Design of columns

1. Reinforced Concrete Design-II
Lecture 01: Slender Columns
Semester – Fall 2020
Dr.Tahir Mehmood

2. Difference between short and Slender or Long column
Strength Control Criteria Stiffness Control Criteria

3. Slender column
• If a column is relatively slender, it may
deflect laterally and failed by bending
(Buckling) under axial loading rather than
by compression.
• A slender column failed by buckling at so-
called critical buckling load (Pc) or
Eurler’s buckling load.
𝑃𝑐 =
𝜋2𝐸𝐼
(𝑘𝑙)2

4. Braced Vs un-braced frames
In summary:
• Strength of concentrically loaded
column decreases with increasing
kl/r
• If a columns is braced then
1/2<kl<1
• If a not braced then effective length
is always larger than l

5. Compression plus bending
Deflection due to bending alone
Deflection under combined action of P and M
𝑀𝑚𝑎𝑥 = 𝑀0 + 𝑃∆ 1

6. Compression plus bending
𝑀𝑚𝑎𝑥 = 𝑀0 + 𝑃∆ ∆= ∆0
1
1 −
𝑃
𝑃𝑐
where
𝑀𝑚𝑎𝑥 = 𝑀0
1
1 −
𝑃
𝑃𝑐
2
𝑀𝑚𝑎𝑥 = 𝑀0 + 𝑃∆0
1
1 −
𝑃
𝑃𝑐
Equation 2 can be written as
3
Moment magnification

7. Moment magnification in single and double curvature bending
End Moment
Magnified moment = End Moment
Magnified moment = Slightly more
than End Moment
Double Curvature

8. Moment magnification in single and double curvature bending
End Moment
Magnified moment = Much more than end Moment
Single Curvature
Magnified moment is greatly
amplified

9. Moment magnification in single and double curvature bending
We can generalize that:
• The moment 𝑀0 will be most strongly amplified at the location where deflection (y) is
maximum and this occurs in the members bent into single curvature by symmetrical loads or
equal end moments.
• If end moments are unequal but have the same sign i.e., produce single curvature, 𝑀0 will
still be strongly magnified but not as much as for equal end moment case.
On the other hand
• There will no or a little magnification of moment 𝑀0 when end moments are of opposite sign
(i.e., producing double curvature).
• Moment magnification depends on relative values of end moments and can be expressed as:
𝑀𝑚𝑎𝑥 = 𝑀0
𝐶𝑚
1 −
𝑃
𝑃

10. Moment magnification in single and double curvature bending
Where 𝐶𝑚
𝐶𝑚 = 0.6 + 0.4
𝑀1
𝑀2
≥ 0.4
• Here 𝑀1 is the numerical smaller and 𝑀2 is numerical larger value of end moment.
• The fraction
𝑀1
𝑀2
is positive when end moment produce single curvature and negative when
they produce double curvature.

11. ACI criteria for slenderness effects in columns
Slenderness effects may be neglected for non-sway (i.e., braced) frames when:
𝑘𝑙𝑢
𝑟
≤ 34 − 12
𝑀1
𝑀2
Where 34 − 12
𝑀1
𝑀2
is not taken greater than 40.
𝑀1 is the smaller factored end moment value and taken as positive when member bent in single
curvature and negative when member bent in double curvature. 𝑀2 is the larger value and
always taken as positive.
Slenderness effects may be neglected for unbraced or sway frames when:
𝑘𝑙𝑢
𝑟
< 22
The radius of gyration r:
r=0.3h (rectangular columns) and r=0.25D for circular columns

12. ACI-Moment Magnifier method for frames braced against sway (i.e.,
non-sway frames)
ACI equation for magnified moment acting with factored axial load Pu is given as :
𝑀𝑐 = 𝛿𝑛𝑠𝑀2
𝛿𝑛𝑠=
𝐶𝑚
1−
𝑃𝑢
0.75𝑃𝑐
≥ 1
𝑃𝑐 =
𝜋2𝐸𝐼
𝑘𝑙𝑢
2
𝐶𝑚 = 0.6 + 0.4
𝑀1
𝑀2
≥ 0.4
The fraction
𝑀1
𝑀2
is positive when end moment produce single curvature and negative when
they produce double curvature.

13. ACI-Moment Magnifier method for frames braced against sway (i.e.,
non-sway frames)
According to ACI 𝑀2 should not be taken less than:
𝑀2,𝑚𝑖𝑛 = 𝑃𝑢 0.6 + 0.03ℎ
Effective stiffness can be calculated as:
𝐸𝐼 =
0.4𝐸𝑐𝐼𝑔
1 + 𝛽𝑑𝑛𝑠
𝛽𝑑𝑛𝑠 =ratio of maximum factored axial sustained load to maximum factored axial load
associated with the same load combination, but not greater than 1.

14. Design Procedure
Step 1: Select a trial section to carry factored axial load (𝑃𝑈) and moment (𝑀𝑈=𝑀2), assuming
short column behavior.
Step 2: Check for consideration of slenderness effect using
𝑘𝑙𝑢
𝑟
≤ 34 − 12
𝑀1
𝑀2
Where 34 − 12
𝑀1
𝑀2
is not taken greater than 40.
Use k=1
Step 3: If slenderness found to be important, then refine the value of k based on alignment
chart with member stiffness EI/l and rotational resistance factor 𝛹.

15. Design Procedure

16. Design Procedure
Step 4: Check Minimum moment
Step 5: Calculate 𝐶𝑚
Step 6: Calculate 𝛽𝑑𝑛𝑠 , and
Step 7: Calculate moment magnification factor 𝛿𝑛𝑠 and
Step 8: Check adequacy of the column to resist axial load and magnified moment, revise the
section dimension and reinforcement if necessary.
Step 9: If column dimensions are revised, determine the revised moment magnification
factor and check adequacy of new design.
𝑀2,𝑚𝑖𝑛 = 𝑃𝑢 0.6 + 0.03ℎ
𝐶𝑚 = 0.6 + 0.4
𝑀1
𝑀2
≥ 0.4
𝐸𝐼 =
0.4𝐸𝑐𝐼𝑔
1 + 𝛽𝑑𝑛𝑠
𝑃𝑐 =
𝜋2
𝐸𝐼
𝑘𝑙𝑢
2
𝑀𝑐 = 𝛿𝑛𝑠𝑀2

17. THANK YOU