http:ihome'ust'hkmech101 - PowerPoint PPT Presentation

Title: http:ihome'ust'hkmech101


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• http//ihome.ust.hk/mech101/
• Web-based Interactive Tool for Visualization of
  Structural Analysis with Finite Element
  Simulation

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4-May-2007
MECH 101 Tutorial 12 zhangms_at_ust.hk
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Example 4) in tutorial 8
wo
x
c
A
B
c
L
Find the shear force and bending diagram at any
cross section of the beam.
wo
woL2/3
Step 1) Find the reaction force
woL/2
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Example 4)
For 0ltxltL
x
c
wox/L
woL2/3
M
V
woL/2
c
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Shear and Bending Moment Diagram (IV)
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Example 1)
qo
c
A
B
c
x
L
Derive the equation of deflection, using
second-order differential equation
c
c
B
M
B
A
M
V
V
L-x
c
c
x
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Example 1)
c
B
B
M
V
L-x
c
In this example, select the right part for
analysis will make the calculation easier
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Example 2)
The beam at the right has a constant E1,I1 and is
support by the fixed wall at B and the elastic
rod AC. If the rod has a cross-sectional area A2
and a Youngs modules E2. Find the reaction
force at B and determine the axial force induced
in the rod AC.
Find the reaction force and moment at fixed
support B
wL1
TAc
MB
L1
By
wx
TAc
M(x)
Derive the bending moment M(x)
x
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Example 2)
Take BC b) into Eq.4)
Take into the second-order differential equation
Take BC b) into Eq.3)
Boundary conditions
Solving Eq.5) and 6) yields
Take BC a) into Eq.4)
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Example 3)
Derive the equations of the deflection curve for
an voerhanging heam ABC subjected to a uniform
load of intensity q action on the overhang. Also,
obtain formulas for the deflection and angle
of roatation at the end of the overhang.
q
Governing equation from 0L
A
C
B
L
(½)L
Boundary condition at A, x0,M0, take into Eq,3)
Boundary condition at C, x(3/2)L,V0, take into
Eq.5)
Governing equation from L(3/2)L
Then Eq.6) becomes
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Example 3)
q
Boundary condition at C, x(3/2)L,M0, take into
Eq.7)
A
C
B
Continue Boundary condition at B
L
(½)L
Then write the first order differential equation
Combine Eq.3) and Eq.7) yields
0ltxltL
Lltxlt(3/2)L
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Example 3)
Also use Continue Boundary condition at B
q
A
C
Combine Eq.8) and Eq.9) yields
B
L
(½)L
For the left part 0ltxltL
Then we have
Boundary condition at A, x0,v0, take into
Eq,10)
For the right part Lltxlt(3/2)L
Boundary condition at B, xL,v0, take into
Eq,10)
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Example 3)
q
Boundary condition at B, xL,v0, take into
Eq,11)
A
C
B
Adjust those previous calculation
L
(½)L
0ltxltL
Lltxlt(3/2)L
The rotation angle at C
The deflection at C
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Example 4)
SFx 0
SFy 0
sin2? 2sin?cos?, cos2? (1 cos2?) / 2,
sin2? (1 - cos2?) / 2
Put ? ? 90o
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In-plane principal stresses and plane stress
transformation
(rotation to obtain principal stresses)
(measured from ve x axis to ve x axis)
(ve in counterclockwise direction)
Mohrs Circle
sx
txy
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Example 5)
200 MPa
Determine the resultant state of stress
represented on the element oriented as shown in
the last figure.
60o
y
350 MPa
x
y

?
x
60o
58 MPa
350 MPa
200 MPa
25o
? - 30o (clockwise direction)

sx -200 MPa, sy -350 MPa, txy 0 MPa
-200 (-350) 2
-200 - (-350) 2
sx
cos(-60o) 0 -237.5 MPa
-200 (-350) 2
-200 - (-350) 2
sy -
cos(-60o) 0 -312.5 MPa
-200 - (-350) 2
txy - sin(-60o) 0
64.95 MPa
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END