Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
= ...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
= ...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432
Answer:
The distance between the station A and B will be:
Explanation:
Let's find the distance that the train traveled during 60 seconds.
We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:
Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.
Then the second distance will be:
The final distance is calculated whit the decelerate value:
The final velocity is zero because it rests at station B. The initial velocity will be v(1).
Therefore, the distance between the station A and B will be:
I hope it helps you!
Answer:
momentum
Explanation:
Mass - Mass is a measurement of how much matter is in an object. It is usually measured in kilograms. Momentum is equal to the mass times the velocity of an object. Momentum is a measurement of mass in motion
Answer:
Heat losses by convection, Qconv = 90W
Heat losses by radiation, Qrad = 5.814W
Explanation:
Heat transfer is defined as the transfer of heat from the heat surface to the object that needs to be heated. There are three types which are:
1. Radiation
2. Conduction
3. Convection
Convection is defined as the transfer of heat through the actual movement of the molecules.
Qconv = hA(Temp.final - Temp.surr)
Where h = 6.4KW/m2K
A, area of a square = L2
= (0.25)2
= 0.0625m2
Temp.final = 250°C
Temp.surr = 25°C
Q = 64 * 0.0625 * (250 - 25)
= 90W
Radiation is a heat transfer method that does not rely upon the contact between the initial heat source and the object to be heated, it can be called thermal radiation.
Qrad = E*S*(Temp.final4 - Temp.surr4)
Where E = emissivity of the surface
S = boltzmann constant
= 5.6703 x 10-8 W/m2K4
Qrad = 5.6703 x 10-8 * 0.42 * 0.0625 * ((250)4 - (25)4)
= 5.814 W