University of Florida/Egm3520/s13.team5.r4

Problem 4.1 (Problem 3.23 in Beer, 2012)
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
Under normal operating conditions a motor exerts a torque of magnitude $$T_F= 1200lbs*in $$ at $$ F$$. Knowing that $$r_D= 8 in, r_C= 3 in $$ and the maximum allowable shearing stress is 10.5 ksi.

Determine the required diameter of member FH.

Given

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$$ \displaystyle T_f=1200\,in\cdot lb $$ (4.1-1)
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$$ \displaystyle r_g=3\, in $$ (4.1-2)
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$$ \displaystyle r_d=8\, in $$ (4.1-3)
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$$ \displaystyle \tau _{allowed}= 10.5\, ksi=10,500 \, \frac{lb}{in^2} $$ (4.1-4)
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Step One: Draw Free Body Diagrams
For part A, by assuming constant velocity for the point of gear contact:

Step Two: Part "A" Analysis
The sum of the forces from the diagram equals zero


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$$ \displaystyle \sum F=0=F_D+F_G=\frac{T_{DE}}{R_D}+\frac{T_{FH}}{R_G} $$ (4.1-5)
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Isolating the torque in CE based on the applied torque,


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$$ \displaystyle T_{CE}=\frac{-T_{FH}\cdot r_D}{r_G} $$ (4.1-6)
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$$\displaystyle \tau _{max}= \frac{2T_E}{\pi \times r_{CE}^3} $$ (4.1-7)
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Manipulating the stress formula the diameter can be determined,


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$$\displaystyle d_{CE}=2(\frac{2T_E}{\pi \tau _{max}})^{1/3} $$ (4.1-8)
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Substituting the values given above, the diameter can be calculated

Step Three: Part "B" Analysis

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$$ \displaystyle \tau_{max}=\frac{T_Hr_{HF}}{J}=\frac{2T_H}{\pi r_{HF}^{3}} $$ (4.1-9)
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Solving for the radius,


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$$ \displaystyle r_{HF}=(\frac{2T_H}{\pi\tau_{max} })^{1/3} $$     (4.1-10)
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To get the diameter, multiply the equation of the radius by 2


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$$ \displaystyle d_{HF}=2(\frac{2T_H}{\pi\tau_{max}})^{1/3} $$     (4.1-11)
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Solving the Equation 4.1-11 with the values given, the diameter can be calculated

Problem 4.2 (Problem 3.25 in Beer, 2012)
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
In the image below, there are two steel shafts, ABC and DEF, for which the maximum allowable shear stress is 8500 psi. They are connected by gears at A and D of given radii 4 in. and 2.5 in., respectively. There is a known applied torque at C, TC of 5 kips•in. and an unknown torque, TF, applied at F.

a) Determine the required diameter of shaft BC b) Determine the required diameter of shaft EF

Given
The magnitude of the torque at C, $$T_C=5\, kip\cdot in$$ Allowable shearing stress in the shafts, $$\tau_{max}=8500\, psi$$ Radius of the gear A, $$r_A=4\, in$$ Radius of the gear D, $$r_D=2.5\, in$$

Step Two: Analysis
From the relation between the torques and the radius of the gears,
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$$ \displaystyle \frac{T_1}{T_2}=\frac{r_1}{r_2} $$ (4.2-1)
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Therefore,
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$$ \displaystyle \frac{T_F}{T_C}=\frac{r_D}{r_A} $$ (4.2-2)
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Now inserting the given values,
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$$ \displaystyle \frac{T_F}{5}=\frac{2.5}{4} $$ (4.2-3)
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$$ \displaystyle T_F=3.125\, kip\cdot in $$ (4.2-4)
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Part A
Allowable shearing stress in the shaft BC,
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$$ \displaystyle \tau_{max}=\frac{T_C*C}{J} $$ (4.2-5)
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Where J for a solid circular shaft is
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$$ \displaystyle J=\frac{\pi}{2} C^4 $$ (4.2-6)
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Insert the values and solve for the radius C.
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$$ \displaystyle 8500=\frac{(5*10^3)C}{\frac{\pi}{2}C^4} $$ (4.2-7)
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$$ \displaystyle C^3=0.3748 $$ (4.2-8)
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$$ \displaystyle C=0.721\, in $$ (4.2-9)
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Diameter of the shaft BC,
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$$ \displaystyle d=2C $$ (4.2-10)
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$$ \displaystyle d=2(0.721) $$ (4.2-11)
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Part B
Allowable shearing stress in the shaft EF,
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$$ \displaystyle \tau=\frac{T_F*C}{J} $$ (4.2-13) Insert the values and solve for the radius C.
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$$ \displaystyle 8500=\frac{(3.125*10^3)C}{\frac{\pi}{2}C^4} $$ (4.2-14)
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$$ \displaystyle C^3=0.234 $$ (4.2-15)
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$$ \displaystyle C=0.616\, in $$ (4.2-16)
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Diameter of the shaft EF,
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$$ \displaystyle d=2C $$ (4.2-17)
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$$ \displaystyle d=2(0.616) $$ (4.2-18)
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Contributors
Team Designee: Daniel Siefman