 3). U = 0.90 as assumed. T d2 = N Fu U A n = 0.75 × 65 × 0.90 × (13.3 - 4 × f × 0.620) = 488 kips

Select a W10×45 of A992 steel.

> 480 kips

O.K. (Ans.)



P7.18.

Chapter 7

page 7-26

A W16 section is used as a tension member to carry a tensile load of 500 kips. The end connection is by fin.-dia. A490-X bolts, in two rows, in each flange, using standard punched holes. The gusset plates are ein.-thick. Find the lightest W16 section and determine the number of bolts required. What is the maximum allowable length of such a member. Solution Factored load, T u = 500 kips W 16 shape: A992 steel 6 F y = 50 ksi; F u = 65 ksi a.


Limit state of fracture in net area requires that:

Diameter of bolts, d = f in. Assume standard punched holes. 6 d e = 1.0 in. Assume U = 0.85 as there will be more than 3 bolts in each line of bolts and the W16s are beam shapes. Assume tf . e in. With two rows of bolts in each flange, there are four bolt holes to be removed from the gross area.

7 controls

A g $ 14.6 in.2 So, tentatively select a W16×50 (A g = 14.7 in.2; tf = 0.63 in.; b f = 7.07 in.; d = 16.3 in.; r x = 6.68 in.; r y= 1.59 in.). b.

Number of bolts required From Table 6.7.1, for a f -in.-dia. A490-X type bolt in single shear, B dv = 33.8 kips ² controls For bolt bearing on member, B dbo = N (2.4Fu) d t = 0.75 × 2.4 × 65 × f × 0.63 = 64.5 kips For bolt bearing on gusset, B dbo = 0.75 × 2.4 × 58 × f × e = 57.1 kips

c.

.

Use 16 bolts;

i.e., 4 bolts in each line

Check section selected T d1 = N F y A g = 0.90 × 50 × 14.7 = 662 kips > 500 kips

O.K.

U = 0.85 T d2 = N FuU A n = 0.75 × 65 × 0.85 × (14.7 - 4 × 1.0 × 0.63) = 505 kips > 500 kips

Select W16×50 section of A992 steel. d.

O.K. (Ans.)

Maximum length of member



Chapter 7

page 7-27

rmin = ry = 1.59 in.

From Eq. 7.10.7, Length of the member should be no more than 39 feet 9 inches.

(Ans.)



P7.19.

Chapter 7

page 7-28

A 16 ft long tension member is subjected to a factored axial load of 100 kips. Select a suitable, single, equalleg angle of 5 in. leg size and an even number of ¾ -in.-dia. bolts of A325-N type. The bolts are to be arranged in two lines, without stagger, with a pitch of 3 in., and an end distance of 1 ½ in. Use standard gages. Solution Factored load, T u = 100 kips Member: Single equal-leg angle of 5-in. leg size; length, L = 16 ft Material: Assume A36 steel Bolts: ¾ -in.-dia. A325-N type a.

Preliminary selection

estimated deduction for bolt holes Assume U = 0.85 and let t = d in.

Try L5×5× d ( A = 3.65 in.2; t = d in.; r z = 0.986 in.; from LRFDM Table 1-7). Provide bolts in two lines, using workable gages g 2 (= 2 in.) and g3(= 1¾ in.) given in Table 6.7.1. This results in a side distance, L s = 1¼ in. Provide a pitch, p = 3 in. and an end distance, L e = 1½ in. From Table 6.7.1, for A325-N type bolt in single shear, B dv = 15.9 kips

Number of bolts required, Consider 8 bolts, 4 bolts in each line without stagger. Using Tables 6.8.1 and 2, show that limit state of bearing doesn’t control the bolt strength. b.

Checks T d1 = Nt1 Fy A g = 0.90 × 36 × 3.65 = 118 kips > 100 kips Net area, An = 3.65 - 2 × f × d = 2.99 in.2 Length of connection, L con = 3 × 3.0 = 9.00 in. Connection eccentricity,

T d2 = Nt2 Fu U A n = 0.75 × 58 × 0.848 × 2.99 = 110 kips > 100 kips Block shear strength There will be one rectangular failure block. Length of block = 3 × 3.0 + 1½ = 10.5 in.; width of block = 1¾ +1¼ in. = 3.00 in.



