University of Florida/Egm3520/s13.team1.r6

R6.1: Problem 4.101

Figure P4.101.

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.

$A=30\cdot 24=720m{m}^2$

• Next calculate the moment of inertia of the rectangular cross section.

$I=\frac{b{d}^3}{12}=\frac{30\cdot 24^3}{12}=34560m{m}^4$

• 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

File:FBD6.1.jpg

• Find the eccentricity

$e=45mm-\frac{24}{2}=33mm$

• Calculate the bending couple using P = 8kN and e = 0.033m

$M=Pe=8\cdot 0.033=264N*m$

• Now we can calculate the stresses

${\sigma }_{centric}=\frac{-P}{A}=\frac{8\cdot 10^3}{7.2\cdot 10^{-4}}=-11.11MPa$

${\sigma }_{bending}=\frac{MC}{I}=\frac{264\cdot 0.012}{34560\cdot 10^{{(-3)}^4}}=90.67MPa$

• Stress induced at point A:

${\sigma }_A={\sigma }_{centric}-{\sigma }_{bending}=-11.11MPa-91.67MPa=-102.78MPa$

• Stress induced at point B:

${\sigma }_B={\sigma }_{centric}+{\sigma }_{bending}=-11.11MPa+91.67MPa=80.56MPa$

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R6.2: Problem 4.103

Figure P4.102.

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):

$t=8mmP=20kN$

• Rectangular cutout is 64 mm x 44 mm

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

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

$c=40mm=0.004m$

$e=200mm+40=240mm⟹0.240m$

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

• Stress induced at point A:

${\sigma }_A=-\frac{20\cdot 10^{3}N}{1.984\cdot 10^{-3}}-\frac{(4.8\cdot 10^{3}N-m)(-0.04m)}{(1.59881\cdot 10^{-6}m)}=130.240\cdot 10^{6}Pa$

• Stress induced at point B:

${\sigma }_B=-\frac{20\cdot 10^{3}N}{1.984\cdot 10^{-3}}-\frac{(4.8\cdot 10^{3}N-m)(0.04m)}{(1.59881\cdot 10^{-6}m)}=-110.0\cdot 10^{6}Pa$

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R6.3: Problem 4.112

Figure P4.112.

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	$d_o$	$0.75in$
Thickness	$t$	$0.08in$
Inner Diameter	$d_i$	$d_i=d_o-t=0.75-0.08(2)=0.59in$
Area	$A$	$\frac{\pi }{4}(d_o^2-d_i^2)=\frac{\pi }{4}(0.75^2-0.59^2)\approx 0.168i{n}^2$
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:

$F=PM=Pd$

(Where F = Force at centroid, P = Line of action load, M = Moment, and d = offset distance.)

• EQ 4.5 on page 271(*) states:

${\sigma }_x=\frac{P}{A}-\frac{M\gamma }{I}$

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

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

$I=\frac{\pi }{64}(d_o^4-d_i^4)=\frac{\pi }{64}(0.75^4-0.59^4)=0.00958i{n}^4$

• 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:

${\sigma }_{all}=4\sigma =4\frac{P}{A}$

${\sigma }_{all}=\frac{P}{A}-\frac{M\gamma }{I}=\frac{P}{A}-\frac{Pd\gamma }{I}$

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

$⟹\frac{3}{0.168}=\frac{d(0.375)}{0.00958}⟹d=+0.456in$

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R6.4: Problem 4.114

Figure P4.114 and P4.115.

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. (*) = 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 $\bar{y}$ as a distance measured from left end of shape.

$\bar{Y}=\frac{\sum \bar{y}A}{\sum A}$

$A_1=(3in)(1in)=3i{n}^2$

$A_2=(3in)(.75in)=2.25i{n}^2$

• Due to same parameters,

$A_3=A_2=2.25i{n}^2$

$\overline{y_1}=\frac{1}{2}in=.5in$

$\overline{y_2}=1in+\frac{3}{2}in=2.5in$

• Due to same parameters,

$\overline{y_3}=\overline{y_2}=2.5in$

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

• Must incorporate the parallel axis theorem to find moment of inertia: $I=\frac{b{h}^3}{12}+A{d}^2$

b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis

$I_1=\frac{b_1{h}_1^3}{12}+A_1{d}_1^{2}where{d}_1=(\overline{y_1}-\bar{Y})⟹I_1=\frac{(3in)(1in{)}^3}{12}+(3i{n}^2)(.5in-1.7in{)}^2=4.57i{n}^4$

$I_2=\frac{b_2{h}_2^3}{12}+A_2{d}_2^{2}where{d}_2=(\overline{y_2}-\bar{Y})⟹I_1=\frac{(.75in)(3in{)}^3}{12}+(2.25i{n}^2)(2.5in-1.7in{)}^2=3.1275i{n}^4$

• Due to same parameters,

$I_3=I_2=3.1275i{n}^4$

$\therefore$ Total moment of inertia is:$I_{Tot}=I_1+I_2+I_3=4.57i{n}^4+3.1275i{n}^4+3.1275i{n}^4⟹10.825i{n}^4$

• The normal stress at point A is due to bending:

${\sigma }_{bending}=-\frac{My}{I}$

• "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(*)):

$M=PdP=F_{max}=maximumforced=$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))

Normal stress due to centric load: ${\sigma }_{centric}=\frac{P}{A}$

• Combine:

$\sigma ={\sigma }_{centric}+{\sigma }_{bending}=\frac{P}{A}-\frac{My}{I}$ (EQ4.50, p.221(*))

$\uparrow =$Total normal stress acting at point A.

