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FromTheMoon [43]
1 year ago
14

A rectangular beam 200 mm deep and 300 mm wide is simply supported over a span of 8 m. What uniformly distributed load per metre

the beam may carry, if the bending stress is not to exceed 120N/mm2.
Physics
1 answer:
Alexandra [31]1 year ago
5 0

The uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

<h3>What is bending stress?</h3>

When an object is subjected to a heavy load at a specific spot, it often experiences bending stress, which causes the object to bend and tire.

The given data in the problem is;  

Bending stress, σ = 120 N/mm2  

Moment of inertia, I = 8.5 × 106 mm⁴

Depth of beam, y = d/2 = 200/2 = 100 mm  

Length of beam, L = 8 m = 8000 mm  

Width of beam, W = 300 mm

The maximum bending moment of the beam with UDL;

\rm W = \frac{wL^2}{8}

From the bending equation;

\rm M = \frac{\sigma I}{y_{max}} \\\\ M = \frac{120 \ N / mm^2 \times 8.5 \times 10^6 }{100 \ mm } \\\\ M = 10.2 \times 10^6 \ N - mm

The maximum bending moment of the beam with UDL;

\rm M = \frac{wL^2}{8} \\\\  10.2 \times 10^6 = \frac{w \times 8^2}{8} \\\\ w = 1.275 \times 10^6 \  N/mm

Hence the uniformly distributed load per meter the beam may carry will be 1.275 × 10 ⁶ N/mm.

To learn more about the bending stress, refer;

brainly.com/question/24227487

#SPJ1

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Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate
maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

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Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

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             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

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            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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