1. Beams Extremely common structural element
In buildings majority of loads are vertical and majority of useable surfaces are horizontal
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2. devices for transferring vertical loads horizontally
Beams
devices for transferring
vertical loads horizontally
action of beams involves combination of
bending and shear
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3. What Beams have to Do Be strong enough for the loads
Not deflect too much
Suit the building for size, material, finish,
fixing etc
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4. Checking a Beam what we are trying to check (test)
stability - will not fall over
adequate strength - will not break
adequate functionality - will not deflect too much
what do we need to know
span - how supported
loads on the beam
material, shape & dimensions of beam
allowable strength & allowable deflection
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5. Designing a Beam what we are trying to do what do we need to know ?
determine shape & dimensions
what do we need to know
span - how supported
loads on the beam
material
allowable strength & allowable deflection
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6. Tributary Areas A beam picks up the load halfway to its neighbours
Each member also carries its own weight
this beam supports the load that comes from this area
span
spacing
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7. Tributary Areas (Cont. 1)
A column generally picks up load from halfway to its neighbours
It also carries the load that comes from the floors above
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8. Dead Loads on Elements Code values per cubic metre or square metre
Multiply by the volume or area supported
Length
Height
Thickness
Load =
Surface area x
Wt per sq m, or
volume x
wt per cu m
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9. Live Loads on Elements Code values per square metre
Multiply by the area supported
Area carried by one beam
Total Load = area x (Live load + Dead load) per sq m
+ self weight
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10. Loads on Beams Point loads, from concentrated loads or other beams
Distributed loads, from anything continuous
Distributed Load
Point Load
Reactions
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11. What the Loads Do
The loads (& reactions) bend the beam, and try to shear through it
Bending
Shear
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12. What the Loads Do (cont.)
Bending C T
Shear
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13. Designing Beams
in architectural structures, bending moment more important
importance increases as span increases
short span structures with heavy loads, shear dominant
e.g. pin connecting engine parts
beams in building
designed for bending
checked for shear
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14. How we Quantify the Effects
First, find ALL the forces (loads and reactions)
Make the beam into a freebody (cut it out and artificially support it)
Find the reactions, using the conditions of equilibrium
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15. Example 1 - Cantilever Beam
Point Load at End
Consider cantilever beam with point load on end W L
R = W
MR = -WL
vertical reaction, R = W
and moment reaction MR = - WL
Use the freebody idea to isolate part of the beam
Add in forces required for equilibrium
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16. Example 1 - Cantilever Beam Point Load at End (cont1.)W
V = W
M = -Wx
Take section anywhere at distance, x from end
Add in forces, V = W and moment M = - Wx
V = W
Shear Force Diagram
Shear V = W constant along length
(X = 0 -> L)
Bending Moment Diagram
BM = WL
BM = Wx
Bending Moment BM = W.x
when x = L BM = WL
when x = BM = 0
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17. Example 2 - Cantilever Beam Uniformly Distributed Load
w /unit length
For maximum shear V and bending moment BM
Total Load W = w.L
L/2
R = W = wL
MR = -WL/2
= -wL2/2
vertical reaction, R = W = wL
and moment reaction MR = - WL/2 = - wL2/2
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18. Example 2 - Cantilever Beam Uniformly Distributed Load (cont.)
