Short Columns
Related NSCP 2015
Provisions
409.7.6.4 Lateral Support of Compression Reinforcement
409.7.6.4.1
Transverse reinforcement shall be provided throughout the distance where
longitudinal compression reinforcement is required. Lateral support of longitudinal
compression reinforcement shall be provided by closed stirrups of hoops in
accordance with Sections 409.7.6.4.2 through 409.7.6.4.4.
409.7.6.4.2
Size of transverse reinforcement shall be at least (a) or (b). Deformed wire or welded
wire reinforcement of equivalent area shall be permitted.
a. 10 mm∅ for longitudinal bars 32 mm∅ and smaller
b. 12 mm∅ for longitudinal bars 36 mm∅ and larger and for longitudinal bundled
bars.
409.7.6.4.3
Spacing of transverse reinforcement shall not exceed the least of (a) through (c):
a. 𝟏𝟔𝒅𝒃 of longitudinal reinforcement;
b. 𝟒𝟖𝒅𝒃 of transverse reinforcement;
c. Least dimension of beam.
409.7.6.4.4
Longitudinal compression reinforcement shall be arranged such that very corner and
alternate compression bar shall be enclosed by the corner of the transverse
reinforcement with an included angle of not more than 135 degrees, and no bar shall
be farther than 150 mm. clear on each side along the transverse reinforcement from
such an enclosed bar.
410 Columns
410.1 Scope
410.1.1
This section shall apply to the design of non-prestressed, prestressed, and composite
columns, including reinforced concrete pedestals.
410.1.2
Design of plain concrete pedestals shall be in accordance with Section 414.
410.2 General
410.2.1 Materials
410.2.1.1
Design properties for concrete shall be selected to be in accordance with Section 419.
410.2.1.2
Design properties for steel reinforcement and structural steel used in composite
columns shall be selected to be in accordance with Section 420.
410.2.1.3
Materials, design, and detailing requirements for embedments in concrete shall be in
accordance with Section 420.7.
410.2.2 Composite Columns
410.2.2.1
If a structural steel shape, pipe, or tubing is used as longitudinal reinforcement, the
columns shall be designed as a composite column.
410.2.3 Connection to Other Members
410.2.3.1
For cast-in-place construction, beam-column and slab-column joints shall satisfy
Section 415.
410.2.3.2
for precast construction, connections shall satisfy the force transfer requirements of
Section 416.2.
410.2.3.3
Connections of columns to foundations shall satisfy Section 416.3.
410.3 Design Limits
410.3.1 Dimensional Limits
410.3.1.1
For columns with a square, octagonal, or other shaped cross sections, it shall be
permitted to base gross area considered, required reinforcement, and design strength
on a circular section with a diameter equal to the least lateral dimension of the actual
shape.
410.3.1.2
For columns with cross sections larger than required by considerations of loading, it
shall be permitted to base gross area considered, required reinforcement, and design
strength on a reduced effective area, not less than one-half the total area. This
provision shall not apply to columns in special moment frames designed in accordance
with Section 418.
410.3.1.3
For columns built monolithically with a concrete wall, the outer limits of the effective
cross section of the column shall not be taken greater than 40 mm. outside the
transverse reinforcement.
410.3.1.4
for columns with two or more interlocking spirals, outer limits of the effective cross
concrete cover.
410.3.1.5
If a reduced effective area is considered according to sections 410.3.1.1 through
410.3.1.4, structural analysis and design of other parts of the structure that interact
with the column shall be based on the actual cross section.
410.3.1.6
For composite columns with a concrete core encased by structural steel, the thickness
of the steel encasement shall be at least (a) or (b):
a.
𝒃
b. 𝒉
𝒇𝒚
for each face of width 𝒃;
𝟑𝑬𝒔
𝒇𝒚
𝟖𝑬𝒔
for circular sections of diamter 𝒉;
410.4.1 Required Strength
410.4.1.1
Required strength shall be calculated in accordance with the factored load
combinations in Section 405.
410.4.1.2
Required strength shall be calculated in accordance with the analysis procedures in
Section 406.
410.4.2 Factored Axial Force and Moment
410.4.2.1
𝑷𝒖 and 𝑴𝒖 occurring simultaneously for each applicable factored load combination
shall be considered.
410.5 Design Strength
410.5.1 General
410.5.1.1
For each applicable factored load combinations, design strength at all sections shall
satisfy ∅𝑺𝒏 ≥ 𝑼, including (a) through (d). Interaction between load effects shall be
considered:
a. ∅𝑃𝑛 ≥ 𝑃𝑢
b. ∅𝑀𝑛 ≥ 𝑀𝑢
c. ∅𝑉𝑛 ≥ 𝑉𝑢
d. ∅𝑇𝑛 ≥ 𝑇𝑢
410.5.1.2
∅ shall be determined in accordance with Section 421.2.
410.5.2 Axial Force and Moment
410.5.2.1
𝑷𝒏 and 𝑴𝒏 shall be calculated in accordance with Section 422.4.
410.5.2.2
For composite columns, forces shall be transferred between the steel section and
concrete by direct bearing, shear connectors, or bond in accordance to the axial
strength assigned to each component.
410.6 Reinforcement Limits
410.6.1 Minimum and Maximum Longitudinal Reinforcement
410.6.1.1
For non-prestressed columns and for prestressed columns with average 𝒇𝒑𝒆 <
1.6 MPa, area of longitudinal reinforcement shall be at least 𝟎. 𝟎𝟏𝑨𝒈 but shall not
exceed 𝟎. 𝟎𝟖𝑨𝒈.
