University of Florida/Egm3520/s13.team1.r4

Report 4

Problem R4.1
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 $$ \displaystyle 10.5 ksi$$.

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:
 * $$ \tau = 10.5 ksi = 10,500 psi $$
 * $$ T_F = 1200 lb*in $$
 * $$ r_D = 8 in $$
 * $$ r_G = 3 in $$


 * $$ \tau = \frac{Tc}{J} = \frac{2T}{\pi c^3} \implies c= \sqrt[3]{\frac{2T}{\pi \tau}} $$

(a) Required diameter of shaft CDE

 * $$ T_E = \frac{r_D}{r_G} T_F = \frac{8 in}{3 in} (1200 lb*in) = 3200 lb*in $$


 * $$ c = \sqrt[3]{\frac{2T_F}{\pi \tau}} = \sqrt[3]{\frac{2(3200 lb*in)}{\pi (10,500 psi)}} = 0.5789 $$


 * $$ \frac{1}{2} d = c \implies d=2c $$


 * $ d_E = 2(0.5789) = 1.157 in $

(b) Required of diameter shaft FGH

 * $$ c = \sqrt[3]{\frac{2T_F}{\pi \tau}} = \sqrt[3]{\frac{2(1200 lb*in)}{\pi (10,500 psi)}} = 0.4175 in $$


 * $$ d_F = 2c = 2(0.4175 in) = 0.8349 in $$

Question 3.25


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:
 * $$ \tau = 8500 psi $$


 * $$ T_C = 5 kip*in = 5000 lb*in $$

$$\tau = \frac{Tc}{J} = \frac{2T}{\pi c^3} \implies c = \sqrt[3]{\frac{2T}{\pi \tau}} $$

(a) Required diameter of shaft BC
$$ c = \sqrt[3]{\frac{2(5000 lb*in)}{\pi(8500 psi)}} = 0.72079 in $$


 * $$ d = 2c \implies d_C = 40442 in $$

(b) Required diameter of shaft EF
$$ T_F = \frac{4 in}{2.5 in} (5000 lb*in) = 8000 lb*in $$

$$ c = \sqrt[3]{\frac{2T_F}{\pi \tau}} = \sqrt[3]{\frac{2(8000 lb*in)}{\pi (8500 psi)}} = 0.84304 in $$


 * $$ d = 2c \implies d_F = 1.686 in $$