Solution to Problem # 01:
Because the section is symmetric, the neutral axis will pass through mid-depth, i.e.,
n=
=
= 15.
= 6.44.
Itr =
+ (n-1) * As *
=
+ (n-1) * As *
+ (6.44-1)*2.37*
+ (6.44-1)*2.3
CEE461
Reinforced Concrete 1
Term Test 1
March 10, 2011
Name: _
(last)
(first)
This test will be graded out of 40 points.
(Question 1: 8 points.)
Consider that cross-section shown below that is 10 inc
CEE461
Reinforced Concrete 1
Term Test 3
Tuesday April 21, 2009
Name: _
(last)
(first)
This test will be initially graded out of 40 points.
(Question 1: 8 points.)
Consider the reinforced concrete cro
Midterm Exam Review
CEE461
Problem 1. Assume the section shown below is subjected to negative bending moment.
Find its bending strength (Mn), reduced bending strength (Mn) and curvature at ultimate
(n
Shurong Lin,
Slim56
REINFORCED CONCRETE DESIGN, I
Spring 2016
Homework Assignment 1
Concrete Construction
Due Tuesday, January 26
1. Describe in a few words what the primary constituents of concrete a
CEE461 Reinforced Concrete I
Gurfinkel and Asem
University of Illinois at Urbana-Champaign
Spring 2017
Name:_
Roster No.:_
Homework Set 8
Follow Homework Assignment requirements in the Class Protocol.
Clarity factor
CEE 4730/6730
Quiz #2, August 28, 2015
Correctness
Quiz Score
Name_
L/2
L/2
wL/2
Uniformly distributed
load w in units of
force per unit length
of beam.
wL/2
Derive the expression for t
CEE 4730/6730 Reinforced Concrete Design
Quiz #3, August 28, 2015
1
What is the sum of 2.34 + 5.6 + 4 + 6.829?
Name _
19
It looks like 18.769, but 4 is only one sig-fig.
Minimum number of decimal plac
Clarity factor
CEE 4730/6730
Quiz #6, September 1, 2015
Correctness
Quiz Score
Name_
Ultimate Strength Moment Capacity
Datum Line
a
d
8.00
C
fc = 4,000 psi
Ec = 57,000 (fc)0.5 psi
Es = 29,000,000 psi
CEE 4730/6730
Quiz #1, Tuesday, August 25, 2015
Name _
Policy on Quizzes, CEE 4730/6730
1. There will be at least 30 quizzes over the semester, meaning that at least one, and
occasionally more than on
\ \xlulxd uumw
g H /
University of Illinois at Urbana-Champaign
C 55461 - Reinforced Concrete I
Gurnkei and Asem Fall 2016 M ,
Name: 914 SIX mun. \n \1.
Roster Nag-"LI.
w: Thad-q!
Homework Set 5
ework
Beam Design for 2016 Semester Project
Flexural design based on moment envelopes
fy (ksi) = 60
f'c (in psi) = 4000
numbers in blue are input
numbers in black are computed
numbers in green are based on
Module 4a:
Floor Loadings
North
1
Up
Floor Plan
N
A
B
C
20
20
Beam 1
10
1
Beam 2W
Beam B
Beam A
10
3
Beam C
Beam 2E
2
Beam 3W
Beam 3E
2
Uniform Floor Loads
Live Load = 50 psf
U = 1.2D + 1.6L = 1.2(75)
CEE 461-NP
Reinforced Concrete Design, I
Spring 2016
Homework Assignment 8
Shear Design of Beams
Due Tuesday, March 29
Design all of the beams for the flooring system shown below so that they will hav
Module 3:
Analysis and Design
for Concentric Axial Loads
CEE 461: Reinforced Concrete Design, slide 3-1
Column Examples
P
P
P
P
CEE 461: Reinforced Concrete Design, slide 3-2
Column Examples
P
P
P
P
C
Module 1:
Concrete Construction and Design
CEE 461: Reinforced Concrete Design, slide 1-1
Introduction
What is concrete?
Early use of concrete: lime mortars 12,000BC.
Advantages of using concrete:
Clarity factor
This quiz introduces the
concept of Cover over the
reinforcing steel
CEE 4730/6730
Quiz #8, September 3, 2015
Correctness
Quiz Score
Name_
fc = 5,000 psi=5.00 ksi=5.00x103 psi
Ultimate
CEE461SolutiontoTest2Spring2011
Question1:FindAvandsforVu=150kips,b=12inches,h=36inches,d=32.5inches,fc=4000psiandfy=60000psi
Vs Vu Vc 150 2 4000 (12)(32.5) / 1000 150.67 kips
Av f y d
s
150.67 kips
CEE461
Final Exam
Spring 2009
Name:_
(last name)
(first name)
This test will be graded out of 50 points. Remember that the numerical grade (# points awarded) on a question
with a qualitative component
CEE461
Reinforced Concrete 1
Term Test 1
Thursday Feb. 19, 2009
Name: _
(last)
(first)
This test will be graded out of 40 points.
(Question 1: 10 points.)
For the cross-section described below, evalua
Materials
2
f c f 'c c c
'c 'c
E
E c 57000 f ' c ; n s
Ec
f cr
fc
1 500 c
2
bal
0.85 f ' c 1 87000
.
87000 f
fy
y
As ,max .75 bal bw d
c ,max
1 /4
2
; k1
; k2
3
3
'c
fy
f'
1 c ;
; R w f
Materials
2
f c f 'c c c
'c 'c
E
E c 57000 f ' c ; n s
Ec
f cr
fc
1 500 c
2
bal
0.85 f ' c 1 87000
.
87000 f
fy
y
As ,max .75 bal bw d
c , max
3
1 / 4
; k1 ; k2
'c
3
3
fy
f'
1 c ;
; R w f