Annexes of Experimental Activities

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Annexes of Experimental

Activities

A) Calculation of the applied force

The real forces applied to the specimens were calculated by the system shown in Figure 11.11.

W

T

P

Tx Ty=F

R

γ Rx Ry

α

Fr L1 _L2 _L3

Figure 11.11: System of loading Considering in the Figure 11.11:

F r is the friction force between the lever and the metal axis,

R is the constrain force,

T is the hanger force,

F is the applied load,

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P = 11.58N is the lever load,

L1= 50mm is the distance between the hinge and the hanger force,

L2= 203mm is the distance between the hinge and the barycenter of the lever,

L3= 366mm is the distance between the hinge and the plates load,

α is the angle between the lever and the horizontal axis,

γ is the angle between the cable force T and the vertical axis.

The equilibrium system is given by X

X = F rx− Rx− Tx= 0 (11.129)

X

Y = Ty− W − P − F ry− Ry= 0 (11.130)

X

M = W L3cosα+ P L2cosα − T L1cos(α + γ) − F r · d = 0 (11.131)

wherein:

F rx= µRy

F ry = µRx

µ is the friction coefficient between the lever and the metal axis

d is the diameter of hinge bar

R=qR2 x+ R2y F r= µR = µqR2 x+ Ry2 Tx= T sinγ Ty= F = T cosγ Substituting, X X = µRy− Rx− T sinγ= 0 (11.132) X Y = T cosγ − W − P − µRx− Ry = 0 (11.133) X

M = W L3cosα+ P L2cosα − T L1cos(α + γ) − µ

q

R2

x+ R2y· d= 0 (11.134)

and solving for Rx and Ry

Rx=

µ(T cosγ − W − P + µT sinγ)

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Ry=

T cosγ − W − P+ µT sinγ

µ2+ 1 (11.136)

From the rotation equilibrium in Equation (11.131) and T = F

cosγ:

W L3cosα+ P L2cosα −_cosγF L1cos(α + γ) −

−µ r _{µ(F −W −P +µF tgγ)} µ2₊₁ − F tgγ 2 +F −W −P +µF tgγ µ2₊₁ 2 · d= 0 (11.137) By interactions of Equation (11.137), the effective force F applied on the specimens can be calculated. 600 1060 60 60 740 370 200

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B) Evaluation of the friction coefficient between the lever

and the axis

The friction coefficient µ between the levers and the metal axes was estimated through the dynamic theory of the simple pendulum with friction in the hinge support, in which the friction causes a certain loss of mechanical energy at every half period done by the pendulum.

Actually, considering the system in Figure 11.13, the equilibrium system is given by

WT = m · aT

WN = T (11.138)

where:

W = m · g is the weight,

WN = cos θ · W is the radial weight,

WT = − sin θ · W is the tangential weight and

aT = L · ∂

2_θ

∂t2 is the tangential acceleration.

Then, the dynamic equation becomes

∂2_θ ∂t2 + L g sin θ = 0 (11.139) T Ty Tx WN WT W

θ

h L m θ θ

Figure 11.13: Pendulum system

For small angle sin θ ≈ θ and cos θ ≈ 1. Therefore Equation (11.138) becomes ∂2θ

∂t2 + L g · θ= 0

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The friction in the hinge is given by F r = µ · T = µ · m · g.

Considering the boundary condition of pendulum with an initial angle θ0, the initial

energy of the overall system is

E0= m · g · h0 (11.141)

where h0= L · (1 − cos θ0).

After the first half period the loss of energy is expressed by

4E = E0− E1= m · g · h0− m · g · h1= m · g(h0− h1) (11.142)

The loss of energy 4E due to the friction in the hinge is given by

4E = F r · θT · R= µ · m · g · θT · R (11.143)

where:

θT is the total angle made by the pendulum, given by θT = θ0+ θ1,

R is the radius of the axes which make up the hinge.

Then, by Equations (11.142)-(11.143), the friction coefficient µ is given by

µ=h0− h1 θT· R =

L(cos θ1−cos θ0)

(θ0+ θ1) · R (11.144)

According to Taylor series expansion

cos θ1−cos θ0= − θ2 1 2 + θ2 0 2 + θ4 1 24− θ4 0 24− θ6 1 720+ θ6 0 720+ O(θ)8 (11.145)

if the terms over fourth order are neglected

cos θ1−cos θ0≈

θ2₀− θ2 1

2 (11.146)

Thus, by substituting Equation (11.146) into (11.144), the friction coefficient can be ex-pressed as

µ=L(θ0− θ1)

2R (11.147)

The friction coefficient was evaluated through a test in which the total number of half

periods made by the pendulum, starting from a little angle θ0, was counted. Therefore,

if the final angle is equal to zero and if n is the total number of half periods, the friction coefficient turns out to be

µ= L · θ0

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C) Influence of friction in the supports on the deflection

Though the large deflection and second order effects can not be taken into account, extensive investigation on the influence of the tension forces on the deflection due to the friction in the supports has been carried out. In Figure 11.14 are shown the experiment models of the first and second order analysis.

