Analysis and Design of Shallow and Deep Foundations
By
Lymon C. Reese,
William M. Isenhower,
Shin-Tower Wang
( All Chapters Included - 100% Verified Solutions )
, CHAPTER 1
Problem 1.1
Obtain a current copy of a magazine on engineering projects and write a brief exposition of the
nature of the structure, the general nature of the soil conditions, and the kind of foundation being
employed for the project. If possible, list the reasons the engineer chose the particular type of
foundation.
NOTES
The problem presents an opportunity for the student to make use of the Engineering Library.
Many journals are probably available, including Civil Engineering from ASCE or Engineering
News-Record. The use of a journal from the past, which may be in a bound volume, is
satisfactory. In any case, the student must cite the issue, the title of the article, the author if
given, and page numbers.
Problem 1.2
In preparation for considering the behavior of a pile under lateral loading with respect to the pile-
head restraint, (a) compute the deflection of a cantilever beam with a load at the free end and
then (b) one with the end fixed against rotation, but free to deflect.
NOTES
Two types of solution are shown here: one employing the area-moment proposition, and the
other employing the difference-equation method. If time allows, you may wish to sketch the
beam, and start the difference-equation solution to the first of the two problems and ask the
students to complete the solution. If you use such a scheme, you could sketch the divisions of
the beam for the solution of the second of the two problems by difference equations, and give
extra credit on homework for the solution of the second problem.
1
2
,SOLUTION
P1.2a
P
Deflection
Yt
0
Shear
P
a b
Moment
PL
1 PL2 2 L PL3
∑M y = Yt =
EI 2
=
3 3 EI
Solution by difference-equation method:
P
-1 0 1 2 3 4 5
Writing equation for moment at Point 0
Y−1 − 2Y0 + Y1 PL
=
(L 5 )2 EI
Boundary conditions
dy
(1) Y0 = 0 ; (2) = 0
dx 0
From (2): Y−1 = Y1 ;
From (1): Y0 = 0
Substituting
PL3 PL3
2Y1 = ; Y1 =
25 EI 50 EI
2
3
, Y0 − 2Y1 + Y2 4 PL
At Point 1 =
( 5)
L
2
5 EI
Substituting
2 PL3 4 PL3
− + Y2 =
50 EI 125EI
3
4 PL 2 PL3 9 PL3
Y2 = + =
125 EI 50 EI 125EI
Y1 − 2Y2 + Y3 3PL
At Point 2 =
(L5 )2 5EI
Substituting
PL3 18 PL3 3PL3
− + Y3 =
50 EI 125 EI 125EI
3
3PL 18 PL 2.5PL3 18.5 PL3
3
Y3 = + − =
125 EI 125EI 125EI 125EI
Y2 − 2Y3 + Y4 2 PL
At Point 3 =
(L 5 )2 5 EI
Substituting
9 PL3 37 PL3 PL3
− + Y4 =
125EI 125 EI 125EI
3
30 PL
Y4 =
125EI
Y3 − 2Y4 + Y5 PL
At Point 4 =
(L5 ) 2
5 EI
Substituting
18.5 PL3 60 PL3 PL3
− + Y5 =
125EI 125 EI 125 EI
3 3
42.5 PL PL
Y5 = =
125 EI 2.94 EI
Error = 2%, a more accurate solution could have been obtained by using a larger number
of increments
3
4