University of Florida/Egm3520/s13.team4.r2

Problem Statement

1)Solve for the reactions at B and at A in the Example 2.04 (see textbook page 72) with the stress strain relation ${\displaystyle \sigma =E\ \epsilon ^{1/2}}$
2)Do the results depend on the length of each segment and the Young's modulus?

Solution

problem 2.04, Beer 2012 p.72

Given:

Length

${\displaystyle L_{AD}=L_{DC}=L=150\,mm}$

${\displaystyle L_{CK}=L_{KB}=L=150\,mm}$

Area

${\displaystyle A_{AD}=A_{DC}=A_{AC}=250\,mm^{2}}$

${\displaystyle A_{CK}=A_{KB}=A_{CB}=400\,mm^{2}}$

Young's modulus

${\displaystyle E_{AD}=E_{DC}=E_{CK}=E_{KB}=E}$

Applied forces

${\displaystyle F_{D}=300\,kN}$

${\displaystyle F_{K}=600\,kN}$

Draw a Free Body Diagram for each section

2.04 problem, Vu-Quoc sec.9 p.9-6

Use ${\displaystyle \sum \mathbf {F=0} }$ on FBD 3

${\displaystyle P_{CK}=R_{B}-F_{K}}$

Knowing that the strain relation is found by:

${\displaystyle \delta =\epsilon \,L={\frac {\sigma ^{2}}{E^{2}}}\,L={\frac {P^{2}\,L}{A^{2}\,E^{2}}}}$

We understand that without a linear strain relation we can't use the superposition principle to separate the reaction as we did in class.
So we'll need to set two FBD's displacement equations equal to each other in order to solve


The displacement equation mentioned previously is applied to FBD 3, which yeilds:

${\displaystyle \delta _{CK}=-{\frac {P_{CK}^{2}L_{CK}}{(A_{CK})^{2}(E_{CK})^{2}}}=-{\frac {(R_{B}-F_{K})^{2}L}{(A_{CB})^{2}E^{2}}}}$

To solve for ${\displaystyle R_{B}}$ we must sum all displacements from each section we cut and set it equal to 0

These equations yield

${\displaystyle \displaystyle \delta _{AD}={\frac {(R_{B}-F_{K}-F_{D})^{2}\,L}{(250)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle \delta _{DC}={\frac {(R_{B}-F_{K})^{2}\,L}{(250)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle \delta _{CK}={\frac {(R_{B}-F_{K})^{2}\,L}{(400)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle \delta _{KB}={\frac {(R_{B})^{2}\,L}{(400)^{2}\,E^{2}}}}$

${\displaystyle \delta _{T}=\sum \delta =0}$

${\displaystyle \delta _{T}=\delta _{AD}+\delta _{DC}+\delta _{CK}+\delta _{KB}=0}$

since all quantities are squared, therefore

${\displaystyle \delta _{AD}=0}$

${\displaystyle \delta _{DC}=0}$

${\displaystyle \delta _{CK}=0}$

${\displaystyle \delta _{KB}=0}$

${\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K}-F_{D})^{2}\,L}{(250)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K})^{2}\,L}{(250)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K})^{2}\,L}{(400)^{2}\,E^{2}}}}$

${\displaystyle \displaystyle 0={\frac {(R_{B})^{2}\,L}{(400)^{2}\,E^{2}}}}$

${\displaystyle R_{B}=0}$

${\displaystyle R_{B}-F_{K}=0}$

${\displaystyle F_{K}=0}$

${\displaystyle R_{B}-F_{K}-F_{D}=0}$

${\displaystyle F_{D}=0}$

Solution 2.04

Problem Statement

A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the required diameter of the thread

Solution

GIVEN:

