### Problem Statement3. The uniformly distributed vertical loads on the surface of a clay layer are as shown in the figure below. Determine the vertical stress increase at points A and B due to the loaded area. Points A and B are located at a depth of 5 meters below the ground surface.### Diagram ExplanationThe diagram illustrates a rectangular plan view of a loaded area on the clay layer surface. The dimensions and load distributions are detailed as follows:- The load is applied over a rectangular area. - The load distribution is divided into two distinct sections: - A central section with a uniform load of \( q = 200 \, \text{kN/m}^2 \), extending horizontally 4 meters (2 meters from the center in each direction). - An additional section to the left, with a uniform load of \( q = 150 \, \text{kN/m}^2 \), extending 2 meters.- The total width of the loaded area is 6 meters (2 meters on the left + 4 meters in the central section).- The loaded area extends 3 meters in the y-direction.Points A and B:- Point A is located at the center, directly below the boundary between the 200 kN/m² and 150 kN/m² sections.- Point B is located at the right boundary of the 200 kN/m² section, 3 meters from the vertical edge and within the horizontal plane containing points A and B. Both points A and B are at a depth of 5 meters beneath the ground surface. The challenge involves determining the increase in vertical stress at these points due to the loads applied above.

For the given loaded area .We will first calculate the radial distance from point of consideration ,…

3. The uniformly distributed vertical loads on the surface of a clay layer as shown in Figure below. Determine the vertical stress increase at A and B due to the loaded area. A and B are located at a depth of 5 m below the ground surface. 2m q-200 kN/m² q-150 kN/m² 2m +-2m B 3 m X

3. The uniformly distributed vertical loads on the surface of a clay layer as shown in Figure below. Determine the vertical stress increase at A and B due to the loaded area. A and B are located at a depth of 5 m below the ground surface. 2m q-200 kN/m² q-150 kN/m² 2m +-2m B 3 m X

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A circular area on the ground surface is subjected to a uniformly distributed load, q = 2200 Ib/ft2.…

A: “Since you have asked multiple question, we will solve the first question for you. If youwant any…

Q: Answer fast

Q: A tensile load of 6 kN at 25 C was applied to a steel rod stretching between two rigid walls as…

A: Given: P=6 KN E=200GPa α=12.2 μm/m°C ∆T=25-(-25)=50°C σ=150 MPa

Q: Using equation 6.14, determine the vertical stress (Aoz) increase below the center of paded area at…

A: Given Area = 6.35*3.35m^2 Depth 2m,4,6,8,10m q= 225kn/m^2 Calculate stress at different different…

Q: A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay…

A: Effective stress: Effective stress is defined as the difference between the total stress and the…

Q: 4- A strip load of q= 100 lb/ft² is applied over a width, B=10 ft. Determine the increase in…

A: This question is related to the Soil Mechanics.

Q: Point loads of magnitude 100,200,and 400kN act at B, C, and D, respectively. Determine the increase…

A: Given :

Q: (c) What will be the change in vertical stress below point E in Figure Q1b at the depth of 5m below…

Q: 5 kN 2 kN 15 kN/m 50 mm P. E A 150 mm 3 m 1 m 3 m 1 m 150 mm

A: Free body diagram - RA + RE = 15×3 + 5 + 2⇒RA + RE = 52 KN .........1 Taking the sum of…

Q: 3. In a saturated soil(5 m) below determine and draw the total, pore water and effective stress at…

Q: The effective stress at a point located at a depth of 5.5 ft in a saturated soil (γm = 108 lb/ft3,…

Q: 2- Point loads of magnitude 100, 200, 300 and 400 lbs. act at A, B, C, and D, respectively.…

A: Given Data: The point loads A, B, C and D are given as: QA = 100 lbsQB = 200 lbsQC = 300 lbsQD = 400…

Q: Refer to Figure 10.47. A flexible rectangular area is subjected to a uniformly distributed load of q…

A: Given data in question Stress 330 kN/m2 Depth 6 m Area 18 m × 9 m To find out Increase in…

Q: The profile of a soil consists of a soil layer 3.1 meters below the ground surface. The weight of…

A: Given Bulk unit weight of soil = 19.3 KN/ m3

Concept explainers

Methods to Determine Vertical Stress

The load applied to the soil is transferred to the underground. The presence of water at soil pores and solid grains above the soil are the primary components that are responsible for the effective load transfer. Due to the load acting, the internal component of the soil or the underground formations undergoes deformations or a tectonic shift. The amount of deformation depends upon the amount of load acting on the soil surface, which indeed, depends on the weight of everything present above the soil. Due to this, the underground soil element develops internal stresses which try to resist the tectonic changes.

