ENG2000 Chapter 7
Beams
ENG2000: R.I. Hornsey
Beam: 1
Overview
• In this chapter, we consider the stresses and
moments present in loaded beams
 shear stress and bending moment diagrams
• We will also look at what happens when a beam is
deformed
 in terms of the axial stress in the deformed beam
 and the flexural rigidity of a beam
• The chapter will finish by considering the failure
of a column due to an axial compressive force
 buckling
ENG2000: R.I. Hornsey
Beam: 2
Beams with concentrated loads
• A beam is defined as a slender structural member
• For trusses we assumed that what happened in
the members was unimportant to the system
 although the nature of the members clearly affects the
strength of a truss
 here we look at what happens when beams are loaded
• We will look at the internal forces – axial and
shear – and moments in the loaded beam
• Three types of beam are statically determinate …
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Beam: 3
simply supported beam
cantilever beam
combination beam
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Beam: 4
Cantilever
Free body diagram of the
cantilever
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Beam: 5
If we cut the beam at an arbitrary point, we
can determine the system of forces and
moments required to maintain equilibrium
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This system must
include axial, shear
and moment
Beam: 6
Sign conventions
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Beam: 7
Shear force & bending
moment diagrams
• In order to determine whether the beam can
support the loads required, we need to determine
the distribution of stress in the beam
 i.e. P, V, and M as a function of x
• Let’s take the following example:
 we first cut the beam at an arbitrary
location between the left end and
the force
 and then the other side of the
force
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Beam: 8
•For 0 < x < 2L/3
P = 0
V = F/3
M = Fx/3
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• For 2L/3 < x < L
 P=0
 V = - 2F/3
 M = (2F/3)(L - x)
Beam: 9
Diagrams
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Beam: 10
General equations
• For a generalised distributed load of w(x) N/m:
dP
(no axial force)
0
dx
dV
(shear increases with x)
 w
dx
dM
V
(moment = shear x distance)
dx
ENG2000: R.I. Hornsey
Beam: 11
Stresses in beams
• A simple beam, such
as this firewood,
snaps when a moment
is applied to each end
• Similarly, structural
members can deform
or fail due to bending
moments
• This will allow us to
calculate the
distribution of axial
stress
ENG2000: R.I. Hornsey
Beam: 12
Geometry of deformation
• We will consider the deformation of an ideal,
isotropic prismatic beam
 the cross section is symmetric about y-axis
• All parts of the beam that were originally aligned
with the longitudinal axis bend into circular arcs
 plane sections of the beam remain plane and perpendicular
to the beam’s curved axis
Note: we will take these
directions for M0 to be
positive. However, they are
in the opposite direction to
our convention (Beam 7),
and we must remember to
account for this at the end.
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Beam: 13
Neutral axis
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Beam: 14
Concrete
• While we are mostly assuming beams made of
steel or other metals, many means are made of
concrete
 and concrete does not support a tensile stress
• For concrete beams, we assume that only the
material on the compressive side of the neutral
axis actually carries a load
ENG2000: R.I. Hornsey
Beam: 15
http://www.uaf.edu/~asce/failure.jpg
• One solution to this is pre-stressed concrete
 where metal bars set within the concrete are pre-stressed to
provide an initial compression to the concrete beam
 so it can withstand some tension, until the pre-stress is
overcome
The yellow guidelines highlight the camber (upward curvature) of a pre-stressed double T. The pre-stressing strands can be
seen protruding from the bottom of the beam, with the larger strands at the bottom edge. The tension is these strands
produces the camber, the beam is straight when cast.
http://urban.arch.virginia.edu/~km6e/tti/tti-summary/half/concrete1-1-01-detail.jpeg
ENG2000: R.I. Hornsey
Beam: 16
Expression for stress
• If we take a small element of
width dx, and deform the
beam…
 r is the radius of the neutral axis
 at y = 0 (neutral axis), dx remains
unchanged
 dx’ is width at y
• Hence we can write a strain
 ex = (dx’ - dx)/dx = (dx’/dx) - 1
• Also
 dx = rd
 dx’ = (r - y)d
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Beam: 17
• Hence
dx
y
 1
leading to
dx
r
y
ex  
r
• Which tells us that the extension parallel to the
beam axis is linearly related to the distance form
the neutral axis
 and the sign indicates compression for positive y, i.e. below
the neutral axis
• We can now calculated the stress
 assuming no additional loads – just the moment
ex  
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Ey
r
Beam: 18
• Hence we can sketch of the stress normal to the
axis of the beam …
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Beam: 19
Location of neutral axis
• The previous analysis used r, but did not relate r
to the applied moment
 and we don’t know where the neutral axis is located either
• Imagine that we cut the beam at some point
• Since the moment M0 does not exert a net force
 the sum total of the stress at the cut section must also be
zero
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Beam: 20
• Expressed mathematically
  x dA  0
A
 where A is the area of the beam’s cross section
• Substituting in for x,  ydA  0
A
• Recall that the centre of mass is given by
 ycom = (∫y dm)/m
• For a geometric shape, the equivalent point is the
centroid, given by
 ydA  ydA
yA
A
A
 dA
A
• Hence the neutral axis (at y = 0) passes through
the beam’s centroid
ENG2000: R.I. Hornsey
Beam: 21
Relating stress to applied moment
• For the free body diagram
of the chopped beam, the
sum of moments must
also be zero for
equilibrium
• For an elemental area of
the cross section, the
moment about the z-axis is
– yxdA
• Hence the total moment
for the segment is
 y x dA  M 0  0
A
ENG2000: R.I. Hornsey
Beam: 22
• Using
ex  
• we find that:
Ey
r
1
M0

