Mechanical and Electrical concepts SOUD1214

James FloundersStudent assignment number 4Static systems

We are given a diagram of a pulley shaft supported at each end in roller bearings. Pulley P1 drives the shaft and pulley P2 is driven by the shaft. The forces on the shaft from each pulley have been resolved into net forces and may be treated as point loads F1 and F2 pulling in opposite directions. The shaft is made from a solid bar diameter Dmm. The total power transmitted from pulley to the other is Pkw at a speed of N rev/min.It may be assumed that the modulus of rigidity G is 900PaAm = 0.25, Bm = 0.15, Cm = 0.1, F1N =2500, F2N =1900, Dm = 0.13 , Pkw = 6.0, Nrev/min = 100Bending and transverse shear1. Draw the free body diagram to scale

A (2500) B C

0.25m 0.15m 0.1m

N RA RB

(1900)

2. Calculate the reaction forces in each bearingMoments about RA; CCw = Positive

∑ M =0

0 = + (-2500x0.25) + (+1900x0.4) + (RBx0.5) => 625 + -760 = RB0.5 => -135 = RB0.5 => RB = -135 = 1270N So RB =-270N

0.5F1 + RB = -2500 + - 270 = -2770F2 + RA = 2770 = 2770 – 1900 = 870 S0 RA = 870N

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3. Draw the shear force (SF) diagram.4. Draw the bending moment (BM) diagram.

Kn 3.0 2.5 2.0 1.5 0.25m 0.1m 1.0 +870n 0.5 -2500 -270

0 x-5 0.1 0.2 0.3 0.4 0.5 -10-15 -1630 +1900-20-25 0.15m NM

250 217.5NM 200 150 100 50 0 0 -50 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 5 -100 -27.00NM -150

At intervals of 0.05:(870x0.05)=43.5+(870x0.05)=87+(870x0.05)=130.5+(870x0.05)=174+(870x0.05)=217.5-(1630x0.05)=136-(1630x0.05)=54.5-(1630x0.05)=-27+(270x0.05)=-13.5+(270x0.05)= 0

5. Determine the position of maximum bending motionThe maximum bending moment is 217.5NM at 0.25m from the left

6. Calculate the maximum bending stress produced.O = MY and M = 217.5NM , Y = D = 0.065m M= Bending moment I 2 I = Moment of inertia I = πD4 = πx0.13⁴ = 1.4x10‾⁵ Y = Distance from neutral axis 64 64so O = 217.5x0.065 = 1009821.429 = 1.0MPa 1.4x10‾⁵

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3.sheer force

4.Bending moment

Torsional shear stress7. Calculate the maximum shear stress produced in the shaft due to torsion

Using Bending formula T = GO = Ϯ we can work through and find Ϯ j L R T = P = 6000 = 573.6Nm N =100rev/min converted into rev/sec = 1.666 2πN 2π1.666 P = 6kw

j = πD⁴ = π0.13⁴ = 2.8x10‾⁵ 32 32we can now find a value for Ϯ Ϯ= TxR = 573.6 x 0.065 = 1.3n/mm₂ j 2.8x10‾⁵

8. Calculate the angle of twist between the two pulleys.

Angle of twist = OO = TL = 573.6 x 0.15 = 2.7radians jG 2.8x10‾⁵x90x10⁹

Now convert radians into degrees 2.7 x 180 = 154.69° πSelection of standard sections9. A beam is to support a bending moment of 350kNm and the maximum bending stress must to exceed 250mPa, determine the section modulus (Z) and select an appropriate universal beam T section using tables of standard steel section.

O = M we need to find Z so change this to Z = M Z O=> Z = 350x10‾⁶ = 1.4x10‾⁹ 250x10ᶟZ = 1400cmUsing the standard steel section tables the minimum size beam to be used should be serial size 267x381, although a serial size 292x419 may be used with a view to extra load being applied in the future.

Serial sizes taken from T section tables on http://www.roymech.co.uk

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10. A steel section is to be used as a column, it is to have a slenderness ratio of 100 and has an overall length of 5m. Determine the radius of gyration and hence select a suitable column T section using tables of standard steel sections. Using this specification of column determine the compressive stress induced for an axial load of 100kN.

