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Principles of Foundation Engineering, SI Edition
8th Edition
ISBN: 9781305446298
Author: Braja M. Das
Publisher: Cengage Learning US
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Chapter 3, Problem 3.16P
To determine
Find the overconsolidation ratio
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Students have asked these similar questions
A simple supported one-way slab spans 24 ft. In addition to its own weight, it supports
an uniformly distributed service live load of 100 psf. A masonry wall, which has a
service dead load of 1500 lb/ft, is also supported by the slab along with an uniform
service floor live load (a line load) of 500 lb/ft loaded as shown below in the figure.
(a)
(b)
Select the depth of the slab using ACI Code's minimum thickness requirements
where deflections are not computed. Design the primary steel and temperature
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has a unit weight of 110 lb/ft³. Assume 3/4 inch cover on the reinforcement aand
that f' 3 ksi and f₁ = 60 ksi. Use #6 bars for the primary steel and #4 bars
for the shrinkage and temperature steel. Assume interior exposure, and use
ACI 318-19.
Repeat part (a) NOT using the ACI Code's minimum thickness requirements for
cases where deflections are not computed. Use #6 bars for the primary steel
and #4 bars…
Water (at 10° C has n = 1.31 x 106 m²/s) flows from reservoir A (surface elevation 100 m) through a 2.25-m-diameter concrete pipe (e = 0.36 mm; f=0.014) to reservoir
B (surface elevation 91.84 m). If the 2 reservoirs are 17 km apart, find the flow velocity V (to 2 decimal places x.xx) ignoring minor losses.
100 m
A
L = 17 km
P₁
h₁ + = +
V²
2g
D=2.25 m
91.84 m
B
+ h₂
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P₂
= =h₂+ +
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The floor system of an apartment building consists of a 4-inch-thick reinforced concrete slab resting on three steel floor beams, which are in turn supported by two steel girders, as shown below. The areas of cross section of the
floor beams and the girders are 18.3 in.2 and 32.7 in.², respectively. Determine the dead load acting on the beam CD.
Steel girder
(A-32.7 in.2)
At
I-
25 ft
B
I
Steel
column
C
D
Steel floor beam
(A =18.3 in.2) I
a. 600.0 lb/ft
b. 662.3 lb/ft
c. 62.3 lb/ft
d. 1347.3 lb/ft
E
4 in.
concrete slab
2 at 12 ft = 24 ft
Chapter 3 Solutions
Principles of Foundation Engineering, SI Edition
Ch. 3 - Prob. 3.1PCh. 3 - Prob. 3.2PCh. 3 - Refer to Figure P3.3. Use Eqs. (3.10) and (3.11)...Ch. 3 - Prob. 3.4PCh. 3 - Prob. 3.5PCh. 3 - Prob. 3.6PCh. 3 - Prob. 3.7PCh. 3 - Prob. 3.8PCh. 3 - Prob. 3.9PCh. 3 - Prob. 3.10P
Ch. 3 - Prob. 3.11PCh. 3 - Following are the standard penetration numbers...Ch. 3 - Prob. 3.13PCh. 3 - Prob. 3.14PCh. 3 - Prob. 3.15PCh. 3 - Prob. 3.16PCh. 3 - Prob. 3.17PCh. 3 - Prob. 3.18PCh. 3 - Prob. 3.19PCh. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - Prob. 3.23PCh. 3 - Prob. 3.24PCh. 3 - Prob. 3.25PCh. 3 - Prob. 3.26PCh. 3 - Prob. 3.27P
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