Reinforcement Design
- Becca Alford (Unlicensed)
Analytics
Reinforcing design is completed in accordance with AASHTO Section 17.4.6. The development of this design method is documented in the references.
Flexural reinforcement is designed in accordance with the following criteria:
Ultimate flexural strength based on yielding of the tensile reinforcement
Minimum reinforcement
Maximum reinforcement based on concrete compression to ensure ductile behavior
Control of cracking at service loads
Ultimate radial tension strength
If maximum reinforcement governs, an informative message is printed, warning the user that concrete compression governs, and the design is halted for that particular design section. To obtain a design, the user must increase the wall thickness, or the concrete strength, or otherwise modify the input. Shear forces at the two design locations are compared against allowable values calculated by AASHTO Equation 17.16. If the shear strength of either section is exceeded, informative messages are printed and the interactive stirrup design routine of Eriksson Pipe is activated.
Users should note that the theories for 0.01 inch crack and diagonal tension strength are semi empirical in nature and were developed from tests on slabs and large diameter pipe (see references). Experience has shown that these equations become increasingly conservative as the pipe diameter decreases below 48 inches.
For circular pipe the user may specify several reinforcing configurations as shown in Figure 5-5.
Indirect Design Method
Eriksson Pipe includes the option of designing reinforcing by the indirect method, also known as the three edge bearing load condition. Reinforced concrete pipe designs in ASTM Standards C 76 (circular pipe), C 506 (arch pipe) and C 507 (horizontal and vertical elliptical pipe) are all based on the three edge bearing load condition which is shown on Figure 5-6. The indirect design module of Eriksson Pipe designs reinforcing for all of these pipe shapes. The ·basis of the indirect method is that the design D-Load is selected to produce a bending moment at the invert that is equal to the bending moment at the invert under installed conditions.
The Eriksson Pipe indirect design module was developed from an existing program. It is completely independent of the remainder of the program and will appear different to the . user, but is easily operated. The module has its own input and output routines. Results are printed initially to the screen and, upon the user's request, to the printer. No files are written to or read from disk storage.
Design of inside reinforcing is based on the same criteria as the direct design module of Eriksson Pipe: ultimate flexure, 0.01 inch crack, diagonal tension and radial tension. The design equations are based on the equations for direct design but are modified for the specific shear and moments of the three edge bearing condition.
The practice of ASTM standards is to size the outside reinforcing as a percentage of the inside reinforcing. This rule is followed in the Eriksson Pipe indirect design method when the inside reinforcing is governed by ultimate flexure or 0.01 inch crack; however, when the inside reinforcing is increased to. improve the diagonal tension strength the outside reinforcing is not increased because the outside reinforcing has no effect on the diagonal tension strength. If the design requires stirrups the indirect module will present a screen that requests input for developable stirrup yield strength and stirrup spacing. Complete stirrup requirements are computed and printed with the circumferential reinforcing requirements.
Users of this module of Eriksson Pipe should remember that the ASTM standards for reinforced concrete pipe have been largely developed from experience and testing. This module uses semi-empirical equations that do not always match that experience. The equivalence between the standards and the Eriksson Pipe indirect design module is obviously affected by the strength reduction factors that are selected for input and experience has shown that method of manufacture and local variations in materials, particularly aggregates have a significant effect on the diagonal and radial tension strength.
As stated for the direct design method, the equations for 0.01 inch crack and diagonal tension become increasingly conservative for diameters below 48 inches.
The design equations used in the indirect method are as follows:
Analysis of Dead Load Capacity
Initially, all the data needed to compute the depth of the reinforcement is available.
Reinforcing Design
The wire diameter is not initially known prior to design. Therefore, the depth of the reinforcement is estimated as,
Weight of Pipe Section
The total weight of structure per unit length, Wp, is estimated in lbs per ft.
Ultimate Flexure
If ultimate flexure strength is to be based on elastic moment distribution and “first yield” in reinforcement (the conventional design basis), use Cmp = 1.0 and the values of the Cm given in the table below for the applicable standard pipe shape.
If the ultimate flexural strength is to be the expected “collapse” load on the pipe in 3-edge loading (the conventional test strength basis), compute Cmp as:
Use values of Cm and Cmo from the table below. The coefficient Cs is the ratio of the amount of reinforcing provided at the springline to the amount of reinforcing provided at the invert. Typical values for Cs from ASTM specifications are listed below. Non-standard reinforcing arrangements (such as circular plus elliptical) may result in other values of Cs.
If the value Cmp computed by the respective equation is less than 1.0, whether first yield or ultimate collapse is selected as the design condition, then reinforcing at the springline may yield before the reinforcing at the invert. If this occurs, the calculation for ultimate flexure and 0.01 inch crack may be in error.
