DESIGN FOR FLEXURE SINGLY REINFORCED RECTANGULAR SECTIONS
The design of singly reinforced rectangular cross-sections for flexure may follow one of
several methods. All of the methods described below are based on satisfying the same set
of fundamental equations of equilibrium. Note that it is assumed that the cross-section is
underreinforced, such that the tension reinforcement has yielded before failure.
b
.003
c
h
.85f 'c
a=1c
C = .85 f 'c a b
NA
d
As
sy
T = As fy
Force Equilibrium ("C=T"):
a 
Asf y
(1)
.85 f 'c b
Moment Equilibrium (assuming flexural strength provided = required strength):
Mu
 M n
a

 A s f y  d  
2

(2)
Design for Cross-Sections of Known Dimensions
In this case, b and d (or h) are assumed to be known prior to design. These dimensions
may have been established by architectural considerations, or for other reasons (such as
repeating beam dimensions across a span for formwork optimization). This is also
typically the case for one-way slab systems, in which the width is assumed to be a 12inch wide strip and the depth is established by shear or deflection criteria.
All three methods involve simultaneously satisfying both of the above equations. For the
case of fixed dimensions, only the parameters a and As are unknown in these equations.
Method 1: Trial and error (iterative) approach
Method 2: Simultaneous algebraic solution
Method 3: Solution by design aids
1
Design Problem
The dimensions of the beam shown below are fixed as 10" x 20" for architectural reasons.
The beam supports an indoor walkway on a 20-foot simply-supported span. The
superimposed dead load is 1000 lb/ft and the superimposed live load is 1000 lb/ft. Select
flexural reinforcement for the beam if the material properties to be used are f 'c = 4000 psi
and fy = 60,000 psi. (Ignore the presence of any top bars used to facilitate stirrup
placement.)
LL = 1.00 k/ft
DL = 1.00 k/ft
h=20"
20 ft.
b=10"
Beam self-weight (assume concrete weighs 145 pcf):
2
 1 ft  
k 
  .145
 10 in .20 in .
  0.20 k / ft
ft 3 
 12 in  
DLsw
Total factored load:
wu
 1.2w D  1.6w L
 1.21.00  0.20 k / ft   1.61.00 k / ft   3.04 k / ft
Total factored moment (design for maximum moment along span):
Mu

w u2
8

3.04 k / ft 20 ft 2
8
 152 ft.  kips
 1824 in .  kips
Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" diameter bars

d = 17.5"
2
Method 1: Trial and Error (Iterative) Approach
For this method, one of the variables a or As must be assumed, and then Equations 1 and
2 are solved iteratively until convergence.
Typically, a is assumed and As is calculated by Equation 1. Then a is calculated by
Equation 2 using the value for As calculated from Equation 1. The computed value of a
is checked against the original assumption and the process is repeated until convergence.
A typical starting assumption is that a = d/4 for beams a = d/6 for slabs. In most cases,
these will yield slightly conservative values for a, and thus slightly conservative values
for As.
An alternative starting assumption is to assume the moment arm jd, where jd = (d - a/2).
In this case, typical starting assumptions are jd = .875d for beams and jd = .925d for
slabs. Using this process, As is first calculated from Equation 2, and then the assumption
is checked using Equation 1.
Note: When using this method, it is not necessarily required that the designer iterate
until convergence is achieved. If the initial assumption is slightly conservative, then the
result will be a slightly conservative design. The designer may check the capacity of the
section based on the assumption, and if adequate and not overly conservative, complete
the design at this stage. This philosophy is used in the solution below.
Solution by Method 1
Assume jd:
jd
a

  d    .875d  .87517.5 in .  15.31 in .
2

Calculate As from Equation 2:
Mu
 M n
As

a

 A s f y  d  
2

Mu
a

f y  d  
2


1824 in .  kips
 2.21 in .2
.90 60 ksi 15.31 in .
At this point, iterate further or assume result is acceptable and analyze section:
*We will not iterate, but if we had, we would get a = 3.83 in. and As = 2.17 in.2
Try 3 #8 bars (As = 2.37 in.2)
3
Analyze section:
Actual d:
d = 20" - 1.5" - 0.5" - 1/2(1.0") = 17.5"
Check reinforcement limits (b from Text Table A-5):
a 
 a

 dt
Asf y
.85 f ' c b

2.37 in . 60 ksi 
2
.854 ksi 10 in .
 4.18 in .
 a
 
 0.239  
 17.5 in .
 dt
As
2.37 in .