F F a = 2'000a 2'000 mm a aa = 2'000 P P S S a = 2'000w(x) x F F a = 2'000 2'000 mm a = 2'000 a a a x a = 2'000 L First order analysis

Second order analysis

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Figure 11.14: First and second order analysis

In the experiment processing, second order analysis is not taken into account because the maximum deflection of the specimens did not exceed 1/10 of the total nominal span L [39]; moreover, horizontal forces due to the inclination of the hanger were also neglected. Indeed, pure bending and shear-bending states in Figure 9.1 can only exist through an experimental model in first order analysis, as in Figure 11.14 (a).

Considering the second order analysis and the inclination of the forces in the model, system in Figure 11.14 (b) must be considered. Let us introduce a simplification into the model thanks to the symmetry, the new experimental model is shown in Figure 11.15.

F

a = 2'000 2'000 mm a a/2

P

S

1 2

S-P

x1 x2

Figure 11.15: Experimental model with second order analysis

In the sistem in Figure 11.15, the bending forces in the two segments are given by

M1(x1) = F · x1− S · w1(x1)

M2(x2) = F · a − (S + P ) · w2(x2) + P · w1(a) (11.149)

Thanks to Euler-Bernoulli’s beam model, the bending moments are also expressed as

(x ) = EJ ·∂2_w _(x ₎

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Thus, we obtain the following differential equations EJ ·∂2w1(x1) ∂x2 1 − S · w1(x1) = F · x EJ · ∂2w2(x2) ∂x2 2 −(S + P ) · w2(x2) = F · a + P · w1(a) (11.151) with the boundary conditions as

w1(0) = 0 ∂w_∂x1(a) =∂w_∂x2(0)

w1(a) = w2(0)

∂w2(a2)

∂x = 0

(11.152) By solving Equations (11.151)-(11.152), the value of S remains unknown. Actually, the extension of the beam produced by the forces S is equal to the difference between the length of the arc along the deflection curve and the initial length L. For small deflection, the extension is given by λ= S · L EA = 1 2    a ˆ 0  ∂w1(x1) ∂x1 2 dx1+ a/2 ˆ 0  ∂w2(x2) ∂x2 2 dx2    (11.153)

By Equation (11.153) and the solutions w1(x1, S), w2(x2, S) of the differential Equations

(11.151)-(11.152), it is possible to obtain the value of S and therefore evaluate the real

deflections w1(x2), w2(x2).

Now, the comparison between the tension forces S on the supports and the maximum forces allowable due to the friction must be compared in order to investigate if the beam can slide or not on the supports. The maximum forces produced by the friction between the beam and metal support with lubricant is given by

SM AX ≈ µ · F (11.154)

with µ the friction coefficient equal to 0.2.

Figure 11.16 shows that the the restraint forces S is always higher than the maximum

friction forces SM AX in the our range of applied load (1460 N÷2589 N). Therefore, the

specimens always slide until the restraint forces S becomes equal to the maximum friction

forces SM AX. Thus, the tension forces at the ends of the specimens are given by the value

of SM AX ≈0.2 · F . 500 1000 1500 2000 2500 3000 5000 10 000 15 000 20 000 F (N) S (N ) SMAX S

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By introducing S = SM AX into the solution of the deflections w1(x1, S), w2(x2, S), the

maximum deflection in the mid-span w2(a/2) of the overall beam can be obtained. Figure

11.17 shows the comparison between the elastic deflection with the restraint forces and the deflection without them. Moreover, Figure 11.17 shows the ratio between the two values of deflection. The maximum discrepancy is about 0.96 at the highest value of the applied force (2589 N). Figure 11.17 also shows the ratio of the deflections versus the friction coefficient for the maximum applied force.

