Atar Derj Statics Final Version Final Project

Done by: Derj Atar Supervised by: Dr.Bentamy

Fall 2013

Introduction:

In the following project, I have aimed atanswering the two exercises of the projectof Statics of Fall 2013.

Throughout this project sheet will be foundall the process I have been going throughto reach the solution desired.

I have tried to make the research completeas far as it goes, and I hope that it isdone successfully.

2 EGR 2301 – Final Project – Derj Atar

Table of contentIntroduction 1

Table of Content 2

Topic 3

First Part: Exercise A 4Statement of the Problem 4Given Data 4Information Required 4Free Body Diagram 4

Solving 5Equation6Plot 7Interpretation of results 8

Conclusion 8

Second Part: Exercise B 9Statement of the Problem 9Given Data 9Information Required 9Free Body Diagram 10

Solving 11Equations 11Plots 12Interpretation of results 12

Conclusion 13

Conclusion 14

Topic:Statics A:

3EGR 2301 – Final Project – Derj Atar

The magnitude of the force P applied to the piston of anengine system during one revolution of crank AB is shownin the figure below. Plot the magnitude of the couple Mrequired to hold the system in equilibrium as a functionof θ for 0 ≤ θ ≤ 2π.

Statics B:

Blocks A and B are supported by an incline which is heldin the position shown. Knowing that the mass of block Ais 10 kg and that the coefficient of static frictionbetween all surfaces of contact is 0.15, usecomputational software to determine the value of θ forwhich motion is impending for values of the mass of blockB from 0 to 50 kg using 5-kg increments.

I- Exercise A:

1- Statement of the Problem


This problem is related to chapter 6 that deals with Analysis of structures.

a. Given Data

We know the length of the crank AB= 90mm and BC=200mm.

b. Information Required

Using the given data, we should plot the magnitude of thecouple M required to hold the system in equilibrium as a function of θ for 0 ≤ θ ≤ 2π.

c. Free body diagram

2. Solving

a. EquationApplying the law of sines, we write:

sin∅AB=sinθBC ↔ sin∅0.09

=sinθ0.2


meaning that: ϕ=arcsin (sin (θ)× 0.090.2

)

→ϕ=arcsin (sin (θ)×0.45)

Using the trigonometry over the triangle BCH:

Sin∅ = BHBC ⟶BH=BC×sin∅

cosθ=AHAB

FBD Piston:

Y-axis:

FBCY=FBC×(BHCH

¿

includes that: FBC=FBCY×(BCCH

) (1)

X-axis:


FBCX= FBC×(BHBC

)

We replace FBC by (1):

FBCX= FBCY×(BCCH )×(BHBC )=FBCY×(BHCH

)

We know that: FBCY=P

Then: FBCX=P×(BHCH

)

FBC is a 2 forces member:

FBD Crank AB:

+ ⇃∑❑

❑

MA=0

FBCX×AH+FBCY×BH−M=0

M= P× BHCH×AH+P×BH


M=(P×tan (ϕ)×AH¿+(P×BH)M=(P×tan (ϕ)×AB×cos (θ)+(P×BC×sin (θ))

M=[P×(tan (arcsin (0.45×sinθ)))×(0.09×cosθ) ]+[0.2×P×sin (arcsin (0.45×sinθ)) ]

M=P×¿

b. PlotUsing excel and plotting the equation obtained in the previous part a. , we get the following graph:

Plot of M in function of the angle θ.

c. Interpretation of results

From the graph obtained above, we observe that it is sinusoidal which means that it is periodic. So the frequency is repetitive during a certain range of θ;¿bemorepreciserepeatsitselfeach6units∨approximativelyeach2Π.


So, equilibrium can be maintained between 0<θ <2Π.

3. Conclusion:

The graph shows that the magnitude varies as the angleθchanges.Wecanextractthe maximum magnitude is 1.12N.m.

II- Exercise B:

1- Statement of the problem

This problem is related to chapter 8 that deals with friction. More specifically, a problem that involves friction of blocks.

a- Given Data:


Here we know that the mass of the block A is 10Kg and that the coefficient of static friction between all surfaces of contact is 0.15.

b- Information Required:In this problem we are asked to find the value of θ for which the motion is impending for values of the mass of block B.

c- Free Body Diagram: Here, we will assume that the impeding motion of the upper block is down the plane while the impending motion of the lower block is up the plane. Thus, the frictional of the lower block is up the plane. Thus, the frictional force F acting on the upper blocks acts up the plane while the friction force FA and FB acting on the lower block act down the plane as indicated on the free body diagram below of the upper and lower blocks. Since both block are required to be on the verge of shipping, then:

FA=μs×NA and FB=μs×NB (since motion impends)

Free body diagram of the block A Free body diagram of the block B


2- Solving

a- Equation

Impending motion for block A:

∑❑

❑

FY=0 : NA=10×cosθ

+↖∑❑

❑

FX=0:T−10×sinθ−μs×NA

then: T= 10×sinθ+0.15× (10×cosθ)

Hence: T= 10×sinθ+1.5×cosθ (1)

Impending motion for block B:


+↗∑❑

❑

FY=0:NB−mB×cosθ−NA=0

NB= mB×cosθ+10×cosθ=(mB+10)×cosθ

FB= μs×NB=0.15 (mB+10)×cosθ

+↖∑❑

❑

FX=0:T−mBsinθ+μs×NA+μs×NB=0

T= mB×sinθ−0.15× (10×cosθ )−0.15× (mB+10)×cosθT= mB×sinθ−1.5× (cosθ)−0.15× (mB+10)×cosθ (2)

(1)-(2) = 10×sinθ+1.5×cosθ -mB×sinθ+1.5× (cosθ)+0.15× (mB+10)×cosθ

=10×sinθ+3×cosθ -mB×sinθ+1.5× (cosθ)+0.15×mB×cosθ =10×sinθ+4.5×cosθ - mB×sinθ+0.15×mB×cosθ

so:

sinθ× (10−mB )+4.5×cosθ+0.15×mB×cosθ=0

sinθ× (10−mB )+4.5×cosθ=−0.15×mB×cosθ

sinθ× (10−mB )=cosθ¿

tanθ=(0.15×mB)−4.510−mB

θ=arctan ( (0.15×mB)−4.510−mB

)

b- Plot


We should plot the value of θ for which motion is impending for values of the mass of block B from 0 to 50 Kg using 5Kg increments.

Plot of θin function of mB

c - I n t e r p r e t a t i o n o f r e s u l t sFrom the graph, we observe that from 0<m<5, the function is in the negative side increasing and in the interval 10<m<50 the function decreases, and at m=10kg it cancels out.


Conclusion:


This project gives us the opportunity tostudy more deeply chapters 6 and 8, aboutanalysis structures and friction. In fact wegain in understanding of this two importantparts of the class by exploring data, andlooking for theoretical and experimentalresults to solve the exercises.

It was a good opportunity to finish the classsince we applied the knowledge gained duringthese two months to solve real life problems.

References:

Mathematica:http://www.wolframalpha.com


Atar Derj Statics Final Version Final Project

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