# Mechanics of materials/Problem set 2

This is problem set 2 from UF EGM3520 spring 2013.

## Problem 2.1 (Pb-9.1 in sec.9.)

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

### Problem Statement
#### Part 1

Using Example 2.04(p.9-1), solve for the reaction forces $\displaystyle R_{A}$ and $\displaystyle R_{B}$ as if the stress-strain relation is modeled by
    $\displaystyle \sigma =E\epsilon ^{1/2}$ (2.1-1)

#### Part 2

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

#### Given

    $\displaystyle A_{AD}=A_{DC}=A_{AC}=250mm^{2}$ (2.1-2)
    $\displaystyle A_{CK}=A_{KB}=A_{CB}=400mm^{2}$ (2.1-3)
    $\displaystyle F_{D}=300\,kN$ (2.1-4)
    $\displaystyle F_{K}=600\,kN$ (2.1-5)
    $\displaystyle L_{AD}=L_{DC}=L_{CK}=L_{KB}=L$ (2.1-6)
    $\displaystyle E_{AD}=E_{DC}=E_{CK}=E_{KB}=E$ (2.1-7)

### Solution
#### Part 1
##### Step 1: Draw the free-Body diagram

Begin by removing the supports and drawing the free-body diagram for the entire bar AB and the four sections as shown below.

##### Step 2: Determine the internal forces

Using $\displaystyle \sum F=0$ the internal force, $\displaystyle P$, can be determined in each segment.
    $\displaystyle P_{AD}=R_{B}-F_{K}-F_{D}$ (2.1-8)
    $\displaystyle P_{DC}=R_{B}-F_{K}$ (2.1-9)
    $\displaystyle P_{CK}=R_{B}-F_{K}$ (2.1-10)
    $\displaystyle P_{KB}=R_{B}$ (2.1-11)

##### Step 3: Calculate the displacement

Rearranging Equation 2.1-1,
    $\displaystyle \epsilon ={\frac {\sigma ^{2}}{E^{2}}}$ (2.1-12)
    $\displaystyle \delta =\epsilon \,L={\frac {\sigma ^{2}}{E^{2}}}L={\frac {P^{2}\,L}{A^{2}\,E^{2}}}$ (2.1-13)
The displacement for the whole bar can be represented as a sum of the displacements of each segment and set equal to 0 since both ends of the bar are fixed,
    $\displaystyle \delta =\sum _{i}\delta _{i}={\frac {(P_{i})^{2}\,L_{i}}{(A_{i})^{2}\,(E_{i})^{2}}}$ (2.1-14)
    $\displaystyle \delta _{AD}={\frac {(P_{AD})^{2}\,L}{(A_{AC})^{2}\,E^{2}}}={\frac {(P_{AD})^{2}\,L}{(250)^{2}\,E^{2}}}$ (2.1-15)
    $\displaystyle \delta _{DC}={\frac {(P_{DC})^{2}\,L}{(A_{AC})^{2}\,E^{2}}}={\frac {(P_{DC})^{2}\,L}{(250)^{2}\,E^{2}}}$ (2.1-16)
    $\displaystyle \delta _{CK}={\frac {(P_{CK})^{2}\,L}{(A_{CB})^{2}\,E^{2}}}={\frac {(P_{CK})^{2}\,L}{(400)^{2}\,E^{2}}}$ (2.1-17)
    $\displaystyle \delta _{KB}={\frac {(P_{KB})^{2}\,L}{(A_{CB})^{2}\,E^{2}}}={\frac {(P_{KB})^{2}\,L}{(400)^{2}\,E^{2}}}$ (2.1-18)
Substituting the expressions for $\displaystyle P_{AD},\,P_{DC},\,P_{CK},\,P_{KB}$ from above and adding the components together, we receive
    $\displaystyle \delta ={\frac {L}{(250)^{2}\,E^{2}}}(2(R_{B})^{2}\,-\,3000\,R_{B}\,+\,1,117,000)\,+\,{\frac {L}{(400)^{2}\,E^{2}}}(2(R_{B})^{2}\,-\,1200\,R_{B}\,+\,360,000)\,=\,0$ (2.1-19)

