Beams. Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are horizontal. 1/39. devices for transferring vertical loads horizontally. Beams. action of beams involves combination of bending and shear. 2/39.
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Beams
1/39
devices for transferring
vertical loads horizontally
Beams
action of beams involves combination of
bending and shear
2/39
What Beams have to Do
fixing etc
3/39
Checking a Beam
4/39
?
?
Designing a Beam
5/39
Tributary Areas
this beam supports the load that comes from this area
span
spacing
6/39
Tributary Areas (Cont. 1)
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Length
Thickness
Load =
Surface area x
Wt per sq m,
or
volume x
wt per cu m
Height
Dead Loads on Elements
8/39
Live Loads on Elements
Area carried by one beam
Total Load =area x (Live load + Dead load) per sq m
+self weight
9/39
Distributed Load
Point Load
Reactions
Loads on Beams
10/39
Bending
Shear
What the Loads Do
11/39
e
e
e
e
C
T
Bending
Shear
What the Loads Do (cont.)
12/39
beams in building
designed for bending
checked for shear
Designing Beams
13/39
How we Quantify the Effects
14/39
W
vertical reaction, R = W
and moment reaction MR =  WL
MR= WL
L
R = W
Example 1  Cantilever Beam
Point Load at End
15/39
W
M = Wx
x
V = W
V = W
Shear Force Diagram
BM = Wx
BM = WL
Bending Moment Diagram
Example 1  Cantilever Beam
Point Load at End (cont1.)
Take section anywhere at distance, x from end
Add in forces, V = W and moment M =  Wx
Shear V = W constant along length
(X = 0 > L)
Bending Moment BM = W.x
when x = L BM = WL
when x = 0 BM = 0
16/39
w /unit length
L
Total Load W = w.L
MR = WL/2
= wL2/2
L/2
L/2
R = W = wL
Example 2  Cantilever Beam
Uniformly Distributed Load
Formaximum shear V and bending moment BM
vertical reaction,R = W = wL
and moment reactionMR =  WL/2 =  wL2/2
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wx
M = wx2/2
X/2
X/2
V = wx
V = wL
= W
Shear Force Diagram
BM = wx /2
2
BM = wL2/2
= WL/2
Bending Moment Diagram
Example 2  Cantilever Beam
Uniformly Distributed Load (cont.)
For distributed V and BM
Take section anywhere at distance, x from end
Add in forces,V = w.xand momentM =  wx.x/2
ShearV = wx
when x = LV = W = wL
when x = 0V = 0
Bending MomentBM = w.x2/2
when x = LBM = wL2/2 = WL/2
when x = 0BM = 0
(parabolic)
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“Positive” shear
“Negative” shear
R.H down
L.H down
L.H up
R.H up
Sign Conventions
Shear Force Diagrams
but this one is common
positive and negative shear forces
19/39
W3
W1
W2
R1
Diagram of loading
R2
W1
R1
W2
R2
W3
Shear Force Diagram
Plotting the
Shear Force Diagram
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Shape of the
Shear Force Diagram
a block diagram
produce triangular diagrams
Diagrams of loading
Shear force diagrams
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Split in timber beam
Reo in concrete beam
What Shear Force does
to the Beam
horizontally along
the grain
22/39
Hogging
NEGATIVE
Sagging
POSITIVE

+
Sign Conventions
Bending Moment Diagrams
23/39
Saggingbending momentisPOSITIVE (happy)
+
Hoggingbending momentisNEGATIVE
(sad)

Sign Conventions
Bending Moment Diagrams (cont.)
24/39


Cantilevers


+
+
Simple beam
Builtin beam
Positive and Negative Moments
25/39
+
+
This way mimics the beam’s deflection
This way is normal coordinate geometry
Where to Draw the
Bending Moment Diagram
(textbooks are divided about this)
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Diagrams of loading
Bending moment diagrams
Shape of the Bending Moment Diagram
27/39
UDL
UDL
Diagrams of loading
Bending moment diagrams
Shape of the Bending Moment Diagram (cont1.)
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Maximum value
Shape of the
Bending Moment Diagram (cont.2)
with the maximum values
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+
+
+
Shape of the
Bending Moment Diagram (cont.3)
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Simply supported
Continuous
Can we Reduce the Maximum BM Values?
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Total load = W
W
(w per metre length)
L
L
Central point load
Max bending moment
= WL/4
Uniformly distributed load
Max bending moment
= WL/8 or wL2/8
where W = wL
Standard BM Coefficients
Simply Supported Beams
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Total load = W
W
(w per metre length)
L
L
End point load
Max bending moment
= WL
Uniformly Distributed Load
Max bending moment
= WL/2 or wL2/2
where W = wL
Standard BM Coefficients Cantilevers
33/39
W
W
W /m
W
W
W
W
W
W
W
W
W /m
W
W
Standard BM Coefficients Simple Beams
Beam
Cable
BMD
34/39
W
W
W = wL
W = wL
L
L
L
L
V = +W/2
V = +W/2
V = +W
V = +W
V = W/2
V = W/2
Mmax = WL/8
= wL2/8
Mmax = WL
Mmax = WL/2
= wL2/2
Mmax = WL/4
SFD & BMD Simply Supported Beams
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How much compressive stress?
How much
tensile stress?
How much deflection?
What the Bending Moment
does to the Beam
34/37
how big & what shape?
How to Calculate the
Bending Stress
section
is the section we are using as a beam
37/39
What to do with the
Bending Stress
38/39
we need to find the
Section Properties
Finding Section Properties
next lecture
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