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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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4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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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4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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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4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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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4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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Procedure for analysis
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)
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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Procedure for analysis
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
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
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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
4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
Assume AEis constant, solve equations
simultaneously,
= 0
= ; =
4 A i l L d
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Procedure for analysis
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
4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
4 A i l L d
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Procedure for analysis
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
4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
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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
4.5 FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS
+=
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.
4.5 FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS
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Procedure for Analysis
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
4.5 FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS
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Procedure for Analysis
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
4.5 FORCE METHOD OF ANALYSIS FOR AXIALLY LOADED MEMBERS
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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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4.7 STRESS CONCENTRATIONS
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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4.7 STRESS CONCENTRATIONS
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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4.7 STRESS CONCENTRATIONS
Stress
Conce
ntration
s
avg
K
max
max= Kt* avg
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4.7 STRESS CONCENTRATIONS
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