EBK FOUNDATION DESIGN
3rd Edition
ISBN: 9780133424478
Author: CODUTO
Publisher: YUZU
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Chapter 4, Problem 4.9QPP
To determine
The angle of internal friction for each test
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Chapter 4 Solutions
EBK FOUNDATION DESIGN
Ch. 4 - Describe a scenario that would require a very...Ch. 4 - How would you go about determining the location of...Ch. 4 - A five-story office building is to be built on a...Ch. 4 - A two-story reinforced concrete building is to be...Ch. 4 - Discuss the advantages of the CPT over the...Ch. 4 - A standard penetration test was performed in a 150...Ch. 4 - Using the CPT data in Figure 4.22, a unit weight...Ch. 4 - Prob. 4.8QPPCh. 4 - Prob. 4.9QPPCh. 4 - A series of vane shear tests have been performed...
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- 8. A prestressed concrete beam is subjected to the following stress distributions: Pi is the initial prestressing force, Pe is the effective prestressing force, M, is the bending moment due to self- weight, Ma and M, are the dead load and live load bending moment, respectively. The concrete has the following properties: fr = 6000 psi and fri = 4200 psi +250 -85 -2500 +550 Pe+ Mo+Ma+Mi P alone P₁+ Mo -2450 -3500 Stress at midspan +210 +250 P, alone Pe alone -2500 -3500 Stress at ends Using Table 22.1, evaluate whether the stresses at the center of the span and the end of the span comply with the permissible stress limits. The beam is classified as U-class. Provide justifications for each condition listed in the table. Note: Calculated stresses are to be taken from the above diagram, and permissible stresses are to be calculated using Table 22.1. Compressive stresses immediately after transfer Tensile stresses immediately after transfer Compressive stresses under sustained and total…arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. (2pts each, 10pt total) Po Pn (1) failure range H 3" 30" Ast 6 No. 10 bars = P 22" I e H 3" (4) e = e small Load path for given e Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb.arrow_forwardNo chatgpt plsarrow_forward
- 11. The prestressed T beam shown below is pretensioned using low relaxation stress-relieved Grade 270 strands. The steel area Aps = 2.5 in². The tensile strength is fpu = 270 ksi, and the concrete compressive strength is fr = 6000 psi. (a) Calculate the nominal moment strength Mn with hr = 6 in. 22" 15" T hf (b) Since this beam is a T-beam, the nominal moment strength M₁ increases with a thicker hf. However, M, stops increasing if he reaches a value. Determine the minimum thickness hy that can achieve the maximum nominal moment strength Mr. Also, calculate the corresponding maximum nominal moment strength Mn with the computed hf.arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward10. A short column is subjected to an eccentric loading. The axial load P = 1000 kips and the eccentricity e = 12 in. The material strengths are fy = 60 ksi and f = 6000 psi. The Young's modulus of steel is 29000 ksi. (a) Fill in the blanks in the interaction diagram shown below. 30" Ast 6 No. 10 bars = Pn (1) Po (4) e = e small Load path for given e failure range Radial lines show constant (2) eb (3) e large failure range Mn (5) e= Mo (b) Compute the balanced failure point, i.e., P and Mb. H 3" P 22" I e H 3"arrow_forward
- 7. Match the given strand profiles with the corresponding loading conditions for a prestressed concrete (PSC) beam. Strand profile (b) (d) (c) (a) Ꮎ Load on a beamarrow_forward4. For serviceability considerations, the effective moment of inertia (Ie) is calculated using the following formula: le 1 - 1cr ((2/3) Mcr) Ma 2 - وا ≥ Note that the upper bound was previously set as Iut in the earlier ACI equation. (a) Arrange the following moment of inertia values in ascending order (from smallest to largest): le, Ier, Ig and lut (b) Mer is the cracking moment. Choose the cross-section that should be used to compute Mcr. NA. h 5. Identify and circle the figure that represents the scenario in which the torsional effect is permitted to be reduced according to the ACI code provisions. (3 pts) mt mi B (b)arrow_forwardI will rate, thanksarrow_forward
- . 9. A reinforced concrete beam is subjected to V/ = 40 kips and Tu/ = 12 ft kips at the critical section. Given conditions: ⚫ Longitudinal reinforcements use No. 8 grade 60 steel with an effective depth d = 20 in. For shear capacity, V = 18 kips and V₂ = 22 kips • For transverse reinforcements, use No. 3 bars with grade 60. • The effective torsional area of A. = 150 in². • Crack angle = 45° ⚫ The minimum stirrup spacing is Smin = 4" and the maximum stirrup spacing is Smax = Find the required stirrup spacing at the critical section. 8".arrow_forward3. The beam shown on the right uses three No. 8 bars made of Grade 60 steel as longitudinal reinforcement. The allowable maximum center-to-center spacing of the longitudinal rebars has been determined to be 10 inches. Now assume that Grade 80 steel will be used instead. Determine whether the beam satisfies the rebar spacing requirements according to the ACI Code. Additional assumptions: • Estimate fs = fy • 20" Clear cover: ? 12" Clear side cover: 1.5" The clear cover depth cc and the clear side cover remain unchanged, regardless of the change in material.arrow_forward6. For the slender columns shown below: a) Determine the effective buckling length factor (k) for each column. b) Circle the column with the largest buckling capacity, assuming all columns have the same length (f) and the same flexural rigidity (E+I) k = (a) (b) (c) (d)arrow_forward
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