Download - Deflection of Beams _ Chapter 12
Chapter Objectives
ü Determine the deflection and slope at specific points on beams and shafts, using various analytical methods including:
Ø The integration method
Ø The use of discontinuity functions
Ø The method of superposition
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READING QUIZ
1) The slope angle θ in flexure equations is
a) Measured in degree
b) Measured in radian
c) Exactly equal to dv/dx
d) None of the above
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APPLICATIONS
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ELASTIC CURVE
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• The deflection diagram of the longitudinal axis that passes through the centroid of each cross-sectional area of the beam is called the elastic curve, which is characterized by the deflection and slope along the curve
ELASTIC CURVE (cont)
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• Moment-curvature relationship:
– Sign convention:
ELASTIC CURVE (cont)
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• Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or
• Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/I
y
ε
ρ−=
1
yEEI
M σ
ρρ−==
1or
1
SLOPE AND DISPLACEMENT BY INTEGRATION
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• Kinematic relationship between radius of curvature ρ and location x:
• Then using the moment curvature equation, we have
( )[ ] 232
22
1
1
dxdv
dvvd
+−=
ρ
( )[ ] 2
2
2/32
22
1
1
dx
vd
dxdv
dxvd
EI
M≈
+==
ρ
SLOPE AND DISPLACEMENT BY INTEGRATION (cont)
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• Sign convention:
SLOPE AND DISPLACEMENT BY INTEGRATION (cont)
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• Boundary Conditions:
– The integration constants can be determined by imposing the boundary conditions, or
– Continuity condition at specific locations
EXAMPLE 1
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The cantilevered beam shown in Fig. 12–10a is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.
EXAMPLE 1 (cont)
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• From the free-body diagram, with M acting in the positive direction, Fig. 12–10b, we have
• Applying Eq. 12–10 and integrating twice yields
Solutions
PxM −=
(1) 6
(2) C2
(1)
21
3
1
2
2
2
CxCPx
EIv
Px
dx
dvEI
Pxdx
vdEI
++−=
+−=
−=
EXAMPLE 1 (cont)
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• Using the boundary conditions dv/dx = 0 at x = L and v = 0 at x = L, equations 2 and 3 become
• Substituting these results, we get
Solutions
3 and
2
60
20
3
2
2
1
21
3
1
2
PLC
PLC
CLCPL
CPL
−==⇒
++−=
+−=
( )
( ) (Ans) 236
2
323
22
LxLxEI
Pv
xLEI
P
−+−=
−=θ
EXAMPLE 1 (cont)
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• Maximum slope and displacement occur at for which A(x =0),
• If this beam was designed without a factor of safety by assuming the allowable normal stress is equal to the yield stress is 250 MPa; then a W310 x 39 would be found to be adequate (I = 84.4(106)mm4)
Solutions
(5) 3
(4) 2
3
2
EI
PLv
EI
PL
A
A
−=
=θ
( ) ( )[ ] ( )[ ]
( ) ( )[ ] ( )[ ] mm 1.74
104.842003
1000530
rad 0222.0104.842002
1000530
6
22
6
22
−=−=
==
A
A
v
θ
EXAMPLE 2
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The simply supported beam shown in Fig. 12–11a supports the triangular distributed loading. Determine its maximum deflection. EI is constant.
EXAMPLE 2 (cont)
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• Due to symmetry only one x coordinate is needed for the solution,
• The equation for the distributed loading is .
• Hence
Solutions
2/0 Lx ≤≤
( )
xLw
L
xwM
xLwx
L
xwMM NA
43
043
;0
02
0
02
0
+−=
=−
+=+ ∑
xL
ww 02
=
EXAMPLE 2 (cont)
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• Integrating twice, we have
• For boundary condition,
Solutions
2,0 and 0,0 Lxdxdvxv ====
213050
12040
0302
2
2460
812
43
CxCxLw
xL
wEIv
CxLw
xL
w
dx
dvEI
xLw
xL
wM
dx
vdEI
+++−=
++−=
+−==
0,192
52
30
1 =−= CLw
C
EXAMPLE 2 (cont)
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• Hence
• For maximum deflection at x = L/2,
Solutions
(Ans) 120
40
maxEI
Lwv −=
xLw
xLw
xL
wEIv
192
5
2460
303050 −+−=