SAMPLE LABORATORY SESSION FOR JAVA MODULE B

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1

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.


2




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 propert
ies are needed:

o


Concrete compressive strength,
'27.5
c
f


MPa

o

Concrete strain corresponding to peak stress (

c
f
),
0.002
co



The default value used by the module for
co


is
0.002
.



Steel P
roperties

o

Steel yield strength,
400
y
f


MPa

o

Steel Young’s Modulus,
200000
s
E


MP
a

The default value used by the module for
s
E

is
200000

MPa.



Section Dimensions


Currently, only rectangular cross
-
sections are allowed by the module.

o

Section height,
60

h

cm

o

Section width,
36

b

cm



Reinforcement

o

Stirrup diameter,
95
.
0

s
d

cm

o

Number of layers of lo
ngitudinal bars,
2

l
n

o

First layer

: bottom layer



Number of longitudinal bars = 4



Diameter of longitudinal bars,
91
.
1

b
d

cm


3



Distance to compression (top) face,
54

d

cm

o

Second layer
: top layer



Number o
f longitudinal bars = 2



Diameter of longitudinal bars,
91
.
1

b
d

cm



Distance to compression (top) face,
6

d

cm


The user input for the first layer of bars is shown in Figure 2.


Figure 2:
R
einforcement for Layer 1.

Selec
ted user
-
specified properties of the section are displayed in Window 1 as shown in
Figure 3.


4


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
(u
p 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
c c st s
f A A f


, where
c
f


is the concrete strength,

5

c
A

is the concrete area,
st
A

is the total steel area, and
s
f

is the steel stress
corresponding to concrete crushing.
Th
e smallest compressive axial force
that can be specified is zero. The module is not designed to consider tensile
forces, i.e., negat
ive axial loads.
The module uses zero axial load as the
default value.

As shown in Figure 4, t
he number of axial forces
spec
ified for

Sample
Cross
-
S
ection

2

is three, namely, 0, 50 and 100
% of the balanced failure
load. O
nly 50% of the balanced load will be considered i
n the sample
calculations below
.


1.3 Strain Condition for P
-
M Interaction

The module requires a user
-
specifie
d maximum compression strain value to
generate the P
-
M interaction diagram
s
.
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.


6

2. Calculations and Equations

The following
sample
calculations are based on the method employe
d by Java Module B.

2.1 Concrete Stress
-
Strain Relationship

The equation for the unconfined concrete stress
-
strain relationship is [1]:

r
c
c
x
r
xr
f
f




1
, where










(1)

co
c
x



,
0
0.002
c














(2)
The tangent modulus of elasticity,
c
E
, is calculated using:

4,741
c c
E f



MPa =
24,862.0

MPa








(3)

sec
E
, 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).

co
c
f
E



sec

=
13,750.0

MPa










(4)
Then,

sec
E
E
E
r
c
c


=
2.24











(
5)

and, the concrete stress
-
strain
( )
c c
f



relationship is given as:

2.24
27.5( )(2.24)
0.002
2.24 1 ( )
0.002
c
c
c
f



 
,









(6)

The above
c c
f



relationship is plotted in Figure 6. It is assumed that crushing of concrete
occurs at a strain of
2 0.004
cu co
 
 
.



7


Figure 6: Concrete stress
-
strain relationship for
Sample Cross
-
S
ection 2.

In Fig. 6,

Circle marker: assumed concrete linear
-
elastic limit at
2
c
c el
f
f f

 

and
1
2
c
c el
c
f
E
 

 
.

Square marker: peak stress at
c c
f f



and
c co
 

.

Diamond marker: assumed ultimate strain at
2
c cu co
  
 
.

2.2 Moment
-
Curvature Relationships

Window 2 generates the moment
-
curvature relationships of the

section for the user
-
specified axial forces. This is an iterative process, in which th
e basic equilibrium
requirement (e.g.,
2 1
c s s t
P C F F C
   

for a section with two layers of reinforcement)

and

8

a linear strain diagram are
used to find the neutral ax
is for a particular

maximum concrete

compressive strain
,
cm

,

selected

(see Figure 7)
.


