A solid steel shaft has to transmit 100 kW at 160 RPM. Taking allowable shear stress at 70 Mpa, find the suitable diameter of th
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Answer:
Total bandwidth: 8 kHz
Explanation:
Data provided:
Transmitter frequency: 3.9 MHz
Modulation up to: 4 kHz
Solution:
For the upper side frequencies:
Upper side frequencies = 3.9 × + 4 × 10³
Upper side frequencies = 3.904 MHz
For the lower side frequencies:
Lower side frequencies = 3.9 × - 4 × 10³
Lower side frequencies = 3.896 MHz
Consequently, the total bandwidth is computed as:
Total bandwidth = upper side frequencies - lower side frequencies
Total bandwidth = 8 kHz
Answer:
Change in length = 0.0913 in
Explanation:
Given data:
Length = 6 ft
Diameter = 0.2 in
Load w = 200 lb/ft
Solution:
We start by applying the equilibrium moment about point C, expressed as
∑M(c) = 0.............1
This can be used to find the force in AB.
10× 200 × ( 5) - (T cos(30)) × 10 = 0
Solving gives us
Tension in wire T(AB) = 1154.7 lb
We also know the modulus of elasticity for A992 is
E = 29000 ksi
And the area will be
Area =
The change in length is expressed as
Change in length = .........2
Substituting values results in
Change in length =
Change in length = 0.0913 in