Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑![F_G_X = W - G_x = 0](https://tex.z-dn.net/?f=F_G_X%20%3D%20W%20-%20G_x%20%3D%200)
![G_X = W = 1500 lb](https://tex.z-dn.net/?f=G_X%20%3D%20W%20%3D%201500%20lb)
∑![F_G_Y = P - G_Y = 0](https://tex.z-dn.net/?f=F_G_Y%20%3D%20P%20-%20G_Y%20%3D%200)
![G_Y = P = -10,800 lb](https://tex.z-dn.net/?f=G_Y%20%3D%20P%20%3D%20-10%2C800%20lb)
∑![F_G_Z = - G_Z = 0](https://tex.z-dn.net/?f=F_G_Z%20%3D%20%20-%20G_Z%20%3D%200)
Where, (
) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
![M_G = r_O_G x F_O](https://tex.z-dn.net/?f=M_G%20%3D%20r_O_G%20x%20F_O)
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
![r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\](https://tex.z-dn.net/?f=r_O_G%20%3D%20%5B%20-%28%203%20%2B%20%5Cfrac%7BH%7D%7B2%7D%20%29%20i%20%2B%20%28%5Cfrac%7Br_o%7D%7B12%7D%29j%20-%20%28%20%5Cfrac%7Br_o%7D%7B12%7D%20%2B%20%5Cfrac%7BL%7D%7B2%7D%29k%20%5D%20%5C%5C%5C%5Cr_O_G%20%3D%20%5B%20-%28%206%20%29%20i%20%2B%20%280.25%29j%20-%20%286%29k%20%5D%20%5C%5C)
![F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb](https://tex.z-dn.net/?f=F_O_G%20%3D%20%5B%20%28%20W%20%29%20i%20%2B%20%28%20P%20%29%20k%20%5D%5C%5C%5C%5CF_O_G%20%3D%20%5B%20%281500%29%20i%20-%20%28%2010%2C800%20%29%20k%20%5D%20lb)
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
![( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k](https://tex.z-dn.net/?f=%28%20M_G_X%20%29i%20%2B%20%28M_G_Y%29j%2B%20%28M_G_Z%29k%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C-6%260.25%26-6%5C%5C1500%26-10%2C800%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%28%20M_G_X%20%29i%20%2B%20%28M_G_Y%29j%2B%20%28M_G_Z%29k%20%3D%20-%28%206%2A10%2C800%20%29%20i%20-%20%28%206%2A1500%20%29%20j%20%2B%20%5B%20%28%2010%2C800%2A6%29%20-%20%28%200.25%2A1500%29%20%5D%20k%5C%5C%5C%5C%28%20M_G_X%20%29i%20%2B%20%28M_G_Y%29j%2B%20%28M_G_Z%29k%20%3D%20-%20%2864%2C800%29i%20-%20%289%2C000%29j%20%2B%20%2864%2C425%29k)
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
![T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft](https://tex.z-dn.net/?f=T%20%3D%20P%2A%5B%5Cfrac%7BL%7D%7B2%7D%5D%20%3D%20%2810%2C800%29%2A%286%29%20%3D%2064%2C800%20lb.ft)
![V = P = -10,800 lb](https://tex.z-dn.net/?f=V%20%3D%20P%20%3D%20-10%2C800%20lb)
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = ![-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}](https://tex.z-dn.net/?f=-%5Cfrac%7BG_x%7D%7BA%7D%20-%20%5Cfrac%7BM_G_Y.%20z%2A%7D%7BI_Y_Y%7D%20%2B%20%20%5Cfrac%7BM_G_Z.%20y%2A%7D%7BI_Z_Z%7D)
Where,
A: The area of pipe cross section
![A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20%2A%20%5B%20%28%20%5Cfrac%7Br_o%7D%7B12%7D%29%5E2%20-%20%20%28%20%5Cfrac%7Br_i%7D%7B12%7D%29%5E2%20%5D%20%3D%20%5Cpi%20%2A%20%5B%20%28%20%5Cfrac%7B3%7D%7B12%7D%29%5E2%20-%20%20%28%20%5Cfrac%7B2.75%7D%7B12%7D%29%5E2%20%5D%20%3D%200.03136%20ft%5E2)
z*: The distance of point "C" along z-direction from central axis ( x )
![z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft](https://tex.z-dn.net/?f=z%2A%3D%20%5B%5Cfrac%7Br_i%7D%7B12%7D%20%5D%20%3D%20%20%20%5B%5Cfrac%7B2.75%7D%7B12%7D%20%5D%20%3D%200.22916%20ft)
I_YY: The second area moment of pipe along and about "y" axis:
![I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4](https://tex.z-dn.net/?f=I_Y_Y%20%3D%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%20%2A%20%5B%20%28%5Cfrac%7Br_o%7D%7B12%7D%29%5E4%20-%20%28%5Cfrac%7Br_i%7D%7B12%7D%29%5E4%20%5D%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%20%2A%20%5B%20%28%5Cfrac%7B3%7D%7B12%7D%29%5E4%20-%20%28%5Cfrac%7B2.75%7D%7B12%7D%29%5E4%20%5D%20%5C%5C%5C%5CI_Y_Y%20%3D%200.00090%20ft%5E4)
y*: The distance of point "C" along y-direction from central axis ( x )
![y* = 0](https://tex.z-dn.net/?f=y%2A%20%3D%200)
- The normal stress ( σ_x ) becomes:
σ_x = ![[-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi](https://tex.z-dn.net/?f=%5B-%5Cfrac%7B1500%7D%7B0.03136%7D%20-%20%5Cfrac%7B-9%2C000%2A0.22916%7D%7B0.00090%7D%20%2B%20%5Cfrac%7B64%2C425%2A0%7D%7B0.00090%7D%20%5D%20%2A%20%28%5Cfrac%7B1%7D%7B12%7D%29%5E2%20%20%20%3D%2015.5%20ksi)
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = ![- \frac{T.c}{J} + \frac{V.Q}{I.t}](https://tex.z-dn.net/?f=-%20%5Cfrac%7BT.c%7D%7BJ%7D%20%2B%20%5Cfrac%7BV.Q%7D%7BI.t%7D)
Where,
c: The radial distance from central axis ( x ) and point "C".