500 1000 1500 2000 2500 3000 10 20 30 40 0 0 50

Deflection in first order analysis Deflection in second order analysis

F (N) w II / w I 500 1000 1500 2000 2500 3000 0 0.6 0.8 1.0 1.2 F (N) w (m m ) Friction coefficient w II / w I 0 0.2 0.4 0.6 0,5 0.8 1.0 0 1 1,5 2

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D) Experimental data

D.1) Pure bending state

TEMPERATURE STRESS 26.08 °C 170.1 MPa Moisture t0 n m Elastic deflection 22.88% 1 min 0.172 0.0118 1.43 mm

26% - 26°C

26% - 32°C

26% - 41°C

35% - 26°C

35% - 32°C

35% - 41°C

45% - 26°C

45% - 32°C

45% - 41°C

TEMPERATURE STRESS 26.08 °C 170.1 MPa Moisture t0 n m Elastic deflection 22.88% 1 min 0.123 0.019 1.72 mm LONG-TERM (26% - 26°C)

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26% - 26°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045

Experimental data (Sp.1) Findley law (Sp.1) Experimental data (Sp.2) Findley law (Sp.2) Findley law (A

verage) 0 0 E la st ic de fl ec ti on: 1.43 mm Figure 11.19: 26% -26 °C (Pure bending state)

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26% - 41°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.1 0.02 0.03 0.04 0.05 0.06 0.07

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35% - 26°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.01 0.02 0.03 0.06 0.05 0.04

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verage) 0 0.09 0.08 0.07 0 E la st ic de fl ec ti on: 3.94 mm Figure 11.24: 35% -41 °C (Pure bending state)

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45% - 26°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.02 0.06 0.04

verage) 0 0.08 0 0.1 0.12 E la st ic de fl ec ti on: 4.40 mm Figure 11.25: 45% -26 °C (Pure bending state)

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verage) 0 0.08 0 0.1 0.12 0.14 0.16 0.18 E la st ic de fl ec ti on: 4.60 mm Figure 11.26: 45% -32 °C (Pure bending state)

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T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.1 0.05

Experimental data (Sp.1) Findley law (Sp.1) Experimental data (Sp.2) Findley law (Sp.2)

0 0.15 0 0.2 0.25 Findley law (A verage) 45% - 41°C E la st ic de fl ec ti on: 4.95 mm Figure 11.27: 45% -41 °C (Pure bending state)

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26% - 26°C Creep de flec tion (m m) 0.01 0.02 0.03 0.06 0.05 0.04

verage) 0 E la st ic de fl ec ti on: 1.72 mm T im e (da ys ) 10 20 30 40 0 0.08 0.07 Figure 11.28: Long-term exp er imen t in pure bending state (26% -26 °C)

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D.2) Shear-bending state

26% - 26°C

26% - 32°C

26% - 41°C

35% - 26°C

35% - 32°C

35% - 41°C

45% - 26°C

45% - 32°C

45% - 41°C

TEMPERATURE STRESS 26.08 °C 170.1 MPa Moisture t0 n m Elastic deflection 22.88% 1 min 0.128 0.185 23.50 mm LONG-TERM (26% - 26°C)

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verage) 0 0 E la st ic de fl ec ti on: 23.32 mm 0.45 Figure 11.30: 26% -26 °C (Shear-b ending state)

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26% - 32°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

verage) 0 0 E la st ic de fl ec ti on: 24.1 1 m m 0.45 Figure 11.31: 26% -32 °C (Shear-b ending state)

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verage) 0 0 E la st ic de fl ec ti on: 34.24 mm 0.7 0.8 0.9 1 Figure 11.35: 35% -41 °C (Shear-b ending state)

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verage) 0 0 E la st ic de fl ec ti on: 42.50 mm 0.8 1 1.2 1.4 Figure 11.36: 45% -26 °C (Shear-b ending state)

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verage) 0 0 E la st ic de fl ec ti on: 42.96 mm 0.8 1 1.2 1.4 1.6 Figure 11.37: 45% -32 °C (Shear-b ending state)

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45% - 41°C T im e (m in) Creep de flec tion (m m) 100 200 300 400 500 600 700 800 900 1000 1 100 1200 1300 1400 0.5 1

verage) 0 0 E la st ic de fl ec ti on: 43.92 mm 1.5 2 2.5 Figure 11.38: 45% -41 °C (Shear-b ending state)

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26% - 26°C T im e (da ys ) Creep de flec tion (m m) 10 20 30 40 0,05 0,15 0,25 0,35 0,45 0,65 0,85

verage) 0 0 E la st ic de fl ec ti on: 23,50 mm 0,75 0,55 Figure 11.39: Long-term exp er imen t in shear-b ending state (26% -26 °C)(26% -26 °C)

Annexes of Experimental Activities