- #1
Double A
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Homework Statement
See attachment.
Homework Equations
[tex]A = \pi r^2[/tex]
[tex]I_x = I_z = \frac{\pi}{4}r^4[/tex]
[tex]J = \frac{\pi}{2}r^4 = 2I_x[/tex]
[tex]Q = \Sigma A_i y_i[/tex]
[tex]\sigma_{Axial} = \frac{P}{A}[/tex]
[tex]\sigma_{Bending} = \frac{Mc}{I}[/tex]
[tex]\tau_{Shear} = \frac{VQ}{Tt}[/tex]
[tex]\tau_{Torque} = \frac{Tr}{J}[/tex]
The Attempt at a Solution
I moved the applied forces to the location at point A with the corresponding moments and torques caused by the applied loads giving me the following values:
[tex]A = \pi(0.75in)^2 = 1.77 in^2[/tex]
[tex]I_x = I_z = \frac{\pi}{4}(0.75 in)^4 = 0.249 in^4[/tex]
[tex]J = 2(0.249 in^4) = 0.497 in^4[/tex]
[tex]M_x = (800lbs)(10in) = 8 kips-in[/tex]
[tex]M_y = T = (800lbs)(14in) = 11.2 kips-in[/tex]
[tex]M_z = (500lbs)(14in) = 7 kips-in[/tex]
[tex]\sigma_{Axial} = \frac{500 lbs}{1.77 in^2} = 282.9 psi[/tex]
[tex]\sigma_{Bending} = \frac{(7 kips-in)(0.75 in)}{0.249 in^4} = 21.1 ksi[/tex]
[tex]\tau_{Torque} = \frac{(11.2 kips-in)(0.75 in)}{0.497 in^4} = 16.9 ksi[/tex]
Now I am not sure if the shear force has an affect on point A. If it does then how would you go about finding Q? I know that point A is on the x-axis and at the left edge of the rod. At this point I think the shear force does not cause a shear stress. Also the moment produced around the x-axis does not cause a normal stress because point A is on the centroid for the x-axis.
Now assuming what I have stated above is true then the only stresses acting at A are normal stresses in the y-direction and shear stresses in the y-z plane.
So then,
[tex]\sigma_y = \sigma_{Axial} + \sigma_{Bending} = 282.9 psi + 21.1 ksi = 21.4 ksi[/tex]
[tex]\tau_{yz} = \tau_{Torque} = 16.9 ksi[/tex]
So I guess what I want to know is does the shear force cause a shear stress and am I also accounting for all the elements that cause a normal stress or shear stress at point A?
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