ENG2000 Chapter 7 Beams

1
ENG2000 Chapter 7Beams
2
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

3
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

4
(No Transcript)
5
Cantilever
Free body diagram of the cantilever
6
If we cut the beam at an arbitrary point, we can
determine the system of forces and moments
required to maintain equilibrium
This system must include axial, shear and moment
7
Sign conventions
8
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
• Lets take the following example
• we first cut the beam at an arbitrarylocation
  between the left end andthe force
• and then the other side of the force

9
(No Transcript)
10
Diagrams
11
General equations

• For a generalised distributed load of w(x) N/m

(no axial force) (shear increases with
x) (moment shear x distance)
12
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

13
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 beams 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.
14
Neutral axis
15
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

http//www.uaf.edu/asce/failure.jpg
16

• 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-summa
ry/half/concrete1-1-01-detail.jpeg
17
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??

18

• Hence
• 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

19

• Hence we can sketch of the stress normal to the
  axis of the beam

20
Location of neutral axis

• The previous analysis used r, but did not relate
  r to the applied moment
• and we dont 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

21

• Expressed mathematically
• where A is the area of the beams cross section
• Substituting in for sx,
• 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
• Hence the neutral axis (at y 0) passes through
  the beams centroid

22
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 ysxdA
• Hence the total moment for the segment is

23

• Using
• we find that
• where
• I is a very important figure of merit for a
  beams shape, known as the moment of inertia
• Hence

24
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
• 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

25
Interpretation of moment of inertia

• What does the moment of inertia
  mean?
• and flexural rigidity, EI
• 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

http//www.tricelcorp.com/images/ibeam.jpg
26
(No Transcript)
27
(No Transcript)
28
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
• where Ix,y are the moments about one set of
  axes and dx,y are the distances to the new set of
  axes

y
y
A
dx
x
dy
x
29
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

30
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

31
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

32

• Here M Pv
• where v is the beams deflection
• It turns out that
• so

33

• The general solution to such a differential
  equation is
• 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 lL 0, or sin lL 0, if the beam is
  deformed (v ? 0)
• The condition is now satisfied if
• where n is an integer

34

• Hence
• 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

35

• 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 Eulers 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

http//chronomath.irem.univ-mrs.fr/jpeg/Euler.jpg
36

• e.g. Cortlandt Street station in New York
  destroyed on 11 September 2001

http//www.nycsubway.org/irt/westside/wtc-damage/i
w-cortlandt-damage-09.jpg
37
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