Peter Smith & Mike Rosenman

Extremely common structural element

In buildings majority of loads are vertical and
majority of useable surfaces are horizontal
University of Sydney – Structures
BEAMS 1/39
Peter Smith & Mike Rosenman
devices for transferring
vertical loads horizontally
action of beams involves combination of
bending and shear
University of Sydney – Structures
BEAMS 2/39
Peter Smith & Mike Rosenman

Be strong enough for the loads

Not deflect too much

Suit the building for size, material, finish,
fixing etc
University of Sydney – Structures
BEAMS 3/39
Peter Smith & Mike Rosenman
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
University of Sydney – Structures
BEAMS 4/39
Peter Smith & Mike Rosenman
Designing a Beam

what we are trying to do


determine shape & dimensions
?
?
what do we need to know

span - how supported

loads on the beam

material

allowable strength & allowable deflection
University of Sydney – Structures
BEAMS 5/39
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 6/39
Peter Smith & Mike Rosenman

A column generally picks up load from
halfway to its neighbours
 It also carries the load that comes from the
floors above
University of Sydney – Structures
BEAMS 7/39
Peter Smith & Mike Rosenman

Code values per cubic metre or square
metre
 Multiply by the volume or area supported
Length
Height
Thickness
University of Sydney – Structures
Load =
Surface area x
Wt per sq m,
or
volume x
wt per cu m
BEAMS 8/39
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 9/39
Peter Smith & Mike Rosenman

Point loads, from concentrated loads or other
beams

Distributed loads, from anything continuous
Distributed
Load
Point Load
Reactions
University of Sydney – Structures
BEAMS 10/39
Peter Smith & Mike Rosenman

The loads (& reactions) bend the beam,
and try to shear through it
Bending
Shear
University of Sydney – Structures
BEAMS 11/39
Peter Smith & Mike Rosenman
e
e
e
e
C
T
Bending
Shear
University of Sydney – Structures
BEAMS 12/39
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 13/39
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 14/39
Peter Smith & Mike Rosenman

Consider cantilever beam with point load on end
W
vertical reaction,
R=W
and moment reaction MR = - WL
MR = -WL
L
R=W

Use the freebody idea to isolate part of the beam

Add in forces required for equilibrium
University of Sydney – Structures
BEAMS 15/39
Peter Smith & Mike Rosenman
W
M = -Wx
Take section anywhere at distance, x from end
x
Add in forces, V = W and moment M = - Wx
V=W
Shear V = W constant along length
(X = 0 -> L)
V=W
Shear Force Diagram
BM = Wx
Bending Moment BM = W.x
when x = L
BM = WL
when x = 0
BM = 0
University of Sydney – Structures
BM = WL
Bending Moment Diagram
BEAMS 16/39
Peter Smith & Mike Rosenman
w /unit length
L
For maximum shear V and bending moment BM
Total Load W = w.L
MR = -WL/2
= -wL2/2
L/2
L/2
R = W = wL
vertical reaction,
and moment reaction
University of Sydney – Structures
R=W
= wL
MR = - WL/2 = - wL2/2
BEAMS 17/39
Peter Smith & Mike Rosenman
wx
For distributed V and BM
M = -wx2/2
Take section anywhere at distance, x from end
Add in forces, V = w.x and moment M = - wx.x/2
X/2 X/2
V = wx
Shear
when x = L
when x = 0
V = wx
V = W = wL
V=0
V = wL
=W
Shear Force Diagram
w.x2/2
BM = wx2 /2
Bending Moment BM =
2/2
BM
=
wL
2
when x = L
BM = wL /2 = WL/2
= WL/2
when x = 0
BM = 0
(parabolic)
Bending Moment Diagram
University of Sydney – Structures
BEAMS 18/39
Peter Smith & Mike Rosenman

To plot a diagram, we need a sign convention
“Positive” shear
“Negative” shear
R.H down L.H down
L.H up
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
University of Sydney – Structures
BEAMS 19/39
Peter Smith & Mike Rosenman
W1
R1

W3
W2
Diagram of loading
R2
Starting at the left hand end, imitate each force
you meet (up or down)
W1
R1
W2
W3
R2
Shear Force Diagram
University of Sydney – Structures
BEAMS 2039
Peter Smith & Mike Rosenman