Figure
7
: Section strains, stresses, and stress resultants.

The calculation of the foll
owing four points on
a

moment
-
curvature curve will be s
hown in
this example:



cu
cm


25
.
0




0.5
cm cu
 




cu
cm


75
.
0




cu
cm



(concrete crushing)

Axial
Load, P

The axial load considered for these sample calculations is 50% of the balanced failure load.

The balance
d load,
b
P
, is computed as follows:

The neutral axis,
y
cu
cu
b
d
c
c






, where
y
y
s
f
E




(7)


With
0.004
cu



and
54

d

cm, this gives a value of
36.0
b
c


cm.

The concrete compressive resultant,
c
C
, is determined by numerically integrating under the
concrete stress distribution curve.


9

0 0
2,795.6
b cu
c
c c c c
cm
c
C f bdx f b d



  
 

kN, where
c
f

is from Eq. (6
)



(8)
The
top

and
bottom

steel forces,
2
s
F

and
1
s
F
, respectively, are calculated using similarity
to find the strains in the layers.

B
alanced failure condition, by definition, has strain values
0.004
cm cu
 
 

for concrete and
1
0.002
s y
 
 

for bottom steel. For the top steel,

c
d
d
c
s
s



'
1
2


=0.
0033









(9)

which implies that the
top

steel is also
at yield stress.

Hence,

1 1 1 1
458.4
s s s s y s
F E A f A

  

kN

(tension)







(10
)

2 2 2 2
229.2
s s s s y s
F E A f A

  

kN (compression)






(11)

where
,

1
s
A
and
2
s
A

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

0.62
r c
f f




MPa









(12)

for normal weight concrete.

Thus, for
27.5
c
f



MPa,
3.3
r
f



MPa.

The

concrete tension force
1
6.96
2
t r ct
C f A

 

kN






(13)

w
here
,

ct
A
is the area of concrete in tension

calculated based on the linear strain diagram
.

The
n, the

balanced
axial
load is found
from equilibrium
as

2 1
2,559.4
b c s s t
P C F F C
    

kN.








(14
)

Therefore, 50% of the balanced load used in the example is
1279.7
P


kN.



10

Instant Centroid

The module assumes that the axial load acts at an “instant” centroi
d 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 conditio
n
produces

a

uniform compression
strain
distribution
throughout the cross
-
section.

Let the uniform compression strain be equal to

ci

. Then

1 2
s s ci
  
 

and
1 2
si s s s ci y
f f f E f

   







(15)

From equilibr
ium,

1 1 2 2
ci c s s s s
f A f A f A P
  











(16
)

1 2
2.24
27.5( )(2.24)
0.002
( ) ( )( ) 1279.7
2.24 1 ( )
0.002
ci
c si s s
ci
A f A A P


    
 

kN




(17)

where
1 2
( ) 2143.0
c s s
A bh A A
   

cm
2


(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:

0.000227 5.6
ci ci
f

  

MPa

and
45.4
si
f


MPa

(bars not yielding)
.

Then,

th
e location of th
e instant centr
o
i
d,
x
,

from the top compression face is determined
as

1 2
1 2
/2'
30.48
ci c s si s si
ci c s si s si
f A h A f d A f d
x
f A A f A f
 
 
 
cm








(19
)





11



Point

1

The calculation
of the first sample point on the moment
-
curvature relationship of the
section

can be

summarized as follows:

1.
001
.
0
25
.
0


cu
cm



2. Assume the neutral axis depth, a distance
15

c

cm.

3. From the
linear
strain diagram geometry

1
( ) 0.0026
cm
s
d c
c


  

(at yield stress) and






(20)

2
( ) 0.0006
cm
s
c d
c



  

(below yield stress)





(21)

4. The steel stress resultants are

1 1
458.4
s y s
F f A
 

kN (tension) and







(22)

2 2 2
68.8
s s s s
F E A

 

kN
(
compression).







(23)

5. Determine
c
C

by integrat
ing numerically under the concrete stress distribution curve.

