# 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
    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.
       $A=30\cdot 24=720mm^{2}$
* Next calculate the moment of inertia of the rectangular cross section.
       $I={\frac {bd^{3}}{12}}={\frac {30\cdot 24^{3}}{12}}=34560mm^{4}$
* Next calculate the centroid of the rectangle
       $c={\frac {h}{2}}=12mm$
* Next we show a free body diagram of the forces present on the bracket
* Find the eccentricity
       $e=45mm-{\frac {24}{2}}=33mm$
* Calculate the bending couple using P = 8kN and e = 0.033m
       $M={\color {blue}Pe}=8\cdot 0.033=264N*m$
* Now we can calculate the stresses
       $\sigma _{centric}={\color {blue}{\frac {-P}{A}}}={\frac {8\cdot 10^{3}}{7.2\cdot 10^{-4}}}=-11.11MPa$
       $\sigma _{bending}={\color {blue}{\frac {MC}{I}}}={\frac {264\cdot 0.012}{34560\cdot 10^{\left(-3\right)^{4}}}}=90.67MPa$
* Stress induced at point A:
       $\sigma _{A}=\sigma _{centric}-\sigma _{bending}=-11.11MPa-91.67MPa={\color {red}-102.78MPa}$
* Stress induced at point B:
       $\sigma _{B}=\sigma _{centric}+\sigma _{bending}=-11.11MPa+91.67MPa={\color {red}80.56MPa}$

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

## 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. (*) = 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 | $d_{o}$ | $0.75in$ |
| Thickness | $t$ | $0.08in$ |
| Inner Diameter | $d_{i}$ | $d_{i}=d_{o}-t=0.75-0.08(2)=0.59in$ |
| Area | $A$ | ${\frac {\pi }{4}}\left(d_{o}^{2}-d_{i}^{2}\right)={\frac {\pi }{4}}\left(0.75^{2}-0.59^{2}\right)\approx 0.168in^{2}$ |
| Stress | $\sigma$ | $\sigma ={\frac {P}{A}}$ |
* 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:
       ${\color {blue}F=P\,\,\,\,\,\,\,\,\,\,M=Pd}$
       (Where F = Force at centroid, P = Line of action load, M = Moment, and d = offset distance.)
* EQ 4.5 on page 271(*) states:
       ${\color {blue}\sigma _{x}={\frac {P}{A}}-{\frac {M\gamma }{I}}}$
       $\gamma ={\frac {d_{o}}{2}}=0.375in$ (Distance from centroid)
* The Moment of Inertia of a Hollowed Cylindrical Cross-Section:
       ${\color {blue}I={\frac {\pi }{64}}\left(d_{o}^{4}-d_{i}^{4}\right)}={\frac {\pi }{64}}\left(0.75^{4}-0.59^{4}\right)=0.00958in^{4}$
* 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:
       $\sigma _{all}=4\sigma =4{\frac {P}{A}}$
       $\sigma _{all}={\frac {P}{A}}-{\frac {M\gamma }{I}}={\frac {P}{A}}-{\frac {Pd\gamma }{I}}$
          $\implies 4{\frac {P}{A}}={\frac {P}{A}}+{\frac {Pd\gamma }{I}}\implies {\frac {4}{A}}={\frac {1}{A}}+{\frac {d\gamma }{I}}\implies {\frac {3}{A}}={\frac {d\gamma }{I}}$
          $\implies {\frac {3}{0.168}}={\frac {d(0.375)}{0.00958}}\implies d={\color {red}+0.456in}$

