University of Florida/Egm3520/s13.team5.r4

Given

${\displaystyle \displaystyle T_{f}=1200\,in\cdot lb}$

(4.1-1)

${\displaystyle \displaystyle r_{g}=3\,in}$

(4.1-2)

${\displaystyle \displaystyle r_{d}=8\,in}$

(4.1-3)

${\displaystyle \displaystyle \tau _{allowed}=10.5\,ksi=10,500\,{\frac {lb}{in^{2}}}}$

(4.1-4)

Step One: Draw Free Body Diagrams

FBD of shafts

FBD of gears

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

${\displaystyle \displaystyle \sum F=0=F_{D}+F_{G}={\frac {T_{DE}}{R_{D}}}+{\frac {T_{FH}}{R_{G}}}}$

(4.1-5)

Isolating the torque in CE based on the applied torque,

${\displaystyle \displaystyle T_{CE}={\frac {-T_{FH}\cdot r_{D}}{r_{G}}}}$

(4.1-6)

${\displaystyle \displaystyle \tau _{max}={\frac {2T_{E}}{\pi \times r_{CE}^{3}}}}$

(4.1-7)

Manipulating the stress formula the diameter can be determined,

${\displaystyle \displaystyle d_{CE}=2({\frac {2T_{E}}{\pi \tau _{max}}})^{1/3}}$

(4.1-8)

Substituting the values given above, the diameter can be calculated

${\displaystyle \displaystyle d_{CF}=1.158in}$

Step Three: Part "B" Analysis

${\displaystyle \displaystyle \tau _{max}={\frac {T_{H}r_{HF}}{J}}={\frac {2T_{H}}{\pi r_{HF}^{3}}}}$

(4.1-9)

Solving for the radius,

${\displaystyle \displaystyle r_{HF}=({\frac {2T_{H}}{\pi \tau _{max}}})^{1/3}}$

(4.1-10)

To get the diameter, multiply the equation of the radius by 2

${\displaystyle \displaystyle d_{HF}=2({\frac {2T_{H}}{\pi \tau _{max}}})^{1/3}}$

(4.1-11)

Solving the Equation 4.1-11 with the values given, the diameter can be calculated

${\displaystyle \displaystyle d_{HF}=0.836in}$

Given

The magnitude of the torque at C, ${\displaystyle T_{C}=5\,kip\cdot in}$
Allowable shearing stress in the shafts, ${\displaystyle \tau _{max}=8500\,psi}$
Radius of the gear A, ${\displaystyle r_{A}=4\,in}$
Radius of the gear D, ${\displaystyle r_{D}=2.5\,in}$

Step One: Draw Free Body Diagrams

Bars ABC and DEF

Gears A and D

Step Two: Analysis

From the relation between the torques and the radius of the gears,

${\displaystyle \displaystyle {\frac {T_{1}}{T_{2}}}={\frac {r_{1}}{r_{2}}}}$

(4.2-1)

Therefore,

${\displaystyle \displaystyle {\frac {T_{F}}{T_{C}}}={\frac {r_{D}}{r_{A}}}}$

(4.2-2)

Now inserting the given values,

${\displaystyle \displaystyle {\frac {T_{F}}{5}}={\frac {2.5}{4}}}$

(4.2-3)

${\displaystyle \displaystyle T_{F}=3.125\,kip\cdot in}$

(4.2-4)

Step Three: Application

Part A

Allowable shearing stress in the shaft BC,

${\displaystyle \displaystyle \tau _{max}={\frac {T_{C}*C}{J}}}$

(4.2-5)

Where J for a solid circular shaft is

${\displaystyle \displaystyle J={\frac {\pi }{2}}C^{4}}$

(4.2-6)

Insert the values and solve for the radius C.

${\displaystyle \displaystyle 8500={\frac {(5*10^{3})C}{{\frac {\pi }{2}}C^{4}}}}$

(4.2-7)

${\displaystyle \displaystyle C^{3}=0.3748}$

(4.2-8)

${\displaystyle \displaystyle C=0.721\,in}$

(4.2-9)

Diameter of the shaft BC,

${\displaystyle \displaystyle d=2C}$

(4.2-10)

${\displaystyle \displaystyle d=2(0.721)}$

(4.2-11)

${\displaystyle \displaystyle d=1.442\,in}$

(4.2-12)

Part B

Allowable shearing stress in the shaft EF,

${\displaystyle \displaystyle \tau ={\frac {T_{F}*C}{J}}}$

(4.2-13)

Insert the values and solve for the radius C.

${\displaystyle \displaystyle 8500={\frac {(3.125*10^{3})C}{{\frac {\pi }{2}}C^{4}}}}$

(4.2-14)

${\displaystyle \displaystyle C^{3}=0.234}$

(4.2-15)

${\displaystyle \displaystyle C=0.616\,in}$

(4.2-16)

Diameter of the shaft EF,

${\displaystyle \displaystyle d=2C}$

(4.2-17)

${\displaystyle \displaystyle d=2(0.616)}$

(4.2-18)

${\displaystyle \displaystyle d=1.233\,in}$

(4.2-19)