Chapter 7

A gt = 3.00 × d = 1.13 in.2; A gv = 10.5 × d = 3.94 in.2; Tension rupture component, N Fu A nt = Shear rupture component, N (0.6F u) A nv = Tension yielding component, N Fy A gt = Shear yielding component, N (0.6F y) A gv =

page 7-29

A nt = 1.13 - 1½ × f × d = 0.633 in.2 A nv = 3.94 - 3½ × f × d = 2.79 in.2 0.75 × 58 × 0.633 = 27.5 kips 0.75 × 0.6 × 58 × 2.79 = 72.8 kips 7 controls 0.75 × 36 × 1.13 = 30.5 kips 0.75 × 0.6 × 36 × 3.94 = 63.8 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 72.8 + min [30.5; 27.5] = 100 kips O.K. O.K.

Select a L5×5× d of A36 steel. Provide 8 bolts, four in each row along workable gage lines. (Ans.)



P7.20.

Chapter 7

page 7-30

Design the tension diagonal of an all welded Warren truss for a pedestrian bridge, in which the chords are made from WT7x19 sections. The factored load in the diagonal member is 125 kips and its length is 15 ft. Use 2Ls of A36 steel LLBB, with their 4-in.-long legs welded to the stem of the WT. The weld consists of a transverse weld and two longitudinal welds.

Solution Factored load, T u = 125 kips A36 steel 6 Fy = 36 ksi; F u = 58 ksi Member: Double angles with their 4 -in.-long legs welded to the stem of a tee. Limit state of yield in the gross section requires that:

Assume U = 0.85 Limit state of fracture in net area requires that:

A g $ 3.86 in.2 Or, $ 1.93 in.2 /angle Provide two L4×3× d LLBB. From LRFDM Table 1-7, for a L4×3× d : A = 2.49 in.2; t = d in.; So, A g = 4.98 in.2 > 3.86 in.2 O.K. (Ans.)

Note: As no information on the inclination of the diagonal and the type of weld is given, the connection cannot be designed.



P7.21.

Chapter 7

page 7-31

Design a single-angle tension member 13 ft long and its connection at each end to a d in. gusset plate. The tension is 24 kips dead load and 52 kips live load. The angle cannot extend more than 15 in. on the gusset. Use LRFD specification, A36 steel, f in. dia. bolts, slip critical connection. Include block shear rupture strength. Solution Service loads: D = 24 kips; L = 52 kips Factored load, T u = max [1.4D; 1.2D + 1.6L] = max[1.4×24; 1.2×24 + 1.6×52] = max[33.6; 112] = 112 kips Member: Material: Bolts: Connection: Restriction: a.

Single angle. A36 steel f-in.-dia. Slip-critical. Angle cannot

Length, L = 13 ft

6 d e = 1 in. assuming standard punched holes Assume A325-SC type bolts. extend more than 15 in. on the gusset.


estimated deduction for bolt holes Assume U = 0.85 and let t = d in.

From Table 6.7.1, for A325-SC type bolts in single shear, design resistance to shear at service load using factored loads is, B dsf = 14.5 kips

Number of bolts required,

say 8

As the length of the connection is restricted, provide the bolts in two rows. This requires angles with connected leg lengths $ 5 in. For higher efficiency, restrict the selection to unequal leg angles, with the long leg connected to the gusset. Try L7×4× d ( A = 4.00 in.2; t = d in.; r z = 0.873 in.;

from LRFDM Table 1-7).

Provide bolts 8 bolts, 4 bolts in each line without stagger, using workable gages g 2 (= 2½ in.) and g 3(= 3 in.) given in Table 6.7.1. This results in a side distance, L s = 1½ in. Provide a pitch, p = 3 in. and an end distance, L e = 2 in. Overlap of the angle over the gusset equals 14 in. ( 3 × 3.0 + 2 × 2.0 ). b.

Checks T d1 = Nt1 Fy A g = 0.90 × 36 × 4.00 = 130 kips > 112 kips

O.K.