${\displaystyle P=10N}$

${\displaystyle E=3.3GPa}$

${\displaystyle \sigma =40MPa}$

WE KNOW THAT

${\displaystyle \delta ={\frac {PL}{AE}}}$

AND IT IS REQUIRED THAT

${\displaystyle \delta <0.001L}$

${\displaystyle \Rightarrow {\frac {PL}{AE}}<0.001L}$

${\displaystyle \Rightarrow {\frac {AE}{P}}<100}$

${\displaystyle \Rightarrow A>{\frac {100P}{E}}}$

AND WE ALSO KNOW THAT ${\displaystyle A={\frac {\pi }{4}}\times d^{2}}$

SO ${\displaystyle {\frac {\pi }{4}}\times d^{2}<{\frac {100P}{E}}}$

${\displaystyle \Rightarrow d^{2}<{\frac {400P}{\pi E}}}$

${\displaystyle \Rightarrow d>20{\sqrt {\frac {p}{\pi E}}}}$

${\displaystyle \Rightarrow d>20{\sqrt {\frac {10}{\pi \times 3.2\times 10^{9}}}}}$

${\displaystyle d>0.63mm}$

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

Problem Statement

The brass tube AB (E = 105 GPa) has a cross-sectional area of 140 mm^2 and is fitted with a plug at A. The tube is attached at B to a rigid plate that is itself attached at C to the bottom of an aluminum cylinder (E = 72 GPa) with a cross-sectional area of 250 mm^2. The cylinder is then hung from a support at D. In order to close the cylinder, the plug must move down through 1 mm. Determine the force P that must be applied to the cylinder.

Solution

Given: Brass Tube AB: ${\displaystyle E_{AB}=105*10^{9}Pa}$

${\displaystyle A_{AB}=140mm^{2}}$

${\displaystyle L_{AB}=1+375=376mm}$

Aluminum Cylinder DC:

${\displaystyle E_{DC}=72*10^{9}Pa}$

${\displaystyle A_{DC}=250mm^{2}}$

${\displaystyle L_{DC}=375mm}$

Equations: Compression of Brass Tube AB: ${\displaystyle \delta _{AB}={\frac {P*L_{AB}}{A_{AB}*E_{AB}}}={\frac {(376)P}{(105*10^{9})(140)}}=(2.56*10^{-}11)*P}$

Tension of Aluminum Cylinder DC: ${\displaystyle \delta _{DC}={\frac {P*L_{DC}}{A_{DC}*E_{DC}}}={\frac {(375)P}{(72*10^{9})(250)}}=(2.1*10^{-}11)*P}$

We know the total deflection:

${\displaystyle \delta _{T}=1mm}$

${\displaystyle \delta _{T}=\delta _{AB}+\delta _{DC}\Rightarrow 1=(2.56*10^{-}11)P+(2.1*10^{-}11)P=(4.66*10^{-}11)P}$

${\displaystyle \Rightarrow P=2.15*10^{1}0mN,or21.5kN}$

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

Problem Statement

For the steel truss (E = 29 x 10^6 psi) and loading shown, determine the deformations of members BD and DE, knowing that their cross-sectional areas are 2 in^2 and 3 in^2, respectively.

Solution

Free Body Diagrams

${\displaystyle \sum F_{x}=0=-F_{x}+90kips\Rightarrow F_{x}=90kips}$

${\displaystyle \sum M_{F}=0=-30kips*16ft-30kips*8ft-24ft*30kips+G_{y}*15ft\Rightarrow G_{y}=96kips(tension)}$

${\displaystyle \sum F_{y}=0=F_{y}+G_{y}\Rightarrow F_{y}=-G_{y}=-96kips}$

${\displaystyle \sum F_{x}=0=-F_{x}+30kips+F_{DE}\Rightarrow F_{DE}=90kips-30kips=60kips}$

${\displaystyle \sum M_{G}=0=-30kips*8ft+F_{y}*15ft-F_{BD}*15ft-F_{DE}*8ft\Rightarrow F_{BD}={\frac {60kips*8ft+30kips*8ft-96kips*15ft}{15ft}}=-48kips(tension)}$

${\displaystyle \delta _{BD}={\frac {PL}{AE}}={\frac {48kips*8ft*12{\frac {in}{ft}}}{2in^{2}*29*10^{3}{\frac {kips}{in^{2}}}}}=0.0794in}$

${\displaystyle \delta _{DE}={\frac {PL}{AE}}={\frac {60kips*15ft*12{\frac {in}{ft}}}{3in^{2}*29*10^{3}{\frac {kips}{in^{2}}}}}=0.124in}$

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

Problem Statement

A polystyrene rod consisting of two cylindrical portions AB and BC is restrained at both ends and supports two 6-kip loads as shown. Knowing that E = 0.45 X 10^6 psi, determine (a) the reactions at A and C, (b) the normal stress in each portion of the rod.