410.6.1.2
For composite columns with a structural steel core, area of longitudinal bars located
within the transverse reinforcement shall be at least 𝟎. 𝟎𝟏 𝑨𝒈 − 𝑨𝒔𝒙 , but shall not
exceed 𝟎. 𝟎𝟖 𝑨𝒈 − 𝑨𝒔𝒙 .
410.7 Reinforcement Detailing
410.7.1 General
410.7.1.1
Concrete cover for reinforcement shall be in accordance with Section 420.6.1
410.7.1.2
Development lengths of deformed and prestressed reinforcement shall be in
accordance with Section 425.4.
410.7.1.3
Bundled bars shall be in accordance with Section 425.6.
410.7.2 Reinforcement Spacing
410.7.2.1
Minimum spacing s shall be in accordance with Section 425.2.
410.7.3 Longitudinal Reinforcement
410.7.3.1
For non-prestressed columns and for prestressed columns with average 𝑓𝑝𝑒 <1.60
MPa, the minimum number of longitudinal bars shall be (a), (b), or (c):
a.) Three within triangular ties;
b.) Four within rectangular or circular ties;
c.) Six enclosed by spirals or for columns of special moment frames enclosed by
circular hoops
414 Plain Concrete
414.1 Scope
414.1.1
This section shall apply to the design of plain concrete members, including (a) and
(b):
a. Members in building structures;
b. Members in non-building structures such as arches, underground utility
structures, gravity walls, and shielding walls.
414.1.3
Plain concrete shall be permitted only in cases (a) through (d):
a. Members that are continuously supported by soil or supported by other structural
members capable of providing continuous vertical support;
b. Members for which arch action provides compression under all conditions of
loading;
c. Walls
d. Pedestals
414.1.4
Plain concrete shall be permitted for a structure assigned to seismic zone 4, only in
cases (a) and (b):
a. Footings supporting cast-in-place reinforced concrete or reinforced masonry walls
provided the footings are reinforced longitudinally with at least two continuous
reinforcing bars. Bars shall be at least two continuous reinforcing bars. Bars shall be
at least 12 mm∅ and have a total area of not less than 0.002 times the gross crosssectional area of the footing. Continuity of reinforcement shall be provided at
corners and intersections;
b. Foundation elements (i) through (iii) for detached one-and two-family dwellings not
exceeding three storeys and constructed with stud bearing walls:
i. Footing supporting walls;
ii. Isolated footings supporting columns or pedestals
iii. Foundation or basement walls not less than 190 mm. thick and retaining no more
than 1.2 m of unbalanced fill.
414.1.5
Plain concrete shall not be permitted for columns and pile caps.
414.2 General
414.2.1.1
Design properties for concrete shall be selected to be in accordance with Section 419.
414.2.1.2
Steel reinforcement, if required, shall be selected to be in accordance with Section
420.
414.2.1.3
Materials, design, and detailing requirements for embedments in concrete shall be in
accordance with Section 420.7.
414.3.3 Pedestals
414.3.3.1
Ratio of unsupported height to average least lateral dimension shall not exceed 3.
Section 415 Beam-Concrete and Slab-Column Joints
415.1 Scope
415.1.1
This section shall apply to the design and detailing of cast-in-place beam-column and
slab-column joints.
415.2 General
415.2.1
Beam-column and slab-column joints shall satisfy Section 415.3 for transfer of
column axial force through the floor system.
415.2.2
If gravity load, wind, earthquake, or other lateral forces cause transfer of moment at
beam-column or slab-column joints, the shear resulting from moment transfer shall
be considered in the design of the joint.
415.2.3
Beam-column and slab-column joints that transfer moment to columns shall satisfy
the detailing provisions in Section 415.4. beam-column joints within special moment
frames, slab-column joints within intermediate moment frames, and beam-column and
slab-column joints using frames not designated as part of the seismic-force-resisting
systems in structures assigned to seismic zone 4, shall satisfy Section 418.
415.2.4
A beam-column joint shall be considered to be restrained if the joint is laterally
supported on four sides by beams of approximately equal depth.
415.2.5
A slab-column joint shall be considered to be restrained if the joint is laterally
supported on four sides by the slab
415.3 Transfer of Column Axial Force through the Floor System
415.3.1
If 𝒇′𝒄 of a column is greater than 1.4 times that of the floor system, transmission of
axial force through the floor system shall in accordance with (a), (b), or (c):
a. Concrete of compressive strength specified for the column shall be place in the
floor at the column location. Column concrete shall extend outward at least 600
mm into the floor slab from face of column for the full depth of the slab and be
integrated with floor concrete.
b. The design strength of a column through a floor system shall be calculated using
the lower value of concrete strength with vertical dowels and spirals as required
to achieve adequate strength;
c. For beam-column and slab-column joints that are restrained in accordance with
Section 415.2.4 or 415.2.5, respectively, it shall be permitted to calculate the
design strength of the column on an assumed concrete strength in the column
joint equal to 75 percent of column concrete strength plus 35 percent of floor
concrete strength, where the value of column concrete strength shall not exceed
2.5 times the floor concrete strength.
415.4 Detailing of Joints
415.4.1
Beam-column and slab-column joints that are restrained in accordance with Section
415.2.4 or 415.2.5, respectively, and are not part of a seismic-force-resisting system
need not satisfy the provisions for transverse reinforcement of Section 415.4.2.
415.4.2
The area of all legs of transverse reinforcement in each principal direction of beamcolumn and slab-column joints shall be at least the greater of (a) and (b):
𝑏𝑠
a. 0.062 𝑓′𝑐
𝑓𝑦𝑡
𝑏𝑠
b. 0.35
𝑓𝑦𝑡
Where 𝒃 is the dimension of the column section perpendicular to the direction under
consideration.
