Engineering, 11.06.2020 03:57 jeff7259

The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe. The uniform sign has a weight of 1500 lb and is supported bythe pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in.. If the face of the sign issubjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at pointsC and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.

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Answer from: juliannabartra

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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

G_X = W = 1500 lb

∑ F_G_Y = P - G_Y = 0

G_Y = P = -10,800 lb

∑ F_G_Z = - G_Z = 0

Where, ( G_X, G_Y, G_Z ) 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

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:

$r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\$

$F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb$

- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):

$( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k$

- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:

$T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft$

V = P = -10,800 lb

- 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 = $-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}$

Where,

A: The area of pipe cross section

$A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2$

z*: The distance of point "C" along z-direction from central axis ( x )

$z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft$

I_YY: The second area moment of pipe along and about "y" axis:

$I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4$

y*: The distance of point "C" along y-direction from central axis ( x )

y* = 0

- The normal stress ( σ_x ) becomes:

σ_x = $[-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi$

- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):

τ_xy = $- \frac{T.c}{J} + \frac{V.Q}{I.t}$

Where,

c: The radial distance from central axis ( x ) and point "C".

$c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft$

J: The polar moment of inertia of the annular cross section of pipe:

$J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4$

Q: The first moment of area for point "C" = semi-circle

$Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3$

I: The second area moment of pipe along and about "y" axis:

$I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4$

t: The effective thickness of thin walled pipe:

$t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft$

- The planar stress is ( τ_xy ) becomes:

τ_xy = $[ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi$

- 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 = $\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }$

σ_2 = $\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }$

σ_1 = $\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi$

σ_2 = $-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi$

- The angle of maximum plane stress ( θ ):

θ = $0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg$

Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at , i'm unable to post here.

Answer from: Quest

answer:

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Answer from: Quest

Yes, yes they do sir

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