10 in .17.5 in .
bd

 min

3 f c'
fy

 4.18 in .


 0.375 1  0.375 (.85)  0.319 (OK)
 TCL
 .0135
3 4000 psi 
 .0032 
60000 psi
200
 .0033 (OK)
fy
Check section capacity (Equations 1 and 2):
M n


4.18 in . 
a


 A s f y  d    .90  2.37 in .2 60 ksi 17.5 in . 

2
2 


 1972 in .  kips  1818 in .  kips (OK)
Check spacing and cover:

Minimum spacing = 1" (ACI 7.6.1)

Minimum cover = 1.5" (assume concrete is not exposed to weather)

Assume minimum clear spacing and check cover.
Cover

= [b - 2(stirrup dia.) - 3(bar dia.) - 2(spacing)] / 2 sides
= [10" - 2(0.5") - 3(1.0") - 2(1.0")] / 2
= 2" clear cover each side > 1.5 " minimum (OK)
Note: several design aids are available for this step, such as Text
Table A-6.
4
Method 2: Simultaneous Algebraic Solution
For this method, Equations 1 and 2 are solved simultaneously to yield a closed form
solution:
Substitute Equation 1 into Equation 2:
Mu

A s f y 
 A s f y  d 

1.7 f c' b 

(3)
Rearrange Equation 3 into a quadratic form:
Mu

As 2f y 2
 Asf yd 
 fy2
0  
1.7 f c'
1.7 f c' b

As 2
b 

f y dA s
M 
  u
  
Solve the quadratic equation for As (keeping only the root that makes physical
sense):
f yd 
As
f y 2d 2

 f y 2  M
 u
 4
 1.7 f ' b  
c 




fy2
.85 f c' b
As


f yd  1 



 M
2
 u
1  
 .85 f ' bd 2  
c







fy2
.85 f c' b
As


 .85 f '  
c 
 bd 
1 
 f y 




  Mu   
 2
 
  bd 2   
1  

'
 .85 f c  



 
(4)
5
or, in alternative form:


 .85 f '  
c 
  
1 
 f y 




  Mu  
 2
 
  bd 2   
1  

'
 .85 f c  



 
(5)
Either Equation 4 or Equation 5 may be used to solve directly for the required amount of
reinforcement.
Solution by Method 2
Solve for required As using Equation 4:
Mu
bd
As

2

1824 in .  kips
.90 10 in .17.5 in .2
 .662 ksi  662 psi

10 in .17.5 in . .854000 psi   1
 60000 psi   


 2662 psi   
   2.17 in .2
1  
 .85  4000 psi   
Try 3 #8 bars (As = 2.37 in.2)
Check reinforcement limits, analyze section, check cover & spacing as before…
Method 3: Solution by Design Aids
This method is identical to the method outlined above (Method 2), except design aids are
used for the solution of the closed form equation.
Repeating Equation 5 from above:


 .85 f '  
c 
  
1 
 f y 




  Mu  
 2
 
  bd 2   
1  

'
 .85 f c  



 
(5)
6
Let us define the resistance coefficient Rn:
Rn

Mu
bd 2
Several design aids have been published which tabulate Rn as a function of  for various
combinations of f 'c and fy. In using these design aids, Rn is typically stated in units of psi.
An example of such a design aid is shown below:
From: McCormac, Design of Reinforced Concrete
Note: MacGregor Text has a similar set of design aid tables (Table A-3), except for one
important difference. MacGregor defines the resistance coefficient as kn, which is
identical to Rn. However, values are tabulated for kn, not kn. Therefore, one must
essentially enter the tables after calculating Mu/bd2 (in units of psi) rather than Mu/bd2.
The MacGregor design aid is shown on the next page:
7
From: MacGregor, Reinforced Concrete - Mechanics and Design
8
To simplify the process, a similar design aid has been published by several authors. This
alternate design aid is a nondimensionalized design aid based on a modified form of
Equation 5. Only one table is needed, regardless of the choice of f 'c and fy.
Define the reinforcement index :
 
fy
f c'
Now rewrite Equation 5 in terms of :