0 0
644.3
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








(24
)

where
c
f

is given
by

Eq. (6).

The concrete
that has not cracked
below the neutral axis contributes to the tension force
.

1
11.1
2
t r ct
C f A

 

kN with
3.3
r
f



MPa from Eq. (12)





(25
)

6. Check to see if
2 1
c s s t
P C F F C
   








(26
)

But

1271
P


kN

243.5

kN

2 1
c s s t
C F F C
   


12

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
32.92
c


cm gives:

1
0.00064
s



(below yield stress) and
2
0.00082
s



(below yield stress).

1
146.7
s
F


kN (tension) and
2
93.7
s
F


kN (compression).

0 0
1,369.2
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.







(27
)

1
26.0
2
t r ct
C f A

 

kN.









(28
)

1279
P


kN


1285 kN
2 1
c s s t
C F F C
   

O.K.

Section c
urvature can then be found from:

5
0.001
3.04 10
32.92
cm
c



   

1
cm











(29
)


The internal lever arm
s

for the resultant compression and tension force
s

of the concrete
measured from the instant
centroid

are

19.15
c
z


cm and

4.39
ct
z


cm, respectively.
Then
,

the
section
moment can be calculated as

1 1 2 2
318.5
c c s s s s t ct
M C z F z F z C z
    

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.
0.5 0.002
cm cu
 
 

2. Assume the neutral axis depth, a distance
18
c


cm.

3. From the
linear
strain diagram geometry


13

1
( ) 0.004
cm
s
d c
c


  

(at yield stress) and







2
( ) 0.0013
cm
s
c d
c



  

(below yield stress)







4. The steel stress resultants are

1 1
458.4
s y s
F f A
 

kN (tension) and








2 2 2
152.8
s s s s
F E A

 

kN (compression).








5. Determine
c
C

by integrating numerically under the concrete stress distribution cu
rve.

0 0
1200.7
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








where
c
f

is given

by

Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force.
1
7
2
t r ct
C f A

 

kN with
3.3
r
f



MPa from Eq. (12)






6. Check to see if
2 1
c s s t
P C F F C
   









But
1271
P


kN

894.4

kN
2 1
c s s t
C F F C
   

So, the neutral axis must be adjusted downwa
rd, 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
23.76
c


cm gives:

1
0.0025
s



(at yield stress) and
2
0.0015
s



(below yield stress).

1
458.4
s
F


kN (tension) and
2
171.3
s
F


kN (compression).

0 0
1,585.1
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.









14

1
9.4
2
t r ct
C f A

 

kN.











1279
P


kN


1288

kN
2 1
c s s t
C F F C
   

O.K.

Section curvature can then be found from:

5
0.002
8.42 10
23.76
cm
c



   

1
cm









The internal lever arms for the resultant compr
ession and tension forces of the concrete
measured from the instant centroid are
21.6
c
z


cm and
5.7
ct
z


cm, respectively. Then,
the section moment can be calculated as

1 1 2 2
491.6
c c s s s s t ct
M C z F z F z C z
    

kN
-
m.










Point

3

The calculation of sample point
three

on the moment
-
curvature relationship of the section
can be summarized as follows:

1.
0.75 0.003
cm cu
 
 

2. Assume the neutral axis depth, a distance
20
c


cm.

3. From the
linear
strain diagram geometry

1
( ) 0.0051
cm
s
d c
c


  

(at yield stress) and







2
( ) 0.0021
cm
s
c d
c



  

(at yield stress)






4. The steel stress resultants are

1 1
458.4
s y s
F f A
 

kN (tension) and








2 2
229.2
s y s
F f A
 

kN (compression).








5. Determine
c
C

by integrating numerically under the concrete stress distribution curve.


15

0 0
1525.3
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








where
c
f

is given
by

Eq.

(6).

The concrete that has not cracked below the neutral axis contributes to the tension force.
1
5.3
2
t r ct
C f A

 

kN with
3.3
r
f



MPa from Eq. (12)






6. Check to see if
2 1
c s s t
P C F F C
   









But

1271
P


kN

1290.8

kN
2 1
c s s t
C F F C
   

So, the neutral axis must be adjusted
upward
, for the particular maximum concrete strain
that was selected in Step 1, until equilibrium is sati
sfied. This iterative process determines
the correct value of
c
.