![c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft](https://tex.z-dn.net/?f=c%20%3D%20%5Cfrac%7Br_i%7D%7B12%7D%20%3D%20%5Cfrac%7B2.75%7D%7B12%7D%20%3D%200.22916%20ft)
J: The polar moment of inertia of the annular cross section of pipe:
![J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%2A%20%5B%20%28%20%5Cfrac%7Br_o%7D%7B12%7D%29%5E4%20-%20%28%20%5Cfrac%7Br_i%7D%7B12%7D%29%5E4%20%5D%20%3D%20%20%5Cfrac%7B%5Cpi%20%7D%7B2%7D%2A%20%5B%20%28%20%5Cfrac%7B3%7D%7B12%7D%29%5E4%20-%20%28%20%5Cfrac%7B2.75%7D%7B12%7D%29%5E4%20%5D%20%3D%20%200.00180%20ft%5E4)
Q: The first moment of area for point "C" = semi-circle
![Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3](https://tex.z-dn.net/?f=Q%20%3D%20Y_c%2AA_c%20%3D%20%5Cfrac%7B4%2A%28%20r_m%29%7D%7B3%5Cpi%20%7D%20%2A%20%5Cfrac%7B%5Cpi%2A%28%20r_m%29%5E2%20%7D%7B2%7D%20%3D%20%5Cfrac%7B2.%20%28%20r_m%29%5E3%7D%7B3%7D%20%5C%5C%5C%5CQ%20%3D%20%5Cfrac%7B2.%20%5B%20%28%20%5Cfrac%7Br_o%7D%7B12%7D%29%5E3%20%20%20-%20%28%20%5Cfrac%7Br_i%7D%7B12%7D%29%5E3%5D%20%7D%7B3%7D%20%3D%20%5Cfrac%7B2.%20%5B%20%28%20%5Cfrac%7B3%7D%7B12%7D%29%5E3%20%20%20-%20%28%20%5Cfrac%7B2.75%7D%7B12%7D%29%5E3%5D%20%7D%7B3%7D%20%3D%200.00239ft%5E3)
I: The second area moment of pipe along and about "y" axis:
t: The effective thickness of thin walled pipe:
![t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft](https://tex.z-dn.net/?f=t%20%3D%202%2A%20%5B%20%5Cfrac%7Br_o%7D%7B12%7D%20-%20%5Cfrac%7Br_i%7D%7B12%7D%20%5D%20%3D%20%202%2A%20%5B%20%5Cfrac%7B3%7D%7B12%7D%20-%20%5Cfrac%7B2.75%7D%7B12%7D%20%5D%20%3D%200.04166%20ft)
- The planar stress is ( τ_xy ) becomes:
τ_xy = ![[ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi](https://tex.z-dn.net/?f=%5B%20-%20%5Cfrac%7B-64%2C800%2A0.22916%7D%7B0.0018%7D%20%2B%20%5Cfrac%7B-10%2C800%2A0.00239%7D%7B0.0009%2A0.04166%7D%20%5D%20%2A%20%5B%20%5Cfrac%7B1%7D%7B12%7D%5D%5E2%20%20%3D%2052.4%20ksi)
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =![\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }](https://tex.z-dn.net/?f=%5Cfrac%7Bsigma_x%2Bsigma_y%7D%7B2%7D%20%2B%20%5Csqrt%7B%28%5Cfrac%7Bsigma_x%2Bsigma_y%7D%7B2%7D%29%5E2%20%2B%20%28tow_x_y%29%5E2%20%7D)
σ_2 = ![\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }](https://tex.z-dn.net/?f=%5Cfrac%7Bsigma_x%2Bsigma_y%7D%7B2%7D%20-%20%5Csqrt%7B%28%5Cfrac%7Bsigma_x%2Bsigma_y%7D%7B2%7D%29%5E2%20%2B%20%28tow_x_y%29%5E2%20%7D)
σ_1 = ![\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi](https://tex.z-dn.net/?f=%5Cfrac%7B15.55%2B0%7D%7B2%7D%20%2B%20%5Csqrt%7B%28%5Cfrac%7B15.55%2B0%7D%7B2%7D%29%5E2%20%2B%20%2852.4%29%5E2%20%7D%20%3D%2060.75%20ksi)
σ_2 =![-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi](https://tex.z-dn.net/?f=-%5Csqrt%7B%5Cleft%28%5Cfrac%7B15.55%2B0%7D%7B2%7D%5Cright%29%5E2%5C%3A%2B%5C%3A%5Cleft%2852.4%5Cright%29%5E2%5C%3A%7D%2B%5Cfrac%7B15.55%2B0%7D%7B2%7D%20%3D%20-45.20%20ksi)
- The angle of maximum plane stress ( θ ):
θ = ![0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg](https://tex.z-dn.net/?f=0.5%2Aarctan%20%28%20%5Cfrac%7Btow_x_y%7D%7B%5Cfrac%7Bsigma_x-sigma_y%7D%7B2%7D%20%7D%20%29%3D%200.5%2Aarctan%2A%28%20%5Cfrac%7B52.4%7D%7B7.8%7D%20%29%20%3D%2040.8%20deg)
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.