Answer:
R min = 28.173 ohm
R max = 1.55 × ohm
Explanation:
given data
capacitor = 0.227 μF
charged to 5.03 V
potential difference across the plates = 0.833 V
handled effectively = 11.5 μs to 6.33 ms
solution
we know that resistance range of the resistor is express as
V(t) = ...........1
so R will be
R = ....................2
put here value
so for t min 11.5 μs
R =
R min = 28.173 ohm
and
for t max 6.33 ms
R max =
R max = 1.55 × ohm
Answer:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.
Explanation:
The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"
Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:
Where:
- Safety factor, dimensionless.
- Ultimate shear strength, measured in pascals.
- Maximum allowable shear stress, measured in pascals.
The maximum allowable shear stress is consequently cleared and computed: (,
)
Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:
Where:
- Maximum allowable shear stress, measured in pascals.
- Shear force, measured in kilonewtons.
- Cross section area, measured in square meters.
As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:
The minimum allowable cross section area is cleared in the shearing stress equation:
If and
, the minimum allowable cross section area is:
The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:
The diameter is now cleared and computed:
The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.