Chapter 7

page 7-32

Net area, An = 4.00 - 2 × 1.00 × d = 3.25 in.2 Length of connection, L con = 3 × 3.0 = 9.00 in. Connection eccentricity,

T d2 = Nt2 Fu U A n = 0.75 × 58 × 0.900 × 3.25 = 127 kips > 112 kips Block shear strength There will be one rectangular failure block. Length of block = 3 × 3.0 + 2 = 11.0 in.; width of block = 2½ + 1½ in. = 4.00 in. A gt = 4.00 × d = 1.50 in.2; A nt = 1.50 - 1½ × 1.00 × d = 0.938 in.2 2 A gv = 11.0 × d = 4.12 in. ; A nv = 4.12 - 3½ × 1.00 × d = 2.81 in.2 Tension rupture component, N Fu A nt = 0.75 × 58 × 0.938 = 40.8 kips Shear rupture component, N (0.6F u) A nv = 0.75 × 0.6 × 58 × 2.81 = 72.3 kips 7 controls Tension yielding component, N Fy A gt = 0.75 × 36 × 1.50 = 40.5 kips Shear yielding component, N (0.6F y) A gv = 0.75 × 0.6 × 36 × 4.12 = 66.7 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 72.3 + min [40.5; 40.8] = 113 kips O.K. O.K.

Strength of connectors Shear resistance of connectors, C dv = N B d sf = 8 × 14.5 = 116 kips There are six interior bolts with pitch, p = 3 in., and two end bolts with L e = 2 in. Using the values given in Tables 6.8.1 and 6.8.2 for Fu = 58 ksi plate material and f-in.-dia. bolts, the design bearing strength of the connectors is obtained as C db = [6×91.3 + 2×79.9]× d = 265 kips Strength of connectors, C d = min [116, 265] = 116 kips > 112 kips O.K.

Select a L7×4× d of A36 steel. Provide 8 bolts, four in each row without stagger, along workable gage lines of the long leg. Use a pitch, p = 3 in. and an end distance, L e = 2 in. (Ans.)



P7.22.

Chapter 7

page 7-33

A tension member is composed of two C12x30 channels placed 12 in. back-to-back and tied together as shown in Fig. 7.2.2c (ii). The member is connected at ends to ½ -in.-thick gusset plates, by twenty four f -in.-dia. A325-N bolts provided in three rows in the webs. If A572 Grade 42 steel is used, find the design strength as per LRFDS. Assume a gage of 3 in., a pitch of 3 in., and an end distance of 2 in. If the force in the member from dead load on the structure is 200 kips, what live load and wind load forces can be applied on the member as per ASCES? What is the maximum recommended length of such a member as per LRFDS? See Figs. 7.2.2c (ii) and 7.11.1b of the text book. Solution a. Section properties A572 Gr 42 steel: Fy= 42 ksi; F u = 60 ksi For the C12x30, from LRFDM Table 1-5: A C = 8.81 in.2, d C = 12.0 in., twC = 0.510 in., b fC = 3.17 in., tfC = 0.501 in., r xC = 4.29 in. , r yC = 0.762 in. = r min C = 0.674 in., IyC = 5.12 in.4 Also from LRFDM Table 1-5, the workable gage, g, for the flange of a C12×25 is 1¾ in. For the built-up section: A = 2 × 8.81 = 17.6 in.2 Iy = = 2 [ 5.12 + 8.81 × (6.0 - 0.674)2] = 510 in.4

b.

ry

=

rx r min

= =

r xC = 4.29 in. r x = 4.29 in.

Yield and fracture strengths Gross area of the section, A g = A = 17.6 in.2 Tie plates connect the flanges of the two channels at the ends, and at intervals along the length of the member. Assume, conservatively, that the holes in the flanges connecting the end tie plate to the member, and the holes in the webs connecting the member to the gussets lie in the same cross section. Net area, A n = 17.6 - 2×3×1.0×0.510 - 2×2×1×0.501 = 12.5 in.2 From Eq. 7.7.2, reduction coefficient,

Use U = 0.90 Design strength corresponding to the limit state of tension yielding, T d1 = 0.9 × 42 × 17.6 = 665 kips tension fracture, T d2 = 0.75 × 60 × 0.900 × 12.5 = 506 kips c.

Block shear rupture strength Gross area in tension, A gt = 2 × 6.0 × 0.510

= 6.12 in.2



Chapter 7

page 7-34

Net area in tension, A nt = 6.12 - 2 (2 × 1.0×0.510) = 4.08 in.2 Gross area in shear, A gv = 2 × 2 × 11.0 × 0.510 = 22.4 in.2 Net area in shear, A nv = 22.4 - 2× 2 (3½ × 1.0× 0.510) = 15.3 in.2 N Fu A nt = 0.75×60× 4.08 = 184 kips N Fuv A nv = 0.75×0.6×60× 15.3 = 413 kips 7 controls N Fy A gt = 0.75×42×6.12 = 193 kips N Fyv A gv = 0.75×0.6×42× 22.4 = 423 kips So, the design strength of the member corresponding to block shear failure is, T d3 = N Fuv A nv + min [ N Fy A gt, N Fu A nt ] = 413 + min [193; 184 ] = 597 kips d.

