SAMPLE LABORATORY SESSION FOR JAVA MODULE B
Calculations for Sample Cross-Section 2
1. User Input
1.1 Section Properties
The properties of Sample Cross-Section 2 are shown in Figure 1 and are summarized below.
Figure 1: Properties of Sample Cross-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
The default value used by the module for  co is 0.002 .

Steel Properties
o Steel yield strength, f y  400 MPa
o Steel Young’s Modulus, Es  200000 MPa
The default value used by the module for Es is 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

Diameter of longitudinal bars, d b  1.91 cm
2

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: 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 Sample Cross-Section 2.
1.2 Axial Forces
The user input for the axial forces is shown in Figure 4. These axial forces
(up to five forces) are used to generate moment-curvature relationships for
Figure 4: Axial forces for Sample Cross-Section 2.
the section. The largest compressive axial force that can be specified for a
cross-section is equal to 0.85 f c Ac  Ast f s , where f c is the concrete strength,
4
Ac is the concrete area, Ast is the total steel area, and f s is the steel stress
corresponding to concrete crushing. The smallest compressive axial force
that can be specified is zero. The module is not designed to consider tensile
forces, i.e., negative axial loads. The module uses zero axial load as the
default value.
As shown in Figure 4, the number of axial forces specified for Sample
Cross-Section 2 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.
1.3 Strain Condition for P-M Interaction
The module requires a user-specified maximum compression strain value to
generate the P-M interaction diagrams. 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.
A user-specified strain value of 0.003 is specified to generate the P-M
interaction diagrams for Sample Cross-Section 2 as shown in Figure 5.
Figure 5: Strain condition for P-M interaction.
5
2. Calculations and Equations
The following sample 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]:

f c xr
fc 
, where
r 1 xr
x
(1)
c
,  c 0  0.002
 co
(2)
The tangent modulus of elasticity, E c , is calculated using:
Ec  4,741 fc MPa = 24,862.0 MPa
(3)
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).
Esec 
fc

 co
= 13, 750.0 MPa
(4)
Then,
r
Ec
= 2.24
Ec  Esec
(5)
and, the concrete stress-strain ( fc   c ) relationship is given as:
fc 
27.5(
c
0.002
2.24  1  (
)(2.24)
c
0.002
,
)
(6)
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 .
6
Figure 6: Concrete stress-strain relationship for Sample Cross-Section 2.
In Fig. 6,
Circle marker: assumed concrete linear-elastic limit at f c  f el 
f c
1 f c
and  c   el 
.
2
2 Ec
Square marker: peak stress at f c  f c and  c   co .
Diamond marker: assumed ultimate strain at  c   cu  2 co .
2.2 Moment-Curvature Relationships
Window 2 generates the moment-curvature relationships of the section for the userspecified axial forces. This is an iterative process, in which the basic equilibrium
requirement (e.g., P  Cc  Fs 2  Fs1  Ct for a section with two layers of reinforcement) and
7
a linear strain diagram are used to find the neutral axis for a particular maximum concrete
compressive strain,  cm , selected (see Figure 7).
Figure 7: Section strains, stresses, and stress resultants.
The calculation of the following four points on a 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)
Axial Load, P
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 
fy
Es
(7)
With  cu  0.004 and d  54 cm, this gives a value of cb  36.0 cm.
The concrete compressive resultant, C c , is determined by numerically integrating under the
concrete stress distribution curve.
8
cb
Cc   f cbdx 
0
 cu

