strength design of reinforced concrete hydraulic
TRANSCRIPT
STRENGTH DESIGN OF REINFORCED CONCRETE HYDRAULIC STRUCTURES
Report I
PRELIMINARY STRENGTH DESIGN CRITERIAby
Tony C. Liu
Structures LaboratoryU. S. Army Engineer Waterways Experiment Station
P. O. Box 631, Vicksburg, Miss. 39160
July 1980
Report I of a Series
Approved For Public Release; Distribution Unlimited
Prepared for Office, Chief of Engineers, U. S. Army Washington, D. C. 20314
Under C W IS 31623
libraryLureou of Reclamation Denver, Colorado
Destroy this report when no longer needed. Do not return it to the originator.
The findings in this report are not to be construed as an officia Department of the Army position unless so designated,
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The contents of this report are not to be used for advertising, publication, or promotional purposes. Citation of trade names does not constitute an official endorsement or approval of the use of
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BUREAU OF RECLAMATION DENVER LIBRARY
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REPO RT DOCUMENTATION PAGE R E A D IN S T R U C T IO N S B E F O R E C O M P L E T IN G FO R M
1. R E P O R T N U M BER 2. G O V T ACCESSION NO.
T e c h n ic a l R ep o rt SL-80-1+3. R E C IP IE N T ’ S C A T A L O G NU M BER
4. T IT L E (and S ub title )STRENGTH DESIGN OF REINFORCED CONCRETE HYDRAULIC STRUCTURES; R ep o rt 1 , PRELIMINARY STRENGTH DESIGN CRITERIA
5. T Y P E O F R E P O R T ft P E R IO D C O V E R E D
R ep ort 1 o f a s e r i e s6. P E R FO R M IN G ORG. R E P O R T NU M BER
7. A U T H O R *»
Tony C. L iu8. C O N T R A C T O R G R A N T N U M B E R *»
9 . P E R F O R M IN G O R G A N IZ A T IO N NA M E AND ADDRESS
U. S . Army E n g in e e r W aterways E xperim ent S ta t io n P. 0 . Box 631 V ick sb u rg , M iss. 39180
10. PROGRAM E L E M E N T , P R O J E C T , TASK A R EA ft WORK U N IT NUM BERS
cw is 31623
11. C O N T R O L L IN G O F F IC E NA M E AND ADDRESS
O f f ic e , C h ie f o f E n g in e e rs , U. S . Army W ash in g to n , Do C. 2031^+
12. R E P O R T O A T E
J u ly 198013. NU M BER O F PA G ES
9714. M O N ITO R IN G AG EN C Y NAM E ft ADDRESS*?/ d ifferen t from Controlling O ffice) 15. S E C U R IT Y CLASS, (o f thia report)
U n c la s s i f ie d15«. D E C L ASSI FI C ATI O N / DOWN GR ADI N G
S C H E D U L E
16. D IS T R IB U T IO N S T A T E M E N T (o f th ie R eport)
A pproved f o r p u b l ic r e l e a s e ; d i s t r i b u t i o n u n l im i te d .
17. D IS T R IB U T IO N S T A T E M E N T (o f the abatract en tered in B lo c k 20, i t d iffe ren t from Report)
18. S U P P L E M E N T A R Y N O TE S
19. K E Y WORDS (C ontinue on reveree a ide i f neceaaary and id en tify by b lock number)C o n cre te s t r e n g th C o n cre te s t r u c t u r e s H y d ra u lic s t r u c t u r e s R e in fo rc e d c o n c re te
20«. A B S T R A C T CContfmim o n reveree aidla f f neceaaary and Id e n tify by b lock number)The o v e r a l l o b je c t iv e o f t h i s s tu d y was t o d ev e lo p a r e a l i s t i c s t r e n g th
d e s ig n m ethodo logy f o r r e in f o r c e d c o n c re te h y d r a u l ic s t r u c t u r e s and to d e v is e an a c c u ra te and e f f i c i e n t d e s ig n p ro c e d u re f o r im p lem en ting th e s e s t r e n g th d e s ig n m eth ods. The f i r s t ph ase o f t h i s s tu d y d ev e lo p ed p r e l im in a r y s t r e n g th d e s ig n c r i t e r i a t h a t w i l l y i e l d d e s ig n s e q u iv a le n t t o th o s e d e s ig n e d by th e w o rk in g - s t r e s s m ethod f o r h y d r a u l ic s t r u c t u r e s . The r e s u l t s o f t h i s f i r s t p h ase s tu d y a re g iv e n in t h i s r e p o r t . ^
U n c la s s i f ie dDD U73 E D IT IO N O F I N O V 6 5 IS O B S O L E T E
S E C U R IT Y C L A S S IF IC A T IO N O F T H IS P A G E (W feit D ata Entered)
92059214
Unclassified_____________________S E C U R IT Y C L A S S IF IC A T IO N O F TH IS PA G EfW Tin Data Entered)
20. ABSTRACT (Continued).The preliminary strength design criteria have been developed, including
load factors and loading combinations, factored base reactions, design strength for reinforcement, distribution of flexural reinforcement, shrinkage and temperature reinforcement, details of reinforcement, control of deflections, maximum tension reinforcement of flexural members, minimum reinforcement of flexural members, provisions for combined flexure and axial load, and shear strengths of various hydraulic structural members. A comprehensive commentary discussing the considerations and background information used in developing the strength design criteria is also included. Examples for designing typical hydraulic structural members using the strength design method are given. A preliminary study on the use of the probabilistic approach to derive the load factors and strength reduction factors for hydraulic structures is presented.
UnclassifiedS E C U R IT Y C L A S S IF IC A T IO N O F T H IS PA G EfW hen D a ta En tered )
PREFACE
The study reported herein was conducted in the Structures Laboratory (SL), U. S. Army Engineer Waterways Experiment Station (WES), under the sponsorship of the Office, Chief of Engineers (OCE), U. S. Army, as a part of Civil Works Investigation Work Unit 31623. Mr. Donald R. Dressier of the Structures Branch, Engineering Division, OCE, served as technical monitor.
This study was conducted during the period October 1978 to September 1979 under the general supervision of Mr. Bryant Mather, Acting Chief, SL; Mr. John Scanlon, Chief, Engineering Mechanics Division, SL; and Mr. James E. McDonald, Chief, Structures Branch, SL. This study was conducted and this report was prepared by Dr. Tony C. Liu, SL.
The assistance and cooperation of many persons were instrumental in the successful completion of this study. The author wishes to acknowledge Mr. Donald R. Dressier, OCE; Professor Phil M. Ferguson, University of Texas at Austin; Mr. Ervell A. Staab, Missouri River Division; Mr. Chester F. Berryhill, Southwestern Division; Mr. V. M. Agostinelli, Lower Mississippi Valley Division; Mr. Garland E. Young, Fort Worth District; Mr. Marion M. Harter, Kansas City District; Mr. George Henson, Tulsa District; and Dr. Paul Mlakar, Dr. N. Radhakrishnan, and Mr. William A. Price, WES, for their critical review of the manuscript.
The Commanders and Directors of the WES during this study and the preparation and publication of this report were COL John L. Cannon, CE, and COL Nelson P. Conover, CE. The Technical Director was Mr. F. R. Brown.
1
CONTENTSPage
PREFACE .......................................................... 1CONVERSION FACTORS, INCH-POUND TO METRIC (Si)UNITS OF MEASUREMENT.................. . . . .................. 3
PART I: INTRODUCTION . . . . . . . . . .......................... h
Background ...................... . .......................... bObjective............... h
PART II: STRENGTH DESIGN CRITERIA............. 6Introduction ...................................... . . . . . 6Strength Requirements ...................................... 6Serviceability Requirements ...................... . . . . . 8Flexure and Axial Loads .................................... 9Shear Strength Requirements.................................. 11
PART III: COMMENTARY.............................................. 13Strength Requirements........................ 13Serviceability Requirements .................... 18Flexure and Axial L o a d s .................... 20Shear Strength Requirements.................................. 22
REFERENCES...................................................... 2 6APPENDIX A: DESIGN EXAMPLES.................................... A1
Design of Retaining W a l l s .................... A1Design of Members Subject to Combined Flexure andAxial L o a d .............................................. . A15
Shear Design................................................. A25APPENDIX B: COMPARISON OF DESIGN CRITERIA FOR HYDRAULIC
STRUCTURES AND ACI 318-77 FOR BUILDINGS.............. B1APPENDIX C: DERIVATION OF LOAD FACTORS AND STRENGTH
REDUCTION FACTORS USING PROBABILISTIC APPROACH . . . . ClIntroduction ................................................ ClBasic Theory............................... ClChoice of Acceptable Probability of Failure ................ C5Derivation of Load Factors and Strength Reduction
F a c t o r s ................. C7Discussion........................................ .. • • • • Cl8Shortcomings of First-Order Probabilistic Procedure ........ Cl8Recommended Future Studies .................................. C19
APPENDIX D: DERIVATION OF MAXIMUM TENSION REINFORCEMENTRATIO FOR FLEXURAL MEMBERS............................ DL
2
CONVERSION FACTORS, INCH-POUND TO METRIC (Si) UNITS OF MEASUREMENT
Inch-pound units of measurement used in this report to metric (Si) units as follows:
Multiplyfeetincheskips (force)kips (force) per footkips (force) per inchkips (force) per square footkips (force) per square inchkip (force)-feetkip (force)-inchespounds (force)pounds (force) per square inch pounds (mass) per cubic foot
_______ Bjr_______0.3048
25.44448.222 1459-3904
175126.8 1+7 - 88026
6.894757 1355.818
112.9848 4.448222
0.006894757 16.01846
square inches 6.U516
can be converted
To Obtain metres millimetres newtonsnewtons per metrenewtons per metrekilopascalsmegapascalsnewton-metresnewt on-met resnewtonsmegapascalskilograms per cubic
metresquare centimetres
3
STRENGTH DESIGN OF REINFORCED CONCRETEHYDRAULIC STRUCTURES
PRELIMINARY STRENGTH DESIGN CRITERIA
PART I: INTRODUCTION
Background
1. Since 1963 the structural engineering profession has been gradually adopting the strength design (SD) approach in lieu of the working- stress method that is the basis of current Engineer Manual EM 1110-1-2101, "Working Stresses for Structural Design" (Office, Chief of Engineers 1963). This EM permits the use of the SD method, but does not provide adequate guidance for proportioning structural members for strength and service requirements. The SD criteria for reinforced concrete hydraulic structures (RCHS) need to be developed for several reasons:
si. The American Concrete Institute (ACl) Building* Code and the American Association of State Highway and Transportation Officials (AASHTO) Bridge Code are not directly applicable to hydraulic structures.
b. The SD approach is more realistic and is potentially capable of producing more economical structures without compromising safety requirements (Winter and Nilson 1972).Structural engineering research in the United States and abroad will only be updated in terms of the SD approach.
d. Engineering schools are only emphasizing the SD approach in their design courses; recent and future engineering graduates will not be familiar with the working-stress method.
2. Consequently, a comprehensive study was initiated at WES in 1978 to develop a realistic SD methodology for RCHS and to devise an accurate and efficient design procedure for implementing these SD methods.
Objective3. The purpose of the first phase of study was to develop general
k
SD criteria that will yield designs (i.e. member dimensions and reinforcements) equivalent to those designed by the working-stress method for RCHS. The results of this first phase study are reported herein.The second phase of this study, which will be initiated in FY 80, will develop a realistic SD methodology and practical procedure that accounts for the special loading and service characteristics of particular RCHS.