Bolt strength From Table 6.7.1, design strength of a single f -in.-dia. A325-N bolt in single shear is B dv = 21.6 kips Design strength of bolts, T d4 = N B d = 2×12×21.6 = 518 kips

e.

Member strength Design strength of member, T d = min [T d1, T d2, T d3, T d4] = min [665, 506, 597, 518] = 506 kips (Ans.) Dead load on the member, D = 200 kips If L is the live load, and W is the wind load on the structure, we have from load combination LC-2: 1.2D + 1.6L # 506 kips 6 L # 166 kips (Ans.) Also, from load combination LC-4, 1.2D + 0.5L + 1.6 W # 506 6 139 kips (Ans.)

Maximum recommended length of the member,



P7.23.

Chapter 7

page 7-35

A tension member in a covered pedestrian bridge is subjected to a dead load of 50 kips, live load of 70 kips, and snow load of 38 kips. Select two angles with long legs back-to-back and separated by d in. for end connections to gusset plates. Assume a single line of holes for ¾ -in.-dia. A325-X bolts. Assume A242 Grade 50 steel. The member is 20 ft long. Determine the spacing of spacer plates, if necessary. Solution Member: Two angles with long legs back-to-back. Material: A242 Grade 50 steel. 6 Fy = 50 ksi; F u = 70 ksi Service loads: D = 50 kips; L = 70 kips; S = 38 kips Load combinations: 1.4D = 70 kips 1.2D + 1.6L + 0.5S = 191 kips ² controls 1.2D + 1.6S + 0.5L = 156 kips Factored load, T u = 191 kips a.

Number of bolts needed From Table 6.7.1, for ¾ -in.-dia. A325-X type bolts in double shear, B dv = 39.8 kips


Say 5

b.

Select a section A g1 $ 191/(0.90 × 50) = 4.24 in.2 A g2 $ (191/NFuU ) + estimated deduction for bolt holes Assume U = 0.85, and t . 0.4 in. A g2 $ 191/(0.75 × 70 × 0.85) + 2 × f × 0.4 = 4.98 in.2 Tentatively select 2L 4×3× d. [From LRFDM Table1-7 for single angles: A L = 2.49 in.2; r xL = 1.26 in.; r zL = 0.636 in. From LRFDM Table 1-14 for double angles: r y = 1.30 in. corresponding to s = d in. ]

c.

Check the section selected Gross area, A g = 4.98 in.2 T d1 = N Fy A g = 0.90 × 50 × 4.98 = 224 kips > 191 kips O.K. T d2 = N Fu U A n = 0.75 × 70 × 0.85 × (4.98 - 2 × f × 0.375) = 193 kips > 191 kips

O.K.

O.K.

d.

Spacer plates For the single angle L4×3× d , r min = r zL = 0.636 in. Spacer plate spacing, S sp # 300 r min = 300×0.636 = 191 in. = 15.9 ft Provide three spacer plates; one at each end of the member, and the third at mid-length of the member, so that S sp = 10 ft < 15.9 ft O.K. So, select 2L 4×3× d of 242 Grade 50 steel, with long legs back to back, connected to d -in.-thick gusset plates by five ¾ -in.-dia. A325-X bolts, and provided with a spacer plate at mid length. (Ans.)



P7.24.

Chapter 7

page 7-36

Select a pair of American Standard channels for a tension member subjected to a dead load of 68 kips and a live load of 140 kips. The channels are placed back-to-back and connected to a ½ in. gusset plate by f -in.dia. A490-N bolts [see Fig. 7.2.2c (i)]. Assume A588 Grade 50 steel for the member and the gusset. The member is 20 ft long. The bolts are arranged in two lines parallel to the length of the member. See Fig. 7.2.2c (i) of the text book. Solution Member: two channels placed back to back. Steel: A588 Gr 50 6 Fy = 50 ksi; F u = 70 ksi Service loads: D = 68 kips; L = 140 kips Factored load, T u = max [1.4D; 1.2D + 1.6L ] = [95.2; 306] = 306 kips a.

