# University of Florida/Egm3520/s13.team1.r6

TEAM 1: REPORT

## R6.1: Problem 4.101

Contents taken from Page 274 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Contents taken from Page 274 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Knowing that the magnitude of the horizontal force P is 8 kN, determine the stress at (a) point A, (b) point B.

Honor Pledge: On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### R6.1 Solution

• We must first analyze the cross sectional area.
${\displaystyle A=30\cdot 24=720mm^{2}}$
• Next calculate the moment of inertia of the rectangular cross section.
${\displaystyle I={\frac {bd^{3}}{12}}={\frac {30\cdot 24^{3}}{12}}=34560mm^{4}}$
• Next calculate the centroid of the rectangle
${\displaystyle c={\frac {h}{2}}=12mm}$
• Next we show a free body diagram of the forces present on the bracket

• Find the eccentricity
${\displaystyle e=45mm-{\frac {24}{2}}=33mm}$
• Calculate the bending couple using P = 8kN and e = 0.033m
${\displaystyle M={\color {blue}Pe}=8\cdot 0.033=264N*m}$
• Now we can calculate the stresses
${\displaystyle \sigma _{centric}={\color {blue}{\frac {-P}{A}}}={\frac {8\cdot 10^{3}}{7.2\cdot 10^{-4}}}=-11.11MPa}$
${\displaystyle \sigma _{bending}={\color {blue}{\frac {MC}{I}}}={\frac {264\cdot 0.012}{34560\cdot 10^{\left(-3\right)^{4}}}}=90.67MPa}$
• Stress induced at point A:
${\displaystyle \sigma _{A}=\sigma _{centric}-\sigma _{bending}=-11.11MPa-91.67MPa={\color {red}-102.78MPa}}$
• Stress induced at point B:
${\displaystyle \sigma _{B}=\sigma _{centric}+\sigma _{bending}=-11.11MPa+91.67MPa={\color {red}80.56MPa}}$

## R6.2: Problem 4.103

Contents taken from Page 274 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Contents taken from Page 274 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

The vertical portion of the press of the press shown consists of a rectangular tube of wall thickness t = 8 mm. Knowing that the press has been tightened on wooden planks being glued together until P = 20 kN, determine the stress at (a) point A, (b) point B.

Honor Pledge: On my honor, I have neither given nor received unauthorized aid in doing this assignment.

## R6.2 Solution

Given(s):

${\displaystyle t=8mm\,\,\,\,\,\,\,P=20kN}$
• Rectangular cutout is 64 mm x 44 mm
${\displaystyle A=(80mm)(60mm)-(61mm)(49mm)=1.984\cdot 10^{3}mm^{2}}$
${\displaystyle I={\frac {(60mm)(80mm)^{3}}{12}}-{\frac {(44mm)(64mm)^{3}}{12}}=1.59881\cdot 10^{6}mm^{2}=1.599\cdot 10^{-6}m^{4}}$
${\displaystyle c=40mm=0.004m}$
${\displaystyle e=200mm+40=240mm\implies 0.240m}$
${\displaystyle {\color {blue}M=P\cdot e}=(20\cdot 10^{3}N)(0.240m)=4.8\cdot 10^{3}N-m}$
• Stress induced at point A:
${\displaystyle \sigma _{A}=-{\frac {20\cdot 10^{3}N}{1.984\cdot 10^{-3}}}-{\frac {(4.8\cdot 10^{3}N-m)(-0.04m)}{(1.59881\cdot 10^{-6}m)}}={\color {red}130.240\cdot 10^{6}Pa}}$
• Stress induced at point B:
${\displaystyle \sigma _{B}=-{\frac {20\cdot 10^{3}N}{1.984\cdot 10^{-3}}}-{\frac {(4.8\cdot 10^{3}N-m)(0.04m)}{(1.59881\cdot 10^{-6}m)}}\,={\color {red}-110.0\cdot 10^{6}Pa}}$

## R6.3: Problem 4.112

Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

* Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664. (*) = Reference to listed textbook

An offset h must be introduced into a metal tube of 0.75 in outer diameter and 0.08 in wall thickness. Knowing the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used.

