西南交通大学基础工程课程设计.doc

Foundation Design
姓名 马 德 林
学号 20100193
班级 2010级土木茅1班
西南交通大学土木工程学院
2013年5月
CONTENTS
Problem-—---———--—-—--———--—-——-----—————-———-----—---——-——--———————-———————-----—--—---2
Design————-——-—-—-——--—---————-—-——--——---———-——----—-----—--—--—-—-———--—-----——————-———4
Chapter one Unit conversion--—--———--—--—-—-—————-—----—--------————————-——--———4
Chapter two Design load calculation---———-———------—---—-—-—-——-——---—-———-----4
Chapter three Geotechnical designing————--——---—————-——-——-———---——————-—---——-4
Step 1 to Step 9 Allowable bearing pressure method-—-————---————-———-—----4
Step 10 Checking moment load—-—--————————-—-—--———--———--—--—--—--———-—-----—8
Chapter four Structural designing--—-—-—-—-————-—----—--—----—-—--—----—--—--—-—-8
Part A Determine required thickness based on a two—way shear analysis——8
Check one—way shear-—--—-———-——-—————---—-—--—-----—-—------—————————-——9
Part B Design the flexural steel-—--—---———-—-——————--—————-———--—-—--------———---9
Chapter five Sketch of the designed footing----—-—-—-——--—-——-—----——-—--——-—-11
Figure 2 Thickness and effective depth——-—-——--——-——--—--—-—--——---—--—---——-—-11
Figure 3 Dimensions and reinforcing steel-—-——-——-—————-——---——-—---——-———----11
Figure 4 Inner block and outer block—-—-—--—-————-———-—---—---———---——-—————---—12
Figure 5 Structural show-—-—-——----------—-—-—---------——-——--————--—---—---———-——12
Problem:
Design for Practice
A proposed office building is to be constructed at the site with a geologic profile showed in figure 1。 The ground table is at 5。5 ft. The shallow strata are very soft。 The data for these strata maybe used in foundation design were obtained from a series in-situ tests and laboratory tests, and showed in table 1。 In the table, Su is undrained shear strength； is preconsolidation stress.

Medium Sand (Dr=60%)
0-ft
12-ft
23-ft
27-ft
GWT @ 5.5-ft
High Plastic Clay (CH)

Silty Clay (CL)
Very Stiff Silty Clay (CL)

Fig。 1 geologic section for the construction site
The design columns of the proposed office building will carry the following loads： dead vertical load range 30-100 k, live vertical load range 20-75 k, and dead load moment range 0-50 ft-k. These columns are to be supported on spreading footings.
A sketch of an interior column and its spreading footing is given in figure 2。 If such an interior carrying a 50k dead vertical load, a 50k live vertical load， and a dead load moment 20 ft—k. Try to determine the spreading footing of this column。 The design task should including followings,
Table 1 data for different strata(ft)

(1) Unit conversion
Before beginning your design, please convert the data in the figures and tables from English to SI， and please use SI in your designing.
(2) Design load calculation
There are two methods of expressing and working with design loads： the allowable stress design (ASD) and resistance factor design (LRFD）. Calculate both of them.
(3) Geotechnical designing
Select a suitable type of the spreading footing， determine the footing depth, determine allowable bearing pressure， and determine the required base dimensions for the footings of the column in Figure 2.
Fig. 2 A sketch of the typical interior column and its footing
(4) Structural designing
Determine the materials using in the designed footings, determine the thickness of the footing, and determine the reinforcing steel of the footing.
(5) Sketch of the designed footing
Show your design in a sketch.
A proposed design
Chapter one Unit conversion
To Convert
To
Multiply by
ft
m
0.3048
psf
kPa
0.04787
pcf
kN/m3
0。1571
Data for different strata （SI）
Depth
Range（m)
Soil
Description

0-3。657
CH
16.495
61。274
0.15
0。02
124。462
3。657—7。010
CL
17.595
86.166
0。11
0.015
143。610
7。010-8。229
Med.Sand
18。852
—
0。006
0。002
-
>8.229
CL
18。538
124。462
0.08
0.01
287.220
Tips: to convenient, English units will be used in the designing.
Chapter two Design load calculation
The allowable stress design (ASD）:
( ）
()
Resistance factor design (LRFD):
Chapter three Geotechnical designing
Step 1—--- per TABLE 8。1 (）
Use an estimated D of 2 ft (24 in)
Step 2—-—- The groundwater table is at 5.5 ft ，and is not a concern at this site
Step 3——-- per Figure 6.11 （Soil Type： Clay Design F with Typical Range）
Use F=3.5
step 4—-—— For high clay, if saturated undrained conditions exist( as same as the problem statement）， we may conduct a stress analysis with the shear strength defined as and 。 In this case， (per TABLE 6.1)
Hence，,. Using square foundation(B=L）.
Using the BEARING。XLS spreadsheet with ，the computed allowable bearing pressure，
Step 5---— per TABLE 2.2 （ Typical commercial and residential buildings）
Per TABLE2。1 for office building, use ( in order to control differential settlement here)
Step 6———- using TABLE 7。5 for clayey natural soil，assuming the foundation is a ”rigid" structure, the design value of is 0。5
Step 7---— ， so the total settlement requirement controls the settlement analysis
Step 8—-—- using classical method to compute total settlement of shallow foundation， which is based on Terzaghi's theory of consolidation
Because of the assumption that all of the soils are over-consolidated， the equation of the total settlement is:
Case1()：
Case 2（):
Where ：
=rigidity factor (per TABLE 7.1, for spread footings， )
=initial vertical effective stress at midpoint of the soil layer
=final vertical effective stress at midpoint of the soil layer
and
could be computed by simplified method equation:
(for square foundation）
In which ,
=bearing pressure
=depth from bottom of foundation to point
=vertical effective stress at a depth D below the ground surface
（in this case， ）
=pre-consolidation stress at midpoint of the soil layer, and
Now， we can compute the consolidation settlement after dividing the soil beneath the footing into layers.
Figure 1 Dividing the soil beneath the footing into five layers
Try with
，
At midpoint of soil layer

