1. ENG2000 Chapter 7 Beams

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. simply supported beam
cantilever beam
combination beam

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
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

9. For 2L/3 < x < L For 0 < x < 2L/3 P = 0 P = 0 V = - 2F/3
M = Fx/3
For 2L/3 < x < L
P = 0
V = - 2F/3
M = (2F/3)(L - x)

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 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.

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

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.

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. We can now calculated the stress
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 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

21. Expressed mathematically
where A is the area of the beam’s 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 beam’s 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 Hence
I is a very important figure of merit for a beam’s 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

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 dx A dy x’ 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 beam’s 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. We are looking for the minimum force to cause deformation, i.e. n = 1
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 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 m4
E = 200 GPa & L = 300mm
hence P = 2.74N ≈ 300 grammes

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

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