Number of bolts From Table 6.7.1, for a f -in.-dia. A490-N bolt in double shear, B dv = 54.1 kips. N $ T u / B dv = 306/54.1 = 5.66 (say six bolts, 3 in each line)

b.

Select a section A g1 $ 306/(0.90 × 50) = 6.80 in.2 Assume U = 0.85, and tw . 0.40 in. A g2 $

6

4.23 in.2 per channel.

Select 2C9×15 [A C = 4.41 in.2; twC = 0.285 in.; IxC = 51 in.4; IyC = 1.91 in.4; ryC = 0.659 in.,

c.

Check the selected section T d1 = N Fy A g = 0.90 × 50 × 8.82 = 397 kips > 306 kips O.K. T d2 = N Fu U A n = 0.75 × 70 × 0.85 × (8.82 - 4 × 1.0 × 0.285) = 343 kips > 306 kips O.K. For the built-up section: Ix = 2 × 51.0 = 102 in.4; Iy = 2 × [ 1.91 + 4.41 × (0.25 + 0.586)2] = 9.98 in.4

O.K.

So, select 2C 9×15 of A588 Grade 50 steel with webs connected to a ½ -in.-thick gusset plate in between. (Ans.)



Chapter 7

page 7-37

P7.25. A steel truss is used to transfer loads from the upper floor columns of an office building across an arcade at street level. The lower chord of the truss is made of two angles, with their vertical legs 8 in. back-to-back, connected to gussets as shown in Fig. P7.2.2a (ii). In addition, the two angles are connected by tie plates. The factored load to be carried by the member is 240 kips and the member length is 18 ft. Use A36 steel, ¾ -in.-dia. A325-N bolts. Select suitable angles. Show the end connection detail, including tie plates. Consider block shear failure. See Figure 7.2.2a (ii) of the text book. Solution Factored load, T u = 240 kips a.


From Table 6.7.1, for a ¾ -in.-dia. A325-N type bolt in single shear, B dv = 15.9 kips Number of bolts required, N $ 240/15.9 = 15.1, say 16 Assume 2 lines of bolts in each angle, and 4 bolts in each line. estimated deduction for bolt holes Assume U = 0.85, and t . 0.5 in.

Tentatively select 2L6×3½×½ with A = 8.96 in.2 b.

Checks T d1 = N Fy A g = 0.90 × 36 × 8.96 = 290 kips > 240 kips T d2 = N Fu U A n = 0.75 × 58 × 0.85 × (8.96 - 4 × 0.5 × f) = 267 kips So, select 2L6×3½×½ .

>

240 kips

O.K. (Ans.)



P7.26.

Chapter 7

page 7-38

A pinned member is to consist of four equal leg angles arranged as shown in Fig. 7.2.2b(i). The factored tensile load is 324 kips. Two e in. bolts will be used in each angle. The member length is 40 ft. For architectural reasons the cross-sectional dimensions should not exceed 12 in.× 12 in. Include the design of tie plates. See Fig. 7.2.2b (i) of the text book. Solution Factored load, T u = 324 kips Member: Four equal-leg angles to form a box shape; length, L = 40 ft Width of the box limited to 12 in. Material: Assume A36 steel Bolts: 5/8 -in.-dia. a.


estimated deduction for bolt holes Assume U = 0.85 and let t = d in. ( = 2.75 in.2/angle).

Try L4×4× d From LRFDM Table 1-7: A L = 2.86 in.2; tL = d in.; r zL = 0.779 in.; Provide bolts in one line, using workable gage g1 (= 2½ in.) given in Table 6.7.1. This results in a side distance, L s = 1½ in. From Table 6.7.1, for 5/8-in.-dia. A325-X type bolt in single shear, B dv = 13.8 kips


say 24. Provide six bolts in each line.

Provide a pitch, p = 3 in. and an end distance, L e = 2 in. Using Tables 6.8.1 and 2, show that limit state of bearing doesn’t control the bolt strength. b.