  .85  1 





.85
1 

  M
 2
u
  bd 2 f '
c
1   

.85



  M
 2
u
  bd 2 f '
c
1   

.85



 
 
1  2
  

.
85


 .85 
 


 .85 
2
1.7
2
 
 2

 .85 
 
 M
u

 bd 2 f '
c





2








  M
 2
u
  bd 2 f '
c
 1   

.85



 M
u
2
 bd 2 f '
c

.85

 M
u
 
 bd 2 f '
c

 
 
 
 



 
















   1  0.59 
(6)
9
A single table provides values for the term on the left side of the equation as a function of
the reinforcement index w. This table is found in several publications, including PCA's
Notes of 318-99, and is shown below:
From: PCA, Notes on ACI 318-99
10
Solution by Method 3
Solve for required  using design aids for Equation 5:
Rn

Mu
bd 2


Mu
bd 2

1824 in .  kips
.90 10 in .17.5 in .2
 .662 ksi  662 psi
or
k n
1824 in .  kips
10 in .17.5 in .2
 .596 ksi  596 psi
Entering table in MacGregor text (Table A-3) with kn = 596 psi, f 'c = 4 ksi, fy = 60 ksi,
and interpolating linearly:
 required
 .0124
  requiredbd 
A s, required
.0124 10 in .17.5 in .
 2.17 in .2
Try 3 #8 bars (As = 2.37 in.2)
Check reinforcement limits, analyze section, check cover & spacing as before…
Alternatively, using the design aid based on Equation 6:
Mu
bd 2 f c'

1824 in .  kips
.90 10 in .17.5 in .2 4 ksi 
 .1654
Entering PCA Notes table with value of 0.1654, and interpolating linearly:
 required  .1858
 required   required
A s, required
f c'
fy

  required bd 
.1858  4 ksi   .0124
60 ksi 
.0124 10 in .17.5 in . 
2.17 in .2
Try 3 #8 bars (As = 2.37 in.2)
Check reinforcement limits, analyze section, check cover & spacing as before…
11
Design for Cross-Sections of Unknown Dimensions
In this case, b and d (or h) are not known prior to design. Therefore, we have three
parameters that the designer must choose - b, h, and As (assuming f 'c and fy are
established). These three parameters must only satisfy two equilibrium equations, so
there are mathematically an infinite number of possible design combinations. Typically,
an assumption is made regarding one of the parameters, usually As (. Examples of
common design assumptions are:



 = 0.5b
 = 0.375b (this corresponds to 0.50max for previous editions of ACI 318)
 = .01 = 1%
The design at this stage is also complicated by the fact that the self-weight of the beam is
unknown.
Once a design assumption is made, the process of beam design generally relies on one of
the design aids presented in the previous section. It is again emphasized that the design
aids are identical to the equations derived in the previous section (Equations 3, 4, 5, & 6),
such that solution by these equations is essentially identical to solution using design aids.
Design Problem (same problem as before with b and d unknown)
The beam below supports an indoor walkway on a 20-foot simply-supported span. The
superimposed dead load is 1000 lb/ft and the superimposed live load is 1000 lb/ft. Select
flexural reinforcement for the beam if the material properties to be used are f 'c = 4000 psi
and fy = 60,000 psi. (Ignore the presence of any top bars used to facilitate stirrup
placement.)
LL = 1.00 k/ft
DL = 1.00 k/ft
h
20 ft.
b
12
Beam self-weight (assume concrete weighs 145 pcf; assume 12"x24" cross-section):
2
 1 ft  
k 
  .145
  0.29 k / ft
ft 3 
 12 in  
12 in .24 in .

