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

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Assignment 1

TASK 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

Task 2

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.

Task 3

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

=26.21×106×1.675×109×0.025 = 0.0223 radian

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

Task 3

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

Torsional loads

= (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

=1.0175×10-4 = 0.01954 radian

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

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- Writing Assignment Answer All Questions On Engineering Description: 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....6 pages/≈1650 words| 1 Source | Harvard | Literature & Language | Math Problem |