Cantilever beam test

Graeme J. Kennedy and Julian J. Rimoli

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Background

A beam is an idealization of the 3D continuum when one of thecharacteristic dimensions of a given body is much larger than theothers.

Beams are critical components of almost every aerospace structure.

There are different ways of idealizing beams, e.g. Euler beam,Timoshenko beam, etc, and each of them works very well in manypractical situations.

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Lab Objectives

Review basic beam theory

Learn about micrometer measurements for beam deflection

Learn about load control vs. displacement control

Apply strain transformation theory

Learn about principal stresses and strains

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Procedures

Two experiments will be conducted: in one a load is applied at the tip ofthe beam (load control) and in the other one a displacement is appliedinstead (displacement control).

Pre-test: measure beam dimensions and distances between points ofapplication of load and displacement and strain gauge.

Test A: apply tip loads and measure rosette strains.

Test B: apply tip deflection and measure rosette strains.

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Experimental set-up

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Experimental set up

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Beam analysis

From Euler-Bernoulli beam theory we know that the stress on the beam isrelated to the moment at a given cross-section through the equation:

σx =Myz

I

At the location of the rosette, the moment is My = P(L− xg ).Considering that the moment of inertia is I = bh3/12 we get that at thelocation of the rosette (z = h/2)

σx =6P(L− xg )

bh2

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Strains

In the experimental setup we measure strains and not stress!

To compare with the theory, we need a stress-strain relation.

In the elastic regime, we adopt Hooke’s Law under the plane-stressassumption:

εX =σXE

εY = −ν σXE

εZ = −ν σXE

γYZ = 0 γXZ = 0 γXY = 0

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Strain transformation

There is one last struggle: we derived strains in the beam’s referenceframe, e.g., εX , εY , and γXY

The strain gauge is not necessarily aligned with the beam!

In summary: we’ll measure εA, εB , and εC and we want to find εx , εy ,and γxy !

This requires a transformation from one frame to another!

We will review a similar transformation we are familiar with first(stresses) and will end with the needed expressions for strains.

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Stress transformation equations: Derivation

X

Y

σx

σY

σX

τxy

τXY

τXYx

y

X

Y

A0σx sec θ

A0σY tan θ

A0σX

A0τxy sec θ

A0τXY tan θ

A0τXY

Stress components Force components

σx = σX cos2 θ + σY sin2 θ + 2τXY sin θ cos θτxy = τXY (cos

2 θ − sin2 θ)− (σX − σY ) sin θ cos θ

Sum of forces for equilibrium:

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Mohr’s circle for stress

σ

τ

2θ

C

R

C = 12(σX + σY )

R = 12

√(σX − σY )2 + 4τ 2XY

(σX , τXY )

(σx , τxy)

(σy ,−τxy)

(σY ,−τXY )

2θp

σ1σ2

tan 2θp =2τXY

σX − σY

σ1,2 = C ± R

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Stress transformation equation

The transformation for stress can also be written using atransformation matrix, T(θ)

This form is more useful in many calculations

σxσyτxy

=

cos2 θ sin2 θ 2 sin θ cos θsin2 θ cos2 θ −2 sin θ cos θ

− sin θ cos θ sin θ cos θ cos2 θ − sin2 θ

σXσYτXY

The transformation matrix is given as follows

T(θ) =

cos2 θ sin2 θ 2 sin θ cos θsin2 θ cos2 θ −2 sin θ cos θ

− sin θ cos θ sin θ cos θ cos2 θ − sin2 θ

The mathematical representation of stresses are tensors.Thistransformation is the same for all tensors, regardless of what theyrepresent!

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Transformations for strains

In this lab we need to transform strains

Recall: σx , σy , and τxy are components of a tensor

εx , εy , and γxy ARE NOT components of a tensor

The components of the strain tensor, E, are εx , εy , and γxy/2:

E =

[εx γxy/2

γxy/2 εy

]

All work on stress tensor can now be applied to the strain tensor

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Mohr’s circle for strain

ε

γ/2

2θ

C

R

C = 12(εX + εY )

R = 12

√(εX − εY )2 + γ2XY

(εX , γXY /2)

(εx , γxy/2)

(εy ,−γxy/2)

(εY ,−γXY /2)

2θp

ε1ε2

tan 2θp =γXY

εX − εY

ε1,2 = C ± R

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General strain transformations

For 2D strain cases, we can also express the transformation of straincomponents as follows:

εxεy

γxy/2

= T(θ)

εXεY

γXY /2

This is often written as follows, in terms of the matrixR = diag{1, 1, 2}

εxεyγxy

= RT(θ)R−1

εXεYγXY

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Orthotropic materials

In this class, we will stick to isotropicmaterials, which is often a good modelfor metallic structures

For orthotropic materials, likecomposites, the properties vary stronglywith directionObtain the stress-strain relationship in the X − Y − Z frame

σXσYτXY

= T(θ)−1CmatRT(θ)R−1

εXεYγXY

= C(θ)

εXεYγXY

For isotropic materials C(θ) = Cmat

Isotropic

Cmat =E

1− ν2

1 ν 0ν 1 00 0 1

2(1− ν)

Orthotropic

Cmat =

Q11 Q12 0Q12 Q22 0

0 0 Q66

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Strain transformation

Back to the strain gauge rosette

The rosette will measure normal strains along three independentdirections given by θA, θB , and θC

We can use these known angles (that you can measure in the lab) andthe strain measurements to determine the desired strain components

X

εA

εBεCY

θC

θB

θA

εA = εX cos2 θA + εY sin2 θA + γXY sin θA cos θA

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Strain transformation

Writing an equation for each gauge in the rectangular rosette givesthe following linear system:

εAεBεC

=

cos2 θA sin2 θA sin θA cos θAcos2 θB sin2 θB sin θB cos θBcos2 θC sin2 θC sin θC cos θC

εXεYγXY

Since we know θA, θB , and θC and we measure εA, εB and εC (usingthe Vishay 7000) we can find εX , εY , and γXY :

εXεYγXY

=

cos2 θA sin2 θA sin θA cos θAcos2 θB sin2 θB sin θB cos θBcos2 θC sin2 θC sin θC cos θC

−1 εAεBεC

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