DLsw
Total factored load:
 1.2w D  1.6w L
wu
 1.21.00  0.29 k / ft   1.61.00 k / ft   3.15 k / ft
Total factored moment (design for maximum moment along span):

Mu
w u2
8

3.15 k / ft 20 ft 2
8
 157 .5 ft.  kips
 1890 in .  kips
Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" diameter bars

d = h - 2.5"
Design assumption:
  0.375  b
 
 fy

f c'
 0.375.0285   .0107
.0107 60 ksi 
4 ksi 
 .1605
Enter design aid ( design aid used here):
Mu
bd 2 f c'
 .1453
Solve for required bd2:
bd 
2
required

Mu

 .1453 
f c'
1890 in .  kips
 3613 in .3
.90 4 ksi .1453 
Possible combinations of b and d:



b = 10", d = 19"
b = 12", d = 17.5"
b = 14", d = 16"



bd2 = 3610 in.3
bd2 = 3675 in.3
bd2 = 3584 in.3
Note: This calculation gets us "in the ballpark". Selection of a b-d combination that
gives us less than the calculated requirement of 3625 in.2 will simply mean that we need a
slightly higher reinforcement ratio. Any of these three combinations will therefore work.
13
Try b = 12", h = 20" (d = 17.5")
Now, proceed as for a design with b and d known.
Beam self-weight (assume concrete weighs 145 pcf):
 1 ft 

 12 in 
12 in .20 in .

DLsw
2
k 

 .145 3   0.242 k / ft
ft 

Total factored load:
wu
 1.2w D  1.6w L
 1.21.00  0.242 k / ft   1.61.00 k / ft   3.09 k / ft
Total factored moment (design for maximum moment along span):
Mu

w u2
8
3.09 k / ft 20 ft 2

8
 154 .5 ft.  kips
Assume 1.5 in. clear cover, 0.5 in. stirrup, and 1.0" bars

 1854 in .  kips
d = 17.5"
Using the design aid based on Equation 6:
Mu
bd 2 f c'

1854 in .  kips
.90 12 in .17.5 in .2 4 ksi 
 .1401
Entering PCA Notes table with value of 0.1401, and interpolating linearly:
 required
 .1541
 required
  required
A s, required
f c'
fy

  required bd 
.1541 4 ksi   .0103
60 ksi 
.0103 12 in .17.5 in . 
2.16 in .2
Try 3 #8 bars (As = 2.37 in.2)
Check reinforcement limits, analyze section, check cover & spacing as before…
(This section works)
14
Design by Simplified Methods
Simplified design methods involve the same basic approach as the methods outlined
previously, but with further simplifications (i.e. assumptions) that reduce the effort
involved in the design process. The method outlined in this section is adapted from the
following references:
Fanella, D. and Ghosh, S.K. (editors), Simplified Design - Reinforced Concrete
Buildings of Moderate Size and Height, Portland Cement Association, Second
Edition, 1993.
Fanella, D., "Time-saving design aids for reinforced concrete," Structural
Engineer, August 2001, pp. 38-41.
Consider the simplifying assumption that
  .375  b
(7A)
A set of design aids (Tables 1A through 5A) have been developed based on this
simplifying assumption.
Note: A second set of design aids (Tables 1B through 5B) have been developed based on
a similar simplifying assumption:
1
b
2
 
(7B)
Using the assumption of a given  as a multiple of b, the values of  and  are tabulated
in Table 1A and 2A (or 1B and 2B) for various combinations of f 'c and fy.
A constant C1 can then be calculated where:
bd 
2
required
 C1 M u
(8)
where b and d are in units of inches , and M u is in units of ft  kips
Equation 8 can be used to help size the cross-section when dimensions have not been
established prior to design.
15
The constant C1 can be determined by manipulation of Equation 6:
 M
u

 bd 2 f '
c

bd 2




Mu

C1 
   1  0.59 
f c' 
1
 0.59 
1
f c' 
1 
 C1 M u
(9)
0.59 
Values of C1 are tabulated in Tables 3A and 3B.
The value of j, where jd is the length of the moment arm between the resultant internal
tensile and compressive forces, can also be computed directly as a function of f 'c and fy:
jd
j
 d 
a
2d
 1 
j
 1 
j
 1 
a
2
 Asf y 


 .85 f ' b 
c


2d

1.7
(10)
Values of j are tabulated in Tables 4A and 4B.
Finally, a required quantity of reinforcement can be computed based on the following
expression:
As,
required

