University of Florida/Egm3520/s13.team1.r2

Report 2

Problem R2.1

Egm3520.s13.team1.stevenchiu (discuss • contribs) 15:28, 6 February 2013 (UTC)

Egm 3520.s13.team1.wcs (discuss • contribs) 15:52, 6 February 2013 (UTC)

Pb-9.1 sec.9 p.9-8. Contents taken from the notes of Dr. Loc Vu-Quoc

Lecture Notes

Text can be viewed here

Problem Statement

Question 1

Solve for the reactions at B and at A in the Example 2.04 (see p.9-1) with the stress-strain relation (1) p.9-5.

Diagram

Solution 1

Figure from Lecture Notes via textbook Mechanics of Materials - Authors: Beer, Johnston, DeWolf, Mazurek

Given
Applied Forces	Area	Length	Young's Modulus
$F_{D}=300kN$	$A_{CK}=A_{KB}=A_{CB}=400mm^{2}={\color {red}A_{1}}$	$L_{AD}=L_{DC}=L_{CK}=L_{KB}=L=150mm$	$E_{AD}=E_{DC}=E_{CK}=E_{KB}=E$
$F_{K}=600kN$	$A_{AD}=A_{DC}=A_{AC}=250mm^{2}={\color {red}A_{2}}$

Non-Linear Stress-Strain Relation

\sigma =E\epsilon ^{1/2}

\implies \delta =\epsilon L={\frac {\sigma ^{2}}{E^{2}}}L={\frac {P^{2}L}{A^{2}E^{2}}}

Free Body Diagrams

FBD from Section 9 Lecture notes. Author Professor Loc Vu-Quoc

FBD 3:

\sum F=-P_{CK}-F_{K}+R_{B}=0

\implies P_{CK}={\color {red}P_{1}}=R_{B}-F_{K}

FBD 4:

\sum F=-P_{DC}-F_{K}=0

\implies P_{DC}={\color {red}P_{2}}=R_{B}-F_{K}

FBD 5:

\sum F=-P_{AD}-F_{D}-F_{K}+R_{B}=0

\implies P_{AD}={\color {red}P_{3}}=R_{B}-F_{D}-F_{K}

Deformation Equations

Left

\delta _{L}=\sum {\frac {P_{i}^{2}L_{i}}{A_{i}^{2}E_{i}^{2}}}=\left(0+{\frac {P_{1}^{2}L}{A_{1}^{2}E^{2}}}+{\frac {P_{2}^{2}L}{A_{2}^{2}E^{2}}}+{\frac {P_{3}^{2}L}{A_{2}^{2}E^{2}}}\right)

Sub:

{\color {blue}P_{1}=P_{2}}

Right

\delta _{R}={\frac {P_{1}^{2}L}{A_{1}^{2}E^{2}}}+{\frac {P_{2}^{2}L}{A_{1}^{2}E^{2}}}

Sub:

{\color {blue}P_{1}=P_{2}=-R_{B}}

Total

\delta _{\text{Total}}=\delta _{L}+\delta _{R}={\frac {P_{1}^{2}L}{A_{1}^{2}E^{2}}}+{\frac {P_{2}^{2}L}{A_{2}^{2}E^{2}}}+{\frac {P_{3}^{2}L}{A_{2}^{2}E^{2}}}-{\frac {R_{B}^{2}L}{A_{1}^{2}E^{2}}}-{\frac {R_{B}^{2}L}{A_{2}^{2}E^{2}}}=0

ALL E's and L's cancel out

$\implies {\frac {P_{1}^{2}}{A_{1}^{2}}}+{\frac {P_{1}^{2}}{A_{1}^{2}}}+{\frac {P_{3}^{2}}{A_{2}^{2}}}-{\frac {R_{B}^{2}}{A_{1}^{2}}}-{\frac {R_{B}^{2}}{A_{2}^{2}}}=0$

$\implies -{\frac {1}{A_{1}^{2}}}R_{B}^{2}+{\frac {1}{A_{1}^{2}}}(1.2*10^{6})R_{B}+{\frac {(6*10^{6})^{2}}{A_{1}^{2}}}-{\frac {1}{A_{2}^{2}}}R_{B}^{2}+{\frac {1}{A_{2}^{2}}}(1.2*10^{6})R_{B}+{\frac {(9*10^{6})^{2}}{A_{2}^{2}}}-{\frac {1}{A_{2}^{2}}}R_{B}^{2}+{\frac {1}{A_{2}^{2}}}(1.8*10^{6})R_{B}+{\frac {(9*10^{6})^{2}}{A_{2}^{2}}}-{\frac {1}{A_{1}^{2}}}R_{B}^{2}-{\frac {1}{A_{2}^{2}}}R_{B}^{2}$

$\implies \left(-{\frac {2}{A_{1}^{2}}}-{\frac {3}{A_{2}^{2}}}\right)R_{B}^{2}+(4*10^{6})\left({\frac {1}{A_{2}^{2}}}+{\frac {2}{A_{2}^{2}}}\right)R_{B}+{\frac {(6*10^{6})^{2}}{A_{1}^{2}}}+{\frac {(9*10^{6})^{2}}{A_{2}^{2}}}+{(9*10^{6})^{2}}{A_{2}^{2}}$

$\implies -6.05*10^{7}R_{B}^{2}+1.6065*10^{14}R_{B}+2.097*10^{21}$

$R_{B}\approx {\color {blue}200kN}$

\sum F=R_{A}-F_{D}-F_{K}+R_{B}=0

\implies R_{A}=300+600-200=0

\implies R_{A}={\color {blue}700kN}

Question 2

Do the results depend on the length of each segment and the Young's modulus?

Solution 2

While working to determine $\delta _{Total}$ , which was an equilibrium equation, the resultants were found not to be dependent on segment length or the Young's modulus because both were constants and were subsequently cancelled out. Refer to process above.

Problem R2.2

Problem Number 2.12, p.73. Contents taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Problem Statement

A nylon thread is to be subjected to a 10-N tension. Knowing that E=3.2 GPa,that the maximum allowable normal stress is 40 MPa, and the length of the thread must not increase by more than 1%, determine the required diameter thread.

Solution

Determine maximum stress allowable with the limit in displacement being $\displaystyle \delta ={\frac {L}{100}}$ :

$\displaystyle \sigma =E{\frac {\epsilon }{L}}=32MPa$

Derive formula relating area of nylon string with ratio of displacement and original length of the string using these three relations:

$\displaystyle \epsilon ={\frac {\delta }{L}}$

$\displaystyle \sigma =E\epsilon$

$\displaystyle \sigma ={\frac {P}{A}}$

Write Area in terms of diameter:

$\displaystyle \sigma ={\frac {P}{A}}={\frac {P}{\pi ({\frac {d}{2}})^{2}}}$

Solve equation for diameter (d):

$\displaystyle d={\sqrt {\frac {4P}{\pi \sigma }}}=.631mm$

Problem R2.3

Problem Number 2.16, p.74. Contents taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Problem Statement

The brass tube AB ( $E=105{\text{GPa}}$ ) has a cross-sectional area of $140{\text{mm}}^{2}$ and is fitted with a plug at A. The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder (E=72 GPa) with a cross-sectional area of 250mm^2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1 mm. Determine the force P that must be applied to the cylinder.

Figure 2.16 from Mechanics of Materials textbook 6th edition Authors Beer, Johnston, DeWolf, and Mazurek

Solution

Given(s):

$\displaystyle E_{AB}=105GPa\geq 105*10^{9}Pa$

$\displaystyle X_{AB}=104mm^{2}\geq 1.4*10^{-4}m$

$\displaystyle E_{C}D=72GPa\geq 72*10^{9}Pa$

$\displaystyle X_{C}D=250mm^{2}\geq 2.5*10^{-4}m$

Required

Must be moved down 1 mm to find P (force or load required).

Assumptions

$\delta ={\frac {PL}{EA}}$

To Brass Tube AB

$L_{AB}=375{\text{mm}}+1{\text{mm}}=376{\text{mm}}$

$\delta _{AB}={\frac {PL_{AB}}{E_{AB}A_{AB}}}\implies {\frac {P(0.376m)}{(105*10^{9}Pa)(1.40*10^{-4}m^{2})}}=2.558*10^{-8}P$

To Aluminum cylinder CD

$\delta _{CD}={\frac {PL_{CD}}{E_{CD}A_{CD}}}\implies {\frac {P(0.375)}{(72*10^{9}Pa)(2.5*10^{-4}m^{2})}}=2.0833*10^{-8}Pa$

Total Deflection

$\delta _{A}=\delta _{AB}+\delta _{CD}\implies 1*10^{-3}m=(2.558*10^{-8}+2.0833*10^{-8})P$

$P={\frac {1*10^{-3}m}{2.558*10^{-8}+2.0833*10^{-8}}}=21546.35N\implies 21.5kN$

Problem R2.4

Problem Number 2.24, p.75. Contents taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Problem Statement

For the steel truss (E=29*10^6 psi) and loading shown, determine the deformations of members BD and DE, knowin that their cross-sectional areas are 2 in^2 and 3 in^2, respectively.

Solution

Done by method of sections. Cut off the top half of the truss by cutting midway through sections BD, DE and EG.

It is assumed that the structure is in equilibrium.