Checks For the built-up section: A g = 4× 2.86 = 11.4 in.2 Ix = 4× [ 4.32 + 2.86 × (6.0 - 1.13)2 ] = 287 in.4 rx = 5.02 in. T d1 = Nt1 Fy A g = 0.90 × 36 × 11.4 = 369 kips > 324 kips Net area, An = 11.4 - 4 × 2 × 3/4 × d = 9.15 in.2 Length of connection, L con = 5 × 3.0 = 15.0 in. Connection eccentricity,



Chapter 7

page 7-39

T d2 = Nt2 Fu U A n = 0.75 × 58 × 0.900 × 9.15 = 358 kips > 324 kips Block shear strength There will be four rectangular failure blocks. Length of each block = 5 × 3.0 + 2.0 = 17.0 in.; width of each block = 1½ in. = 1.50 in. A gt = 1.50 × d× 4 = 2.25 in.2; A nt = 2.25 - ½ × 3/4 × d × 4 = 1.69 in.2 2 A gv = 17.0 × d × 4 = 25.5 in. ; A nv = 25.5 - 5½ × 3/4 × d × 4 = 19.3 in.2 Tension rupture component, N Fu A nt = 0.75 × 58 × 1.69 = 73.5 kips Shear rupture component, N (0.6F u) A nv = 0.75 × 0.6 × 58 × 19.3 = 504 kips 7 controls Tension yielding component, N Fy A gt = 0.75 × 36 × 2.25 = 60.8 kips Shear yielding component, N (0.6F y) A gv = 0.75 × 0.6 × 36 × 25.5 = 413 kips

N (0.6Fu) A nv > N Fu A nt, from Eq. 7.8.2, T bs = N (0.6Fu) A nv + min [N Fy A gt; N Fu A nt] = 504 + min [60.8; 73.5] = 565 kips O.K. O.K.

Select four L4×4× d of A36 steel. Provide twenty-four 5/8-in.-dia. A325-X bolts, six in each

d.

row along workable gage lines to connect the vertical legs to gussets. Use a pitch, p = 3 in. and an end distance, L e = 2 in. (Ans.) Design of tie plates Distance between bolt lines, G = 12.0 - 2 × 2½ = 7.00 in. Length of tie plate, L tp $ b × 7.00 = 4.67 in. (say, 5 in.) Thickness of tie plate,

(say, ¼ in.)

Width of plate, W tp = G + 2 L eh = 7.00 + 2×1½ = 10 in. Tie plates are 10×5×¼ in. Spacing of the plates, S tp # 300 r zL = 300×0.779 = 19.5 ft Provide 4 sets of tie plates, one at each end of the member and the other two at a and b of the length. Note that at the ends, there will be two (horizontal) tie plates only. At the intermediate locations there will b four. Two vertical and two horizontal. S tp =

O.K.

So, select two C12x25 of A572 Grade 50 steel with their webs back to back. Provide four tie plates 12×7×¼ in. of A36 steel. (Ans.)



Chapter 7

page 7-40

P7.27. Two channels with their webs 6 in. back to back as shown in Fig. 7.2.2c (iii), are used as a tension member to carry a factored load of 480 kips. The member is 40 ft long. At the ends, the flanges are connected by tie plates, while the webs are connected to gusset plates. Using A572 Grade 50 steel, LRFD Specification, and ¾ -in.-dia. bolts, select the channels. Design the tie plates. See Figure 7.2.2c (iii) of the text book. Solution Member: Two channels back to back separated by 6 in. Member length, L = 40 ft Factored load, T u = 480 kips a.

Preliminary selection A g1 $ 480/ (0.90×50) = 10.7 in.2 Note that the webs are connected to gusset plates at the ends; while the two sets of flanges are interconnected by tie plates at intervals. Only the gussets transmit load from the member to the rest of the structure (U < 1.0). Assume, conservatively, that holes in the flange (connecting it to tie plate) line up with the holes in the web (connecting it to gusset plate). Assume N R > 3 6 U = 0.85. Let tw . 0.5 in. and t f . 0.5 in.

6 7.15 in.2 per channel. Tentatively select C12×25 shapes (A = 7.34 in.2; tw = 0.387 in.; tf = 0.501 in.; b f = 3.05 in.; Ixx = 144 in.4; Iyy = 4.45 in.4; r y = 0.779 in.) b.

Strength checks T d1 = N Fy A g = 0.9 × 50 × 2 × 7.34 = 661 kips > 480 kips O.K. T d2 = N Fu U A n = 0.75 × 70 × 0.85 × 2 × [7.34 - 2 × f × 0.387 - 2 × f × 0.501] = 516 kips > 480 kips O.K.

c.

