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BRAE433 - Chapter 2b
Course: BRAE 433, Fall 2011School: Cal Poly
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Word Count: 432
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2 Chapter RC Design Evaluation and Design RC Rectangular Beam Problems Part 2 ITRC Cal Poly Ultimate Moment, Mn Fig 1-1 As = Area of steel Tables A-1 and A-2 Also called the nominal moment strength Point just prior to failure of the beam BRAE/ITRC Cal Poly 1 Ultimate Moment Theoretical internal resultant forces ,NC and NT internal resultant forces Parallel, equal and opposite Separated by distance Z...
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2 Chapter RC Design
Evaluation and Design RC
Rectangular Beam Problems
Part 2
ITRC
Cal Poly
Ultimate Moment, Mn
Fig 1-1
As = Area of steel
Tables A-1 and A-2
Also called the nominal moment strength
Point just prior to failure of the beam
BRAE/ITRC
Cal Poly
1
Ultimate Moment
Theoretical internal resultant forces ,NC and NT
internal resultant forces
Parallel, equal and opposite
Separated by distance Z
internal resisting couple = ultimate moment (or nominal
moment strength)
BRAE/ITRC
Cal Poly
Ultimate Moment
To solve,
need to calculate the resultant force NC in the concrete
determine its location in the beam
BRAE/ITRC
Cal Poly
2
Equivalent Stress Distribution
BRAE/ITRC
Cal Poly
Beta vs. fc'
0.900
1 = 0.85 for fc 4000 psi
0.850
1 = 0.85 0.05 (fc-4000)/1000
(fc
for fc > 4001 psi to 0.65
Beta
0.800
0.750
0.700
0.650
0.600
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
fc' (psi)
BRAE/ITRC
Cal Poly
3
Isometric View of the Internal Couple
Ready to Solve for Mn
BRAE/ITRC
Cal Poly
BRAE/ITRC
Cal Poly
4
Ultimate Moment (nominal
moment strength) - Mn
Example 2-1ss
Given: fc = 5,000 psi, A615 grade 75 steel
12
24
4 - #6
BRAE/ITRC
Cal Poly
Steps to Solve
Ultimate Moment, Mn
1.
NC = NT Nc = 0.85 fc a b,
NT = As fy
2.
0.85 fc a b = As fy
fc
3.
solve for a = As x fy/ (0.85 x fc x b)
4.
Z = d a/2
5.
solve Mn for = NC (d a/2)
Strain Check (did steel yield?)
Check (did steel yield?)
1.
solve for c = a / 1
2.
solve actual steel strain t = (d c) x 0.003 / c
3.
ck against yield strain y = fy/Es = fy / 29,000,000
BRAE/ITRC
Cal Poly
5
Did the Steel Yield?
Perform a strain check...
Is actual strain in steel less than
yield strain?
If concrete were to fail (exceed 0.003
strain). Not good...
BRAE/ITRC
Cal Poly
BRAE/ITRC
Cal Poly
6
Values in Table A-1
0.002
10
BRAE/ITRC
Cal Poly
Limiting the steel strain and
determining the required steel
Example 2-2ss
Given: fc = 5,000 psi, A615 grade 75 steel
12
0.003
24
4 - #6
0.005
BRAE/ITRC
Cal Poly
7
Solving for Required As
Example 2-2ss
1. steel strain t = 0.005
st
2. solve for c = 0.003 (d) / 0.008
3. solve for NC = 0.85 fc (c 1) b
4. NC = NT = As fy
5. solve for As required = NC / fy
for
fy
6. Check As maximum against As supplied
7. Comment on adequacy
BRAE/ITRC
Cal Poly
Reinforcement Ratio
Design As must be lowered
to ensure failure in steel
t limit at 0.005
Balanced
As
=
bd
BRAE/ITRC
Cal Poly
8
Minimum Steel
Balanced
As , min =
3 f 'c
200
bwd
bwd
fy
fy
BRAE/ITRC
Cal Poly
Minimum Steel
Need to make sure there is enough steel
Rectangular beams: bw = b
Table A5 Values for As,min and
recommended
As , min =
3 f 'c
200
bwd
bwd
fy
fy
BRAE/ITRC
Cal Poly
9
End
ITRC
Cal Poly
10
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