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15 Νοε 2013 (πριν από 4 χρόνια και 7 μήνες)

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Indeterminate beams Copyright Prof Schierle 2011 1
Indeterminat
e
beams
Indeterminate beams Copyright Prof Schierle 2011 2
Indeterminate beams
1 Simple beam (determinate)
2 Fixed end beam (indeterminate)
M = WL/24 (mid-span)
M = -WL/12 (supports)
3 Continuous beam (2-bays)
4 Continuous beam (3-bays)
Indeterminate beams cannot be analyzed
by static analysis.
However, bending coefficients are available for
common conditions, such as beams with
and supported by equally spaced supports.
Using coefficients, bending moments are:
M = C W L
M = bending moment
C = Coefficient
W = w L (total load)
L = span between supports
Indeterminate beams Copyright Prof Schierle 2011 3
Fixed end beams
Usually have moment resistant
Connections to columns
Fixed ends create negative
bending at the supports that
reduces the positive bending
between supports
The combined bending is equal
to that of an equivalent simple beam
but the separate smaller bending
moments allow smaller beams
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Indeterminate beams Copyright Prof Schierle 2011 4
Bending coefficients for fixed end beam
\
Examples
Assume: L = 30’, W = 60 k = 2klf x 30’
1 Fixed end beam with uniform load
M = -.083 (60) 30 M = -149 k’
M = .042 (60) 30 M = 76 k’
2 Fixed end beam with 1 point load
M = -.125 (60) 30 M = -225 k’
M = .125 (60) 30 M = 225 k’
3 Fixed end beam with 2 point loads
M = -.111 (60) 30 M = -200 k’
M = .056 (60) 30 M = 101 k’
4 Fixed end beam with 3 point loads
M = -.104 (60) 30 M = -187 k’
M = .063 (60) 30 M = 113 k’
Indeterminate beams Copyright Prof Schierle 2011 5
Bending coefficients for continuous beams
live load (LL) may or may not be applied:
• Beams at left are fully loaded with DL
• Beams at right have LL on all or alternate
bays, whichever gives the greatest results
Indeterminate beams Copyright Prof Schierle 2011 6
Assume:
L = 30’, W = 60k / bay (W = w L)
1 Beam with uniform DL
2-bay beam
Bay M = .070 (60) 30 M = 126 k’
Support M = -.125 (60) 30 M = -225 k’
3-bay beam
End bay M = .080 (60) 30 M = 144 k’
Mid bay M = .025 (60) 30 M = 45 k’
Support M = -.100 (60) 30 M = -180 k’
2 Beam with 1 point DL per bay
3-bay beam
End bay M = .175 (60) 30 M = 315 k’
Mid bay M = .100 (60) 30 M = 180 k’
Support M = -.150 (60) 30 M = -270 k’
3 Beam with 2 point DL per bay
3-bay beam
End bay M = .122 (60) 30 M = 220 k’
Mid bay M = .033 (60) 30 M = 59 k’
Support M = -.134 (60) 30 M = -241 k’
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Indeterminate beams Copyright Prof Schierle 2011 7
Assume:
L = 30’, W = 60k / bay (W = w L)
4 Beam with uniform LL
2-bay beam
Bay M = .096 (60) 30 M = 173 k’
Support M = -.125 (60) 30 M = -225 k’
3-bay beam
End bay M = .101 (60) 30 M = 182 k’
Mid bay M = .075 (60) 30 M = 135 k’
Support M = -.117 (60) 30 M = -211 k’
5 Beam with 1 point LL per bay
3-bay beam
End bay M = .213 (60) 30 M = 383 k’
Mid bay M = .175 (60) 30 M = 315 k’
Support M = -.175 (60) 30 M = -315 k’
6 Beam with 2 point LL per bay
3-bay beam
End bay M = .144 (60) 30 M = 259 k’
Mid bay M = .100 (60) 30 M = 180 k’
Support M = -.156 (60) 30 M = -281 k’
Indeterminate beams Copyright Prof Schierle 2011 8
GerberBeam
The Gerber beam,named after its 19
th
century
inventor, the German engineer Gerber, was in
response to railroad bridge failures caused by
uneven support settlements.
1 Large bending moment of simple beams
2 Reduced bending of continuous beam
3 Increased settlement bending causes failure
4 Gerber beam with hinges at inflection points
allows settlement without bending increase
Indeterminate beams Copyright Prof Schierle 2011 9
Gerber beams are used for glue-lam
warehouse roof beams to reduce
bending moments and beam size
Hinge detail
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Indeterminate beams Copyright Prof Schierle 2011 10
Beamdiagrams
• Typical V M  diagrams without computing
• Visualize deflection as a flexible ruler
• Draw shear and bending diagrams left to right;
starting and ending with zero beyond the beam
• Uniform load cause downward sloping shear
• Point loads cause downward shear offset
• Upward reactions cause upward shear offset
• Estimate shear area to draw bending diagrams
1 Cantilever beam with point load
2 Cantilever beam with uniform load
3 Cantilever beam with mixed load
4 Simple beam with point loads
5 Simple beam with uniform load
6 Simple beam with mixed load
7 Beam with 1 overhang and point load
8 Beam with 1 overhang and uniform load
9 Beam with 1 overhang and mixed load
10 Beam with 2 overhangs and point loads
11 Beam with 2 overhangs and uniform load
12 Beam with 2 overhangs and mixed load
Indeterminate beams Copyright Prof Schierle 2011 11
Shear effect
1 Beam with square mark to study stress
2 Shear stress on square
3 Equivalent split shear stress
4 Shear stress as tension/compression stress
5 Equivalent tension/compression stress
cause diagonal tension cracks at beam supports
Isostatic lines combine bending and shear stress
(compressive ”arch” lines and tensile “cable” lines)
Indeterminate beams Copyright Prof Schierle 2011 12
Shear and bending distribution
1 Beam diagram
2 Shear diagram
3 Bending diagram
4 Shear stress
5 Bending stress
A Best location of possible pipe hole:
• Zero shear force at mid-span
• Zero bending stress at mid-depth
The diagrams reveal an interesting paradox:
• Linear shear force over beam length
• Parabolic shear stress over beam depth
• Parabolic bending moment over beam length
• Linear bending stress over beam depth
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Indeterminate beams Copyright Prof Schierle 2011 13
Girder optimization
1
Stepped bending diagram used to optimize
:
4
4
5
6
2 Steel girder with plates welded outside flanges
3 Steel girder with plates welded inside flanges
4 Concrete girder with rebar lengths as required
5 Parabolic girder reflecting bending moment
6 Tapered girder approximates bending moment
Indeterminate beams Copyright Prof Schierle 2011 14
Overhang effect
1 Simple beam
2 Beam overhangs reduce bending moment
~1/3 overhangs equalize positive and negative bending
Overhangs can provide synergy with architectural design
Overhangs reduce bending up to
~ 600%
Indeterminate beams Copyright Prof Schierle 2011 15
Overhang/span ratio
Beams with overhangs are most efficient if positive and
negative bending are equal (optimal cross section use)
Find ratios C/L for equal positive and negative bending
1 Beam with uniform load and two overhangs
+M = Abs (-M)
Considering the Area Method
Positive shear must be 2 times negative shear
+V = 2-V
L/2 = 2
1/2
C = 1.414 C L = 2.828 C
2 Beam with uniform load and one overhang
X = 2
1/2
C = 1.414 C
L = C+X = 1+1.414 C L = 2.414 C
1
L-x= C
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Indeterminate beams Copyright Prof Schierle 2011 16
happy end