Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 6, Problem 6.4P
Redo Problem 6.2 using the general bearing capacity equation [Eq. (6.28)].
A 5.0 ft wide square footing is placed at 3.0 ft depth within the ground where c′ = 200 lb/ft2, ф′ = 25°, and γ = 115.0 lb/ft3. Determine the ultimate bearing capacity of the footing using Terzaghi’s bearing capacity equation and the bearing capacity factors from Table 6.1. What is the maximum column load that can be allowed with a factor of safety of 3.0?
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==
H.W.5: A beam section is limited to a width of 300 mm, and total depth h=500 mm, and has to resist
a factored moment 405 kN.m.Calculate the required reinforcement, given f'c=28 MPa,
fy=420 MPa.
(As=6425mm, As'=2425 mm Ans)
Chapter 6 Solutions
Principles of Foundation Engineering (MindTap Course List)
Ch. 6 - For the following cases, determine the allowable...Ch. 6 - A 5.0 ft wide square footing is placed at 3.0 ft...Ch. 6 - Prob. 6.3PCh. 6 - Redo Problem 6.2 using the general bearing...Ch. 6 - The applied load on a shallow square foundation...Ch. 6 - A 2.0 m wide continuous foundation carries a wall...Ch. 6 - Determine the maximum column load that can be...Ch. 6 - A 2.0 m wide strip foundation is placed in sand at...Ch. 6 - A column foundation (Figure P6.9) is 3 m × 2 m in...Ch. 6 - For the design of a shallow foundation, given the...
Ch. 6 - An eccentrically loaded foundation is shown in...Ch. 6 - Prob. 6.12PCh. 6 - For an eccentrically loaded continuous foundation...Ch. 6 - A 2 m 3 m spread footing placed at a depth of 2 m...Ch. 6 - Prob. 6.15PCh. 6 - A tall cylindrical silo carrying flour is to be...Ch. 6 - A 2.0 m 2.0 m square pad footing will be placed...Ch. 6 - An eccentrically loaded continuous foundation is...Ch. 6 - A square foundation is shown in Figure P6.19. Use...Ch. 6 - The shallow foundation shown in Figure 6.25...Ch. 6 - Consider a continuous foundation of width B = 1.4...
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- H.W.1: A rectangular beam has a width of 300 mm, and effective depth d=570 mm to centroid of tension steel bars. Tension steel reinforcement consist of 6628mm in two rows, compression reinforcement of 2022mm. Calculate the design moment strength of the beam, where f'c=28 MPa, fy=420 MPa. (Mu=822 kN.m Ans)arrow_forwardH.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 resistarrow_forwardH.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)arrow_forward
- 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)arrow_forwardH.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_forward
- A 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_forwardAn 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_forward
- I really need help on barrow_forwardWU Example 6 For the exterior transverse frame of the flat slab floor shown in figure, and by using the Direct Design Method, find: a. Longitudinal distribution of the total static moment at factored loads. b. Lateral distribution of moment at exterior panel (column and middle strip moments at exterior support) D= 6.5 kN/m² L= 5.0 kN/m² تفكر وکھل flat slap ما لا يوجد bon حامل . 3000 1000 5000 160 + 2000+ +2000+ 5000 2608 300 2000 Drop Panal السعفarrow_forwardExample 4 For the transverse interior frame (Frame C) of the flat plate floor with edge beams shown in Figure, by using the Direct Design Method, find: 1) Longitudinal distribution of total static moment at factored loads. 2) Lateral distribution of moment at interior panel (column and middle strip moments atnegative and positive moments). 3) Lateral distribution of moment at exterior panel (column and middle strip moments atnegative and positive moments). Plat 5000-5000 5000 -Frame C لا بوجود deen 0009 0009 Slab thickness = 180 mm, d = 150 mm q₁ = 16.0 kN/m² All edge beams = 250x 500 mm All columns = 500x 500 mm 6000arrow_forward
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CE 414 Lecture 02: LRFD Load Combinations (2021.01.22); Author: Gregory Michaelson;https://www.youtube.com/watch?v=6npEyQ-2T5w;License: Standard Youtube License