University of Florida/Egm3520/s13.team1.r4

Report 4

R4.1

Problem R4.1

Figure from Mechanics of Materials textbook.

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 $T_f$ to shaft FGH attached to a gear with radius $r_G$ which in turn applies a torque $T_d$ to a gear with radius $r_D$ attached to shaft CDE. No slipping occurs between the gears. The Allowable shearing stress in each shaft is $10.5ksi$.

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:

$\tau =10.5ksi=10,500psi$

$T_F=1200lb*in$

$r_D=8in$

$r_G=3in$

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

(a) Required diameter of shaft CDE

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

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

$\frac{1}{2}d=c⟹d=2c$

Template:Color

(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

Figure from Mechanics of Materials.

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:

$\tau =8500psi$

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

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

(a) Required diameter of shaft BC

$c=\sqrt[3]{\frac{2(5000lb*in)}{\pi (8500psi)}}=0.72079in$

$d=2c⟹d_C=40442in$

(b) Required diameter of shaft EF

$T_F=\frac{4in}{2.5in}(5000lb*in)=8000lb*in$

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

$d=2c⟹d_F=1.686in$