# University of Florida/Egm3520/s13.team1.r4

Report 4

## R4.1

### Problem R4.1 Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Problem statement: A motor exerts a torque $\displaystyle T_{f}$ to shaft FGH attached to a gear with radius $\displaystyle r_{G}$ which in turn applies a torque $\displaystyle T_{d}$ to a gear with radius $\displaystyle r_{D}$ attached to shaft CDE. No slipping occurs between the gears. The Allowable shearing stress in each shaft is $\displaystyle 10.5ksi$ .

Question statement:

What is required diameter of shaft CDE?
What is required diameter of shaft FGH?

Contents taken from Page 157 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Givens:
$\tau =10.5ksi=10,500psi$ $T_{F}=1200lb*in$ $r_{D}=8in$ $r_{G}=3in$ $\tau ={\frac {Tc}{J}}={\frac {2T}{\pi c^{3}}}\implies c={\sqrt[{3}]{\frac {2T}{\pi \tau }}}$ ### (a) Required diameter of shaft CDE

$T_{E}={\frac {r_{D}}{r_{G}}}T_{F}={\frac {8in}{3in}}(1200lb*in)=3200lb*in$ $c={\sqrt[{3}]{\frac {2T_{F}}{\pi \tau }}}={\sqrt[{3}]{\frac {2(3200lb*in)}{\pi (10,500psi)}}}=0.5789$ ${\frac {1}{2}}d=c\implies d=2c$ $d_{E}=2(0.5789)=1.157in$ ### (b) Required of diameter shaft FGH

$c={\sqrt[{3}]{\frac {2T_{F}}{\pi \tau }}}={\sqrt[{3}]{\frac {2(1200lb*in)}{\pi (10,500psi)}}}=0.4175in$ $d_{F}=2c=2(0.4175in)=0.8349in$ ## R4.2

### Question 3.25 Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Problem Statement: Given the diagram of gears A and D connected to rods BC and EF. There is a given torque of 5 kip*in for $T_{C}$ and $T_{F}$ is unknown. The shafts of rods ABC and DEF are solid and their diameters are unknown. Each shaft has an allowable shearing stress of 8500psi

Question Statement:

Using the known values find the required minimum diameter a) of shaft BC and b) shaft EF

Contents taken from Page 158 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Givens:
$\tau =8500psi$ $T_{C}=5kip*in=5000lb*in$ $\tau ={\frac {Tc}{J}}={\frac {2T}{\pi c^{3}}}\implies c={\sqrt[{3}]{\frac {2T}{\pi \tau }}}$ ### (a) Required diameter of shaft BC

$c={\sqrt[{3}]{\frac {2(5000lb*in)}{\pi (8500psi)}}}=0.72079in$ $d=2c\implies d_{C}=40442in$ ### (b) Required diameter of shaft EF

$T_{F}={\frac {4in}{2.5in}}(5000lb*in)=8000lb*in$ $c={\sqrt[{3}]{\frac {2T_{F}}{\pi \tau }}}={\sqrt[{3}]{\frac {2(8000lb*in)}{\pi (8500psi)}}}=0.84304in$ $d=2c\implies d_{F}=1.686in$ 