415.4.2.1
At beam-column and slab-column joints, an area of transverse reinforcement
calculated in accordance with Section 415.4.2 shall be distributed within the column
height not less than the deepest beam or slab element framing into the column.
415.4.2.2
For beam-column joints, the spacing of the transverse reinforcement s shall not
exceed one-half the depth of the shallowest beam.
415.4.3
If longitudinal beam or column reinforcement is spliced or terminated in a joint,
closed transverse reinforcement in accordance with Section 410.7.6 shall be provided
in the joint, unless the joint region is restrained in accordance with Section 415.2.4 or
415.2.5.
415.4.4
Development of longitudinal reinforcement terminating in the joint shall be in
accordance with Section 425.4.
422.4.2 Maximum Axial Compressive Strength
422.4.2.1
Nominal axial compressive strength, 𝑷𝒏 , shall not exceed 𝑷𝒏,𝒎𝒂𝒙 , in accordance with
Table 422.4.2.1, where 𝑷𝒐 is calculated by Eq. 422.4.2.2 for non-prestressed members
and composite steel and concrete members, and by Eq.422.4.2.3 for prestressed
members.
Table 422.4.2.1
Maximum Axial Strength
Member
Non-prestressed
Prestressed
Composite steel and
concrete columns in
accordance with
Section 410
Transverse Reinforcement
𝑃𝑛,𝑚𝑎𝑥
Ties conforming to Section 422.4.2.4
0.80𝑃𝑜
(a)
Spirals conforming to Section
422.4.2.5
0.85𝑃𝑜
(b)
Ties
0.80𝑃𝑜
(c)
Spirals
0.85𝑃𝑜
(d)
All
0.85𝑃𝑜
(e)
422.4.2.2
For non-prestressed members and composite steel and concrete members, 𝑷𝒐 shall be
calculated by
𝑷𝒐 = 𝟎. 𝟖𝟓𝒇′𝒄 𝑨𝒈 − 𝑨𝒔𝒕 + 𝒇𝒚 𝑨𝒔𝒕 ⟶ 𝟒𝟐𝟐. 𝟒. 𝟐. 𝟐
Where 𝑨𝒔𝒕 is the total area of non-prestressed longitudinal reinforcement.
422.4.2.3
For prestressed members, 𝑷𝒐 shall be calculated by:
𝑷𝒐 = 𝟎. 𝟖𝟓𝒇′𝒄 𝑨𝒈 − 𝑨𝒔𝒕 − 𝑨𝒑𝒅 + 𝒇𝒚 𝑨𝒔𝒕 − 𝒇𝒔𝒆 − 𝟎. 𝟎𝟎𝟑𝑬𝒑 𝑨𝒑𝒕 ⟶ 𝟒𝟐𝟐. 𝟒. 𝟐. 𝟑
Where 𝑨𝒑𝒕 is the total area of prestressing reinforcement, 𝑨𝒑𝒅 is the total area
occupied by duct, sheathing, and prestressing reinforcement, and the value of 𝒇𝒔𝒆
shall be at least 𝟎. 𝟎𝟎𝟑𝑬𝒑 . For grouted, post-tensioned tendons, it shall be permitted
to assume 𝑨𝒑𝒅 equals 𝑨𝒑𝒕.
425.7.3 Spirals
425.7.3.1
Spirals shall consist of evenly spaced continuous bar or wire with clear spacing
conforming to (a) and (b):
a. At least the greater of 25 mm and 𝟒Τ𝟑 𝒅𝒂𝒈𝒈
b. Not greater than 75 mm
425.7.3.2
For cast-in-place construction, spiral bar or wire diameter shall be at least 10 mm.
425.7.3.3
Volumetric spiral reinforcement ratio 𝝆𝒔 shall satisfy Eq. 425.7.3.3
𝑨𝒈
𝒇′𝒄
𝝆𝒔 ≥ 𝟎. 𝟒𝟓
−𝟏
⟶ 𝟒𝟐𝟓. 𝟕. 𝟑. 𝟑
𝑨𝒄𝒉
𝒇𝒚𝒕
Where the value of 𝒇𝒚𝒕 shall not be taken greater than 700 MPa.
Introduction
Columns are structural elements used primarily to support compressive loads.
Short, stocky columns subjected to small bending moments are often called
“axially loaded”. A reinforced concrete column can be roughly divided into the
following categories.
a.) Short Compression Block/Pedestals
if the height of an upright compression member is less than three times
its least dimensions, it may be considered to be a pedestals. A pedestal may be
designed with unreinforced or plain concrete with a maximum design compressive
stress equal to 0.85∅𝑓′𝑐 𝐴𝑔 .
b.) Short Reinforced Concrete Columns
should a reinforced concrete column fail due to initial material failure, it
is classified as short column. The load that it can support is controlled by the
dimensions of the cross section and the strength of the materials of which it is
constructed. A short column is a stocky member with little flexibility.
c.) Long or Slender Reinforced Concrete Columns
one in which the load capacity is influenced by slenderness which
produces additional bending because of transverse deformations. As slenderness
ratios are increased, bending deformations will increase as will the resulting
secondary moments. If these moments are of such magnitude, the axial load
capacity of columns is significantly reduce.