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PART II: STRENGTH DESIGN CRITERIA
Introduction
U. This part defines the strength design criteria for reinforced concrete hydraulic structures. The considerations and background information used in developing these criteria are given in Part III.* Examples for designing typical hydraulic structural members using the strength design method are given in Appendix A.
5. A hydraulic structure is defined as a structure that will be subjected to submergence, wave action, spray, chemically contaminated atmosphere, and severe climatic conditions (OCE 1963). Typical hydraulic structures are stilling basin slabs and walls, concrete-lined channels, portions of powerhouses, spillway piers, spray walls and training walls, flood walls, intake and outlet structures below maximum high water and wave action, lock walls, guide and guard walls, and retaining walls subject to contact with water (OCE 1963)*
6. Reinforced concrete hydraulic structures may be designed with the strength design method in accordance with the current "Building Code Requirements for Reinforced Concrete," ACI 318-77** (ACI 1977a) except as hereinafter specified. The notations used are the same as those used in ACI 318-77* except those defined in this report.
Strength RequirementsRequired strength
7. Reinforced concrete hydraulic structures and structural members shall be designed to have design strengths at all sections at least equal to required strengths calculated for the factored loads and forces in the following combinations that are applicable:+
* Part III is a commentary on the criteria.** Hereinafter referred to as ACI 318-77.+ Both the full and zero values of L in Equations 3 and b shall be considered to determine the more severe condition.
6
(1)U = 1.5D + 1.9 (L + H + H + F + F + F )w p w p u
U = 0.9D + 1.9 (H + H + F + F + F ) (2)v p w p u
U = 0.75 [l.5D + 1.9 (L + H + H + F + F• w p w p(3)
+ F + p + W + T)1 u -1
U = 0.75 [l.5D + 1.9 (L + H + H + F + F + F + PL w p w p u(1»)
+ E + T)]
whereH = earth mass, or related internal moments and forces wH = lateral earth pressure, or related internal moments and forcesF = water mass, or related moments and forces wF = lateral water pressure or related internal moments and forces F = vertical uplift pressure or related internal moments and forces P = additional pressure due to wave action
Base reactions for hydraulic structures
8. The factored base reactions for most hydraulic structures shall be approximated by applying a load factor of 1.5 to 1.9 to the base reactions obtained from a stability analysis for unfactored normal load cases. Factored base reactions for abnormal load cases shall be approximated by applying reduced load factors to the base reactions from appropriate stability analyses, consistent with paragraph 7.Design strength for reinforcement
9. Design shall be based on a 1+0,000-psi yield strength of reinforcement for Grades 1+0 and 60 steel (American Society for Testing and Materials 1978). The reinforcement with yield strength in excess of Grade 60 shall not be used, except for prestressing tendons.
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Serviceability Requirements
Distribution of flexural reinforcement
10. For reinforced concrete hydraulic structures, spacing of flexural tension reinforcement shall not generally exceed 12 in.*
11. The spacing of flexural tension reinforcement exceeding the limit specified in paragraph 9* but less than 18 in. may be used if it can be justified. In no case shall the flexural tension reinforcement spacing exceed 18 in.Shrinkage and temperature reinforcement
12. Area and spacing of shrinkage and temperature reinforcement shall be in accordance with the requirements specified in EM 1110-2- 2103 (OCE 1971)5 or in appropriate Engineering Manuals for specific structures.Details of reinforcement
13. Bending and splicing of reinforcement, minimum reinforcement spacing, and minimum concrete cover for principal reinforcement shall be in accordance with the requirements specified in EM 1110-2-2103 (OCE 1971).Control of deflections
lH. Deflections at service loads need not be computed if the limits of the reinforcement ratio specified in paragraph 16 are not exceeded.
15. For reinforcement ratios exceeding the limits specified in paragraph 16, deflections shall be computed in accordance with Section 9*5 of ACI 318-775 or other methods that predict deflections in substantial agreement with the results of comprehensive tests.
* A table of factors for converting inch-pound units of measurement to metric (Si) units is given page 3.
8
Flexure and Axial Loads
Maximum tension rein- forcement of flexural members
l6. For flexural members, and for members subjected to combinedflexure and compressive axial load when the design axial load strength<f>P is less than the smaller of either 0.10 f’A or d>P, , the ratio n c g Y bof tension reinforcement p provided shall not generally exceed
IT. Reinforcement ratios exceeding the limits specified in para-
shown to significantly affect the operational characteristics of the structure,
18. Reinforcement ratios in excess of 0.50 shall not be usedbunless a detailed investigation (e.g. laboratory testing, linear or nonlinear finite element analyses) of serviceability requirements, including computation of deflections, is conducted in consultation with higher authority.
Minimum reinforcement of flexural members
19. At any section of a flexural member where tension reinforcement is required by analysis, the minimum reinforcement requirements specified in ACI 318-77 shall apply except that the f shall be in accordance with paragraph 9•Combined flexure and axial load
20. The design axial load strength (JjP^ of compression member shall not be taken greater than the following:
* The reinforcement ratio for conduits or culverts, designed inaccordance with EM 1110-2-2902 (0CE 1969), shall not generallyexceed 0.375 P-. • b
0.25 Pb .*
graph l6 but less than 0.50 p^ may be used if deflections are not
+c f P )y g /
(5)
9
where pgarea9 A
is the ratio of area of reinforcement to the gross concrete
21. The strength of a cross section is controlled by compression if the factored axial load has an eccentricity e no greater thanthat given by Equation 6* and by tension if e exceeds this value.
_ p’mCa - a*) + o.i aeb (p ' - p)m + 0 . 6
(6)
where
fm * o Ifc22. Sections controlled by compression shall be proportioned by
Equation J.*
1 (T)
whereP B = 0.65 (l + P m)f'A A \ Pg ) c gMp = 0.11 f^bh2
23. For sections controlled by tension, the moment strength <f>Mnshall be considered to vary linearly with the axial load strength 4>P from <J>Mq (when the section is in pure flexure) to (when theaxial load strength is equal to (|>P ); and <f>P shall be determined from e. and Equation 7); <f>M from Equation 8. b o
<j»M = <(»pf bd o y - 0.59 P (8)
* For conduits or cubverts, Mp = O.lU f Hah2
10
Shear Strength Requirements
2k. The nominal shear strength V provided by concrete shall beccomputed in accordance with ACI 318-77 except as modified by paragraphs 26-28.
25. Provisions of paragraphs 26 and 27 shall apply to straight members as follows:
a. Members with an ultimate shear strength limited to the load capacity that causes formation of the first inclined crack.
b. Members with beams or frames having rigid, continuous joints or corners.
c_. Members subjected to uniformly distributed loads, or loads closely approximating this condition.
jd. Members subjected to internal shear, flexure, and axial compression, but not axial tension.
e,. Members with rectangular cross-sectional shapes.jf. Members with % /d less than 10 and f! not more than
6000 psi. n c£,. All straight members designed in accordance with para
graphs 26 and 27 shall include the stiffening effects of wide supports and haunches in determining moments, shears, and member properties.
h. The reinforcing details for all straight members designed in accordance with paragraphs 26 and 27 shall meet the following requirements:(1) Straight, full-length reinforcement shall be used.
Flexural reinforcement shall not be terminated even though it is no longer a theoretical requirement.
(2) Reinforcement in the exterior face shall be bent around corners and shall have a vertical lap splice in a region of compressive stress.
(3) Reinforcement in the interior face shall extend into and through the supports.
i. Shear strength for straight members shall not be takengreater than 20/F*" bd when £ /d is between 2 and 10 c n
11
26. At a distance 0.15 straight members with £^/d
£ from the face of the support, nbetween 2 and 6
for
Vc (10)
27. At points of contraflexure, for straight members with £n/d between 6 and 10
Vc 11,000( 0 . 0 1 + 6 + p ) ^ L 2 +
f c1,000 bd (11)
The length V is the distance between the points of contraflexure.28. At points of maximum shear, for uniformly loaded curved cast-
in-place members with R/d > 2.25 where R is the radius of curvature to the centerline of the member
V = c
but the shear strength shall not exceed 1 0 bd .29. Shear strength based on the results of detailed model tests
approved by higher authority shall be considered a valid extension of the provisions in paragraphs 26-28.
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PART III: COMMENTARY
30. This part discusses some of the considerations and background information used in developing the strength design criteria contained in Part II.
31• As an alternative to the working-stress design method for reinforced concrete hydraulic structures, the design provisions of ACI 318-77 (ACI 1977) are generally applicable. However, because of the unique strength and serviceability requirements of the RCHS, some design criteria of ACI 318-77 need to be modified. The specific design criteria that are different from those of ACI 318-77 are given in Part II. A comparison of the design criteria for RCHS and ACI 318-77 is presented, in Appendix B. Most of the notations used in Part II are the same as those used in ACI 318-77 and therefore are not defined. Special notations that are not used in the ACI 318-77 have been defined in Part II.
Strength Requirements
Required strength32. The dead load factor of ACI 318-77 is increased from l.k to
1.5 to account for the greater load uncertainties for hydraulic structures. The live load factor is increased from 1.7 to 1.9* The basis for this modification is discussed in subsequent paragraphs.
33. Considering the loading combination of live and dead loads, the required capacity of a flexural member shall be at least equal to
Mu - V d + h \ (13)
whereMu*D“ d*L
required factored moment capacitydead load factorbending moment due to dead loadlive load factorbending moment due to live load
13
3^. Since designs using the working-stress design method specified in EM 1110-1-2101 (OCE 1963) are satisfactory, the strength design method should yield equivalent designs* Therefore
M = (L )M (lb)U r W
whereL_ = overall load factor FM = required moment capacity derived from working-stress method w
Substituting Equation lU into Equation 13
(V Mv - V d * ( 15)
35» Recent studies indicated that the overall load factor is approximately 1.9 for p < 0.25 P- and <J> = 0.90 (Figures 1 and 2). Substituting 1.9 for L into Equation 15r
!-9 Mw = jy jp + ^
Since M wthen
Md + Ml
1.9 = K ^“ d + **L
*
“d
(16)
(IT)
Let
D**L
a
lU
LOAD
FAC
TOR
L_
L0AD
FAC
TOR
L
STEEL RATIO p
Figure 1 f =
y
Overall load *±0,000 psi and
factor-steel ratio,f f = 3,000 psi c
Figure 2. Overall load factor-steel ratio, f = *±0,000 psi and f^ = *±,000 psi
y ^
15
and
1.9 = k d — + k l 1 + a (18)
or
Kj = 1.9 (1 + a) - oKp (19)
Since
Kd = 1.5
therefore
= 1 . 9 + 0.1m (20)
It can he seen for a given that the value of is not a constanthut a function of a (Figure 3).
36. For many hydraulic structures, the dead load of the concrete structure is much smaller than the live load: a << 1 . To he consistentwith the concept of constant live load factor used in ACI 318-77, the value a is assumed to he zero; and therefore
^ = 1.9 (21)
In special cases where the dead load effects are significant, a larger shall he used, and in these cases should he determined from
Equation 20.37* In paragraph 7, a load factor of 1.9 was chosen for all load
ings due to lateral earth pressure, earth weight, lateral water pressure, vertical uplift pressure, water weight, wave pressure, wind loads,
l6
Figure 3. Relation between *L and adCp = 1.5)earthquake effects, and the effects of temperature, creep, shrinkage, and differential settlement. Load factors, depending upon the degree of uncertainty of loads considered can he determined using a probabilistic approach (Appendix C) and will be refined during future phases of this study.