## 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. (*) = 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 $\sigma _{all}=+5ksi$ and $\sigma _{all}=-12ksi$, 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:
       $\sigma _{allow}=+5ksi$ $F_{max\downarrow }=?$
       $\sigma _{allow}=-12ksi$ $F_{max\uparrow }=?$
* Find centroid:
   * Take ${\overline {y}}$ as a distance measured from left end of shape.
          ${\overline {Y}}={\frac {\sum {\overline {y}}A}{\sum A}}$
          $A_{1}=(3in)(1in)=3in^{2}$
          $A_{2}=(3in)(.75in)=2.25in^{2}$
   * Due to same parameters,
          $A_{3}=A_{2}=2.25in^{2}$
          ${\overline {y_{1}}}={\frac {1}{2}}in=.5in$
          ${\overline {y_{2}}}=1in+{\frac {3}{2}}in=2.5in$
   * Due to same parameters,
          ${\overline {y_{3}}}={\overline {y_{2}}}=2.5in$
          $\therefore {\overline {Y}}={\frac {(.5in)(3in)+(2.5in)(2.25in)+(2.5in)(2.25in)}{7.5in}}=1.7in$
* Must incorporate the parallel axis theorem to find moment of inertia: ${\color {blue}I={\frac {bh^{3}}{12}}+Ad^{2}}$
       b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis
       ${\color {blue}I_{1}={\frac {b_{1}h_{1}^{3}}{12}}+A_{1}d_{1}^{2}}\,\,where\,\,{\color {blue}d_{1}=({\overline {y_{1}}}-{\overline {Y}})}\implies I_{1}={\frac {(3in)(1in)^{3}}{12}}+(3in^{2})(.5in-1.7in)^{2}=4.57in^{4}$
       ${\color {blue}I_{2}={\frac {b_{2}h_{2}^{3}}{12}}+A_{2}d_{2}^{2}}\,\,where\,\,{\color {blue}d_{2}=({\overline {y_{2}}}-{\overline {Y}})}\implies I_{1}={\frac {(.75in)(3in)^{3}}{12}}+(2.25in^{2})(2.5in-1.7in)^{2}=3.1275in^{4}$
* Due to same parameters,
    $I_{3}=I_{2}=3.1275in^{4}$
    $\therefore$ Total moment of inertia is:${\color {blue}I_{Tot}=I_{1}+I_{2}+I_{3}}=4.57in^{4}+3.1275in^{4}+3.1275in^{4}\implies 10.825in^{4}$
* The normal stress at point A is due to bending:
       ${\color {blue}\sigma _{bending}=-{\frac {My}{I}}}$
   * "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(*)):
          ${\color {blue}M=Pd}\,\,\,\,\,{\color {blue}P=F_{max}=maximumforce}\,\,\,\,\,d=$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))
          Normal stress due to centric load: ${\color {blue}\sigma _{centric}={\frac {P}{A}}}$
* Combine:
       ${\color {blue}\sigma =\sigma _{centric}+\sigma _{bending}={\frac {P}{A}}-{\frac {My}{I}}}$ (EQ4.50, p.221(*))
       $\uparrow =$Total normal stress acting at point A.
* Largest downward force:
   * Assuming conventions:$\sigma _{max}=+5ksi,A=7.5in^{2},I=10.825in^{4},y=-1.7in,d$ = distance of acting force from the centroid $=1.5+1.7=3.2in$
       ${\color {blue}\sigma _{max}={\frac {P_{max}}{A}}-{\frac {My}{I}}}={\frac {P_{max}}{A}}-{\frac {P_{max}dy}{I}}=P_{max}({\frac {1}{A}}-{\frac {dy}{I}})$
          $\implies P_{max\downarrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\frac {5ksi}{({\frac {1}{7.5in^{2}}}-{\frac {(3.2in)(-1.7in)}{10.825in^{4}}})}}=7.86kips$
* Largest upward force:
       $\sigma _{all}=-12ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=4-1.7=2.3in,\,d=1.5+1.7=3.2in$
       $P_{max\uparrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}=21.955kips$
* The limiting factor is at 7.86 kips force upward.
* Apply negative sign throughout equation:
       ${\color {blue}\sigma _{max}={\frac {-P_{max}}{A}}-{\frac {-my}{I}}\implies P_{max}={\frac {-\sigma _{all/max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}}$
   * The Downward force becomes:
          $\sigma _{all}=+5ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=2.3in,\,d=3.2in$
             $\implies P_{max\downarrow }=9.15kips$
   * The Upward Force becomes:
          $\sigma _{max}=-12ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=2.3in,\,d=3.2in$
             $\implies P_{max\uparrow }=18.87kips$
$\therefore$ Limit is at ${\color {red}9.15kips}$

## 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
    A vertical rod is attached at point B to the cast iron hanger shown. Knowing that the allowable stress in the hanger are $\sigma _{all}=+5ksi$ and $\sigma _{all}=-12ksi$, 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

