0% found this document useful (0 votes)
24 views38 pages

MoM Ch11

Uploaded by

Amna Omer
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPTX, PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
24 views38 pages

MoM Ch11

Uploaded by

Amna Omer
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPTX, PDF, TXT or read online on Scribd
You are on page 1/ 38

MECHANICS OF

MATERIALS
MECH 305
Dr. Erwin Sulaeman
UAE University
College of Engineering
Department of Mechanical Engineering
Chapter 11: Buckling
11.1 Introduction
11.2 Buckling and Stability
11.3 Columns with Pinned Ends
11.4 Columns with Other Support Conditions
11.5 Columns with Eccentric Axial Loads
11.6 The Secant Formula for Columns
11.7 Elastic and Inelastic Column Behavior
11.8 Inelastic Buckling
11.9 Design Formulas for Columns
Chapter Summary & Review
11.1 Introduction
• Load-carrying structures may fail in a variety of ways, depending
upon the type of structure, the conditions of support, the kinds of
loads, and the materials used.
• For instance, an axle in a vehicle may fracture suddenly from
repeated cycles of loading, or a beam may deflect excessively, so
that the structure is unable to perform its intended functions.
• Another type of failure is called buckling, which is the subject
matter of this chapter.
• If a compression member is relatively slender, it may deflect
laterally and fail by bending rather than failing by direct
compression of the material.
• When lateral bending occurs, we say that the column has buckled.
• Under an increasing axial load, the lateral deflections will increase
too, and eventually the column will collapse completely.
11.1 Introduction
11.2 Buckling and Stability
Point C moved slightly
out of alignment. Consider model with two rigid rods connected at C
by a pin and a torsional spring of constant K.
After a small perturbation,

K 2   restoring moment


L L
P sin   P   destabilizing moment
2 2

Column is stable (tends to return to aligned


orientation) if
L
P   K 2 
2
4K
P  Pcr 
L
Free-body diagram of rod
AC in unaligned position.
11.2 Buckling and Stability
Assume that a load P is applied. After
a perturbation, the system settles to a
new equilibrium configuration at a
finite deflection angle.

L
P sin   K 2 
2
PL P 
 
4 K Pcr sin 

(a) Model column in Noting that sinq < q , the assumed


buckled position, (b) free-
body diagram of rod AC. configuration is only possible if
P > Pcr.
11.2 Buckling and Stability
Critical Load
• Long slender members subjected to
axial compressive force are called
columns.
• The lateral deflection that occurs is
called buckling.
• The maximum axial load a column can
support when it is on the verge of
buckling is called the critical load, Pcr.
11.3 Columns with Pinned Ends
Consider an axially loaded column.
After a small perturbation, the system
reaches an equilibrium configuration
such that
d2y M P
2
   y
dx EI EI

d2y P
2
 y0
dx EI

Solution with assumed configuration can


only be obtained if
 2 EI
P  Pcr 
L2
Free-body diagrams of (a)
buckled column and (b)
P
    cr 
 
 2 E Ar 2

 2E
portion AQ. A 2
L A L r 2
11.3 Columns with Pinned Ends
• A column will buckle about the principal axis of the
x-section having the least moment of inertia
(weakest axis).
• For example, the meter stick shown will
buckle about the a-a axis and not
the b-b axis.
• Thus, circular tubes made excellent
columns, and square tube or those
shapes having Ix ≈ Iy are selected
for columns.
11.3 Columns with Pinned Ends
• Buckling equation for a pin-supported long slender
column,
 2 EI
Pcr  2 11-15
L

Pcr = critical or maximum axial load on column just


before it begins to buckle. This load must not cause
the stress in column to exceed proportional limit.
E = modulus of elasticity of material
I = Least modulus of inertia for column’s x-sectional
area.
L = unsupported length of pinned-end columns.
10
11.3 Columns with Pinned Ends
• Expressing I = Ar2 where A is x-sectional area of column and
r is the radius of gyration of x-sectional area.
 2E
 cr  11-17 
L r 
2

cr = critical stress, an average stress in column just before the


column buckles. This stress is an elastic stress and therefore
cr  Y
E = modulus of elasticity of material
L = unsupported length of pinned-end columns.
r = smallest radius of gyration of column, determined from

where I is least moment of inertia of column’s cross-sectional area A.


