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Lesechka [4]
3 years ago
6

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

he uniform distributed load that can be applied to the beam without risk of causing the strut to buckle. take f.s. = 2 against bucklin

Physics
1 answer:
EastWind [94]3 years ago
3 0

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

Ay = By = \frac{w * 6}{2} = 3w

P_c_r = 3w * F.S = 3w * 2.0 = 6w

I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7

To find the maximum intensity, w, let's take the Pcr formula, we have:

P_c_r = \frac{\pi^2 E I}{(KL)^2}

Let's take k = 1

E = 200*10^9

Substituting figures, we have:

6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}

Solving for w, we have:

w = \frac{67266.84}{6}

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

\sigma _c_r = \frac{w}{A}

\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y. This means it is safe

The maximum intensity w = 11.211KN/m

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b) (7.308 × 10⁻⁷) Wb

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b) Another loop lies in the same plane but has an irregular shape, resembling a starfish. Its area is three times greater than that of the square loop

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Hope this Helps!!!

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Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
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6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


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