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fahraynk
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Homework Statement
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The shaft connecting A to D is rotating at 1200RPM clockwise and has a concentrated unbalanced load of W2 of 8lb at a radius of 2in. in a plane midway between points A and D as shown in the figure. A constant vertical load W1 of 1500 lb is applied by means of two anti-friction bearings B and C. If the shaft material has an ultimate strength of 60,000 PSI and yield strength 45,000 PSI, determine the required diameter of the shaft at the mid section for a design factor of safety N=2. Only consider the mid point.
Homework Equations
W1=1500 lb
W2=8 lb
The Attempt at a Solution
The book gives an answer, and part of the answer is this : "(1)Show that the bending moment due to the rotating load varies from -6540 in-lb to 6540 in-lb. (2) show that the bending moment due to the constant vertical load is 7500 in-lb. (3) The combined bending moment then varies from 960 in-lb to 14,040 in-lb."
I can't figure out how the book got (1) or (2). Here is my work :
To get forces at B and C due to load W1:
$$B+C=-1500\\\\14(1500)-20C=0\\\\B=-450\\\\C=-1050$$
To get forces at A and D:
$$A-450-8-1050+D=0\\\\450*10+8*20+1050*30-D*40=0\\\\A=604\\\\D=904$$
This gives me this moment diagram:
7580 total moment max at the 20" mark.
Okay, so the book says the bending moment due to the rotating load varies from -6540 to 6540 in-lb.
I don't understand where they got this from. Is this ONLY the rotating load? So, that would mean solve a force diagram where W1 is removed? That makes no sense because then you would have A=4 lb B = 4 lb holding up W1=8 lb. The bending moment would be 8*20 = 160
So, I try to take the rotation into account.
It is rotating at w=1200RPM $$w=[1200[RPM]*\frac{2\pi}{60}[\frac{Rads}{RPM * Sec}] = 125.7[\frac{Rads}{Sec}]$$
This would give the mass a velocity of RwCos(wt) = 2*125.7= 251.4
Also an acceleration dv/dt = -Rw^2sin(wt), with a min and max of +-Rw^2.
But... ##R*w^2=2*125.7^2=31601 >> 6540## ... So this has to be wrong. Not even close to 6540.
Here is another failed attempt :
If I look at that section of the beam as a 2D section, it would be a circle with a mass 2 inches from the center. Sum of torque = change in angular momentum :
$$T= dH/dt \\\\
H=I_{zz}w \hat{k} + [-R\hat{i} x (-8Rw\hat{j})]$$
mass of the shaft is 0, so I_zz vanishes. ##H=8R^2w\hat{k}##
dH/dt=0 because w is constant. Sum of Torques = 0. That would mean r cross F = 0, no forces in the tangential direction, nothing that would point down..
What am I doing wrong!? Please help