Honor Pledge: On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### R6.3 Solution

GIVEN
Term Desig Value
Outer Diameter ${\displaystyle d_{o}}$ ${\displaystyle 0.75in}$
Thickness ${\displaystyle t}$ ${\displaystyle 0.08in}$
Inner Diameter ${\displaystyle d_{i}}$ ${\displaystyle d_{i}=d_{o}-t=0.75-0.08(2)=0.59in}$
Area ${\displaystyle A}$ ${\displaystyle {\frac {\pi }{4}}\left(d_{o}^{2}-d_{i}^{2}\right)={\frac {\pi }{4}}\left(0.75^{2}-0.59^{2}\right)\approx 0.168in^{2}}$
Stress ${\displaystyle \sigma }$ ${\displaystyle \sigma ={\frac {P}{A}}}$

• The internal forces in the cross section are equivalent to a centric force P and a bending curve M. (*) Ex:4.07 on pg 272
• EQ 4.49 on page 270(*) states:
${\displaystyle {\color {blue}F=P\,\,\,\,\,\,\,\,\,\,M=Pd}}$
(Where F = Force at centroid, P = Line of action load, M = Moment, and d = offset distance.)
• EQ 4.5 on page 271(*) states:
${\displaystyle {\color {blue}\sigma _{x}={\frac {P}{A}}-{\frac {M\gamma }{I}}}}$
${\displaystyle \gamma ={\frac {d_{o}}{2}}=0.375in}$ (Distance from centroid)
• The Moment of Inertia of a Hollowed Cylindrical Cross-Section:
${\displaystyle {\color {blue}I={\frac {\pi }{64}}\left(d_{o}^{4}-d_{i}^{4}\right)}={\frac {\pi }{64}}\left(0.75^{4}-0.59^{4}\right)=0.00958in^{4}}$
• To ensure the the max stress does not exceed 4 times the stress in the tube and making an assumption that P = 1, we can derive the following solution:
${\displaystyle \sigma _{all}=4\sigma =4{\frac {P}{A}}}$
${\displaystyle \sigma _{all}={\frac {P}{A}}-{\frac {M\gamma }{I}}={\frac {P}{A}}-{\frac {Pd\gamma }{I}}}$
${\displaystyle \implies 4{\frac {P}{A}}={\frac {P}{A}}+{\frac {Pd\gamma }{I}}\implies {\frac {4}{A}}={\frac {1}{A}}+{\frac {d\gamma }{I}}\implies {\frac {3}{A}}={\frac {d\gamma }{I}}}$
${\displaystyle \implies {\frac {3}{0.168}}={\frac {d(0.375)}{0.00958}}\implies d={\color {red}+0.456in}}$

## R6.4: Problem 4.114

Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664. (*) = Reference to listed textbook

A vertical rod is attached at point A to the cast iron hanger shown. Knowing that the allowable stress in the hanger are ${\displaystyle \sigma _{all}=+5ksi}$ and ${\displaystyle \sigma _{all}=-12ksi}$, determine the largest downward force and the largest upward force that can be exerted by the rod.

Honor Pledge: On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### R6.4 Solution

• Max allowable stresses on the hanger:
${\displaystyle \sigma _{allow}=+5ksi}$ ${\displaystyle F_{max\downarrow }=?}$
${\displaystyle \sigma _{allow}=-12ksi}$ ${\displaystyle F_{max\uparrow }=?}$
• Find centroid:
• Take ${\displaystyle {\overline {y}}}$ as a distance measured from left end of shape.
${\displaystyle {\overline {Y}}={\frac {\sum {\overline {y}}A}{\sum A}}}$
${\displaystyle A_{1}=(3in)(1in)=3in^{2}}$
${\displaystyle A_{2}=(3in)(.75in)=2.25in^{2}}$
• Due to same parameters,
${\displaystyle A_{3}=A_{2}=2.25in^{2}}$
${\displaystyle {\overline {y_{1}}}={\frac {1}{2}}in=.5in}$
${\displaystyle {\overline {y_{2}}}=1in+{\frac {3}{2}}in=2.5in}$
• Due to same parameters,
${\displaystyle {\overline {y_{3}}}={\overline {y_{2}}}=2.5in}$

