Answer:
Flow energy is defined as, flow energy is the energy needed to push fluids into control volume and it is the amount of work done required to push the entire fluid. It is also known as flow work. Flow energy is not the fundamental quantities like potential and kinetic energy.
Fluid at state of rest do not possess any flow energy. It is mostly converted into internal energy as, rising in the fluid temperature.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Answer:
The shear strain is 0.05797 rad.
Explanation:
Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.
Step1
Given:
Height of the pad is 1.38 in.
Deformation at the top of the pad is 0.08 in.
Calculation:
Step2
Shear strain is calculated as follows:
![tan\phi=\frac{\bigtriangleup l}{h}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7B%5Cbigtriangleup%20l%7D%7Bh%7D)
![tan\phi=\frac{0.08}{1.38}](https://tex.z-dn.net/?f=tan%5Cphi%3D%5Cfrac%7B0.08%7D%7B1.38%7D)
![tan\phi= 0.05797](https://tex.z-dn.net/?f=tan%5Cphi%3D%200.05797)
For small angle of
,
can take as
.
![\phi = 0.05797 rad.](https://tex.z-dn.net/?f=%5Cphi%20%3D%200.05797%20rad.)
Thus, the shear strain is 0.05797 rad.
Answer:
The value of Modulus of elasticity E = 85.33 ×
![\frac{lbm}{in^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Blbm%7D%7Bin%5E%7B2%7D%20%7D)
Beam deflection is = 0.15 in
Explanation:
Given data
width = 5 in
Length = 60 in
Mass of the person = 125 lb
Load = 125 × 32 = 4000![\frac{ft lbm}{s^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Bft%20lbm%7D%7Bs%5E%7B2%7D%20%7D)
We know that moment of inertia is given as
![I = \frac{bt^{3} }{12}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7Bbt%5E%7B3%7D%20%7D%7B12%7D)
![I = \frac{5 (1.5^{3} )}{12}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B5%20%281.5%5E%7B3%7D%20%29%7D%7B12%7D)
I = 1.40625 ![in^{4}](https://tex.z-dn.net/?f=in%5E%7B4%7D)
Deflection = 0.15 in
We know that deflection of the beam in this case is given as
Δ = ![\frac{PL^{3} }{48EI}](https://tex.z-dn.net/?f=%5Cfrac%7BPL%5E%7B3%7D%20%7D%7B48EI%7D)
![0.15 = \frac{4000(60)^{3} }{48 E (1.40625)}](https://tex.z-dn.net/?f=0.15%20%3D%20%5Cfrac%7B4000%2860%29%5E%7B3%7D%20%7D%7B48%20E%20%281.40625%29%7D)
E = 85.33 ×
![\frac{lbm}{in^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7Blbm%7D%7Bin%5E%7B2%7D%20%7D)
This is the value of Modulus of elasticity.
Beam deflection is = 0.15 in