Sample Laboratory Session for Java Module B

SAMPLE LABORATORY SESSION FOR JAVA MODULE B

Calculations for Sample Cross-Section 2

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.

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:

Concrete compressive strength, MPa
Concrete strain corresponding to peak stress (),

The default value used by the module for is .

Steel Properties
Steel yield strength, MPa
Steel Young’s Modulus, MPa

The default value used by the module for is MPa.

Section Dimensions

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

Section height, cm
Section width, cm

Reinforcement
Stirrup diameter, cm
Number of layers of longitudinal bars,
First layer : bottom layer
Number of longitudinal bars = 4
Diameter of longitudinal bars, cm
Distance to compression (top) face, cm
Second layer: top layer
Number of longitudinal bars = 2
Diameter of longitudinal bars, cm
Distance to compression (top) face, 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.

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 , where is the concrete strength, is the concrete area, is the total steel area, and 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 forSample 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 forSample Cross-Section 2 as shown in Figure 5.

Figure 5: Strain condition for P-M interaction.

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]:

, where (1)

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

MPa = MPa (3)

, the secant modulus of elasticity, is the slope of the line connecting the origin andpeak stress on the compressive stress-strain curve (see Fig. 6).

= MPa (4) Then,

= (5)

and, the concrete stress-strain relationship is given as:

, (6)

The above relationship is plotted in Figure 6. It is assumed that crushing of concrete occurs at a strain of .

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

In Fig. 6,

Circle marker: assumed concrete linear-elastic limit at and.

Square marker: peak stress at and.

Diamond marker: assumed ultimate strain at.

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.,for a section with two layers of reinforcement) and a linear strain diagram are used to find the neutral axis for a particular maximum concrete compressive strain, , 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:

(concrete crushing)

Axial Load, P

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

The balanced load,, is computed as follows:

The neutral axis, , where (7)

With and cm, this gives a value of cm.

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

kN, where is from Eq. (6) (8) The top and bottom steel forces, and , respectively, are calculated using similarity to find the strains in the layers. Balanced failure condition, by definition, has strain values for concrete and for bottom steel. For the top steel,

=0.0033 (9)

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

Hence,

kN (tension) (10)

kN (compression) (11)

where,and 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

MPa (12)

for normal weight concrete. Thus, for MPa, MPa.

The concrete tension force kN (13)

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

Then, the balanced axial load is found from equilibrium as

kN. (14)

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

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. Then

and (15)

From equilibrium,

(16)

kN (17)

where 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 can be calculated as:

MPa and MPa (bars not yielding).

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

cm (19)

Point 1

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

2. Assume the neutral axis depth, a distance cm.

3. From the linear strain diagram geometry

(at yield stress) and (20)

(below yield stress) (21)

4. The steel stress resultants are

kN (tension) and (22)

kN (compression). (23)

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

kN. (24)

where is given by Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force. kN with MPa from Eq. (12) (25)

6. Check to see if (26)

But kN kN

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 .

Trying neutral axis depth cm gives:

(below yield stress) and (below yield stress).

kN (tension) and kN (compression).

kN. (27)

kN. (28)

kN 1285 kN O.K.

Section curvature can then be found from:

(29)

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m. (30)

Point 2

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

2. Assume the neutral axis depth, a distance cm.

3. From the linear strain diagram geometry

(at yield stress) and

(below yield stress)

4. The steel stress resultants are

kN (tension) and

kN (compression).

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

kN.

where is given by Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force. kN with MPa from Eq. (12)

6. Check to see if

But kN kN

Trying neutral axis depth cm gives:

(at yield stress) and (below yield stress).

kN (tension) and kN (compression).

kN.

kN 1288 kN O.K.

Section curvature can then be found from:

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m.

Point 3

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

2. Assume the neutral axis depth, a distance cm.

3. From the linear strain diagram geometry

(at yield stress) and

(at yield stress)

4. The steel stress resultants are

kN (tension) and

kN (compression).

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

kN.

where is given by Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force. kN with MPa from Eq. (12)

6. Check to see if

But kN kN

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 .

Trying neutral axis depth cm gives:

(at yield stress) and (at yield stress).

kN (tension) and kN (compression).

kN.

kN 1280 kN O.K.

Section curvature can then be found from:

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m.

Point 4

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

2. Assume the neutral axis depth, a distance cm.

3. From the linear strain diagram geometry

(at yield stress) and

(at yield stress)

4. The steel stress resultants are

kN (tension) and

kN (compression).

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

kN.

where is given by Eq. (6).

The concrete that has not cracked below the neutral axis contributes to the tension force. kN with MPa from Eq. (12)

6. Check to see if

But kN kN

Trying neutral axis depth cm gives:

(at yield stress) and (at yield stress).

kN (tension) and kN (compression).

kN.

kN 1279.4 kN O.K.

Section curvature can then be found from:

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m.

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 and pairs calculated for the four points above are plotted on the moment-curvature curve for .

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-Minteraction diagramsfor the user-defined cross-section by determining the axial load and moment pairs of the section for a user-specified maximum concrete compression strain, . The P-M interaction diagram for each cross-section is generated by selecting successive choices of the neutral axis distance, , from aninitial small value to a large one that gives a pure axial loading condition. The initial neutral axis value, , corresponds to the pure bending condition (i.e., no axial force) of the cross-section.

The calculation of the following three points on the P-M interaction diagram of Sample Cross-Section 2 will be shown in this example.

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 sectionmoment-curvature relationship. The axial load is assumed to be applied at the instant centroid, whichis determined from a uniform compression strain distribution over the section, equal to the user-specified maximum concrete strain, , for P-M interaction.

For Sample Cross-Section 2, the user-specified maximum concrete compression strain (see Figure 5).

Thus, MPa [from Eq. (6)] and

MPa (at yield).

With cm2 , the location of the instant centroid, ,from the top compression face is determined as

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, , corresponding to pure bending is determined as follows:

and (31)

where the user specified

Then, and (32)

Since ,from equilibrium:

(33)

Substituting for theconcrete and steel stress resultants givesthe following equation

(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 cm.

The internal forces can then be calculated as:

kN (tension, at yield) and kN (compression, below yield).

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m.

Point 2

The calculation of the second sample point,, on the P-M interaction diagram of the section is described below.

and

The neutral axis iscm.

The internal forces can then be calculated as:

kN (tension, at yield) and kN (compression, belowyield).

Then, the section axial force can be calculated as

kN.

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m.

Point 3

The calculation of the third sample point,, on the P-M interaction diagram of the section is described below.

and

The neutral axis is cm.

The internal forces can then be calculated as:

kN (tension, at yield) and kN (compression, at yield).

The internal lever arms for the resultant compression and tension forces of the concrete measured from the instant centroid are cm and cm, respectively. Then, the section moment can be calculated as

kN-m

Plots

The and 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