SAMPLE LABORATORY SESSION FOR JAVA MODULE B …concrete/java/bending/helppage/design1.pdf · SAMPLE LABORATORY SESSION FOR JAVA MODULE B Calculations for Sample Cross-Section 2 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.

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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 properties are needed:

o Concrete compressive strength, ' 27.5cf = MPa

o Concrete strain corresponding to peak stress ( ′cf ), 0.002coε =

The default value used by the module for coε is 0.002 .

• Steel Properties

o Steel yield strength, 400yf = MPa

o Steel Young’s Modulus, 200000sE = MPa

The default value used by the module for sE 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=sd cm

o Number of layers of longitudinal bars, 2=ln

o First layer : bottom layer

Number of longitudinal bars = 4

Diameter of longitudinal bars, 91.1=bd cm

3

Distance to compression (top) face, 54=d cm

o Second layer: top layer

Number of longitudinal bars = 2

Diameter of longitudinal bars, 91.1=bd 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: Reinforcement for Layer 1.

Selected user-specified properties of the section are displayed in Window 1 as shown in

Figure 3.

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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 c c st sf A A f′ + , where cf ′ is the concrete strength,

cA is the concrete area, stA is the total steel area, and sf 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.

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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]:

rc

c xrxrf

f+−

′=

1, where (1)

co

cxεε

= , 0 0.002cε = (2)

The tangent modulus of elasticity, cE , is calculated using:

4,741c cE f ′= MPa = 24,862.0 MPa (3)

secE , 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

cfEε

′=sec =13,750.0 MPa (4)

Then,

secEEE

rc

c

−= = 2.24 (5)

and, the concrete stress-strain ( )c cf ε− relationship is given as:

2.24

27.5( )(2.24)0.002

2.24 1 ( )0.002

c

cc

f

ε

ε=− +

, (6)

The above c cf ε− relationship is plotted in Figure 6. It is assumed that crushing of concrete

occurs at a strain of 2 0.004cu coε ε= = .

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

In Fig. 6,

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

c elff f′

= = and 12

cc el

c

fE

ε ε′

= = .

Square marker: peak stress at c cf f ′= and c coε ε= .

Diamond marker: assumed ultimate strain at 2c cu coε ε ε= = .

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 the basic equilibrium

requirement (e.g., 2 1c s s tP C F F C= + − − for a section with two layers of reinforcement) and

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a linear strain diagram are used to find the neutral axis for a particular maximum concrete

compressive strain, cmε , 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:

• cucm εε 25.0=

• 0.5cm cuε ε=

• cucm εε 75.0=

• cucm εε = (concrete crushing)

Axial Load, P

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

The balanced load, bP , is computed as follows:

The neutral axis, ycu

cub dcc

εεε+

== , where yy

s

fE

ε = (7)

With 0.004cuε = and 54=d cm, this gives a value of 36.0bc = cm.

The concrete compressive resultant, cC , is determined by numerically integrating under the

concrete stress distribution curve.

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0 0

2,795.6b cuc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN, where cf is from Eq. (6) (8)

The top and bottom steel forces, 2sF and 1sF , respectively, are calculated using similarity

to find the strains in the layers. Balanced failure condition, by definition, has strain values

0.004cm cuε ε= = for concrete and 1 0.002s yε ε= = for bottom steel. For the top steel,

cddc

ss −−

='

12 εε =0.0033 (9)

which implies that the top steel is also at yield stress.

Hence,

1 1 1 1 458.4s s s s y sF E A f Aε= = = kN (tension) (10)

2 2 2 2 229.2s s s s y sF E A f Aε= ⇒ = kN (compression) (11)

where, 1sA and 2sA 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

0.62r cf f′ ′= MPa (12)

for normal weight concrete. Thus, for 27.5cf ′ = MPa, 3.3rf ′ = MPa.

The concrete tension force 1 6.962t r ctC f A′= = kN (13)

where, ctA is the area of concrete in tension calculated based on the linear strain diagram.

