Principles of Foundation Engineering (MindTap Course List)
9th Edition
ISBN: 9781337705028
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 13, Problem 13.10P
A free-headed drilled shaft is shown in Figure P13.10. Let Qg = 260 kN, Mg = 0, γ = 17.5 kN/m3, ϕ′ = 35°, c' = 0, and Ep = 22 × 106 kN/m2. Determine
- a. The ground line deflection, xo
- b. The maximum bending moment in the drilled shaft
- c. The maximum tensile stress in the shaft
- d. The minimum penetration of the shaft needed for this analysis
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
When an open-ended square tube is placed vertically into a pool of water, the water rises 4 mm
up inside of the tube. A) Determine the inner length of the square tube.
A solid cylindrical rod is then placed vertically down the center of the open-ended square tube
and the water rises an additional 4 mm up the tube.
B) Determine the diameter of the solid rod that was inserted. 0.073 @ 20N m T C
s = = o .
Please use the following labels in the image such as Va, Vbr, Vbl and etc.
Show step by step solution for each.
Thanks!
The single story building shown in Fig. 2 has an applied uniform load of 300 psf (0.3ksf) including the self weight of the beams and the girders. The roof has a 16 ft x 15 ft opening as shown. 1. Determine the axial loads on Columns C1 and C2 using reactions from the beams supported on the columns. 2. Determine the axial loads on Columns C1 and C2 using the concept of tributary areas.
Chapter 13 Solutions
Principles of Foundation Engineering (MindTap Course List)
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Determine the ultimate load-carrying capacity of...Ch. 13 - For the same data given in Problem 13.4, determine...Ch. 13 - Prob. 13.6PCh. 13 - A 3 ft diameter straight drilled shaft is shown in...Ch. 13 - Prob. 13.8PCh. 13 - Figure P13.9 shows a drilled shaft extending into...Ch. 13 - A free-headed drilled shaft is shown in Figure...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.Similar questions
- A built-up beam section is formed by welding 2xL8x6x1 angles to the bottom flange of W36x210 as shown in Fig. 1. Determine the following section properties of the built-up cross section: a. Cross sectional area, A (in2) b. The location of the centroids CG-X and CG-Y id of the built up section from the bottom of angle ( this value is given, so you need to check that you get this value) c. Moments of inertia ICG-X (in4) and ICG-Y (in4) d. Section modulus Sx (bot) (in3) and Sx (top) (in3) e. Radius of gyration rx (in) and ry (in) f. Weight of the built up section, w (lb/ft) (use density of steel = 490 pcf) g. Surface area of the built-up section, S(ft2/ft)arrow_forward11. Design the main beam of a building supporting concrete floor slab as shown in Fig. 10.61 and with the following data: (i) Beam centres: 6 m (ii) Span (simply supported): 7.4 m (iii) Concrete slab (spanning in two directions): 240-mm thick (iv) Finished screed: 40-mm thick (v) Imposed load: 4 kN/m² (vi) Take weight of concrete slab as 24 kN/m³ and total weight of 40-mm thick screed as 1.0 kN/m² Assume Fe 410 grade steel and take initial weight of beam as 1.0 kN/m. H H- Main beam 7.4 m 6.0 m H 40 mm screed I -H. Fig. 10.61 k 240 mm slab 6.0 m Typical bay of large floor areaarrow_forwardConsider the situation in Question 2. If all the cables are made of the same material and have a maximum tensile force of 500 lb, what is the heaviest load that can be supported by the system?arrow_forward
- A flexible circular area is subjected to a uniformly distributed load of 148 (see the figure below). The diameter of the load area is 2 . Estimate the average stress increase () below the center of the loaded area between depths of 3 and 6 . Use the equations: and (Enter your answer to three significant figures.) =arrow_forwardA square flexible foundation of width B applies a uniform pressure go to the underlying ground. (a) Determine the vertical stress increase at a depth of 0.625B below the center using Aσ beneath the corner of a uniform rectangular load given by Aσ = Variation of Influence Value I qoI. Use the table below. n 0.8 1.0 m 0.2 0.4 0.5 0.6 0.2 0.01790 0.03280 0.03866 0.04348 0.05042 0.05471 0.4 0.03280 0.06024 0.07111 0.08009 0.09314 0.10129 0.5 0.03866 0.07111 0.08403 0.09473 0.11035 0.12018 0.6 0.04348 0.08009 0.09473 0.10688 0.12474 0.13605 0.8 0.05042 0.09314 0.11035 0.12474 0.14607 0.15978 1.0 0.05471 0.10129 0.12018 0.13605 0.15978 0.17522 (Enter your answer to three significant figures.) Ασ/90 = (b) Determine the vertical stress increase at a depth of 0.625B below the center using the 2 : 1 method equation below. 90 x B x L Aσ = (B+ z) (L + z) (Enter your answer to three significant figures.) Ασ/90 = (c) Determine the vertical stress increase at a depth of 0.625B below the center using…arrow_forwardPoint loads of magnitude 100, 200, and 360 act at , , and , respectively (in the figure below). Determine the increase in vertical stress at a depth of 6 below point . Use Boussinesq's equation. (Enter your answer to three significant figures.) =arrow_forward