Types of Reinforced Columns
a.) Tied Column
a tied column has a series of closed ties. These ties are effective in
increasing the column strength. They prevent the longitudinal bars from being
displaced during construction and they resist the tendency of the same bars to
buckle outwards under load, which would cause the outer concrete cover to break
or spall off. Tied columns are ordinarily square or rectangular, but they can be
octagonal, round, L-shaped, etc.
b.) Spiral Column
a spiral column has a continuous helical spiral made from bars or heavy
wire wrapped around the longitudinal bars. Spirals are even more effective than
ties in increasing column strength. The closely spaced spirals do a better job of
holding the longitudinal bars in place, and also confine the concrete inside and
greatly increase its resistance to axial compression. Spiral column are normally
rounds, but they also can be made into rectangular, octagonal, or other shapes.
Spirals are usually used for large heavily loaded columns and in seismic areas
because of their resistance to earthquake loads.
c.) Composite Column
These are concrete columns that are reinforced longitudinally by
structural steel shapes, which may or may not be surrounded by structural bars, or
they may consist of structural tubing filled with concrete (commonly called lally
column).
Short Columns
(𝑃𝑢 ≤ ∅𝑃𝑛 )
𝑷𝒖 =?
203.3 Load Combination using Strength Design or Load
and Resistance Factor Design
203.1 Basic Load Combinations
1.4(D+F)
(203-1)
1.2(D+F+T)+1.6(L+H)+0.5(Lr or R)
(203-2)
1.2D+1.6(Lr or R)+(f1L or 0.5W)
(203-3)
1.2D+1.0W+f1L+0.5(Lr or R)
(203-4)
1.2D+1.0E+f1L
(203-5)
0.9D+1.0W+1.6H
(203-6)
0.9D+1.0E+1.6H
(203-7)
Where
f1
=1.0 for floors in places of public assembly, for live loads in excess of 4.8kPa,
and for garage live load, or
=0.5 for other live loads
203.2 Other Loads
Where “P” is to be considered in design, the applicable load shall be added to Section
203.3.1 factored as 1.2P.
Load Combination (Section 405.3; 2015)
1.) 1.4𝐷
2.) 1.2𝐷 + 1.6𝐿 + 0.5(𝐿𝑟 𝑜𝑟 𝑅)
3.) 1.2𝐷 + 1.6(𝐿𝑟 𝑜𝑟 𝑅) + (1.0𝐿𝑟 𝑜𝑟 0.5𝑅)
4.) 1.2𝐷 + 1.0𝑊 + 1.0𝐿 + 0.5(𝐿𝑟 𝑜𝑟𝑅)
5.) 1.2𝐷 + 1.0𝐸 + 1.0𝐿
6.) 0.9𝐷 + 1.0𝑊
7.) 0.9𝐷 + 1.0𝐸
𝑮𝒓𝒆𝒂𝒕𝒊𝒆𝒔𝒕 𝑮𝒐𝒗𝒆𝒓𝒏𝒔‼!
405.3.3 The load factor on live load L in eq. 3, 4 and 5, shall be permitted to 0.5
except for (a), (b), or (c):
a.) Garages
b.) Areas occupied as places of public assembly
c.) Areas where L is greater than 4.8 kPa.
∅ =?
Strength Reduction Factors
(Section 421; NSCP 2015)
Table 421.2.1
Strength Reduction Factors,𝝓
Action or Structural Element
𝜙
Exceptions
Moment, axial force, or combined moment and axial force
0.65 to 0.90
-Check NSCP
Shear
0.75
-Check NSCP
Torsion
0.75
-
Bearing
0.65
-
Post-tensioned anchorage zones
0.85
-
Brackets and Corbels
0.75
-
0.75
-
0.90
-
Plain concrete elements
0.90
-
Anchors in concrete elements
0.45-0.75 in
accordance with
section 417
Struts, ties, nodal zones, and bearing areas designed in
accordance with strut-and-tie method in Section 423
Components of connections of precast members controlled
by yielding of steel elements in tension
NSCP Table 421.2.2
Strength Reduction Factor, ∅, for Moment, Axial Force, or Combined Moment and
Axial Force
∅
Net tensile
strain 𝜖𝑡
Classification
Type of transverse reinforcement
Spiral conforming to Sect.
425.7.3
𝜖𝑡 ≤ 𝜖𝑡𝑦
Compression
Controlled
𝜖𝑡𝑦 < 𝜖𝑡
< 0.005
Transition
𝜖𝑡 ≥ 0.005
Tension
Controlled
0.75
0.75 + 0.15
𝜖𝑡 − 𝜖𝑦
0.005 − 𝜖𝑡𝑦
0.90
Other
(a)
(c)
(e)
0.65
0.65 + 0.25
𝜖𝑡 − 𝜖𝑦
0.005 − 𝜖𝑡𝑦
0.90
421.2.2.1 for deformed reinforcement, 𝝐𝒕𝒚 shall be 𝑓𝑦 /𝐸𝑠 . For Grade 280 deformed
reinforcement, it shall be permitted to take 𝝐𝒕𝒚 equal to 0.002
(b)
(d)
(f)
𝑷𝒏 =?
Yield Axial Resistance, 𝑃𝑜 =?
By applying equilibrium
Condition: If 𝒆 = 𝟎
Σ𝐹𝑣 = 0
𝑃𝑜
𝑃𝑜 = 𝑃𝑐 + 𝑃𝑠
From the concrete;
𝑃𝑐
𝑃𝑐 = 0.85𝑓′𝑐 𝐴𝑐 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠
Note
𝐴𝑔 = 𝐴𝑐 + 𝐴𝑠 ; 𝐴𝑐 = 𝐴𝑔 − 𝐴𝑠
From the steel;
𝑃𝑠 = 𝑓𝑦 𝐴𝑠
𝑃𝑠
𝑒
Cross
Section
𝑃𝑜
By substitution,
𝑃𝑜 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠 + 𝑓𝑦 𝐴𝑠
By applying Gross Steel Ratio, 𝝆𝒈
𝐴𝑠
; 𝐴𝑠 = 𝜌𝑔 𝐴𝑔
𝜌𝑔 =
𝐴𝑔
𝑃𝑜 = 0.85𝑓′𝑐 𝐴𝑔 − 𝜌𝑔 𝐴𝑔 + 𝑓𝑦 𝜌𝑔 𝐴𝑔
𝑃𝑜 = 𝐴𝑔 0.85𝑓′𝑐 1 − 𝜌𝑔 + 𝑓𝑦 𝜌𝑔
(422.4.2.2)
Kern of Column (Elastic Analysis)
𝑃
𝑀 = 𝑃𝑒
𝑃
Review from M.D.B.
