University of Florida/Egm3520/s13.team1.r6

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:


 * $$ \sigma_{A} = \sigma_{centric} - \sigma_{bending} = -11.11MPa - 91.67MPa = {\color{red}-102.78 MPa} $$


 * Stress induced at point B:


 * $$ \sigma_{B} = \sigma_{centric} + \sigma_{bending} = -11.11MPa + 91.67MPa = {\color{red}80.56 MPa} $$

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):
 * $$ t = 8 mm \,\,\,\,\,\,\,   P = 20 kN $$


 * Rectangular cutout is 64 mm x 44 mm


 * $$ A = (80mm)(60mm)-(61mm)(49mm) =  1.984 \cdot 10^3 mm^2 $$


 * $$ I =  \frac{(60mm)(80mm)^3}{12}  - \frac{(44mm)(64mm)^3}{12}  =  1.59881 \cdot 10^6 mm^2 = 1.599 \cdot 10^{-6} m^4$$


 * $$ c  =  40  mm  =  0.004  m$$


 * $$ e  =  200  mm  +  40  =  240  mm  \implies  0.240  m$$


 * $$ {\color{blue} M =  P \cdot e}  =  (20 \cdot 10^3  N)(0.240  m)  =  4.8 \cdot 10^3  N-m$$


 * 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

 * 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:


 * $$ {\color{blue} \sigma_x = \frac{P}{A} - \frac{M \gamma}{I}} $$


 * $$ \gamma = \frac{d_o}{2} = 0.375 in $$   (Distance from centroid)


 * The Moment of Inertia of a Hollowed Cylindrical Cross-Section:


 * $$ {\color{blue} I = \frac{\pi}{64} \left(d_o^4 - d_i^4 \right)} = \frac{\pi}{64} \left( 0.75^4 - 0.59^4 \right) = 0.00958 in^4 $$


 * 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:


 * $$ \sigma_{all} = 4 \sigma = 4 \frac{P}{A} $$


 * $$ \sigma_{all} = \frac{P}{A} - \frac{M \gamma}{I} = \frac{P}{A} - \frac{Pd \gamma}{I} $$


 * $$ \implies 4 \frac{P}{A} = \frac{P}{A} + \frac{Pd \gamma}{I} \implies \frac{4}{A} = \frac{1}{A} + \frac{d \gamma}{I} \implies \frac{3}{A} = \frac{d \gamma}{I} $$


 * $$ \implies \frac{3}{0.168} = \frac{d(0.375)}{0.00958} \implies d = {\color{red} +0.456 in} $$

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 $$ \sigma_{all} = +5 ksi $$ and $$ \sigma_{all} = - 12 ksi $$, 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:$${\color{blue}I_{Tot}=I_1+I_2+I_3} = 4.57in^4 + 3.1275in^4 + 3.1275in^4 \implies 10.825 in^4 $$


 * 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(*)) :


 * $${\color{blue} M = Pd}\,\,\,\,\, {\color{blue}P = F_{max}= maximum force} \,\,\,\,\, d=$$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:$$\sigma_{max} = +5ksi, A = 7.5in^2, I = 10.825 in^4, y = -1.7 in, d $$ = distance of acting force from the centroid $$ = 1.5 + 1.7 = 3.2 in$$


 * $${\color{blue}\sigma_{max}=\frac{P_{max}}{A}-\frac{My}{I}} = \frac{P_{max}}{A}-\frac{P_{max}dy}{I} = P_{max}(\frac{1}{A}-\frac{dy}{I}) $$


 * $$ \implies P_{max\downarrow}=\frac{\sigma_{max}}{(\frac{1}{A}-\frac{dy}{I})}= \frac{5ksi}{(\frac{1}{7.5in^2}-\frac{(3.2in)(-1.7in)}{10.825in^4})}= 7.86 kips$$


 * Largest upward force:


 * $$\sigma_{all}= -12 ksi, \,A = 7.5 in^2, \, I = 10.825 in^4, \, y = 4 - 1.7 = 2.3 in, \, d = 1.5 + 1.7 = 3.2 in$$


 * $$P_{max\uparrow}=\frac{\sigma_{max}}{(\frac{1}{A}-\frac{dy}{I})}= 21.955 kips$$


 * 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:


 * $$\sigma_{all}=+5 ksi, \, A = 7.5 in^2, \, I = 10.825 in^4, \, y = 2.3 in,\, d = 3.2 in $$


 * $$ \implies P_{max\downarrow}=9.15 kips$$


 * The Upward Force becomes:


 * $$\sigma_{max}= -12 ksi, \, A = 7.5 in^2, \, I = 10.825 in^4, \, y = 2.3 in,\, d = 3.2 in $$


 * $$ \implies P_{max\uparrow}= 18.87 kips$$

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

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 $$ \sigma_{all} = +5 ksi $$ and $$ \sigma_{all} = - 12 ksi $$, 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:$${\color{blue}I_{Tot}=I_1+I_2+I_3} = 4.57in^4 + 3.1275in^4 + 3.1275in^4 \implies 10.825 in^4 $$


 * 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(*)) :


 * $${\color{blue} M = Pd}\,\,\,\,\, {\color{blue}P = F_{max}= maximum force} \,\,\,\,\, d=$$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:$$\sigma_{max} = +5ksi, A = 7.5in^2, I = 10.825 in^4, y = +2.3in, d $$ = distance of acting force from the centroid $$ = 3.2 in$$


 * $${\color{blue}\sigma_{max}=\frac{P_{max}}{A}-\frac{My}{I}} = \frac{P_{max}}{A}-\frac{P_{max}dy}{I} = P_{max}(\frac{1}{A}-\frac{dy}{I}) $$


 * $$ \implies P_{max\downarrow}=\frac{\sigma_{max}}{(\frac{1}{A}-\frac{dy}{I})}= \frac{5ksi}{(\frac{1}{7.5in^2}-\frac{(3.8in)(-2.3in)}{10.825in^4})}= {\color{Red}6.15 kips}$$


 * Largest upward force:
 * Apply negative sign throughout equation:


 * $${\color{blue}\sigma_{max}=\frac{-P_{max}}{A}-\frac{-my}{I} \implies P_{max}=\frac{-\sigma_{all/max}}{(\frac{1}{A}-\frac{dy}{I})}} $$


 * $$\sigma_{all}= +5 ksi, \,A = 7.5 in^2, \, I = 10.825 in^4, \, y = -1.7 in, \, d = 3.2 in$$


 * $$P_{max\uparrow}=\frac{\sigma_{max}}{(\frac{1}{A}-\frac{dy}{I})}= {\color{Red}13.54 kips}$$

Egm3520.s13.Jeandona (discuss • contribs) 12:47, 10 April 2013 (UTC)