CONCRETE STRUCTURES
CONCRETE STRUCTURES 
PROBLEM STATEMENT: Design a short circular column to carry Pu = 424 kips and Mu = 127.2 ft kips only 4,000 psi concrete and Grade 60 rebar is available. The steel is to be contained within a 3/8 in diameter spiral. Allow 2.5 in concrete coverage from face of column to steel.
A. GIVEN: ========= Short Column h<10 cross section secondary moment <5% primary moment Pu = 424 kips Mu = 127.2 ft kips fc = 4,000 psi = 4 kips/sqin fy = 6,000 psi = 6 kips/sqin concrete coverage 2.5 in spiral rebar diameter #3 rebar = 3/8 in
B. ASKED: ========= Required steel area number of longitudinal bars IIII size of longitudinal bars # pitch of spiral s diameter of column (diameter
C. GRAPH: =========
Step (#) by Step (#)
===============
D. SOLUTION:
============
#1. Estimate the average concrete stress at
fc average = 1/2 fc
fc average = 1/2 fc = 4 kips/sq inch / 2
fc average = 2 kips/ sq inch
=====================
#2. Compute a Trial Area Ag
A g = Pu
---------------
f c average
= 424 kips
-------------------
2 kips/ sq inch
Ag = 212 sq inch
#3. Selected column diameter 16"
for the column
based on a guesstimated cross sectional area
#4. Ag for a 16" diameter concrete column
Ag = phi * d * d
-----------------
4
Ag = 3.14 * 16 * 16
----------------------------
4
= 200.96
Ag = 201 sq inches
===============
#5. Concrete Coverage
2.5 inches
from face of column to outside diameter
of steel spiral
given - otherwise refer to
Table 7.1 Leet p. 314
Table 7.1 Leet p. 314
Minimum Concrete Cover for
cast-in-place column reinforcement
============================
Condition Reinforcement Min. Cover
cast against
earth all sizes 3"
surfaces formed
but exposed to
weather or earth # 6 - 8 2"
# 5 - 1 1.5"
no exposure to
weather or earth main reinforcement 1.5"
ties 1.5"
spirals 1.5"
#6. Calculate Diameter of Spiral
concrete coverage left 2.5"
concrete coverage rifght 2.5"
-----------------------------------------------'
total concrete coverage 5.0"
column diameter 16"
- concrete coverage 5"
-----------------------------------------------
core diameter 11" hg
#7. gama=
Ratio of distance between rows of reinforcement on opposite sides of
column to
depth of column in direction of bending
used in ACI Design Aids
gamma = hc/hg
= 11"/16"
( gamma= 0.69
=======
#8. Variables for ACI Interactive Curves
#9. e = eccentricity of load
e = Mu
---------------------
Pu
e = 127.12 * 12 (kips ft * 12)
--------------------------------------
424 kips
e = 3.6 inches
===========
#10.
e = 3.6"
h 16"
e = 0.225
h
=========
#11. Pu = Factored Load
----> read "phi P sub u"
phi Pu = Pu
Pu = 424 kips given in problem
===========
#12.
phiPu = 424 kips
Ag 201 sqin
= 2.11 kips/squint
===============
#13.
phi Pn * e
Ag h
2.11kips/sqinch * 0.225
= 0.475 kips/sqinch
===============
#14. Interpolating between
fig. 7.38 p. 328 Leet
p = 2.11
e = 0.225
Ag
h
2.0
pg = 0.041
(read)
Pu * e = 0.475
Ag h
#15. Reinforcement Ration pg
pg = 0.03
pg = 0.05
pg = 0.041
===========
#16. Required Steel
Ast = pg * Ag
= 0.041 * 201 sqinch
Ast = 8.25 sqinch
==============
#16. Steel Area
Table B1 p. 614 Leet
7 = number of bars
I
V
#10 8.86 sq in
#17. #10 bars selected #18. As sup = 8.86 sq inch > Ast = 8.24 sq in Table B1 Leet 614 Bar Area and Spacing in Slabs Area of groups of standard bars (sq inches) # 2 3 4 5 6 7 8 4 0.39 0.58 0.78 0.98 1.18 1.37 1.57 5 0.61 0.91 1.23 1.53 1.84 2.15 2.45 6 0.88 1.32 1.77 2.21 2.65 3.09 3.53 7 1.20 1.80 2.41 3.01 3.61 4.21 4.81 8 1.57 2.35 3.14 3.93 4.71 5.50 6.28 9 2.00 3.00 4.00 5.00 6.00 7.00 8.00 10 2.53 3.79 5.06 6.33 7.59 8.8610.12 11 3.12 4.68 6.25 7.81 9.37 10.9412.50 14S4.50 6.75 9.00 11.25 13.50 15.7518.00 18S8.00 12.00 16.00 20.00 24.00 28.0032.00
#19. Answer 7 (#10) vertical evenly distributed #3 spiral