38. The reduction factor of 0.75 used in Equations 3 and b is consistent with the provision of increasing the allowable stresses by 33-1/3 percent for Group II loadings (OCE 1963) for the working-stress designs.Base reactions for hydraulic structures
39. Theoretically, the base reactions for hydraulic structures should be evaluated from the factored loads such as earth pressures, water pressures, and structural weights. However, since the bearing pressures are caused by loadings with different load factors, such a procedure will cause relocation of the resulting eccentricities and lead to base reactions that are in principle different from those obtained
17
under service load conditions. For the purposes of design, paragraph 8 specifies that a load factor between 1.5 and 1.9 be applied to the base reactions obtained from the investigation of the service load conditions. A weighted average method, as given in Equation 22, may be used for determining the load factor for base reaction for strength design:
(22)
factor for bearing pressure factor for dead load factor for earth weight factor for lateral earth pressure factor for water weight factor for lateral water pressure factor for uplift pressure ute value of the load considered
Justification for selecting an appropriate load factor should be submitted to the higher authority.Design strength for reinforcement
UO. Paragraph 9 limits the yield strength of reinforcement for design to H0,000 psi. This limit is more restrictive than ACI 318-77, which placed the upper limit at 80,000 psi. Since the maximum width of crack due to load is approximately proportional to stress in reinforcement, this limit is necessary to minimize the crack width in the hydraulic structures.
where= load = load = load
% = lo a d
Kpv = lo a dK^p = loadKpu = load I I = absol
Serviceability Requirements
Distribution of flexural reinforcement
hi. Control of cracking in RCHS is particularly important. RCHS
18
designed by the working-stress method have low concrete and steel stresses and have served their intended functions with very limited flexural cracking.
1+2. The ACI 318-77 criteria for distribution of flexural reinforcement in beams and one-way slabs is based on the results of small beam tests (Gergely and Lutz 1968). The present ACI criteria may not be applicable for large RCHS members.
1+3. Extensive investigation of the cracking phenomenon in reinforced concrete structures revealed that smaller bar sizes placed at closer spacings are more effective in controlling flexural cracking than a few larger bars of equivalent area placed at wider spacing between bars (ACI 1972). The best crack control is obtained when the reinforcement is well distributed over the zone of maximum concrete tension. The provisions of paragraph 10 limiting the bar spacing to 12 in. are empirical but have been used satisfactorily for hydraulic structures.Shrinkage and temperature reinforcement
1+1+. Volume changes in concrete due to drying shrinkage, heat of hydration of cement, and seasonal variations in temperature in restrained members can produce excessive tensile stresses and thus cause wide cracks. Therefore, shrinkage and temperature reinforcement is required at right angles to the principal reinforcement to prevent excessive cracking.
1+5. EM 1110-2-2103 (OCE 1971) classifies three degrees of restraint for concrete hydraulic structural members: unrestrained, restrained,or partially restrained. The areas of reinforcement required vary from 0.2 percent to 0.1+ percent depending on the degree of restraint. The shrinkage and temperature reinforcement requirements for conduits or culverts given in EM 1110-2-2902 (OCE 1969) shall be satisfied.Details of reinforcement
1+6. According to EM 1110-2-2103 (OCE 1971)» the bending and splicing of reinforcement shall conform to ACI 318-77.
1+7• The maximum size of coarse aggregate used in hydraulic structures is generally larger than that used in building structures.
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Therefore, the bar spacing requirements specified in EM 1110-2-2103 (OCE 1971) are generally larger than those of ACI 318-77-
U8. Thick concrete covers (from k to 6 in.) for hydraulic structures are generally required not only for protection of reinforcement against corrosion but also for erosion protection against abrasion and cavitation.Control of deflections
U9 . The members designed by strength-design methods are oftenmore slender than those designed by working-stress methods. Reasons forthis include (a) the use of large steel ratios and (b) the use of high-strength steel and concrete. Because the deflection of slender membersmay in some cases exceed desirable limits, paragraph 15 specifies thatdeflection shall be checked when the reinforcement ratio exceeds 0.25P,b(or 0.375P-, for conduits or culverts), b
Flexure and Axial Loads
Maximum reinforcement of flexural members
50. The limitation on the amount of tensile reinforcement that may be used in a flexural member is to ensure that flexural members designed by the strength criteria will have ductile behavior.
51. The maximum tension reinforcement of flexural members for hydraulic structures is limited to 0.25p, and to 0.375P-U ^or conduitsb bor culverts. The derivation of this limitation is given in Appendix D.This limit is much smaller than 0.75P-. allowed in ACI 318-77- Theblow reinforcement limit will result in a stiffer structure, which is considered desirable for hydraulic structures.
52. For unusual situations when a larger amount of reinforcement is necessary because of physical constraints on member dimensions due to functional or esthetic requirements, paragraph 17 limits the reinforcement ratio to 0.50p^ provided that the deflection criteria specified in paragraphs ih and 15 are met.
20
Minimum reinforce- ment of flexural members
53. This provision applies to members that for functional or other reasons are much larger in cross section than required by strength consideration. The computed moment strength as a reinforced concrete section -with very little tensile reinforcement becomes less than that of the corresponding plain concrete section computed from its modulus of rupture. Failure in such a case could be quite sudden. A minimum percentage of reinforcement should be provided to prevent such a mode of failure•Combined flexural and axial load
5^. The strength design procedure for members subjected to combined flexure and axial load can be summarized in Figure k .
Figure U. Typical interaction diagram
21
55* For zero or small eccentricities (interval ac ), the strength of a section is that for concentric compression (Equation 5). Equation 5 is derived to yield ^ n(max) approximately 1.9 times the maximum allowable axial load calculated by the working-stress method using Section lU03 of ACI 318-63. The calculated by Equation 5 isabout .10 percent less than that computed by Equation (10-2) of ACI318-77.
56. For moderate eccentricities, when compression governs in theinterval cb , the interaction Equation 7 applies. It is representedby the straight line connecting P^ (when M = 0) and (when P = 0).In that range, one determines <|>P and <|>M from Equation 7 and com-n npares (¡>P with <f>P , % as calculated from Equation 5- The smallern T n(max)of the two is the design strength.
57. Point b determines the boundary between members governed by compression and those governed by tension. It can be determined by calculating e^ from Equation 6 and then (j)P from Equation 7.
58. For large eccentricities, when tension governs, a linearvariation is assumed between the moment (j)M at the balance point band d>M for simple flexure, shown by straight line bd . o
59« It can be seen from Figure b that the general shape of the figure acbd is very similar to the interaction diagram for working- stress design specified in ACI 318-63 (ACI 1963). The similarity of the figures comes from the fact that the SD criteria are devised to provide comparable design with the conventional working-stress design methods.
Shear Strength Requirements
Shear strength provided by concrete60. In general, for retaining walls or flood walls, the shall
be computed in accordance with Section 11.3.2 of ACI 318-77- The maximum shear for horseshoe-shaped conduits or culverts will generally occur in the base slab near the walls at a point of high moment, shear, and axial thrust. In this case, the shall also be computed in accordance with the provisions of Section 11.3.2 of ACI 318-77« For thick
22
horseshoe-shaped conduits or culverts with £ /d less than 5, thenspecial provisions for deep flexural members given in Section 11.8 of ACI 318-77 may be used, if appropriate. Paragraphs 26 and 27 shall apply.
61. The walls of an intake structure generally have high uniform loads caused by either water pressure or earth embankment fill that result in high axial compression thrusts on the members. Section 11.3 of ACI 318-77 shall be used to compute . When the structural conditions of intake structure walls are clearly similar to the box conduits investigated in the laboratory (Diaz de Cossio and Siess 1969 and Ruzickaet al. 1976), Equations 10 and 11 may be used. The special provisions for deep flexural members given in ACI 318-77 may apply to some intake structure walls; however, since there is no provision for axial loads, the results will usually be too conservative. Paragraph 28 is generally applicable to uniformly loaded circular or oblong conduits or culverts.
62. For hydraulic structures, the factored shear force willnot generally exceed the shear strength provided by concrete ; andtherefore, shear reinforcement is not normally required. However, incertain areas where V does exceed d>V , shear reinforcement shallu Y cbe provided in accordance with Section 11.5.6 of ACI 318-77*
63. The limitations for using Equations 10 and 11 are specified in paragraph 25. These requirements are necessary to ensure that the actual structural performance is consistent with the modes of failure observed in comprehensive tests performed at the University of Illinois by Diazde Cossio and Siess (1969), Gamble (1977), and Ruzicka (Ruzicka et al. 1976). Special attention should be given to the following:
a. In frames of normal proportions, the supporting member can be idealized as a line structure in determining moments and shears. However, as the widths of the supporting members become larger relative to the span (e.g., in a thick-walled multiple opening conduit), it becomes necessary to take these widths into account if the distribution of moments is to be adequately predicted. As the supports become wider, one must also reconsider the definition of the span (clear span or center-to-center span), and the effects of the growing joint areas on the flexural stiffness factors. The recommended idealization of thick box
23
con d u its w ith and w ithout haunches can he found in Diaz de C ossio (1969) and Gamble (1 9 7 7 ).
b . The r e in fo r c in g d e t a i l s , in c lu d in g th e le n g th s o f b a r s , th e lo c a t io n s o f b ar te rm in a tio n s , and th e lo c a t io n s o f bends should be con sid ered both to ensure th e adequate beh av ior o f th e s t ru c tu r e and to minimize m a te r ia l and f a b r ic a t io n c o s t s * The o u ts id e re in forcem en t should be bent around th e corn ers in a th ic k -w a lle d m u ltip le opening co n d u it , and th e v e r t i c a l b a r s should then be la p - s p l ic e d in a c co rdance w ith ACI 318-77. A ll o th er b a r s a re s t r a ig h t and f u l l le n g th . No b a r s should be cut o f f sh o r t , even though in some c a se s th e shape o f th e bending moment diagram s might appear to a llo w t h i s to be done. The e x te n s io n s o f th e in s id e b a r s (bottom b a r s in top member, top b a r s in bottom member, and in s id e b a r s in e x te r io r v e r t i c a l memb e r s ) in to and n e a r ly through th e su p p o rts are recommended.
6k . The bent b a r s have been used in some box c o n d u its , but th e se
become q u ite in e f f i c i e n t and in e f f e c t iv e when members w ith sm all i^ /d
v a lu e s are c o n sid e red . The u se o f a l l s t r a ig h t b a r s ra th e r than bent
b a r s i s recommended fo r th ic k -w a lle d co n d u its .
65 . For th ic k -w a lle d re in fo rc e d con crete box con d u its or c u lv e r t s ,
th e low er bound o f th e nominal sh ear s tre n g th g iven in E quation 10 i s
ob ta in ed from Ruzicka e t a l . (1 9 7 6 ). Equation 10 i s only a p p lic a b le
when th e te c h n ic a l requ irem ents s p e c i f ie d in p aragraph 2k a re met.
66. Equation 11 i s ob ta in ed from Diaz de C o ssio and S i e s s ( 1969) ,
and i s g e n e ra lly a p p lic a b le to box con d u its o r c u lv e r t s w ith iW d 'be-
tween 6 and 10* The requ irem ents s p e c i f ie d in paragraph 25 should a ls o
be met. For box conduit se c t io n s o f o rd in ary d im en sion s, th e c r i t i c a l
member fo r sh ear i s u su a l ly th e h o r iz o n ta l member.
