University of Florida/Egm3520/s13.team1.r1

Problem R1.1
Pb-1.1 sec.1 p.1-10. Contents taken from the notes of Dr. Loc Vu-Quoc Text can be viewed here

Problem Statement
The textbook by Beer et. al. 2012 as listed on amazon.com has dimensions 8.2 in x 1.2 in x9.9 in, and weighs 3.6 lbs. Consider a bookshelf lined up with the above books, from support to support. the bookshelf, made of Ponderosa Pine (p.1-4), has a rectangular cross section (p.1-5):

Given: L= 100 in, b=9 in, h=0.5 in

1) Find the vertical mid-span deflection v under the weight of the books and the shelf itself.

2) Increase the shelf thickness to 1 in, find the vertical mid-span deflection.

3) Repeat part 1 and 2 with the shelf made from structural steel ASTM-A36.

4) Next, consider reinforcing the bookshelf with 2 side strips so to have the H cross section (p.1-6), with the following geometry:

Solution
1)	Using equation (1) from section 1 notes.
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$$ \displaystyle \nu = \frac{5qL^4}{384EI} $$   (1)
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Given(s):
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$$ q=\frac{3.6 lb}{100 in}=0.036\frac{lb}{in} $$
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$$ L=100 in $$
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$$ \displaystyle I=\left(\frac{bh^3}{12}\right)=\left(\frac{9in*0.5in)^3}{12} \right) =0.09375in^{4} $$
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$$ \displaystyle E_{pine}=1.3*10^{6} psi $$
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Plugging these values into the equation yields
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$$ \displaystyle \nu=0.385 in $$
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2) Increasing the thickness to 1in will change the moment of inertia
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$$ \displaystyle I=\left(\frac{bh^3}{12}\right)=\left(\frac{9in*(1in)^3}{12}\right)=0.75in^{4} $$ By plugging in this new value of I into equation (1) we get:
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$$ \nu=0.48in $$
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3) This section ask to use the Young's modulus for ASTM A-36 steel instead of the Young's modulus for Pine. Replacing:
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$$ \displaystyle E_{pine}=1.3*10^6 psi $$ with $$ \displaystyle E_{ASTM A-36 Steel}=29*10^6 psi $$ then plug this new E value into equation (1)
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$$ \displaystyle Thickness=0.5in $$ $$ \nu=0.0172 in $$
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$$ \displaystyle Thickness=1.0 in $$ \nu=0.00216 in $$
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4) Converting the original bookshelf into an H-cross section changes the moment of inertia.
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$$ \displaystyle I_{x_1}+A^{2} dy=\frac{bh^3}{12}+A^{2} dy=\frac{9in*(0.5in)^3}{12}+(9*0.5)^{2} (0)=0.09375in^4 $$
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$$ \displaystyle I_{x_2}+A^{2} dy=\frac{bh^3}{12}+A^{2} dy=\frac{9in*(0.3in)^3}{12}+(9*0.3)^{2} (1)=7.31025 in^4 $$
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Therefore the moment of inertia is equal to the sum of
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$$ \displaystyle I_{x_1}+I_{x_2}=7.404 in^4 $$
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Plugging in this new value for the moment of inertia into equation one yields:
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$$ \displaystyle \nu_{pine}=0.00487in $$
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$$ \displaystyle \nu_{steel}=2.18*10^{-4} in $$
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Problem R1.2
Problem Number 1.5, p.20. 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
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 200 MPa in the bolts and 130 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.

Figure displayed here

Solution
We can infer that "the most economical and safe design" is one in which the bolts and spacers exceed their maximum average normal stress at the same time; weaker spacers would weaken the design, and stronger spacers would add cost without resulting in a stronger design.

The normal stress exerted on each spacer is equal in magnitude to the force on each bolt. Max stress divided by bolt cross sectional area equals 200 MPa, and max stress divided by spacer cross sectional area equals 130 MPa. Therefore, we can determine that the cross sectional areas of the spacers and bolts on the plane normal to the stress vector have a ratio of 200 to 130, or 20/13, respectively.

The bolt's cross section has area $$ \pi(8mm)^2 $$ . The spacer's cross section with respect to OD (outer diameter) is


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$$ \displaystyle \pi\left(\frac{OD}{2}\right)^{2}-\pi(8mm)^{2} $$
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$$ \displaystyle \frac{\pi\left(\frac{OD}{2}\right)^{2}-\pi(8mm)^2)}{\pi(8mm)^2}=\frac{20}{13}\implies OD=2\sqrt{\frac{\frac{20}{13}\pi(8mm)^{2}+\pi(8mm)^{2}}{\pi}} $$ $$ OD=25.49mm $$
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Problem R1.3
Problem Number 1.11 page 22. 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 frame shown consists of four wooden members, ABC, DEF, BE, and CF. Knowing that each member has a 2 3 4-in. rectangular cross section and that each pin has a 12-in. diameter, determine the maximum value of the average normal stress (a) in member BE, (b) in member CF.

Diagram available here

Solution
The first step in the process is to create a free body diagram of the pieces of the structure

Free Body Diagram

Now that we have 4 unknowns we much try to solve for as many of them as possible. Since this is a static problem we know that the moment about any point will equal zero.


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$$ \displaystyle \sum M_A =0
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40in*D_x-(45+30)in*480lb=0

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$$ \displaystyle 40in*D_x-(45+30)in*480lb=0 $$
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$$ \displaystyle D_x=900lb $$
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Next we must separate the components of the system. Since we already solved for Dx, using the bottom member will be the best choice to eliminate more variables.

Bottom Member FBD

Force D must be parallel to FBE and FCF


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$$ \displaystyle D_y=\frac{4}{3} D_x=1200lb $$
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From that we chose points E and F and take the moment at each


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$$ \displaystyle \sum M_E = 0 \rightarrow 30 * \frac{4}{3} F_{CF}-15D_y $$
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$$ \displaystyle F_{CF} = 750lb $$
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$$ \displaystyle \sum M_F = 0 = -30 * \frac{4}{3} F_{BE}-(30+15)D_y $$
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$$ \displaystyle F_{BE} = -2250lb $$
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Now that each of the forces on the bottom member we can calculate stresses. We are looking for maximum value of the stress. Since the force is equal all along the member the maximum occurs where the area is the smallest. This location is where the pin is in the member.


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$$ \displaystyle A_{member} = 2in * 4in = 8in^2 $$
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$$ \displaystyle A_{pin} = 2in * (4 - 0.5)in = 7in^2 $$
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$$ \displaystyle \sigma_{BE}=\frac{F_{BE}}{A}= \frac{2250lb}{7in^{2}}=321.psi $$
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$$ \displaystyle \sigma_{CF}=\frac{F_{CF}}{A}= \frac{750lb}{7in^{2}}=107psi $$
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Problem R1.4
Problem Number 1.20, p.24. 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 axial force in the column supporting the timber beam shown is P 5 20 kips. Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 400 psi.

[http://s13.postimage.org/c05qkxs47/1_20.png Diagram available here. ]

Solution
Using equation 1.5 from the textbook we know the relationship between normal stress, the magnitude of the load and the cross sectional area.
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$$ \sigma = \frac{P}{A} = \frac{P}{LW} $$ σ = normal stress, P = magnitude of the load, A =  cross sectional area
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Rearranging the formula we are able to determine the minimum length required for the bearing stress not to exceed 400 psi


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$$ \frac{P}{LW} < 400psi $$
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Reorganizing the formula allows us to solve directly for L


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$$ L > \frac{P}{W\sigma} = \frac{20kips}{(6in)(400psi)} $$
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Given that 1kip = 1000lbs and l psi = 1 lb/in2


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$$ L > \frac{20000lb}{(6in)(\frac{400lb}{in^2})} $$
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$$ L > 8.33in $$ The smallest possible length for the given stress is 8.33inches
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Problem R1.5
Problem Number 1.33, p.36. Contents, including images, 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 steel pipe of 12in outer diameter is fabricated from 1/4in thick plate by welding along a helix that forms an angle of 25° with a plane perpendicular to the axis of the pipe. Knowing that the maximum allowable normal and shearing stresses in the directions respectively normal and tangential to the weld are σ = 12ksi and Ί = 7.2ksi, determine the magnitude P of the largest axial force that can be applied to the pipe.

Diagram

Solution:
Find cross sectional area of pipe parallel with weld $$ (A_{\theta}) $$ <ul> $$A_{\circ}$$: cross sectional area perpendicular to axis $$A_{\circ} = A_\theta cos(\theta) = \pi (r_\circ^{2}-r_i^{2})$$

$$\theta = 25^\circ$$ $$r_\circ= 6 in$$ $$r_i= 6 in - .25 in = 5.75$$ $$A_{\circ} = 9.2 in^2 $$ $$A_\theta = \frac{A_0}{cos(\theta)}$$ </ul> Find tangential force (V) and force normal (F) to weld in terms of P <ul> $$P_{max}$$ : magnitude of largest axial force $$F = P_{max} cos(\theta)$$ $$V = P_{max} sin(\theta)$$ </ul> $$ \sigma $$ is the limiting factor for choosing the maximum $$P$$ <ul> Normal stress: $$ \sigma = \frac{F}{A_\theta} = \frac{P_{max} (cos(\theta))^{2}}{A_{\circ}} $$ <ul>$$P_{max} = \frac{\sigma A_{\circ}}{(cos(\theta))^{2}} = 134.4 kips$$</ul> Tangential stress: $$ \tau =\frac{V}{A_\theta} =\frac {P_{max} sin(2\theta)}{2A_\circ} $$ <ul>$$P_{max} = \frac{\tau 2A_\circ}{sin(2\theta)} = 172.9 kips$$</ul> </ul>

Problem R1.6
Problem Number 1.42, p.38. Contents and equations taken from Mechanics of Materials Textbook 6th edition. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664

The links are drawn using the program paint and taken from the the textbook posted using postimage.org

Problem Statement
A steel loop ABCD of length 1.2 m and of 10-mm diameter is placed as shown around a 24-mm-diameter aluminum rod AC. Cables BE and DF, each of 12-mm diameter, are used to apply the load Q. Knowing that the ultimate strength of the steel used for the loop and the cables is 480 MPa and that the ultimate strength of the aluminum used for the rod is 260 MPa, determine the largest load Q that can be applied if an overall factor of safety of 3 is desired.

Solution
Knowns

Steel loop ABCD has

$$L$$ (length) = 1.2 meters

$$d$$ (diameter) = 10 millimeters

The cables BE and DF are used to apply $$Q$$.

Ultimate strength of steel for loop and cables is 480MPa.

Ultimate strength of aluminum used for the rod is 260MPa.

Assumptions

Assuming symmetry, utilize the Method of Joints

Joint B


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$$ 2 \left(\frac{3}{5}\right) F_{AB} - Q = 0 $$

$$ Q = \frac{6}{5} F_{AB}  \leftrightarrow   F_{AB} = \frac{5}{6}Q $$

Joint A

$$F_{AB} = F_{AD} $$

$$ 2 (\frac{4}{5}) F_{AB} - F_{AC} = 0 $$

$$ (\frac{8}{5}) (\frac{5}{6}) Q - F_{AC} = 0 $$

$$(\frac{4}{3}) Q = F_{AC}   \leftrightarrow   Q = (\frac{3}{4})  F_{AC} $$

$$ \sigma =\frac{P}{A} $$ from book, however in this case $$ \sigma = \frac{Q}{A} \leftrightarrow \sigma A = Q $$

Solve

For Cable BE

$$ Q' = \sigma A = \frac{\pi \sigma d^2}{4}= \frac{\pi (480 * 10^6) (0.012)^2}{4} = 54286.72 N $$

For Steel loop

$$ Q'' = (\frac{6}{5}) F_{AB} \implies (\frac{6}{5}) \sigma A \implies \frac{6 \pi \sigma d^2}{(5)(4)} \implies \frac{6 \pi (480*10^6)(0.010)^2}{(5)(4)}= 45238.93 N $$

For Rod AC

$$Q''' = (\frac{3}{4}) F_{A}C = (\frac{3}{4}) \sigma A \implies \frac{3\pi\sigma d^2}{(4)(4)} \implies \frac{3\pi (260*10^6)(0.024)^2}{(4)(4)}= 88215.92 N $$

The smallest $$Q$$ will be the 1 with the largest load $$ \implies Q'' $$

Allowable load $$ Q = \frac{Q''}{Factor of Safety} = \frac{45238.93}{3} = 15079.64 N $$

So the largest load that can be applied if the overall factor of safety is 3 is 15 kN.

Nomenclature

m = meter

mm = millimeter

d = diameter

L = length

M = mega

Pa = pascals

A = area

Q or P = load

σ = stress

N = Newtons