Axial Stress Alone Bending Stress Alone
𝑃
𝑀𝑐
𝜎1 =
𝜎2 =
𝐴
𝐼
Combined Axial & Bending Stress
𝑃
𝑀𝑐
𝜎𝑡𝑜𝑡𝑎𝑙 = ±
𝐴
𝐼
Combined Stress Diagram:
Assumed 𝜎1 ≥ 𝜎2 𝑠𝑎𝑦 𝜎1 = 𝜎2
𝑒
L
R
L
𝜎1
Axial Stress
Diagram
C
𝜎2
T
Bending Stress
Diagram
R
𝜎1
𝜎2
Combined Stress
at Side L
0
𝑃
𝑀𝑐
𝜎𝑡𝑜𝑡𝑎𝑙 = −
𝐴 ℎ 𝐼
𝑃𝑒
2
𝑃 𝑀𝑐
=
𝐴
𝐼 𝑏ℎ3
𝑏ℎ
12
𝒉
𝒆=
𝟔
Permissible Eccentricity
Tied Column:
Spiral Column:
𝑒 = 0.10ℎ
𝑒 = 0.05ℎ
Nominal Axial Load, 𝑷𝒏 =?
𝑃𝑛 = 𝛼𝑃𝑜
Table 422.4.2.1 NSCP 2015
Tied Column:
Spiral Column:
𝛼 = 0.80
𝛼 = 0.85
Limiting Gross Steel Ratio
410.6.1 NSCP 2015
𝜌𝑚𝑖𝑛 = 0.01
𝜌𝑚𝑎𝑥 = 0.08
Transverse Reinforcement
Required Diameter
409.7.6.4.2
Size of transverse reinforcement shall be at least (a) or (b). Deformed wire or
welded wire reinforcement of equivalent area shall be permitted.
a. 10 mm∅ for longitudinal bars 32 mm∅ and smaller
b. 12 mm∅ for longitudinal bars 36 mm∅ and larger and for longitudinal bundled
bars.
Tied Column:
Maximum Spacing - 409.7.6.4.3
Spacing of transverse reinforcement shall not exceed the least of (a)
through (c):
a. 𝟏𝟔𝒅𝒃 of longitudinal reinforcement;
b. 𝟒𝟖𝒅𝒃 of transverse reinforcement;
c. Least dimension of beam.
Spiral Column:
425.7.3.3
Volumetric spiral reinforcement ratio 𝝆𝒔 shall satisfy
𝑨𝒈
𝒇′𝒄
𝝆𝒔 ≥ 𝟎. 𝟒𝟓
−𝟏
Note 𝒇𝒚𝒕 ≤ 700 MPa
𝑨𝒄𝒉
𝒇𝒚𝒕
Minimum & Maximum Spacing - 425.7.3.1
a. At least the greater of 25 mm and 𝟒Τ𝟑 𝒅𝒂𝒈𝒈
b. Not greater than 75 mm
Plastic Centroid
Is the centroid of resistance of the
section if all the concrete is compressed to the
maximum stress and all the steel is
compressed to the yield stress with uniform
strain over the section.
Problem
The T-shaped column shown is reinforced with 4-28 mm
diameter bar with fy=414 MPa. Determine the plastic
centroid of the column measured from the 400 mm side,
f ’c=24 MPa.
75mm
75mm
100mm
200mm
400mm
100mm
150
200
275
250
75
𝑻𝟏 = 𝟓𝟎𝟗. 𝟖𝟒𝐤𝐍
75mm
𝑻𝟏
𝑻𝟐 = 𝟓𝟎𝟗. 𝟖𝟒𝐤𝐍
𝑻𝟐
75mm
𝑪𝟏
𝑪𝟏 = 1198.87
400mm
𝑪𝟐 = 790.88
200mm
𝑪𝟐
𝑥ҧ
150
100mm
𝑃𝑡𝑜𝑡𝑎𝑙
200
100mm
Computing 𝑻𝟏 & 𝑻𝟐 =?
414
𝑇1 = 𝑓𝑦 𝐴𝑠,1 = 509,842.79 𝑁 = 509.84 𝑘𝑁 ; Note 𝑇2 = 𝑇1 = 509.84 𝑘𝑁
2
𝜋 2
𝑑 𝑁
4 𝑏
28
28 2
Computing 𝑪𝟏 & 𝑪𝟐 =? 𝜋
𝑑𝑏 2 𝑁
24
4
𝐶1 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠,1
= 1,198,877.31 𝑁 = 1198.87 𝑘𝑁
150
𝑏ℎ
28
2
400
𝜋 2
𝑑 𝑁
24
4 𝑏
𝐶2 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠,2
= 790,877.31 𝑁 = 790.88 𝑘𝑁
200
𝑏ℎ
200
Computing 𝑷𝒕𝒐𝒕𝒂𝒍 =?
509.84 509.84 1198.87790.88
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑇1 + 𝑇2 + 𝐶1 + 𝐶2 = 3009.43 𝑘𝑁
ഥ =?