r EI
• where
I   y 2 dA
A
 I is a very important figure of merit for a beam’s shape,
known as the moment of inertia
• Hence
x  
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M 0y
I
Beam: 23
Conventional directions
• We have to revert to our convention for the
positive direction of moments
 see Beam 7 and Beam 13
 by letting M = -M0
• Hence, our equations are as follows
• Radius of curvature 1   M
r
EI
 positive r means the positive y-axis is on the concave side
of the neutral axis
 EI is known as the flexural rigidity of the beam
• Extensional strain
• Normal stress
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y
My
ex   
r EI
My
x 
I
Beam: 24
Interpretation of moment of inertia
• What does the moment of inertia I   y 2 dA mean?
 and flexural rigidity, EI
A
• Essentially, this says that the beam is stiffer if the
material in the beam is located further away from
the centroid (neutral axis)
 so any area, dA, is more effective at stiffening the beam
depending on the square of the distance
 hence, if you want to make a strong beam with little material,
make sure that the material is as far as possible from the
centroid
 hence, ‘I’ beams
Beam: 25
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http://www.tricelcorp.com/images/ibeam.jpg
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Beam: 26
ENG2000: R.I. Hornsey
Beam: 27
Parallel axis theorem
• If we know I for a particular axis, we can calculate
the value for a different parallel axis in a
straightforward way
y’
y
I x  I x d y2 A
I y  I y d x2 A
A
dx
x’
dy
x
• where Ix’,y’ are the moments about one set of axes
and dx,y are the distances to the new set of axes
ENG2000: R.I. Hornsey
Beam: 28
Summary so far
• So we can now calculate stresses and moments
within a beam, and we know how the beam shape
effects the stress
• Calculation of the deformed shape of the beam is
possible, but beyond the scope of this course
 essentially, the ways in which the ends of the beam are ixed
determines the deformed shape
 by providing boundary conditions to the differential equations
relating deformation to load
 see chapter 9 of Riley
• However, we did say earlier that buckling of long
slender beams is also important
 which is why we need trusses in the first place
ENG2000: R.I. Hornsey
Beam: 29
Buckling of columns
• Imagine a hacksaw blade
 sy = 520Mpa
 cross section is 12mm x 0.5mm
 so the blade should withstand a compression of 3120N ≈
300kg
• However it is easy to cause the blade to fail
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Beam: 30
Euler buckling
• Buckling is a geometric instability that causes a
structural column to fail well before its ultimate
load
• Let us assume a column has already deformed
due to an axial load
 and we will determine the force we need to maintain
equilibrium
• Cutting the beam at an arbitrary point gives us …
ENG2000: R.I. Hornsey
Beam: 31
• Here M = Pv
 where v is the beam’s
deflection
• It turns out that
d 2v
M
2 
EI
dv
• so
d 2v 2
 v 0
2
dv
where
P
2
 
EI
ENG2000: R.I. Hornsey
Beam: 32
• The general solution to such a differential
equation is
v  Bsin x  C cos x
 where B and C are determined by the boundary conditions
• At x = 0, v = 0
 so C = 0
• Also v = 0 at x = L
 so Bsin L = 0, or sin L = 0, if the beam is deformed (v ≠ 0)
• The condition is now satisfied if
n

L
 where n is an integer
ENG2000: R.I. Hornsey
Beam: 33
• Hence
n 2  2 EI
P
L2
nx 
v  B sin 
 L 
• This represents a number
of sinusoidal deformations
with different wavelengths
 buckling modes
• We are looking for the
minimum force to cause
deformation, i.e. n = 1
P
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 2 EI
L2
Beam: 34
• Note that the deformation is dependent on “B”
 but we consider the buckling load to be the force to cause an
arbitrarily small deformation
• This is known as the Euler buckling load
 after Leonhard Euler’s 1744 formulation
• We can increase the load required to cause
buckling by restricting deflection at the
appropriate places along the beam
 which is essentially what a truss does
• For the hacksaw blade
 I = bh3/12 (a rectangular section) = 1.25 x 10-13 m4
 E = 200 GPa & L = 300mm
 hence P = 2.74N ≈ 300 grammes
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http://chronomath.irem.univ-mrs.fr/jpeg/Euler.jpg
Beam: 35
• e.g. Cortlandt Street station in New York
destroyed on 11 September 2001
http://www.nycsubway.org/irt/westside/wtc-damage/iw-cortlandt-damage-09.jpg
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Beam: 36
Summary
• This completes our brief look at structural
mechanics and statics
• Based on our knowledge of materials science, we
pursued a course that explained mechanical
properties of materials through to structures
• We will now return to materials science to explore
other macroscopic properties arise from atomic
bonding
 e.g. electronic, thermal, optical, magnetic
ENG2000: R.I. Hornsey
Beam: 37