First

$\displaystyle \sum M_{E}=(-30kips)(0)+(-30kips)(8)+(-30kips)(16)+F_{BD}*15=0\implies$

$F_{BD}=48kips$

Then

$\displaystyle \delta _{BD}={\frac {F_{BD}L_{BD}}{A_{BD}E_{BD}}}={\frac {48kips*96in}{2in*(29E-6)psi}}\implies$

$\delta _{BD}=0.0794in$

$\displaystyle \delta _{DE}={\frac {F_{DE}L_{DE}}{A_{DE}*E_{DE}}}={\frac {60kips*180in}{3in*(29E-6)psi}}\implies$

$\delta _{DE}=0.124in$

Problem R2.5

Problem Number 2.40, p.89. Contents taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Problem Statement

A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at both ends and supports and supports two 6-kip loads as shown. Knowing that E=0.45*10^6 psi, determine (a) the reactions at A and C, (b) the normal stress in each portion of the rod.

Solution

We begin by first sketching the free body diagram of the system From this diagram we can see that:

$\displaystyle R_{A}+R_{B}=12kips$

(1)

Next, we consider the deformation of the rod to be zero. Using equation 2.15 from the text book we see:

$\displaystyle \delta ={\frac {P_{AB}L_{AB}}{A_{AB}E}}+{\frac {P_{BC}L_{BC}}{A_{BCE}}}=0$

(Text 2.15)

Having been given values for all variables in the equation except P we must solve for those two forces. Substituting into the original equation we find:

$\displaystyle {\frac {R_{A}L_{AB}}{A_{AB}}}-{\frac {R_{C}L_{BC}}{A_{BC}}}=0$

Calculating the areas gives us:

$\displaystyle A_{AB}=\pi (0.625)^{2}=1.22in^{2}$

$\displaystyle A_{BC}=\pi (1)^{2}=3.14in^{2}$

$\displaystyle R_{C}={\frac {L_{AB}A_{BC}}{L_{BC}A_{AB}}}*R_{A}=4.29R_{A}$

(2)

Using the equations (1) and (2) we can solve for $R_{A}$ and $R_{B}$

$R_{A}=2.3kips$

$R_{B}=9.7kips$

To find the normal stress we simply use formula 1.5 from the first chapter of the text:

$\displaystyle \sigma ={\frac {P}{A}}$

(Text 1.5)

$\displaystyle \sigma _{AB}={\frac {R_{A}}{A_{AB}}}=1.88ksi$

$\displaystyle \sigma _{BC}={\frac {-R_{C}}{A_{BC}}}=-3.09ksi$

Problem R2.6

Problem Number 2.44, p.90. Contents taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664 Egm3520.s13.team1.scheppegrell.jas (discuss • contribs) 16:57, 6 February 2013 (UTC)

Problem Statement

Figure from Lecture Notes via textbook Mechanics of Materials - Authors: Beer, Johnston, DeWolf, Mazurek

The rigid bar AD is supported by two steel wires of 1/16in diameter (E=29*10^6 psi) and a pin and bracket at D. Knowing that the wires were initially taut, determine (a) the additional tension in each wire when a 120-lb load P is P is applied at B, (b) the corresponding deflection of point B.

Solution

Given
Applied Force	Area	Length	Young's Modulus
P=120lb	$A_{AE}=A_{CF}={\frac {\pi }{4}}{\frac {1}{256}}in^{2}=A$	$L_{AE}=15in=L_{A}$	$E_{AE}=E_{CF}=29*10^{6}psi$
		$L_{CF}=8in=L_{C}$

$\delta _{i}={\frac {P_{i}L_{i}}{AE}}$

$\delta _{A}={\frac {P_{A}15}{{\frac {\pi }{4}}{\frac {1}{256}}(29*10^{6})}}$

$\delta _{C}={\frac {P_{C}8}{{\frac {\pi }{4}}{\frac {1}{256}}(29*10^{6})}}$

$\delta _{A}={\frac {3}{2}}\delta _{B}$

$\delta _{C}={\frac {1}{2}}\delta _{B}$

${\frac {2}{3}}{\frac {P_{A}15}{{\frac {\pi }{4}}{\frac {1}{256}}(29*10^{6})}}=2{\frac {P_{C}8}{{\frac {\pi }{4}}{\frac {1}{256}}(29*10^{6})}}$

${\frac {2}{3}}P_{A}15=2P_{C}8$

$P_{A}={\frac {8}{5}}P_{C}$

${\frac {8}{5}}P_{C}{\frac {3}{2}}+P_{C}{\frac {1}{2}}=120$

$P_{C}{\frac {29}{10}}=120$

$P_{C}={\frac {1200}{29}}\implies P_{A}={\frac {1920}{29}}$

Solution (a)

$Tension_{C}=41.4lb$

$Tension_{A}=66.2lb$

Solution (b)

$\delta _{C}={\frac {8*41.4}{{\frac {\pi }{4*256}}29*10^{6}}}=3.72*10^{-3}$

$\delta _{B}=\delta _{B}*2=3.72*10^{-3})*2=7.44*10^{-3}inches$

Deflection of point B is approximately $7.44*10^{-3}inches$

Nomenclature

m = meter

mm = millimeter

d = diameter

G = giga

Pa = Pascal

A = area

Q/P = load

N = Newton

δ = deflection

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