##### Step 4: Solve for the reaction at B

$\displaystyle {\frac {L}{E^{2}}}$ can be cancelled out and like-terms can be grouped to give us
    $\displaystyle 0\,=\,({\frac {2}{250^{2}}}+{\frac {2}{400^{2}}})(R_{B})^{2}-({\frac {3000}{250^{2}}}+{\frac {1200}{400^{2}}})R_{B}+({\frac {1,117,000}{250^{2}}}+{\frac {360,000}{400^{2}}})$ (2.1-20)
    $\displaystyle 0\,=\,(4.45*10^{-5})(R_{B})^{2}-(.0555)R_{B}+(20.122)$ (2.1-21)
Since $\displaystyle 4.45*10^{-5}\approx 0$, Equation 2.1-21 can be rewritten as
    $\displaystyle (.0555)R_{B}=(20.122)$ (2.1-22)
| $\displaystyle R_{B}=362.559\,kN$ | (2.40-13) |

##### Step 5: Solve for the reaction at A

Referring back the free-body diagram of the entire bar,
    $\displaystyle \sum F=0=R_{A}+R_{B}-F_{K}-F_{D}$ (2.1-23)
    $\displaystyle R_{A}=300\,kN+600\,kN-362.559$ (2.1-24)
| $\displaystyle R_{A}=537.441\,kN$ | (2.1-25) |

#### Part 2

Referring to Equation 2.1-19, $\displaystyle {\frac {L}{E^{2}}}$ can be divided out of both terms and set equal to zero. This confirms that the results do not depend on the length of each segment nor Young's modulus, so long as these values are the same in every segment of the bar.

## Problem 2.2 (P2.12, Beer 2012)

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 thread stretching from B to C is subjected to a tension of 10 N. The maximum allowable normal stress is 10 MPa and $E$= 200GPa. The length of the thread may not increase by more than 1%. What is the required diameter of the thread?

### Solution
#### Step One: Determine a relationship between diameter and normal stress and diameter and elongation

The normal stress in the string, $\sigma$, is given by Equation 2.2-1, where $P$ is the applied load and $A$ is the area.
    $\displaystyle \sigma ={\frac {P}{A}}$ (2.2-1)
The elongation of the string is given by Equation 2.2-2, where $\delta$ is the elongation of the string, $L$ is the length of the string and $E$ is the Young's modulus given in the problem statement.
    $\displaystyle \delta ={\frac {PL}{AE}}$ (2.2-2)
The diameter, $d$, can later be inserted into Equation 2.2-1 by equating it to the area, as shown in Equation 2.2-3
    $\displaystyle A={\frac {\pi }{4}}\times (d^{2})$ (2.2-3)

#### Step Two: Calculate the diameter if the maximum normal stress is limiting

Inserting Equation 2.2-3 in to 2.2-1 and solving for $d$ gives
    $\displaystyle d={\sqrt {\frac {4P}{\pi \sigma }}}$ (2.2-4)
Plugging in the given values from the problem statement,
    $\displaystyle d={\sqrt {\frac {4\times 10N}{\pi \times 10\times 10^{6}Pa}}}=0.01128m=1.13mm$ (2.2-5)

#### Step Three: Calculate the diameter if the elongation is limiting

First it was assumed that the string had a length $L$ of 1 meter, meaning that a 1% elongation would be 0.01 meters. The proportionality of these values is what is important, because their units will cancel later.
Inserting Equation 2.2-3 in to 2.2-2 and solving for d gives
    $\displaystyle d={\sqrt {\frac {4PL}{\pi E\delta }}}$ (2.2-6)
Plugging in the given values from the problem statement and the assumed values,
    $\displaystyle d={\sqrt {\frac {4\times 10N\times 1m}{\pi \times 200\times 10^{9}Pa\times 0.01m}}}=7.98\times 10^{-5}m=0.798mm$ (2.2-7)

