ENG2000: R.I. Hornsey Beam: 1 ENG2000 Chapter 7 Beams.

ENG2000: R.I. Hornsey Beam: 1

ENG2000 Chapter 7Beams


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


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 …


simply supported beam

cantilever beam

combination beam


Cantilever

Free body diagram of the cantilever


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


Sign conventions


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 andthe force

and then the other side of the force


•For 0 < x < 2L/3P = 0V = F/3M = Fx/3

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


Diagrams


General equations

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

€

dPdx

=0

dVdx

=−w

dMdx

=V

(no axial force)

(shear increases with x)

(moment = shear x distance)


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


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.


Neutral axis


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


• 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


Expression for stress• If we take a small element of

width dx, and deform the beam… 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 x = (dx’ - dx)/dx = (dx’/dx) - 1

• Also dx = d dx’ = ( - y)d


• 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

€

d ′ x dx

=1−yρ

leading to

εx =−yρ

€

εx =−Eyρ


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


Location of neutral axis• The previous analysis used , but did not relate

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


€

σxA∫ dA=0

€

yA∫ dA=0

• Expressed mathematically

where A is the area of the beam’s cross section

• Substituting in for x,

• 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

€

y =ydA

A∫

A=

ydAA∫

dAA∫


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 – yxdA

• Hence the total moment for the segment is

€

−yσxA∫ dA−M0 =0


• Using

• we find that:

• where

I is a very important figure of merit for a beam’s shape, known as the moment of inertia

• Hence

€

εx =−Eyρ

€

1ρ

=M0

EI

€

I = y2dAA∫

€

σx =−M0y

I


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 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

€

1ρ

=−MEI

€

εx =−yρ

=MyEI

€

σx =MyI


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

€

I = y2dAA∫

http://www.tricelcorp.com/images/ibeam.jpg


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

€

Ix =I ′ x +dy2A

Iy =I ′ y +dx2A

A

x

y

dy

dx

x’

y’


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


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


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 …


• Here M = Pv where v is the beam’s

deflection

• It turns out that

• so

€

d2v

dv2=−

MEI

€

d2v

dv2+λ2v=0

where

λ2 =PEI


• 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 L = 0, or sin L = 0, if the beam is deformed (v ≠ 0)

• The condition is now satisfied if

where n is an integer

€

v=Bsinλx+Ccosλx

€

λ =nπL


• 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

€

P =n2π2EI

L2

v=BsinnπxL

⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

P =π2EI

L2


• 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

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


• 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


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: 1 ENG2000 Chapter 7 Beams.

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Transcript of ENG2000: R.I. Hornsey Beam: 1 ENG2000 Chapter 7 Beams.