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:
Explanation:
From the information given:
Life requirement = 40 kh = 40 
Speed (N) = 520 rev/min
Reliability goal 
= 0.9
Radial load 
= 2600 lbf
To find C10 value by using the formula:
where;
The Weibull parameters include:
∴
Using the above formula:
![C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}](https://tex.z-dn.net/?f=C_%7B10%7D%20%3D%203640%20%5Ctimes%20%5Cbigg%5B%5Cdfrac%7B1248%7D%7B0.9933481582%7D%5Cbigg%5D%5E%7B%5Cdfrac%7B3%7D%7B10%7D%7D)
Recall that:
1 kN = 225 lbf
∴
Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
\
We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v = 
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s
Answer:
In engineering and materials science, a stress–strain curve for a material gives the relationship between stress and strain. It is obtained by gradually applying load to a test coupon and measuring the deformation, from which the stress and strain can be determined (see tensile testing).
Explanation:
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