#### Step Four: Choose the smallest diameter

The smallest allowable diameter, and the minimum that is required, occurred when the one-percent elongation was the limiting factor.
* $\displaystyle d=0.798mm$

## Problem 2.3 ( P2.16, Beer 2012)

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 brass tube AB connected at point A, is being applied a load P on top of the tube. At the bottom of the structure is a rigid plate C attached at point B. An aluminum cylinder is hung downwards from point D, and attached to the rigid plate at point C. With the given values, what is the force load, $P$, that is being applied to the brass tube?
Given:
    $\displaystyle E_{AB}=150GPa$ , $\displaystyle E_{CD}=72GPa$ $\displaystyle A_{AB}=140mm^{2}$ , $\displaystyle A_{DC}=250mm^{2}$ $\displaystyle L_{AB}=376m$ , $\displaystyle L_{DC}=375m$

### Solution
#### Step One:Determine the deflection on the brass tube

First step is to analyze the brass tube at point A and B.
At the segment AB, the brass tube experiences a state of compression when the load $P$ is applied at point A.
    $\displaystyle L_{AB}=375mm+1mm$ $\displaystyle L_{AB}=376mm$ (2.3-1)
In order to find the load applied to the brass tube, a relation between the cylinder and the tube has to be made through the equations of deflection. For the brass tube the deflection is:
    $\displaystyle \delta _{AB}={\frac {PL_{AB}}{E_{AB}A_{AB}}}$ (2.3-2)
After substituting the given values from the problem statement:
    $\displaystyle \delta _{AB}={\frac {376m\times P}{105\times 10^{-9}Pa\times 1.40\times 10^{-4}m^{2}}}$ (2.3-3)
    $\displaystyle \delta _{AB}=2.55\times 10^{-5}\times P({\frac {m}{N}})$ (2.3-4)

#### Step Two:Determine the deflection on the aluminum cylinder

Next, is to find the deflection that is created by the load at point A, on the aluminum cylinder collected at point D. The deflection of the aluminum cylinder hanging at point D.
The deflection of the aluminum cylinder is measured by the following equation:
    $\displaystyle \delta _{CD}={\frac {PL_{CD}}{E_{CD}A_{CD}}}$ (2.3-5)
After substituting the values given in the above problem statement, the equations looks like the following:
    $\displaystyle \delta _{CD}={\frac {375m\times P}{72\times 10^{9}Pa\times 2.5\times 10^{-4}m^{2}}}$ (2.3-6)
    $\displaystyle \delta _{CD}=2.08\times 10^{-5}P({\frac {m}{N}})$ (2.3-7)

#### Step 3 (Finding the stress load)

Total deflection is as followed:
    $\displaystyle \delta =\delta _{AB}+\delta _{CD}$ (2.3-8)
    $\displaystyle 0.001=2.55\times 10^{-5}\times P+2.08\times 10^{-5}\times P$ (2.3-9)
* $\displaystyle P=21.598kN$

## Problem 2.4 ( P2.24, Beer 2012)

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 deformations of members BD and DE in the steel truss ($E$ = 29 E6 psi) shown in Figue 2.4-1. Their cross-sectional areas of BD and DE are 2 in^2 and 3 in^2 respectively. The steel truss is shown in Figure 2.4-1
Given:
$E=29\times 10^{6}psi$
$P=30kips$
$A_{BD}=2in^{2}$
$A_{DE}=2in^{2}$

### Solution
#### Step One:Draw a free-body diagram of the system

The free-body diagram is shown in Figure 2.4-2.

#### Step Two: Use the equilibrium of moments and forces at point F

The sum of moments around point F is zero and the counterclockwise direction is taken as positive.
    $\displaystyle \Sigma M_{F}=0=G_{y}\times 15ft-30kips\times [8ft+16ft+24ft]$ (2.4-1)
Solving for $G_{y}$ gives,
    $\displaystyle G_{y}=96kips$
The sum of forces in the y-direction is also zero.
    $\displaystyle \Sigma F_{y}=0=G_{y}-F_{y}$ (2.4-2)
Therefore, $F_{x}$ is
    $\displaystyle F_{y}=96kips$
The sum of forces in the x-direction is also zero.
    $\displaystyle \Sigma F_{x}=0=30kips+30kips+30kips-F_{x}$ (2.4-3)
Therefore, $F_{y}$ is
    $\displaystyle F_{x}=90kips$

#### Step Three: Isolate the truss at FDG and solve for the forces

First, a free-body diagram must be made for the isolated truss FDG.
To solve for $F_{DE}$, take the sum of the forces in the x-direction, which is zero.
    $\displaystyle \Sigma F_{x}=0=F_{DE}-F_{x}$ (2.4-4)
Knowing $F_{x}$ from Step Two, $F_{DE}$ is
    $\displaystyle F_{DE}=90kips$
Next, the sum of the moments about G is take, which also equals zero.
    $\displaystyle \Sigma M_{G}=0=(F_{y}\times 15ft)-(F_{DE}\times 8ft)-(F_{DB}\times 15ft)$ (2.4-5)
Therefore, $F_{DB}$ is
    $\displaystyle F_{DB}=48kips$

#### Step Four: Solve for the deformation of members BD and DE

The deformation of members BD and DE are given by Equations 2.4-6 and 2.4-7, respectively.
    $\displaystyle \delta _{BD}={\frac {F_{BD}L_{BD}}{A_{BD}E}}$ (2.4-6)
    $\displaystyle \delta _{DE}={\frac {F_{DE}L_{DE}}{A_{DE}E}}$ (2.4-7)
Solving for $\delta _{BD}$ and $\delta _{DE}$ with the given and calculated values gives
* $\displaystyle \delta _{BD}={\frac {48\times 10^{3}kips\times 8ft\times 12{\frac {in}{ft}}}{2in^{2}\times 29\times 10^{6}kips}}=0.079in$
* $\displaystyle \delta _{DE}={\frac {60\times 10^{3}kips\times 15ft\times 12{\frac {in}{ft}}}{3in^{2}\times 29\times 10^{6}kips}}=0.124in$

## Problem 2.5 ( P2.40, Beer 2012)

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 Polystyrene rod consisting of two cylindrical portion AB and BC is restrained at both ends ans supports two 6-kips loads as shown. Knowing that $E=.04x10^{6}psi$, determine the reactions at A and C, and the normal stress in each portion of the rod.

### Solution
#### Step One:Find the reactions at A and C

The elongation of the rod is zero.
    $\displaystyle \delta ={\frac {P_{AB}+L_{AB}}{{\frac {\pi }{4}}d_{AB}^{2}E}}+{\frac {P_{BC}+L_{BC}}{{\frac {\pi }{4}}d_{BC}^{2}E}}=0$ (2.40-1)
The forces are equal in opposite directions.
    $\displaystyle P_{AB}=+R_{A}$ (2.40-2)
    $\displaystyle P_{BC}=-R_{C}$ (2.40-3)
By substitution and simplifying the elongation of the rod can be rearranged to solve for $\displaystyle R_{C}$ in terms of $A_{c}$
    $\displaystyle {\frac {R_{A}L_{AB}}{d_{AB}^{2}}}-{\frac {R_{C}L_{BC}}{d_{BC}^{2}}}=0$ (2.40-4)
    $\displaystyle R_{C}={\frac {L_{AB}}{L_{BC}}}({\frac {d_{BC}}{d_{AB}}})^{2}R_{A}={\frac {25}{15}}({\frac {2}{1.25}})^{2}R_{A}$ (2.40-5)
    $\displaystyle R_{C}=4.2667R_{A}$ (2.40-6)
From the free body diagram the sum of
$\displaystyle R_{A}+R_{C}$ totals to 12 kips.
    $\displaystyle R_{A}+R_{C}=12kips$ (2.40-7)
After substitution and simplifying the reactions can be calculated.
    $\displaystyle 1R_{A}+4.2667R_{A}=12kips$ (2.40-8)
    $\displaystyle 5.2667R_{A}=12$ (2.40-9)
| $\displaystyle R_{A}=12/5.2667=2.2785kips$ | (2.40-10) |
| $\displaystyle R_{C}=4.2667(2.2785)=9.7217kips$ | (2.40-11) |

