**Title: Analysis of a Retaining Wall in Saturated Soft Clay****Diagram Description:**The diagram shows a 4.5-meter high retaining wall. The wall is homogenous and constructed in saturated soft clay.**Given Data:**- Height of Retaining Wall: 4.5 meters- Material: Homogeneous saturated soft clay- Saturated Unit Weight (\( \gamma_{\text{sat}} \)): 19.6 kN/m³- Cohesion (\( c_u \)): 19.3 kN/m² (undrained shear strength)**Objectives:**1. **Rankine's Active Pressure Calculation:** - Calculate the active pressure from a water-filled tension crack to the failure plane.2. **Determine Total Active Force:** - Calculate the total active force from water-filled tension crack per unit length of the wall after tensile cracks have occurred. - Ignore the location of tension stresses and include the force from the water-filled crack in your drawings.3. **Graphical Representation:** - Illustrate the variation of pressure along the wall with depth. - Show the depth at which a tensile crack can occur. - Determine depth before tensile crack and ensure maximum stresses exist and do not cause cracks.**Detailed Calculations and Steps:**- Understand the properties of the soft clay and calculate the pressure distribution along the wall.- Apply Rankine's theory to determine the potential failure surfaces and lateral earth pressure distribution.**Important Considerations:**- Analyze the conditions leading to tensile cracking.- Include water pressure acting on the wall due to the tension crack being water-filled. This analysis is crucial for designing safe retaining walls in areas with similar geological conditions. Proper consideration of tension cracks and shear strength is essential to prevent structural failures.

Given Saturated unit weight= 19.6 KN/m3 Undrained Shear strength= 19.3 KN/m3 Note According to…

Soft Clay sat- 19.6 kN/m² Cu= 19.3 kN/m3 4.5-m Find: → Make the necessary calcu lation ard draw the varialion Rankine's active pressure on the wall with depth. nclude the pressure from a waker-filled crack in Find the depth Delermink the total active wall bepore the tensile crack çcurs. Assome tension stresses exist ard do not include he force from waler - filled Crack. - Delermine the total active force force per wall after the tensile of the resultant. Negect tersion shesses and include the of your drawings unit length of the up to which a tensile crack can occur. -> force -> per unit legth of the occurs. Also, find the location gack 아 force from the waer- pilled crack.

Soft Clay sat- 19.6 kN/m² Cu= 19.3 kN/m3 4.5-m Find: → Make the necessary calcu lation ard draw the varialion Rankine's active pressure on the wall with depth. nclude the pressure from a waker-filled crack in Find the depth Delermink the total active wall bepore the tensile crack çcurs. Assome tension stresses exist ard do not include he force from waler - filled Crack. - Delermine the total active force force per wall after the tensile of the resultant. Negect tersion shesses and include the of your drawings unit length of the up to which a tensile crack can occur. -> force -> per unit legth of the occurs. Also, find the location gack 아 force from the waer- pilled crack.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Determine the absolute maximum shear stress for each of the following plane stress states for which the principal stresses are given. (a) op1 = 2700 psi, op2 = -2470 psi, Op3 = 0 (b) op1 = 44 MPa, op2 = 25 MPa, op3 = 0
For the beams shown in Figs. 1 and 2, determine: Plane stress st ate Princip al stresses Maximum in-plane she ar stress state with corresponding Plot ALL stre ss st ates a bove Plot Mohr Circle for ALL stress st ates a bove 12 kN/m Line 1 Line 2 Line 3 1.4 m 1.9 m 2.8 m Fig. 1 Simply-Supported Beam 12 kN/m Line 1 Line 2 Line 3 10 1.4 m 1.9 m 2.8 m Fig. 2 Cantilever Beam 200 mm 25 mm Line 1 125 mm Line 2 -250 mm 30 mm Line 3 25 mm 100 mm Fig. 3 Cross-Section CS Scanned with CamScanner
A circular aluminum alloy [E = 70 GPa; a = 22.5 × 10-6/°C; v = 0.33] pipe has an outside diameter of 215 mm, a wall thickness of 12 mm, and a length of 4.3 m. The pipe supports a compressive load of 695 kN. After the temperature of the pipe drops 43°C, determine: (a) the axial deformation of the pipe. (b) the change in diameter of the pipe. Answer: (a) ō = i AD = i mm mm
Find the normal stress value on the new x-axis obtained by turning 53 ° clockwise of the stressed element in the figure.
A spherical pressure vessel has an inside diameter of 5 m and a wall thickness of 10 mm. The vessel will be constructed from a special steel [E = 231 GPa; v = 0.32] that has a yield strength of 440 MPa. If the internal pressure in the vessel is 1020 kPa, determine (a) the normal stress in the vessel wall, (b) the factor of safety with respect to the yield strength, (c) the normal strain in the sphere and (d) the increase in the outside diameter of the vessel. Answers: (a) o = i МРа. (b) FS = (c) e = i 106. (d) AD = i mm.
Which of the following is NOT TRUE for a state of stress at point for a thin spherical pressure vessel? O The in-plane shear stress is equal to zero. O The minimum principal stress is equal to pr/2t. The maximum principal stress is equal to pr/2t. The in-plane shear stress is equal to pr/2t
Please try to solve in 30 minute
Problem-3 b. A water main 160 cm dia. Contains water at a pressure head of 200 m. Take weight of water to be 1000 kg/m³. The thickness of the metal shell required for the water main given that the maximum permissible stress in the metal is 400 kg/cm², will be...
1
During the tensile test, experiment on a sample of mild steel, data represented in the figure below was obtained with an initial diameter (do) of 0.505 inch. At failure, the reduced diameter (d;) of the sample was 0.305 inch. Gauge Length (Lo) is 2 inches and final Gauge Length (L) is 2.625 inches. Answer the questions below: Us TR 70- 60 - Lo LyP *IR 50 - EL PL: proportimal limit EL Elastic limit UXP = 4pper yield point Lyp : Lower us : uiti mated streagth IR : Indicanteed Strengty Rupture TR: True strength Ruptue 40 AL or 8t P 30 **PL (proportimal limt) Direct strain E strain (in/in) aStress Iblinz (esi) 0.003. ogo o-200
A pipe of diameter 30 cm carries water in turbulent flow regime. The average height of the surface roughness is 0.25 mm. If the shear stress developed is 5.5 N/m? then the thickness of laminar sublayer, in mm is - (Kinematics viscosity of water, v = 0.01 stokes)
All part plzz