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ECCENTRIC LOADING of
COLUMNS
B T venkata jagadeesh
15131D1603
Given By
A seminar on
INTRODUCTION
• Bars are categorized as columns and beams.
• Column is geometrically looks like beam but loading is different when
compared.
• Beam takes up load perpendicularly where as column takes axial.
• Again columns divided into SHORT and LONG columns.
• Both the types differentiated by ratio of length(a) to lateral
dimensions(b).
• Long columns have more (a/b) ratio comparted to short columns.
• Short columns, generally withstand higher compressive loads.
• Long columns are used in axial load applications.
Cntd..
Beam Types
• Simple
• Continuous
• Cantilever
Moment
(fixed at one end)
Beam Types
• Fixed
Moments at each end
• Propped- Fixed at one end supported at other
• Overhang
Forces and Supports
Supports are translated into forces and moments in a free body diagrams.
The following are three common supports and the forces and moments used to
replace them.
Roller:
Pin
Connection:
Fixed
Support:
Fy
Fy
Fx
Fx
Fy
Mo
Centric loading:
The load is applied at the centroid of the cross section. The limiting
allowable stress is determined from strength (P/A) or buckling.
Eccentric loading:
The load is offset from the centroid of the cross section because of how the
beam load comes into the column. This offset introduces bending along
with axial stress.
P
𝜎 = 𝑃/𝐴
P
e
This causes Direct stress
e
No problem with this set
e
This set forms couple
M = P * e
P
P
P
Eccentric loading :
Combined force and couple
Application of Eccentric load using short column :
P
e
P
M = P * e
This causes bending stress in
addition to direct stress.
P
e
P
M = P * e
P
𝜎 = 𝑃/𝐴
M = P * e
Max. Bending Stress
Max. Bending Stress
Direct stress𝜎 = 𝑃/𝐴
b= M ∗ C2/I
C1C2
b= M ∗ C1/I
𝜎min 𝜎max
𝛔min = P/A - M*C2/I
𝛔max = P/A + M*C1/I
Representation of stress distribution
ECCENTRIC LOADING ON LONG COLUMNS
=
Take sec - AQ
Taking sec – A Q
Algebraic sum of moments about Q be zero
We know
let
differential equation
EQUATION SOLVING
General solution
Boundary conditions
Put in y(max)
By solving
Important NOTE
Time to replace with next slide
Than QThan QThan Q

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Ppt on ecdcentric loading of short column

  • 1. ECCENTRIC LOADING of COLUMNS B T venkata jagadeesh 15131D1603 Given By A seminar on
  • 2. INTRODUCTION • Bars are categorized as columns and beams. • Column is geometrically looks like beam but loading is different when compared. • Beam takes up load perpendicularly where as column takes axial. • Again columns divided into SHORT and LONG columns. • Both the types differentiated by ratio of length(a) to lateral dimensions(b). • Long columns have more (a/b) ratio comparted to short columns. • Short columns, generally withstand higher compressive loads. • Long columns are used in axial load applications. Cntd..
  • 3. Beam Types • Simple • Continuous • Cantilever Moment (fixed at one end)
  • 4. Beam Types • Fixed Moments at each end • Propped- Fixed at one end supported at other • Overhang
  • 5. Forces and Supports Supports are translated into forces and moments in a free body diagrams. The following are three common supports and the forces and moments used to replace them. Roller: Pin Connection: Fixed Support: Fy Fy Fx Fx Fy Mo
  • 6. Centric loading: The load is applied at the centroid of the cross section. The limiting allowable stress is determined from strength (P/A) or buckling. Eccentric loading: The load is offset from the centroid of the cross section because of how the beam load comes into the column. This offset introduces bending along with axial stress. P 𝜎 = 𝑃/𝐴 P e This causes Direct stress
  • 7. e No problem with this set e This set forms couple M = P * e P P P Eccentric loading : Combined force and couple
  • 8. Application of Eccentric load using short column : P e P M = P * e This causes bending stress in addition to direct stress.
  • 9. P e P M = P * e P 𝜎 = 𝑃/𝐴 M = P * e Max. Bending Stress Max. Bending Stress Direct stress𝜎 = 𝑃/𝐴 b= M ∗ C2/I C1C2 b= M ∗ C1/I
  • 10. 𝜎min 𝜎max 𝛔min = P/A - M*C2/I 𝛔max = P/A + M*C1/I Representation of stress distribution
  • 11. ECCENTRIC LOADING ON LONG COLUMNS = Take sec - AQ
  • 12. Taking sec – A Q Algebraic sum of moments about Q be zero We know let differential equation
  • 13. EQUATION SOLVING General solution Boundary conditions Put in y(max) By solving
  • 14. Important NOTE Time to replace with next slide