Axial Load Lecture

7/25/2019 Axial Load Lecture

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Cha ter 4

Axial Load

Elastic Deformation of an Axially Loaded

Component (4.2)

Statically Indeterminate Axially Loaded

Member 4.4

Principle of Superposition (4.3)

The Force Method of Analysis for Axially

.


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E = constant

A(x)

P x

d

L

P(x)

Generally, internal axial load P(x) varies with position x,

dx)(

)(

xA

x

dx

)(and x


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)()( xEx Now apply Hookes law:

dxE

)(

xA

Solve for the elemental displacement d

:)( dxxP

)(xEA

The total dis lacement is :

LLdxxP P(x): internal axial force

00

)(xEAq. - : oung s mo u us

A(x): cross-sectional area

This is the GENERAL equation for deflection of abody subjected to an axial applied force.


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SPECIAL CASE 1: The axial displacement

subjected to a constant axial force:

PP

A

LL

dxEAxEAxx

00)(

PL Eq. (4-2)

EA


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Example 1:

The shi is ushed throu h the water usin an A-36 steel ro eller shaft that is 8 m lon measuredfrom the propeller to the thrust bearingD at the engine. The shaft has an outer diameter of 400 mm and

a wall thickness of 50 mm. The force exerted on the shaft from the propeller can be assumed as an axial

force. When this force is 5 kN, determine the amount of axial contraction of the shaft. The bearings atB

and C are ournal bearin s.


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SPECIAL CASE 2: The total axial displacement

of a series components connected in series:

Each component has different axial internal loading or

Each component is made of different material or

Each has different constant cross section or

a bc d

All the above conditions are satisfied (a general example below)

I II III PdPaP

b

Pc

deflection of end d w.r.t. end a:

abbccdad ////

E . 4-3IIIIIIIIIIII LPLPLPn

iiLP

IIIIIIIIIIII

adAEAEAE

/i ii

AE1


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CAUTION:

n or er o app y q. . o e erm ne e sp acemen o

one end of the bar to the other end, the sign convention

MUST be followed

Sign Convention: Tension (elongation): POSITIVE

ompress on con rac on :

Tips: It is convenient to assume that the internal normal loading

is always tensile when draw the FBD of the sectioned segment.

If the result calculated from e uilibrium e uations is ositive

it is real positive (tensile) according to the sign convention for

the normal loading; if the result is negative, it is real

.

DIRECTLY into Eq. (4.3) to calculate displacement.


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Calculation of the axial deformation of a tube assembly

Given: The assembly shown below consists of an aluminum tube AB having a

cross-sectional area of 500 mm2. A steel rod having a cross-sectional area

of 80 mm2is attached to a rigid collar at B and passes through the tube. A

tensile load of 100 kN is applied to the rod.

Example 2:

BA

C

100 kN

Al tube

Steel rod

400 mm

600 mm

Find: The displacement of the end C of the rod. Take Esteel = 200 GPa and Eal =70 GPa.


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Example 3: Calculation of the axial deformation of a rigid beam

Given: A rigid beam AF rests on two posts. Post AC is made of steel and has a diameter of 20 mm.

ost s ma e o a um num an as a ameter o mm.

Find: The displacement of point F if a vertical load of 90 kN is applied downward at this point.

= =steel al .


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How to calculate F:

A/C B F

. .

A

B/DF

B

A/C

'''' AAFFAABB

..

/ / /B D A C F A C

0.3 0.7

7.03.0

Attention: Each length component has absolute value


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# of equilibrium equations < # of unknown

Otherwise, reactions can be determined from equilibrium

equations-Statically Determinate Problem

**These problems cant be solved with simple

deflection of the components.


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Example 4: Calculation of the axial deformation of a rigid beam using the compatibility method

Given: A rigid beam AG rests upon three 20 mm diameter posts (AC, BD, and EF).

osts an are ma e o stee w e post s ma e o a um num.

Find: The displacement of point G if a vertical load of 90 kN is applied downward at this point.

Take Esteel = 200 GPa and Eal = 70 GPa.


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90 kNFBD:

A B GE

Ay ByX Ey


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Methods for solving statically indeterminate problems

Method 1 The compatibility method:

1

Draw a Free-Body Diagram of the member to identify all the forces that act on it.

2

Write the equations of equilibrium for the member.

3 Add additional equations by describing the displacement of the various axially

loaded components of the member keeping in mind the condition that, at points

where the sections meet the displacements must be equal (i.e. compatibility must.

4

These additional equations are used to determine the number of unknown reaction

forces.

Compatibility (kinematic) condition is introduced,

which specifies the conditions for displacement


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F = 100 kNFBD:

YA B

X A C

Z

Ay ByCy = k(- 0.02)

woo

woo1 m


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Example 6: Calculation of the reaction forces in a

member using the force superposition method

Given: Rod AB is fixed rigidly at the top and attached to rod BC at its other end.

Rod BC is fixed ri idl at the bottom. Both are solid rods made of steeland are each 0.5 m long. Rod AB has diameter of 30 mm while rod BC hasa diameter of 15 mm. A 400 kN load is applied to point B.

A

C

0.5 m

B

0.5 m400 kN

Find: The reaction forces at ends A and C. Take Esteel= 200 GPa.


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Example 7:

e s r u e oa ng s suppor e y e ree suspen er ars. an are ma e o a um num anCD is made of steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum

intensity w of the distributed loading so that an allowable stress in the steel and in the

aluminum is not exceeded. AssumeACE is rigid.Est

= 200 GPa,Eal

= 70 GPa. 180allow st MPa

94allow al

MPa

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Axial Load Lecture