EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 10 Columns

Columns

Introduction

Columns are vertical prismatic members subjected to

compressive forces

Goals:

1. Study the stability of columns: resistance to buckling

under axial loadings

2. Determine critical load, stress, and/or column

dimension information (length/cross-section area)

Column Subjected to Vertical Load

Example: a vertical column with pin connections at both

ends subject to centric axial compressive load P

Concerns addressed before (chapters 1-2):

• (Internal compressive) stress is within

allowable limits

• (Axial or normal) deformation is within

the allowable limit

Loads/length should not be too large while

transverse section not too small

𝜎 =𝑃

𝐴< 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

𝛿 =𝑃𝐿

𝐴𝐸< 𝛿𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

Concern with Stability for a Column

Additional concern for columns under compression:

Due to unavoidable imperfect

alignment for loads and/or

environmental disturbance,

axial compressive force/stress

could lead to transverse

deformation or buckling,

often in a sudden way, causing

loss of stability for a column.

• Consideration of a column’s

stability puts additional

constraints on compressive

load and part dimension

Pin & Spring Connected Rigid Rods (1)Consider two rigid rods of AC and BC pin connected at A and

B and also connected at C by a pin and a torsional spring of

constant K Perfect

alignment

After

slight disturbance

Implication:

• When actual load P < Pcr, system will return to straight

system stable

To determine whether the system is stable or

not against a small disturbance , consider

the top rod AC at an angle and draw FBD

If reaching equilibrium, balance of moment:

We have

Pin & Spring Connected Rigid Rods (2)

When disturbance

~ 0 (very small)

)2(sin2

KL

Pcr

LKPcr /4

𝑠𝑖𝑛∆𝜃 ≈ ∆𝜃

When P > Pcr, system

moves further away from

straight/vertical state, if it

settles into a new

“equilibrium” state

with a finite angle ,

Because not close to 0,

equation cannot be

simplified, and we have

Pin & Spring Connected Rigid Rods (3)

)2(sin2

KL

P

sin4

K

PL

For in the range of 0 - /2,

> sin Therefore:

• For P < Pcr

No such exist system

stable and remains in the

straight state

• For P > Pcr (but not too large)

may be found numerically

However, system unstable

because disturbance for slight

increase in P necessitates

further increase in , which

eventually lead to collapse or

loss of stability

Pin & Spring Connected Rigid Rods (4)

P < Pcr P > Pcr

1

sin/44

crP

P

LK

P

K

PL

Euler’s Formula for Pin-Ended Columns (1)

Similarly, to determine critical load Pcr for a pin-ended

column AB, the FBD for the entire column is drawn

For part AQ near top

draw FBD:

Bending moment

M = -Py

From chapter 9

Rewrite, we have

𝑑2𝑦

𝑑𝑥2=𝑀

𝐸𝐼= −

𝑃

𝐸𝐼𝑦

𝑑2𝑦

𝑑𝑥2+𝑃

𝐸𝐼𝑦 = 0

Define

The general solution to this

differential equation is:

Consider boundary conditions:

• x = 0, y = 0 B = 0

• x = L, y = 0

Euler’s Formula for Pin-Ended Columns (2)

Therefore,

From previous𝑑2𝑦

𝑑𝑥2+𝑃

𝐸𝐼𝑦 = 0 𝑝2 =

𝑃

𝐸𝐼

𝑑2𝑦

𝑑𝑥2+ 𝑝2𝑦 = 0

𝑦 = 𝐴𝑠𝑖𝑛𝑝𝑥 + 𝐵𝑐𝑜𝑠𝑝𝑥

𝐴𝑠𝑖𝑛𝑝𝐿 = 0

1. A = 0 y 0 the column remains straight! or

2. sin pL = 0 pL = n

Recall

Therefore, P satisfy

Smallest Pcr

occurs when n = 1

Euler’s formula

Euler’s Formula for Pin-Ended Columns (3)

Two possibilities:

2

2

L

EIPcr

We have,

𝐴𝑠𝑖𝑛𝑝𝐿 = 0

𝑝 =𝑛𝜋

𝐿

𝑝2 =𝑃

𝐸𝐼

𝑃 =𝑛2𝜋2𝐸𝐼

𝐿2

For previous, if A ≠ 0

When P = Pcr

Hence

This is the elastic curve when

P = Pcr

Euler’s Formula for Pin-Ended Columns (4)

2

2

L

EIPcr

Since B = 0, the general solution

becomes

𝑝2 =𝑃𝑐𝑟𝐸𝐼

=𝜋2𝐸𝐼

𝐿2𝐸𝐼=𝜋2

𝐿2

𝑝 =𝜋

𝐿

𝑦 = 𝐴𝑠𝑖𝑛𝑝𝑥 + 𝐵𝑐𝑜𝑠𝑝𝑥

𝑦 = 𝐴𝑠𝑖𝑛𝜋𝑥

𝐿

• If 0 < P < Pcr

Euler’s Formula for Pin-Ended Columns (5)

To satisfy boundary condition of

x = L, y = Asin (x/L) = 0

A = 0 and y = Asin (x/L) ≡ 0,

The system remains stable (i.e.,

column remains straight and

stable against small disturbance)

• If P ≥ Pcr the system becomes

unstable: a small disturbance

(e.g., misalignment of force)

may cause the system to

suddenly change from straight to

a buckled state

0 ≤ P < Pcr P ≥ Pcr

pL < sin pL = 0 cannot be satisfied

Critical load

Euler’s Formula for Pin-Ended Columns (4)

2

2

L

EIPcr

Notes:

If a column has circular or square cross-

section, it may buckle in different

directions

If a column has other geometry, it tends

to buckle in the orientation for which it

has the smallest I. Therefore, to

determine critical load Pcr , I = Imin for

the cross-section should be used.

For the rectangle as illustrated, Imin = Iy

x

y

Iy < Ix

Critical load

Setting where rg is radius of gyration

L/rg Slenderness ratio of the column

Euler’s Formula for Pin-Ended Columns (5)

2

2

L

EIPcr

Critical Stress

2

2

)/( g

crrL

E We have

If a column has different directions with different moment

of inertia, the minimum radius of gyration should be used

2

gArI

𝜎𝑐𝑟 =𝑃𝑐𝑟𝐴

=𝜋2𝐸𝐼

𝐿2𝐴

Critical load

Euler’s Formula for Pin-Ended Columns (6)

2

2

L

EIPcr

Critical Stress 2

2

)/( g

crrL

E

Notes about critical stress

• Proportional to E

• Decreases as

slenderness ratio L/rg

increases

• Ceases to be meaningful

when

cr > yield strength Y

Column w/ One End Fixed & One End FreeFor column AB with B side fixed and A side free & subjected

to compressive load P:

It can be treated as top half

of a pin connected vertical

column as discussed before,

with effective length Le = 2L

Applied force/stress

has to be even lower to avoid

loss of stability

2

2

e

crL

EIP

2

2

)/( ge

crrL

E

Column w/ Both Ends Fixed (1)For column AB with both

A and B side fixed

subjected to compressive

load P.

Draw FBD for AB and

top half AC

Symmetry of loads &

supports dictates

• Zero horizontal reactions

at A & B Zero shear at

C

• Center symmetric for AC around D

D is an inflection point (i.e., infinite curvature) with

zero bending moment

Le = L/2

Column w/ Both Ends Fixed (2)For column AB with both A and B sides fixed and subjected

to compressive load P, symmetry considerations enable to

treat DE section in the center as pin-ended column, with

effective length

2

2

e

crL

EIP

2

2

)/( ge

crrL

E

Applied force/stress

can be a bit higher due to

way of fixing at two ends

Summary

2

2

e

crL

EIP

2

2

)/( ge

crrL

E

More stableLess stable

1. Long columns: obey Euler’s Equation

2. Short columns: dominated by y

3. Intermediate columns: mixed behavior

Empirical Test Data for Critical Stress

Design of Columns under an Eccentric Load

Two approaches:

Eccentric loads are treated as a

centric load P superposed with

bending moment M = Pe

Maximum compressive stress

• Allowable Stress Method

• Interaction Method

Over

conservative

𝜎 = 𝜎𝑐𝑒𝑛𝑡𝑟𝑖𝑐 + 𝜎𝑏𝑒𝑛𝑑𝑖𝑛𝑔

𝜎𝑚𝑎𝑥 =𝑃

𝐴+𝑀𝑐

𝐼

𝑃

𝐴+𝑀𝑐

𝐼≤ 𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒

𝑃/𝐴

𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒(𝑐𝑒𝑛𝑡𝑟𝑖𝑐)+

𝑀𝑐/𝐼

𝜎𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒(𝑏𝑒𝑛𝑑𝑖𝑛𝑔)≤ 1

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Homework 10.0

Read section 10.1, 10.3 and 10.4 and give a statement confirm

the reading

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Homework 10.1

Knowing the spring constant at D is of constant k and the column DE is rigid. Please calculate the critical load Pcr

E

L

Dk

P

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Homework 10.2

A column EH is hang horizontally by two springs with equal length and

spring constant k symmetrically at F and G, as illustrated. Please

calculate the critical load Pcr

E

k

P’P

F G Hb

L

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Homework 10.3

A long cylinder shaped solid column has length L and outside

diameter (OD) d. It is made of a material with elastic modulus E. If a

separate cylinder-shaped hollow column is to be made with the same

material and same length L and O.D. d, but has inner diameter of 0.5d,

please calculate (a) the saving in weight for the hollow column versus

the original solid column, and (b) the relative reduction in critical load

for the hollow column versus the original solid cylinder.

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 1 Introduction

Homework 10.4

Members EF and GH are 32 mm diameter

and 4 m long steel rod and members of EH

and FG are 24 mm diameter and 3 m long

steel road. When rope between EG and FH

are tightened, and if the factor of safety is

2, please calculate the largest allowable

tension force in EG. Knowing E = 200 GPa

and consider buckling only in the plane of

the structure.

E H

F G

4 m

3 m