67 . E quation 12 i s d eriv ed b ased on a p r in c ip a l s t r e s s a n a ly s is o f
an e l a s t i c member su b je c te d to a x ia l com pression and sh e a r , and in
which i t i s assumed th a t f a i lu r e w i l l occur when the p r in c ip a l t e n s i l e
s t r e s s reach es a l im it in g v a lu e o f • E quation 12 i s only a p p l ic a
b le a t p o in ts o f zero bending moment. The nominal sh ear s tre n g th fo r
o th er s e c t io n s su b je c te d to combined sh e a r , f le x u r e , and a x i a l th r u s t
should be computed in accordance w ith th e a p p ro p r ia te se c t io n s o f Chap
t e r 11 o f ACI 318-77.
68. In g e n e ra l, th e R/d o f th e c i r c u la r o r oblong con d u its i s
2k
greater than 2.25* For unusually thick curved members with R/d ^ 2.25 ? the effects of member curvature on the stress distributions should be considered.
25
REFERENCES
American Concrete Institute. 1963. "Building Code Requirements for Reinforced Concrete," ACI Manual of Concrete Practice, ACI 318-63,Detroit, Mich.
1965. "Reinforced Concrete Design Handbook - Working Stress Method," Special Publication SP-3, Detroit, Mich.__________. 1972. "Control of Cracking in Concrete Structures,"Proceedings, Vol 69, No. 12, pp 717-753._________. 1977a. "Building Code Requirements for Reinforced Concrete,"ACI Manual of Concrete Practice, ACI 318-77, Detroit, Mich.__________. 1977b. "Commentary on Building Code Requirements for Reinforced Concrete," ACI Manual of Concrete Practice, ACI 318-77> Detroit, Mich.American Society for Testing and Materials. 1978. "Standard Specifications for Deformed and Plain Billet-Steel Bars for Concrete Reinforcement," Designation: A 615-78, 1978 Book of ASTM Standards, Part U,Philadelphia, Pa.Cornell, C. A. 1969* "A Probability-Based Structural Code," American Concrete Institute Proceedings, Vol 66, No. 12, pp 97^-985.Diaz de Cossio, R., and Siess, C. P. 1969. "Development of Design Criteria for Reinforced Concrete Box Culverts; Recommendations for Design," Civil Engineering Studies, Structural Research Series No. l6U, Part II, University of Illinois, Urbana, 111.Ellingwood, B. 1978. "Reliability Basis of Load and Resistance Factors for Reinforced Concrete Design," NBS Building Science Series 110,National Bureau of Standards, Washington, D. C.Ferguson, P. M. 1965« Reinforced Concrete Fundamentals with Emphasis on Ultimate Strength, 2d ed., Wiley, New York.Gamble, W. L. 1977. "Design of Thick-Walled Multiple Opening Conduits to Resist Large Distributed Loads," Civil Engineering Studies, Structural Research Series No. b k 0 9 University of Illinois, Urbana, 111.Gergely, P., and Lutz, L. A. 1968. "Maximum Crack Width in Reinforced Concrete Flexural Members," Cause» Mechanism, and Control of Cracking in Concrete, Special Publication SP-20, American Concrete Institute, Detroit, Mich., pp 1-17•Lind, N. C. 1971. "Consistent Partial Safety Factors," Journal, Structural Division, American Society of Civil Engineers, Vol 97, No. St 6,pp 1651-1670.MacGregor, J. G. 1976. "Safety and Limit States Design for Reinforced Concrete," Canadian Journal of Civil Engineering, Vol 3, No. k 9
pp U8U—513.
26
Office* Chief of Engineers* Department of the Army. 1963. "Working Stresses for Structural Design," Engineer Manual EM 1110-1-2101, Washington* D. C.__________. 1969* "Conduits* Culverts* and Pipes*" Engineer ManualEM 1110-2-2902* Washington* D. C.__________• 19T1* "Details of Reinforcement - Hydraulic Structures,"Engineer Manual EM 1110-2-2103* Washington, D. C.Ravindra* M. K., Heaney* A. C., and Lind* N. C. 1969« "Probabilistic Evaluation of Safety Factors*" Final Report* Symposium on Concepts of Safety of Structure and Methods of Design* International Association for Bridge and Structural Engineering* London, pp 35-1*6.Rosenblueth* E., and Esteva* L. 1972. "Reliability Basis for Some Mexican Codes," Probablistic Design of Reinforced Concrete Buildings, Special Publication SP-31, American Concrete Institute, pp 1-1*2.Ruzicka, G. C., et al. 1976. "Strength and Behavior of Thick Walled Reinforced Concrete Conduits," Civil Engineering Studies* Structural Research Series No. 1*22, UILU-ENG-76-2001*, University of Illinois, Urbana, 111.Siu* W. W. C.* Parioni* S. R.* and Lind* N. C. 1975* "Practical Approach to Code Calibration," Proceedings, American Society of Civil Engineers, Vol 101* No. ST 7, PP ll*69-ll*80.Winter* G., and Nilson* A. H. 1972. Design of Concrete Structures,8th ed.* McGraw-Hill, New York.
27
APPENDIX A: DESIGN EXAMPLES
Design of Retaining Walls
Design data1. Design information for retaining walls is presented in the
following paragraphs.¡a. Soil data.
Unit mass of sand (submerged) = 65*5 lb/cu ft Unit mass of clay (submerged) = 57*5 lb/cu ft Active earth pressure coefficient for sand = 0.33 Passive earth pressure coefficient for clay = 2.0^1
b. Material data.Concrete compressive strength, f¿ = 3,000 psi Yield strength of steel, f y = 1+0,000 psi
Loading diagrams and dimensions2. The critical loading diagrams and structural dimensions result
ing from stability analysis of service load condition are shown in Figure Al.Design of stem
3. Working-stress design. Referring to Figure A2, the moment at the base of the stem is
M = (P1 x 8.5) + (P2 x 8.5) - (P3 x 1.5)
whereP1 = “ X 0.0625
= 20.3 kips p2 = \ * 0.33 x
= 7.0 kipsP3 = 2 X ° ‘ o625
0.0655 x (25.5)2
x (25.5)2
x (U.5) 2= 0.6 kips
Thus
M = (20.3 x 8.5) + (T.O x 8.5) - (0.6 x 1.5)
= 231.2 kip-ft
Al
Figure Al. Critical loading diagrams and structural dimensions
A2
7777?4 1.5 F T
WATER ONLYFigure A2* Loading diagram for item design
U . The effective depth d required for a given moment can heI
calculated by
d =
for f ' c = 3,000 psi , f = 0.35 f* = 1,050 psi , and f = 20,000 psi
R = I52*
Thus
231,200 x 12152 x 12
= 39 in.
* Obtained from Ferguson 1965*
A3
26.5
FT
The required steel area can be computed by
AsM_ad
where a = for f = 20,000 psiTherefore
As231 >2
l.kb x 39
= H.ll sq in.
Use No. 11 at h-l/2 in. (Ag = b.l6 sq in.).
Check shear at d from top of base.
V = + 0.33 x 0.
= 20.8 kips
v V_bd
20,800 12 x 39 = kk.k psi < 1 .1 f^ = 60 psi
5. Strength design. The factored moment at the base of the stem is
M = 1.9M u= 1.9 x 231.2
= ^39-3 kip-ft
Ak
The effective depth d can he determined from the following equation
Mu ♦Pfy bel l - o.59P^rj
Use p = 0.25
= 0.25 0.853fc 87,000
1 fy y87,000 + f
= 0.25 O .85 x 0.85 x 87,000kO 1 8 7 ,0 0 0 + f
= 0.0093
Thus
U39.3 x 12 = 0.9 x 0.0093 x UO x 12 x d2 (l - 0.59 x 0.0093 x
d = 37-6 in.
Use d = 38 in.
A = pbds
= 0.0093 x 12 X 38
= b.2h sq in.
Use No. 11 at b-l/2 in. (As = 1+.16 sq in.)
Check shear at d from the base.
A5
Vu |l.9(0.0625 + 0.33 x 0
- [0.0625 (4.5 - f§f *
0655) X (25.5
12
= 39.8 kips
According to Equation (ll-3) of the ACI 318-77
Vc
= 2/3000 X 12 X 38
= 50 kips
<(>Vc = 0.85 x 50
1+2.5 kips > V u
Therefore, shear reinforcement is not required.Design of heel
6. Working-stress design. Referring to Figure A3, the moment at the face of the support* can be computed as
* Under most loadings, the critical heel moment is at the center line of the stem steel.
A 6
hI cm
hIcm
(29.u + 28.1 X + 0.525 x (1 6 .55)22
. ' 0 \ t S t-t- \ 16.55(2 X 1.8l + 0.98)(0.98 + 1 . 8 l ) ( I 6 . 5 5 ) X 3(1.81 + 0.98)
(2.8U)(15.72) x I x 15.72
1*75.8 + 71.9 - 210.0 - 117.0
220.7 kip-ftH =29.4 KIPS w
F =28.1 KIPS1116.55 F T
CONCRETE D - 0.15 X 3.5 - 0.525 KIPS
Figure A3. Loading diagram for heel design
AT
The & required
d =
d J220,700 x 12 152 x 12
d = 38.1 in.
Use d = 38 in.7. Using k-in. concrete cover, the total thickness of the heel is
T = 38 + U.5 = U2.5 in. = 3.5b ft
This is satisfactorily close to the assumed thickness of 3-5 ft used in the calculation of concrete weight.
As ad
- 220.7“ l.UU x 38
= U.03 sq in.
Use No. 11 at k-l/2 in. (A = U.l6 sq in.)s
Check shear at the face of the stem.
V = (29.U + 28.1) + (0.525 x 16.55)
- I (0.98 + 1 .8 1H 16 .55)
- | (2 .810(1 5.7 2)
= 20.8 kips
A8
V 20,800 _ ^v - M ' I SV 38 k5'6 psl * 60 psl
8. Strength design.*
M = 1.9 x 1*75.8 + 1.5 x 71.9 - 1.9 x 210.0 - 1.9 x 117.0 u
=_ 390.6 kip-ft
Mu = (f.pfybd |l - 0.59P^f
Use p = 0.25p, = 0.0093D
390.6 X 12 = 0.9 * 0.0093 X 1*0 X 12 X a2 X (1
a = 35.5 in.
0.59 x 0.0093 X
Use a = 36 in. and
A = pbd s
= 0.0093 x 12 x 36
= 1+.02 sq in.
Use No. 11 at U-l/2 in. (A = U.l6 sq in.)s
Check shear at the face of the stem.
* A load factor of 1.9 is used for factored base reactions.
A9
vu = i.9 (2 9.U + 28.1) + 1 .5(0 .5 2 5 x 16.55)
‘ 1.9 X | X (1.81 + 0.98) X 16.55
- 1.9 x | x (2.8U x 15.72)
= 36 kips
♦Vc = 0.85 2^fJ
= 0 .8 5 x 2 x ¿3000 x 12 x 36
= U0 . 2 kips > V uNo shear reinforcement is required.Design of toe
9 . Working-stress design. Referring to Figure AU, the moment at the face of the support is
M = | X 3 .0 6 X i X (5.7 5)2 + \ x 3.80 X | X (5 .7 5)2
= 58.7 kip-ft
Therefore
d =/¡T _ 158,700 x 12>Rb ~ y 152 x 12
d = 19.7 in.
Use a = 20 in.Using k-in. concrete cover, the total thickness requirea for the toe at the face of the support is
T = 20 + it. 5 = 2k . 5 in.
A10
0.54 KIPSEARTH + WATER
0.72 KIPS
+
+
0.79 KIPS1
UPLIFT i 0.50 KIPS
+
3.06 KIPS
3.80 KIPS
Figure Ak. Loading diagram for toe design - working-stress design
10. This thickness is only about 60 percent of the assumed thick
ness of b2 in. used in the calculation of concrete weight. A recalcula
tion of M and d may be necessary. However, in this design example,
the recalculation is not performed.
11. The steel area required can be determined by
a = ^s ad
58.71.1+U x 20
= 2.0k sq in.
Use No. 11 at 9 in. (A = 2.08 sq in.)s
Check shear at d from the face of the stem.
V = | (3.^5 + 3.80) X (5.75 - 3.0)
= 10 kips
V_ _ 10,000bd 12 x 20 Ul.7 psi < 60 psi
No shear reinforcement is necessary.
12. Strength design. Referring to Figure A5,* the factored moment
at the support is
Mu = | x 6.01 x i x (5.75)2 + \ x 7.3U x | x (5.75)2
= llU.O kip-ft
M = <J>pf bd2 u y 1 - 0.59P JL
fc
* A load factor of 1.9 is used for factored base reaction.
A12
1.03 KIPS EARTH + WATER1.37 KIPS
+
0.45 KIPS
0.95 KIPS
+
8.21 KIPS
n
Figure A5- Loading diagram for toe design - strength design
Use p = 0.25P, = 0.0093D
llU.O X 12 = 0.9 x 0.0093 X 1(0 X 12 X d2 | l - 0.59 x
d = 19.2 in.
Use d = 19.5 in.
A = pbd s
= 0.0093 x 12 x 19.5
= 2.l8 sq in.
Use No. 11 at 8-1/2 in. (A = 2.20 sq in.)S
Check shear at d from the face of the stem.
V « ! (6.70 + 7.3*0 x 2.75
=19.3 kips
4>vc = 0.85 x bd
= 0.85 x 2 x v/3000 x 12 x 19.5
= 21.8 kips > V u
No shear reinforcement is necessary.
AlU
0.0093 x
Summary13. The retaining wall design using both working-stress design
and strength design methods is summarized in the following tabulation. The depths of the sections designed by the strength design method are approximately 5 percent less than those designed by the working-stress design and the steel areas required by the strength design are approximately 3 percent more than those required by the working-stress design.
Working-StressDesign Strength Design
A Ad s d sArea in. sq in. in. sq in.Stem 39.0 1+.11 37.6 k.2k
Heel 38.1 i*.03 35.5 H.02
Toe 19. 7 2.0U 19.2 2.18
Design of Members Subject to CombinedFlexure and Axial !Load
lU. The following design information is given: au Overall depth of section* h = 2k in.bu Distance from extreme compression fiber to centroid of
tension reinforcement* d = 20 in.c_. Width of the section, b = 12 in.cl. Compressive strength of concrete, f^ = 3,000 psi£. Yield strength of reinforcement, f = U0,000 psi
Y.f. Ratio of area of tension reinforcement to effective area of concrete, p = 0.00769Ratio of area of reinforcement to the gross concrete area,p = 0.006Ug
h_. Plot interaction diagrams for (l) working stress design per ACI 318-63 (ACI 1963)* (2) strength design per ACI 318-77 (ACI 1977a), and (3) strength design per Part II
Working-stress design15. According to Equation lU-10 of ACI 318-63*
A15
P = 0.3^1 + p mV'Aa \ g / c g
= 0. 3b(l + 0.006U B78 ~ ) (3)(12)(2U)
= 323.3 kips
1 6 . The bending moment that could be permitted for bending alone,
S , = the section modulus of the transformed, uncracked section ut
= 1 1 5 .8 kip-ft17. According to Equation (lU-8) of ACI 318-63
= P'm(d - d') + O.ldeb (p* - p)m + 0.6
For this design example, p' = 0 , and therefore
O.ld
where
eb 0.6 - pm
0.1 x 200.6 - 0.00769 * 0
(0.85 x 3)
= It. 17 in.
Al6
The maximum allowable axial load PmaxSection lU03 of the ACT 318-63
can be calculated according to
Pmax = 0.85A (0.25f' + f P ) g c s g
= 0.85 x 12 x 2M0.25 x 3 + 0.1+ x 1+0 x 0 .006H)
= 208.7 kips
18. The allowable bending moment without axial load can be calcu
lated according to Equation (ll+-13) of the ACI 318-63:
M = 0.1+0A f jd o s y°
= 0.1+0 x 0.00769 x 12 x 20 x 1+0 x 0.891 x 20
= 526.2 kip-in.
= 1+3.9 kip-ft
Based on the above calculation, the interaction diagram can be plotted
in Figure A6.
Strength design, ACI 318-77
19. = 0. The design moment strength without axial load <j>Mo
can be determined by the following equation:
<J>Mo d>pf bdy
1 - 0 .59^
= 0 .9 x 0.00769 X l+o X 12 X 20"
x ^1 - 0.59 x 0.00769 x(8 bis)
= 121+8.1+ kip-in.
= 10l+ kip-ft
A17
KIPS
p
Figure A6. Interaction diagram
A18
20. <j>P = O.lOf^A^ = 86 .k kips. Referring to Figure AT, the depthof equivalent rectangular stress block a can he determined from the following equation:
P“ = 0.85f,ab - A f $ c s y
86 . 1+
0.7 = 0.85 x 3 x a x 12 - 0.00769 x 12 x 20 x 1+0
a = 6.1+5 in.
Figure AT. Stress-strain distribution for member subject to combined flexure and axial load
21. The eccentricity measured from the centroid of tension reinforcement e f can be determined from the following equation (see Figure AT).
A19
P e 1 u = 0 .8 5 f£ a b (d -
8 6 .1+e1O.T = 0 .8 5 x 3 x 6.1+5 x 12(20 -
e ! = 26.8 in .
The v a lu e fo r d>M can be determ ined from th e fo llo w in g : u
cJ>M /d - d >
e' =^ + l“^
06 8 - ^ + ( 20 ~ ^26,8 " 86.1+ + ( ■)<|>M = 135.1+ k ip - f t
22. B alanced co n d itio n . The b a lan ced lo a d stre n g th <fiP and
b alan ced moment s tre n g th (JiM can be computed by *
d)P = é (0 .8 5 f 'b a . - A f \ b T \ c b s y)
and
= + 0'®?*b l - r) * Vyfl - 0where
. = (— 8 li2 2 2 ___ V d% ( 87,000 + f P r
S u b s t itu t in g th e g iven d esig n in fo rm ation
* O btained from ACI (1977b).
A20
% = ( sT.ooo’ fto .ooo ) * ° ‘85 * 20
= 1 1 .6 5 in .
<(>Pb = 0 . 7 ( 0 . 8 5 x 3 x 12 x 1 1 .6 5 - 0 .00769 x 12 x 20 x 1+0)
= 197.9 kips
^ = 0 . 7 ^ 0 . 8 5 x 3 x 12 x 1 1 .6 5 - 1~ 2~ )
+ 0 .0 0 7 6 9 x 12 x 20 x 1+0 - 1+ J
= 1 9 5 ^ .3 kip-in.
= 162.9 kip-ft
2 3. <|>M = 0 . According to Equation (10-2 ) of ACI 318-77
4>P , \ = 0 .80(j) |0 .85f' /a - A A + f A , 1mmax) |_ c\ g sty y s t l
= 0.80 x 0.7^0.85 x 3(12 x 2l+ - 0.00769 x 12 x 20)
+ 0 .0 0 7 6 9 x 12 x 20 x l+o]
= ^50 kips
- 5?jf - ?62-5 ^PS
The interaction diagram for strength design according to ACI 318-77 is plotted in Figure A6.
A21
Strength design, hydraulic structures, f = 0.35f'_______________ c________ c_
2k. According to Equation 7
P B = 0.65/1 + P m\f'A A V g / c g
= 0.65(1 + 0.006k x -0■ '3) X 3 X 12
= 6l8,0 kips
and
O.llf'hh2c
= 0.11 X 3 x 12 X 2h2
= 2281 kip-in.
= 190.1 kip-ft
25. The eccentricity at balanced condition e_bfrom Equation 6
p!m(d - d*) + O.ld eb (p1 - p)m + 0.6
__ O.ld0.6 - pm
= U.17 in.
26. The <()P and can be determined from
n
A
4>Mn < 1
x 2k
can be determined
(6 bis)
e^ and Equation 7
(7 his)
A22
Substituting
<f>P = <{>P, n b
*Mn ' +pb <eb>
= lt.l7*P b
P^ = 6l8 kips
Mp = 190.1 x 12
into Equation 7
2281.2 kip-in.
(|)P = 290*2 kips
and
^ = 290.2 x U.17 -5- 12
= 100.8 kip-ft
The maximum design axial load strength <j>P should he computed innaccordance with Equation 5.
<f>Pn(max) 0.70*Ag(0.85r + f/ g )
= 396.0 kips(5 bis)
The design moment strength for bending alone <|>M can be computed by Equation 8
A23
(8, bis)
<J>M d>pf bdYK y “ 0.59P -£$•
= lOU kip-ft
27. Based on the above data, the interaction diagram for strength design of hydraulic structures is also plotted in Figure A6.
Strength design, hydraulic structures, f = O.U5f'__________ _____ c_________ç
28. It can be shown that P. , e, , <|>P , \ , and <f>M are theA b n(max) osame as those calculated in paragraphs 21+-27.
Mp, = O.lUfMDh2
2= 0.ll( X 3 X 12 X 2l)
= 2U1..9 k ip -f t
and d)P, can be calculated by b
<f>Pb e= 1
Substituting into the above equation
= 6l8 kips
e = U.17 in.
Mp = 2U1.9 x 12
= 2902.8 kip-in.
A2k
then
(|)P = 327.1+ kips
and therefore
4 ^ = cj>Pb e
= 113.8 kip-ft
29. The interaction diagram for hydraulic structures withf = O.l^f1 is shown in Figure A8. The interaction diagrams for ACI c c318-63 (WSD) (ACI 1963) and ACI 318-77 (ACI 1977a) are also plotted in Figure A8 for comparison.
Shear Design
Circular conduits30. The stress analysis of a typical circular conduit (Figure A9)
indicates that the maximum shear force and thrust at a section 1+5 deg from the crown (where moment is zero) are 81.25 kips and 162.5 kips, respectively, under a service load condition. Determine whether shear reinforcement is required. A 1+000-psi concrete is to be used. The thickness of the conduit as shown in Figure A6 is 1+8 in. and has a U-in. concrete cover.
