SAMPLE LABORATORY SESSION FOR JAVA MODULE B
Calculations for Sample Cross-Section 2
1. User Input
1.1 Section Properties
The properties of Section 2 are shown in Figure 1 and are summarized below.
Figure 1: Properties of Section 2.
1

Concrete Properties
Currently the entire cross-section is assumed to be unconfined. The
compressive stress-strain relationship of the unconfined concrete is
determined using a method developed by Mander et al. [1]. The
following user specified properties are needed:

o
Concrete compressive strength, f c '  27.5 MPa
o

Concrete strain corresponding to peak stress ( f c ),  co  0.002
Steel properties
o Steel yield strength, f y  400 MPa
o Steel Young’s Modulus, Es  200000 MPa

Section Dimensions
Currently, only rectangular cross-sections are allowed by the module.
o Section height, h  60 cm
o Section width, b  36 cm

Reinforcement
o Stirrup diameter, d s  0.95 cm
o Number of layers of longitudinal bars, nl  2
o First layer : bottom layer

Number of longitudinal bars = 4
2

Diameter of longitudinal bars, d b  1.91 cm

Distance to compression (top) face, d  54 cm
o Second layer: top layer

Number of longitudinal bars = 2

Diameter of longitudinal bars, d b  1.91 cm

Distance to compression (top) face, d  6 cm
The user input for the first layer of bars is shown in Figure 2.
Figure 2: Input reinforcement for Layer 1.
Selected user-specified properties of the section are displayed in Window 1 as shown in
Figure 3.
3
Figure 3: Window 1 representation of Section 2.
1.2 Axial Forces
The user input for the axial forces is shown in Figure 4. The number of
axial forces acting on the cross-section is three, namely, 0, 50 and 100% of
the balanced failure load (only 50% of the balanced load will be
considered in the sample calculations below).
Figure 4: Axial forces input.
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1.3 Strain Condition for P-M Interaction
A strain value less than or equal to the concrete crushing strain may be
used. The module uses the concrete crushing strain as the default value.
Section 2 uses a user-specified strain value of 0.003 as shown in Figure 4.
Figure 4: Strain condition for P-M interaction.
2. Calculations and Equations
The following calculations are based on the method employed by Java Module B.
2.1 Concrete Stress-Strain Relationship
The equation for the unconfined concrete stress-strain relationship is [1]:
(1) f c 
(2) x 

f c xr
, where
r 1 xr
c
,  c 0  0.002
 co
The tangent modulus of elasticity, E c , is calculated using:
(3) Ec  4,741 fc MPa = 24,862.0 MPa
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Esec , the secant modulus of elasticity, is the slope of the line connecting the origin and
peak stress on the compressive stress-strain curve (see Fig. 6).
(4) Esec 
fc

 co
= 13, 750.0 MPa
Then,
(5) r 
Ec
= 2.24
Ec  Esec
and, the concrete stress-strain relationship is given as:
fc 
27.5(
c
0.002
2.24  1  (
)(2.24)
c
0.002
,
)
2.24
The above fc   c relationship is plotted in Figure 6. It is assumed that crushing of
concrete occurs at a strain of  cu  2 co  0.004
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Figure 6: Concrete stress-strain relationship for cross-section 2.
In Fig. 6,
Circle marker: assumed concrete linear-elastic limit at f c  f el 
Square marker: peak stress at f c  f c and  c   co .
Diamond marker: assumed ultimate strain at  c   cu  2 co .
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f c
1 f c
and  c   el 
.
2
2 Ec
2.2 Moment-Curvature Relationships
Window 2 generates the moment-curvature relationships of the user-defined section for
the user-specified axial forces. This is an iterative process, in which the basic equilibrium
requirement, P  Cc  Fs 2  Fs1  Ct , can be used to find the neutral axis for a particular
compressive strain selected.
Figure 3: Section strains, stresses, and stress resultants.
The calculation of the following four points on the moment curvature curve will be
shown in this example:

 cm  0.25 cu

 cm  0.5 cu

 cm  0.75 cu

 cm   cu (concrete crushing)
The axial load considered for these sample calculations is 50% of the balanced failure
load.
The balanced load, Pb , is computed as follows:
The neutral axis, c  cb  d
 cu
 cu   y
, where  y 
8
fy
Es
With  cu  0.004 and d  54 cm, this gives a value of cb  36 cm.
The concrete compressive resultant, C c , is determined by numerically integrating under
the concrete stress distribution curve.
cb
Cc   ( fcb)dx  2, 795.6 kN.
0
For balanced failure condition, by definition,  s1   y and  cm   cu . The compression
and tension steel forces, Fs 2 and Fs1 , respectively, are calculated using similarity to find
the strains in the layers.
 s 2   s1
c  d'
=0.0033, which implies that the compression steel is also at yield stress.
d c
Hence,
Fs1   s1 Es As1  f y As1  458.4 kN
Fs 2   s 2 Es As 2  f y As 2  229.2 kN, where As1 and As 2 are the total reinforcing steel
areas in each layer.
The module considers the concrete tensile strength in the tension region. ACI-318 (1)
recommends the modulus of rupture to be taken as f r  0.62 fc MPa for normal weight
concrete. Thus, for f c  27.5 MPa, f r  3.3 MPa.
The, concrete tension force Ct 
1 
f r Act  25.6 kN, where Act is the area of concrete in
2
tension calculated based on the linear strain diagram.
The balanced load is found as Pb  Cc  Fs 2  Fs1  Ct  2,540.8 kN.
Therefore, 50% of the balanced load used in the example is P  1271 kN.
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The module determines an “instant” centroid by assuming an initial condition where only
the user-selected axial load acts on the section without moment, such that it produces
uniform compression strain throughout the cross-section.
Let the uniform compression strain be equal to  ci . Then  s1   s 2   ci , and
f si  f s1  f s 2  Es ci  f y
From equilibrium,
fci Ac  f s1 As1  f s 2 As 2  P ,

27.5(
 ci
0.002
2.24  1  (
)(2.24)
 ci
0.002
)
( Ac )  ( f si )( As1  As 2 )  P  1271 kN
2.24
where Ac  bh  ( As1  As 2 )  2143 cm2
 ci  0.000185  fci  4.83 MPa and f si  37 MPa.
Then, the location of the instant centroid, x , from the top compression face is determined
as
f ci bh 2 / 2  As1 f si d  As 2 f si d '
x
 30.48 cm
f ci bh  As1 f si  As 2 f si

Point 1
The calculation process can be summarized as follows:
1.  cm  0.25 cu  0.001
2. Assume the neutral axis depth, a distance c  15 cm.
3. From the strain diagram geometry,  s1  0.0026 (at yield stress) and  s 2  0.0006
(below yield stress).
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4. The steel stress resultants are Fs1  458.4 kN (tension) and Fs 2  68.8 kN
(compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
Cc  644.3 kN. The concrete below the neutral axis that has not cracked contributes to
the tension force Ct 
1 
f r Act  11.1 kN.
2
6. Check to see if P  Cc  Fs 2  Fs1  Ct
But P  1271 kN  243.5 kN  Cc  Fs 2  Fs1  Ct
So, the neutral axis must be adjusted downward, for the particular concrete strain that was
selected in Step 1, until equilibrium is satisfied. This determines the correct value of c .
Trying neutral axis depth c  32.92 cm gives:
 s1  0.00064 (below yield stress) and  s 2  0.00082 (below yield stress).
Fs1  146.7 kN (tension) and Fs 2  93.7 kN (compression).
c
Cc   ( fcb)dx  1,363.8 kN.
0
CT 
1 
f r Act  26.0 kN.
2
P  1271 kN  1285 kN  Cc  Fs 2  Fs1  CT
O.K.
Curvature can then be found from:

 cm
c

0.001
1
 3.04  105
32.92
cm
(1)
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  19.15 cm and zct  4.39 cm, respectively.
Then, the moment can be calculated as
11
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  319.8 kN-m.

(2)
Point 2
The calculation process can be summarized as follows:
1.  cm  0.5 cu  0.002
2. Assume the neutral axis depth, a distance c  18 cm.
3. From the strain diagram geometry,  s1  0.004 (at yield stress) and  s 2  0.00133
(below yield stress).
4. The steel stress resultants are Fs1  458.440 kN (tension) and Fs 2  152.8 kN
(compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
Cc  1220.179 kN.
6. Check to see if P  Cc  Fs 2  Fs1 .
But P  1304  914.54  Cc  Fs 2  Fs1
So, the neutral axis must be adjusted upward or downward, for the particular concrete
strain that was selected in Step 1, until equilibrium is satisfied. This determines the
correct value of c .
Curvature can then be found from:

1
(1)
c1
The internal lever arm, z , from the centroid of the concrete stress distribution to the
tensile resultant is calculated after which
M  Cz  Tz

(2)
Point 3
12
The calculation process can be summarized as follows:
1.  cm  0.75 cu  0.003
2. Assume the neutral axis depth, a distance c  20 cm.
3. From the strain diagram geometry,  s1  0.0051 (at yield stress) and  s 2  0.0021 (at
yield stress).
4. The steel stress resultants are Fs1  458.440 kN (tension) and Fs 2  229.2 kN
(compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
Cc  1417.77 kN.
6. Check to see if P  Cc  Fs 2  Fs1 .
But P  1304  1188.53  Cc  Fs 2  Fs1
So, the neutral axis must be adjusted upward or downward, for the particular concrete
strain that was selected in Step 1, until equilibrium is satisfied. This determines the
correct value of c .
Curvature can then be found from:

1
(1)
c1
The internal lever arm, z , from the centroid of the concrete stress distribution to the
tensile resultant is calculated after which
M  Cz  Tz

(2)
Point 4
The calculation process can be summarized as follows:
1.  cm   cu  0.004
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2. Assume the neutral axis depth, a distance c  24 cm.
3. From the strain diagram geometry,  s1  0.005 (at yield stress) and  s 2  0.003 (at
yield stress).
4. The steel stress resultants are Fs1  458.440 kN (tension) and Fs 2  229.2 kN
(compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
Cc  1804.665 kN.
6. Check to see if P  Cc  Fs 2  Fs1 .
But P  1304  1575.43  Cc  Fs 2  Fs1
So, the neutral axis must be adjusted upward or downward, for the particular concrete
strain that was selected in Step 1, until equilibrium is satisfied. This determines the
correct value of c .
Curvature can then be found from:

1
(1)
c1
The internal lever arm, z , from the centroid of the concrete stress distribution to the
tensile resultant is calculated after which
M  Cz  Tz
(2)
The three axial forces specified to generate moment-curvature relationship for section 2
are represented as P1, P2 and P3 in window 2 as shown below.
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Figure 4: Section 2 moment-curvature relationship shown in window 2.
2.2 Axial-Force-Bending-Moment Interaction Diagram
Window 3 generates P-M interaction diagrams for the user-defined cross-sections. It
determines the axial load and moment pairs of the cross-section for user specified
concrete compression strain,  cm , values. This diagram is generated by selecting
successive choices of neutral axis distance, c , from a small value to a large one that gives
a pure axial loading condition. The initial neutral axis value corresponds to pure bending
condition (no axial force) of the cross section determined by the module.
The calculation of the following three points on the P-M interaction diagram will be
shown in this example.

c  cinitial

c  1.5cinitial

c  3cinitial
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Figure 5: Section strains, stresses, and stress resultants for P-M interaction.