Solution

Splitting the diagram into two parts we get part 1 and part 2 So we are given:

Part 1:

${\displaystyle d_{1}=1.25in}$

${\displaystyle L_{1}=25in}$

Part 2:

${\displaystyle d_{2}=2in}$

${\displaystyle L_{2}=15in}$

${\displaystyle E=0.45\times 10^{6}psi}$

${\displaystyle A_{1}={\frac {\pi d_{1}^{2}}{4}}=1.227in^{2}}$

${\displaystyle A_{2}={\frac {\pi d_{2}^{2}}{4}}=3.14in^{2}}$

Looking at the free body diagram we see that

${\displaystyle R_{1}+R_{2}=12kip}$

The total ${\displaystyle \delta _{total}}$ equals zero so

${\displaystyle \delta _{total}=0=\delta _{1}+\delta _{2}}$

Using the formula

${\displaystyle \delta ={\frac {PL}{AE}}}$

We find that

${\displaystyle {\frac {P_{1}L_{1}}{A_{1}E_{1}}}+{\frac {P_{2}L_{2}}{A_{2}E_{2}}}=0}$

Substituting in the reaction forces ${\displaystyle R_{1}}$ and ${\displaystyle R_{2}}$

${\displaystyle (R_{2}}$ will show negative because the force is facing opposite to ${\displaystyle R_{1})}$

${\displaystyle {\frac {R_{1}L_{1}}{A_{1}E_{1}}}-{\frac {R_{2}L_{2}}{A_{2}E_{2}}}=0}$

Solving for ${\displaystyle R_{1}}$ we get

${\displaystyle R_{1}={\frac {R_{2}A_{1}L_{2}}{L_{1}A_{2}}}}$

So

${\displaystyle R_{1}=.2344R_{2}....(1)}$ and solving a system of equations with

${\displaystyle R_{1}+R_{2}=12kip....(2)}$ Gives

${\displaystyle R_{1}=2.27kips}$
${\displaystyle R_{2}=9.72kips}$

(b) Using the formula ${\displaystyle \sigma ={\frac {P}{A}}}$

So

${\displaystyle \sigma _{1}={\frac {R_{1}}{A_{1}}}=1.85ksi}$
${\displaystyle \sigma _{2}={\frac {R_{2}}{A_{2}}}=3.09ksi}$

On our honor, we did this assignment 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 1 16-in. diameter (E = 29 X 10^6 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

${\displaystyle \delta _{AE}={\frac {P_{AE}L_{AE}}{AE}}}$

${\displaystyle \Rightarrow P_{AE}={\frac {\delta _{AE}*AE}{L_{AE}}}={\frac {24in*\theta *{\frac {\pi }{4}}*({\frac {1}{16}}in)^{2}*29*10^{6}psi}{15in}}=142.4kips*\theta }$

${\displaystyle \delta _{CF}={\frac {P_{CF}*L_{CF}}{AE}}}$

${\displaystyle \Rightarrow P_{CF}={\frac {\delta _{CF}AE}{L_{CF}}}={\frac {8in*{\frac {\pi }{4}}*({\frac {1}{16}}in)^{2}*29*10^{6}psi}{8in}}=88.97kips*\theta }$

${\displaystyle \sum M_{D}=0=-24in*P_{AE}+16in*P-8in*P_{CF}=-24in*142.4*10^{3}lb*\theta +16in*120lb-8in*88.97*10^{3}lb*\theta }$

${\displaystyle \Rightarrow \theta =465*10^{-6}}$

${\displaystyle P_{AE}=142.4*10^{3}lb*465*10^{-6}=66.21lb}$

${\displaystyle P_{CF}=88.97*10^{3}lb*465*10^{-6}=41.37lb}$

${\displaystyle \delta _{B}=16in*\theta =16in*465*10^{-6}=7.44*10^{-3}(downward)}$

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