（in)
Layer
No。
H
(ft)
zf
(ft)
psf
psf
psf
psf
Case
①
3.5
1.75
289
1290
1579
2889
OC-I
0.15
0。02
0.53
②
6。5
6.75
716
837
1553
3316
OC—I
0.15
0。02
0。44
③
5
12。5
978
399
1377
3978
OC—I
0。11
0.015
0.11
④
6
18
1251
221
1472
4251
OC-I
0.11
0.015
0.06
⑤
4
23
1515
143
1658
1515
0C-II
0。006
0.002
0.01

Σ= 1.15
， so the settlement criterion has not been satisfied.
Try with
,
At midpoint of soil layer

（in)
Layer
No。
H
(ft)
zf
（ft)
psf
psf
psf
psf
Case
①
3。5
1.75
289
976
1265
2889
OC—I
0.15
0.02
0.46
②
6。5
6.75
716
711
1427
3316
OC-I
0.15
0。02
0.40
③
5
12。5
978
375
1353
3978
OC-I
0。11
0.015
0。11
④
6
18
1251
216
1467
4251
OC—I
0。11
0。015
0。06
⑤
4
23
1515
142
1657
1515
0C-II
0.006
0.002
0。01

Σ= 1.04
, so the settlement criterion has not been satisfied。
Try with
，
At midpoint of soil layer

(in）
Layer
No.
H
(ft）
zf
（ft)
psf
psf
psf
psf
Case
①
3.5
1。75
289
863
1152
2889
OC—I
0.15
0。02
0。43
②
6.5
6.75
716
657
1373
3316
OC-I
0.15
0。02
0。37
③
5
12.5
978
363
1341
3978
OC—I
0.11
0。015
0.10
④
6
18
1251
213
1464
4251
OC—I
0。11
0.015
0。06
⑤
4
23
1515
141
1656
1515
0C—II
0。006
0。002
0。01

Σ= 0.97
， so the settlement criterion has been satisfied。
（Tips: using the SETTLEMENT。XLS with and can also produce )
Step 9——-- 1078〈2770， so the settlement controls the design。 Rounding to a multiple of 500 psf gives：
For a 100k column load（a 50k dead vertical load and a 50k live vertical load），
use
Step 10——-- checking moment load
Using Equation 5。5：
， so OK for eccentric loading
Since （in practical engineering, 5％ error is allowed）, this design is satisfactory.
Hence， a suitable type of the spreading footing may be chosen as：
The footing depth：
Allowable bearing pressure：
Base dimension for the footing: （square foundation)
Chapter four Structural designing
Now， a 21-inch square reinforced concrete column（c=21in) carries a vertical dead load of 50k, a vertical live load of 50k， and a dead load moment 20 ft—k( that’s to say， , ）。
Because of the small applied load and inspection costs of high strength concrete， we will use and 。
Part A—-—- Determine required thickness based on a two—way shear analysis:
Try T=12 in：
Not acceptable!
Try T=15 in:
OK！
However， moment load is present here， one—way shear needs to be checked， too.
Check one—way shear, using Equation 9.8：
Using Equation 9。9:
OK!
Therefore， T=15 in; d=11 in.
Part B———— Design the flexural steel
Find the required steel area
Check minimum steel (for grade 60 steel）
Use
Use 7 ＃7 bars each way ()
(Tips: The final value of (＃7 bars） is determined as a part of the flexural analysis, and is different from 1 in assumed before. However, this difference is small compared to the construction tolerances, so there is no need to repeat the shear analysis with revised 。)
and Acceptable!
Clear space between bars=144/8-0。875=17 in< 3T=45 in or 18 in (ACI 10.5。4）
Check development length
For spread footings, use ， which is conservation。
c=spacing or cover dimension=3+0.875/2=3。44 in
=reinforcement location factor=1。0 for all other cases
=coating factor=1。0 for uncoated bars
=reinforcement factor=1。0 for #7 and larger bars
=lightweight concrete factor=1。0 for normal concrete
use 2.5
，so the development length is OK.
（Tips： It's easy to compute that using 9＃6 bars each way （) is also acceptable。)
To sum up， the spread footing structure can be designed as：
Materials：
Concrete:
Reinforcing steel: grade 60 steel ()
7 #7bars each way ( )
Structure:
The footing thickness： effective depth：
Chapter five Sketch of the designed footing
Figure 2 Thickness and effective depth
Figure 3 Dimensions and reinforcing steel
Figure 4 Inner block and outer block
Figure 5 Structural show
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