For pipe with wall thicknesses 5” or less, 1” nominal cover concrete, and outer cage reinforcement equal to six-tenths of inner cage reinforcement, tests show that both the inner and outer cages will develop their full tension strengths at the crown and invert prior to collapse. Thus, to account for this behavior when flexural strength is to be based on collapse load rather than on first yield.
Reinforcement required at invert, and crown, to obtain required Dlu
Minimum reinforcement area
Anchorage of discontinuous reinforcement and splices
Where reinforcement is terminated at sections where it is no longer required, it must be extended sufficiently to develop at least 0.5 fy As.
Reinforcement splices made in zones of flexural tension must develop fy As. Use fsu and Cmp = 1.1 if reinforcement design is based on collapse condition.
Inner reinforcement splices made in zones of flexural compression must develop fy (min Asi).
0.01 inch crack
The following equations are for pipes 48 inches in diameter and larger. They may be used for smaller pipe but become increasingly more conservative as pipe diameter decreases from 48 inches.
Predicted dead load at 0.01 inch crack
B0.01 and C0.01 are as follows:
Reinforcement required at invert to obtain required DL0.01
Method for using “proof of design” tests to modify estimate of 0.01 inch crack strength for particular manufacturing and materials conditions
The equations given above for 0.01 inch crack DL strength, are semi-empirical with design constraints determined by analyzing a large number of tests on pipe having many different manufacturing and materials conditions. Sometimes, process or material characteristics at a particular plant, or locality, may result in pipe with increases (or occasionally decreased) 0.01 inch strength. In such cases, the designer may obtain improved estimates of 0.01 inch crack strength by applying modification factors obtained as described below to the design equations given previously in this sub-section.
1.) Determine a series of maximum 0.01 inch crack index limits, using “proof of design” tests of specific designs tested to 0.01 inch DL strength:
The tested designs, should cover the range of variables expected to occur in the design practice with particular manufacturing and materials conditions. Average measured values of d, tb, and fc for the test specimens should be used.
2.) Select either the lowest value of FcrT, or the arithmetic mean less 1.07 standard deviations, as defined in 9.1.1 of ASTM C 655, as the 0.01 inch crack strength modification factor.
3.) Multiply the right side of the equation by the value of FcrT selected above to obtain estimated 0.01 inch crack strengths, based on “proof of design” tests of pipe having specific manufacturing and materials conditions. Alternately the required (DL0.01 + 9 Wp/Si) may be divided by FcrT.
Reinforcement required at springline
Outside reinforcing requirements at the springline are computed by the following:
Values for Cs based on ASTM specifications may be taken from the table above.
Asi is the maximum reinforcing required for ultimate flexure or 0.01 inch crack.
Ultimate diagonal tension without stirrups
Predicted DL at diagonal tension failure
Area of reinforcement for required ultimate and DL based on diagonal tension strength
FMV is greater than or equal to 1.0, then the M/Vd ratio is greater than 3.0 and use FMV = 1.0.
If Asi is greater than 0.24d, stirrups must be used, with Asi governed by flexure or 0.01 inch crack.
Method for using “proof of design” tests to modify estimate of ultimate diagonal tension strength for particular manufacturing and materials conditions
The equations given above for ultimate diagonal tension strength are semi-empirical, with design constants determined by analyzing a large number of tests on pipe having many different manufacturing and materials conditions. Sometimes, process or materials characteristics at a particular plant or locality may result in pipe with increased (or occasionally decreased) diagonal tension strength. In such cases, the designer may obtain improved estimates of diagonal tension strength by applying modification factors obtained as described below to the design to the design equations given previously in this sub-section.
Determine a series of maximum diagonal tension strength index limits, Rdtm, using proof of design tests (ASTM C 655) of specific designs that fall in diagonal tension:
2. Select either the lowest value of Rdtm, or the arithmetic mean Rdtm less 1.07 standard deviations, as defined in paragraph 9.1.1 of ASTM C 655, as the diagonal tension strength modification factor.
3. To preclude an excessive correction caused by some local distortion in test results, limit the maximum Rdtm to less than or equal to 1.20.
4. Multiply the right side by the value of Rdtm selected above to obtain estimated diagonal tension strengths, based on proof of design tests of pipe having specific manufacturing and materials conditions. Alternatively, the required (DLu + 11 Cw Wp / Si) may be divided by Rdtm.
Ultimate Tension Without Stirrups
Predicted DL at radial tension failure
Obtain rs by deducting (0.5h – tb) from the values of r given in the table.
Stirrup design for diagonal and radial tension strength
If the required DLu exceeds the DLu provided by the diagonal tension strength of the concrete, stirrups may be used at the crown and invert regions to increase the ultimate diagonal strength of the pipe.
Maximum stirrup spacing:
Required stirrup reinforcement area per foot of pipe per line of stirrups:
LS should be determined based on field load and bedding conditions. In the absence of a design for field loading conditions, provide stirrups over minimum lengths in the crown and invert region: