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Chapter : Structural, On-shore For additional information on this subject, contact
File Reference: CSE10803 C.C. Baldwin on 873-1567
Engineering EncyclopediaSaudi Aramco DeskTop Standards
Design Of Reinforced Concrete Two-Way
Slabs and Columns; Frame Analysis Techniques
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Contents Pages
INTRODUCTION TO DESIGN OF TWO-WAY SLABS ............................................................... 1
Direct Design Method......................................................................................................... 1
Equivalent Frame Method................................................................................................... 6
Yield Line Method.............................................................................................................. 8
General .................................................................................................................. 8
Yield Line Patterns................................................................................................ 8
Analysis Methods ............................................................................................................... 9
Slab Reinforcement ............................................................................................. 11
COLUMN DESIGN ....................................................................................................................... 12
Column Design Basis ....................................................................................................... 12
Determining Slenderness Effects......................................................................... 13
Using Axial Load-Moment Interaction Diagrams............................................................. 19
DETERMINING REINFORCING STEEL REQUIREMENTS ..................................................... 25
FRAME ANALYSIS TECHNIQUES ............................................................................................ 29
Beam Formulas ................................................................................................................. 30
ONE-STORY FRAMES................................................................................................................. 33
Computer Programs .......................................................................................................... 36
WORK AID 1: RECTANGULAR COLUMN: LOAD-MOMENT INTERACTION
DIAGRAMS .................................................................................................... 37
COLUMNS 7.4.3Load-moment strength interaction diagram for R4-60.75
columns ....................................................................................... 37
COLUMNS 7.4.4Load-moment strength interaction diagram for R4-60.90
columns ....................................................................................... 38
WORK AID 2: SPIRALLY REINFORCED COLUMN: LOAD-MOMENTINTERACTION DIAGRAMS......................................................................... 39
COLUMNS 7.23.1Load-moment strength interaction diagram for C5-
60.45 spirally reinforced columns ............................................. 39
COLUMNS 7.23.1Load-moment strength interaction diagram for C5-
60.60 spirally reinforced columns ............................................. 40
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WORK AID 3: MINIMUM FACE DIMENSIONS OF RECTANGULAR TIED
COLUMNS ACCOMMODATING VARIOUS NUMBERS OF BARS
PER FACE....................................................................................................... 41
WORK AID 4: FRAME 39, SYMMETRICAL RECTANGULAR TWO-HINGED FRAME, CASE #1 .......................................................................... 42
WORK AID 5: FRAME 39, SYMMETRICAL RECTANGULAR TWO-
HINGED FRAME, CASE #2-4 ....................................................................... 43
WORK AID 6: FRAME 39, SYMMETRICAL RECTANGULAR TWO-
HINGED FRAME, CASE #5-7 ....................................................................... 44
WORK AID 7: FRAME 39, SYMMETRICAL RECTANGULAR TWO-
HINGED FRAME, SYMMETRICAL LOADING - CASE #8-
10 ..................................................................................................................... 45
WORK AID 8: FRAME 41, SYMMETRICAL RECTANGULAR FULLY-FIXED FRAME, CASE #1 .............................................................................. 46
WORK AID 9: FRAME 41, SYMMETRICAL RECTANGULAR FULLY-
FIXED FRAME, CASE #2-3........................................................................... 47
WORK AID 10: FRAME 41, SYMMETRICAL RECTANGULAR FULLY-
FIXED FRAME, CASE #4-5........................................................................... 48
WORK AID 11: FRAME 41, SYMMETRICAL RECTANGULAR FULLY-
FIXED FRAME, CASE #6-7 ....................................................................... 49
WORK AID 12: FRAME 41, SYMMETRICAL RECTANGULAR FULLY-
FIXED FRAME, CASE #8-9 ....................................................................... 50
WORK AID 13: APPENDIX LOAD TERMS; GENERAL NOTATIONS............................. 51
WORK AID 14: APPENDIX LOAD TERMS; NOTATIONS, CASE 1-3.............................. 52
WORK AID 15: APPENDIX LOAD TERMS; GENERAL NOTATIONS, CASE 4-6........... 53
WORK AID 16: APPENDIX LOAD TERMS; NOTATIONS, CASE 7-11............................ 54
GLOSSARY................................................................................................................................... 55
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Introduction to Design of Two-Way Slabs
In CSE10802, different types of slabs were identified and the methodology for design of one-way slabspresented. For two-way slabs, however, the methods for determining the distribution of moments throughout
the slab are considerably more complex than for beams or one-way slabs. Consequently, the analysis and
design of two-way slab systems is usually performed using computer programs.
The following three methods for performing two-way slab analysis and design will be outlined in this module.
The first two methods are specified in the ACI 318 Code and are based on elastic analysis; the third is a
collapse method of analysis. An introduction to some of the concepts employed in the methods is presented
below. Details of all three methods, however, are considered beyond the scope of this course:
Direct Design Method
Equivalent Frame Method
Yield Line Method
Code-specified minimum thickness of slabs and deflection limitations will also be presented, as will some
details of slab reinforcement.
The designer must understand the concepts and provisions governing the analysis and design of two-way slabs.
This understanding is necessary to develop the necessary input data for computer programs and to be able to
hand-check the validity of the output. For a few limited cases, the Direct Design Method (DDM) per the Code
can be used to analyze and design the slab without a computer. Of equal importance, the designer can use thesteps and concepts contained in the DDM to evaluate computer-aided designs.
Direct Design Method
This semiempirical method consists of a set of rules for the proportioning of slab and beam sections to resist
flexural stresses.
Step 1 - Limitations Check
The first step checks the validity of certain assumptions stated in the Code. These critical limitations are stated
in Figure 1. If any of these assumptions are not true, the designer must resort to an alternative approach.
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Step 2 - Thickness/Shear Check
The second step selects a trial thickness and checks the perimeter or punching shearcapacity at the critical
section near each column. Refer to ACI Code for details. For two-way slabs, Section 9.5.3 of the Code
provides guidelines defining the minimum thickness for deflection and serviceability requirements; theguidelines are also summarized in
Figure 2. For slabs without drop panels, h 120 mm; for slabs with drop panels, h 100 mm per ACI 318 M.
For two-way slabs, a good first estimate is: h/l -- 1/20 to 1/30
The trial thickness can be based on guidelines in Figure 2 (serviceability requirements) or on experience.
Usually, a uniform slab thickness is selected and checked for shear. Note that an allowance should be made for
the slab-to-column moment transfer capacity, which is to be checked.
Step 3 - Detailed Design
Steps in the detailed design procedure are given in the ACI Code. Detailed design is beyond the scope of this
course.
Slabs with thicknesses less than the minimum given in Figure 2 may be used if it can be shown that deflections
will not exceed the limits specified in Figure 3. Deflections shall be computed taking into account the size and
shape of the panel, the conditions of support, and the nature of restraint at panel edges.
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DIRECT DESIGN METHOD LIMITATIONS
ACI 318-89, Section 13.6.1
Step 1 Determine whether slab geometry and loading satisfy conditions for use of
Direct Design Method.
A) Minimum of three continuous spans in each direction.
B) Panels shall be rectangular with ratio of longer to shorter span within a
panel not greater than 2.0.
C) Successive span lengths in each direction must not differ by more than
one-third of longer span.
D) No column offset more than 10% of span in the direction of offset.
E) All loads are due to gravity loading only and are uniformly distributed.
The services live loads are not greater than three times total dead load.
F) If the panel is supported by beams on all sides, the relative stiffness of the
beams in two perpendicular directions shall not be less than 0.2 nor greater
than 5.0.
G) No negative moment redistribution is permitted.
H) Variations to the above are permitted if:
- Conditions of equilibrium and compatibility are met.
- Design strength is at least equal to required strength.
- Serviceability conditions are met.
FIGURE 1
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TABLE 9.5 (c) MINIMUM THICKNESS OF SLABS WITHOUT INTERIOR BEAMS
Without drop panels With drop panels
Yield Stress
fy,
Exterior panels Interior
panels
Exterior panels Interior
panels
psi Without
edge beams
With edge
beams
Without
edge beams
With edge
beams
40,000 l n
33
l n
36
l n
36
l n
36
l n
40
l n
40
60,000 l n
30
l n
33
l n
33
l n
33
l n
36
l n
36
Min.
Thk.
9.5.3.2
5 in. (120 mm) 4 in. (100 mm)
Authorized reprint from ACI 318-89, Table 9-5(c), Page 99, with permission from the American Concrete Institute.
FIGURE 2
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TABLE 9.5(b) MAXIMUM PERMISSIBLE COMPUTED DEFLECTIONS
Type of member Deflection to be considered Deflection limitationFlat roofs not supporting or
attached to nonstructural
elements likely to be damaged
by large deflections.
Immediate deflection due to live
load L l
180
*
Floors not supporting or
attached to nonstructural
elements likely to be damaged
by large deflections.
Immediate deflection due to live
load L l
380
Roof or floor construction
supporting or attached to
nonstructural elements likely tobe damaged by large
deflections.
That part of the total deflection
occurring after attachment of
nonstructural elements (sum ofthe long-time deflection due to
all sustained loads and the
immediate deflection due to any
additional live load)
l
480
Roof or floor construction
supporting or attached to
nonstructural elements not
likely to be damaged by large
deflections.
l *
240
*
* Limit not intended to safeguard against ponding. Ponding should be checked by suitablecalculations of deflection, included added deflections due to ponded water,
and considering long-term effects of all sustained loads, camber, construction
tolerances, and availability of provisions for drainage.
** But not greater than tolerance provided for nonstructural elements. Limit may be
exceeded if camber is provided so that total deflection minus camber does not
exceed limit.
Authorized reprint from ACI 318-89, Table 9-5(b), Page 98, with permission from the American Concrete Institute.
FIGURE 3
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Equivalent Frame Method
The Equivalent Frame Method involves the representation of the three-dimensional slab system by a series of
two-dimensional frames which are then analyzed for loads acting in the plane of the frames. The negative and
positive moments so determined at the critical design sections of the frame are distributed to the slab sections in
accordance with rules in the ACI Code governing column strips, beams, and middle strips.
Application of the equivalent frame to a regular structure is illustrated in Figure 4. The three-dimensional
building is divided into a series of two-dimensional frame bents (equivalent frames) centered on column or
support centerlines with each frame extending the full height of the building. The width of each equivalent
frame is bounded by the centerlines of the adjacent panels. The complete analysis of a slab system for a
building consists of analyzing a series of equivalent (interior and exterior) frames spanning longitudinally and
transversely through the building.
The equivalent frame comprises three parts: (1) the horizontal slab strip, including any beams spanning in the
direction of the frame; (2) the columns or other vertical supporting members, extending above and below the
slab; and (3) the elements of the structure that provide moment transfer between the horizontal and vertical
members.
Details of the method are provided in Section 13.7 of the ACI Code.
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DEFINITION OF EQUIVALENT FRAME
l 1
l 1
l 1
l 2l 2l 2
l 2
l 22
SlabBeamStrip 2
Centerlineof Panel
Interior Equivalent Frame
ExteriorEquivalentFrame
Edge
CenterlineAdjacent Panel
Column Strip
One-Half Middle Strip
Authorized reprint from ACI 318-89, Figure 13.7.2.1, Page 231, with permission from the American Concrete Institute.
FIGURE 4
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Yield Line Method
General
In the conventional methods of slab design considered previously, the design moments and shears are
determined by simplified procedures which are based mainly on elastic concepts of slab behavior, with the final
proportioning of the slab to determine the reinforcement requirements carried out by means of the ultimate
strength method. An alternative approach to reinforced concrete two-way slab design is to consider the
inelastic distribution of moments and shears in the statically indeterminate structure just prior to collapse.
The proportion of flexural reinforcement which is used in the construction of a slab is typically much lower
than in beams and girders. This ensures a high degree of ductility, and since a slab is a highly indeterminate
structural system, an extensive redistribution of moments can thus occur prior to collapse. This very
satisfactory aspect of the overload behavior of slabs has led to the application of the basic concepts of simpleplastic theory to their analysis and design.
The collapse methods of slab analysis and design were originally developed in the Scandinavian countries, and
have obtained greater acceptance in Europe than in the
United States. One such method, the yield line theory, is an acceptable basis for slab design and is briefly
discussed below.
One of the important practical advantages in using plastic methods is that they can be applied to slabs of
irregular and complex shape. Therefore, they can be used to design slabs which cannot be designed by the
"elastic" methods. However, it must be emphasized that the collapse load methods are concerned solely with
the design objective of adequate strength; problems of serviceability, such as excessive deflection, needseparate and careful consideration. Although not included in the ACI Code, slab analysis by yield line theory
may be useful in providing the needed information for understanding the behavior of irregular or single-panel
slabs with various boundary conditions.
Yield Line Patterns
Yield line theory for two-way slabs requires a different treatment to that of limit (plastic) analysis of continuous
beams, because in this case the yield lines will generally not be parallel to each other, but will instead form a
yield line pattern. The entire slab area will be divided into several segments which can rotate along the yield
lines as rigid bodies at the condition of collapse or unstable equilibrium. Some yield line patterns for typicalsituations are shown in Figure 5.
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The slab of Figure 5(a) has nonparallel supports. At the collapse condition, this slab will break into two
segments; one segment will have an edge rotating about I and the other will have an edge rotating about II. The
positive moment yield line must then intersect lines I and II at their intersection, point 0. The exact position of
yield line III will depend on the reinforcement amount and direction, both in the positive and negative moment
regions.
For the case of Figure 5(b) where a rectangular panel is either simply supported or continuous over four linear
supports, the collapse mechanism consists of four slab segments. The exact locations of points aand bwill
depend on the moment strengths at the supports and the positive moment reinforcement in each direction.
The slab in Figure 5(c) is supported along two edges and, in addition, is supported by two isolated columns.
The rotational axes for the slab segments at collapse must occur along the supports (lines I and II), and
additional rotational axes must pass through the isolated columns. The critical position of the positive moment
yield lines, a, b, c, d and eis a function of the reinforcement amount and direction; in the meantime,
compatibility of deflection along the yield lines must be maintained during the rigid body rotations of the slab
segments.
For a concentrated load at a significant distance from a supported edge, the yield line pattern will be circular as
shown in Figure 5(d). The circular pattern will be a yield line of negative bending moment while the radial
yield lines are due to positive bending moment. For concentrated loads near a free edge, a fan or partial circular
pattern is typical.
Analysis Methods
There are two methods of yield line analysis of slabs: the virtual work method and the equilibrium method.
Based on the same fundamental assumptions, the two methods should give exactly the same results. In either
method, a yield line pattern must be first assumed so that a collapse mechanism is produced. For a collapse
mechanism, rigid body movements of the slab segments are possible by rotation along the yield lines while
maintaining deflection compatibility at the yield lines between slab segments. There may be more than one
possible yield line pattern, in which case solutions to all possible yield line patterns must be sought and the one
giving the smallest ultimate load would actually happen and thus should be used in design.
Most texts on advanced reinforced concrete design contain a discussion on the application of yield line
methods.
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TYPICAL YIELD LINE PATTERNS
III
O
(a)
Free Edge
Supported
Edges
Free Edge
III
e
a
b
c
d
II
IV
IIII
Supported Edges
Columns
(c)
(d)(b)
Negative
Moment
Y ield Line
Positive
Moment
Y ield Line
Four Supported Edges
a b
P
FIGURE 5
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Slab Reinforcement
When the shear design is satisfactory and the slab design moments and column moment transfer aredetermined, the designer selects the slab reinforcement. (Details are provided in the ACI Code).
The area of flexural reinforcement required to resist the factored moments in each strip is computed using the
flexure formulas/work aids for beams and one-way slabs. A uniform bar spacing is usually used within a given
strip. Note that reinforcement for any required code-specified additional column moment transfer must be
placed within a width not exceeding the column width plus one and one-half times the slab thickness on each
side of the column. Bar diameters and spacing are selected using the same considerations as for one-way slabs,
covering minimum reinforcement ratio for shrinkage and temperature, crack control, and minimum bar spacing.
Maximum bar spacing for two-way slabs shall not exceed two times the slab thickness. Bar development for
two-way slabs shall follow the provisions stated in Section 13.4 of the code (see Figure 13.4.8).
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Column Design
Column Design Basis
The design of columns and other structural members subjected to compressive loads is based on the same
assumptions covered in Module 2 in the discussion of ultimate strength design of beams. The procedures
described herein are based on the provisions in Section 10 of the Code. The design of compression members is
based on forces and moments determined from an elastic analysis. The designer needs to consider the influence
of axial loads and variable moment of inertia on member stiffness, the effect of deflections on moments and
forces, and the effects of the duration of loads (creep).
Column design notation is presented in Figure 6. Also, refer to Work Aid 1 in Module 2 for notation specific to
flexure and shear design.
For a short column without bending moments, the design axial load strength, Pn,is given by
For spiral reinforced:
Pn (max)= 0.85 0.85 fc' ( Ag- Ast) + fyAst
For tie reinforced:
Pn ( max )= 0.80 0.85 fc' ( Ag- Ast ) + fyAst
For members with spiral reinforcement conforming to ACI Code Section 10.9.3, is 0.75. For tie reinforcedmembers, is 0.7. See Section 9.3.2 of ACI Code.
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Determining Slenderness Effects
The Code defines a column as "short" based on a maximum value of klu/r as given below:
For columns braced against sidesway,
klu/r < 34 - 12 M1/M2
and for columns not braced against sidesway,
klu/r < 22
M1and M2are the smaller and larger factored end moments on the column, respectively. M1is positive if the
column is bent in single curvature and negative if the column is bent in double curvature. M2is always
positive.
The effective length factor, k, may be obtained from the nomograph shown on Figure 7. The terms Aand Bare the ratios of the sum of column stiffnesses, EI/lc, for each end joint divided by the sum of the beam
stiffnesses, EI/lb.
The unsupported length, lu, shall be taken as the clear distance between slabs, beams, or other members
providing lateral support. The radius of gyration, r, may be set equal to 0.3 times the overall dimension of the
rectangular member in the direction stability is considered, and equal to 0.25 times the diameter for circular
members. Note that if the moments at both ends of the column, M1and M2, are zero, the ratio M1/M2shall be
set equal to one.
If the value of klu/r exceeds that given above, column slenderness effects must be considered. Columns are
typically sized such that klu/r values are less than 50. Although permitted by the Code, experience indicates that
columns should not be sized such that klu/r values exceed 100.
The Code accounts for the effects of column slenderness by a moment magnification procedure provided in
Section 10.11. Columns are designed using the factored axial load and a magnified factored moment. Details
of this method are beyond the scope of this course.
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COLUMN DESIGN NOTATION
Ab
Ac
Ag
Ast
b
Cm
d
db
e
ex
ey
Ec
EI
=
=
=
=
=
=
=
=
=
=
=
=
=
Area of an individual bar.
Area of core of spirally reinforced columnmeasured to outside diameter of spiral.
Gross area of column cross section.
Total area of longitudinal reinforcement in
a column cross section.
Overall cross-section dimension of a
rectangular column.
A factor relating actual moment diagram to
an equivalent uniform moment diagram.
(For members braced against sidesway and
without transverse loads between supports,
Cm= 0.6 + 0.4 [Mlb/M2b] but not less
than 0.4. For all other cases, Cmshall be
taken as 1.0).
Depth to reinforcement.
Nominal diameter of bar.
Eccentricity of axial load at end of
member, measured from the centroid of the
tension reinforcement, calculated byconventional methods of frame analysis.
Eccentricity e along x-axis.
Eccentricity e along y-axis.
Modulus of elasticity of concrete, psi. (For
normal weight concrete. Ecmay be taken
as 57,000 fc'psi.)
Flexural stiffness term.
c
fs
fy
h
he
I
Ig
Ise
k
l
lc
ldb
ldc
lu
=
=
=
=
=
=
=
=
=
=
=
=
=
=
Specified compressive strength of
concrete.
Stress in reinforcement.
Specified yield strength of reinforcement.
Diameter of a round column or side of a
rectangular column.
Effective thickness of a column for
slenderness considerations.
Moment of inertia of section resisting
externally applied factored loads.
Moment of inertia of gross concrete
section about centroidal axis, neglecting
reinforcement.
Moment of inertia of reinforcement about
centroidal axis of member cross-section.
Effective length factor for compression
members.
Span length of beam or slab, as defined in
ACI 318-89, Section 8.7.
Height of column, center-to-center of
floors or roof.
Development length.
Development length for bars in
compression.
Unsupported length of member.
Authorized reprint from ACI, SP-17(91), Vol. 2, Notation, Page xi, with permission from the American
Concrete Institute.
FIGURE 6
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COLUMN DESIGN NOTATION (CONT'D)
Mc
Md
Mn
Mu
M1
M2
M2s
Pc
Pn
Pnx
Pny
Pno
=
=
=
=
=
=
=
=
=
=
=
=
Magnified factored moment to be used for
design of column.
Moment due to dead load.
Nominal moment strength at section.
Factored moment at section.
Value of smaller factored end moment on
compression member due to the loads that
result in no appreciable sidesway,
calculated by conventional elastic frame
analysis. Positive if member is bent in
single curvature, negative if bent in
double curvature.
Value of larger factored end moment on
compression member due to the loads that
result in no appreciable sidesway,
calculated by conventional elastic frame
analysis. Always positive.
Value of larger factored end moment on
compression member due to the loads that
result in appreciable sidesway, calculated
by conventional elastic frame analysis.Always positive.
Critical axial load.
Nominal axial load strength at given
eccentricity.
Nominal axial load strength for
eccentricity ey, along y-axis only,
x-axis being axis of bending.
Nominal axial load strength for
eccentricity ex along x-axis only.
y-axis being axis of bending.
Nominal axial load strength at zero
eccentricity, kips.
Pu
r
s
S
Sb
d
(gamma)
b(delta)
s
(xi)
g (rho)
(phi)
(psi)
=
=
=
=
=
=
=
=
=
=
=
=
=
Factored axial load at given eccentricity
Pn.
Radius of gyration of cross section of a
compression member.
Center-to-center spacing of bars.
Pitch of spiral, center-to-center of bar.
Clear spacing between bars.
Ratio of maximum factored dead load
moment to maximum factored total load
moment. Always positive.
Ratio of distance between centroid of
outer rows of bars and thickness of cross
section, in the direction of bending.
Moment magnification factor for columns
braced against sidesway.
Moment magnification factor for frames
not braced against sidesway, to reflect
lateral drift resulting from lateral and
gravity loads.
Dimensionless constant used in
computing Igand Ise.
Ast/Ag= ratio of total reinforcement area
to cross-sectional area of column.
Strength reduction factor as defined in
Section 9.3 of ACI 318-89.
Ratio of sum of stiffness _ (EI/l) of
compression members to _(EI/l) of
flexural members in a plane at one end of
a compression member.
FIGURE 6 (CONT'D)
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EFFECTIVE LENGTH FACTORS
A k B
A k B
50.0
10.0
5.0
3.0
2.0
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1.0
0.9
0.8
0.7
0.6
0.5
50.0
10.0
5.0
3.0
2.0
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
50.0
10.0
5.0
3.0
2.0
1.0
9.0
8.0
7.0
0 1.0
20.0
3.0
2.0
1.5
100.0
30.0
20.0
4.0
6.0
10.0
5.0
4.0
50.0
10.0
5.0
3.0
2.0
1.0
9.0
8.0
7.0
0
100.0
20.0
4.0
6.0
(a) (b)
Braced Frames Unbraced Frames
= Ratio of (EI/lc) of compression members to (EI/lb) of flexural members in a plane at one end of acompression member.
k = Effective length factor.
For pinned support, = 10
For fixed support, = 1
FIGURE 7
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Example Problem 1
For a rectangular tied column with bars equally distributed along four faces, determine whether slenderness
effects can be ignored.
Given: Loading
Factored axial loadPu= 560 kips
Factored momentMu= 3920 kip-in.
Materials
Compressive strength of concrete f'c = 4 ksi
Yield strength of reinforcement fy= 60 ksi
Nominal maximum size of aggregate is 1 in.
Design conditions
Unsupported length of columns lu= 10 ft
Column is braced against sidesway
b = 1 6 I n .
h
h = 20 In.
e
P
u
= 5 6 0 k i p
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Example Problem 1 (Cont' d)
ACI 318-89
Reference Procedure Calculation
Handbook
Reference
Step 1--Determine column section size
Given: h= 20 in. b= 16 in.
10.11.4.1
10.11.5.1
10.11.3
10.11.4.1
10.11.2.1
10.11.4.1
Step 2--Check whether
slenderness ratio klu/ris
less than critical value. If
so, slenderness effects may
be neglected.
If not, slenderness effects
must be considered.
A) Compute M1/M2and
read critical value of klu/r
B) Compute klu/r and
compare with critical value;
determine whether slender-
ness effects must be
considered.
In this case, assume
conservatively that M1= M2=
3920 kip-in. Hence, upper limitof klu/r is 22.
For columns braced against
sidesway: k = 1.0
Given:
l u =10 ft =120 in.
r =0.3 h = 0.3 x 20 = 6.0 in .
kl u
r=
(1.0)(120 )
6 .0= 20
< 22
Slenderness effects may be neglected
Authorized reprint from ACI, SP-17(91), Vol. 2, Columns Example 1, Pages 1-3, with permission from the
American Concrete Institute.
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Using Axial Load-Moment Interaction Diagrams
As discussed previously, the strength of a member under axial load only, P n, and the strength of a member
under flexure only, Mn, are given by:
For tied columns
Pn = 0.80 0. 85 f c Ag Ast( )+ fyAst
For beams
Mn = fybd2
1 0.59fy
fc
For members subjected to both axial load and moment, the strength of the member depends on the interaction
between load and moment. As the applied moment increases, the axial load strength decreases. Due to the
nonlinear stress-strain behavior of concrete, the relationship between axial load strength and moment capacity is
nonlinear. Solutions to the tedious calculation for determining the axial load strength for a given moment
capacity are presented in the form of interaction diagrams or charts in the ACI SP-17 Design Handbook for
Columns. The charts cover both rectangular and circular columns, with varying 28-day strengths, reinforcement
arrangement, and yield strength. Note that these charts are applicable to the design of columns subjected to
moment about one axis or direction only.
A typical chart (from Work Aid 1) is shown on Figure 8. This chart is applicable to rectangular columns with
equal reinforcing in each face, fc' = 4 ksi, fy = 60 ksi, and = 0.75 referring to the ratio of the distance betweenreinforcing layers in opposite faces to the total thickness or dimension of the column.
Figure 8 relates values of the axial load strength Pn to the design moment strength (capacity)Mn, based onthe ratio e/h of load eccentricity to column thickness and ratio of the area of longitudinal reinforcing to the
column cross-section area, g. Generally the axial load and moment have been previously computed andtherefore are "known". The designer will assume a column shape and size, and reinforcing pattern, and will use
the charts to find the ratio .
Knowing the design moment and axial loads, the designer computes the abscissa and ordinate based on the
column size and enters the chart to obtain the reinforcing steel ratio required (g). Note that for a successfuldesign, the point on the chart corresponding to the combination of load and moment must fall below the solid
(or dashed) line corresponding to the strength curve for the steel ratio to be used.
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LOAD MOMENT STRENGTH INTERACTION DIAGRAM
FOR R4-60.75 COLUMNS
Columns 7.4.3 Load-moment strength interaction diagram for R4-60.75 columns
References: ACI 318-89 Sections 9.3.2.2., 10.2, and 10.3: ACI Publication SP-7. pp 152-182
PnAg
eh
= MnA gh
,ksi
Authorized reprint from ACI, SP-17(90) Vol. 2, Columns 7.4.3, Page 82, with permission from the
American Concrete Institute.
FIGURE 8
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Depending on the value of gobtained, the designer will revise his assumptions and reuse the chart until adesired value is found. The desired value of gis discussed in the following section.
A table of he/h values is provided on each chart. The value for hereflects the additional flexural stiffness
provided by the reinforcement and is used as an adjustment factor when computing slenderness ratios and
moment magnification factors in the event slenderness effects need to be considered.
Figure 9 illustrates some additional information regarding development and use of these charts. Example
Problem 2 indicates use of the charts to further design the column as defined in Example Problem 1.
For columns subjected to biaxial bending or moments about both column axes, such as corner columns in a
building, the axial load strength follows the form of the interaction surface. See the ACI Code and Design
Handbook for details.
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INTERACTION DIAGRAMS
P
o
Maximum Design Axial
Load Strength,
P
n m ax
Changes
Balanced Design
Design Moment Strength Index
Authorized reprint from ACI, SP-17(91) Vol. 2, Figure 1, page 206, with permission from the
American Concrete Institute.
FIGURE 9
1. The coordinates have been reduced by the appropriate values of the strength reduction factor inaccordance with Section 9.3.2.2 of ACI 318-89. Over most of the curve, is 0.75 for spirally reinforcedcolumns and 0.70 for tied columns. But increases toward 0.90 at low values of axial load, as providedin Section 9.3.2.2 of ACI 318-89, which accounts for the discontinuity at the low end of curves.
2. Curves of fs= 0 and fs = fyare given. The first of these two conditions represents the point at which the
stress in tension face steel changes from compression to tension as moment increases. The second
condition represents the point at which a longitudinal bar first reaches yield stress; that is, the point of
balanced load Pnb where simultaneously the concrete reaches a 0.3% strain and the steel reaches itstensile yield stress level. These values are useful in selecting splice lengths for bars in columns.
3. Dashed lines are used for curves for reinforcement ratios gof 0.05, 0.06, 0.07, and 0.08 to call attentionto the likelihood of difficulty at splices due to congestion of bars where ratios greater than 0.04 are used.
4. At the low end of the moment range, as moment decreases, design axial load strength Pnreaches amaximum and levels off at that value, in accordance with Section 10.3.5 of ACI 318-89.
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Example Problem 2
This is the same column as presented in Example Problem 1.
For a rectangular tied column with bars equally distributed along four faces, select the
required area of reinforcement.
Given: Loading
Factored axial loadPu= 560 kips
Factored momentMu= 3920 kip-in.
Materials
Compressive strength of concretef'c= 4 ksi
Yield strength of reinforcement fy= 60 ksi
Nominal maximum size of aggregate is 1 in.
Design conditions
Unsupported length of columns lu= 10 ft
Column is braced against sidesway
b = 1 6 I n .
h
h = 20 In.
e
P
u
= 5 6 0
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Example Problem 2 (Cont' d)
ACI 318-89
Reference Procedure Calculation
Handbook
Reference
Step 1--See Example Problem 1.
Step 2--See Example Problem 1.
9.3.2.2(b)
10.2.7
9.3.2.2(b)10.2
10.3
Step 3--Determine
reinforcement ratio using known values of
variables on appropriate
interaction diagram(s) and
compute required cross
section areaAst of
longitudinal reinforcement.
A) Compute PuAg
B) ComputeMu
Agh
C) Estimate h 5
h
D) Determine appropriate interaction diagram(s).
E) Read gforPu/Ag (
< Pn/Ag) andMu/Agh (< Mn/Agh)
F) Compute requiredAst from
Ast = gAg
Given:
Pu= 560 kips
Mu= 3920 kip-in
h= 20 in.
b= 16 in.
Ag=bx h= 20 x 16 = 320 in2
PuAg
= 560320 = 1.75 ksi
Mu
Agh=
3920
320 20= 0.61 ksi
20 5
20= 0.75
For a rectangular tied column with bars along
four faces, f'c= 4 ksi,
fy= 60 ksi, and an
estimated of 0.75, use R4-60.75.
ForPu/Ag= 1.75 from
Step 3A andMu/Agh=
0.61 from Step 3B:
g= 0.040
RequiredAst = 0.040 x 320
= 12.8 in.2
Work Aid 1
Authorized reprint from ACI, SP-17(91), Vol. 2, Columns Example 1, Pages 1-3, with permission from the
American Concrete Institute.
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Determining Reinforcing Steel Requirements
Requirements for reinforcing steel in columns are covered in sections 7.10, 10.9, and 12 of the Code. For
longitudinal reinforcing, the ratio gmust be within the range of 0.01 to 0.08. Based on typical columndimensions and loading and bar congestion, which occurs at high steel ratios, usually falls within the range of0.01 to 0.04. The Code does not specify a minimum or maximum diameter for longitudinal bars. Generally, #5
to #11 size bars are used. The minimum number of bars is 4 for columns with lateral ties and 6 for columns
with spirals. The maximum number of bars that can be placed in each face of the column depends on the
column dimension, cover, longitudinal bar diameter, tie or spiral bar diameter, a clear spacing between bars of
1.5db, but not less than 1.5 inches, and the type of column splicing.
Often, due to the height of the column and concrete placing/forming sequence, splices of longitudinal bars are
necessary. Column reinforcing bars may be spliced by welding, mechanical connectors, or lapping of bars.
Welding and mechanical connectors are not often used due to quality control and bar placement difficulties.
Note that the Code limits the size of bars that are lap spliced to #11 maximum. The length of the lap dependswhether the bars are in tension or compression and the percentage of bars to be spliced for bars in tension.
Minimum length of lap for tension lap splices shall be as required for Class A or B splices, but not less than 12
inches, where:
Class A splice. . . . . . . . . . . . . . . . . . . . . .1.0 ldb
Class B splice. . . . . . . . . . . . . . . . . . . . . .1.3 ldb
where ldbis the tensile development length for the specified yield strength fy,in accordance with Work Aids 6
through 8 of Module 2 adjusted to SAES-Q-001.
For bars in compression, use ldb, using Work Aid 9 of Module 2.
It is preferred to splice no more than 50% of the bars at one splice location. Splices should be spaced at least 24
inches apart. Each splice shall involve at least one bar in each face. The 1.5dbspacing requirement also applies
to the clear distance between sets of lapped bars. Work Aid 3 contains minimum face dimensions of
rectangular tied columns accommodating various numbers of bars per face.
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For lateral ties and spirals, the Code sets the minimum diameter at 3/8 inch. For spirals, the ratio of spiral
reinforcement, s(defined as the volume of spiral reinforcement to total volume of core (out-to-out of spirals)),shall not be less than:
s= 0.45AgAc
- 1fc'
fy
where Agis gross column area and Acis area of core measured to the outside diameter of the spirals.
TENSION LAP SPLICES
Maximum percent of Asspliced within required lap length
Asprovided*
Asrequired
50 100
Equal to or greater than 2 Class A Class B
Less than 2 Class B Class B
*Ratio of area of reinforcement provided to area of reinforcement required by analysis at splice location.
Authorized reprint from ACI 318-89, Table 12.15.2, Page 204, with permission from the American Concrete
Institute.
FIGURE 10
The spacing between ties shall not exceed 16 longitudinal bar diameters, 48 tie bar diameters, or the least
dimension of the column. As mentioned in CSE 108.01, every corner and alternate longitudinal bar shall be
laterally supported by a corner of a tie, with no more than a 135 included angle. The spacing between supported
longitudinal bars shall not exceed 6 inches. The clear spacing between spirals must fall within the range of 1 to 3
inches. Lap splices for spiral reinforcement shall be 48db, but not less than 12 inches.
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Example Problem 3
This is the same column as presented in Example Problem 1 and 2.
For a rectangular tied column with bars equally distributed along four faces, select the
reinforcement.
Given: Loading
Factored axial loadPu= 560 kips
Factored momentMu= 3920 kip-in.
Materials
Compressive strength of concrete f'c= 4 ksi
Yield strength of reinforcement fy= 60 ksi
Nominal maximum size of aggregate is 1 in.
Design conditions
Unsupported length of columns lu= 10 ft
Column is braced against sidesway
b = 1 6 I n .
h
h = 20 In.
e
P
u
= 5 6 0 k
ACI 318-89
Reference Procedure Calculation
Handbook
Reference
Step 1--See Example Problem 1
Step 2--See Example Problem 1
Step 3--See Example Problem 2 Required area of steel = 12.8 in2
10.9.2
Step 4--Select optimum
reinforcement
A) Assume trial bar quantities
4
(2/face)
8
(3/face)
12
(4/face)
16
(5/face)
10.9.1
7.10.5.1
B) Determine smallest bar size
to provide Ast list resulting Ast in.2
compute resulting g= Ast/Ag and check that 0.01
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Example Problem 3 (Cont' d)
ACI 318-89
Reference Procedure Calculation
Handbook
Reference
3.3.3(c)
7.6.3
7.6.4
7.7.1
7.10.5.1
12.14.2.1
E) Check whether reinforcement can be accommodated along smaller
face with
--Bearing splices
--Normal lap splices
--Tangential lap splices
OK
Not per-
mitted
Not per-
mitted
OK
Not per-
mitted
Not per-
mitted
OK
NO
NO
NO
NO
NO
Work Aid 3
Work Aid 3
Work Aid 3
7.10.5.2
7.8
7.9
7.10
F)Determine tie spacing as least of 16 longitudinal bar diameters, in.
--48 tie bar diameters, in.
--Least dimension of column, in.
G)Consider special design details
H)Select most cost-efficient
reinforcement
36
24
16
Omitted
27
24
16
in this
20
18
16
example
(Probable)
first choice
Solution Use 12 #10 bars with bearing splices
and #3 ties spaced not more than 16 in.
apart. (Choice is based on minimum
steel requirement, use #3 ties instead of#4, and ease of handling #10 bars
instead of larger bars.)
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Frame Analysis Techniques
In this section of CSE 108.03, the participant will learn the use of beam formulas, as well as those for simple
one-story frames.
In previous sections in CSE 108.02 and CSE 108.03 covering the design of structural components (e.g., beams,
columns, slabs), the design moments, shears, and axial loads were given. As shown in Figure 11, the designer
must first determine the moments, shears, and axial loads from a structural analysis (Step 4), as well as compute
reactions and displacements. To conduct the analysis, the dimensions and sizes of the individual structure
members are assumed, based on experience and preliminary sizing guidelines.
STRUCTURAL DESIGN FLOWCHART
No
Obtain AllStructure
Requirements
SelectConstruction
Materials
DetermineDesign Loads
and LoadCombinations
StructuralAnalysis
StructuralMemberDesign
DoesDesign Match
AnalysisModel?
Develop
Drawings/Specifications
Review/MonitorConstruction
Owner / ClientOther Engineering Disciplines
Saudi Aramco Standards
Saudi Aramco Building Code
Uniform Buidling Code
SAES-Q-001ACI 318M
Saudi Aramco Standards
ACI Codes / Reports
ASTM / SASO Standards
FIGURE 11
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Using an assumed structure, the structural analysis is performed to convert the design loadings covered in CSE
108.01 (dead, live, wind, etc.) to moments and shears for each member in the structure.
Structural analysis is based on two principles:
1. Equilibrium of Forces.The structure as a whole must be in static equilibrium under the action of the
applied loads and reactions. Each member of the structure alone must also be in equilibrium with applied
loads and boundary forces. This means that the resultant of all forces and moments acting at a given
point in the structure or member must be zero.
2. Compatibility of Displacements. For any loading, the displacements of all the members of the structure
due to their respective stress-strain relationships must be consistent with respect to each other; that is, the
continuity of the structure must be satisfied. Continuity applies to deflections and rotations in the
member and structure.
For all but simple beams and one-story frames, hand calculation methods for structural analysis have been
replaced with computer analysis programs. For beams and one-story frames, published solutions may be
directly used. These solutions are also helpful in performing quick rough checks of computer-generated output.
Computer structural analysis is beyond the scope of this course.
Beam Formulas
Diagrams and formulas for moments, shears, and deflections covering a wide variety of loading patterns and
end restraint conditions are provided in standard texts, including the American Institute for Steel Construction
(AISC) Manual.
Diagrams for a uniformly distributed load for different end restraint conditions are shown in Figure 12, which
also shows the effect of end restraint on moments and deflections. Maximum positive moments in the beam
vary between 1/24 and 1/8 wl2. Maximum negative moments vary from 0 to 1/12 wl2. The maximum beam
deflection varies from 1/384 to 5/384 wl4/EI. Diagrams for a concentrated load at midspan are shown in
Figure 13. Similar variations in moment and deflection occur for the different end restraint conditions.
Note that when Figures 12 and 13 are compared by setting wl= P, moments for the concentrated load are 1.5 to
3.0 times higher than for the uniform load. As expected, deflections are higher for the concentrated load. These
diagrams can be used to check the upper and lower bounds to solutions for beams with more complicated
loadings or partial end restraint.
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BEAM DIAGRAMS AND FORMULAS
Beam Fixed At One End, Supported At Other - Uniformly Distributed Load
Beam Fixed At Both Ends Uniformly Distributed Load
R = V . . . . . . . . . . . . . . . . . . . . . . =w
2
. . . . . . . . . . . . =
. . . . . . . . . . =
Simple Beam Uniformly Distributed Load
w 2
12
w 2
24
w 4
384EI
3w8
. . . . . . . . . . . . . . . . . . . . . . =
M1 (at x = 38
) . . . . . . . . . . . . . =
R1 = V1
. . . . . . . . . =
. . . . . . . . . . . . . . . . . =
. . . . . . . . . . . . . . . . . . . . . =
w 4
185EI
9128
w 2
R2 = V2 max
Mmax
max (At x = 0.4215 )
R
V
V
Moment
Shear
2 2
M1
R
M maxM max
R = V . . . . . . . . . . . . . . . . . . . . . . . =
. . . . . . . . . . . . =
w
2
. . . . . . . . . . . . = 5w4
384EI
w 2
8M max (At Center)
max (At Center)
max (At Center)
M m ax (At Ends) . . . . . . . . . . . . =
M1(At Center)
Moment
38
4
R1
V1
V2
M1
MmaxMoment
w
x
2 2
R Rw
V
V
M max
Moment
w 2
8
R2
w
5w8
FIGURE 12
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BEAM DIAGRAMS AND FORMULAS
Simple Beam Concentrated Load at Center
R = V . . . . . . . . . . . . . . . . . . . . . . . . . =
. . . . . . . . =
. . . . . . . . . =
P2
P4
P
2 2
R
VShear
Moment
P 3
48EI
Beams Fixed at Both Ends ConcentratedLoad at Center
R = V . . . . . . . . . . . . . . . . . . . . . . =
. . . =
. . . . . . . . . . . . =
P2
P8
P
2 2
Shear
Moment
R
V
R
V
4
P3
192E
R 1 = V1
R 2 = V 2 max
. . . . . . . . . . . =
5P
16
11P16
3P16
5P
32
. . . . . . . . . . =
. . . . . . . . . . =
. . . . . . . . . . =
. . . . . . . . . . . . . . . . . . . =
. . . . . . . . . . . . . . . . . . . . . . =
P
2 2
Shear
Moment
P 3
48EI 5
7P 3
768EI
Beam Fixed at One End, Supported at Other Concentrated Load at Center
M max (At Point of Load)
max (At Point of Load)
M max (At Fixed End)
M 1 (At Point of Load)
x (At Point of Load)
R 1 R 2
V1
V 2
M 1
M max311
x
M max
M max
M max (At Center and Ends)
max (At Center)
Mmax
max (At x = 0.4472 )
FIGURE 13
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One-Story Frames
Published solutions to moments, shears, and axial forces for simple one-story frames subjected to certain
loading patterns are provided in the book entitled "Rigid Frame Formulas," by Kleinlogel, and are reproducedin part in Work Aids 4-16. Work Aids 4-12 cover the generalized solutions for different frames, with load terms
defined in Work Aids 13-16.
Figure 14 illustrates a fully fixed symmetrical rectangular frame. The solutions for this frame depend on
coefficients k, N1, and N2, and on load factors:
, ; r, 1; S, W;Mx
,
My
The coefficients are determined based on the ratio of the moment of inertia of the girder (beam), J2, to that of
the column, J1
. The load factors are determined using Work Aids 8-12 for the specific loading pattern. The
generalized solutions for two common load cases are shown in Figure 15.
For the special case of uniformly distributed load in Figure 15, note that the negative moment at the end of the
girder is given by the same expression as on the beam diagrams, except that it is modified by the factor 2/N 1.
Using the coefficients from Figure 14, based on a frame with height, h, equal to the span l, and similar moments
of inertia for the beam and column (J2/J1= 1), the value of 2/N1is 2/3. Therefore, the negative end moment for
the girder is 2/3 of that for a fully fixed restraint condition. The resulting positive moment in the span is equal
to (ql2/8 - 2*ql 2/(3*12) = 5ql2/72, or 5/3 the value for a fully-fixed beam, or 5/9 the value of a simply
supported beam
These frame solutions are directly appropriate for analyzing one-story building frames and pipe racks. Analysis
of more complicated frames can often be simplified and approximated using these published solutions.
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FRAME 41
Fully fixed symmetrical rectangular frame
B C
DA
h
Frame Dimensions andNotations
B Cx x'
Coefficients:
y'
HA
MA
HD
VDVA
MD
J1 J1
J2
k =J2J1
. h N1 = k + 2 N2= 6k + 1
This Sketch Shows the PositiveDirection of the Reactions and theCoordinates Assigned to Any Point. ForSymmetrical Loading of the Frame Usey and y'. Positive Bending MomentsCause Tension at the Face Marked by aDashed Line.
y
Source: Kleinlogel "Rigid Frame Formulas", Frederick Ungar Publishing Co. New York, 1980, Frame 41, Page
147.
FIGURE 14
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GENERALIZED SOLUTION FOR TWO COMMON LOAD CASES
-M
J1
J1
J2
P
h
A D
2B C
y2
-M y1
2M
x
-HA
-MA-V
A-V
D
MD
HD
MAMD
= + Ph2. 3k + 1
N2M BM C
= Ph2
. 3kN 2
;
H D= HA=P2
; VD= V A=2 M B ;
Case 41/8: Horizontal Concentrated Load at the Girder
Case 41/3: Girder Loaded by Any Type of Vert ical Load , Acting Symmetricall y.
B C
S
A D
J1 J1
+
B C
-My
HA HD
MA MDVA
VD
y
+
2
MA= M D= 3N 1
;
My= MA HA
Special Case: Uniformly Distributed Load S = q l
HA= HD=3M A
hVA= VD= S
2;
MB= MC= - 2MA
MA= MD= +q 2
12 N1V A= VD=
q
2
M x= M xo+ M B
max M x=q 2
8+ M B
y
-M
J2
+
+ +
+-
Source: Kleinlogel, "Rigid Frame Formulas", Frame 41, Pages 148 and 151.
FIGURE 15
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Computer Programs
During the past ten years, numerous computer programs have been developed for public use. These programs
are used for traditional structural analysis and design of structural components. The analysis and design often
use different programs.
The designer must develop the input data covering materials, structure layout, framing, and preliminary
member sizes and determine the applicable design loads and load combinations. The designer must be aware of
any limitations of the program. The use of consistent units is essential. The accuracy of the program output
depends directly on the quality of the input prepared by the designer and whether the assumptions/limitations
of the program have been checked and satisfied. After the analysis is performed and validated, the analysis
output (moments, shears, forces, reactions, and deflections) is used as input to the component design program.
The same checks regarding consistent units, program limitations, and assumptions must be made. Computer
structural analysis and design is beyond the scope of this course.
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Work Aid 1: Rectangular Column: Load-Moment Interaction Diagrams
COLUMNS 7.4.3Load-moment strength interaction diagram for R4-60.75 columns
References: ACI 318-89 Sections 9.3.2.2., 10.2, and 10.3: ACI Publication SP-7. pp 152-182
Authorized reprint from ACI, SP-17(90) Vol. 2, Columns 7.4.3 , Page 82 with permission from the American
Concrete Institute.
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COLUMNS 7.4.4Load-moment strength interaction diagram for R4-60.90 columns
References: ACI 318-89 Sections 9.3.2.2., 10.2, and 10.3: ACI Publication SP-7. pp 152-182
Authorized reprint from ACI, SP-17(90) Vol. 2, Columns 7.4.4, Page 82 with permission from the American
Concrete Institute.
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Work Aid 2: Spirally Reinforced Column: Load-Moment Interaction
Diagrams
COLUMNS 7.23.1Load-moment strength interaction diagram for C5-60.45 spirallyreinforced columns
References: ACI 318-89 Sections 9.3.2.2. 10.2. and 10.3; ACI Publication SP-7, pp. 152-182
Authorized reprint from ACI, SP-17(90) Vol. 2, Columns 7.23.1 , Page 119 with permission from the American
Concrete Institute.
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COLUMNS 7.23.1Load-moment strength interaction diagram for C5-60.60 spirally
reinforced columns
References: ACI 318-89 Sections 9.3.2.2. 10.2. and 10.3; ACI Publication SP-7, pp. 152-182
Authorized reprint from ACI, SP-17(90) Vol. 2, Columns 7.23.2, Page 119 with permission from the American
Concrete Institute.
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Work Aid 3: Minimum Face Dimensions of Rectangular Tied Columns
Accommodating Various Numbers of Bars per Face
Note: This Work Aid needs to be adjusted for specific Saudi Aramco cover requirements per SAES-Q-001.
Authorized reprint from ACI, SP-17 (90) Vol. 2, Reinforcement 2, Page 165 with permission from the
American Concrete Institute.
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Work Aid 4: Frame 39, Symmetrical Rectangular Two-Hinged Frame, Case
#1
Source: Kleinlogel, "Rigid Frame Formulas," Page 138, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 5: Frame 39, Symmetrical Rectangular Two-Hinged Frame, Case
#2-4
Source: Kleinlogel, "Rigid Frame Formulas," Page 139, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 6: Frame 39, Symmetrical Rectangular Two-Hinged Frame, Case
#5-7
Source: Kleinlogel, "Rigid Frame Formulas," Page 140 ,Frederick Ungar Publishing Company, New York,
1980
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Work Aid 7: Frame 39, Symmetrical Rectangular Two-Hinged Frame,
Symmetrical Loading - Case #8-10
Source: Kleinlogel, "Rigid Frame Formulas," Page 141, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 9: Frame 41, Symmetrical Rectangular Fully-Fixed Frame, Case
#2-3
Source: Kleinlogel, "Rigid Frame Formulas," Page 148, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 10: Frame 41, Symmetrical Rectangular Fully-Fixed Frame, Case
#4-5
Source: Kleinlogel, "Rigid Frame Formulas," Page 149, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 11: Frame 41, Symmetrical Rectangular Fully-Fixed Frame, Case
#6-7
Source: Kleinlogel, "Rigid Frame Formulas," Page 150, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 12: Frame 41, Symmetrical Rectangular Fully-Fixed Frame, Case
#8-9
Source: Kleinlogel, "Rigid Frame Formulas," Page 151, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 13: Appendix Load Terms; General Notations
Source: Kleinlogel, "Rigid Frame Formulas," Page 440, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 14: Appendix Load Terms; Notations, Case 1-3
Source: Kleinlogel, "Rigid Frame Formulas," Page 441, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 15: Appendix Load Terms; General Notations, Case 4-6
Source: Kleinlogel, "Rigid Frame Formulas," Page 442, Frederick Ungar Publishing Company, New York,
1980
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Work Aid 16: Appendix Load Terms; Notations, Case 7-11
Source: Kleinlogel, "Rigid Frame Formulas," Page 443, Frederick Ungar Publishing Company, New York,
1980
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GLOSSARY
biaxial bending Bending moments about two perpendicular axes.
column strip Area of slab adjacent to column line used to analyze and design two-way
slabs.
compatibility of
displacements
Defected shape of frame or member. Must be smooth (no kinks) unless
pinned joints are used.
drop panels Areas of thickened slab surrounding columns for enhancing shear transfer.
equilibrium of forces All forces (internal and external) must equate to zero.
inflection point Point along axis of member where moments change sign.
lap spliced Describes how load transfer occurs from one rebar to another, bars are
overlapped.
middle strip Strip outside of column strip used for analysis and design of two-way slabs.
one-way slabs Reinforcement and loads carried in one principal direction.
orthogonal direction Perpendicular directions (two).
perimeter or punchingshear
Shear failure mode where column or load breaks through slab on four sidesor circle.
sidesway Lateral sway or deflection of frame due to lateral or unsymmetric gravity
loads.
spiral reinforced Lateral reinforcing in columns, made by a continuous vertical spiral.
stiffness matrix
approach
Method used to analyze frames and structures using matrix operations using
member stiffness.
tie reinforced Lateral reinforcing in columns by one or more discrete enclosure bars.
two-way slab Reinforcement and loads carried in both orthogonal directions.