#### Step Two:(Calculate the normal stress for both members)

The normal stress for the member equals the stress divided by the area of the members. After substitution these normal stresses can be calculated.
    $\displaystyle \sigma _{AB}={\frac {P_{AB}}{A_{AB}}}={\frac {+R_{A}}{A_{AB}}}={\frac {2.2785}{{\frac {\pi }{4}}(1.25)^{2}}}$
(2.40-12)
| $\displaystyle \sigma _{AB}=+1.875ksi$ | (2.40-13) |
    $\displaystyle \sigma _{BC}={\frac {P_{BC}}{A_{BC}}}={\frac {-R_{C}}{A_{BC}}}={\frac {-9.7217}{{\frac {\pi }{4}}(2)^{2}}}$
(2.40-14)
| $\displaystyle \sigma _{BC}=-3.09ksi$ | (2.40-15) |

## Problem 2.6 ( P2.44, Beer 2012)

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

### Problem Statement

The rigid bar AD is supported by two steel wires of ${\frac {1}{16}}$ in. diameter $(E=29x10^{6}$ 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 applied at B, (b) the corresponding deflection of point B.

### Solution
#### Step One:(description of step here)

Let $\theta$ be the rotation of the bar ABCD.
    $\displaystyle \delta _{A}={\frac {24}{\theta }}$ (2.44-1)
    $\displaystyle \delta _{C}={\frac {8}{\theta }}$ (2.44-2)
    $\displaystyle \delta _{A}={\frac {P_{AE}L_{AE}}{AE}}$ (2.44-3)
    $\displaystyle P_{AE}={\frac {EA\delta _{A}}{L_{AE}}}={\frac {(29x10^{6}){\frac {\pi }{4}}({\frac {1}{16}})^{2}(24\theta )}{15}}$ (2.44-4)
    $\displaystyle =142.353x10^{3}\theta$ (2.44-5)
    $\displaystyle \delta _{C}={\frac {P_{CF}L_{CF}}{AE}}$ (2.44-6)
    $\displaystyle P_{CF}={\frac {EA\delta _{C}}{L_{CF}}}={\frac {(29x10^{6}){\frac {\pi }{4}}({\frac {1}{16}})^{2}(8\theta )}{8}}$ (2.44-7)
    $\displaystyle =88.97x10^{3}\theta$ (2.44-8)
Using the free body diagram ABCD the sum of the momentum is equal to 0.
    $\displaystyle -24P_{AE}+16P-8P_{CF}=0$ (2.44-9)
    $\displaystyle -24(142.353x10^{3}\theta )+16(120)-8(88.971x10^{3}\theta )=0$ (2.44-10)
    $\displaystyle \theta =.46510x10^{-}3rad$ (2.44-11)
Substituting into these equation give the tensions in each wire.
| $\displaystyle P_{AE}=(142.353x10^{3})(.46510x10^{-}3)=66.2lb$ | (2.44-12) |
| $\displaystyle 1P_{CF}=(88.971x10^{3})(.46510x10^{-}3)=41.4lb$ | (2.44-13) |
The deflection of the beam is the angle of the moment $\theta$ multiplied by the length of the beam (16in).
| $\displaystyle 1\delta _{B}=16\theta =16(.46510x10^{3})=7.44x10^{3}in$ | (2.44-14) |