An operating gear box (transmission) has 350 hp at its input shaft while 250. hp are delivered to the output shaft. The gear box
1 answer:
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Answer:
γ =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
Part a: The yield moment is 400 k.in.
Part b: The strain is
Part c: The plastic moment is 600 ksi.
Explanation:
Part a:
As per bending equation
Here
- M is the moment which is to be calculated
- I is the moment of inertia given as
Here
- b is the breath given as 0.75"
- d is the depth which is given as 8"
- y is given as
- Force is 50 ksi
The yield moment is 400 k.in.
Part b:
The strain is given as
The stress at the station 2" down from the top is estimated by ratio of triangles as
Now the steel has the elastic modulus of E=29000 ksi
So the strain is
Part c:
For a rectangular shape the shape factor is given as 1.5.
Now the plastic moment is given as
The plastic moment is 600 ksi.