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.

Shafts CE and FH with gears

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

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

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

(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.

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, $T_{C}=5\,kip\cdot in$
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 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,

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

(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

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