eng2000 chapter 7 beams - york university chapter 7 beams. ... moments present in loaded 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 isdeformed§ in terms of the axial stress in the deformed beam

§ and the flexural rigidity of a beam

• The chapter will finish by considering the failureof 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 inthe 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 andshear – 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 thecantilever


If we cut the beam at an arbitrary point, wecan determine the system of forces andmoments required to maintain equilibrium

This system mustinclude axial, shearand moment


Sign conventions


Shear force & bendingmoment diagrams

• In order to determine whether the beam cansupport the loads required, we need to determinethe 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 theforce


•For 0 < x < 2L/3§P = 0

§V = F/3

§M = 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 momentis applied to each end

• Similarly, structuralmembers can deformor fail due to bendingmoments

• This will allow us tocalculate thedistribution of axialstress


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 alignedwith 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 thesedirections for M0 to bepositive. However, they arein the opposite direction toour convention (Beam 7),and we must remember toaccount for this at the end.


Neutral axis


Concrete• While we are mostly assuming beams made of

steel or other metals, many means are made ofconcrete§ and concrete does not support a tensile stress

• For concrete beams, we assume that only thematerial on the compressive side of the neutralaxis 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 isovercome

The yellow guidelines highlight the camber (upward curvature) of a pre-stressed double T. The pre-stressing strands can beseen protruding from the bottom of the beam, with the larger strands at the bottom edge. The tension is these strandsproduces 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 thebeam…§ ρ is the radius of the neutral axis

§ at y = 0 (neutral axis), dx remainsunchanged

§ dx’ is width at y

• Hence we can write a strain§ εx = (dx’ - dx)/dx = (dx’/dx) - 1

• Also§ dx = ρdθ and

§ dx’ = (ρ - y)dθ


• Hence

• Which tells us that the extension parallel to thebeam axis is linearly related to the distance formthe 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− y leading to

x = − y

x = − Ey


• Hence we can sketch of the stress normal to theaxis 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


xA∫ 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 thecentroid, given by

• Hence the neutral axis (at y = 0) passes throughthe beam’s centroid

y =ydA

A∫

A=

ydAA∫

dAA∫


Relating stress to applied moment• For the free body diagram

of the chopped beam, thesum of moments mustalso be zero forequilibrium

• For an elemental area ofthe cross section, themoment about the z-axis is– y xdA

• Hence the total momentfor the segment is

−y xA∫ 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

x = − Ey

1 = M0

EI

I = y2dAA∫

x = − M 0yI


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 ρ means the positive y-axis is on the concave sideof the neutral axis

§ EI is known as the flexural rigidity of the beam

• Extensional strain

• Normal stress

1 = − MEI

x = − y = MyEI

x = 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 thematerial in the beam is located further away fromthe 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 thecentroid

§ 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 astraightforward way

• where Ix’,y’ are the moments about one set of axesand 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 shapeeffects the stress

• Calculation of the deformed shape of the beam ispossible, 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 equationsrelating deformation to load

§ see chapter 9 of Riley

• However, we did say earlier that buckling of longslender 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 ultimateload

• Let us assume a column has already deformeddue 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 + 2v = 0

where

2 = PEI


• The general solution to such a differentialequation 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 λL = 0, or sin λL = 0, if the beam is deformed (v ≠ 0)

• The condition is now satisfied if

§ where n is an integer

v = Bsin x + C cos x

= nL


• Hence

• This represents a numberof sinusoidal deformationswith different wavelengths§ buckling modes

• We are looking for theminimum force to causedeformation, i.e. n = 1

P = n2 2EI

L2

v = Bsinn xL

P =2EI

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 causebuckling by restricting deflection at theappropriate 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 Yorkdestroyed 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, wepursued a course that explained mechanicalproperties of materials through to structures

• We will now return to materials science to exploreother macroscopic properties arise from atomicbonding§ e.g. electronic, thermal, optical, magnetic

eng2000 chapter 7 beams - york university chapter 7 beams. ... moments present in loaded beams...

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