Point loads produce
a block diagram

Uniformly distributed loads
produce triangular diagrams
Diagrams of loading
Shear force diagrams
University of Sydney – Structures
BEAMS 21/39
Peter Smith & Mike Rosenman

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
Concrete
needs
special shear reinforcement
(45o or stirrups)
University of Sydney – Structures
Reo in concrete beam
BEAMS 22/39
Peter Smith & Mike Rosenman

To plot a diagram, we need a sign convention
Hogging
NEGATIVE
Sagging
POSITIVE
+

This convention is almost universally agreed
University of Sydney – Structures
BEAMS 23/39
Peter Smith & Mike Rosenman
Sagging bending moment is POSITIVE
(happy)
+
Hogging bending moment is NEGATIVE
(sad)
University of Sydney – Structures
BEAMS 24/39
Peter Smith & Mike Rosenman

Cantilevers produce negative moments
-
Cantilevers

Simple beams produce positive moments
-
+
Simple beam

-
+
Built-in beam
Built-in & continuous beams have both, with
negative over the supports
University of Sydney – Structures
BEAMS 25/39
Peter Smith & Mike Rosenman

Positive moments are drawn downwards
(textbooks are divided about this)
+
+
This way mimics the
beam’s deflection
University of Sydney – Structures
This way is normal
coordinate geometry
BEAMS 26/39
Peter Smith & Mike Rosenman

Point loads produce triangular diagrams
Diagrams of loading
Bending moment diagrams
University of Sydney – Structures
BEAMS 27/39
Peter Smith & Mike Rosenman

Distributed loads produce parabolic diagrams
UDL
UDL
Diagrams of loading
Bending moment diagrams
University of Sydney – Structures
BEAMS 28/39
Peter Smith & Mike Rosenman

We are mainly concerned
with the maximum values
Maximum value
University of Sydney – Structures
BEAMS 29/39
Peter Smith & Mike Rosenman

Draw the Deflected Shape (exaggerate)

Use the Deflected shape as a guide to where the
sagging (+) and hogging (-) moments are
-
-
-
+
University of Sydney – Structures
-
+
+
BEAMS 30/39
Peter Smith & Mike Rosenman

Cantilevered ends reduce the positive bending
moment

Built-in and continuous beams also have lower
maximum BMs and less deflection
Simply supported
Continuous
University of Sydney – Structures
BEAMS 31/39
Peter Smith & Mike Rosenman

Use the standard formulas where you can
W
L
Central point load
Max bending moment
= WL/4
University of Sydney – Structures
Total load = W
(w per metre length)
L
Uniformly distributed load
Max bending moment
= WL/8 or wL2/8
where W = wL
BEAMS 32/39
Peter Smith & Mike Rosenman
W
L
End point load
Max bending moment
= -WL
University of Sydney – Structures
Total load = W
(w per metre length)
L
Uniformly Distributed Load
Max bending moment
= -WL/2 or -wL2/2
where W = wL
BEAMS 33/39
Peter Smith & Mike Rosenman
W
W
W
W
W
W
W W W
W /m
W
W /m
Beam
Cable
W
W
BMD
University of Sydney – Structures
BEAMS 34/39
Peter Smith & Mike Rosenman
W
L
W = wL
W
W = wL
L
L
L
V = +W/2
V = +W
Mmax = -WL
University of Sydney – Structures
V = +W
Mmax = -WL/2
= -wL2/2
V = -W/2
Mmax = WL/4
V = +W/2
V = -W/2
Mmax = WL/8
= wL2/8
BEAMS 35/39
Peter Smith & Mike Rosenman

Causes compression on one face and
tension on the other
 Causes the beam to deflect
How much
compressive stress?
How much deflection?
University of Sydney – Structures
How much
tensile stress?
BEAMS 34/37
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 37/39
Peter Smith & Mike Rosenman

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
University of Sydney – Structures
BEAMS 38/39
Peter Smith & Mike Rosenman
we need to find the
Section Properties
next lecture
University of Sydney – Structures
BEAMS 39/39