31. Working-stress design. According to Equation (ih) of EM 1110-2-2902 (OCE 1969),
fc2
wheref^ = principal tensile stress fc = average compressive stress v = average shear stress
A25
KIP
SP
M, KIP-FT
Figure A8. Interaction diagram for hydraulic structureswith f = 0.^5^' c c
A2 6
LOAD = 17.55 KIPS/SQ FT
Figure A9. Circular conduit design
A27
HO
RIZ
. = 1
/3 V
ER
TIC
AL
LOA
D A
T EA
CH
POIN
T
f = 162,500 = 8 c 1+8 x 12 * P
v =
f, =
= 81,2501+3.5 x 12282.1
= 155.7 psi
- ♦ (155.7)"
= 69 psi < 2^f^ = 126.5 psi
Therefore, no shear reinforcement is required.32. Strength design. Since R/d = 11 x 12/U3.5 = 3.03 > 2.25 ,
the nominal shear strength provided by concrete, , shall be computedby Equation 12:
Vc 1 + bd (12 bis)
where= 162.5 x 1.9 = 308.8 kips
A = 12 x 1+8 = 576 sq in.§bd = 12 x 1+3.5 = 522 sq in.f* = 1+000 psi c
Thus,
V = 1+ A 000 c 1 +308,800576
1+Aooo X 522
= 233,227 lb = 233.2 kips
♦V = O .85 x 233.2
= 198.2 kips
A28
V = 1.9 x 81.25u= 15I+.I+ kips
Since V < 4>V , shear reinforcement is not required, u cBox culverts or conduits
33. For a typical rectangular one-cell reinforced concrete box culvert (Figure A10), the following design information for the horizontal member is given:
a. Clear span, £ = 10 ft— nb. Span between points of contraflexure, V = 9 ft
Uniform service load, to = 7.2 kips/ftcL Thrust, N = ih.Q kips_e. Effective depth, d = 2k in._f. Overall thickness, 28 in.£. Steel ratio, p = 0.005h. Concrete compressive strength, f^ = 1+000 psi
Design shear reinforcement for the horizontal member.3l+. Working-stress design. The unit shearing stress of box culvert
can be determined using Equation (15) of EM 1110-2-2902 (0CE 1969) •
vpc 11,000f!c
1+000
whereV
N/V£!/d
Pf1c
total shear at point of contraflexureo ) £ t2
7 = 32.1+ kips
Ik. 8/32.1+ = 0.U6 9 x 12/2U = 1+.5 0.005 1+000 psi
A29
= 2.
4 KI
PS/F
T<0= 7.2 KIPS/FT
Figure AIO. Rectangular one-cell reinforced concrete box culvert section
A30
Therefore,
vpc 11,000 (0.046 + 0.005)(12 + 0.46) I k000(19 + 4.5) >4000
= 297.4 psi
v V_ _ 32,400 bd 12 x 24 = 112.5 psi
The factor of safety in shear for 3 < £'/d < 5 is
FS = 1.25 + 4d
= 1.25 + 9 x 12 4 x 24
= 2.38
v_E£. =FS
297.4 _2.38 125.2 psi > V = 112.5 psi
Shear reinforcement is not required.35* Strength design. Since £ /d = 10/2 = 5 < 6 , Equation 10 is
applicable, and shear should be checked at 0.15£ from the face ofnthe support:
A31
a + bdVc 11•5-r)V^.
= ( 1 1 . 5 - 5 ) A 000I7
83.75 A 000
x 12 x 2l*
(10 bis)= 132.6 kips < 2.lfll.5 - = 2i+8*8 kips
<f>V = 0.85 x 132.6 c
= 112.7 kips
V at 0.15& from the face of the support is u n
V = - 0.15& )u \2 n/
= 1.9 x 7.2 - 0.15 x 10
= U7.9 kips < <|>Vc
Shear reinforcement is also not required at 0.15& from the face ofnthe support.Horseshoe-shaped conduits
36. The elastic analysis of a horseshoe-shaped conduit (Figure All) shows that the maximum shear occurs at Section 10. The moment, shear, and thrust at this section are 37 kips-ft, 19*7 kips, and 13 kips, respectively. The thickness of the Section 10 is 2.18 ft and the effective depth is 1.9k ft. Concrete having f^ = ^000 psi is to be used.
A3 2
SYMMETRY LINE
CROWN
CENTER
INVERT
Figure All. Horseshoe-shaped conduit
The steel ratio p is 0.01 . Determine whether shear reinforcement is required.
37. Working-stress design. According to Equation (B-3) of the ACI 318-77,
V = xff7 + 1300p ^ c y c M
A33
whereV = 19*T kips
va _ 19.T x 1.9U x 12 M 37 x 12
= 1.03p = 0.01
f* = UOOO psi c
v = AOOO + 1300 x 0.01 x 1.03C
= 76.6 psi
V 19,700I.9U X 12 X 12 = 70.5 psi < Vc
No shear reinforcement is required,38. Strength design. Since the maximum shear occurs at Section 10
where moment and thrust also exist, the provision of Section 11.3.2.2 of ACI 318-77 applies:
vc = + 2500p r - ) bd
and
M = M m u - NuUh - d
8From the given information
Mu = 37 x 1.9 = 70.3 kip-ft
N = 13 x 1.9 = 2k.7 kips u
A3k
Vu = 19.7 x 1.9 = 37.^ kips
and
Mm = 70.3 - 2U.7 r- — '
= k9.U kip-ft
V = (1 .9AÒÒO + 2500 X 0.01 ILiL-Liiii) x 12 X I.9U x 12
= 1+3.8 kips
*V = 0.85 x 1+3.8 c
= 37.2 kips
Since this value is almost equal to V (37.1+ kips) , the shear reinforcement is not considered necessary.Intake structure vail
39. For an intake structure wall (Figure A12), the following information is given:
a. Clear span, l = ik ft“ nb. Overall thickness, h = 36 in.c_. Effective depth, d = 33.5 in. = 2.79 ftcL. Moment at face of support, M = 2U9 kip-ftse. Thrust, N = ll6 kips.f. Shear at face of support, V = 8 3 kipss
Uniform load at compression face, a) = 11.8 kips/fth. Concrete compressive strength, f 1 = 3000 psi — cju Reinforcement ratio, p = 0.006
Determine if shear reinforcement is required.1+0. Working-stress method.
a. Check shear at d from the face of the support.
A35
WATER + EARTH LOAD
V i r 1T 11___ 3 ' H_____ 3w = 11.8 KIPS/FT
£ TOWER
3 - 0 " 14'-0" I1 3 '-0"4 '-0".3 /-0".3 '-0" 4 - 0 "
I■=fri
PLAN AT EL 1342.83
Figure A12. Intake structure wall
V , = V - wd d s
= 83 - 11.8 x 2.79
= 50.1 kips
For members subject to axial compression, vc may be computed according to Equation B-U of ACI 318-77.
A36
cV
where
N_ _ 116„000A 12 x 36 g
= 268.5 psi
v = 1.1(1 + 0.006 x 268.5)^3000 c
= 1 5 7 .3 psi
v 50»100 12 x 33.5 = 12U.6 psi < Vc 1 5 7 .3 psi
Shear reinforcement is not required.h. Check shear at points of contraflexure. Since the moment
at any point M along the wall can he computed hy x
2M = M + — ---V xx s 2 s
The point of contraflexure can he determined hy lettingM = 0 and solve for x. x
0 = 2U9 + _ 83x
x = k.3 ft
The span between points of contraflexure can becalculated by
V = l - 2x n
= ]A - 2 x U.3
= 5. ft
A37
The shear force at point of contraflexure V is
V = fa>&' 2
11.8 x 5.U2 31.9 kips
and
_ V_ = 31,900bd 12 x 33.5
= 79-^ psi
Ul. Since the structural conditions of this intake structure wallare similar to the box culverts» Equation (15) of EM 1110-2-2902 (OCE 1969) shall apply
vpc 11,000(0.0U6 + p
19 +c
kooo
whereB/V = = 3.6
V/i * 2 ^ 9 = 1-9l‘P = 0.006
f = 3000 c
pcv 11,000 (0.0U6 + 0.006)(12 + 3.6) f 3000 (19 + 1.9*0 yjkooo
= 369 psi
A38
The factor of safety in shear for £'/d < 3 is 2.0 according to EM 1110-2-2902 (OCE 1969).
h2. Strength design.a. Check shear at d from the face of the support. Since
the intake structure wall is subjected to moment, shear, and axial compression, the provisions of Section 11.3.2.2 of ACT 318-77 shall apply.
= 120.k kip-ft
<Va - x -9l-v s - “d)
= 1.9(83 - 11.8 x 2.79)
= 95 kips
N = 1.9 x 116 = 220.k kips u
According to Equation 11-7 of ACI 318-77
= 18U.5 Psi > T9.U Psi
= -2923.9 kip-ft
A39
Since is negative, Vc shall he computed by Equa-tion (11-8) of ACT 318-77
vc - 3 .5yftl bdN
1 + 5o o r
= 3 . 5/3000 x 12 x 33.5 220,^00 500 x 12 x 36
= 109*5 kips
(¡>V = 0.85 x 109.5 c
= 93 kips
This value is approximately 2 percent less than the at Section d from the face of the support and the design is considered acceptable without shear reinforcement.
b. Check shear at 0,15&n from the face of the support.From the given design information
Z n] n _ lbd “ 2.79 5.0 < 6
Therefore, Equation 10 is applicable:
V =
Nu
l u . 5 - f . W i l .5 Vi bd (10,. bis)
(11.5 - 5) /3000 Jl + ■ x 12 x 36
= 260.2 kips < 2.1
<j)V = 0.85 x 260.2 c= 221.2 kips
5/3000
l
“ •5 - t m bd = 323 kips
A^0
The factored shear force at 0.15&n from the face of the support can be computed:
V = 1.9 fv - o>(0.15* )1u L S n J
= 1.9(83 - 11.8 x 0.15 x lU)
= 110.6 kips
Since vu < <()VC , the shear reinforcement is not required at 0.15&n from the face of the support. Since £n/d = 5 and the intake structure wall is loaded at the compression face, the provisions for deep flexural members given in Section 11.8 of ACI 318-77 may be used. According to Equation (11-29) of ACI 318-77, the nominal shear strength, Vc at 0.15&n from the face of the support, can be computed by
Vc 3-5 " 2.5 “ 1* 1 >9 V ^ + 2500p
where
Mu
Vud
f1cPbd
and
1.9 M +su>(0.15£ )— r - V°
-.9 ^1.9 2^9 + 11.8(0.15 x iU)‘
15* )n
- 83(0.15 X Ik)
191. kip-ft 110.6 kips2.79 ft 3000 psi 0.006 12 in.33.5 in.
AUl
Therefore
M3.5 - 2.5 y j = 2.0 < 2.5
u
V dv c = 2.o(l.9>/F + 2500p ^ j b d
u
= 2 .0 1.9/3000 + 2500 X 0.006
X 110
= 103.1 kips
Comparison of this value with the Vc of 260.2 kips computed “by Equation.7 indicates that the VQ value computed by Equation (11-29) of ACI 318-77 is too conservative because the effects of axial loads on the shear strength of concrete are not considered by the ACI equation. Therefore, the ACI equation shall not be used if the deep flexural member is subjected to significant axial compression, in addition to moment and shear.
Ak2
APPENDIX B: COMPARISON OF DESIGN CRITERIAFOR HYDRAULIC STRUCTURES AND ACI 318-77 FOR BULIDINGS
B1
Table B1Comparison of Design Criteria for Hydraulic Structures
and ACT 318-77 for Buildings
__________ ItemRequired strength
wro
Design strength for reinforcement
Distribution of reinforcement
Hydraulic StructuresU = 1.5D + 1.9/l + H + HV w P
+ F + F + F \w p u JU = 0.9D + 1.9/h + H + F
V w P 1+ F + F \p uy
U = 0.75[l.5D + 1.9^L + + H
+ p + F + F + P + W + TÌw p u )]
U = 0.75 |l • 5D + 1.9^L + Hw +
+ F + F + F + P + E + Tw p u1+0,000 psi
’)]
12 in.