Mu
C2 d
(11)
where d is in units of inches , A s in units of in . , and M u in units of ft  kips
2
16
The constant C2 can be determined by manipulation of Equation 2:
Mu
a

 A s f y  d    A s f y jd
2

As

C2
 f y j
Mu
f y jd

Mu
C2d
(12)
Values of C2 are tabulated in Tables 5A and 5B.
Design Problem (same problem as before with b and d unknown)
Consider the design example solved previously, and use the design assumption:
  0.375  b
Note that as before, we assume a self weight for the beam such that Mu = 157.5 ft.-kips:
From Table 1A 
 = .0107
From Table 2A 
 = .1604
From Table 3A 
C1 = 23.0
(bd2)required = 23.0(157.5) = 3623 in.2
Try b = 12", d = 17.5" (h = 20")
Correcting our self-weight assumption to the values for the selected beams size, we get
Mu = 154.5 ft.-kips:
From Table 4A 
j = 0.906
From Table 5A 
C2 = 4.08
(As)required = (154.5)/[(4.08)(17.5)] = 2.16 in.2
Try 3 #8 bars (As = 2.37 in.2)
Check reinforcement limits, analyze section, check cover &
spacing as before… (This section works)
Technically, the designer only needs to look up values in Table 3A and Table 5A. The
values in tables 1A, 2A, and 4A are only provided for reference.
17
TABLE 1a
0.375b
bd 
 C1 M u
2
required
where
As,
required
b and d are in units of inches , and M u is in units of ft  kips

Mu
C2 d
where d is in units of inches , A s in units of in .2 , and M u in units of ft  kips
TABLE 2a
fy (psi)
(corresponding to
0.375b)
40000
50000
60000
75000
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
0.1856
0.1856
0.1856
0.1801
0.1747
0.1638
0.1528
0.1419
0.1419
0.1419
0.1419
0.1419
0.1721
0.1721
0.1721
0.1670
0.1619
0.1518
0.1417
0.1316
0.1316
0.1316
0.1316
0.1316
0.1604
0.1604
0.1604
0.1556
0.1509
0.1415
0.1321
0.1226
0.1226
0.1226
0.1226
0.1226
0.1455
0.1455
0.1455
0.1412
0.1369
0.1284
0.1198
0.1113
0.1113
0.1113
0.1113
0.1113
TABLE 4a
60000
75000
0.0139
0.0162
0.0186
0.0203
0.0218
0.0246
0.0267
0.0284
0.0319
0.0355
0.0390
0.0426
0.0103
0.0120
0.0138
0.0150
0.0162
0.0182
0.0198
0.0211
0.0237
0.0263
0.0289
0.0316
0.0080
0.0094
0.0107
0.0117
0.0126
0.0141
0.0154
0.0163
0.0184
0.0204
0.0225
0.0245
0.0058
0.0068
0.0078
0.0085
0.0091
0.0103
0.0112
0.0119
0.0134
0.0148
0.0163
0.0178
C1
40000
50000
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
60000
75000
26.9
23.0
20.2
18.4
17.0
15.0
13.7
12.8
11.4
10.3
9.3
8.5
28.7
24.6
21.6
19.7
18.2
16.1
14.7
13.7
12.2
11.0
10.0
9.2
30.6
26.2
23.0
21.0
19.4
17.1
15.6
14.7
13.0
11.7
10.7
9.8
33.4
28.6
25.1
22.9
21.2
18.7
17.1
16.0
14.3
12.8
11.7
10.7
fy (psi)
TABLE 5a
fy (psi)
j
f 'c
(psi)
50000
TABLE 3a