Computing 𝒙
Using Varignon’s Theorem
509.84
3009.43
509.84
1198.87
790.88
𝑃𝑡𝑜𝑡𝑎𝑙 𝑥ҧ = 𝑇1 75𝑚𝑚 + 𝑇2 275𝑚𝑚 + 𝐶1 75𝑚𝑚 + 𝐶2 250𝑚𝑚
ഥ = 𝟏𝟓𝟒. 𝟖𝟕 𝒎𝒎
𝒙
Problem
Calculate the ultimate axial load capacity of a rectangular column
350 mm x 450 mm reinforced with six 25 mm bars. Use fy=276 MPa and
f ’c=28 MPa.
Solution 0.65
𝑃𝑢 ≤ ∅𝛼𝑃𝑜
0.80
Therefore
28
157,500 276 2945.24
; 𝑃𝑜 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠 + 𝑓𝑦 𝐴𝑠
2945.24
𝑃𝑜 = 4,491,289.52 𝑁
𝑃𝑢 = 0.65 0.80 4,491,289.52
25
350
; 𝐴𝑔 = 𝑏ℎ
; 𝐴𝑠 =
450
𝐴𝑔 = 157,500
𝜋 2
𝑑 𝑁
4 𝑏
6
𝐴𝑠 = 2945.24
= 2,335,470.55 𝑁 = 2335.47 𝑘𝑁
Problem
Design a square tied column to support an axial dead load of 850
kN and an axial live load of 1040 kN. Use fy=414 MPa and f ’c=27.6
MPa. Assume 2 % steel ratio and use 28 mm main bars and 10 mm ties.
Solution
850
𝑃𝑢 ≤ ∅𝛼𝑃𝑜
Therefore
0.65
1040
; 𝑃𝑢 = 1.2𝑃𝐷𝐿 + 1.6𝑃𝐿𝐿
27.6
0.02 414
𝑃𝑢 = ∅𝛼 𝐴𝑔 0.85𝑓′𝑐 1 − 𝜌𝑔 + 𝑓𝑦 𝜌𝑔
2684 1000
= 2684 𝑘𝑁
0.02
; 𝐴𝑔 = 165,059.37 𝑚𝑚2
0.80
𝒙
Computing Column Width, 𝒙 =?
165,059.37
𝐴𝑔 = 𝑥 2 ; 𝑥 = 406.27 𝑚𝑚
𝒙
Safe Width, 𝒙𝒔𝒂𝒇𝒆 =?
𝑥𝑠𝑎𝑓𝑒 = 450 𝑚𝑚
Computing the number of rebar, 𝑵 =?
4050
0.02
𝐴𝑠
; 𝐴𝑠 = 𝜌𝑔 𝐴𝑔 = 4050 𝑚𝑚2
𝑁=𝜋
𝑑2
4
450 450
28
𝑁 = 6.58 ≈ 7 𝑝𝑐𝑠
Computing Lateral Tie Spacing, 𝒔 =?
From Section 409.7.6.4.3 (NSCP 2015)
28
1. ) 16𝑑𝑏 = 448 𝑚𝑚
10
2. ) 48𝑑𝑡
= 480 𝑚𝑚
3. ) 𝑙𝑒𝑎𝑠𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 = 450 𝑚𝑚
Problem
Calculate the ultimate axial load capacity of a 480 mm round spiral
column reinforced with six 22 mm bars. Use fy=345 Mpa and f ’c=34
MPa.
Ans: 3793.50 kN
Solution
0.75
𝑃𝑢 ≤ ∅𝛼𝑃𝑜
0.85
180,955.74
2280.80
345
34
480
; 𝑃𝑜 = 0.85𝑓′𝑐 𝐴𝑔 − 𝐴𝑠 + 𝑓𝑦 𝐴𝑠 ; 𝐴 = 𝜋 𝐷 2
𝑔
4
2280.80
𝑃𝑜 = 5,950,581.77 𝑁
22
; 𝐴𝑠 =
𝐴𝑔 = 180,955.74 𝐴𝑠 = 2280.80
Therefore
𝑃𝑢 = 0.75 0.85 5,950,581.77
𝜋 2
𝑑 𝑁
4 𝑏
6
= 3,793,495.88 𝑁 = 3793.50 𝑘𝑁
Problem
Design a round spiral column to support an axial dead load of
950 kN and an axial live load of 1210 kN. Use fy=276 MPa and
f ’c=27.6 MPa. Assume 2 % steel ratio and use 28 mm bars and 10 mm
spiral. Use 38 mm steel cover.
Solution
950
𝑃𝑢 ≤ ∅𝛼𝑃𝑜
Therefore
0.75
1210
; 𝑃𝑢 = 1.2𝑃𝐷𝐿 + 1.6𝑃𝐿𝐿
27.6
0.02 276
𝑃𝑢 = ∅𝛼 𝐴𝑔 0.85𝑓′𝑐 1 − 𝜌𝑔 + 𝑓𝑦 𝜌𝑔
3076 1000
= 3076 𝑘𝑁
0.02
; 𝐴𝑔 = 169,237.55 𝑚𝑚2
0.85
𝑫
𝑺𝒄
𝒅𝒄𝒉
Computing Column Diameter, 𝑫 =?
169,237.55
𝜋
𝐴𝑔 = 𝐷 2 ; 𝐷 = 464.20 𝑚𝑚
4
Safe D, 𝑫𝒔𝒂𝒇𝒆 =?