S.R = L => 100 = 5000 K KRearrange to find KK = L => 5000 = 50mm S.R 100

Looking in the standard steel section tables http://www.roymech.co.uk/Useful_Tables/Sections/TeeA_dimprop.html we use y,y axis in the radius of gyration column and use serial size, 229x305x70 this is the narrowest beam that will not buckle.

we can now find the compressive stress induced for an axial load of 100kN= F The area given on the tables for the beam is 89.1cm₂

A=> 100 = 1.12MPa 89.1

Using lesson notes and information taken from Tooley and Dingle, Engineering science 2012 and

various engineering websites, including MD solids, free study and Roymech. I have completed all

questions in the static systems section of this assignment and where possible backed my findings up

with evidence from other sources, I believe double checking all my findings is essential where

possible to check for calculation errors on my part and to gain a greater understanding of the results.

SOUD1214 Dynamic SystemsFor a charity event, an engineering team is tasked with designing a launch system for a small un-manned glider. The aim is to use no electrical assistance and keep the cost as low as possible. It is decided to use a gravity driven flywheel system to generate the energy required for launch.Our task is to analyse the proposed system as detailed below.Uniform accelerationThe system comprises of a rim type flywheel of mean radius ‘R’ 0.7 meters and mass ‘M’ 90kg. The flywheel is powered by a weight of ‘W’ 22 kg free falling from a height of ‘X’ 4.5meters . the weight ‘W’ is attached to the flywheel by means of a thin cord wrapped around a drum of radius 0.25m.

If the fly wheel starts at rest and gravity is taken as 9.81ms‾1. calculate:

1. The potential energy of the mass when initially stationary

According to Tooley and Dingle (2012, P156) Potential Energy (PE) is energy possessed by a body by virtue of its position, relative to some datum. The change in PE is equal to its weight multiplied by the change in height. Since the weight of a body = MG, then the change in PE may be written as PE = MG∆H

MG∆H can be written as MGZ, so PE = MGZPE = 22 x 9.81 x 4.5 = 971.19j

So the potential Energy of the mass when initially stored is 971.19j

2. Find the linear kinetic energy of the mass as it hits the ground

Tooley and Dingle (2012, P158) state that: Kinetic Energy (KE) is energy possessed in a body by virtue of its motion. Translational KE i.e. the KE of a body travelling in a linear direction (straight line), is: Translational KE (J) = mass (kg) x (velocity)₂ (m/s) 2

We already have the mass of the weight (22kg) but we need to find velocity, so first we have to find inertia (I)I = MK₂ We have M (22kg) and we have K (0.7) so I = 90 x 0.7₂ = 44.1kg/m₂

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The next step is to find V₂V₂ = 2GZ so v₂ = 2 x (9.81x4.5) 88.29 1+ I 1 + 44.1 = 1+ 44.1 = 2.66 MR₂ 22 x 0.25₂ 1.375

V = v₂ = 2.66 = 1.63mp/s

KE = M x V₂ = 22 x 2.66 = 29.26j 2 2So the linear kinetic energy of the mass when it hits the ground is 29.26j

3. Find the angular kinetic energy of the fly wheel when the mass hits the ground.First we need to find the angular velocity (Ѡ)Ѡ = V = 1.63 = 6.52rad/s R 0.25

Now we can find the RKE = IѠ₂ = 44.1 x 6.52₂ = 937.4j 2 2We can now add the values of KE together to find the total KE when the mass hits the ground

29.26j + 937.4j = 966.66j

4. Explain why you think that the kinetic energy is different for the flywheel and the mass, also explain why the theoretical values might be higher than actual values for this system?

The change in potential energy of the falling mass increases the kinetic energy in both the flywheel

and the falling mass but the energy used to increase the KE of the falling mass isn’t available to

increase the KE of the flywheel. The actual values could be subject to exterior influence such as

wind, temperature, human interface and quality of the products used in the experiment.

I have experienced this in the past whilst performing insulation resistance tests on copper cable,

having taken the same length of cable and using the same test kit, reading from one end of the cable

to the other has given back different results when double checked a day later. Slight changes in

temperature and positioning can make small changes to the values taken whilst performing an

experiment or test.

