3. WHAT IS UNIAXIAL COLUMN?
• Uniaxial columns are those whose rotation are strictly bound
about a single axis only as the load eccentricity is pure about
a single axis.
4. Bending moments can occur in
columns because:
• - Unbalance gravity
loads
• - Lateral loads: wind,
earthquake
• - In general a certain
degree of eccentricity is
present due to
inaccuracies in
construction.
7. Column types based on dimensional
property
SHORT – both lex/h and ley/b < 15 for braced columns
< 10 for unbraced columns
BRACED - If lateral stability to structure as a whole is
provided by walls or bracing designed to resist all lateral
forces in that plane, else unbraced.
28. 1. If column controlled by compression
eccentricity, e= M/P <= eb :• For symmetrical tied
column, eb=
(0.67Pgm+0.17)d
• For spiral column, eb=
(0.43PgmDs+0.14t)
• Pg= As/Ag
• m= fy/0.85f’c
29. Selection of column is done using the relation –
(fa/Fa) + (fb/Fb) <=1, (this is also known as interaction
formula.)
Where,
fa= P/Ag
Fa= 0.34(1+Pgm)fc’
fb= M/S
S= I/c (c=0.5h)
Fb= 0.45 fc’
30. 2. If column controlled by tension,
e>eb
Section of column shall be proportioned by using the
relationM= Mo+ P/Pb (Mb-Mo)
Pb & Mb shall be calculated by using the relation(fa/Fa) + (fb/Fb)= 1
=> (Pb/Ag)/Fa + (Pbeb/S)/Fb= 1
For symmetrical tied column, Mo= 0.4 As fy (d-d’), As is
tensile steel.
For spiral column, Mo= 0.12 Ast fy Ds, Ast is total steel.
32. Slender columns are also becoming
increasingly important and popular
because of the following reasons:
(i) the development of high strength materials
(concrete and steel),
(ii) improved methods of dimensioning and
designing with rational and reliable design
procedures,
(iii) innovative structural concepts – specially, the
architect’s expectations for creative structures.
33. HYSTERESIS?
• slender or long columns may fail at a much lower value of the load
compared to short column when sudden lateral displacement of the
member takes place between the ends. Thus, short columns
undergo material failure, while long columns may fail by buckling
(geometric failure) at a critical load or Euler’s load, which is much
less in comparison to that of short columns having equal area of
cross-section.
• The general expression of the critical load Pcr at which a member
will fail by buckling is as follows:
• Pcr = π2EI /(kl)2 or,
• Pcr = π2EA /(kl/r)2
• Thus, Pcr of a particular column depends upon kl/r or slenderness
ratio. It is worth mentioning that kl is termed as effective length le of
the column.
34.
35.
36.
37. Critical loads in terms of Pcr of hinge-hinge column and effective lengths le = kl
of elastic and reinforced concrete columns with different boundary conditions
and for a constant unsupported length l
Sl.
No.
Support conditions
Critical load
Pcr
Effective length
le = kl
(A) Elastic single columns
1.
2.
3.
4.
5.
6.
Hinged at both ends, no
sidesway
Fixed against rotation at
both ends – no sidesway
Partially restrained against
rotation by top and bottom
cross-beams, no sidesway
Fixed at one end and
entirely free at other end –
sidesway not prevented
Rotationally fixed at both
ends – sidesway not
prevented
Partially restrained against
rotation at both ends –
sidesway not prevented
Pcr
l
4Pcr
0.5 l
Between Pcr and 4Pcr
l > kl > l/2
0.25 Pcr
2 l, one PI is on imaginary
extension
Pcr
l, one PI is on imaginary
extension
Between zero and slightly
less than Pcr *
l < kl < α
(B) Reinforced concrete columns
7.
8.
Hinged portal frame – no
sidesway
Hinged portal frame –
> Pcr
kl < l
<< Pcr
kl > 2 l
38. Additional Moment Method of
analysis of short column
1.Braced columns bent
symmetrically in single
curvature and
undergoing balanced
failure
M = Mo + Py
=>Mo + P ea = Mo + Ma
• Where,
• P is the factored design
load
• M are the total factored
design moments
• Mo are the primary
factored moments
• Ma are the additional
moments
• ea are the additional
eccentricities
41. Assuming the column undergoes a balanced failure when Pu=
Pub, the maximum curvature at the mid-height section of
the column can be expressed as given below, assuming
(i) the values of εc = 0.0035, εst = 0.002 and d’/D = 0.1, and
(ii) the additional moment capacities are about eighty per cent
of the total moment. Then,
φ = eighty per cent of {(0.0035 + 0.002)/0.9D}
or φ = 1/200D
Substituting the value of φ , we have
ea = D(le/D)2/2000
42. • Therefore, the additional moment Ma can be
written as,
• Ma = Py = PΔ = Pea = (PD/2000) (le/D)2
43. 2.Braced columns subjected to unequal primary
moments at the two ends
• Mo max = 0.4 M1 + 0.6 M2
• Mo max ≥ 0.4 M2
Where,
• M1 is smaller end moment.
• M2 is larger end moment.