${\displaystyle \therefore {\overline {Y}}={\frac {(.5in)(3in)+(2.5in)(2.25in)+(2.5in)(2.25in)}{7.5in}}=1.7in}$
• Must incorporate the parallel axis theorem to find moment of inertia: ${\displaystyle {\color {blue}I={\frac {bh^{3}}{12}}+Ad^{2}}}$
b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis
${\displaystyle {\color {blue}I_{1}={\frac {b_{1}h_{1}^{3}}{12}}+A_{1}d_{1}^{2}}\,\,where\,\,{\color {blue}d_{1}=({\overline {y_{1}}}-{\overline {Y}})}\implies I_{1}={\frac {(3in)(1in)^{3}}{12}}+(3in^{2})(.5in-1.7in)^{2}=4.57in^{4}}$
${\displaystyle {\color {blue}I_{2}={\frac {b_{2}h_{2}^{3}}{12}}+A_{2}d_{2}^{2}}\,\,where\,\,{\color {blue}d_{2}=({\overline {y_{2}}}-{\overline {Y}})}\implies I_{1}={\frac {(.75in)(3in)^{3}}{12}}+(2.25in^{2})(2.5in-1.7in)^{2}=3.1275in^{4}}$
• Due to same parameters,
${\displaystyle I_{3}=I_{2}=3.1275in^{4}}$
${\displaystyle \therefore }$ Total moment of inertia is:${\displaystyle {\color {blue}I_{Tot}=I_{1}+I_{2}+I_{3}}=4.57in^{4}+3.1275in^{4}+3.1275in^{4}\implies 10.825in^{4}}$
• The normal stress at point A is due to bending:
${\displaystyle {\color {blue}\sigma _{bending}=-{\frac {My}{I}}}}$
• "The internal forces in the cross section are equivalent to a centric force P and a bending couple M " (Example Problem 4.07 page 272(*)):
${\displaystyle {\color {blue}M=Pd}\,\,\,\,\,{\color {blue}P=F_{max}=maximumforce}\,\,\,\,\,d=}$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))
Normal stress due to centric load: ${\displaystyle {\color {blue}\sigma _{centric}={\frac {P}{A}}}}$
• Combine:
${\displaystyle {\color {blue}\sigma =\sigma _{centric}+\sigma _{bending}={\frac {P}{A}}-{\frac {My}{I}}}}$ (EQ4.50, p.221(*))
${\displaystyle \uparrow =}$Total normal stress acting at point A.

• Largest downward force:
• Assuming conventions:${\displaystyle \sigma _{max}=+5ksi,A=7.5in^{2},I=10.825in^{4},y=-1.7in,d}$ = distance of acting force from the centroid ${\displaystyle =1.5+1.7=3.2in}$
${\displaystyle {\color {blue}\sigma _{max}={\frac {P_{max}}{A}}-{\frac {My}{I}}}={\frac {P_{max}}{A}}-{\frac {P_{max}dy}{I}}=P_{max}({\frac {1}{A}}-{\frac {dy}{I}})}$
${\displaystyle \implies P_{max\downarrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\frac {5ksi}{({\frac {1}{7.5in^{2}}}-{\frac {(3.2in)(-1.7in)}{10.825in^{4}}})}}=7.86kips}$
• Largest upward force:
${\displaystyle \sigma _{all}=-12ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=4-1.7=2.3in,\,d=1.5+1.7=3.2in}$
${\displaystyle P_{max\uparrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}=21.955kips}$
• The limiting factor is at 7.86 kips force upward.
• Apply negative sign throughout equation:
${\displaystyle {\color {blue}\sigma _{max}={\frac {-P_{max}}{A}}-{\frac {-my}{I}}\implies P_{max}={\frac {-\sigma _{all/max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}}}$
• The Downward force becomes:
${\displaystyle \sigma _{all}=+5ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=2.3in,\,d=3.2in}$
${\displaystyle \implies P_{max\downarrow }=9.15kips}$
• The Upward Force becomes:
${\displaystyle \sigma _{max}=-12ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=2.3in,\,d=3.2in}$
${\displaystyle \implies P_{max\uparrow }=18.87kips}$

${\displaystyle \therefore }$ Limit is at ${\displaystyle {\color {red}9.15kips}}$

## R6.5: Problem 4.115

Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Contents taken from Page 276 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

A vertical rod is attached at point B to the cast iron hanger shown. Knowing that the allowable stress in the hanger are ${\displaystyle \sigma _{all}=+5ksi}$ and ${\displaystyle \sigma _{all}=-12ksi}$, determine the largest downward force and the largest upward force that can be exerted by the rod.

Honor Pledge: On my honor, I have neither given nor received unauthorized aid in doing this assignment.

