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Beam review
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Beam diagrams
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 causedownward shear offset
Upward reactions causeupward shear offset Estimate shear area to draw bending diagrams
1 Cantilever beamwithpoint load
2 Cantilever beamwithuniformload
3 Cantilever beamwithmixed load4 Simplebeamwithpoint loads
5 Simplebeamwithuniformload
6 Simplebeamwithmixed load
7 Beamwith1overhangandpoint load8 Beamwith1overhanganduniformload
9 Beamwith1overhangandmixed load
10Beamwith2overhangs andpoint loads
11Beamwith2overhangs anduniformload12Beamwith2overhangs andmixed load
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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 linescombine bending and shear stress
(compressive arch lines and tensile cable lines)
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Shear and bending distr ibut ion
1 Beam diagram
2 Shear diagram
3 Bending diagram
4 Shear stress (over beam depth)
5 Bending stress (over beam depth)
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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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
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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%
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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 = 21/2C = 1.414 C L = 2.828 C
2 Beam with uniform load and one overhangX = 21/2C = 1.414 C
L = C+X = 1+1.414 C L = 2.414 C
1
L-x= C
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48
2810 10
Alternate beams
1. Simple beam
Span: L = 48
2. Beam with overhangs
Span L = 28, overhangs C = 10
Note:
Columns for beam with overhang may define circulation
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Deflection formulas
=slopeof the tangent of theelasticcurve
=maximumdeflection.
1 Cantilever beamwithpoint load =(PL)(L/2)/(EI) = 2/3L
= 1/2 PL2/(EI)
= 1/3 PL3/(EI)
2 Cantilever beamwithuniformload
=(WL/2)(L/3)/(EI) = 3/4L
= 1/6 WL2/(EI)
= 1/8 WL3/(EI)
3 Simplebeamwithpoint load
=(PL/4)(L/4)/(EI) = 1/3L
= 1/16 PL2
/(EI) = 1/48 PL3/(EI)
4 Simplebeamwithuniformload
=(WL/8)(2/3L/2)/(EI) = 5/16L
= 1/24 WL2/(EI)
= 5/384 WL3/(EI)
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Beam formulas
The formulas include for common beams:
M = bending moment V = shear
= deflection
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Deflection vs. span
The formulas
V = w L M = w L2/8
= (5/384) wL4/EI
reveal:
V increases linearly with LM increases quadratic with L
increases to the 4th power with L
If L doubles increase16 times !1 Beam with= 1
2 Double span with= 16
3 Short span: shear V governs
4 Medium span: bending M governs5 Long span: deflection governs
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Deflection vs. I (moment of inertia)
641/644 Four boards glued
81/83 Twin board glued
21/22 Twin board
111 Single board
IType of beam