Trying neutral axis depth
19.86
c


cm gives:

1
0.0052
s



(at yield stress) and
2
0.002
s



(
at

yield stress).

1
458.4
s
F


kN (tension) and
2
229.2
s
F


kN (compression).

0 0
1,514.6
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








1
5.2
2
t r ct
C f A

 

kN.










1279
P


kN


128
0

kN
2 1
c s s t
C F F C
   

O.K.

Secti
on curvature can then be found from:

4
0.003
1.51 10
19.86
cm
c



   

1
cm










16

The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are
22.33
c
z


cm and
10.0
ct
z


cm, respectively.
Then, the section moment can be calculated as

1 1 2 2
501.6
c c s s s s t ct
M C z F z F z C z
    

kN
-
m.









Point

4

The calculation of sample point four on the moment
-
curvature relationship of the
section
can be summarized as follows:

1.
1.0 0.004
cm cu
 
 

2. Assume the neutral axis depth, a distance
21
c


cm.

3. From the
linear
strain diagram geometry

1
( ) 0.0063
cm
s
d c
c


  

(at yield stress) and







2
( ) 0.0029
cm
s
c d
c



  

(at yield stress)






4. The steel stress resultants are

1 1
458.4
s y s
F f A
 

kN (tension) and








2 2
229.2
s y s
F f A
 

kN (compression).








5. Determine
c
C

by integrating n
umerically under the concrete stress distribution curve.

0 0
1630.4
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








where
c
f

is given
by

Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force.
1
4.1
2
t r ct
C f A

 

kN with
3.3
r
f



MPa from Eq. (12)







17

6. Check to see if
2 1
c s s t
P C F F C
   








But
1271
P


kN

1396.7

kN
2 1
c s s t
C F F C
   

S
o, 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
19.48
c


cm gives:

1
0.0071
s



(at yield stress) and
2
0.0028
s



(at yield stress).

1
458.4
s
F


kN (tension) and
2
229.2
s
F


kN (compression).

0 0
1,512.4
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN.








1
3.8
2
t r ct
C f A

 

kN.










1279
P


kN


1279.4

kN
2 1
c s s t
C F F C
   

O.K.

Section curvature can then be found from:

4
0.004
2.05 10
19.48
cm
c



   

1
cm









The int
ernal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are
21.82
c
z


cm and
10.6
ct
z


cm, respectively.
Then, the section moment can be calculated as

1 1 2 2
493.5
c c s s s s t ct
M C z F z F z C z
    

kN
-
m.








18

Plots





The three axial forces specified to generate
the
moment
-
curvature relationship
s

for

Sample
Cross
-
S
ection 2 are represented as P1, P2 and P3 in
W
indow 2 as shown below.

The
M
and


pairs calculated for the four points above are plotted on the moment
-
curvature
curve for
2
0.50
b
P P

.


Figure 8
:
Sample Cross
-
Section 2 m
oment
-
c
urvature relationship
s

shown in W
indow 2.

2.2
Axial
-
Fo
rce
-
Bending
-
Moment
Interaction Diagram

Window 3 generates
P
-
M

i
nteraction diagram
s

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 a
n

initial
small

value to a large on
e

that gives a pure axial loading condition.
The
initial neutral axis
value
,
o
c
,

corresponds to
the
pure bending condition (
i.e.,
no axial force) of the cross
-
s
ection
.


19

The calculation of the following three points on the P
-
M interaction diagram

of Sample
Cross
-
Section 2

will be s
hown in th
is

example.



o
c c




1.5
o
c c





3
o
c c



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 t
o the user
-
specified
maximum concrete
strain
,
cm

, for

P
-
M interaction.

For Sample Cross
-
S
ection 2, the user
-
specified
maximum
concr
ete compression strain

0.003
cm



(see Figure 5)
.