fcb
0
c
 cm
d  c  2, 795.6 kN, where f c is from Eq. (6)
(8)
The top and bottom steel forces, Fs 2 and Fs1 , respectively, are calculated using similarity
to find the strains in the layers. Balanced failure condition, by definition, has strain values
 cm   cu  0.004 for concrete and  s1   y  0.002 for bottom steel. For the top steel,
 s 2   s1
c  d'
=0.0033
d c
(9)
which implies that the top steel is also at yield stress.
Hence,
Fs1   s1 Es As1  f y As1  458.4 kN (tension)
(10)
Fs 2   s 2 Es As 2  f y As 2  229.2 kN (compression)
(11)
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
(12)
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  6.96 kN
2
(13)
where, Act is the area of concrete in tension calculated based on the linear strain diagram.
Then, the balanced axial load is found from equilibrium as
Pb  Cc  Fs 2  Fs1  Ct  2,559.4 kN.
(14)
Therefore, 50% of the balanced load used in the example is P  1279.7 kN.
9
Instant Centroid
The module assumes that the axial load acts at an “instant” centroid location for the
calculation of the moment-curvature relationship. The location of the instant centroid is
determined by assuming an initial condition where only the user-selected axial load acts on
the cross-section without moment. This loading condition produces a uniform compression
strain distribution 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
(15)
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
)
(16)
( Ac )  ( f si )( As1  As 2 )  P  1279.7 kN
(17)
2.24
where Ac  bh  ( As1  As 2 )  2143.0 cm2
(18)
A trial-and-error solution on Eq. (17) is needed since it is not known in advance if the bars
are yielding; from which the strain  ci can be calculated as:
 ci  0.000227  fci  5.6 MPa and f si  45.4 MPa (bars not yielding).
Then, the location of the instant centroid, x , from the top compression face is determined
as
x
f ci Ac h / 2  As1 f si d  As 2 f si d '
 30.48 cm
f ci Ac  As1 f si  As 2 f si
(19)
10

Point 1
The calculation of the first sample point on the moment-curvature relationship of the
section 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 linear strain diagram geometry
 s1  (d  c)
 cm
 s 2  (c  d )
c
 0.0026 (at yield stress) and
(20)
 0.0006 (below yield stress)
(21)
 cm
c
4. The steel stress resultants are
Fs1  f y As1  458.4 kN (tension) and
(22)
Fs 2   s 2 Es As 2  68.8 kN (compression).
(23)
5. Determine C c by integrating numerically under the concrete stress distribution curve.
c
Cc   fcbdx 
0
 cm

0
f cb
c
 cm
d  c  644.3 kN.
(24)
where f c is given by Eq. (6).
The concrete that has not cracked below the neutral axis contributes to the tension force.
Ct 
1 
f r Act  11.1 kN with f r  3.3 MPa from Eq. (12)
2
6. Check to see if P  Cc  Fs 2  Fs1  Ct
(25)
(26)
But P  1271 kN  243.5 kN  Cc  Fs 2  Fs1  Ct
11
So, the neutral axis must be adjusted downward, for the particular maximum concrete strain
that was selected in Step 1, until equilibrium is satisfied. This iterative process 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   fcbdx 
0
Ct 
 cm

0
f cb
c
 cm
d  c  1,369.2 kN.
(27)
1 
f r Act  26.0 kN.
2
(28)
P  1279 kN  1285 kN  Cc  Fs 2  Fs1  Ct
O.K.
Section curvature can then be found from:

 cm
c

0.001
1
 3.04  105
32.92
cm
(29)
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 section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  318.5 kN-m.

(30)
Point 2
The calculation of sample point two on the moment-curvature relationship of the section
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 linear strain diagram geometry
12
 s1  (d  c)
 cm
 s 2  (c  d )
 0.004 (at yield stress) and
c
 cm
c
 0.0013 (below yield stress)
4. The steel stress resultants are
Fs1  f y As1  458.4 kN (tension) and
Fs 2   s 2 Es As 2  152.8 kN (compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
c
Cc   fcbdx 
0
 cm

f cb
0
c
 cm
d  c  1200.7 kN.
where f c is given by Eq. (6).
The concrete that has not cracked below the neutral axis contributes to the tension force.
Ct 
1 
f r Act  7 kN with f r  3.3 MPa from Eq. (12)
2
6. Check to see if P  Cc  Fs 2  Fs1  Ct
But P  1271 kN  894.4 kN  Cc  Fs 2  Fs1  Ct
So, the neutral axis must be adjusted downward, for the particular maximum concrete strain
that was selected in Step 1, until equilibrium is satisfied. This iterative process determines
the correct value of c .
Trying neutral axis depth c  23.76 cm gives:
 s1  0.0025 (at yield stress) and  s 2  0.0015 (below yield stress).
Fs1  458.4 kN (tension) and Fs 2  171.3 kN (compression).
c
Cc   fcbdx 
0
 cm

0
fcb
c
 cm
d  c  1,585.1 kN.
13
Ct 
1 
f r Act  9.4 kN.
2
P  1279 kN  1288 kN  Cc  Fs 2  Fs1  Ct
O.K.
Section curvature can then be found from:

 cm
c

0.002
1
 8.42  105
23.76
cm
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  21.6 cm and zct  5.7 cm, respectively. Then,
the section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  491.6 kN-m.

Point 3
The calculation of sample point three on the moment-curvature relationship of the section
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 linear strain diagram geometry
 s1  (d  c)
 cm
 s 2  (c  d )
c
 cm
c
 0.0051 (at yield stress) and
 0.0021 (at yield stress)
4. The steel stress resultants are
Fs1  f y As1  458.4 kN (tension) and
Fs 2  f y As 2  229.2 kN (compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
14
c
Cc   fcbdx 
0
 cm

f cb
0
c
 cm
d  c  1525.3 kN.
where f c is given by Eq. (6).
The concrete that has not cracked below the neutral axis contributes to the tension force.
Ct 
1 
f r Act  5.3 kN with f r  3.3 MPa from Eq. (12)
2
6. Check to see if P  Cc  Fs 2  Fs1  Ct
But P  1271 kN  1290.8 kN  Cc  Fs 2  Fs1  Ct
So, the neutral axis must be adjusted upward, for the particular maximum concrete strain
that was selected in Step 1, until equilibrium is satisfied. This iterative process determines
the correct value of c .
Trying neutral axis depth c  19.86 cm gives:
 s1  0.0052 (at yield stress) and  s 2  0.002 (at yield stress).
Fs1  458.4 kN (tension) and Fs 2  229.2 kN (compression).
c
Cc   fcbdx 
0
Ct 
 cm

0
f cb
c
 cm
d  c  1,514.6 kN.
1 
f r Act  5.2 kN.
2
P  1279 kN  1280 kN  Cc  Fs 2  Fs1  Ct
Section curvature can then be found from:

 cm
c

0.003
1
 1.51104
19.86
cm
15
O.K.
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  22.33 cm and zct  10.0 cm, respectively.
Then, the section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  501.6 kN-m.

Point 4
The calculation of sample point four on the moment-curvature relationship of the section
can be summarized as follows:
1.  cm  1.0 cu  0.004
2. Assume the neutral axis depth, a distance c  21 cm.
3. From the linear strain diagram geometry
 s1  (d  c)
 cm
 s 2  (c  d )
 0.0063 (at yield stress) and
c
 cm
c
 0.0029 (at yield stress)
4. The steel stress resultants are
Fs1  f y As1  458.4 kN (tension) and
Fs 2  f y As 2  229.2 kN (compression).
5. Determine C c by integrating numerically under the concrete stress distribution curve.
c
Cc   fcbdx 
0
 cm

0
f cb
c
 cm
d  c  1630.4 kN.
where f c is given by Eq. (6).
The concrete that has not cracked below the neutral axis contributes to the tension force.
Ct 
1 
f r Act  4.1 kN with f r  3.3 MPa from Eq. (12)
2
16
6. Check to see if P  Cc  Fs 2  Fs1  Ct
But P  1271 kN  1396.7 kN  Cc  Fs 2  Fs1  Ct
So, the neutral axis must be adjusted upward, for the particular maximum concrete strain
that was selected in Step 1, until equilibrium is satisfied. This iterative process determines
the correct value of c .
Trying neutral axis depth c  19.48 cm gives:
 s1  0.0071 (at yield stress) and  s 2  0.0028 (at yield stress).
Fs1  458.4 kN (tension) and Fs 2  229.2 kN (compression).
c
Cc   fcbdx 
0
Ct 
 cm

0
f cb
c
 cm
d  c  1,512.4 kN.
1 
f r Act  3.8 kN.
2
P  1279 kN  1279.4 kN  Cc  Fs 2  Fs1  Ct
O.K.
Section curvature can then be found from:

 cm
c

0.004
1
 2.05  104
19.48
cm
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  21.82 cm and zct  10.6 cm, respectively.
Then, the section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  493.5 kN-m.
17
Plots
The three axial forces specified to generate the moment-curvature relationships for Sample
Cross-Section 2 are represented as P1, P2 and P3 in Window 2 as shown below. The
M and  pairs calculated for the four points above are plotted on the moment-curvature
curve for P2  0.50 Pb .
Figure 8: Sample Cross-Section 2 moment-curvature relationships shown in Window 2.
2.2 Axial-Force-Bending-Moment Interaction Diagram
Window 3 generates P-M interaction diagrams for the user-defined cross-section by
determining the axial load and moment pairs of the section for a user-specified maximum
concrete compression strain,  cm . The P-M interaction diagram for each cross-section is
generated by selecting successive choices of the neutral axis distance, c , from an initial
small value to a large one that gives a pure axial loading condition. The initial neutral axis
value, co , corresponds to the pure bending condition (i.e., no axial force) of the crosssection.
18
The calculation of the following three points on the P-M interaction diagram of Sample
Cross-Section 2 will be shown in this example.

c  co

c  1.5co

c  3co
Figure 9: Section strains, stresses, and stress resultants for P-M interaction.
Instant Centroid
The module calculates an “instant” centroid for P-M interaction, similar to the instant
centroid used for the section moment-curvature relationship. The axial load is assumed to
be applied at the instant centroid, which is determined from a uniform compression strain
distribution over the section, equal to the user-specified maximum concrete strain,  cm , for
P-M interaction.
For Sample Cross-Section 2, the user-specified maximum concrete compression strain
 cm  0.003 (see Figure 5).
Thus,  ci  0.003  f ci  24.83 MPa [from Eq. (6)] and
 si   ci  0.003  f si  400 MPa (at yield).
19
With Ac  bh  ( As1  As 2 )  2143 cm2 , 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
 30.91 cm
f ci bh  As1 f si  As 2 f si

Point 1
The calculation of the first sample point on the P-M interaction diagram of the section is
described below. The initial neutral axis location, co , corresponding to pure bending is
determined as follows:
 s1   cm
d  co
c  d
and  s 2   cm o
co
co
(31)
where the user specified  cm  0.003
Then, f s1   s1 Es  f y and f s 2   s 2 Es  f y
(32)
Since P  0 , from equilibrium:
Cc  f s 2 As 2  f s1 As1  Ct
(33)
Substituting for the concrete and steel stress resultants gives the following equation
 cm

f cb
0
co
 cm
d  c  f s 2 As 2  f s1 As1 
1 
f r Act
2
(34)
An iterative solution of Eq. (34) is needed (since the yielding bars are not known in
advance), from which the value of the neutral axis can be obtained as co  6.01 cm.
The internal forces can then be calculated as:
Fs1  458.4 kN (tension, at yield) and Fs 2  0.57 kN (compression, below yield).
co
Cc   fcbdx 
0
 cm

0
f cb
co
 cm
d  c  458.4 kN
20
Ct 
1 
f r Act  1.6 kN
2
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  28.46 cm and zct  24.72 cm, respectively.
Then, the section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  236.1 kN-m.

Point 2
The calculation of the second sample point, c  1.5co , on the P-M interaction diagram of
the section is described below.
 s1   cm
d c
c  d
and  s 2   cm
c
c
The neutral axis is c  9.02 cm.
The internal forces can then be calculated as:
Fs1  f y As1  458.4 kN (tension, at yield) and Fs 2   s 2 Es As 2  115.1 kN (compression,
below yield).
c
Cc   fcbdx 
0
Ct 
 cm

0
f cb
c
 cm
d  c  687.5 kN
1 
f r Act  2.4 kN
2
Then, the section axial force can be calculated as
P  Cc  Fs 2  Fs1  Ct  341.8 kN.
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  27.21 cm and zct  21.6 cm, respectively.
Then, the section moment can be calculated as
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M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  321.1 kN-m.

Point 3
The calculation of the third sample point, c  3.0co , on the P-M interaction diagram of the
section is described below.
 s1   cm
d c
c  d
and  s 2   cm
c
c
The neutral axis is c  18.03 cm.
The internal forces can then be calculated as:
Fs1  f y As1  458.4 kN (tension, at yield) and Fs 2  f y As 2  229.2 kN (compression, at
yield).
c
Cc   fcbdx 
0
Ct 
 cm

0
fcb
c
 cm
d  c  1,375.1 kN
1 
f r Act  4.7 kN
2
P  Cc  Fs 2  Fs1  Ct  1141.2 kN
The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are zc  23.52 cm and zct  12.35 cm, respectively.
Then, the section moment can be calculated as
M  Cc zc  Fs1 zs1  Fs 2 zs 2  Ct zct  485.8 kN-m
Plots
The M and P pairs calculated for the three points above are plotted on the interaction
diagram shown in Figure 10 below.
22
Figure 10: Sample Cross-Section 2 interaction diagram shown in Window 3
23