Slenderness check Ix = 2×144.0 = 288 in.4 Iy = 2× [4.45 + 7.34×(3.0 + 0.674)2 ] = 207 in.4

;

d.

6

Design of tie plates When using C12 × 25 channel with b f . 3 in., the workable gage distance on the flange is 1¾ in. Distance between bolt lines, G = 6.0 + 2 × 1.75 = 9.5 in. Length of tie plate, L tp $ b × 9.5 = 6.33 in. (say, 7 in.) Thickness of tie plate,

(say, ¼ in.)

Width of plate, W tp = G + 2 L eh = 9.5 + 2×1¼ = 12 in. PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 7

page 7-41

Tie plates are 12×7×¼ in. Spacing of the plates, S tp # 300 ryC = 300×0.779 = 19.5 ft Provide 4 sets of tie plates, one at each end of the member and the other two at a and b of the length. S tp =

O.K.

So, select two C12x25 of A572 Grade 50 steel with their webs back to back. Provide four tie plates 12×7×¼ in. of A36 steel. (Ans.)



P7.28.

Chapter 7

page 7-42

A floor beam hanger is subjected to tensile forces of 30 kips dead load and 48 kips live load. Select a suitable round steel threaded rod. Assume A36 steel (Note: Use impact factor from ASCES 7-98 Section 4.7.2). Solution Member: Threaded rod Material: A36 steel Service loads: D = 30 kips; L = 48 kips For hangers supporting floor beams, the impact factor is 33% from ASCES Section 4.7.2. Factored load, T u = 1.2D + 1.6(L ×1.33) = 1.2×30 + 1.6×48×1.33 = 138 kips From Eq. 7.12.8, the cross-sectional area of the threaded rod is,

Use 2½ -in.-dia. A 36 steel rod.

(Ans.)



P7.29.

Chapter 7

page 7-43

Roof trusses of an industrial building, with a span of 80 ft and a rise of 20 ft are spaced on 27 ft centers and have purlins spaced at 6 ft intervals (see Fig. 3.4.2). To support the purlins sag rods are provided at a points between the trusses. The estimated weight of the roof covering and of the purlins is 7 psf and 3 psf of the roof surface respectively. Uses 35 psf of ground snow load. Design sag rods using A36 steel and LRFDS. (Assume: exposure factor, C e = 1.0; thermal factor, C t = 1.0; and importance factor I = 1.0, in ASCES Eq. 7-1.)

Solution Span of roof truss, L = 80 ft Rise, h = 20 ft Inclination of the roof, 2 = tan -1 Spacing of trusses, S T = 27 ft Spacing of sag rods, S S = 27/3 a.

= 26.6 o

= 9 ft

Loads Tributary area of the roof surface per line of sag rods,

Weight of roofing and purlins = 10 psf Dead load, D = 403× 10 ÷ 1000 = 4.03 kips Horizontal projection of the roof area, A H = 40 × 9 = 360 ft2 Ground snow load, p g = 35 psf Exposure factor, C e = 1.0 Thermal factor, C t = 1.0 Importance factor, I = 1.0 From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as, p f = 0.7 C e C t I p g = 0.7 × 1.0 × 1.0 × 1.0 × 35 = 24.5 psf From Fig. 7-2 of the ASCES, the roof slope factor C s is 1.0 for warm roofs with roof slope 2 = 26.6 o. From Eq. 7.2 of the ASCES, the sloped roof snow load is p s = C s p f = 24.5 psf Snow load, D = 400× 24.5 ÷ 1000 = 9.80 kips As there are only dead loads and snow loads, combination LC-3 controls. Factored vertical load on the roof, V = 1.2D + 1.6S = 1.2×4.03 + 1.6×9.80 = 20.5 kips b.

Top sag rod The top sag rod parallel to the roof carries only that component of the vertical load, V, parallel to the roof. Thus, tension in the top sag rod: T S = V sin 2 = 20.5 sin 26.6 o = 9.18 kips Required area of the threaded rod for the sag rod (Eq. 7.12.10):

From Table 1-21 of the LRFDM, select a 5/8 in. diameter threaded rod having a cross-sectioanraela , 0.307 in.2 > 0.282 in.2 O.K. (Ans.) PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by M cGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.