Shrinkage and temperature Per EM 1110-2-2103 (0CE 1971), reinforcement depending on degree of restraint
_____________ Buildings_______U = 1.1+D + 1.7L
U = 0.75 (l.l+D + 1.7L + 1.7W)
U = 0.9D + 1.3W
U = 0.75 (l.l+D + 1.7L + 1.9E)
U = 1.1+D + 1.7L + 1.7H
U = 1.1+D + 1.7L + 1.1+F
U = 0.75 (1.1+D + 1.1+T + 1.7L)
U = 1.1+ (D + T)
1+0,000 to 80,000 psi
Z = 175 kips/in. - Interior exposure
= ll+5 kips/in. - Exterior exposure
0.2$ for GRl+0 or 50 0.18$ for GR600.1 8$ x 60,000
fy
for fy
> 60,000
(Continued) (Sheet 1 of 3)
Table B1 (Continued)
__________ Item_______
Control of deflection
Maximum reinforcement ratio
Minimum reinforcement ratio
Maximum axial load strength
Combined flexure and axial load
Nominal shear strength provided by concrete
Hydraulic Structures Buildings
Deflections need not be computed if Check deflectionsp < 0.25pb
°.25Pb 0.75Pb
0.5 percent
(J)P X X = 0.70<t>A /o.85f1 + f p \ n(max) g\ c y g )
_ p'm(a - a*) + o.iaeb ( p' - p)m + 0.6
npa “f
Straight members with ¡¿q/d > 6
200fy
4>P / v = 0.80<t>lo.85f'/A - A A n(max) Y[_ c\ g sty
+ f A . 1 y stj
Based on stress and strain compatability
(0.0U6
V = c 11,000 bd None
(Continued)
(Sheet 2 of 3)
Table B1 (Concluded)
Item
bd-F-
_____ Hydraulic StructuresStraight members with 2 < Jl /d < 6
Curved cast-in-place members with R/d > 2.25
Buildings
None
None
(Sheet 3 of 3)
APPENDIX C: DERIVATION OF LOAD FACTORS AND STRENGTHREDUCTION FACTORS USING PROBABILISTIC APPROACH
Introduction
1. The purpose of this appendix is to show how the load factors and strength reduction factors can be determined in a rational manner, taking into account the inherent uncertainties of the resistances and load effects.
2. The "first-order" or "second-moment" probabilistic methoddeveloped by Cornell (1969), Lind (l97l)> Rosenblueth and Esteva (1972), and others (Ravindra, Heaney, and Lind 1969) was used. This is a simplified method that uses only two statistical parameters: meanvalues and the coefficient of variation of the relevant variables.
Basic Theory
3. If R and U represent the distributions of strengths and loads, respectively, any given structure will fail if U > R . Thus the probability of failure P^ is the probability that U > R , or
Pf = P [(H - U) < 0] (Cl)
or alternatively
Pf = p(| < l.o) (C2)
Since £n 1.0 = 0 , Equation C2 can be expressed as
Pf = P < 0] (C3)
b . In other words, the probability of failure can be expressed as the probability that £n(R/U) is less than zero. This probability is represented by the shaded area in Figure Cl* As shown, the distance of
Cl
I
Figure Cl. Definition of safety index
the mean of iln(R/U) , [£n(R/U)] , with respect to the origin canmconveniently he measured as a number £ times the standard deviaton a of £n(R/U) .
5. If the actual distribution of £n(R/U) were known, and if a value of the probability of failure could be agreed upon, a complete probability-based set of design criteria could be established. Unfortunately, so much information is not known* The distribution shape of each of the many variables (e.g. material, load) has an influence on the shape of the distribution of £n(R/U) . At best only the means and the standard deviations of the many variables involved in the make-up of the resistance and the load effect can be estimated. However, this information is enough to build a first-order approximate design criterion by stipulating the following design condition.
( C U )
C2
6. Equation can be simplified by using first-order probability theory as follows:
KL • KE £n( mumy
(C5)
and
2 „ (Ra 9,n{~9 in(|)
2a2 +
3 KDl3R °R +
m3U U
m
2 2°R °'~ + .R Um m
HU.2
(C 6)
Since a^/R = V_, and aTT/U = VTT , the coefficients of variation of R m R U m Uthe resistance R and the load effect U , Equation CH becomes
“0W v 2 + v 2R U (C7)
7. To separate the resistance and load terms, Lind (1971) proposed a linear approximation to the square root term in Equation C7
W + VU2 ‘ “(VR + Vu) (C8)
where a is a constant equal to 0.75- This approximation is good towithin +6 percent for the range 1/3 < V„/V < 3 . By this lineariza-i\ Ution, it is possible to write Equation C7 as
Rto ^ > gaVR + 3aVu
m(C9)
or
C3
+(CIO)
RmU ~m
(s«V> e x
Rearranging this gives
Rmf -BaVR>
> U — mg a V u
(Cll)
This resembles the current ACI 318-77 (ACI 1977a) format in that theaverage strength R^ is multiplied by a factor less than 1.0 and theaverage load U is multiplied by a factor greater than 1.0. However, mwhen the designer uses the code design equations and the specified strengths, he computes the design strength R rather than the mean strength R^ . Similarly, the design is based on values of specified U rather than the mean load U . Therefore, y^ and yTT are defined
Rm Ryr (C12)
and
U = Uyttm U (C13)
then Equation Cll can he rewritten as
-eqvr gaV.RYr (e > U U le U' (ClU)
or
R <J> _> U A (C15)
where <|> is the strength reduction factor and A is the load factor. Thus
-3otV(ci 6)
and
gaVUX = Y e u (C1T)
Before values of <|> and X can be derived, an appropriate level of safety defined by the safety index 3 must be defined, and y^ , V_ ,rv nYy , and V estimated. The choice of 3 and the calculation of these terms will be given in the following sections.
8. There are two ways in which guidance in selecting 3 can be obtained. Relationships given in Equation Cl6 and C17 can be used to calculate the value of 3 corresponding to the load factors and strength reduction factors in the current codes. If these 3 values and the related probabilities of failure are felt to be realistic, they can be used to derive new values of <|> and X for use in the new criteria.If not, more appropriate target values of 3 can be selected on the basis of the performance of the current code or engineering judgment.This approach is called "calibration" since the new criteria are calibrated or made to agree with a target established by a study of the old code. This assumes that the load factors in the old code have been developed over a long period and represent a good engineering estimate of the required safety. Siu, Parioni, and Lind (1975) used this technique to estimate the weighted average 3 values in ACI 318-77•
9. The shortcoming of calibration as the sole means of setting the value of 3 for new criteria is that the level of safety in the current code will generally vary widely from one type of member to
Choice of Acceptable Probability of Failure
Flexure 3=U.2, P^ 1.3 x 10 ^1 -7Tied columns 3 = 5*22, P 2 x io
XShear 3 = 3.6U, ~ 1.3 x 10
C5
anotherj as shorn by the values listed in the previous paragraph. A uniform value of 3 for all structural members is more desirable because it will provide a more consistent reliability for the new criteria.
10. An alternative to calibration is to select a probability of failure comparable to the risks people are prepared to accept in other activities. The risks involved in a number of activities are presented in Table Cl.
Table ClRisk of Death for Various Activities*
Activity__________Motorcycle racing Mountain climbing Mining SwimmingAutomobile travel Airplane travel Fire in buildings Poisoning LightningVaccinations and inoculations Structural collapse
during construction All others
Death Rate per PersonThose
Concerned—
5 * 10 Z 5 x io"rT x io:1 X 10
1 x 10
3 x 10"5
per year___________Total
Population
23.6x 10ZlX 10
x 10 x 10 x 10 x 10
-5-5-7-8
—62 x 10
* From MacGregor (1976).
This method has the advantage of obtaining a desired degree of reliability for all structural members. This approach is chosen herein.
11. Table Cl suggests that the probability of failure of an RCHS should be about 10 per year. This corresponds to about 3 x 10~^ during the 30-year life of an RCHS. Assuming normal distributions of R and U , the probability of failure of 3 x 10 will yield a 3 value of approximately J+.O (MacGregor 1976). A value of 3 = +.0 will be used.
C6
Derivation of Load Factors and Strength Reduction Factors
Selection of statisti- cal properties of variables
12. The properties assumed in the calculations are listed in Table C2.
Table C2Statistical Distributions Assumed in Calculations
SpecifiedMean
in situ Percent O VMaterial Strengths M
Concrete strength 4000 psi 3800 psi 0.95 — 0 .I8Steel yield strength 1+0 ksi Hi ksi 1.03 — O . 0 7
Dimensions Eb-beam^ column* in. 1 2 12.05 l.OOU 0.3 0.05d-beam, in. 1 8 17.85 0.992 0.1+5 0.05h-column, in. 1 2 12.05 l.OOU 0.3 0.05Ag-beam, in.2 2 . 0 0 2 . 0 0 1 . 0 0 — O .06As-column, in.2 2 . 1 6 2 . 1 6 1 . 0 0 — O.O6
Accuracy of Code Equations PMu-under-reinforced beams — — 1 . 0 6 — 0.04Pu-axially loaded columns — — 0 . 9 8 — 0.05Vc-shear carried by concrete — — 1 . 1 0 — 0.15
Loadings SDead load — — 0.9 — 0.15Live load — — 0 . 8 0 . 3 0
Structural Analysis EDead load effects — --- 1 . 0 -- 0 . 1 0Live load effects
~
1 . 0 0 . 2 0
Computation of $ for flex- ure of a reinforced- concrete beam
13. Design strength R . The assumptions in ACI 318-77 could beused to calculate the beam strength (neglecting $ ) as
C7
R = M = A f (à -u s y V 2 y
= 2 .0 x 1+0 2 .0 x 1+Q2 x 0 . 8 5 x 1+ x 12 12
= 113.5 kip-ft
ll+. Mean strength Rm . The mean strength can he computed using the mean strengths and dimensions in the equation for M .
where
a _ ________ 2 . 0 x l+i________2 2 x 0.85 x 3.8 x 1 2 .0 5 1.05
M = 2 . 0 x 1+1(17.85 - 1 . 0 5 ) * 12u
= llH.8 kip-ft
This value of M must he corrected to allow for errors in the equation uitself. Thus
R = M x P = 111+.8 x 1 .0 6m u
= 121.7 kip-ft
RmYR R
121.7113.5 1.07
(C12 his)
15. Coefficient of variation . The coefficient of variation
C8
will be calculated in a number of stagessi. Va / 2 “ Since a/2 is the product of a number of variables,
the coefficient of variation of a/2 can be computed as follows.
VSi2
+ \2
- 40.062 + 0.0T2 + 0.182 + 0.05
= 0.21
9 #b. V - Since d - — is the sum of two variables, it is j a dd“2necessary to compute the standard deviation of this term
and then convert that to a coefficient of variation.Thus
where
°a ‘ (fX - ^ * °-212 2
= 0.22
a = Jo.k52 + 0 . 222 = 0.50a-f
and
C 9
V
d È.2
0 .50I T .85 - 1 .05 0 .03
* VMu - Because Mu is calculated as the product of As , fy , and (d - a/2), V]^ can he calculated as
y = y 2 + y 2 + y 2M A vf vu \ s y H )
yjo.062 + 0.0T2 + 0 .0 3 2
= 0.10
d. V-p - It is necessary to include the effect of the accuracy of the design equation in the coefficient of variation
V = fv 2 + V 2 R \ Mu p
= yjo.102 + O.Oit2
= 0.10
16. Computation of j> . According to Equation Cl6
-$aV_♦ = V (Cl6 his)
CIO
As discussed in paragraph 12, g = U.O . The term a -will be taken as 0.75 as explained in paragraph 5. Thus
4> = 1.07 e-U.0 x 0.75 x 0.10
= 0.78
The value of <|> for this particular problem is 0.78. This value of (j> will correspond to the load factors to be developed in paragraph 25. By calculating <p values for a range of different properties, a weighted average value can be obtained
Computation of cf> for an axially loaded tied column
17. Design strength R . The design strength can be calculated by the following equation.
Again, this value must be corrected to allow for errors in the equation itself. Thus
= 0.85 x U (lUU - 2.16) + (Uo x 2.16)
= 568.7 kips
18. Mean strength . Based on the mean strengths and dimen-sions, the mean axial load capacity is
P = 0.85 x 3.8 (12.05 x 12.05 - 2.16) + (kl x 2.16)o
= 550.6 kips
R = P x P = 550.6 x 0.98m o
= 539-6 kips
Oil
and
y = JE = W AtR R 568.T
= 0.95
19. Coefficient of variation VD . The strength of the axially______________________________ nloaded column is the sum of the load carried by the concrete P andcthe steel P . The coefficient of variation of P and P will be s c sevaluated separately.
