Find the settlements of mat foundation (S corner & S center ) & check angular rotation using empirical formula.Here Id =30 2mnvm2 +n² +1 (m² + n2 + 24n m2 +n2 +m²n² + 1 m2 + n² + 1Use the following figures for the calculation.Given, Each Corner Column = 400 + ID * 0.001 kips2mnvm2 +n2 +1\m2 +n2 -m2n² + 1/+ tan-1Each Exterior Column = 340 + ID * 0.001 kipsEach Interior Column = 380 + ID * 0.001 kips65 + ID * 0.00135 + ID * 0.001 6 ftSilt y = 80 pcf8 ft6 ftSand y = 100 pcf20 ftSand y = 110 pcfClay y = 120 pefCe = 0.4e, = 1.215 ftRock

Solution: Given Data: Here ID=30Each corner column=400+ID×0.001 kips=400+30×0.001=400.03 kipsEach…

Answered: Use the following figures for the…

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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Question

Find the settlements of mat foundation (S corner & S center ) & check angular rotation using empirical formula .Here Id =30

2mnvm2 +n² +1 (m² + n2 + 2
4n m2 +n2 +m²n² + 1 m2 + n² + 1
Use the following figures for the calculation.
Given, Each Corner Column = 400 + ID * 0.001 kips
2mnvm2 +n2 +1
\m2 +n2 -m2n² + 1/
+ tan-1
Each Exterior Column = 340 + ID * 0.001 kips
Each Interior Column = 380 + ID * 0.001 kips
65 + ID * 0.001
35 + ID * 0.001

6 ft
Silt y = 80 pcf
8 ft
6 ft
Sand y = 100 pcf
20 ft
Sand y = 110 pcf
Clay y = 120 pef
Ce = 0.4
e, = 1.2
15 ft
Rock

Expert Solution

Step 1: Answer

Solution:

Given Data:

$H e r e I D = 30 E a c h c o r n e r c o l u m n = 400 + I D \times 0.001 k i p s = 400 + 30 \times 0.001 = 400.03 k i p s E a c h E x t e r i o r C o l u m n = 340 + I D \times 0.001 k i p s = 340 + 30 \times 0.001 = 340.03 k i p s E a c h i n t e r i o r C o l u m n = 380 + I D \times 0.001 k i p s = 380 + 30 \times 0.001 = 380.03 k i p s L e n t h o f f r a m e = 65 + I D \times 0.001 = 65 + 30 \times 0.001 = 65.03 k i p s B r e a d t h o f f r a m e = 35 + I D \times 0.001 = 35 + 30 \times 0.001 = 35.03 k i p s$