Analysis of Various Forces Stresses Assignment

Introduction
Analysis of various forces, stresses and strains acting on components to be used in daily
operations of human life is an important aspect since it gives the engineer time to design and
analyze the possible failure of the component. From the design and analyzes he/she is able to
design a component whose failure rate is low thus minimizing risks that leads to accidents for
instance in the helicopter. It also helps in minimizing possible losses that might occur if the
designed and fabricated component fails or its failure to withstand forces leading to the need
of redesigning and developing a new component. Torque, bending moments, induced stresses
and strains should always be analyzed before fabrication of the component. (Johnstonn,
2006) This also raises the need for a safety factor to cater for extra forces that might be
induced during normal operations of the component.
(a) Calculation of direct and shear stress acting on the 2-D element
Numerical data
M- Bending moments is 3.6KNm and 2.5KNm = 6.1KNm
e- eccentricity is 15mm
T- torque
F-force is 50KN
61 and 62 - stress
F= √T 2−M 2
1 +F1
T = MJ
I =
6.1 × 106∗π
32 ( D4−d4 )
π
64 ( D4 −d4 )
=12.2 ×106 Nmm
Therefore: F= √ (12.2× 106 )2−(6.1 ×106 )2
1 + 5.0×106 = 13.65 ×106
Where I is the moment of inertia and J is the polar moment of inertia
P= F
A = 13.6 5× 106
π
4 ¿ 402
=10860.91 Nm m2
Where P isthe pressure∧ A is the areaacted upon by pressure
Then 6Ѳ= PD
2 t =10860.91 × 40
2× 5 =¿ 43443.64N/mm

6z = PD
4 t =10860.91 × 40
4 × 5 =¿ 21721.82N/mm
Ʈxy= 6.1× 106 × 25
π
64 ( 504−404 )
=841.816 N /mm2
61,62 =
6 Ѳ+ 6 z
2 ± 1
2
√ (6 z−6 Ѳ)2+ 4 Ʈxy2
1 =32582.73+ 43443.64
2 ± 1
2
√ (32582.73−4 3443.64)2 +4 (841.816)2
1
Thus 61 = 38013.185 + 5495.32 = 43508.505N/mm
62 = 38013.185 – 5495.32 = 32517.865N/mm
Then maximum shear stress(Ʈmax) =
1
2
√ (6 z−6 Ѳ)2+ 4 Ʈxy2
1 =1
2
√ (32582.73−3443.64)2+ 4 (841.816)2
1 =5495.32 N / mm
CITATION DrT12 \l 1033 (MEGSON, 2012)
(b) Mohr’s circle

(Ibrahim, 2013)
From the diagram, principal stresses are
Max principal stress = N = 59.4N
Min principal stress = M = -13N
Max shear stress = Ʈmax = 3.78
Angle = 1
2 × φ= 1
2 × 47=23.5°
(c) From the formulae principal stress are
Max principal = 61 =
6 Ѳ+ 6 z
2 + 1
2
√ (6 z−6 Ѳ)2 + 4 Ʈxy2
1 = 50+3.6
2 + 1
2
√ (50−3.6)2 + 4(2.5)2
1 =50.1343 KNm
CITATION RCH09 \l 1033 (Hibbeler, 2009)