- no chatgpt, show solution. thx^^arrow_forwardP 20 mm 15 mm 100 mm 200 mm 100 mm 15 mm -450.0 mm- -300.0 mm 300.0 mm -300.0 mm -450.0 mm- 300.0 mm The butt connection shows 8-24mm bolts. Spacing and dimensions are shown in figure PSAD-020. The plate is made of A36 steel. Allowable tensile stress on gross area = 0.6Fy Allowable tensile stress on net area = 0.5Fu Allowable shear stress on net area = 0.3Fu Bolt hole diameter = 26mm Determine the allowable tensile load P based on net area rupture. Determine the allowable tensile load P based on gross area yielding. Determine the allowable tensile load P based on block shear strength. Parrow_forward300 mm P 75 mm -75 mm P I 75 mm 100 mm -125 mm Shown in figure PSAD-022 is a splice connection made of 8-22mm A325 bolts and a PL 300 x 12 mm made of A36 steel, connected back-to-back. The connection joins together two A992 W410X67 (Fy = 50 ksi, Fu = 65 ksi) carrying a tension load P. Assume U = 0.70. Ignore bolt shear failure. Determine the design capacity of the connection based on yielding of the plate. Determine the design capacity of the section based on yielding of the wide flange section. Determine the design capacity of the section based on rupture of the plate. Determine the design capacity of the section based on block shear of the plate.arrow_forward
- Properties of L 152 x 152 x 11.1 A = 3280 mm² d = 152 mm t = 11.1 mm x = 41.9 mm Ix=7.33 x 10 mm⭑ Iy = 7.33 x 10 mm Zx=119 x 103 mm³ Sx = 66.5 x 103 mm³ P ° ° ° ° ° ° ° ° P Shown in figure PSAD-021 is a truss joint detail. It makes use of two rows of 22-mm bolts. The members are L 152 x 152 x 11.1 made of A36. The plate connecting the members is a 9mm-thick A36 plate (Fy = 36 ksi, Fu = 58 ksi). Assume that U = 0.85, ignore 85% limit of net area. Determine the effective net area of the section. Determine the allowable capacity of the connection.arrow_forwardA floor consists of 8 steel beams/girders supporting three 1-way slab panels. The beams are supported on 6 columns around the perimeter of the roof. The roof is subjected to a uniform pressure of 150 psf . All beams and girders weigh 90lb//ft. Use free-body diagrams and statics equations to determine the load and reactions on all the beams (Beam 1, 2, 3 and 4), girders (Girders 1 and 2), and columns (Columns 1, 2 and 3). (we only focused on one-way slabs in the class + pls include FBDs)arrow_forwardDesign an intake tower with gates meet the following requirement: • Normal water surface elevation = 100 m mean sea level • Max. reservoir elevation = 106 m msl • Min. reservoir elevation = 90 m msl • Bottom elevation = 81m msl • Flow rate=57369.6 m³/day.. Velocity = 0.083 m/s ⚫c=0.6, Density for water =1000 kg/m³. Density for concrete =2310 kg/m³ Estimate water elevation that make safety factor =1 ร 4 m Uppr gate 2m 1 m Lower gate 2m Gate 6m 2m 4 m ร 2 m wzarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Principles of Foundation Engineering (MindTap Cou...Civil EngineeringISBN:9781337705028Author:Braja M. Das, Nagaratnam SivakuganPublisher:Cengage Learning
Principles of Foundation Engineering (MindTap Cou...Civil EngineeringISBN:9781305081550Author:Braja M. DasPublisher:Cengage Learning
Fundamentals of Geotechnical Engineering (MindTap...Civil EngineeringISBN:9781305635180Author:Braja M. Das, Nagaratnam SivakuganPublisher:Cengage Learning
Principles of Geotechnical Engineering (MindTap C...Civil EngineeringISBN:9781305970939Author:Braja M. Das, Khaled SobhanPublisher:Cengage Learning
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781337705028
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Principles of Foundation Engineering (MindTap Cou...
Civil Engineering
ISBN:9781305081550
Author:Braja M. Das
Publisher:Cengage Learning
Fundamentals of Geotechnical Engineering (MindTap...
Civil Engineering
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning
Principles of Geotechnical Engineering (MindTap C...
Civil Engineering
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Cengage Learning
Difference between Direct and Bending stress || Combined stresses; Author: Civil Engineering;https://www.youtube.com/watch?v=ZXGSSddI5ew;License: Standard YouTube License, CC-BY