For distributed V and BM
X/2 wx
V = wx
M = -wx2/2
Take section anywhere at distance, x from end
Add in forces, V = w.x and moment M = - wx.x/2
Shear V = wx
when x = L V = W = wL
when x = 0 V = 0
V = wL
= W
Shear Force Diagram
Bending Moment Diagram
BM = wL2/2
= WL/2
BM = wx /2 2
Bending Moment BM = w.x2/2
when x = L BM = wL2/2 = WL/2
when x = BM = 0
(parabolic)
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19. Sign Conventions Shear Force Diagrams
To plot a diagram, we need a sign convention
“Positive” shear
“Negative” shear
L.H up
R.H down
L.H down
R.H up
The opposite convention is equally valid,
but this one is common
There is no difference in effect between
positive and negative shear forces
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20. Plotting the Shear Force Diagram Diagram of loading R1 R2 W1 W2 W3
Starting at the left hand end, imitate each force you meet (up or down) R1
Shear Force Diagram W1 W2 R2 W3
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21. Shape of the Shear Force Diagram Point loads produce a block diagram
Uniformly distributed loads
produce triangular diagrams
Diagrams of loading
Shear force diagrams
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22. What Shear Force does to the Beam
Although the shear forces are vertical, shear stresses are both horizontal and vertical
Timber may split
horizontally along
the grain
Split in timber beam
Shear is seldom critical for steel
Reo in concrete beam
Concrete needs
special shear reinforcement
(45o or stirrups)
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23. Bending Moment Diagrams
Sign Conventions
Bending Moment Diagrams
To plot a diagram, we need a sign convention
Sagging
POSITIVE + -
Hogging
NEGATIVE
This convention is almost universally agreed
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24. Bending Moment Diagrams (cont.)
Sign Conventions
Bending Moment Diagrams (cont.)
Sagging bending moment is POSITIVE (happy) +
Hogging bending moment is NEGATIVE
(sad) -
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25. Positive and Negative Moments
Cantilevers -
Cantilevers produce negative moments
Simple beam +
Simple beams produce positive moments
Built-in & continuous beams have both, with negative over the supports
Built-in beam - +
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26. Bending Moment Diagram
Where to Draw the
Bending Moment Diagram
Positive moments are drawn downwards
(textbooks are divided about this)
This way mimics the beam’s deflection +
This way is normal coordinate geometry +
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27. Shape of the Bending Moment Diagram
Point loads produce triangular diagrams
Bending moment diagrams
Diagrams of loading
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28. Shape of the Bending Moment Diagram (cont1.)
Distributed loads produce parabolic diagrams
UDL
Diagrams of loading
Bending moment diagrams
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29. Bending Moment Diagram (cont.2)
Shape of the
Bending Moment Diagram (cont.2)
We are mainly concerned
with the maximum values
Maximum value
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30. Bending Moment Diagram (cont.3)
Shape of the
Bending Moment Diagram (cont.3)
Draw the Deflected Shape (exaggerate)
Use the Deflected shape as a guide to where the sagging (+) and hogging (-) moments are - - + + -
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31. Can we Reduce the Maximum BM Values?
Cantilevered ends reduce the positive bending moment
Built-in and continuous beams also have lower maximum BMs and less deflection
Simply supported
Continuous
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32. Standard BM Coefficients Simply Supported Beams
Use the standard formulas where you can
Total load = W
(w per metre length) W L
Central point load
Max bending moment
= WL/4
Uniformly distributed load
Max bending moment
= WL/8 or wL2/8
where W = wL L
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33. Standard BM Coefficients CantileversW L
(w per metre length)
Total load = W L
Uniformly Distributed Load
Max bending moment
= -WL/2 or -wL2/2
where W = wL
End point load
Max bending moment
= -WL
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34. Standard BM Coefficients Simple BeamsW W W
W /m
Beam W W W
W /m
Cable
BMD
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35. SFD & BMD Simply Supported Beams
V = +W
Mmax = -WL L W L
W = wL
V = +W
Mmax = -WL/2
= -wL2/2 W L
V = +W/2
V = -W/2
Mmax = WL/4 L
W = wL
V = +W/2
V = -W/2
Mmax = WL/8
= wL2/8
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36. What the Bending Moment
does to the Beam
Causes compression on one face and tension on the other
Causes the beam to deflect
How much deflection?
How much compressive stress?
How much
tensile stress?
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37. How to Calculate the Bending Stress
It depends on the beam cross-section
We need some particular properties of the
section
how big & what shape?
is the section we are using as a beam
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38. What to do with the Bending Stress
Codes give maximum allowable stresses
Timber, depending on grade, can take 5 to 20 MPa
Steel can take around 165 MPa
Use of Codes comes later in the course
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39. Finding Section Properties
we need to find the
Section Properties
next lecture
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