Chapter 7

page 7-44

c. Horizontal tie rod The horizontal tie rod between ridge purlins has to carry a tensile force, T T :

Required area of the horizontal tie rod :

From Table 1-21 of the LRFDM, select a 3/4 in. diameter threaded rod of A36 steel having a crosssectional area, 0.442 in.2 > 0.316 in.2 O.K. (Ans.)



P7.30.

Chapter 7

page 7-45

Steel trusses with a span of 100 ft and a pitch of 1:5 are spaced at 30 ft on centers to support the roof of an industrial building. The roof covering is light weight cement boards on S-shape purlins, 6 ft on centers. The dead load of the roofing, purlins and the attachments may be assumed to be 13 psf. For the snow load recommended by ASCES 7-98 for Green Bay, Wisconsin, design sag rods at the a points between trusses. Use LRFDS. (Assume: exposure factor, C e = 1.0; thermal factor, C t = 1.0; and importance factor I = 1.0, in ASCES Eq. 7-1.)

Solution Span of roof truss, L = 100 ft Pitch of the roof: 1 in 5 Inclination of the roof, 2 = tan -1 (1/5) = 11.3 o Spacing of trusses, S T = 30 ft Spacing of sag rods, S S = 30/3 = 10 ft a.

Loads Tributary area of the roof surface per line of sag rods,

Weight of roofing and purlins = 13 psf Dead load, D = 510× 13 ÷ 1000 = 6.63 kips Horizontal projection of the roof area, A H = 50 × 10 = 500 ft2 Ground snow load, p g = 40 psf Exposure factor, C e = 1.0 Thermal factor, C t = 1.0 Importance factor, I = 1.0 From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as, p f = 0.7 C e C t I p g = 0.7 × 1.0 × 1.0 × 1.0 × 40 = 28.0 psf From Fig. 7-2 of the ASCES, the roof slope factor C s is 1.0 for warm roofs with roof slope 2 = 11.3 o. From Eq. 7.2 of the ASCES, the sloped roof snow load is p s = C s p f = 28.0 psf Snow load, D = 500× 28.0 ÷ 1000 = 14.0 kips As there are only dead loads and snow loads, combination LC-3 controls. Factored vertical load on the roof, V = 1.2D + 1.6S = 1.2×6.63 + 1.6×14.0 = 30.4 kips b.

Top sag rod The top sag rod parallel to the roof carries only that component of the vertical load, V, parallel to the roof. Thus, tension in the top sag rod: T S = V sin 2 = 30.4 sin 11.3 o = 5.96 kips Required area of the threaded rod for the sag rod (Eq. 7.12.10):

From Table 1-21 of the LRFDM, select a ½ in. diameter threaded rod having a cross-sectional area, 0.196 in.2 > 0.183 in.2 O.K. (Ans.)



Chapter 7

page 7-46

c. Horizontal tie rod The horizontal tie rod between ridge purlins has to carry a tensile force, T T :

Required area of the horizontal tie rod :

From Table 1-21 of the LRFDM, select a ½ in. diameter threaded rod of A36 steel having a crosssectional area, 0.196 in.2 > 0.186 in.2 O.K. (Ans.)



Chapter 7

page 7-47

P7.31.

The height to eaves of an industrial building is 32 ft. The columns are spaced 24 ft on centers. The side wall is covered with light weight cement boards weighing 5 psf. Aluminum and glass windows, weighing 8 psf, occupy 50% of the wall area. Girts are C5×9 spaced 5 ft on centers. Design sag rods located at the ¼ points between columns (see Fig. 3.4.3). Use A36 steel and LRFDS.

P7.32.

A three hinged parabolic tied-arch has a span of 60 ft and a rise of 20 ft. It carries a uniform horizontally distributed vertical load of 3 klf. Determine the diameter of an A36 steel tie rod and select suitable clevises and turn buckle. The load given is a factored load.

Solution Structure: Three hinged parabolic tied arch. Span, L = 60 ft Rise, h = 20 ft Factored uniformly distributed vertical load, q u = 3 klf over the entire span of 60 ft Tension in the tie, T u = (q u L 2) ÷ (8 h) = (3.00 × 60.0 2) ÷ (8 × 20.0) = 67.5 kips Required area of the tie rod :

From Table 1-21 of the LRFDM, select a 1¾ in. diameter threaded rod of A36 steel having a crosssectional area, 2.41 in.2 > 2.07 in.2 O.K. (Ans.)


solucionario-capitulo-7

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