Question

### Problem Statement

3. The uniformly distributed vertical loads on the surface of a clay layer are as shown in the figure below. Determine the vertical stress increase at points A and B due to the loaded area. Points A and B are located at a depth of 5 meters below the ground surface.

### Diagram Explanation

The diagram illustrates a rectangular plan view of a loaded area on the clay layer surface. The dimensions and load distributions are detailed as follows:

- The load is applied over a rectangular area.
- The load distribution is divided into two distinct sections:
- A central section with a uniform load of \( q = 200 \, \text{kN/m}^2 \), extending horizontally 4 meters (2 meters from the center in each direction).
- An additional section to the left, with a uniform load of \( q = 150 \, \text{kN/m}^2 \), extending 2 meters.
- The total width of the loaded area is 6 meters (2 meters on the left + 4 meters in the central section).
- The loaded area extends 3 meters in the y-direction.

Points A and B:
- Point A is located at the center, directly below the boundary between the 200 kN/m² and 150 kN/m² sections.
- Point B is located at the right boundary of the 200 kN/m² section, 3 meters from the vertical edge and within the horizontal plane containing points A and B.

Both points A and B are at a depth of 5 meters beneath the ground surface. The challenge involves determining the increase in vertical stress at these points due to the loads applied above.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1: Given

VIEW

Step 2: Given data

VIEW

Step 3: Load calculation

VIEW

Step 4: Vertical Stress increment at point A

VIEW

Step 5: Vertical Stress increment at point B

VIEW

Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 6 steps with 4 images

SEE SOLUTION Check out a sample Q&A here

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

4- Determine a suitable W profile to limit the maximum stress σd=200 MPa
3- Due to application of line loads q1 and q2, the vertical stress increase at point A is 42 lb/ft?. Determine the magnitude of q2 q.=292 Ib/ft 92 45° 4.5 ft 3 ft 3 ft A
the profile of on 2 m from the load. 5-11 A uniform surface load of 220 kPa is applied over the area shown below. Estimate the stress increase in soil due to the surface load at a depth of 15 m below point A. 10 m 20 m IT 10 m A
The center of a rectangular area at the ground surface has Cartesian coordinates (0, 0), and one corner has coordinates (7, 18). All dimensions are in meters. The area carries a uniform pressure of 135 kPa. Estimate the stresses at a depth of 15 m below the ground surface at each of the following locations: (0, 0), (0, 18), (7, 0), (7, 18), and (12, 28).a. Obtain the values by the Boussinesq methodb. Compare the results with those of the 2:1 methodc. Comment on the resultsHint: In the Boussinesq method, the equations attached can be used.
6- A flexible rectangular area is subjected to a uniformly distributed load of q=250 lb/ft?. Determine the increase in vertical stress at a depth of z=6 ft under points A, B, and C. 6 ft + 2.5 ft 3 ft 1.5 ft A C q=250 Ib/ft? 1.8 ft
The cross-sectional area of bar ABCD is 500 mm^2 . Determine the stress in segment BC
1. The bar has a cross-sectional area of 400*106 m². If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads, determine the average normal stress in the bar as a function of x, for 0.5 m < x≤ 1.25 m. -X 0.5 m w = 8 kN/m -6 kN 0.75 m 3 kN
Prob. 3 The plan of a flexible rectangular loaded area is shown in Figure below. The uniformly distributed load on the flexible area, q, is 100 kN/m². Determine the increase in the vertical stress, Aoz, at a depth of z = 2 m below a. Point A b. Point B c. Point C 4 m 1.6 m 2 m 0.8 m q = 100 kN/m² A 1.2 m-
A concentrated load of 2200 kN is applied to the ground surface. Detemine the vertical stress increment due to this load at a point 6 m. below the ground surface and 4.8 m horizontally from the line of the concentrated load. Use Westergaard equation. #32 P=2200 kN z=6m A r-4.8m
2. Find the induced stress at point B at depth 20 ft below the surface. The bearing pressure on the area is 400 psf. 40 ft 35 ft 70 ft