Thus,
83
.
24
003
.
0



ci
ci
f


M
Pa
[from Eq. (6)]
and

0.003 400
si ci si
f
 
   

MPa

(at yield).


20

W
ith
1 2
( ) 2143
c s s
A bh A A
   

cm
2

, t
he location of the instant centroid,
x
,

from the top
compression face is determined as

2
1 2
1 2
/2'
30.91
ci s si s si
ci s si s si
f bh A f d A f d
x
f bh A f A f
 
 
 

cm



Point 1

T
he calculation of the first sample point on the P
-
M interaction diagram of the section is
described below.
The initial neutral axis

location,
o
c
,

corresponding to pure bending

is
determined as follows:

1
o
s cm
o
d c
c
 



and
2
o
s cm
o
c d
c
 






(
31
)

where the user specified
0.003
cm



Then,
1 1
s s s y
f E f

 

and
2 2
s s s y
f E f

 





(32
)

S
ince
0
P

,

from equilibrium:

2 2 1 1
c s s s s t
C f A f A C
  


(33)

S
ubstituting

for
the

concrete and
steel

stress resultants

gives

the
following

equation

2 2 1 1
0
1
2
cm
o
c c s s s s r ct
cm
c
f b d f A f A f A




  









(34
)

An iterative solution

of
Eq
.
(
3
4
)

is needed (since the yielding bars are not known in
advance),

from which the value of the neutral a
xis can be
obtained as
6.01
o
c


cm.

The internal forces
can

then

be calculated as:

1
458.4
s
F


kN (tension
, at yield
) and
2
0.57
s
F


kN (compression
, below yield
).

0 0
458.4
o cm
c
o
c c c c
cm
c
C f bdx f b d



  
 

kN


21

1
1.6
2
t r ct
C f A

 

kN

The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are
28.46
c
z


cm and
24.72
ct
z


cm, respectively.
Then, the
section
moment can be calcu
lated as

1 1 2 2
236.1
c c s s s s t ct
M C z F z F z C z
    

kN
-
m.



Point 2

The calculation of the second sample point,

o
c
c
5
.
1

, on the P
-
M interaction diagram of
the section is described below
.


1
s cm
d c
c
 



and
2
s cm
c d
c
 




T
he neutr
al axis
is

9.02
c


cm.

The internal forces can then be calculated as:

1 1
458.4
s y s
F f A
 

kN (tension, at yield) and
2 2 2
115.1
s s s s
F E A

 

kN (compression,
below

yield).

0 0
687.5
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN

1
2.4
2
t r ct
C f A

 

kN

Th
en, the section axial force can be calculated as

2 1
341.8
c s s t
P C F F C
    

kN.

The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are
27.21
c
z


cm and
21.6
ct
z


cm, respectively.
Then, the section moment can be calculated as


22

1 1 2 2
321.1
c c s s s s t ct
M C z F z F z C z
    

kN
-
m.



Point
3

The calculation of the third

sample point,

o
c
c
0
.
3

, on the P
-
M interaction diagram of the
section is described below.

1
s cm
d c
c
 



and
2
s cm
c d
c
 




The neutral axis is
18.03
c


cm.

The internal forces can then be calculated as:

1 1
458.4
s y s
F f A
 

kN (tension, at yield) and
2 2
229.2
s y s
F f A
 

kN (compression, at
yield)
.

0 0
1,375.1
cm
c
c c c c
cm
c
C f bdx f b d



  
 

kN

1
4.7
2
t r ct
C f A

 

kN

2 1
1141.2
c s s t
P C F F C
    

kN

The internal lever arms for the resultant compression and tension forces of the concrete
measured from the instant centroid are
23.52
c
z


cm and
12.35
ct
z


cm, respectively.
Then, the section moment can be calculated as

1 1 2 2
485.8
c c s s s s t ct
M C z F z F z C z
    

kN
-
m

Plots

The
M
and
P

pairs calculated for the three points above are plotted on the interaction
d
iagram shown in Figure 10 below.


23


Figure 10
:
Sample Cross
-
Section
2 interaction diagram shown in W
indow 3