𝐷𝑠𝑎𝑓𝑒 = 500 𝑚𝑚
Center
Spacing of Spiral Reinforcement
𝒅𝒄𝒉
𝐴𝑠 𝐿
𝜌𝑠 =
𝐴𝑐ℎ 𝐿
𝑉𝑠
=
𝑉𝑐ℎ
𝜋 𝑑𝑐ℎ − 𝑑𝑠
𝜋 2 𝑑𝑠 2 𝑑𝑐ℎ − 𝑑𝑠
𝑉𝑠 = 𝐴𝑠 𝐿 =
4
𝜋 2
𝑑
4 𝑠
𝑠
𝑽 =?
𝑽𝒔 =?
𝒔
𝒄𝒉
𝑉𝑐ℎ = 𝐴𝑐ℎ 𝐿
𝜋
𝑑𝑐ℎ 2
4
𝜋𝑑𝑐ℎ 2
=
𝑠
4
By Substitution:
𝜋 2 𝑑𝑠 2 𝑑𝑐ℎ − 𝑑𝑠
2
𝜋𝑑
𝑑𝑐ℎ − 𝑑𝑠
𝑠
4
;
𝑠
=
𝜌𝑠 =
𝜌𝑠 𝑑𝑐ℎ 2
𝜋𝑑𝑐ℎ 2
𝑠
4
Computing the number of rebar, 𝑵 =?
3926.99
0.02
𝐴𝑠
; 𝐴𝑠 = 𝜌𝑔 𝐴𝑔 = 3926.99 𝑚𝑚2
𝑁=𝜋
𝑑2
500
4
𝜋
2
𝐷
28
4
𝑁 = 6.38 ≈ 7 𝑝𝑐𝑠
Computing Spacing of Spiral, 𝒔 =?
10 434 10
38 10
500
𝜋𝑑𝑠 2 𝑑𝑐ℎ − 𝑑𝑠
𝑑𝑠
𝑠=
= 434 𝑚𝑚
;
𝑑
=
𝐷
−
2
𝑆
−
𝑐ℎ
𝑐
2
𝜌𝑠 𝑑𝑐ℎ 2
500
𝜋
27.6
2
0.015
𝐷
434
𝐴𝑔 4
𝑓′𝑐 276
; 𝜌𝑠 = 0.45
−1
= 0.015
𝐴
𝑓
𝑐ℎ
25𝑚𝑚 < 𝑠 = 47.15 𝑚𝑚 < 75𝑚𝑚
𝑺𝒂𝒇𝒆
Safe Spacing, 𝒔𝒔𝒂𝒇𝒆 =?
𝑠𝑠𝑎𝑓𝑒 = 45 𝑚𝑚
𝑦𝑡
434
𝜋
2
𝑑
4 𝑐ℎ
Eccentrically Loaded Column with Uniaxial Bending
(Elastic)
Nominal Moment
𝑀𝑛 = 𝑃𝑛 𝑒
𝑀𝑛
𝑒=
𝑃𝑛
Eccentricity
𝑃𝑛
𝒆
Cross
Section
𝒆
𝑃𝑛
𝒆
Cross
Section
C
T
𝑃𝑛
𝒆
𝑃𝑛
𝒆
𝑃𝑛
𝑃𝑛
𝒆
Cross
Section
C
C
T
T
Failures of Eccentrically Loaded Column
a.) Balance Failure
A balance failure occurs when tension steel just reaches the yield
strength and the extreme fiber concrete compressive strain reaches 0.003 at the
same time.
b.) Tension Failure
A tension failure occurs if 𝑃𝑢 < 𝑃𝑏 since the smaller column load means
that 𝑐 < 𝑐𝑏 , and in reference to the strain, 𝜖𝑠 > 𝜖𝑦 .
NOTE:
in tension failure, the tension steel will yield while the compression steel
is not yielding except at 𝑃 ≅ 0.
c.) Compression Failure
A compression failure occurs if 𝑃𝑢 > 𝑃𝑏 , since the larger column load
means that 𝑐 > 𝑐𝑏 and in terms of strain 𝜖𝑠 < 𝜖𝑦 , or tension steel is not yielding.
NOTE:
in compression failure, the tension steel does not reach the yield strain
while the compression steel may or may not yield.
d.) Pure Flexure
the section in this case is subjected to bending moment where axial load
is zero.
Analysis of Eccentrically Loaded Column
𝑃𝑛
𝒉
Applying Equilibrium Equation:
Σ𝐹𝑣 = 0
𝑃𝑛 = 𝐶𝑐 + 𝐶𝑠 − 𝑇
e
𝒃
Applying Varignon’s Theorem:
𝑃𝑛 𝑒 = 𝐶𝑐
𝒅
𝒂
Stress
Diagram
𝑇
𝐶𝑐
𝜖𝑠
Neutral Depth:
0.003𝑑
𝑐=
𝜖𝑠 + 0.003
𝐶𝑠
Strain Diagram
𝜖𝑐 = 0.003
𝒄
ℎ 𝑎
ℎ
−
+ 𝐶𝑠
− 𝑑′
2 2
2
ℎ
+𝑇 𝑑 −
2
Problem
A 300 x 500 mm column is reinforced with 4-28 mm∅, one in each corner.
The concrete cylinder strength is f ’c=24 MPa and the steel yield
strength is 345 MPa. Cover to centroid of bar is 65 mm. Calculate
a.) load P for balanced failure and the corresponding eccentricity e.
b.) What nominal force can be safely applied at an eccentricity of ?
