Loading in 2 Seconds...
Loading in 2 Seconds...
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.
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
1/39
vertical loads horizontally
Beams
action of beams involves combination of
bending and shear
2/39
fixing etc
3/39
4/39
?
Designing a Beam
5/39
this beam supports the load that comes from this area
span
spacing
6/39
7/39
Thickness
Load =
Surface area x
Wt per sq m,
or
volume x
wt per cu m
Height
Dead Loads on Elements
8/39
Area carried by one beam
Total Load =area x (Live load + Dead load) per sq m
+self weight
9/39
Point Load
Reactions
Loads on Beams
10/39
designed for bending
checked for shear
Designing Beams
13/39
14/39
vertical reaction, R = W
and moment reaction MR =  WL
MR= WL
L
R = W
Example 1  Cantilever Beam
Point Load at End
15/39
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
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
17/39
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)
18/39
“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
W1
W2
R1
Diagram of loading
R2
W1
R1
W2
R2
W3
Shear Force Diagram
Plotting the
Shear Force Diagram
2039
Shear Force Diagram
a block diagram
produce triangular diagrams
Diagrams of loading
Shear force diagrams
21/39
Reo in concrete beam
What Shear Force does
to the Beam
horizontally along
the grain
22/39
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 produce negative moments


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)
26/39
Bending moment diagrams
Shape of the Bending Moment Diagram
27/39
UDL
Diagrams of loading
Bending moment diagrams
Shape of the Bending Moment Diagram (cont1.)
28/39
Shape of the
Bending Moment Diagram (cont.2)
with the maximum values
29/39



+
+
+
Shape of the
Bending Moment Diagram (cont.3)
30/39
Continuous
Can we Reduce the Maximum BM Values?
31/39
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
32/39
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 = 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
35/39
How much
tensile stress?
How much deflection?
What the Bending Moment
does to the Beam
34/37
How to Calculate the
Bending Stress
section
is the section we are using as a beam
37/39
Bending Stress
38/39