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Rudiy27
2 years ago
3

For a cantilever beam subjected to a triangular distributed load, what is the order of the internal moment expression of the bea

m as a function of the location

Physics
1 answer:
bearhunter [10]2 years ago
6 0

Answer:

The Order is 4th order

Explanation:

Let take the diagram on the first uploaded image as an example of triangular distributed load  

Let x denote the location

To obtain the vertical force acting at point a we integrate the function  

          \frac{dA_y}{dx} = w

    where A_y is the vertical force

    dx is the change in location

     w is a single component of force acting downwards

             A_y = \int\limits^L_0 {w} \, dx

           A_y = \int\limits^8_0 {\frac{x^2}{8} } \, dx

                 =  \frac{1}{8}  [\frac{8^3}{3} ]

                  = 21.33N

To obtain the moment of of inertia \

              M_A = \int\limits^L_0 {wx} \, dx

                     =\int\limits^8_0 {[\frac{x^2}{8} ]x} \, dx

                     = \frac{8^4}{32}

                   = 128 Nm

To obtain the shear force acting at a distance x from A

             V = A_y -\int\limits^x_0 {[\frac{x^2}{8} ]} \, dx

                  = 21.33 - \frac{x^3}{24}

   To obtain the moment about the triangular distributed load section that is at a distance x from A as shown on the diagram

       = >    dM = \int\limits^x_0 {V} \, dx

          =>        M_x -M_A = \int\limits^x_0 {[21.33- \frac{x^3}{24}]} \, dx

      =>           M_x = 128 + 21.33x -\frac{x^4}{96}

Looking at this equation as a polynomial we see that the order is 4 i.e x^4

             

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A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s
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Read 2 more answers
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Answer:

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<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

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c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

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