Explanation:
Outer di ameter [tex]d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}[/tex]
[tex]\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10[/tex]
[tex]d_{1}=380 \mathrm{mm}[/tex]
Given loading on the cylinder [tex]P=300 \mathrm{kN}[/tex] Helix an gle of the weld form [tex]\theta=20^{\circ}[/tex]
(i) Normal stress on the plane at angle [tex]\theta=20^{\circ}[/tex] is
[tex]\sigma=\frac{P \cos ^{2} \theta}{A_{0}}[/tex]
[tex]\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)[/tex]
[tex]\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)[/tex]
[tex]=12252.21 \mathrm{mm}^{2}[/tex]
[tex]=12.25221 \times 10^{-9} \mathrm{m}^{2}[/tex]
[tex]\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}[/tex]
[tex]=-21.6 \mathrm{MPa}[/tex]
(ii) Shear stress along an angle of [tex]\theta=20^{\circ}[/tex] is [tex]\tau=\frac{P}{A_{0}} \cos \theta \sin \theta[/tex]
[tex]=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}[/tex]
[tex]=-7.86 \mathrm{MPa}[/tex]
The normal and shearing stresses are equal to -21.63 MPa and -7.84 Mpa respectively.
Given the following data:
Conversion:
Mathematically, the normal stress is given by this formula:
[tex]\sigma = \frac{P}{A} cos^2 \theta[/tex] ....equation 1.
Where:
Note: The steel pipe is hollow and as such its area would be calculated as that of a disk.
Generally, a disk can be described as a large outer circle with its smaller-inner circle expunged. Thus, the area of a disk is given by:
[tex]Area = \pi (r_o^2-r_i^2)[/tex] .....equation 2.
Where:
For the inner radius:
[tex]r_i = r_o-t\\\\r_i=0.2-0.01\\\\r_i=0.19\;m[/tex]
Substituting the parameters into eqn. 2, we have:
[tex]Area = 3.142 (0.2^2-0.19^2)\\\\Area = 3.142(0.04-0.0361)\\\\Area = 3.142 \times 0.0039\\\\Area = 0.0123\;m^2[/tex]
For the normal stress:
[tex]\sigma = \frac{-300 \times 10^3}{0.0123} \times cos^2 20\\\\\sigma =-24.49 \times 10^{6} \times 0.8830\\\\\sigma =-21.63 \times 10^{6}\;Pa[/tex]
Note: 1 MPa = [tex]1\times 10^6\;Pa[/tex]
Normal stress = -21.63 MPa.
We would derive the formula for shearing stress from the normal stress formula by applying trigonometric functions.
Note: [tex]sin2 \theta = 2sin \theta cos \theta[/tex]
[tex]\tau = \frac{P}{A} sin \theta cos \theta\\\\\tau = \sigma \times \frac{sin \theta}{cos \theta}\\\\\tau = \frac{P}{A} cos^2 \theta \times \frac{sin \theta}{cos \theta}\\\\\tau = \frac{P}{2A} sin2\theta[/tex]
Substituting the parameters into the formula, we have;
[tex]\tau = \frac{-300 \times 10^3}{2 \times 0.0123} \times sin 2(20)\\\\\tau = \frac{-300 \times 10^3}{0.0246} \times sin (40)\\\\\tau = -12.2 \times 10^6 \times 0.6428\\\\\tau = -7.84 \times 10^6\;Pa[/tex]
Note: 1 MPa = [tex]1\times 10^6\;Pa[/tex]
Shearing stress = -7.84 Mpa.
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