### R6.5 Solution

3D Representation of Figure 4.115
• Max allowable stresses on the hanger:
${\displaystyle \sigma _{allow}=+5ksi}$ ${\displaystyle F_{max\downarrow }=?}$
${\displaystyle \sigma _{allow}=-12ksi}$ ${\displaystyle F_{max\uparrow }=?}$
• Find centroid:
• Take ${\displaystyle {\overline {y}}}$ as a distance measured from left end of shape.
${\displaystyle {\overline {Y}}={\frac {\sum {\overline {y}}A}{\sum A}}}$
${\displaystyle A_{1}=(3in)(1in)=3in^{2}}$
${\displaystyle A_{2}=(3in)(.75in)=2.25in^{2}}$
• Due to same parameters,
${\displaystyle A_{3}=A_{2}=2.25in^{2}}$
${\displaystyle {\overline {y_{1}}}={\frac {1}{2}}in=.5in}$
${\displaystyle {\overline {y_{2}}}=1in+{\frac {3}{2}}in=2.5in}$
• Due to same parameters,
${\displaystyle {\overline {y_{3}}}={\overline {y_{2}}}=2.5in}$

${\displaystyle \therefore {\overline {Y}}={\frac {(.5in)(3in)+(2.5in)(2.25in)+(2.5in)(2.25in)}{7.5in}}=1.7in}$

• Must incorporate the parallel axis theorem to find moment of inertia: ${\displaystyle {\color {blue}I={\frac {bh^{3}}{12}}+Ad^{2}}}$
b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis
${\displaystyle {\color {blue}I_{1}={\frac {b_{1}h_{1}^{3}}{12}}+A_{1}d_{1}^{2}}\,\,where\,\,{\color {blue}d_{1}=({\overline {y_{1}}}-{\overline {Y}})}\implies I_{1}={\frac {(3in)(1in)^{3}}{12}}+(3in^{2})(.5in-1.7in)^{2}=4.57in^{4}}$
${\displaystyle {\color {blue}I_{2}={\frac {b_{2}h_{2}^{3}}{12}}+A_{2}d_{2}^{2}}\,\,where\,\,{\color {blue}d_{2}=({\overline {y_{2}}}-{\overline {Y}})}\implies I_{1}={\frac {(.75in)(3in)^{3}}{12}}+(2.25in^{2})(2.5in-1.7in)^{2}=3.1275in^{4}}$
• Due to same parameters,
${\displaystyle I_{3}=I_{2}=3.1275in^{4}}$
${\displaystyle \therefore }$ Total moment of inertia is:${\displaystyle {\color {blue}I_{Tot}=I_{1}+I_{2}+I_{3}}=4.57in^{4}+3.1275in^{4}+3.1275in^{4}\implies 10.825in^{4}}$
• The normal stress at point A is due to bending:
${\displaystyle {\color {blue}\sigma _{bending}=-{\frac {My}{I}}}}$
• "The internal forces in the cross section are equivalent to a centric force P and a bending couple M " (Example Problem 4.07 page 272(*)):
${\displaystyle {\color {blue}M=Pd}\,\,\,\,\,{\color {blue}P=F_{max}=maximumforce}\,\,\,\,\,d=}$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))
Normal stress due to centric load: ${\displaystyle {\color {blue}\sigma _{centric}={\frac {P}{A}}}}$
• Combine:
${\displaystyle {\color {blue}\sigma =\sigma _{centric}+\sigma _{bending}={\frac {P}{A}}-{\frac {My}{I}}}}$ (EQ4.50, p.221(*))
${\displaystyle \uparrow =}$Total normal stress acting at point A.

• Largest downward force:
• Assuming conventions:${\displaystyle \sigma _{max}=+5ksi,A=7.5in^{2},I=10.825in^{4},y=+2.3in,d}$ = distance of acting force from the centroid ${\displaystyle =3.2in}$
${\displaystyle {\color {blue}\sigma _{max}={\frac {P_{max}}{A}}-{\frac {My}{I}}}={\frac {P_{max}}{A}}-{\frac {P_{max}dy}{I}}=P_{max}({\frac {1}{A}}-{\frac {dy}{I}})}$
${\displaystyle \implies P_{max\downarrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\frac {5ksi}{({\frac {1}{7.5in^{2}}}-{\frac {(3.8in)(-2.3in)}{10.825in^{4}}})}}={\color {Red}6.15kips}}$
• Largest upward force:
• Apply negative sign throughout equation:
${\displaystyle {\color {blue}\sigma _{max}={\frac {-P_{max}}{A}}-{\frac {-my}{I}}\implies P_{max}={\frac {-\sigma _{all/max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}}}$
${\displaystyle \sigma _{all}=+5ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=-1.7in,\,d=3.2in}$
${\displaystyle P_{max\uparrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\color {Red}13.54kips}}$

Egm3520.s13.Jeandona (discusscontribs) 12:47, 10 April 2013 (UTC)