• Largest downward force:

• Assuming conventions:${\sigma }_{max}=+5ksi,A=7.5i{n}^2,I=10.825i{n}^4,y=-1.7in,d$ = distance of acting force from the centroid $=1.5+1.7=3.2in$

${\sigma }_{max}=\frac{P_{max}}{A}-\frac{My}{I}=\frac{P_{max}}{A}-\frac{P_{max}dy}{I}=P_{max}(\frac{1}{A}-\frac{dy}{I})$

$⟹P_{max\downarrow }=\frac{{\sigma }_{max}}{(\frac{1}{A}-\frac{dy}{I})}=\frac{5ksi}{(\frac{1}{7.5i{n}^2}-\frac{(3.2in)(-1.7in)}{10.825i{n}^4})}=7.86kips$

• Largest upward force:

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

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

• The limiting factor is at 7.86 kips force upward.

• Apply negative sign throughout equation:

${\sigma }_{max}=\frac{-P_{max}}{A}-\frac{-my}{I}⟹P_{max}=\frac{-{\sigma }_{all/max}}{(\frac{1}{A}-\frac{dy}{I})}$

• The Downward force becomes:

${\sigma }_{all}=+5ksi,A=7.5i{n}^2,I=10.825i{n}^4,y=2.3in,d=3.2in$

$⟹P_{max\downarrow }=9.15kips$

• The Upward Force becomes:

${\sigma }_{max}=-12ksi,A=7.5i{n}^2,I=10.825i{n}^4,y=2.3in,d=3.2in$

$⟹P_{max\uparrow }=18.87kips$

$\therefore$ Limit is at $9.15kips$

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R6.5: Problem 4.115

Figure P4.114 and P4.115.

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 ${\sigma }_{all}=+5ksi$ 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

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Figure P4.114 and P4.115.

3D Representation of Figure 4.115

• Max allowable stresses on the hanger:

${\sigma }_{allow}=+5ksi$ $F_{max\downarrow }=?$

${\sigma }_{allow}=-12ksi$ $F_{max\uparrow }=?$

• Find centroid:

• Take $\bar{y}$ as a distance measured from left end of shape.

$\bar{Y}=\frac{\sum \bar{y}A}{\sum A}$

$A_1=(3in)(1in)=3i{n}^2$

$A_2=(3in)(.75in)=2.25i{n}^2$

• Due to same parameters,

$A_3=A_2=2.25i{n}^2$

$\overline{y_1}=\frac{1}{2}in=.5in$

$\overline{y_2}=1in+\frac{3}{2}in=2.5in$

• Due to same parameters,

$\overline{y_3}=\overline{y_2}=2.5in$

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

• Must incorporate the parallel axis theorem to find moment of inertia: $I=\frac{b{h}^3}{12}+A{d}^2$

b = base, h = height, A = area, and d = perpendicular distance between centroidal axis and parallel axis

$I_1=\frac{b_1{h}_1^3}{12}+A_1{d}_1^{2}where{d}_1=(\overline{y_1}-\bar{Y})⟹I_1=\frac{(3in)(1in{)}^3}{12}+(3i{n}^2)(.5in-1.7in{)}^2=4.57i{n}^4$

$I_2=\frac{b_2{h}_2^3}{12}+A_2{d}_2^{2}where{d}_2=(\overline{y_2}-\bar{Y})⟹I_1=\frac{(.75in)(3in{)}^3}{12}+(2.25i{n}^2)(2.5in-1.7in{)}^2=3.1275i{n}^4$

• Due to same parameters,

$I_3=I_2=3.1275i{n}^4$

$\therefore$ Total moment of inertia is:$I_{Tot}=I_1+I_2+I_3=4.57i{n}^4+3.1275i{n}^4+3.1275i{n}^4⟹10.825i{n}^4$

• The normal stress at point A is due to bending:

${\sigma }_{bending}=-\frac{My}{I}$

• "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(*)):

$M=PdP=F_{max}=maximumforced=$distance of the force from the centroid of the cross section. (EQ 4.49 page 270 (*))

Normal stress due to centric load: ${\sigma }_{centric}=\frac{P}{A}$

• Combine:

$\sigma ={\sigma }_{centric}+{\sigma }_{bending}=\frac{P}{A}-\frac{My}{I}$ (EQ4.50, p.221(*))

$\uparrow =$Total normal stress acting at point A.

• Largest downward force:

• Assuming conventions:${\sigma }_{max}=+5ksi,A=7.5i{n}^2,I=10.825i{n}^4,y=+2.3in,d$ = distance of acting force from the centroid $=3.2in$

${\sigma }_{max}=\frac{P_{max}}{A}-\frac{My}{I}=\frac{P_{max}}{A}-\frac{P_{max}dy}{I}=P_{max}(\frac{1}{A}-\frac{dy}{I})$

$⟹P_{max\downarrow }=\frac{{\sigma }_{max}}{(\frac{1}{A}-\frac{dy}{I})}=\frac{5ksi}{(\frac{1}{7.5i{n}^2}-\frac{(3.8in)(-2.3in)}{10.825i{n}^4})}=6.15kips$

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

${\sigma }_{max}=\frac{-P_{max}}{A}-\frac{-my}{I}⟹P_{max}=\frac{-{\sigma }_{all/max}}{(\frac{1}{A}-\frac{dy}{I})}$

${\sigma }_{all}=+5ksi,A=7.5i{n}^2,I=10.825i{n}^4,y=-1.7in,d=3.2in$

$P_{max\uparrow }=\frac{{\sigma }_{max}}{(\frac{1}{A}-\frac{dy}{I})}=13.54kips$

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

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