Column Design(4.11-4.13)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

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Real world examples

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Continue…

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Jack

Column failure

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Quebec Bridge designed by Theodore Cooper. It collapsed in 1907.

Introduction

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“Long” columns fail structurally (buckling failure)

“Short” columns fail materially (yielding failure)

We will examine buckling failure first, and then transition to yielding failure.

P

P

“Long” 2

2crEIP

l

P

P

“Short”AP crcr

Buckling of Columns (4.12)

For a simply supported beam (pin-pin)

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y, v

x

F.B.D

y, v

v(x)

0

)(

)()(

2

2

2

2

Pvdx

vdEI

PvxMdx

vdEI

xPvxM

The solution to the above O.D.E is:

xEIPBx

EIPAxv cossin)(

Buckling of Columns (4.12)

Apply boundary conditions: v(0)=0 & v(l)=0

Either A = 0 (trivial solution, no displacement) or

The critical load is then

The lowest critical load (called the Euler buckling load) occurs when n=1.

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xEIPBx

EIPAxv cossin)(

0 & sin 0PB A lEI

sin 0P Pl l nEI EI

where n = 1, 2, …

22( )cr

EIP nl

2

2crEIP

l

Buckling of Columns (4.12)

The corresponding deflection at the critical load is

The beam deflects in a half sine wave.

The amplitude A of the buckled beam cannotbe determined by this analysis. It is finite, butnon-linear analysis is required to proceedfurther

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( ) sin xv x Al

Pcr

A

Buckling of Columns (4.12)

Define the radius of gyration as

Pcr can be written as

If l/k is large, the beam is long and slender. If l/k is small, the beam is short an stumpy.

To account for different boundary conditions (other than pin-pin), use the end-condition constant, C.

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2 IkA

2

2( / )crP EA l k

2

2( / )cr

crP C EA l k

Buckling of Columns (4.12)

Theoretical end condition constants for various boundary conditions (Figure 4-18 in textbook) (a) pin-pin (b) fixed-fixed (c) fixed-free (d) fixed-pin

Values of C for design are given in Table 4-2 in the textbook

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Buckling of Columns (4.12)

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1.Stability

(3) Neutral equilibrium

(1) Unstable equilibrium

(2) Stable equilibrium

Buckling of Columns (4.12)

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crP P crP P crP P

FF

Example 4-16

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Example

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Column AB carries a centric load P of magnitude 15 kips. Cables BC and BD are taut and prevent motion of point B in the xz-plane. Using Euler’s formula and a factor of safety of 2.2, and neglecting the tension in the cables, determine the maximum allowable length L.Use E = 29 x 106 psi.

How to prevent buckling? 1) Choose proper cross-sections

2) End conditions 3) Change compression to tension

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(cable stayed bridges)

Critical Stress (4.12)

Plot σcr=Pcr/A vs. l/k Long column: buckling occurs elastically before the yield stress

is reached. Short column: material failure occurs inelastically beyond the

yield stress. A “Johnson Curve” can be used to determine the stress for an

intermediate column

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?

No failureFailure

Sy

Johnson Curve

Note:Sy = yield strength = σy

l/k

l/k

Critical Stress (4.13)

Johnson equation

Where

Notice as l/k 0, Pcr/A Sy

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2crP la bA k

21

2y

y

Sa S and b

CE

Critical Stress (4.13)

To find the transition point (l/k)1 between the Johnson and Euler regions

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2 22

2

2

1

12

2

yy

y

Euler Johnson

SC E lSCE kl

k

l CEk S

Design of Columns (4.13)

Procedure Calculate (l/k)1 and l/k If l/k ≥ (l/k)1 use Euler’s equation

If l/k ≤ (l/k)1 use Johnson’s equation

Notice we haven’t included factor of safety yet…Column Design20

2

2( / )crP C EA l k

2crP la bA k

Example 4-19

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ExampleFind the required outer diameter, with F.S.=3, for the column shown below. Assume P = 15 kN, L = 50 mm, t = 5 mm, Sy = 372 MPa, E = 207,000 MPa, and the material is 1018 cold-drawn steel.

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t

do

ExampleFind the maximum allowable force, Fmax, that can be applied without causing pipe to buckle. Assume L = 12 ft, b = 5 ft, do = 2 in, t = 0.5 in, and the material is low carbon steel.

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t

do

F

L

b

Struts or short compression members

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Strut - short member loaded in compression

If eccentricity exists, maximum stress is at B with axial compression and bending.

Note that it is not a function of length

If bending deflection is limited to 1 percent of e, then from Eq. (4-44), the limiting slenderness ratio for strut is