Then, the balanced axial load is found from equilibrium as

2 1 2,559.4b c s s tP C F F C= + − − = kN. (14)

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

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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 ciε . Then

1 2s s ciε ε ε= = and 1 2si s s s ci yf f f E fε= = = ≤ (15)

From equilibrium,

1 1 2 2ci c s s s sf A f A f A P+ + = (16)

1 22.24

27.5( )(2.24)0.002 ( ) ( )( ) 1279.7

2.24 1 ( )0.002

ci

c si s sci

A f A A P

ε

ε⇒ + + = =− +

kN (17)

where 1 2( ) 2143.0c s sA bh A A= − + = 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 ciε can be calculated as:

0.000227 5.6ci cifε = ⇒ = MPa and 45.4sif = MPa (bars not yielding).

Then, the location of the instant centroid, x , from the top compression face is determined

as

1 2

1 2

/ 2 ' 30.48ci c s si s si

ci c s si s si

f A h A f d A f dxf A A f A f

+ += =

+ +cm (19)

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• Point 1

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

section can be summarized as follows:

1. 001.025.0 == cucm εε

2. Assume the neutral axis depth, a distance 15=c cm.

3. From the linear strain diagram geometry

1 ( ) 0.0026cms d c

cεε = − = (at yield stress) and (20)

2 ( ) 0.0006cms c d

cεε ′= − = (below yield stress) (21)

4. The steel stress resultants are

1 1 458.4s y sF f A= = kN (tension) and (22)

2 2 2 68.8s s s sF E Aε= = kN (compression). (23)

5. Determine cC by integrating numerically under the concrete stress distribution curve.

0 0

644.3cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN. (24)

where cf is given by Eq. (6).

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

1 11.12t r ctC f A′= = kN with 3.3rf ′ = MPa from Eq. (12) (25)

6. Check to see if 2 1c s s tP C F F C= + − − (26)

But 1271P = kN≠ 243.5 kN 2 1c s s tC F F C= + − −

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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 32.92c = cm gives:

1 0.00064sε = (below yield stress) and 2 0.00082sε = (below yield stress).

1 146.7sF = kN (tension) and 2 93.7sF = kN (compression).

0 0

1,369.2cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN. (27)

1 26.02t r ctC f A′= = kN. (28)

1279P = kN ≈ 1285 kN 2 1c s s tC F F C= + − − O.K.

Section curvature can then be found from:

50.001 3.04 1032.92

cm

cεψ −= = = × 1

cm (29)

The internal lever arms for the resultant compression and tension forces of the concrete

measured from the instant centroid are 19.15cz = cm and 4.39ctz = cm, respectively.

Then, the section moment can be calculated as

1 1 2 2 318.5c c s s s s t ctM 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 0.002cm cuε ε= =

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


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1 ( ) 0.004cms d c

cεε = − = (at yield stress) and

2 ( ) 0.0013cms c d

cεε ′= − = (below yield stress)


1 1 458.4s y sF f A= = kN (tension) and

2 2 2 152.8s s s sF E Aε= = kN (compression).


0 0

1200.7cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.



1 72t r ctC f A′= = kN with 3.3rf ′ = MPa from Eq. (12)

6. Check to see if 2 1c s s tP C F F C= + − −


So, the neutral axis must be adjusted downward, for the particular maximum concrete strain




1 0.0025sε = (at yield stress) and 2 0.0015sε = (below yield stress).


0 0

1,585.1cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.

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1 9.42t r ctC f A′= = kN.



50.002 8.42 1023.76

cm

cεψ −= = = × 1

cm


measured from the instant centroid are 21.6cz = cm and 5.7ctz = cm, respectively. Then,

the section moment can be calculated as

1 1 2 2 491.6c c s s s s t ctM 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


1. 0.75 0.003cm cuε ε= =



1 ( ) 0.0051cms d c


2 ( ) 0.0021cms c d

cεε ′= − = (at yield stress)



2 2 229.2s y sF f A= = kN (compression).


15

0 0

1525.3cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.



1 5.32t r ctC f A′= = kN with 3.3rf ′ = MPa from Eq. (12)



So, the neutral axis must be adjusted upward, for the particular maximum concrete strain




1 0.0052sε = (at yield stress) and 2 0.002sε = (at yield stress).


0 0

1,514.6cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.

1 5.22t r ctC f A′= = kN.