* Max allowable stresses on the hanger:
       $\sigma _{allow}=+5ksi$ $F_{max\downarrow }=?$
       $\sigma _{allow}=-12ksi$ $F_{max\uparrow }=?$
* Find centroid:
   * Take ${\overline {y}}$ as a distance measured from left end of shape.
          ${\overline {Y}}={\frac {\sum {\overline {y}}A}{\sum A}}$
          $A_{1}=(3in)(1in)=3in^{2}$
          $A_{2}=(3in)(.75in)=2.25in^{2}$
   * Due to same parameters,
          $A_{3}=A_{2}=2.25in^{2}$
          ${\overline {y_{1}}}={\frac {1}{2}}in=.5in$
          ${\overline {y_{2}}}=1in+{\frac {3}{2}}in=2.5in$
   * Due to same parameters,
          ${\overline {y_{3}}}={\overline {y_{2}}}=2.5in$
          $\therefore {\overline {Y}}={\frac {(.5in)(3in)+(2.5in)(2.25in)+(2.5in)(2.25in)}{7.5in}}=1.7in$
* Must incorporate the parallel axis theorem to find moment of inertia: ${\color {blue}I={\frac {bh^{3}}{12}}+Ad^{2}}$
       b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis
       ${\color {blue}I_{1}={\frac {b_{1}h_{1}^{3}}{12}}+A_{1}d_{1}^{2}}\,\,where\,\,{\color {blue}d_{1}=({\overline {y_{1}}}-{\overline {Y}})}\implies I_{1}={\frac {(3in)(1in)^{3}}{12}}+(3in^{2})(.5in-1.7in)^{2}=4.57in^{4}$
       ${\color {blue}I_{2}={\frac {b_{2}h_{2}^{3}}{12}}+A_{2}d_{2}^{2}}\,\,where\,\,{\color {blue}d_{2}=({\overline {y_{2}}}-{\overline {Y}})}\implies I_{1}={\frac {(.75in)(3in)^{3}}{12}}+(2.25in^{2})(2.5in-1.7in)^{2}=3.1275in^{4}$
* Due to same parameters,
    $I_{3}=I_{2}=3.1275in^{4}$
    $\therefore$ Total moment of inertia is:${\color {blue}I_{Tot}=I_{1}+I_{2}+I_{3}}=4.57in^{4}+3.1275in^{4}+3.1275in^{4}\implies 10.825in^{4}$
* The normal stress at point A is due to bending:
       ${\color {blue}\sigma _{bending}=-{\frac {My}{I}}}$
   * "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(*)):
          ${\color {blue}M=Pd}\,\,\,\,\,{\color {blue}P=F_{max}=maximumforce}\,\,\,\,\,d=$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))
          Normal stress due to centric load: ${\color {blue}\sigma _{centric}={\frac {P}{A}}}$
* Combine:
       ${\color {blue}\sigma =\sigma _{centric}+\sigma _{bending}={\frac {P}{A}}-{\frac {My}{I}}}$ (EQ4.50, p.221(*))
       $\uparrow =$Total normal stress acting at point A.
* Largest downward force:
   * Assuming conventions:$\sigma _{max}=+5ksi,A=7.5in^{2},I=10.825in^{4},y=+2.3in,d$ = distance of acting force from the centroid $=3.2in$
       ${\color {blue}\sigma _{max}={\frac {P_{max}}{A}}-{\frac {My}{I}}}={\frac {P_{max}}{A}}-{\frac {P_{max}dy}{I}}=P_{max}({\frac {1}{A}}-{\frac {dy}{I}})$
          $\implies P_{max\downarrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\frac {5ksi}{({\frac {1}{7.5in^{2}}}-{\frac {(3.8in)(-2.3in)}{10.825in^{4}}})}}={\color {Red}6.15kips}$
* Largest upward force:
* Apply negative sign throughout equation:
       ${\color {blue}\sigma _{max}={\frac {-P_{max}}{A}}-{\frac {-my}{I}}\implies P_{max}={\frac {-\sigma _{all/max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}}$
       $\sigma _{all}=+5ksi,\,A=7.5in^{2},\,I=10.825in^{4},\,y=-1.7in,\,d=3.2in$
       $P_{max\uparrow }={\frac {\sigma _{max}}{({\frac {1}{A}}-{\frac {dy}{I}})}}={\color {Red}13.54kips}$
Egm3520.s13.Jeandona (discuss • contribs) 12:47, 10 April 2013 (UTC)