Point 1
The initial neutral axis corresponding to pure bending is determined as follows:
 s1   cm
d c
c  d
and  s 2   cm
c
c
(1)
Since P  0 ,
from equilibrium:
Cc  Fs 2  Fs1  Ct
substituting the steel strain values with eq (1) above gives the second order equation
 cm

0
fcb
c
 cm
d   Es ( cm
c  d
d c
1
) As 2  Es ( cm
) As1  f r Act , from which the value of the
c
c
2
neutral axis can be obtained by checking f s1  Es s1  f y and f s 2  Es s 2  f y .
Therefore, cinitial  6.01 cm.
The module calculates the instant centroid for both P-M interaction and momentcurvature relationship in the same way. The axial load acting on the section in this case,
however, is the lesser of the user-specified concrete crushing load or P  0.85 f c Ac . This
16
load is the load assumed to be acting with no moment on the cross-section, i.e., pure
thrust load.
The module determines an “instant” centroid by assuming an initial condition where only
the user-selected axial load acts on the section without moment, such that it produces
uniform compression strain throughout the cross-section.
The user-specified concrete crushing strain value ( 0.003 ) gives higher value than
P  0.85 f c Ac , the axial load in this calculation becomes P  0.85 f c Ac  5009.3 kN.
Let the uniform compression strain be equal to  ci . Then  s1   s 2   ci , and
f si  f s1  f s 2  Es ci  f y
From equilibrium,
fci Ac  f s1 As1  f s 2 As 2  P ,
27.5(

 ci
0.002
2.24  1  (
)(2.24)
 ci
0.002
)
( Ac )  ( f si )( As1  As 2 )  P  5009.3 kN
2.24
where Ac  bh  ( As1  As 2 )  2143 cm2
 ci  0.00853  fci  4.83 MPa and f si  37 MPa.
Then, the location of the instant centroid, x , from the top compression face is determined
as
x
f ci bh 2 / 2  As1 f si d  As 2 f si d '
 30.48 cm
f ci bh  As1 f si  As 2 f si
M  Cc zc  Fs1 zs1  Fs 2 zs 2
Calculation of load and moment for balanced point:
17
Computing neutral axis, c for balanced point,
c  cb  d
u
u   y
, where  y 
fy
(1)
Es
With d  54 cm, and a user-defined concrete crushing strain  cu  0.003 , cb  32.4 cm
The concrete compressive resultant, C is given by
cb
Cc   ( fcb)dx  2, 499 kN
0
When concrete reaches its maximum strain of 0.003 at the compression face, similarity of
triangles can be used and the strain in each bar is calculated as follows:
Figure 3: Strain locations
 s1   cu
d c
, for tension steel
c
and  s 2   cu
(4)
cd'
, for the compression steel
c
so  s 2  0.0024 and  s1  0.002 that give the stresses when multiplied with Es .
Hence, the stress values become f s1  400 MPa (tension) and f s 2  480
MPa (compression).
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The balanced load is calculated as follows:
Pb  C  As 2 f s 2  As1 f s1 ,
(5)
where A is the area of steel. This gives the axial load value for balanced failure
Pb  2316 kN.
For unsymmetrical reinforcement such as section 2, a point known as the plastic centroid,
x is of importance. It coincides with the geometric center when both the section and
reinforcement are symmetrical.
x
0.85 fc ' bh2 / 2  As f y d  As ' f y d '
(6)
0.85 fc ' bh  As f y  As ' f y
According to equation 6, x  31 cm .
This plastic centroid becomes the axis about which the balanced moment is measured.
a
M b  C ( x  )  As 2 f s 2 ( x  d ')  As1 f s1 (d  x )
2
(7)
This gives M b  605 kNm .
Using the same method, the interaction diagram can be generated and most easily done
by selecting successive neutral axis values using computer programs like matlab.
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Figure 4: Section 2 interaction diagram shown in window 3
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