Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Chapter 3, Problem 3.29PFS
To determine
The LRFD design strength and the ASD allowable strength of the given section,
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Determine the LRFD design strength and the ASD allowable strength . Neglect block shear.
A36 steel and 3/4-in Ø bolts
PL 5/8 x 12
3/4-in-diameter bolts
Determine the LRFD design tensile strength of the bolted connection shown. The angles are connected
through the longer legs to a % x 11 in plate. Holes are made for % in bolts. Use steel F,-50 ksi F-65 ksi.
L5x3x% (4, = 4.92in,
x%30.947in, y=1.69in). Consider block shear in your calculations.
%3D
3/4" plate
2L5 x 3 % x 5/8
3"
Pr
3"
3 @ 3"
12" -
3"
P,
Chapter 3 Solutions
Structural Steel Design (6th Edition)
Ch. 3 - Prob. 3.1PFSCh. 3 - Prob. 3.2PFSCh. 3 - Prob. 3.3PFSCh. 3 - Prob. 3.4PFSCh. 3 - Prob. 3.5PFSCh. 3 - Prob. 3.6PFSCh. 3 - Prob. 3.7PFSCh. 3 - Prob. 3.8PFSCh. 3 - Prob. 3.9PFSCh. 3 - Prob. 3.10PFS
Ch. 3 - Prob. 3.11PFSCh. 3 - Prob. 3.12PFSCh. 3 - Prob. 3.13PFSCh. 3 - Prob. 3.14PFSCh. 3 - Prob. 3.15PFSCh. 3 - Prob. 3.16PFSCh. 3 - Prob. 3.17PFSCh. 3 - Prob. 3.18PFSCh. 3 - Prob. 3.19PFSCh. 3 - Prob. 3.20PFSCh. 3 - Prob. 3.21PFSCh. 3 - Prob. 3.22PFSCh. 3 - Prob. 3.23PFSCh. 3 - Prob. 3.24PFSCh. 3 - Prob. 3.25PFSCh. 3 - Prob. 3.26PFSCh. 3 - Prob. 3.27PFSCh. 3 - Prob. 3.28PFSCh. 3 - Prob. 3.29PFSCh. 3 - Prob. 3.30PFSCh. 3 - Prob. 3.31PFSCh. 3 - Prob. 3.32PFSCh. 3 - Prob. 3.33PFSCh. 3 - Prob. 3.34PFSCh. 3 - Determine the LRFD design strength and the ASD...Ch. 3 - Prob. 3.36PFSCh. 3 - Prob. 3.37PFS
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- Determine the LRFD design strength and the ASD allowable strength . Neglect block shear. A36 steel and 7/8-in Ø boltsarrow_forwardProblem 1 Given the double splice connection shown made up of 6mm x 150mm steel plates with double row of 19mm diameter bolts. O = 552.93 MPa , t= 207MPA and op = 552.93 MPa a. Compute the tensile force of the steel plates ANS b. Compute the number of 19mm diameter bolts from shear ANS c. Compute the number of 19mm diameter bolts from bearing ANS d. Compute the number of 19mm diameter bolts for the connection ANS 150mm 6mmarrow_forwardProblem # 8.0 The tension member shown is a C12 x 20.7 of A572 Grade 50 steel. Will it safely support a service dead load of 60 kips and a service live load of 125 kips? Use Equation 3.1 for U. by O O 2½" 22" | 2½" 7/8-in.-diameter bolts. оооо оооо m 000 U=0.90 Single and double angles m W10x 19 = 0.394 (for parent shape) 0000 by Single and double angles … W8 x 24 -=0.820 > U=0.90 O O O 00 U=0.80 W shape U=0.70arrow_forward
- 1. The tension member shown in Figure 3.4-2 is a PL 58 x 10, and the steel is A36. The bolts are 3/4-inch in diameter. a. Draw the different potential failure lines. b. Compute the effective net area. 22" 2½" 2" 4 3" *₁ 3" ✓ 2" 2. A double-channel shape, 2C10 x 20 steel, Fy = 50 ksi and Fu = 70 ksi, is used for a built-up tension member as shown in the figure. The holes are for 12-inch-diameter bolts. PNA a. Determine the design strength for LRFD. b. Determine the allowable strength for ASD. YD Shape A 2" 2" -X Area, Depth, d | 3²²-|-1 3²² | 00 "There is no elevator to success, you have to take the stairs." - Zig Ziglar God bless. Ⓒ O O O Web Thickness, tw in. in. O 4" C10-30 x25 x20 5.87 10.0 10 0.379 % x15.3 4.48 10.0 10 0.240 4 4" Table 1-5 C-Shapes Dimensions Flange Width, b₁ 2 in. in. in.2 in. C15x50 14.7 15.0 15 0.716 1168 3.72 3% 0.650 % x40 11.8 15.0 15 0.520 x33.9 10.0 15.0 15 0.400 % 14 3.52 32 0.650 % %/16 3.40 3% 0.650 % C12-30 8.81 12.0 12 0.510 4 3.17 3% x25 7.34 12.0 12…arrow_forward3.4-2 The tension member shown in Figure 3.4-2 is a PL 5 × 10, and the steel is A36. The bolts are 7/8-inch in diameter. a. Determine the design strength for LRFD. b. Determine the allowable strength for ASD. 2" 3" 2" |27|27|--|--| 3" О 2" FIGURE P3.4-2arrow_forwardBottom chord of truss is composed of two angle bars each having a dimension of 185 mm x 146 mm x 8.7 mm. Between them is a gusset plate, 9.5mm thick. At each end of joints, four 19mm-dia. bolts are fastened along the gage line, having an edge distance of 44 mm and 59mm pitch. Use A36 steel Fy 248 MPa Fu = 400 MPa %3D spacing of bolt = 68.9 mm %3D 9 5mm thick gunset plate edge dist Compute the capacity of the bottom chord based on block shear strength, in KN. uadarrow_forward
- Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|Yarrow_forwardDetermine the controlling (lowest) net area of the member. X = 2.69 Y = (6-2.69)/2 Z=0.7 Y" -X Y" 5 @ 1/2" OD O a Do -OHD -(5)-z" bolts -L6 X 4 X 7/16arrow_forwardA W16 x 45 of A992 steel is connected to a plate at each flange as shown. Determine the nominal strength based on the net section using alternative value of shear lag factor. * -2¼" 8-in.-diameter bolts W16 x 45arrow_forward
- (2) The truss diagonal member consists of a pair of angles L5x3/2 x 1/2 that are loaded in tension. The bolts to be used are 7/8" A 490-x bolts. The bolt design Shear strength for is angle 37.9 kips per bolt. one 2/4 23/4" 11 o 11 2 10 0 0 0 = L 3 32 33 3 The design strength of one A36 steel member. angle?arrow_forwardPlease solve with complete detailed solution. Thank you.arrow_forwardProblem 1: The tension member is a PL 12 × 6. It is connected to a 38-inch-thick gusset plate with78-inch-diameter bolts. Both components are of A36 steel. a. Check all spacing and edge-distance requirements. b. Compute the nominal strength in bearing W' PL PL ½x 6 24" 24" 1"arrow_forward
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