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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