chapter 4 - axial loading

5/28/2018 Chapter 4 - Axial Loading

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xialLoading


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4. Axial Load

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CHAPTER OBJECTIVES

Determine deformationofaxially loadedmembers

Develop a method to find

support reactions when itcannot be determined from

equilibrium equations

Analyze the effects of thermal stress, stress

concentrations.


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CHAPTER OUTLINE

1. Saint-VenantsPrinciple

2. Elastic Deformation of an Axially Loaded Member

3. Principle of Superposition

4. Statically Indeterminate Axially Loaded Member

5. Force Method of Analysis for Axially Loaded Member

6. Thermal Stress

7. Stress Concentrations


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4.1 SAINT-VENANTS PRINCIPLE

Saint-Venantsprinciple states that both localizeddeformationandstress tend to even out at a distance

sufficiently removed from these regions.

=

Section a-a Section b-b Section c-c


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Relative displacement () of one end of bar with respect toother end caused by this loading

Applying Saint-VenantsPrinciple, ignore localized

deformations at points of concentrated loading and where

x-section suddenly changes

4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER

x dx


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Use method of sections, and draw free-body diagram

Assume proportional limit not exceeded, thus apply

Hookes Law

P(x)P(x)

dx

d = ()

() =

and

= E ()() = .

= .


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Eqn. 4-1

= displacement of one point relative to another point

L = distance between the two points

P(x) = internal axial force at the section, located a distance xfrom one end

A(x) = x-sectional area of the bar, expressed as a function of x

E = modulus of elasticity for material

= . .

P(x)P(x)

dx

d


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Constant load and X-sectional area


Eqn. 4-2

For constant x-sectional area A, and homogenous material,E is constant

With constant external force P, applied at each end, theninternal force P throughout length of bar is constant

= .

=


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Constant load and X-sectional area

If bar subjected to several different axial forces, or x-sectionalarea or E is not constant, then the equation can be applied to

each segment of the bar and added algebraically to get

=

B


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Sign convention

Sign Forces Displacement

Positive (+) Tension Elongation

Negative () Compression Contraction


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Procedure for analysis

Internal force

Use method of sections to determine internal axialforce Pin the member

If the force varies along members strength, sectionmade at the arbitrary location xfrom one end ofmember and force represented as a function of x,i.e., P(x)

If several constant external forcesact on member,internal force in each segment, between twoexternal forces, must then be determined



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Internal force

For any segment, internal tensile force is positive and

internal compressive force is negative. Results of

loading can be shown graphically by constructing thenormal-force diagram

Displacement

When members x-sectional area variesalong its axis,

the area should be expressed as a function of its

position x, i.e., A(x)



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Displacement

If x-sectional area, modulus of elasticity, or internal

loading suddenly changes, then Eqn 4-2 should be

applied to each segment for which the quantity isconstant

When substituting data into equations, account for

proper sign for P, tensile loadings +ve, compressive

ve. Use consistent set of units. If result is +ve,elongation occurs, ve means its a contraction



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EXAMPLE 4.1

Composite A-36 steel bar shownmade from two segments AB and

BD. Area AAB= 600 mm2and ABD

= 1200 mm2.

Determine the vertical

displacement of end Aand

displacement of Brelative to C.


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EXAMPLE 4.1 (SOLN)

Internal force

Due to external loadings, internal axial forces in regions

AB, BCand CDare different.

Apply the method of

sections and

equation of vertical

force equilibrium as

shown. Variation is

also plotted.


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EXAMPLE 4.1 (SOLN)

Displacement

From tables, Est= 210(103) MPa.

Use sign convention, vertical displacement of Arelative

to fixed support Dis

A=PL

AE [+75 kN](1 m)(10

6)

[600 mm2(210)(103) kN/m2]=

[+35 kN](0.75 m)(106)

[1200 mm2(210)(103) kN/m2]

+

[45 kN](0.5 m)(106)

[1200 mm2(210)(103) kN/m2]+

= +0.61 mm


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EXAMPLE 4.1 (SOLN)

Displacement

Since result is positive, the bar elongates and so

displacement at Ais upward

Apply Equation 4-2 between Band C,

A=PBCLBC

ABCE

[+35 kN](0.75 m)(106)

[1200 mm2(210)(103) kN/m2]=

= +0.104 mm

Here, Bmoves away from C, since segment elongates

4 A i l L d


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After subdividing the load into components, the principle

of superpositionstates that the resultant stress or

displacement at the point can be determined by first

finding the stress or displacement caused by each

component load acting separatelyon the member.

