Calibration values

Weights(KG) Force (N) Measurement A(mV) Measurement B(mV)

0 0 0.05 0

1.14 11.1834 2.9 2.4

1.42 13.9302 3.8 3.2

1.98 19.4238 5.3 4.7

5.02 49.2462 13.8 13.1

ф of the beam 5 mm

Length of the beam 46.6 mm

Gage A Gage B3.26 2.46 175 94 199 1420

A1A2

A3

to find the forces at A & B A(mm) F (N)FA 94 13.9302

FB 344 19.4238

beam DataArea 1.96428571428571E-05 m^2inertai 4.91071428571429E-10 m^4d 0.005 mE 200 Gpa 30000000 lbf/in² (psi)

B1 (mm) B2 (mm) B3 (mm) m 1 (g)

σb

B1 B2

A1

A2

A B

Ra

a

B1 0.175B2 0.095B3 0.199F1 13.9302F2 19.4238

Finding the reactions

ΣMa=0

ΣMb=0Ra = 17.0832939914163 NRb = 16.4854313304721 NC1 0.4M1 0.4 VC2 0.475M2 0.475Vmax.moment 3.898 N.m

Rb= (F1A1+F2(A1+A2))/L

Ra=(F1(A2+A3)+F2*A3)/L

Calibration values

Measurement B(mV)

0

2.4

3.2

4.7

13.1

1980 4.2 3.3m B (g) RA (mV) RB (mV)

B3

A3

Rb

b

B1 B2

A1

A2

A

Ra

a

B1 B2

A1

A2

A

Ra

a

B2B3

A3B

Rb

b

B2B3

A3B

Rb

b

calibrationmass(g) volts A(mV) volts B(mV)

0 0 0.051140 2.9 2.41420 3.8 3.21980 5.3 4.75020 13.8 13.1

Measurement for Am=c1*vC1= 361.9047619 RA= 12.070971 RA=(C1*V*9.81)/1000Measurements for B RB= 11.92896 RB=m=c2*vC2= 357.6470588

φ of the beam 0.006 m

Beam length 0.466 m

Gage B Gage A B1 mm B2 B3mm M1 (g) M2 (g) Ra (mV)

3.1 2.33 19 9.8 17.8 1150 1420 3.4

0.19 0.098 0.178 11.2815 13.9302

W1= 11.2815 N

W2= 13.9302 N

Ra= 12.29323519 N

Rb= 9.920730901 N

E 2E+11 Pa

I 6.36429E-11

E*I 12.72857143

First Moment equation ( 0<x<0.288)M=RaX-W1(X-0.190)

EI(dY/dX)=((Ra-W1)/2)X^2+(0.190*W1*X)+C1EIY=((Ra-W1)/6)X^3+((0.190*W1)/2)X^2)+C1X+C2

Using initial conditions

X=0 Y=0 C2=0

dY/dX=0 X=0 C1=0EIY=((Ra-W1)/6)X^3+((0.190*W1)/2)X^2)

Now To Calculate The Deflection at the first Mass

m4

EI (d2Y/dX2)=(Ra-W1)X+0.190W1

X=0 0.19EIY 0.039846486Y 0.003130476Now To Calculate The Deflection at the first GageX 0.094EIY 0.009609972Y 0.000754992

Second Moment Equation ( 0.288<X<0.466)M=RaX-(W1(X-0.190))-(W2(X-0.288))

EI(dY/dX)=((Ra-W1-W2)/2)X^2+0.190W1X+0.288W2X+C1EIY=((Ra-W1-W2)/6)X^3+((0.190W1)/2)X^2+((0.288W2)/2)X^2+C1X+C2

using the initial conditionsX=L Y=0X=L dY/dX=0C1 -1.46574722C2 0.519956611

Now to calculate the deflection at the second massX 0.288EIY 0.301665Y 0.023699832 0.047399663

Now for deflection at the second gageX 0.344EIY 0.338451099Y 0.026589873

To find the maximum bending momentX 0 M 0 0X 0.19 M 2.335714687 2335.7147X 0.288 M 2.434864736 2434.8647X 0.466 M 0.135378 135.378

To find the bending stressC 0.003

M 0 σ 0

M 2.335715 σ 110101029

M 2.434865 σ 114774769

M 0.135378 σ 6381454.545

X 0 σ 0

X 0.19 σ 110101029

X 0.288 σ 114774769

EI (d2Y/dX2)=(Ra-W1-W2)X+0.190W1+0.288W2

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.5

1

1.5

2

2.5

3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

20000000

40000000

60000000

80000000

100000000

120000000

140000000

X 0.466 σ 6381454.545

B1 B2B3

A1

A2

A3A B

Ra Rb

a b

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

20000000

40000000

60000000

80000000

100000000

120000000

140000000

Rb (mv)

3.4

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.5

1

1.5

2

2.5

3

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

20000000

40000000

60000000

80000000

100000000

120000000

140000000

the deflection of the beamX 0 Y 0 0X 0.094 Y 0.00151 -0.00151X 0.19 Y 0.00313 -0.00313X 0.288 Y 0.02659 -0.02659X 0.344 Y 0.0237 -0.0237X 0.466 Y 0.022577 -0.02258

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

20000000

40000000

60000000

80000000

100000000

120000000

140000000