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# Writing Assignment Answer All Questions On Engineering (Math Problem Sample)

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Assignment 1
1 Figure 1 represents a light beam ABCD of length 10m which is simply supported at the points A and C. The beam carries a uniformly distributed load of 30kN/m between A and B and point loads of 100kN and 70kN at points B & D, as shown.
Show that the reactions at A & C are 120kN and 230kN respectively
8Rav +2×70 = 180×5+ 100×2
8Rav = 900+200-140
8Rav =960
Rav is therefore 120 kN
8Rc =180×3 + (100×6) + (70×10)
8Rc = 1840
Rc = 230 Kn

29199273297002919937855300259720316307825892631707623088597155394
SFD(KN) 6-X
11756575319129967696867
33809744551
18902083265700187441316896900 X
176672517191400
2604887433800BMD(KN)
3991702675910039957129159300 (-)
(+)
X/120 = (6-X)/160
X= 18
Max shear
= ½ x 960 x 18
=8.640KN
1 Suitable I-beam from table
The suitable beam will be of 200.9 kg/m
2 v1 = 120
v2 = 120-180 = -60
v3 = -60-100 =-160
v4 = -160= 230 = 70
therefore max moment = 120KNM at point A
i = 3 x 160/(2 ) = 240/100
2.4MPa
Which is more than there it will fail.
3. The standard beams such as the I-beams resists shear forces, while the flanges resist most of the bending moment experienced by the beam. Beam theory shows that the I-shaped section is a very efficient form for carrying both bending and shear loads in the plane of the web. On the other hand, the cross-section has a reduced capacity in the transverse direction, and is also inefficient in carrying torsion, for which hollow structural sections are often preferred. Therefore, the design and structure of the I beam makes it uniquely capable of handling a variety of loads.
1 A hollow circular shaft has an external diameter of 80mm and an internal diameter of 50mm. The shaft is designed to transmit 450kW of power at 1200rpm.
Taking G = 75GPa, answer the following questions:
* Find the maximum and minimum shear stress acting on the shaft.
I=π32D4-d4
external diameter = 0.08 m
internal diameter = 0.05 m
= π36 (0.084-x0.054) =π363.471×10-5
=3.41×10-6Nxm4
Torque
T = p×62πN = 450×1000×602×π×1200
=3579.55 Nm
Shear maximum = TRJ = 3575.55×0.043.41×10-6
= 4.1954×107 Pa change to mPa
= 4.1954×101 mPa
Minimum shear
=3575.55×0.0253.41×10-6
=26.21 mPa
(b) angle of twist
θ= ἰ LGR =41.95×1.6×10675×109×0.04 =0.022 radians
Converting to degrees = 180π×0.0223 = 1.28°
Angle of twist for minimum
Converting to degrees = 180π× 0.0223
=1.28°
Hence the angles are the same
angle of twist = 1.28°.
(c) Find the diameter of a solid circular shaft that would perform the same operation as above.
* Polar moment
Torque =3575.55
Power =450 KW
RPm =1200
G =75 Gpa
L =1.6 M
J = πD432 hence
3.41×10-6 = π32D4
D =0.07676 m =0.077 m
=77 mm
Question two
2) A hollow steel shaft with a diameter ratio of 0.8 and a length of 2.5 m is required to transmit 1250 kW at 200rpm. The maximum shear stress is not to exceed 60 MPa. Determine the following:
1 The diameters of the shaft
Diameter ratio = 0.8
P = 1.25 x 106 watts
Polar moments =
J = pi/32(D4 -d4)
Pi/32 (D4 – 0.8D4)
Pi/32 x 0.5904
= 0.05799D4
= (p x 60)/2 x pi x N
(1.250 X 106 X 60)/( 2 X 3.142 X 200)
=59659.091 Nm
TJ = ἰΥ =59659.091Nm0.05799D2 = 2×60×106D
=2×60×106×0.05799D459659.09D Nm =116.64D4D
D3 =1116.64
D3 =8.5732×10-3
D =∛8.5732×10-3 = 0.2047 meters
Larger diameter= 204.7 mm
But d= 0.8D4 = (0.8×204.7) mm
The smaller diameter = 163.73 mm
2 The maximum angle of twist
θ=TLJ = 59659.091×2.5GJ
Given that J=0.05799×0.20474
G=75×109
Converting to degrees = 0.01954×180π
=1.12°
(c) Hollow shafts are much better to take torsional loads compared to solid shafts. shear stress in a shaft subjected to torsion varies linearly from zero at the center to the maximum at the boundary. Inside a solid shaft, most of the material carries a shear stress whose value is much below the maximum shear stress which is the Interior portion of the shaft. But at the same they are adding to the weight, without contributing much to the capability of the shaft to carry torsional load. Similarly, a Hollow shaft has a greater Strength to weight ratio.
Assignment 2
1 An Airbus A 380-800, with the specification given in table 1, shown below, sits on a runway ready for take-off. Assuming it has undercarriage wheels of diameter 1.4m, is operating at maximum take-off weight (MTOW) and accelerates uniformly from rest to a take-off speed of 300 km/h in the distance given in table. Determine:
* The take-off speed in ms-1
300 x 1000 =83.33 m/s
3600
* The linear acceleration of the aircraft from rest to take-off speed
311 x 4 x 1000
560 000
=2.221m/s2
* The time taken to reach take-off speed
V= u + 2at
83.33 = 2 x 2.221x t
t= 18.76
* The angular velocity of the undercarriage wheels at take-off speed
=V/r
= 83.33/ 0.7
* The angular acceleration of the undercarriage wheels
Angular velocity/ time
=119.04/18.76
* The number of revolutions made by each wheel during the take-off run
½( angular accelaration)time
=½ x 119.04 x 18.76
Therefore revolution per wheel
=1116.5952/(2 x 3.142)
=177.64 revs
* The force required to produce the linear acceleration
T= Mαr2
=560000 x 2.23 x 0.7
= 874.160N
1 In a manufacturing plant as part of a certain process, a cylindrical drum of diameter 0.7 m and mass 15kg is rolled down an incline from rest. The radius of gyration of each drum is 0.3m. If the drum rolls down the slope without slipping and descends a height of 0.5m, calculate the following:
a.The moment of inertia, I, of the drum
I = Mk2 where M is mass and K is the radius of gyration
I = 15 x 0.32
=1.35kg m2
b. The angular velocity of the drum
W2 = (2 x 9.81 x 0.5)/ (0.352 x 0.32)
c. The linear velocity of the drum
v = wr
= 29.83 x 0.35
=10.44m/s
* The potential energy lost by the drum
P.E = mgz
=
15 x 9.81 x 5
= 735.75 J
* The kinetic energy gained by the drum
K.E = mv2/2
15 x 10.442/2
= 817.45J
2 Derive the equation for ω
= MGZ = MV2/2 + Iw2/2
W2 = 2gz/r2 + k2