University of Florida/Egm3520/s13.team1.r5

Team 1: Report 5

R5.1: Problem 4.7

Contents taken from Page 238 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 238 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Two W4 x 13 rolled sections are welded together as shown. Knowing that for the steel alloy used, $\sigma _{\gamma }=36ksi$ and $\sigma _{\upsilon }=58ksi$ and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis.

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

R5.1 Solution

**Appendix C:Page A18**
Term	Desig	Value
Area	A	3.83 in²
Depth	d	4.16 in
Width	w	4.06 in
Moment of Inertia_x	I_x	11.3 in⁴
Moment of Inertia_y	I_y	3.86 in⁴

**Problem Given**
Yield Strength	$\sigma _{\gamma }=36ksi$
Ultimate Stress	$\sigma _{\upsilon }=58ksi$
Factor of Safety	3

Determine Allowable Stress: $\sigma _{all}={\frac {\sigma _{\upsilon }}{F.S.}}={\frac {58ksi}{3}}=19.33ksi$

Use Parallel Axis Theorem to determine the Polar Moment of Inertia: ${\color {blue}I_{z}=\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]+\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]}\implies {\color {blue}2\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]}$

\implies 2\left[11.3in^{4}+\left(3.83in^{2}\right)\left(2.08in\right)^{2}\right]=55.74in^{4}

Determine the Couple Moment: ${\color {blue}\sigma _{all}={\frac {Mc}{I}}}\implies {\color {blue}M={\frac {\left(\sigma _{all}\right)\left(I\right)}{c}}}$

\implies M={\frac {\left(19.33ksi\right)\left(55.74in^{4}\right)}{4.16in}}={\color {red}259kip*in}

R5.2: Problem 4.8

Egm 3520.s13.team1.wcs (discuss • contribs) 07:27, 27 March 2013 (UTC)

Contents taken from Page 238 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 238 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

Two W4 x 13 rolled sections are welded together as shown. Knowing that for the steel alloy used, $\sigma _{\gamma }=36ksi$ and $\sigma _{\upsilon }=58ksi$ and using a factor of safety of 3.0, determine the largest couple that can be applied when the assembly is bent about the z axis.

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

R5.2 Solution

**Appendix C:Page A18**
Term	Desig	Value
Area	A	3.83 in²
Depth	d	4.16 in
Width	w	4.06 in
Moment of Inertia_x	I_x	11.3 in⁴
Moment of Inertia_y	I_y	3.86 in⁴

**Problem Given**
Yield Strength	$\sigma _{\gamma }=36ksi$
Ultimate Stress	$\sigma _{\upsilon }=58ksi$
Factor of Safety	3

Determine Allowable Stress: $\sigma _{all}={\frac {\sigma _{\upsilon }}{F.S.}}={\frac {58ksi}{3}}=19.33ksi$

Use Parallel Axis Theorem to determine the Polar Moment of Inertia: ${\color {blue}I_{z}=\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]+\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]}\implies {\color {blue}2\left[I_{x}+A\left({\frac {d}{2}}\right)^{2}\right]}$

\implies 2\left[3.86in^{4}+\left(3.83in^{2}\right)\left(2.03in\right)^{2}\right]=39.29in^{4}

Determine the Couple Moment: ${\color {blue}\sigma _{all}={\frac {Mc}{I}}}\implies {\color {blue}M={\frac {\left(\sigma _{all}\right)\left(I\right)}{c}}}$

\implies M={\frac {\left(19.33ksi\right)\left(39.29in^{4}\right)}{4.06in}}={\color {red}187.1kip*in}

R5.3: Problem 4.13

Contents taken from Page 238 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 238 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 a beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN*m, determine the total force acting on the shaded portion of the web.

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

R5.3 Solution

The first step of the problem is to split the shaded portions into two separate portions.

Calculate each moment of inertia about the horizontal axis at the centroid.

Thomas Burley

For section A the moment of inertia is :

I_{XA}={\frac {1}{12}}bh^{3}+Ad^{2}

I_{XA}={\frac {1}{12}}*216*36^{3}+(216*36)*36^{2}

I_{XA}=10.9*10^{6}mm^{4}=10.9*10^{-6}m^{4}

For section B the moment of inertia is :

I_{XB}={\frac {1}{12}}bh^{3}+Ad^{2}

I_{XB}={\frac {1}{12}}*76*90^{3}+(72*90)*36^{2}

I_{XB}=17.5*10^{6}mm^{4}=17.5*10^{-6}m^{4}

Next add both moments of inertia to calculate the total moment of inertia

I_{XTotal}=I_{XA}+I_{XB}

I_{XTotal}=28.4*10^{-6}m^{4}

Next we need to calculate the stress using the pure bending equation

$\sigma ={\frac {My}{I_{XTotal}}}={\frac {6*10^{3}y}{28.4*10^{-6}}}=0.211*10^{9}y$

The final step is to calculate the force on the beam's cross section. We know that force is equal to stress on an area.

$F=\int _{y_{1}}^{y_{2}}\sigma bdy$

Now input the values for the cross section of the beam into the equation

$y_{1}=0,y_{2}=0.09m,b=0.072$

$F=\int _{0}^{0.09}0.211*10^{9}*0.072*y*dy$

$F=7.59*10^{6}(0.09^{2}-0)=61.5kN$

R5.4: Problem 4.16

Contents taken from Page 239 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 "T" shapes beam that is made of nylon, the allowable stress is 24 MPa in tension and 30 MPa in compression. The largest couple M that can be applied to the beam is?

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

R5.4 Solution

1) $\displaystyle A=600mm^{2}$

\displaystyle y=7.5mm

\displaystyle A*y=4500mm^{3}

2) $\displaystyle A=500mm^{2}$

\displaystyle y=20mm

\displaystyle A*y=10000mm^{3}

1+2) A = 1100 mm^2 A*y = 14500 mm^3 y = (14500)/(1100) = 13.18mm

I = (1/12)*b* h^3 + A * d^3

I_1 = (1/12) (40) * (15^3) + 600*(20^2) = 251250 mm^4

I_2 = (1/12) (20)* (25^3) + 500*(10^2) = 76041.66 mm^4

I = I_1 + I_2 = 327291.67 mm^4 = 3.27 *10^-7 m^4

M = σ* I/y

Top M_1 = (24 * 10^6)(3.27*10^-7) / 0.02682 = 292.617 N-m

Bottom M_2 = (30*10^6) (3.27 * 10^-7)/0.01318 = 744.973 N-m

the smaller value is correct couple... so M_1

R5.5: Problem 4.20

Contents taken from Page 239 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 239 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

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

The cantilever beam with the trapezoidal cross-section has an allowable tension stress threshold, $\displaystyle \sigma _{T}$ and allowable compressive stress, $\displaystyle \sigma _{C}$ .

R5.5 Solution

What is largest couple moment $\displaystyle M$ that can be applied?

Given(s):

\displaystyle \sigma _{T}=120\,MPa

\displaystyle \sigma _{C}=-150\,MPa

Relation between cross-sectional area, stress, and moment

\displaystyle M_{max}={\frac {\sigma _{m}I}{c}}

Determine position of centroid in y direction

\displaystyle h

: height

\displaystyle {\bar {y}}={\frac {1}{\sum {A_{i}}}}\int \mathrm {y\,\mathrm {d} } A_{i}

\displaystyle \sum {A_{i}}={\frac {h(b_{1}+b_{2})}{2}}=3240\,mm^{2}

Using composite area method

\displaystyle {\bar {y_{1}}}={\bar {y_{3}}}

\displaystyle A_{1}=A_{3}

\displaystyle {\bar {y}}={\frac {2{\bar {y_{1}}}A_{1}+{\bar {y_{2}}}A_{2}}{\sum {A_{i}}}}

\displaystyle {\color {blue}Area\,1:}

By similar triangles

\displaystyle u={\frac {by}{h}}

\displaystyle \mathrm {d} A=u\mathrm {d} y={\frac {by}{h}}\mathrm {d} y

\displaystyle {\bar {y_{1}}}={\frac {b}{A_{1}h}}\int _{0}^{h}\,\mathrm {y^{2}\,\mathrm {d} } y={\frac {bh^{2}}{3A_{1}}}=36\,mm

\displaystyle {\color {blue}Area\,2:}

Because of symmetry

\displaystyle {\bar {y_{2}}}={\frac {h}{2}}=27\,mm

Therefore, $\displaystyle {\bar {y}}=30\,mm$ and $\displaystyle c\,=30\,mm$

Determine 2nd moment of inertia

\displaystyle I\,=\,\int {\mathrm {y^{2}\,\mathrm {d} } A}\,=\,2\int {\mathrm {y^{2}\,\mathrm {d} } A_{1}}+\int {\mathrm {y^{2}\,\mathrm {d} } A_{2}}\,=\,{\color {Red}2Y+Z}

\displaystyle {\color {Red}Y}\,=\,\int {\mathrm {y^{2}\,\mathrm {d} } A_{1}}\,=\,{\frac {b}{h}}\int _{0}^{h}{\mathrm {y^{3}\,\mathrm {d} } y}\,=\,{\frac {bh^{3}}{4}}

\displaystyle {\color {Red}Z}\,=\,\int {\mathrm {y^{2}\,\mathrm {d} } A_{2}}\,=\,b\int _{0}^{h}{\mathrm {y^{2}\,\mathrm {d} } y}\,=\,{\frac {bh^{3}}{3}}

\displaystyle I=2({\frac {bh^{3}}{4}})+{\frac {bh^{3}}{3}}=3674160\,mm^{4}

Find limiting constraint to the maximum couple moment, $\displaystyle \sigma _{T}$ or $\displaystyle \sigma _{C}$

\displaystyle M_{max}={\frac {\sigma _{T}I}{c}}={\color {blue}-1469.664\,Nm}

\displaystyle M_{max}={\frac {\sigma _{C}I}{c}}={1837.08\,Nm}

R5.6: Problem 3.53

Contents taken from Page 174 in the Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664.

Rod AB is made of aluminum, which has a Young's modulus of 3,700,000 pounds per square inch. Rod BC is made of brass, which has a Young's modulus of 5,600,000 pounds per square inch.

A torque T equal to 12,500 pound inches is applied at point B along the axis AC.

The figure shows that AB is 12 inches in length and 1.5 inches in diameter, and that BC is 18 inches long and 2 inches in diameter.

Find a) The max shear stress in rod AB and b) the max shear stress BC.

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

R5.6 solution

Formulas used: $R={\frac {D}{2}}$

$\tau ={\frac {TL}{JG}}$

$J={\frac {\pi }{2}}R^{4}$

$\tau _{max}={\frac {TR}{J}}$

Applied torque T is equal to the sum of the reaction forces at points A and C:

$T=T_{AB}+T_{BC}=12.5kip*in$

Total rod twist angle is equal to the sum of section AB and BC's twist angles; total twist angle is equal to zero because points A and C are held stationary:

$\phi =\phi _{AB}+\phi _{BC}=0$

Insert formula to relate rod twist to variables T, L, J, and G for their respective sections:

$\phi =\phi _{AB}+\phi _{BC}={\frac {T_{AB}*L_{AB}}{J_{AB}*G_{AB}}}-{\frac {T_{BC}*L_{BC}}{J_{BC}*G_{BC}}}=0$

Rearrange, insert numerical values, and simplify to solve for T in section BC relative to A:

$T_{BC}={\frac {T_{AB}*L_{AB}}{J_{AB}*G_{AB}}}*{\frac {J_{BC}*G_{BC}}{L_{BC}}}$

$T_{BC}=T_{AB}*{\frac {12}{.497*3.7E6}}*{\frac {1.57*5.6E6}{18}}$

$T_{BC}=T_{AB}*3.19$

Solve for $T_{AB}$ relative to applied torque $T$ , solve for numerical values of $T_{AB}$ and $T_{BC}$

$T=T_{AB}+T_{AB}*3.19=12.5kip*in$

$T=T_{AB}*4.19=12.5kip*in$

$T_{AB}={\frac {12.5kip*in}{4.19}}$

$T_{AB}=3kip*in\implies T_{BC}=9.5kip*in$

Find max shear stress in each section of rod using their respective torques:

$\tau _{ABmax}={\frac {T_{AB}*{\frac {D_{AB}}{2}}}{J_{AB}}}={\frac {3*.75}{.497}}=4.53E3psi$

$\tau _{BCmax}={\frac {T_{BC}*{\frac {D_{BC}}{2}}}{J_{BC}}}={\frac {9.5*1}{1.57}}=6.05E3psi$

Egm3520.s13.team1.scheppegrell.jas (discuss • contribs) 19:18, 27 March 2013 (UTC)