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Sergeu [11.5K]
3 years ago
14

What is the maximum value of the bending stress at the critical cross-section?

Engineering
1 answer:
Greeley [361]3 years ago
8 0

Answer:

The bending stress increases linearly away from the neutral axis until the maximum values at the extreme fibers at the top and bottom of the beam. The maximum bending stress is given by: where c is the centroidal distance of the cross section (the distance from the centroid to the extreme fiber).

Explanation:

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A crankcase heater is often used to prevent refrigerant from mixing with compressor oil during periods of:
stira [4]

Answer:

Low ambient temperature

Explanation:

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3 years ago
Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.
Vlad1618 [11]

The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>

The sequence of operation is A - E - D - C - B - A - F

The given parameters;

  • <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
  • <em>initial unit load specified in the first machine, L₁ = 50</em>
  • <em>final unit load, L₂ = 100 </em>
  • <em>the capacity of the vehicle = 1 unit load</em>

<em />

The given sequence of operation of the vehicle;

A - E - D - C - B - A - F

<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>

The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.

100 unit loads ----------------- 6 trips

2000 unit loads --------------- ?

= \frac{2000}{100} \times 6\\\\= 120 \ trips

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

Learn more here:brainly.com/question/21468592

6 0
2 years ago
Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper
Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

\frac{w}{Q1} = \frac{35}{100}

W = 1.2*\frac{35}{100}*1000kj/s

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

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