# University of Florida/Egm3520/s13.team1.r1

Report 1

## 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.
    $\displaystyle \nu ={\frac {5qL^{4}}{384EI}}$ (1)
Given(s):
    $q={\frac {3.6lb}{100in}}=0.036{\frac {lb}{in}}$
    $L=100in$
    $\displaystyle I=\left({\frac {bh^{3}}{12}}\right)=\left({\frac {9in*0.5in)^{3}}{12}}\right)=0.09375in^{4}$
    $\displaystyle E_{pine}=1.3*10^{6}psi$
Plugging these values into the equation yields
    $\displaystyle \nu =0.385in$
2) Increasing the thickness to 1in will change the moment of inertia
    $\displaystyle I=\left({\frac {bh^{3}}{12}}\right)=\left({\frac {9in*(1in)^{3}}{12}}\right)=0.75in^{4}$
By plugging in this new value of I into equation (1) we get:
    $\nu =0.48in$
3) This section ask to use the Young's modulus for ASTM A-36 steel instead of the Young's modulus for Pine.
Replacing:
    $\displaystyle E_{pine}=1.3*10^{6}psi$ with $\displaystyle E_{ASTMA-36Steel}=29*10^{6}psi$
then plug this new E value into equation (1)
    $\displaystyle Thickness=0.5in$ $\nu =0.0172in$
    $\displaystyle Thickness=1.0in$ $\nu =0.00216in$
4) Converting the original bookshelf into an H-cross section changes the moment of inertia.
    $\displaystyle I_{x_{1}}+A^{2}dy={\frac {bh^{3}}{12}}+A^{2}dy={\frac {9in*(0.5in)^{3}}{12}}+(9*0.5)^{2}(0)=0.09375in^{4}$
    $\displaystyle I_{x_{2}}+A^{2}dy={\frac {bh^{3}}{12}}+A^{2}dy={\frac {9in*(0.3in)^{3}}{12}}+(9*0.3)^{2}(1)=7.31025in^{4}$
Therefore the moment of inertia is equal to the sum of
    $\displaystyle I_{x_{1}}+I_{x_{2}}=7.404in^{4}$
Plugging in this new value for the moment of inertia into equation one yields:
    $\displaystyle \nu _{pine}=0.00487in$
    $\displaystyle \nu _{steel}=2.18*10^{-4}in$

## 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
    $\displaystyle \pi \left({\frac {OD}{2}}\right)^{2}-\pi (8mm)^{2}$
    $\displaystyle {\frac {\pi \left({\frac {OD}{2}}\right)^{2}-\pi (8mm)^{2})}{\pi (8mm)^{2}}}={\frac {20}{13}}\implies OD=2{\sqrt {\frac {{\frac {20}{13}}\pi (8mm)^{2}+\pi (8mm)^{2}}{\pi }}}$ $OD=25.49mm$

## 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.
    $\displaystyle \sum M_{A}=040in*D_{x}-(45+30)in*480lb=0$
    $\displaystyle 40in*D_{x}-(45+30)in*480lb=0$
    $\displaystyle D_{x}=900lb$
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
    $\displaystyle D_{y}={\frac {4}{3}}D_{x}=1200lb$
From that we chose points E and F and take the moment at each
    $\displaystyle \sum M_{E}=0\rightarrow 30*{\frac {4}{3}}F_{CF}-15D_{y}$
    $\displaystyle F_{CF}=750lb$
    $\displaystyle \sum M_{F}=0=-30*{\frac {4}{3}}F_{BE}-(30+15)D_{y}$
    $\displaystyle F_{BE}=-2250lb$
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.
    $\displaystyle A_{member}=2in*4in=8in^{2}$
    $\displaystyle A_{pin}=2in*(4-0.5)in=7in^{2}$
    $\displaystyle \sigma _{BE}={\frac {F_{BE}}{A}}={\frac {2250lb}{7in^{2}}}=321.psi$
    $\displaystyle \sigma _{CF}={\frac {F_{CF}}{A}}={\frac {750lb}{7in^{2}}}=107psi$

## 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.
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.
    $\sigma ={\frac {P}{A}}={\frac {P}{LW}}$
σ = normal stress, P = magnitude of the load, A = cross sectional area
Rearranging the formula we are able to determine the minimum length required for the bearing stress not to exceed 400 psi
    ${\frac {P}{LW}}<400psi$
Reorganizing the formula allows us to solve directly for L
    $L>{\frac {P}{W\sigma }}={\frac {20kips}{(6in)(400psi)}}$
Given that 1kip = 1000lbs and l psi = 1 lb/in2
    $L>{\frac {20000lb}{(6in)({\frac {400lb}{in^{2}}})}}$
    $L>8.33in$
The smallest possible length for the given stress is 8.33inches

## 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 $(A_{\theta })$
Find tangential force (V) and force normal (F) to weld in terms of P
$\sigma$ is the limiting factor for choosing the maximum $P$

## 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
    $F_{AB}=F_{BC}$
$2\left({\frac {3}{5}}\right)F_{AB}-Q=0$
$Q={\frac {6}{5}}F_{AB}\leftrightarrow F_{AB}={\frac {5}{6}}Q$
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
$Q'=\sigma A={\frac {\pi \sigma d^{2}}{4}}={\frac {\pi (480*10^{6})(0.012)^{2}}{4}}=54286.72N$
For Steel loop
$Q''=({\frac {6}{5}})F_{AB}\implies ({\frac {6}{5}})\sigma A\implies {\frac {6\pi \sigma d^{2}}{(5)(4)}}\implies {\frac {6\pi (480*10^{6})(0.010)^{2}}{(5)(4)}}=45238.93N$
For Rod AC
$Q'''=({\frac {3}{4}})F_{A}C=({\frac {3}{4}})\sigma A\implies {\frac {3\pi \sigma d^{2}}{(4)(4)}}\implies {\frac {3\pi (260*10^{6})(0.024)^{2}}{(4)(4)}}=88215.92N$
The smallest $Q$ will be the 1 with the largest load $\implies Q''$
Allowable load $Q={\frac {Q''}{FactorofSafety}}={\frac {45238.93}{3}}=15079.64N$
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