¡a. Vp andc c
/ p p P= J o .18 + 0.05 + 0.05
= 0.19
The standard deviation of the load carried by the concrete is
oPcX p c
= 0.19 x b 62 = 87.8 kips
b. Vp and 0p . s_________ s
C12
2 2s
+ VAst
= yjo.Ol2 + 0. 062
= 0.09
and
0p = 0.09 x 88.6 = 8.2 kips s
V_ and a_ . Since P is a sum. a values must be P P oo_________ ocombined to get
o
°p % / oP2to p 2c s
= ■ Q'T.Q2 + 8.22 = 88.2 kips
Thus
V? sso T = °*16550.6
Again, this must he adjusted equation used to compute Po
to allow for errors in the
20.05 0.17
Cl 3
20. Computation of <j> . According to Equation Cl6
-BoV.R
= 0.95 e-U.0 x 0.75 x 0.17 (Cl6 bis)
= 0.57
The value of <|> for this particular example is 0.57. Before a final value can be chosen, a number of different column sections, including eccentrically loaded columns, must be studied.Computation of 4> for shear
21. Design Strength R .
V = 2 bdu
= 2 x À000 x 12 x 181000
=27.3 kips
22. Mean Strength Rm
V = 2 x 1/3800 x 12.05 xu17.851000
= 26.5 kips
R = V x p = 26.5 x 1.10m u
= 29.2 kips
and
ClU
23. Coefficient of variation V .n
v = /v2 + V 2 + V 2V J Vf’ Vb vdc > c
= y/CKl82 + 0.052 + 0.052 = 0.19
V = jv 2 + v 2R ^ V P
= yjo.192 + 0.152 = 0.2^
2U. Computation of <j> . Referring again to Equation Cl6
♦ V-6aVR
, ft<7 -U.o x 0.T5 x 0.2U1 .Of e
= 0.52
Computation of load fac- tors for dead and live loads
25. The derivation of X will he based on Equation C17
$aB« _ u
(Cl6 bis)
(C17 bis)
But, U = D + L where both D and L are separate variables. Because the coefficient of variation of D is much smaller than that of
L , it is desirable to separate these Since
U = Uy m u (C13 bis)
Equation C17 can be rewritten asBaV
yU = U e ’ mu (Cl8)
XUsing the first two terms of the series expansion for e gives
BaVU e u = (D + L ) ( 1 + $aV ) m m m u
= (D + L ) ll +m m 14Ba x/ D2V2 + L2V2’ m D m L
D + Lm m
(D + L ) ll +m m 1ga2D + 6a2L V m D m b
D '+ L m m
or
gaVu / 2 \ / 2 ^V “ Dm 1 + 60 VD + Lm 1 + 301 VL , (C19)
The parenthetical terms in Equation C19 can be rewritten in exponential form for consistency, giving separate load factors for D and L .
2Ba V.XD = V D (C20)
2£a V_XL = YLe (C21)
Cl6
(C20 bis)
26 . Derive X^ .
The term V is affected by variations in the load V and variationsD o J J
due to standard analysis V • Thusjiil)
V = .jv 2 + v 2 SD ED
= yjo.152 + 0.102 = 0.18Therefore
XD = 0.9 ek.O x 0.75 x 0.18
= 1.3527. Derive X .L
2got VTXL " V
where
V = »/V 2 + V 2L \ SL EL
= Jo^32 + 0.22 = 0.36
X = 0.8 e1*-0 * °-75 * °'36L
= 1.80
(C21 bis)
C17
28. Other load combinations. The above derivation was limited tothe combination of dead plus live loads. Similar analyses are required for other loading combinations.
Discussion29. Load factors and strength reduction factors derived from these
particular examples compared to those specified in Part II are summarized in the following tabulation.
FactorCalculatedValues
Calculated Values x 1.11
Specified Values for Design
Dead load 1.35 1.50 1.50Live load 1.80 2.00 1.90
<J> for flexure 0.78 0.87 0.90*<f> for tied columns 0.57 0.63 0.70*<f> for shear 0.52 0.58 0.85*
* Same as ACI 318-77 (ACI 1977a).The calculated values, except for shear, are very close to those specified in Part II. The strength reduction factor for shear given in ACI 318-77 appears to be inadequate; the O .85 value is insufficient to account for the large coefficient of variation in shear resistance.This deficiency has also been noted by other investigators (Ellingwood 1978).
30. Much more extensive study of various sizes of members, reinforcing ratios, and live to dead load ratios is required before final load factors and <|> factors can be determined. However, the final values are expected to be similar to the ones derived in this appendix.
Shortcomings of First-Order Probabilistic Procedure
31. According to MacGregor (1976), there are four major shortcomings of the first-order probabilistic procedure used to calculate load and strength reduction factors.
Cl8
a. The procedure is only as good as the data used in the solution, This is a problem to all procedures. Statistical data of the type required are not widely available.
b. The procedure assumes that specific probabilities of failure can be evaluated. Since this calculation depends on a knowledge of the extreme ranges of the strength distributions that are not adequately known, the computed probabilities of failure could differ from the actual values by as much as a factor of 10.
c_. Failure of the structure is assumed to occur when one cross section or element reaches its capacity. For a weakest-link structure such as a truss, failure will occur when the weakest of many elements is overloaded. For a ductile indeterminate structure, loads will be distributed from section to section before the entire structure fails. The solution will tend to overestimate the safety of the first case and underestimate the second.
<1. Only failure by known overloads has been considered.Causes of failure such as gross error, fire, or explosion have not been considered.
32. In spite of these shortcomings, the first-order probabilistic procedure used in this appendix provides a rational procedure for estimating safety factors and should be used to derive load and strength reduction factors for RCHS.
Recommended Future Studies
33. It is envisaged that the following efforts are required before a rational procedure for derivation of load and strength reduction factors can be realized:
a>. Collect data on statistical distribution of variables.Extensive data of the type summarized in Table C2 must be collected for each component affecting the strength of RCHS. In addition* data must be obtained from designers about typical dead, live, and wind loads, and earth and water pressures in RCHS, as well as typical concrete strengths and steel percentages, so that optimal factors can be used for the most common cases.
b. Determine theoretical and design strength equations.Procedures for calculating the theoretical member strengths must be selected and compared to available tests. The design equations should be those in the code. The
C19
theoretical equations should be as definitive as necessary to accurately estimate the true member strength.
c_. Calculate Yr and Vr for various structural actions. Values of Yr and Vr must be calculated for flexure, column cross sections in combined bending and axial load, slender columns, shear, bond, and deflections, and possibly for cracking. For some members these terms can be computed using direct statistical methods similar to those outlined in this appendix. For other members such as column cross sections or slender columns, the interaction of the variables is more complex and a Monte-Carlo simulation technique must be employed to estimate Yr» and V_ .n n
C20
APPENDIX D: DERIVATION OF MAXIMUM TENSION REINFORCEMENTRATIO FOR FLEXURAL MEMBERS
1. With reference to Figure Dl, the equilibrium between external and internal bending moments M at working stress, can be expressed as
M = | fc kjbd2 (Dl)
Figure Dl. Stress and strain distribution, working-stress design
The minimum d required for working-stress design (WSD) can be obtained from Equation Dl
* - ^ <D2>
where k and j are the values for the balanced WSD stress condition.2. From Figure D2, the factored moment strength M can be written
as
Mu (D3)
Dl
Figure D2. Stress and strain distribution, strength design
and
aA f. s ,y. „0.85f'bc
pf d. y .0.85f'c
(DU)
By substituting the value of a obtained from Equation Db- and substi-tuting pbd for A into Equation D3s
Mu = <(>pfybd2il - 0.59P 1 (D5)
Thus
Md = u
<|)pf (l - 0.59P ^ J b(D6)
3. For an equivalent design, Equation D2 should be equal to Equation D6, and therefore
D2
For hydraulic structures
M = 1.9 M u (d8)
f = 0.35 f (D9)c c
<j) = 0.9 (D10)
f = Uo ksi (Dll)
Substituting Equations D8-D11 into Equation D7
72p|l - 0.59P j f j = 0.67f^kj (D12)
The tension reinforcement ratio for equivalent design p can be obtained from Equation D12 for given f f , k , and j . Table D1 showscthe p and p/p. values for various ff . It can be seen that p/p_ b e bvaries from 0.23 to 0.29. The value of 0.25 is recommended for design of hydraulic structures. For conduits or culverts, the allowable concrete stress f is O.U5f! . Substituting this value together with c cEquations D8, DIO, and Dll into Equation D7
f^ P 1 - (D13)
The p values for conduits or culverts for various f^ are given inTable D2. The p/pn varies from 0.36 to 0.^2. The recommended valuebof p for conduits or culverts is 0.375P-, •b
Table DITension Reinforcement Ratio for Equivalent
Design of Hydraulic Structures
f 'cksi
0 .35ffcksi k* J* P Pb** p/pb
3 1.050 0.326 0 .891 0.0086 0 .0371 0 .2 3
k l.HOO 0 .359 0 .880 0 .0128 0.0U90 0.265 1.750 0 .3 8 3 0 .872 0 .0169 0 .0582 0 .2 9
* Obtained from American Concrete Institute (1965). ** Obtained from American Concrete Institute (1977b):
n = n fiRft C 87,000 pb 0 , ° 561 f 87 ,000 + fy y
Table D2
of Conduits or Culverts
f’cksi
0 .1+5f ’ cksi k* j * P V * p/pb
3 1.350 0.383 0.872 0.013U 0.0371 0.360k 1.800 0.U19 0.860 0.0195 0.0U95 0.3935 2.250 o.kkk 0.852 0.0257 7.0582 0.UU2
Pb = 0 .856! ffc 87,000
87,000 + fy y
* Obtained from American Concrete Institute (1965). ** Obtained from American Concrete Institute (1977b)
DU
In accordance with letter from DAEN-RDC, DAEN-ASI dated 22 July 1977, Subject: Facsimile Catalog Cards forLaboratory Technical Publications, a facsimile catalog card in Library of Congress MARC format is reproduced below.
Liu, Tony CStrength design of reinforced concrete hydraulic struc
tures; Report 1: Preliminary strength design criteria / by Tony C. Liu. Vicksburg, Miss. : U. S. Waterways Experiment Station ; Springfield, Va. : available from National Technical Information Service, 1980.
27, [70] p. : ill. ; 27 cm. (Technical report - U. S. Army Engineer Waterways Experiment Station ; SL-80-4, Report 1) Prepared for Office, Chief of Engineers, U. S. Army, Wash
ington, D. C., under CWIS 31623.References: p. 26-27.
1. Concrete strength. 2. Concrete structures. 3. Hydraulic structures. 4. Reinforced concrete. I. United States. Army. Corps of Engineers. II. Series: United States. Waterways Experiment Station, Vicksburg, Miss. Technical report ; SL-80-4, Report 1.TA7.W34 no.SL-80-4 Report 1