# 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 $r_{D}=8in,r_{C}=3in$ and the maximum allowable shearing stress is 10.5 ksi.
Determine the required diameter of member FH.

#### Given

    $\displaystyle T_{f}=1200\,in\cdot lb$ (4.1-1)
    $\displaystyle r_{g}=3\,in$ (4.1-2)
    $\displaystyle r_{d}=8\,in$ (4.1-3)
    $\displaystyle \tau _{allowed}=10.5\,ksi=10,500\,{\frac {lb}{in^{2}}}$ (4.1-4)

### Soultion
#### 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
    $\displaystyle \sum F=0=F_{D}+F_{G}={\frac {T_{DE}}{R_{D}}}+{\frac {T_{FH}}{R_{G}}}$ (4.1-5)
Isolating the torque in CE based on the applied torque,
    $\displaystyle T_{CE}={\frac {-T_{FH}\cdot r_{D}}{r_{G}}}$ (4.1-6)
    $\displaystyle \tau _{max}={\frac {2T_{E}}{\pi \times r_{CE}^{3}}}$ (4.1-7)
Manipulating the stress formula the diameter can be determined,
    $\displaystyle d_{CE}=2({\frac {2T_{E}}{\pi \tau _{max}}})^{1/3}$ (4.1-8)
Substituting the values given above, the diameter can be calculated
* $\displaystyle d_{CF}=1.158in$

#### Step Three: Part "B" Analysis

    $\displaystyle \tau _{max}={\frac {T_{H}r_{HF}}{J}}={\frac {2T_{H}}{\pi r_{HF}^{3}}}$ (4.1-9)
Solving for the radius,
    $\displaystyle r_{HF}=({\frac {2T_{H}}{\pi \tau _{max}}})^{1/3}$ (4.1-10)
To get the diameter, multiply the equation of the radius by 2
    $\displaystyle d_{HF}=2({\frac {2T_{H}}{\pi \tau _{max}}})^{1/3}$ (4.1-11)
Solving the Equation 4.1-11 with the values given, the diameter can be calculated
* $\displaystyle d_{HF}=0.836in$

## 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, T$_C$ of 5 kips•in. and an unknown torque, T$_F$, 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, $r_{D}=2.5\,in$

### Solution
#### Step Two: Analysis

From the relation between the torques and the radius of the gears,
    $\displaystyle {\frac {T_{1}}{T_{2}}}={\frac {r_{1}}{r_{2}}}$ (4.2-1)
Therefore,
    $\displaystyle {\frac {T_{F}}{T_{C}}}={\frac {r_{D}}{r_{A}}}$ (4.2-2)
Now inserting the given values,
    $\displaystyle {\frac {T_{F}}{5}}={\frac {2.5}{4}}$ (4.2-3)
    $\displaystyle T_{F}=3.125\,kip\cdot in$ (4.2-4)

#### Step Three: Application
##### Part A

Allowable shearing stress in the shaft BC,
    $\displaystyle \tau _{max}={\frac {T_{C}*C}{J}}$ (4.2-5)
Where J for a solid circular shaft is
    $\displaystyle J={\frac {\pi }{2}}C^{4}$ (4.2-6)
Insert the values and solve for the radius C.
    $\displaystyle 8500={\frac {(5*10^{3})C}{{\frac {\pi }{2}}C^{4}}}$ (4.2-7)
    $\displaystyle C^{3}=0.3748$ (4.2-8)
    $\displaystyle C=0.721\,in$ (4.2-9)
Diameter of the shaft BC,
    $\displaystyle d=2C$ (4.2-10)
    $\displaystyle d=2(0.721)$ (4.2-11)
| $\displaystyle d=1.442\,in$ | (4.2-12) |

##### Part B

Allowable shearing stress in the shaft EF,
    $\displaystyle \tau ={\frac {T_{F}*C}{J}}$ (4.2-13)
Insert the values and solve for the radius C.
    $\displaystyle 8500={\frac {(3.125*10^{3})C}{{\frac {\pi }{2}}C^{4}}}$ (4.2-14)
    $\displaystyle C^{3}=0.234$ (4.2-15)
    $\displaystyle C=0.616\,in$ (4.2-16)
Diameter of the shaft EF,
    $\displaystyle d=2C$ (4.2-17)
    $\displaystyle d=2(0.616)$ (4.2-18)
| $\displaystyle d=1.233\,in$ | (4.2-19) |

## Contributors

Team Designee: Daniel Siefman