on the same scale for stress, the tensile true stress-true strain curve is higher than the engineeringstress-engineering strain
1 answer:
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Answer:
I. Tension (cable A) ≈ 6939 lbf
II. Tension (cable B) ≈ 17199 lbf
Explanation:
Let's begin by listing out the data that we were given:
mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,
g = 32.17405 ft/s²
The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.
Mathematically represented thus:
T = mg + ma
where:
T = tension, m = mass, g = gravitational force,
a = acceleration
I. For Cable A, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-20)]
T = 18339.2085 - 11400 = 6939.2085
T ≈ 6939 lbf
II. For Cable B, we have:
T = mg + ma = (570 * 32.17405) + [570 * (-2)]
T = 18339.2085 - 1140 = 17199.2085
T ≈ 17199 lbf
Explanation:
Temperature range → 0 to 80'c
respective voltage output → 0.2v to 0.5v
required temperature range 20'c to 40'c
Where T = 20'c respective voltage
Therefore, Sensor output changes from 0.275v to 0.35volts for the ADC the required i/p should cover the dynamic range of ADC (ie - 0v to 3v)
so we have to design a circuit which transfers input voltage 0.275volts - 0.35v to 0 - 3v
Therefore, the formula for the circuit will be
The simplest circuit will be a op-amp
NOTE: Refer the figure attached
Vs is sensor output
Vr is the reference volt, Vr = 0.275v
choose R2, R1 such that it will maintain required ratio
The output Vo can be connected to voltage buffer if you required better isolation.