40.003 1.51 1019.86

cm

cεψ −= = = × 1

cm

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• Point 4

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


1. 1.0 0.004cm cuε ε= =



1 ( ) 0.0063cms d c


2 ( ) 0.0029cms c d

cεε ′= − = (at yield stress)



2 2 229.2s y sF f A= = kN (compression).


0 0

1630.4cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.



1 4.12t r ctC f A′= = kN with 3.3rf ′ = MPa from Eq. (12)

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So, the neutral axis must be adjusted upward, for the particular maximum concrete strain




1 0.0071sε = (at yield stress) and 2 0.0028sε = (at yield stress).


0 0

1,512.4cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN.

1 3.82t r ctC f A′= = kN.

1279P = kN ≈ 1279.4 kN 2 1c s s tC F F C= + − − O.K.


40.004 2.05 1019.48

cm

cεψ −= = = × 1

cm





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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 ψ pairs calculated for the four points above are plotted on the moment-curvature

curve for 2 0.50 bP P= .

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

section.

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

Cross-Section 2 will be shown in this example.

• oc c=

• 1.5 oc c=

• 3 oc 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 to the user-specified maximum concrete strain, cmε , for

P-M interaction.

For Sample Cross-Section 2, the user-specified maximum concrete compression strain

0.003cmε = (see Figure 5).

Thus, 83.24003.0 =⇒= cici fε MPa [from Eq. (6)] and

0.003 400si ci sifε ε= = ⇒ = MPa (at yield).

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With 1 2( ) 2143c s sA bh A A= − + = cm2 , the location of the instant centroid, x , from the top

compression face is determined as

21 2

1 2

/ 2 ' 30.91ci s si s si

ci s si s si

f bh A f d A f dxf bh A f A f

+ += =

+ + cm

• 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, oc , corresponding to pure bending is

determined as follows:

1o

s cmo

d cc

ε ε −= and 2

os cm

o

c dc

ε ε′−

= (31)

where the user specified 0.003cmε =

Then, 1 1s s s yf E fε= ≤ and 2 2s s s yf E fε= ≤ (32)

Since 0P = , from equilibrium:

2 2 1 1c s s s s tC f A f A C+ = + (33)

Substituting for the concrete and steel stress resultants gives the following equation

2 2 1 10

12

cmo

c c s s s s r ctcm

cf b d f A f A f Aε

εε

′+ = +∫ (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 6.01oc = cm.

The internal forces can then be calculated as:

1 458.4sF = kN (tension, at yield) and 2 0.57sF = kN (compression, below yield).

0 0

458.4o cmc

oc c c c

cm

cC f bdx f b dε

εε

= = =∫ ∫ kN

21

1 1.62t r ctC f A′= = kN





• Point 2

The calculation of the second sample point, occ 5.1= , on the P-M interaction diagram of

the section is described below.

1s cmd cc

ε ε −= and 2s cm

c dc

ε ε′−

=

The neutral axis is 9.02c = cm.


1 1 458.4s y sF f A= = kN (tension, at yield) and 2 2 2 115.1s s s sF E Aε= = kN (compression,

below yield).

0 0

687.5cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN

1 2.42t r ctC f A′= = kN

Then, the section axial force can be calculated as

2 1 341.8c s s tP C F F C= + − − = kN.




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• Point 3

The calculation of the third sample point, occ 0.3= , on the P-M interaction diagram of the

section is described below.

1s cmd cc

ε ε −= and 2s cm

c dc

ε ε′−

=

The neutral axis is 18.03c = cm.


1 1 458.4s y sF f A= = kN (tension, at yield) and 2 2 229.2s y sF f A= = kN (compression, at

yield).

0 0

1,375.1cmc

c c c ccm

cC f bdx f b dε

εε

= = =∫ ∫ kN

1 4.72t r ctC f A′= = kN

2 1 1141.2c s s tP C F F C= + − − = kN




1 1 2 2 485.8c c s s s s t ctM 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

diagram shown in Figure 10 below.

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

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SAMPLE LABORATORY SESSION FOR JAVA MODULE B …concrete/java/bending/helppage/design1.pdf · SAMPLE LABORATORY SESSION FOR JAVA MODULE B Calculations for Sample Cross-Section 2 1