f 'c
(psi)
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
fy (psi)
40000
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
40000
50000
60000
75000
0.891
0.891
0.891
0.894
0.897
0.904
0.910
0.917
0.917
0.917
0.917
0.917
0.899
0.899
0.899
0.902
0.905
0.911
0.917
0.923
0.923
0.923
0.923
0.923
0.906
0.906
0.906
0.908
0.911
0.917
0.922
0.928
0.928
0.928
0.928
0.928
0.914
0.914
0.914
0.917
0.919
0.924
0.930
0.935
0.935
0.935
0.935
0.935
C2
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
fy (psi)
40000
50000
60000
75000
2.67
2.67
2.67
2.68
2.69
2.71
2.73
2.75
2.75
2.75
2.75
2.75
3.37
3.37
3.37
3.38
3.39
3.42
3.44
3.46
3.46
3.46
3.46
3.46
4.08
4.08
4.08
4.09
4.10
4.13
4.15
4.18
4.18
4.18
4.18
4.18
5.14
5.14
5.14
5.16
5.17
5.20
5.23
5.26
5.26
5.26
5.26
5.26
18
TABLE 1b
0.5b
bd 
 C1 M u
2
required
where
As,
required
b and d are in units of inches , and M u is in units of ft  kips

Mu
C2 d
where d is in units of inches , A s in units of in .2 , and M u in units of ft  kips
TABLE 2b
fy (psi)
(corresponding to
0.5b)
40000
50000
60000
75000
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
0.2475
0.2475
0.2475
0.2402
0.2329
0.2184
0.2038
0.1892
0.1892
0.1892
0.1892
0.1892
0.2294
0.2294
0.2294
0.2227
0.2159
0.2024
0.1889
0.1754
0.1754
0.1754
0.1754
0.1754
0.2138
0.2138
0.2138
0.2075
0.2012
0.1886
0.1761
0.1635
0.1635
0.1635
0.1635
0.1635
0.1940
0.1940
0.1940
0.1883
0.1826
0.1712
0.1598
0.1484
0.1484
0.1484
0.1484
0.1484
TABLE 4b
60000
75000
0.0186
0.0217
0.0247
0.0270
0.0291
0.0328
0.0357
0.0378
0.0426
0.0473
0.0520
0.0568
0.0138
0.0161
0.0184
0.0200
0.0216
0.0243
0.0264
0.0281
0.0316
0.0351
0.0386
0.0421
0.0107
0.0125
0.0143
0.0156
0.0168
0.0189
0.0205
0.0218
0.0245
0.0272
0.0300
0.0327
0.0078
0.0091
0.0103
0.0113
0.0122
0.0137
0.0149
0.0158
0.0178
0.0198
0.0218
0.0237
C1
40000
50000
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
60000
75000
21.0
18.0
15.8
14.4
13.3
11.7
10.6
9.9
8.8
7.9
7.2
6.6
22.4
19.2
16.8
15.3
14.2
12.5
11.3
10.6
9.4
8.5
7.7
7.1
23.8
20.4
17.8
16.3
15.0
13.3
12.1
11.3
10.0
9.0
8.2
7.5
25.9
22.2
19.4
17.7
16.4
14.4
13.2
12.3
10.9
9.8
9.0
8.2
40000
50000
60000
75000
2.56
2.56
2.56
2.58
2.59
2.61
2.64
2.67
2.67
2.67
2.67
2.67
3.24
3.24
3.24
3.26
3.27
3.30
3.33
3.36
3.36
3.36
3.36
3.36
3.93
3.93
3.93
3.95
3.97
4.00
4.03
4.07
4.07
4.07
4.07
4.07
4.98
4.98
4.98
5.00
5.02
5.06
5.10
5.13
5.13
5.13
5.13
5.13
fy (psi)
TABLE 5b
fy (psi)
j
f 'c
(psi)
50000
TABLE 3b

f 'c
(psi)
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
fy (psi)
40000
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
40000
50000
60000
75000
0.854
0.854
0.854
0.859
0.863
0.872
0.880
0.889
0.889
0.889
0.889
0.889
0.865
0.865
0.865
0.869
0.873
0.881
0.889
0.897
0.897
0.897
0.897
0.897
0.874
0.874
0.874
0.878
0.882
0.889
0.896
0.904
0.904
0.904
0.904
0.904
0.886
0.886
0.886
0.889
0.893
0.899
0.906
0.913
0.913
0.913
0.913
0.913
C2
f 'c
(psi)
3000
3500
4000
4500
5000
6000
7000
8000
9000
10000
11000
12000
fy (psi)
19