axial load lecture
TRANSCRIPT
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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