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