Cantilever beam test - gatech.edu
Transcript of Cantilever beam test - gatech.edu
Cantilever beam test
Graeme J. Kennedy and Julian J. Rimoli
1 / 18
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.
2 / 18
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
3 / 18
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.
4 / 18
Experimental set-up
5 / 18
Experimental set up
6 / 18
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
7 / 18
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
8 / 18
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.
9 / 18
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:
10 / 18
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
11 / 18
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!
12 / 18
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
13 / 18
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
14 / 18
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
15 / 18
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
16 / 18
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
17 / 18
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
18 / 18