# 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 ${\displaystyle T_{F}=1200lbs*in}$ at ${\displaystyle F}$. Knowing that ${\displaystyle r_{D}=8in,r_{C}=3in}$
and the maximum allowable shearing stress is 10.5 ksi.

Determine the required diameter of member FH.

Shafts CE and FH with gears

#### 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)

### Soultion

#### 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)

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

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

## 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.

shafts AC and DF

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

#### 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, ${\displaystyle r_{D}=2.5\,in}$

### Solution

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)

## Contributors

Team Designee: Daniel Siefman

 Table of Assignments Problem Number Solved by Reviewed by 4.1 María José Carrasquilla﻿, Joshua Herrera﻿, Gregory Grannell﻿, and Phil D Mauro﻿ All 4.2 Tim Shankwitz﻿, Andrew Moffatt﻿, Michael Lindsay﻿, and Daniel Siefman﻿ All