SOUD1214 Energy Transfer.The energy stored in the flywheel from part 1 is to be transferred to the glider by means of a clutch and tow-rope arrangement . The flywheel is instantaneously coupled to the stationary clutch (at the moment the mass hits the ground). The mass of the clutch is 20kg and it has an effective diameter of 300mm and can be considered to be a solid disk.

1. Determine the angular velocity of the combination.

Final Angular Velocity (Ѡf) = I1xѠ1 + I2xѠ2 I1+I2

First we need to find the inertia (I2) D ₂ 0.3 ₂I2 = M 2 => 20 2 = 0.225kgm/s₂ 2 2The clutch itself has no angular velocity so we can now work out Ѡf

Ѡf = (44.1 x 6.52) + (0.225 x 0) = 287.532 = 6.48rad/s 44.1 + 0.225 44.325So the angular velocity of the combination is 6.48rad/s

2. Calculate the loss in kinetic energy as the clutch engages.

We need to add the two values of Inertia together 44.1 + 0.225 =44.325.

Now we can find the kinetic energy in the clutchKe = IѠ₂ = 44.325 x 6.48₂ = 930.6 2 2

We want to find the loss in kinetic energy so we need to take our value of rotational KE (RKE), used in uniform acceleration question 3. This value is 937.4j

To find the loss we need to subtract the KE in the clutch from the RKE

937.4 – 930.6 = 6.8jSo the total loss in Kinetic energy as the clutch engages is 6.8J

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Oscillating mechanical systems

1. A spring of stiffness 15kN/m supports a mass of 5kg. The mass is pulled downwards by 12mm and released to produce linear oscillations. Calculate the frequency of oscillation and the periodic time. Hence determine the maximum linear velocity and maximum linear acceleration for the system.

First we can find the angular frequency Ѡ = k = 15000 = 3000 = 54.7723rad m 5we can now find the frequency F = Ѡ = 54.7723 = 8.717hz 2π 2π

Time period (T) = 1 = 1 = 0.115 seconds F 8.717

t = T = 0.115 = 0.0287 Xᴑ = -12 4 4

Velocity = dX = ѠXᴑSin(Ѡt) and acceleration = dV = Ѡ₂XᴑCos(Ѡt) dt dt

X = -12Cos(Ѡt) = -12Cos(54.7723x0.0287) = 0.014

dX = (54.7723x-12) x (Sin54.7723x0.0287)dt ( -657.26) x (0.999) = -656.60mmp/s

dV = (54.7723₂x-12) x (Cos54.7723x0.0287)dt (-36000) x (-1.17) = 42120mm/s

Maximum Acceleration = T = 0.115 = 0.0575mm/s 2 2

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For the spring mass system describe the effects of damping, particularly critical damping and describe the principle of simple harmonic motion (SHM). Explain what happens when the system vibrates at its natural frequency, when it vibrates at a forced frequency and when it’s forced to vibrate at the same frequency as its natural frequency.

Damping is an effect that reduces the amplitude of oscillations in an oscillatory system. The rate that

a spring mass system is bought to rest depends on the degree of damping. Three types of damping

are known as light, heavy and critical damping. We get critical damping when the time taken for the

displacement to become zero is a minimum without oscillation. A good example of critical damping

can be found in the recoil mechanism in a gun, after firing a weapon, the sooner the mechanism is

returned to its original position the sooner it can be used again.

The diagram to the left taken from Tooley and Dingle, (2012,P212), clearly shows the difference

between light, heavy and critical damping. With light damping, the number of oscillations which

occur before the displacement of the motion is reduced to zero is high and heavy damping occurs

when the displacement is reduced very quickly.

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any motion that repeats itself with equal time intervals is called

periodic motion, within this periodic motion find simple harmonic

motion (SHM). SHM is a type of periodic motion in which the

acceleration of the body is always directed down a fixed path and

the magnitude of the acceleration of the body is proportional to its

distance from the fixed point.

The natural frequency of the system is found when the system oscillates freely once set into motion

without any external force. Damping and resonance can also affect the natural frequency, resonant

frequency is where an external forces vibration matches the natural frequency of the spring, this

increases the amplitude of the oscillation and therefore increases the overall frequency.

The above diagram taken from hyperphysics.phy shows the simple harmonic motion of a spring

mass system.