Bending and extension of beams

The elastic field problem is solved by satisfying a set of 15 equations in 15 unknowns.Out of these 9 are differential equations. The geometric shapes and boundaryconditions for most practical structures are so complex that exact theory of elasticitysolutions are either imposible to obtain or too complex to be useful in the field.Approximate solutions using strength of materials techniques are accurate enoughfor engineering purposes and are commonly used.

Stress Resultants

A beam is defined to be a member capable of resisting bending moments and whoselength is large compared to it's cross sectional area. It is assumed that the beamlength is along the x axis and that the cross sectional area A of the beam variessmoothly in x. In beam theory, it is convenient to work with force and momentresultants of the stresses that act upon the cross section of the beam.

Consider a beam withexternally applied loadings asshown below. The externallyapplied forces per unit lengthare px, py and pz and momentsper unit length are mx,my andmz. The two loadings areshown on two separate figuresbelow. Note that externallyforces and moments areassumed to be positive if theyhave the same sense as theirrespective coordinatedirections.

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Now suppose that a cuttingplane is passed through thebeam at a distance x from theorigin and a free body diagramis constructed as shown. Ingeneral, six internal resultantscan be constructed at theintersection of the x axis withthe cross section defined bythe cutting plane.

On element area dA, Force=dF

dF= σxxdA ix + σxydA iy +σxzdA iz

Total force on face = ∫ (σxxdAix + σxydA iy + σxzdA iz)

or, F=(P ix+ Vy iy+ Vz iz)

Due to element are dA, themoment dM is

dM= r × dF

The total moment on the faceis the integral of this value.

M=(Mx ix+ My iy+ Mz iz)

The six stress resultants cannow be written as:

Axial Force P = ∫ σxxdA

Shear Force Vy = ∫ σxydA

Shear Force Vz = ∫ σxzdA

Torsional Moment Mx = ∫ (yσxz-zσxy)dA

Bending Moment My = ∫ zσxxdA

Bending Moment Mz = ∫ (-yσxx)dA

The face on which these internal resultants act is called a positive face because the xcoordinate axis is coming out of it. Internal resultants are defined to be positive when

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they act in their respective coordinate axis directions on a positive face.

Linear Differential equations of equilibrium

For the beam shown in the above figure subjected to externally applied forces perunit length px, py and pz and moments per unit length mx,my and mz, the differentialequations of equilibrium can be written as:

dP/dx = -px

dVy/dx = -py

dVz/dx = -pz

dMx/dx = -mx

dMy/dx = -my + Vz(x)

dMz/dx = -mz - Vy(x)

The above equations satisfy the equilibrium equations only in an average sense overthe cross section. The above equations can be solved if the boundary conditions areknown.

Stresses due to extension and bending

Normal stresses σxx are developed in the x coordinate direction due to the axialloading P and bending moments My and Mz. We will be making the followingassumptions (called Bernoulli Euler assumptions):

The transversecomponents of normalstress σyy and σzz areassumed to be negligiblecompared to the axialstress σxx.Cross sections areassumed to remainplanar and normal to the

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centroidal axis ofdeformation.

It can be seen that the axial displacement u(x,y,z) is the sum of the contributions dueto the axial extension u0=u(x,0,0), rotation due to bending about the z and y axesrespectively:

u(x,y,z)=u0 - θz(x)y + θy(x)z

The axial strain εxx is given by:

εxx= ∂u/∂x = du0/dx -y dθz/dx + z dθy/dx

Consider a beam whose material properties are hetrogenous. Substitution in linearelastic stress strain equations gives:

σxx= E (du0/dx -y dθz/dx + z dθy/dx)

Substitution into resultant equations gives:

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P = ∫E (du0/dx -y dθz/dx + z dθy/dx) dA

My = ∫E (du0/dx -y dθz/dx + z dθy/dx) z dA

Mz = - ∫E (du0/dx -y dθz/dx + z dθy/dx) ydA

Now define the modulus weighted section properties as follows:

A* = dA

y* = y dA

z* = z dA

I*yy = z2 dA

I*yz = yz dA

I*zz = y2 dA

In the case of a homogeneous beam, the reference modulus E1 may be chosen to bethe same as actual material modulus. In heterogeneous beams, the modulus weightedcentroidal axes(x*,y*,z*) should be used. If the modulus weighted centroidal axes arechosen, then y* and z* defined above will be identically zero. With such a choice, theequations become:

du0/dx = P/ (E1 A*)

dθz/dx = (MzI*yy+MyI*

yz)/ (E1(I*yyI*

zz- I*yz

2))

dθy/dx = (MyI*zz+MzI*

yz)/ (E1(I*yyI*

zz- I*yz

2))

Substituting these results into the strain equations gives:εxx = P/ (E1 A*) -y (MzI*

yy+MyI*yz)/ (E1(I*

yyI*zz- I*

yz2))

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+z (MyI*zz+MzI*

yz)/ (E1(I*yyI*

zz- I*yz

2))

σxx = E P/ (E1 A*) -y E(MzI*yy+MyI*

yz)/ (E1(I*yyI*

zz- I*yz

2))

+z E(MyI*zz+MzI*

yz)/ (E1(I*yyI*

zz- I*yz

2))

Determination of modulus weighted section properties

Consider a beam composed of n discrete portions of homogeneous make up. In thiscase the equations for A* etc. reduce to:

A* = Ei/E1 dA

or, A* = dA

or, A*= Ai

Also,

y* = y dA

or, y'* = yi'Ai

Similarly z* = z dA

or, z'* = zi'Ai

where yi' and zi' are the coordinates of the ith portion taken with respect to thearbitrary y' and z' axes.The modulus weighted moments of inertia may be determined about the modulusweighted centroid, or alternatively, we can determine these properties about thearbitrary y' and z' axes and then use the parallel axis theorem.

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I*yy = I*

y'y' - (z'*)2 A*

I*yz = I*

y'z' - (y'*)(z*') A*

I*zz = I*

z'z' - (y'*)2 A*

Example Problem 1

The figure below shows anunsymmetrical beam sectioncomposed of four stringersa,b,c,d of equal area 0.1 sq.inch connected by a thin web.Determine stress and total loadon each stringer if themoments applied on thesection are My=10,000 lb-inand Mz=5000 lb-in.

Example Problem 2

An airfoil is idealized as prismaticwith average cross section shownbelow. All stringers are made ofstructural steel (E= 30 X 106 psi) withdimensions shown below. All skinsare Aluminum(E=10 X 106 psi).determine the modulus weightedsection properties assuming E1= 10 X

106 psi.

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Example Problem 3

The idealized airfoil describedin the previous example issubjected to the loadingshown. Determine the stressesin the stringers at x=0.

Deformations due to bending and extension

It is possible to determine u(x,y,z), v(x,y,z) and w(x,y,z) at various points in the beam.It is of practical importance to determine the three displacement components alongthe centroidal axes i.e. u(x,0,0),v(x,0,0) and w(x,0,0).The axial component u0= u(x,0,0)can be ontained as:du0/dx = P/(E1 A*)To obtain v0 and w0, we use the relations :

dθz/dx = d2v0/dx2

dθy/dx = d2w0/dx2

The above equations containing the variables u0,v0,w0 can be solved by integratingdirectly after applying displacement boundary conditions to account for integrationconstants.

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