Resultant stress/displacement determined algebraically

by addingthe contributions of each component

4.3 PRINCIPLE OF SUPERPOSITION

4 A i l L d


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For a bar fixed-supported at one end, equilibriumequations is sufficient to find the reaction at the

support. Such a problem is statically determinate

If bar is fixed at both ends, then two unknown axialreactions occur, and the bar is statically indeterminate

4.4 STATICALLY INDETERMINATEAXIALLY LOADED MEMBER

4 A i l L d


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4.4 STATICALLY INDETERMINATEAXIALLY LOADED MEMBER

A member is statically indeterminate whenequations of equilibrium are not sufficient to

determine the reactions on a member.

+ = 0

+ = 0

4 A i l L d


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To establish addition equation, consider geometry of

deformation.

4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER

This equation can be expressed in terms of applied

loads using a load-displacement relationship, which

depends on the material behavior

/ = 0 Since relative displacement of one end of bar to theother end is equal to zero, (end supports fixed),

Such an equation is referred to as a

compatibility/kinematic condition.

4 A i l L d


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For linear elastic behavior, compatibility equation can

be written as


Assume AEis constant, solve equations

simultaneously,

= 0

= ; =

4 A i l L d


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Equilibrium

Draw a free-body diagram of member to identigy all

forces acting on it

If unknown reactions on free-body diagram greater

than no. of equations, then problem is statically

indeterminate

Write the equations of equilibrium for the member


4 A i l L d


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Compatibility

Draw a diagram to investigate elongation or contraction

of loaded member

Express compatibility conditions in terms ofdisplacements caused by forces

Use load-displacement relations (=PL/AE) to relate

unknown displacements to reactions

Solve the equations. If result is negative, this meansthe force acts in opposite direction of that indicated on

free-body diagram


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EXAMPLE 4.5

Steel rod shown has diameter of 5 mm. Attached to fixed

wall atA, and before it is loaded, there is a gap betweenthe wall at Band the rod of 1 mm.

Determine reactions atAand B if rod is subjected to axialforce of P= 20 kN.

Neglect size of collar at C. Take Est= 200 GPa

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EXAMPLE 4.5 (SOLN)

Equilibrium

Assume force Plarge enough to cause rods end B to

contact wall at B. Equilibrium requires

B/A= 0.001 m

FA FB+ 20(103) N = 0+ F= 0;

Compatibility

Compatibility equation:

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EXAMPLE 4.5 (SOLN)

FA(0.4 m) FB (0.8 m) = 3927.0 Nm

Compatibility

Use load-displacement equations (Eqn 4-2), apply to

ACand CB

B/A= 0.001 m =FALAC

AE

FBLCB

AE

Solving simultaneously,

FA= 16.6 kN FB= 3.39 kN

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Used to also solve statically indeterminate problems by

using superposition of the forces acting on the free-bodydiagram

First, choose any one of the two supports as redundant

and remove its effect on the bar

Thus, the bar becomes statically determinate

Apply principle of superposition and solve the equationssimultaneously

4.5 FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS

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From free-body diagram, we can determine the

reaction at A


+=

No displacement

at B

Displacement at B

When redundant

force at B is removed

Displacement at B

when only the

redundant force at B

is applied

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Procedure for Analysis

Compatibility

Choose one of the supports as redundantand write the

equation of compatibility.

Known displacement at redundant support (usually zero),

equated to displacement at support caused onlyby

external loads acting on the member plus the

displacement at the support caused only by theredundant reaction acting on the member.


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Compatibility

Express external load and redundant displacements in

terms of the loadings using load-displacement

relationship

Use compatibility equation to solve for magnitude of

redundant force


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Equilibrium

Draw a free-body diagram and write appropriate

equations of equilibrium for member using

calculated result for redundant force.

