Mechanics of Solids (VDB1063)

Column Stability

Lecturer: Dr. Montasir O. Ahmed

LEARNING OUTCOMES

•To determine the critical load on a column with various types of supports

LECTURE OUTLINES

• Critical Load

• Ideal Column with Pin Supports

• Columns Having Various End-Conditions

• Long slender members subjected to an axial compressive force are called columns,

and the lateral deflection that occurs is called buckling.

• The maximum axial load that a column can support when it is on the verge of buckling

is called the critical load.

• Additional loading will cause the column to buckle .

CRITICAL LOAD

1. Stable equilibrium: Restoring force is greater than the applied load (F > 2×P tanθ) OR (k Δ > 2P tanθ)

OR (k θ L/2 > 2P tanθ) OR (P < k L/4).

2. Unstable equilibrium: Restoring force is less than the applied load (P > k L/4).

3. Neutral equilibrium: Restoring force is equal to the applied load (P = k L/4)

• Three different states for the column members:

CRITICAL LOAD

IDEAL COLUMN WITH PIN SUPPORTS

• Ideal column characteristics:

It is perfectly straight before loading

Both ends are pin-supported

Loads are applied throughout the centroid of the cross section

Euler load

𝑃𝑐𝑟 =𝜋2𝐸𝐼

𝐿2where

Pcr = Critical or max. axial load on the column just before it begins to buckle. This load must not cause the

stress in the column to exceed the proportional limit.

E = Modulus of elasticity

I = least moment of inertia for column’s cross sectional area

L = Unsupported length of the column, whose ends are pinned.

Note: The column will buckle about the principal axis of the cross section having the least I

• For purpose of design, Euler formula can be written in a more useful form by

expressing I = Ar2, where A is the cross sectional area and r is the radius of gyration

of the cross sectional area. Thus

𝑃𝑐𝑟 =𝜋2𝐸𝐼

𝐿2𝑃𝑐𝑟 =

𝜋2𝐸(𝐴𝑟2)

𝐿2𝑃

𝐴 𝑐𝑟=

𝜋2𝐸

𝐿/𝑟 2 𝜎𝑐𝑟 =𝜋2𝐸

𝐿/𝑟 2

• where

• 𝜎𝑐𝑟 = critical stress, which is the average normal stress in the column just before the

column buckles. This stress is an elastic stress and therefore 𝜎𝑐𝑟 ≤ 𝜎𝑌• E = Modulus of elasticity

• L = Unsupported length of the column, whose ends are pinned.

• r = smallest radius of gyration of the column, determined from r = 𝐼/𝐴, where I is

the least moment of inertia of the column’s cross sectional area A.

• (L/r) = slenderness ratio.

IDEAL COLUMN WITH PIN SUPPORTS

The critical-stress curves are hyperbolic, valid only for σcr is below yield stress

IDEAL COLUMN WITH PIN SUPPORTS

EXAMPLE 1

Copyright © 2011 Pearson Education South Asia Pte Ltd

The A-36 steel W200 46 member shown in Fig. 13–8 is to be used as

a pin-connected column. Determine the largest axial load it can

support before it either begins to buckle or the steel yields.

EXAMPLE 1 (cont)

Copyright © 2011 Pearson Education South Asia Pte Ltd

• From Appendix B,

• By inspection, buckling will occur about the y–y axis.

• When fully loaded, the average compressive stress in the column is

• Since this stress exceeds the yield stress,

Solutions

2N/mm 5.3205890

10006.1887

A

Pcrcr

46462 mm 103.15,mm 105.45,mm 5890 yx IIA

kN 6.1887

4

1000/1103.15102002

4462

2

2

L

EIPcr

(Ans) MN 47.1kN 5.14725890

250 PP

COLUMNS HAVING VARIOUS END-CONDITIONS

Le = KL. where Le is the effective length, K is the effective length factor, L is the column length

• The Euler formula for column's with various end conditions is

• (KL/r) is called the column’s effective slenderness ratio 2

2

2

2

/,

rKL

E

KL

EIP crcr

EXAMPLE 2

Copyright © 2011 Pearson Education South Asia Pte Ltd

A W150 24 steel column is 8 m long and is fixed at its ends as shown

in Fig. 13–11a. Its load-carrying capacity is increased by bracing it

about the y–y (weak) axis using struts that are assumed to be pin

connected to its mid-height. Determine the load it can support so that

the column does not buckle nor the material exceed the yield stress.

Take Est = 200 GPa and σY = 410 MPa.

EXAMPLE 2 (cont)

Copyright © 2011 Pearson Education South Asia Pte Ltd

• Effective length for buckling about the x–x and y–y axis is

• From the table in Appendix B,

• Applying Eq. 13–11,

Solutions

mm 2800m 8.22/87.0

mm 4000m 485.0

y

x

KL

KL

46

46

mm 1083.1

mm 104.13

y

x

I

I

(2) kN 8.460

2800

1083.1200

(1) kN 2.16534000

104.13200

2

62

2

2

2

62

2

2

y

xcr

x

xcr

KL

EIP

KL

EIP

EXAMPLE 2 (cont)

Copyright © 2011 Pearson Education South Asia Pte Ltd

• By comparison, buckling will occur about the y–y axis.

• The average compressive stress in the column is

• Since this stress is less than the yield stress,

buckling will occur before the material yields.

• Thus,

• From Eq. 13–12 it can be seen that buckling will always

occur about the column axis having the largest

slenderness ratio, since it will give a small critical stress

Solutions

MPa 6.150N/mm 6.150

3060

108.460 23

A

Percr

(Ans) kN 461crP

Important Points in this Lecture

• The three states for the column member are:

1. Stable equilibrium: P < k L/4

2. Unstable equilibrium: P ˃ k L/4

3. Neutral equilibrium: P = k L/4

• The Euler formula for column's with various end conditions is

• σcr ≤ σY

22

2

2

/,

rKL

E

KL

EIP crcr

Next Class

FINAL EXAM