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 3.34PFS
To determine
The LRFD design strength and the ASD allowable strength of the given section,
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H.W.4: A beam section is limited to a width of 250 mm, and total depth h=550 mm, and has to resist
a factored moment 307 kN.m.Calculate the required reinforcement, given f'c=21 MPa,
fy=350 MPa, d'=65mm.
(As=5425mm, As'=2022 mm Ans)
HW5 A beam section is limited to a width of 300 mm and total denth h-500 mm and has to resist
H.W.3: A rectangular beam has a width of 400 mm, and effective depth d=700 mm to centroid of
tension steel bars. Tension steel reinforcement consist of 4036mm in two rows,
compression reinforcement of 2622mm. Calculate the design moment strength of the
beam, where f'c=21 MPa, fy=420 MPa, d'=65mm.
(Mu=927 kN.m Ans)
H.W.4: A beam section is limited to a width of 250 mm, and total depth h=550 mm, and has to resist
a factored moment 307 kN.m.Calculate the required reinforcement, given f'c=21 MPa,
fy=350 MPa, d'=65mm.
(As=5425mm, As'=2422 mm Ans)
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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- H.W.6: Design the steel reinforcement for flexural for the beam shown in the fig. below. Given f'c=28 MPa, fy=420 MPa. D.L 100 kN/m L.L=200 kN 3 m 3 m 600 mm- A =? 300 mm.arrow_forwardThe tension member shown in the figure below must resist a service dead load of 60 kips and a service live load of 45 kips. Does the member have enough strength? The steel is A588: Fy = 50 ksi, F₁ = 70 ksi; and the bolts are 11/8 inches in diameter. Assume that A = An PL 38 x 72 оо a. Use LRFD. Determine the design strength and the factored load. Make a conclusion about the member. (Express your answers to three significant figures.) +Pn Pu = = kips kips -Select- b. Use ASD. Determine the allowable strength and required strength. Make a conclusion about the member. (Express your answers to three significant figures.) Ft Ae = P₁ = -Select- kips kipsarrow_forwardA single-angle tension member of A36 steel must resist a dead load of 49 kips and a live load of 84 kips. The length of the member is 18 feet, and it will be connected with a single line of 1-inch-diameter bolts, as shown in the figure below. There will be four or more bolts in this line. For the steel Fy = 36 ksi and F₁ = 58 ksi. Try the tension members given in the table below. Tension member 4, (in.) rz (in.) 7 L6 × 6 × 9.75 1.17 8 L 5 × 3 × 4.93 0.746 L 5 × 3 × 5 2.56 0.758 16 7 L5 × 3 × 3.31 0.644 16 Bolt line a. Select a single-angle tension member to resist the loads. Use LRFD. A) L 6 × 6 × B) L 5 × 3 × CL5×3× D) L 5 × 3 × -Select- 5 16 7 16 What is the required gross area? (Express your answer to three significant figures.) A₁ = in.² What is the required effective area? (Express your answer to three significant figures.) A = in.2 What is the minimum radius of gyration? (Express your answer to three significant figures.) "min = in. b. Select a single-angle tension member to…arrow_forward
- An L6 × 4 × 5/8 tension member of A36 steel is connected to a gusset plate with 1-inch-diameter bolts, as shown in the figure below. It is subjected to the following service loads: dead load = 40 kips, live load = 100 kips, and wind load = 45 kips. Use the equation for U: U = 1 − For A36 steel: Fy = 36 ksi, F = 58 ksi. x l For L6 × 4×5/8: Ag = 5.86 in.², x = 1.03 in. 21/4" L6 × 4 × 5/8 11/2" 21/2" 11/2" a. Determine whether this member is adequate using LRFD. -Select- What is the design strength for LFRD? (Express your answer to three significant figures.) Φι Ρη - = kips Which AISC load combination controls? -Select- What is the controlling AISC load combination? (Express your answer to three significant figures.) Pu = kips b. Determine whether this member is adequate using ASD. -Select- What is the allowable strength for ASD? (Express your answer to three significant figures.) Pn Sit kips Which ASD load combination controls? -Select- What is the controlling ASD load combination?…arrow_forwardA double-angle shape, 2L7 × 4 × 5/8, is used as a tension member. The two angles are connected to a gusset plate with 7/8-inch-diameter bolts through the 7-inch legs, as shown in the figure below. A572 Grade 50 steel is used: Fy Fu = 65 ksi. Suppose that t = 5/8 in. = 50 ksi, For L7 x 4 x 5/8: Ag = 6.5 in.², x = 0.958 in. 21/2" оо 11/2" 2L7 x 4 x t a. Compute the design strength. (Express your answer to three significant figures.) ФЕРП = kips b. Compute the allowable strength. (Express your answer to three significant figures.) 'n Sit = kipsarrow_forwardI really need help on barrow_forward
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