The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :
![$J=\frac{\pi}{32}d^4$](https://tex.z-dn.net/?f=%24J%3D%5Cfrac%7B%5Cpi%7D%7B32%7Dd%5E4%24)
![$J=\frac{\pi}{32}\times (46)^4$](https://tex.z-dn.net/?f=%24J%3D%5Cfrac%7B%5Cpi%7D%7B32%7D%5Ctimes%20%2846%29%5E4%24)
J = 207.6 ![mm^4](https://tex.z-dn.net/?f=mm%5E4)
So the shear stress at point A is :
![$\tau_A =\frac{Tc_A}{J}$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%5Cfrac%7BTc_A%7D%7BJ%7D%24)
![$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%5Cfrac%7B85%20%5Ctimes%2010%5E3%5Ctimes%2012%20%7D%7B207.6%7D%24)
![$\tau_A = 4913.29 \ MPa$](https://tex.z-dn.net/?f=%24%5Ctau_A%20%3D%204913.29%20%5C%20MPa%24)
Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
![d_1=50 mm,f_1=0.0048](https://tex.z-dn.net/?f=d_1%3D50%20mm%2Cf_1%3D0.0048)
For pipe 2
![d_2=75 mm,f_2=0.0058](https://tex.z-dn.net/?f=d_2%3D75%20mm%2Cf_2%3D0.0058)
Q=2.8 l/s
![Q=2.8\times 10^{-3]](https://tex.z-dn.net/?f=Q%3D2.8%5Ctimes%2010%5E%7B-3%5D)
We know that Q=AV
![Q=A_1V_1=A_2V_2](https://tex.z-dn.net/?f=Q%3DA_1V_1%3DA_2V_2)
![A_1=1.95\times 10^{-3}m^2](https://tex.z-dn.net/?f=A_1%3D1.95%5Ctimes%2010%5E%7B-3%7Dm%5E2)
![A_2=4.38\times 10^{-3} m^2](https://tex.z-dn.net/?f=A_2%3D4.38%5Ctimes%2010%5E%7B-3%7D%20m%5E2)
![So V_2=0.63 m/s,V_1=1.43 m/s](https://tex.z-dn.net/?f=So%20V_2%3D0.63%20m%2Fs%2CV_1%3D1.43%20m%2Fs)
head loss (h)
![h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7Bf_1L_1V_1%5E2%7D%7B2gd_1%7D%2B%5Cdfrac%7Bf_2L_2V_2%5E2%7D%7B2gd_2%7D%2B0.5%5Cdfrac%7BV_1%5E2%7D%7B2g%7D)
Now putting the all values
![h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B0.0048%5Ctimes%2015%5Ctimes%201.43%5E2%7D%7B2%5Ctimes%209.81%5Ctimes%200.05%7D%2B%5Cdfrac%7B0.0058%5Ctimes%2024%5Ctimes%200.63%5E2%7D%7B2%5Ctimes%209.81%5Ctimes%200.075%7D%2B0.5%5Cdfrac%7B1.43%5E2%7D%7B2%5Ctimes%209.81%7D)
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.
Answer:
15,000 psi
Explanation:
The solution / solving is attach below.
Explanation:
Conduction:
Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally, phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.
Convection:
In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.