EXAMPLES 1SolutionA uniform stress of 250 KPa is applied to the loaded area shown below:(Dimensions in meters)Using Newmark chart, compute the stress at a depth of 80 m below the groundsurface due to the loaded area under point O'.Uniform lead, 250 KPScale I em-20 mot4060Draw the loaded area such that the length of the line OQ is scaled to 80m. Forexanple. the ilist ance ĀB in Fig. Ex. 8.3a is 1.5 times the distance OQ.0Q = 80 mand AB - 120 m. Next, place point O', the point where the stress is required, overthe center of the influence chart (as shown in Fig. Ex. 8.3b to a slightly smaller scale).The number of blocks (and partial blocks) are counted under the loaded area. In thiscase, about eight blocks are found. The vertical stress at 80 m is then indicated by:(a)Influenceo,- qdxNo. of blocksbleck-0.02Where q.- surface or contact stress, and1 - influence value per block (0.02 in Fig. Ex. 8.3b)Therefore,o,- 250 KPax0.02x8 blocks = 40 KPaDephpreTo compute the stress at other depths, the process is repeated by making otherdrawings for the different depths, changing the scale each time to correspond to thedistance OQ on the influence chart.(b)Figure Ex 83 (ler Newmark, 1942)

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A uniform stress of 250 KPa is applied to the loaded area shown below: Using Newmark chart, compute the stress at a depth of 80 m below the ground surface due to the loaded area under point O'.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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

EXAMPLES 1
Solution
A uniform stress of 250 KPa is applied to the loaded area shown below:
(Dimensions in meters)
Using Newmark chart, compute the stress at a depth of 80 m below the ground
surface due to the loaded area under point O'.
Uniform lead, 250 KP
Scale I em-20 m
ot
40
60
Draw the loaded area such that the length of the line OQ is scaled to 80m. For
exanple. the ilist ance ĀB in Fig. Ex. 8.3a is 1.5 times the distance OQ.0Q = 80 m
and AB - 120 m. Next, place point O', the point where the stress is required, over
the center of the influence chart (as shown in Fig. Ex. 8.3b to a slightly smaller scale).
The number of blocks (and partial blocks) are counted under the loaded area. In this
case, about eight blocks are found. The vertical stress at 80 m is then indicated by:
(a)
Influence
o,- qdxNo. of blocks
bleck-0.02
Where q.- surface or contact stress, and
1 - influence value per block (0.02 in Fig. Ex. 8.3b)
Therefore,
o,- 250 KPax0.02x8 blocks = 40 KPa
Deph
pre
To compute the stress at other depths, the process is repeated by making other
drawings for the different depths, changing the scale each time to correspond to the
distance OQ on the influence chart.
(b)
Figure Ex 83 (ler Newmark, 1942)

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