Solve the equations for other reactions


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EXAMPLE 4.9

A-36 steel rod shown has diameter of 5 mm. Its

attached to fixed wall at A, and before it is loaded,theres a gap between wall at Band rod of 1 mm.

Determine reactions at Aand B.

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EXAMPLE 4.9 (SOLN)

Compatibility

Consider support at Bas redundant.Use principle of superposition,

0.001 m = PB Equation 1( + )

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EXAMPLE 4.9 (SOLN)

Compatibility

Deflections Pand Bare determined from Eqn. 4-2

P=PLAC

AE= = 0.002037 m

B=FBLAB

AE= = 0.3056(10-6)FB

Substituting into Equation 1, we get0.001 m = 0.002037 m 0.3056(10-6)FB

FB= 3.40(103) N = 3.40 kN

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EXAMPLE 4.9 (SOLN)

Equilibrium

From free-body diagram

FA+ 20 kN 3.40 kN = 0

FA= 16.6 kN

+ Fx= 0;

3.40 kN

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4.6 THERMAL STRESS

Expansion or contraction of material is linearly related to

temperature increase or decrease that occurs (forhomogenous and isotropic material)

From experiment, deformation of a member having length

Lis

= linear coefficient of thermal expansion. Unit measure strain per degree

of temperature: 1/oC(Celsius) or 1/

oK(Kelvin)

T= algebraic change in temperature of member

T = algebraic change in length of member

= . .

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4.6 THERMAL STRESS

For a statically indeterminatemember, the thermal

displacements can be constrained by the supports,

producing thermal stresses that must be

considered in design.

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EXAMPLE 4.10

A-36 steel bar shown is constrainedto just fit between two fixed supportswhen T1= 30

oC.

If temperature is raised to T2= 60oC,

determine the average normalthermal stress developed in the bar.

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EXAMPLE 4.10 (SOLN)

Equilibrium

As shown in free-body diagram,

+ Fy= 0;

Problem is statically indeterminate since the

force cannot be determined from equilibrium.

CompatibilitySince A/B=0, thermal displacement TatA

occur. Thus compatibility condition atAbecomes

+ A/B=0 = TF

=

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EXAMPLE 4.10 (SOLN)

From magnitude ofF, its clear that changes

in temperature causes large reaction forces

in statically indeterminate members.

Average normal compressive stress is

Compatibility

Apply thermal and load-displacement relationship,

0 = TL FL

AL

F=

TAE= = 7.2 kN

F

A= = = 72 MPa

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4.7 STRESS CONCENTRATIONS

Stress concentrations occur when cross-sectional areachanges.

Maximum stress is determined using a stress

concentration factor, K, which is a function of geometry.

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In engineering practice, actual stress distributionnot needed, only maximum stressat thesesections must be known. Member is designed toresist this stress when axial load Pis applied.

Kis defined as a ratio of the maximum stress tothe average stress acting at the smallest crosssection:

=

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Kis independent of the bars geometry and the type ofdiscontinuity, only on the bars geometry and the typeof discontinuity.

As size rof the discontinuity is decreased, stress

concentration is increased.

It is important to use stress-concentration factors indesign when using brittle materials, but not necessaryfor ductile materials

Stress concentrations also cause failure structuralmembers or mechanical elements subjected to fatigueloadings

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Stress

Conce

ntration

s

avg

K

max

max= Kt* avg

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avg

K

max

Stres

sConc

entratio

ns

max= Kt* avg

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EXAMPLE 4.13

Steel bar shown below has allowable stress,

allow= 115 MPa. Determine largest axial force Pthat

the bar can carry.

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EXAMPLE 4.13 (SOLN)

Because there is a shoulder fillet, stress-

concentrating factor determined using the graphbelow

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EXAMPLE 4.13 (SOLN)

Calculating the necessary geometric parameters

yields

Thus, from the graph, K= 1.4

Average normal stress at smallestx-section,

r

n=

10 mm

20 mm= 0.50

w

h

40 mm

20 mm= 2=

P

(20 mm)(10 mm)avg= = 0.005PN/mm2

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EXAMPLE 4.13 (SOLN)

Applying Eqn 4-7 with allow= maxyields

allow= K max

115 N/mm2= 1.4(0.005P)

P= 16.43(103) N = 16.43 kN

chapter 4 - axial loading

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