Ans:Pb=1436.73kN;eb=242.04;Pn=1327.29kN260 mm
e
500mm
300
435mm
𝐏
𝒆 =e
𝟐𝟒𝟐. 𝟎𝟔
𝟏𝟖𝟓𝒎𝒎
𝐏
500mm
𝑷𝒃 = 𝟏𝟒𝟑𝟔. 𝟔𝟏kN
300
435mm
a
Stress Diagram
𝑻
𝑻 = 𝟒𝟐𝟒. 𝟖𝟕𝐤𝐍
𝑪𝟏 = 𝟏𝟒𝟑𝟔. 𝟔𝟏kN
𝑪
𝟏
𝑪𝑪𝟐 = 𝟒𝟐𝟒. 𝟖𝟕kN
𝟐
Solution: Balance Failure
Tension Side:
28
345 1231.50
𝜋 2
; 𝐴𝑠 = 𝑑𝑏 𝑁 = 1231.50 𝑚𝑚2
𝑇 = 𝑓𝑦 𝐴𝑠
4
2
𝑇 = 424,867.5 𝑁 = 424.87 𝑘𝑁
Compression Side: Assumed Steel Yields
435
0.85
300
65
Concrete
24
600𝑑
𝐶1 = 0.85𝑓′𝑐 𝑎𝑏 ; 𝑎 = 𝛽1 𝑐𝑏 ; 𝑐𝑏 =
; 𝑑 = 500 − 𝑆𝑐 = 435 𝑚𝑚
𝑓𝑦 + 600
65
276.19
276.19
234.76
𝑐 − 𝑑′
345
𝐶1 = 1436.73 𝑘𝑁 𝑎 = 234.76 𝑐𝑏 = 276.19 ; 𝑓′𝑠 = 600
𝑐
276.19
28
1231.50 345
24
𝑓′𝑠 = 458.79 > 𝑓𝑦
Steel
𝜋 2
𝐶2 = 𝐴′𝑠 𝑓𝑦 − 0.85𝑓′𝑐 ; 𝐴′𝑠 = 𝑑𝑏 𝑁 = 1231.50 𝑚𝑚2
4
2
𝐶2 = 399,744.9 𝑁 = 399.74 𝑘𝑁
Total Axial Load
𝑃𝑏 = 𝐶1 + 𝐶2 − 𝑇 = 1436.73 + 424.87 − 399.74 = 1461.86 𝑘𝑁
↻ +𝚺𝑴𝒄𝒆𝒏𝒕𝒓𝒐𝒊𝒅 = 𝟎 500 234.76 500
424.87 500
65
435
399.74
1436.73 1436.73
ℎ 𝑎
ℎ
ℎ
; 𝑒 = 238.80 𝑚𝑚
𝑃𝑏 𝑒 = 𝐶1
−
+ 𝐶2
− 𝑑′ + 𝑇 𝑑 −
2 2
2
2
𝒆 =e
𝟐𝟔𝟎𝒎𝒎
𝟏𝟖𝟓𝒎𝒎
𝐏
500mm
𝑷 = 𝟏𝟑𝟐𝟕. 𝟑𝟏kN
300
435mm
a
Stress Diagram
𝑻
𝑻 = 𝟒𝟐𝟒. 𝟖𝟕𝐤𝐍
𝑪𝟏 = 5.202𝑐 𝑘𝑁
𝑪𝟏
𝑪𝑪𝟐 = 𝟒𝟐𝟒. 𝟖𝟕kN
𝟐
Solution
Tension Side: Assume Steel yields
28
345 1231.50
𝜋 2
; 𝐴𝑠 = 𝑑𝑏 𝑁 = 1231.50 𝑚𝑚2
𝑇 = 𝑓𝑦 𝐴𝑠
4
2
𝑇 = 424,867.5 𝑁 = 424.87 𝑘𝑁
Compression Side: Assume Steel yields
0.85
24
300
𝐶1 = 0.85𝑓′𝑐 𝑎𝑏 ; 𝑎 = 𝛽1 𝑐 ∴ 𝐶1 = 0.85𝑓′𝑐 𝛽1 𝑐𝑏 = 5202𝑐 𝑁 = 5.202𝑐 𝑘𝑁
1231.50 345
24
28
𝜋 2
; 𝐴′𝑠 = 𝑑𝑏 𝑁 = 1231.50 𝑚𝑚2
4
2
𝐶2 = 399,744.9 𝑁 = 399.74 𝑘𝑁
𝐶2 = 𝐴′𝑠 𝑓𝑦 − 0.85𝑓′𝑐
Total Axial Load
𝑃 = 𝐶1 + 𝐶2 − 𝑇 = 5.202𝑐 + 399.74 − 424.87 = 5.202𝑐 − 25.13 𝑘𝑁
0.85
↻ +𝚺𝑴𝒄𝒆𝒏𝒕𝒓𝒐𝒊𝒅 = 𝟎
5.202𝑐
500
424.87 500
500 𝛽1 𝑐
65
5.202𝑐 − 25.13
435
399.74
ℎ 𝑎
ℎ
ℎ
; 𝑐 = 256.74 𝑚𝑚
𝑃 𝑒 = 𝐶1
−
+ 𝐶2
− 𝑑′ + 𝑇 𝑑 −
2 2
2
2
260
Check Assumptions
256.74
0.85
𝑎 = 𝛽1 𝑐 = 218.23 𝑚𝑚
435
256.74
𝑑−𝑐
𝑓𝑠 = 600
= 416.59 𝑀𝑃𝑎 > 𝑓𝑦 Correct!!
𝑐
256.74 65
𝑐 − 𝑑′
𝑓′𝑠 = 600
𝑐
= 448.1 𝑀𝑃𝑎 > 𝑓𝑦 Correct!!
256.74
Therefore
256.74
𝑃 = 5.202𝑐 − 25.13 𝑘𝑁 = 1310.43 𝑘𝑁