11.3 Columns with Pinned Ends
• The geometric ratio L/r in Equation 11-17 is known
as the slenderness ratio.

• It is a measure of the column’s flexibility and will be


used to classify columns as long, intermediate or
short.
11.3 Columns with Pinned Ends
IMPORTANT
• Columns are long slender members that are subjected to
axial loads.
• Critical load is the maximum axial load that a column can
support when it is on the verge of buckling.
• This loading represents a case of neutral equilibrium.
• An ideal column is initially perfectly straight, made of
homogeneous material, and the load is applied through the
centroid of the x-section.
• A pin-connected column will buckle about the principal axis
of the x-section having the least moment of inertia.
• The slenderness ratio L/r, where r is the smallest radius of
gyration of cross-section. Buckling will occur about the axis
where this ratio gives the greatest value.
11.3 Columns with Pinned Ends
The value of stress corresponding to the
critical load,

 2 EI
P  Pcr 
L2
P P
    cr  cr
A A

 cr 
 
 2 E Ar 2
L2 A
 2E
  critical stress
L r 2

L
 slenderness ratio
r

Preceding analysis is limited to


Plot of critical stress. centric loadings.

10 - 14
11.3 Columns with Pinned Ends
Example A:

A 7.2-m long A-36 steel tube


having the x-section shown is to
be used a pin-ended column.
Determine the maximum
allowable axial load the column
can support so that it does not
buckle.
11.3 Columns with Pinned Ends
Example A: Solution
Use equations for critical load: Est = 200 GPa.
 2 EI
Pcr 
L2
1
 2  200 106  kN/m 2     70  1 m /1000 mm 
4 4

 4  228.2 kN
7.2 m 
2

This force creates an average compressive stress in the


column of
Pcr 228.2 kN 1000 N/kN 
 cr    100.2 N/mm 2
 100 MPa
A   75     70   mm
2 2 2
 

Since cr < Y = 250 MPa, application of Euler’s equation is


appropriate.
11.3 Columns with Pinned Ends
Example B:

The A-36 steel W20046 member


shown 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.

17
11.3 Columns with Pinned Ends
Example B: (Solution)
From tables, column’s cross-sectional area and moments of
inertia are A = 5890 mm2,
Ix = 45.5106 mm4,and Iy = 15.3106 mm4.
By inspection, buckling will occur about the y-y axis.
Applying Equation 11-15, we have

 2 EI
Pcr 
L2


 2
200106  kN/m2 15.3104  mm4 1 m /1000 mm4
4 m 2
 1887.6 kN

18
11.3 Columns with Pinned Ends
Example B: (Solution)
When fully loaded, average compressive stress in column is

Pcr 1887.6 kN1000 N/kN 


 cr  
A 5890 mm 2
 320.5 N/mm 2

Since this stress exceeds yield stress (250 N/mm2), the load P
is determined from simple compression:

P
250 N/mm 2 
5890 mm2
P  1472.5 kN

19
11.3 Columns with Pinned Ends
Example C:
The hoisting arrangement for lifting a large pipe is shown in the figure. The spreader is a
steel tubular section with outer diameter 70 mm and inner diameter 57 mm. Its length is
2.6 m and its modulus of elasticity is 200 GPa.
Based upon a factor of safety of 2.25 with respect to Euler buckling of the spreader, what
is the maximum weight of pipe that can be lifted? (Assume pinned conditions at the ends
of the spreader.)
11.3 Columns with Pinned Ends
Example C: Solution T
Given:
P

W/2
11.4 Columns with Other Support
Conditions
A column with one fixed and one
free end, will behave as the upper-
half of a pin-connected column.

The critical loading is calculated


from Euler’s formula,
 2 EI
Pcr 
L2e

 2E
 cr 
Le r 2
Le  2 L  equivalent length
Effective length of a fixed-free
column of length L is equivalent to
a pin-ended column of length 2L.
11.4 Columns with Other Support
Conditions
Effective length
• If a column is not supported by pinned-ends, then
Euler’s formula can also be used to determine the
critical load.
• “L” must then represent the distance between the
zero-moment points.
• This distance is called the columns’ effective length,
Le.
11.4 Columns with Other Support Conditions

Fig. 11-20 Critical loads, effective lengths, and effective-length factors for ideal
columns.
11.4 Columns with Other Support Conditions
Example 11.2

25
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support Conditions
Example 11.2 continued
11.4 Columns with Other Support
Conditions
Example 11.3
11.4 Columns with Other Support
Conditions
Example 11.3: continued
11.4 Columns with Other Support
Conditions
Example 11.3: continued
11.4 Columns with Other Support
Conditions
Example 11.3: continued
11.4 Columns with Other Support
Conditions
Example 11.3: continued
11.4 Columns with Other Support
Conditions
Example 11.3: continued
11.4 Columns with Other Support
Conditions
Example 11.3: continued

You might also like