Respuesta :
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
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
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:
[tex]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 ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( 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[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- 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 = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]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[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]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[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]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[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]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[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]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[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- 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 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
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.
