Which one?? please someone quick!
2 answers:
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Here is the complete question
The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod
The missing image which is the remaining part of this question is attached in the image below.
Answer:
The horizontal shear stress at point H is
Explanation:
Given that :
The internal shear force V = 80 kN = 80 × 10³ N
The moment of inertia = 64,900,000
The length = 20 mm below the centriod
The horizontal shear stress can be calculated by using the equation:
where;
Q = moment of area above or below the point H
b = thickness of the beam = 10 mm
From the centroid ;
Q =
Q =
Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³
Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³
Q = ( 38500 + 307125 ) mm³
Q = 345625 mm³
The horizontal shear stress at point H is
Answer:
acceleration will be tripled.
Explanation:
We know, when an object is performing Simple harmonic motion, the force
experience by it is directly proportional to its displacement from its mean position.
Also, F = ma , therefore, acceleration is also proportional to its displacement .
Now, F = kx
Therefore,
If we triple the displacement i.e, 3x.
Acceleration
Therefore, acceleration is also tripled.
Hence, this is the required solution.