A transmission line with an imperfect dielectric is connected to an ideal time-invariant voltage generator. The other end of the
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Answer:
axial stress in bar B = 25Mpa.
Deformation of bar A = 0.4mm.
Explanation:
PS: Kindly check the attached picture for the diagram showing the two bars that is to say the bar A and the bar B.
So, we are given the following data or information or parameters which we are going to use in solving this particular question or problem. Here they are;
The cross-sectional areas of Bars A and B = 400 mm2, the modulus of elasticity of bar A and bar B = 200 GPa, applied force = 10kN.
STEP ONE: The first step is to determine or calculate the axial stress in bar B. Therefore,
Axial stress in bar B = 10 × 10³ ÷ 400 × 10⁻⁶ = 25 Mpa.
STEP TWO: The second step here is to determine or calculate the deformation of bar A. Therefore,
The deformation of bar A = 20 × 10³ ×1.5 ÷ 400 × 10⁻⁶ × 200 × 10³ = 0.375 mm.
The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A
The <em>equivalent force couple</em> system at O due to force <em>F</em> are;
Force, F = (<u>8.65·i - 4.6·j</u>) KN
Couple, M₀ ≈ <u>40.9 </u>k kN·m
The reason the above values are correct is as follows:
The known values for the <em>cantilever</em> are;
The <em>height </em>of the beam = 0.65 m
The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN
The <em>length </em>of the beam = 4.9 m
The <em>angle </em>away from the vertical the force is applied = 26°
The required parameter:
The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever
Solution:
The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;
The <em>equivalent force</em> = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j
Which gives;
The <em>equivalent force</em> ≈ (<u>8.65·i - 4.6·j</u>) KN
The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;
The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>
The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>
The sum of the moments = Anticlockwise moment - Clockwise moments
∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;
M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em> ≈ 40.9 kN·m
The couple acting at O, due to F, M₀ ≈ <u>40.9 kN·m</u>
The equivalent force couple system acting at point O due the force, F, is as follows
F = (8.65·i - 4.6·j) KN
M₀ ≈ <u>40.9 </u>k kN·m
Learn more about equivalent force systems here: