User:Egm3520.s13/Report table/R1

--Siefman (discuss • contribs) 20:32, 6 February 2013 (UTC)=Problem 1.1=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
A bookshelf is lined up with books( 8.2" x 1.2" x 9.9", 3.6 lbs each) from end to end. The bookshelf has a rectangular cross-section, L = 100 in, b = 9 in, and h = 0.5 in.



1) Find $$v$$, the vertical mid-span deflection, under the weight of the books and shelf itself. 2) Increase the shelf thickness to 1 in and recalculate $$v$$. 3) Repeat parts 1 and 2 with the shelf made from structural steel ASTM-A36. 4) Reinforce the bookshelf with two side strips so as to have an H cross-section, such as: where H= 1 in, h= 0.3 in, b= 9 in

Find $$v$$ for both materials, Pondorosa Pine and A36 steel.

Solution
The mid-span deflection is given by:


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$$ \displaystyle \frac{5qL^4}{384EI} $$      (1.1-1) where $$q$$ is uniform distributed load / unit beam length, $$L$$ is the beam length, $$E$$ is the Young's modulus, and $$I$$ is the second moment of inertia of the cross section. The Young's modulus of the ponderosa pine and the A36 steel are 1.3E6 psi and 2.9E7 psi respectively.
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For each part 1-4 of the problem statement, the procedures are similar.

The first step is to calculate the values for the variables $$q$$ and $$I$$ in Equation (1.1-1). For the variable $$q$$ one has to calculate the total weight of the books that the shelf can support and the weight of the shelf itself per unit length. In order to calculate the load the books apply to the shelf, the maximum number of books that can be loaded on the shelf must be calculated.


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$$ \displaystyle \frac{L_{shelf}}{W_{book}}= \frac{100 in}{1.2 in}=83.33 $$      (1.1-2)
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Each books weighs a total of 3.6 lb, thus the total weight of the books that is supported is 288 lb. The weight of the shelf varies depending on its dimensions and material. For the shelf constructed of Ponderosa Pine, whose density is 0.015 lb/in3 and dimensions are 100" x 9" x 0.5", the total weight is, $$W_{shelf,p}= 6.75 lb.$$ For the same shelf but with dimensions 100" x 9" x 1" the total weight is, $$W_{shelf,p}= 13.5 lb.$$ For the shelf made of structural steel ASTM-A36, whose density is 0.284 lb/in3 and dimensions are 100" x 9" x 0.5", the total weight is, $$W_{shelf,s} = 127.8 lb.$$ For the same shelf but with dimensions 100" x 9" x 1"  the total weight is, $$W_{shelf,s}= 255.6 lb.$$


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$$ \displaystyle \frac{W_{shelf}+W_{books}}{L_{beam}} $$      (1.1-3)
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The second moment of inertia with respect to the z-axis is given by,


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$$ \displaystyle I_{z}=\frac{bh^3}{12} $$      (1.1-4)
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Then, it is just a matter of substituting in the values of the base and the height of the cross-section of the beam into Equation (1.1-4) to obtain the moment of inertia.

Once $$q$$ and $$I$$ are calculated,the subsequent step is to substitute these values into Equation (1.1-1) which yields the answers to parts 1, 2, and 3 of the problem. Part 1:

Part 2:

Part 3:

Part 4 requires further calculation due to its H-shaped cross-section. The variable $$q$$ is still obtained in the same manner as in the preceding steps, however, to obtain the moment of inertia of this H-shape requires one to divide the composite area into simple geometric shapes, calculate the moment of inertia for each segment, and then add them together. The values obtained from doing so are then substituted into Equation (1.1-1). Resulting in :

=Problem 1.5=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
Two steel plates are bolted together 16-mm-diameter steel bolts ecapsulated inside cylindrical brass spacers. The average normal stress must not exceed 200 MPa in the bolts and 130 MPa in the spacers. Find the minimum outer diameter of the spacers so that the system does not fail.



Solution
A free body diagram of the system is shown in Figure 1.5-1.



The magnitude of the force, $$ P $$, is the same for the bolt as it is for the spacer, so that,


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$$ \displaystyle {P=P_b=P_s} $$      (1.5-1)
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Where $$ P_b $$ is the force on the bolt, and $$ P_s $$ is the force on the spacer.

The normal stress for the bolt $$ \sigma_b $$ and the spacer $$ \sigma_s $$ are calculated as in Equations 1.5-2 and 1.5-3.


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$$ \displaystyle \sigma_b=\frac{P}{A_b}=200MPa $$      (1.5-2)
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$$ \displaystyle \sigma_s=\frac{P}{A_s}=130 MPa $$      (1.5-3)
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Equations 1.5-2 and 1.5-3 can be used to solve for the outer diameter of the spacer, $$ d_s $$, knowing that the bolt diameter, $$ d_b $$, is 16-mm.


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$$ \displaystyle \sigma_s=\frac{4P}{\pi(d_s^2-d_b^2)} $$      (1.5-4)
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$$ \displaystyle \sigma_b=\frac{4P}{(\pi)d_b^2} $$      (1.5-5)
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Solving for $$P $$ in Equations 1.5-4 and 1.5-5 gives,


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$$ \displaystyle P=\frac{\pi}{4}\sigma_s(d_s^2-d_b^2) $$      (1.5-6)
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$$ \displaystyle P=\frac{\pi}{4}\sigma_b(d_b^2) $$      (1.5-7)
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This allows Equations 1.5-6 and 1.5-7 to be set equal to eachother, and for $$ d_s $$ to be solved for.


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$$ \displaystyle d_s=\sqrt{\frac{\sigma_b(d_b^2)}{\sigma_s}+d_b^2} $$      (1.5-8)
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The final value for $$\displaystyle {d_s}$$ is

=Problem 1.11=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
There is a frame, that consists of four members, ABC, DEF, BE, and CF. The frame is secured to a wall via pin connections at A and D. Each member has a 2-in. X 4-in. cross section and every pin has a diameter, d=0.5 in. Find the maximum value for the average normal stress in members BE and CF.



Solution
First, remove the supports and add support reactions to the figure as shown.

Setting up the entire frame as a free-body diagram, we can use an equation of equilibrium
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$$ \displaystyle \sum M_A=0 $$      (1.11-1) so that,
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$$ \displaystyle (40\, in.)D_x-(45\, in.+30\, in.)(480\, lb.)=0 $$      (1.11-2) Solving Equation 1.11-2 for $$D_x$$ yields $$D_x=900\, lb$$
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From there, draw a free-body diagram for member $$ \displaystyle DEF $$



The line of action of reaction force $$ \displaystyle D $$ is parallel to $$ \displaystyle BE $$ and $$ \displaystyle CF $$ so,


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$$ \displaystyle \sum F=0 $$      (1.11-3)
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$$ \displaystyle \frac{3}{5}D_y-\frac{4}{5}D_x=0 $$      (1.11-4)
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$$ \displaystyle D_y=\frac{4}{3}D_x $$      (1.11-5) From (1.11-5) we can determine $$ \displaystyle D_y=1200\, lb $$
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Now, taking the sum of the moments about point F
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$$ \displaystyle \sum M_F=0 $$      (1.11-6)
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$$ \displaystyle (-30)(\frac{4}{5}F_{BE})-(30+15)(D_y)=0 $$      (1.11-7)
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Solving equation (1.11-7) for $$ \displaystyle F_{BE} $$ yields


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$$ \displaystyle F_{BE}=-\frac{45}{24}D_y $$      (1.11-7) Taking the sum of the moments about point E gives us
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$$ \displaystyle \sum M_E=0 $$     (1.11-8)
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$$ \displaystyle (30)(\frac{4}{5}F_{CE})-(30+15)(D_y)=0 $$     (1.11-9) and solving (1.11-9) for $$ \displaystyle F_{CF} $$
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$$ \displaystyle F_{CF}=\frac{5}{8}D_y $$     (1.11-10)
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Using (1.11-7) and (1.11-10), respectively, we can get $$ \displaystyle F_{BE}=-2250\, lb $$ and $$ \displaystyle F_{CF}=750\, lb $$

Normal stress for each beam is defined as
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$$ \displaystyle \sigma=P/A $$     (1.11-11) where $$ \displaystyle P $$ is the force in each member and $$ \displaystyle A  $$ is the cross-sectional area of the member.
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The maximum normal stress occurs at the pin because that is where the beams' cross-sectional area is the smallest. Since the two beams $$ \displaystyle BE,CF $$ are identical, their cross-sectional areas are equal to the area highlighted in yellow in the graphic below




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$$ \displaystyle A_{BE,CF}=(2\, in.)(3.5\, in.)=7\, in.^2 $$     (1.11-12) So now we can solve for the maximum stress in each member
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$$ \displaystyle \sigma_{BE}=\frac {F_{BE}}{A_{BE}} $$     (1.11-13)
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$$ \displaystyle \sigma_{BE}=\frac {2250\, lb.}{7\, in^2}=321.4\, psi. $$     (1.11-14)
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$$ \displaystyle \sigma_{CF}=\frac {F_{CF}}{A_{CF}} $$     (1.11-15)
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$$ \displaystyle \sigma_{CF}=\frac {750\, lb.}{7\, in^2}=107.1\, psi. $$     (1.11-16)
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=Problem 1.20=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
Determine the smallest allowable length, L, of the bearing plate if the bearing stress in the timber is not to exceed 400 psi. The axial force in the column, P, supporting the timber beam shown is 5.20 kips



Solution
The bearing stress, $$\sigma_{b}$$, is given by,
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$$ \displaystyle \sigma_{b} = \frac{P}{A}=\frac{P}{L w} $$ (1.20-1) Where $$P$$ is pressure, $$A$$ is area, $$L$$ is length, and $$w$$ is width.
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Rearranging Equation 1.20-1 to solve for $$L$$,


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$$ \displaystyle L = \frac{P}{A}=\frac{P}{\sigma_{b}w} $$     (1.20-2)
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From the problem statement, it is known that $$w$$ is 6 inches and $$P$$ is 5.20 kips, or 5,200 psi. To solve for the minimum allowable length, the bearing stress must be set to the maximum allowable value, or 400 psi.

Solving for $$L$$ gives,

=Problem 1.33=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
Determine the maximum axial force, $$ P_{max} $$ ,that can be applied to a 12 inch. outer diameter pipe if the maximum normal stress, $$\sigma$$, and shear stress, $$\tau$$, is 12 ksi and 7.2 ksi respectively. The pipe is made from a 0.25 inch thick plate, welded along a helix to form a 25 degree angle with a plane normal to the axis of the pipe.



Solution
The normal stress is given by,


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$$ \displaystyle \sigma = \frac{P}{A}cos^2\theta $$     (1.33-1)
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And the shear stress is given by,


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$$ \displaystyle \tau = \frac{P}{2A}sin2\theta $$     (1.33-2)
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From the problem statement, it is known that $$ \theta $$ is 25 degrees, $$\sigma$$ is 12 ksi, and $$\tau$$ is 7.2 ksi. $$ P $$ must be solved for, so the area of the pipe, $$ A $$, must be calculated from the given geometry.

The area of the pipe is calculated,


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$$ \displaystyle A = \pi(r_o^2-r_i^2) $$     (1.33-3)
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Where $$ r_o $$ is the outer radius of the pipe, or half the outer diameter, $$ D $$,


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$$ \displaystyle r_0 = \frac{D}{2} = \frac{12inch}{2}= 6 inch $$     (1.33-4)
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and $$ r_i $$ is the inner radius calculated by subtracting the outer radius from the pipe thickness, $$ t $$.


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$$ \displaystyle r_i = \frac{D}{2}-t = \frac{12inch}{2} - 0.25inch = 5.75inch $$     (1.33-5)
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Therefore the area of the pipe is calculated,


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$$ \displaystyle A = \pi((6inch)^2-(5.75inch)^2) = 9.22inch^2 $$     (1.33-6)
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Now that the area is known, the maximum applied for $$ P $$ can be calculated by rearranging Equations 1.33-1 and 1.33-2 and substituting the known variables.

Normal Stress:


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$$ \displaystyle P = \frac{\sigma A}{cos^2(\theta)}=\frac{12ksi \times 9.22inch^2}{cos^2(25^{\circ})} = 136,690lbs $$     (1.33-7)
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Shear Stress:


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$$ \displaystyle P = \frac{\tau A}{cos^2(\theta)}=\frac{7.2ksi \times 9.22inch^2}{cos^2(25^{\circ})} = 173,310 lbs $$     (1.33-8)
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Therefore, largest magnitude of the axial force, $$ P_{max} $$ is given by the normal stress,

=Problem 1.42=

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement
A steel structure with 1.2m equal length sides ABCD and 10-mm diameter is put around a 24-mm-diameter aluminum rod AC. Load Q is applied through cables BE and DF, both of which are 12mm in diameter. The ultimate strength of the steel used for the structure and the cables is 480 MPa. The ultimate strength of the aluminum used for the rod is 260 MPa. Find the largest load Q that can be applied to acheive an overall factor of safety of 3.



Solution
The angle $$ \theta $$ needs to be calculated for the junctions in structure. The angle is shown in Figure 2 and 3.









Using a common (3,4,5) triangle,its possible to calculate $$ \theta $$ with respect to $$ Q $$ the applied force.


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$$ \displaystyle sin(\theta )=\frac{3}{5} $$     (1.42-1)
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$$ \displaystyle Q=2Tsin\theta $$     (1.42-2)
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Solving for T,


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$$ \displaystyle T=\frac{Q}{2sin(\theta)} $$     (1.42-3)
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$$ \displaystyle T=\frac{5Q}{6} $$     (1.42-4)
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From Free Body Diagram 3 the compressible force P in the aluminum rod can be expressed in terms of T and the angle $$ \theta $$


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$$ \displaystyle P=2Tcos(\theta) $$     (1.42-5)
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$$ \displaystyle P=2*\frac{5Q}{6}*\frac{4}{5}=\frac{4Q}{3} $$     (1.42-6) From the given ultimate strength of the aluminum and steel are as follows
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$$ \displaystyle {\sigma_{u,st}=480 MPA} $$     (1.42-7)
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$$ \displaystyle {\sigma_{u,Al}=430 MPA} $$     (1.42-8)
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With a Factor of saftey of:
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$$ \displaystyle {f_s=3} $$     (1.42-9)
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The allowable stress in the members can be calculated by taking the strength of the materials and divide by the factor of safety


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$$ \displaystyle {\sigma_{st}=\frac{\sigma_{u,steel}}{3}=\frac{480}{3}=160MPa} $$     (1.42-10)
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$$ \displaystyle {\sigma_{Al}=\frac{\sigma_{uAl}}{3}} $$     (1.42-11)
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$$ \displaystyle {\frac{260}{3}=86.66MPa} $$     (1.42-12)
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The aluminum rod compressible stress is calculated by taking the compressible force and dividing by the area


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$$ \displaystyle {\sigma_{Al}=\frac{P}{A}} $$     (1.42-13)
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$$ \displaystyle {=\frac{\frac{4Q}{3}}{\frac{\pi }{4}{d}^2}} $$     (1.42-14)
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$$ \displaystyle {PQ_{Al}=\frac{\pi }{4}*\sigma_{Al}*d^2*\frac{3}{4}} $$     (1.42-15)
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$$ \displaystyle {PQ_{Al}=\frac{\pi }{4}*(86.66*10^6)*(24*10^{-3})^2*\frac{3}{4}=29.405kN} $$     (1.42-16)
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Steel length tensile stress can be calculated by taking the force of the loop divided by it's area.


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$$ \displaystyle {\sigma_{st loop}=\frac{P_{loop}}{A_{loop}}} $$     (1.42-17)
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$$ \displaystyle {P_{loop}=\sigma_{st}*{A_{loop}}} $$     (1.42-18)
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$$ \displaystyle {=160*10^6*\frac{\pi }{4}*10^2*10^-6} $$     (1.42-19)
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$$ \displaystyle {T=400\pi} $$     (1.42-20)
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$$ \displaystyle {\frac{5Q_{loop}}{6}=400\pi=15.079kN} $$     (1.42-21)
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$$ \displaystyle {\sigma=\frac{P_{cable}}{A_{cable}}} $$     (1.42-22)
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$$ \displaystyle {Q_{cable}=\frac{\pi }{4}*\sigma_{st}*d^2} $$     (1.42-23)
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$$ \displaystyle {\frac{\pi }{4}*160*10^6*12^2*10^{10}=18.095kN} $$     (1.42-24)
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Applied load is the minimum force of the rod, cable, and loop.

=Contributing Team Members= = Contributors = Team Designee: Daniel Siefman

=References=