ENGINEERING COLLEGE

What is the deflection equation for a simply supported beam with a uniformly distributed load?

Answers

Answer 1

Answer:

Answer:

$\Delta _(max)=(5wL^4)/(384EI)$

Explanation:

Given that

Load is uniformly distributed load and beam is simply supported.

Ra + Rb= wL

Ra = Rb =wL / 2

Lets x is measured from left side,then the deflection of beam at any distance x is given as

$\Delta _x=(wx)/(24EI)(L^3-2Lx^2+x^3)$

The maximum deflection of beam will at x = L/2 (mid point )

$\Delta _(max)=(w* (L)/(2))/(24EI)(L^3-2L* \left ((L)/(2)\right)^2+\left((L)/(2)\right)^3)$

$\Delta _(max)=(5wL^4)/(384EI)$

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The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Answers

Answer:

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

$W=p* \Delta V$

Where

W is the work done by the system

p is the constant pressure

$\Delta V$ is the change in volume

So, the correct option is (c) " isobaric process ".

A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.

Answers

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

A Coca Cola can with diameter 62 mm and wall thickness 300 um has an internal pressure of 100 kPa. Calculate the principal stresses at a point on the cylindrical surface of the can far from its ends.

Answers

Answer:

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

Explanation:

Given that

Diameter(d)=62 mm

Thickness(t)= 300 μm=0.3 mm

Internal pressure(P)=100 KPa

Actually there is no any shear stress so normal stress will become principle stress.This is the case of thin cylinder.The stress in thin cylinder are hoop stress and longitudinal stress .

The hoop stress

$\sigma _h=(Pd)/(2t)$

Longitudinal stress

$\sigma _l=(Pd)/(4t)$

Now by putting the values

$\sigma _h=(Pd)/(2t)$

$\sigma _h=(100* 62)/(2* 0.3)$

$\sigma _h=10.33MPa$

$\sigma _l=5.16MPa$

So the principle stress are

$\sigma _1=10.33MPa$

$\sigma _2=5.16MPa$

_________ items are similar to the free issue items, but their access is limited. (CLO5) a)-Bin stock items free issue b)-Bin stock controlled issue c)-Critical or insurance spares d)-Rebuildable spares e)-consumables

Answers

Answer:

a)-Bin stock items free issue

Explanation:

Bin stock items free issue items are similar to the free issue items, but their access is limited.

Bin stock items free issue items are similar to the free issue items, but their access is limited.

An escalator with 35° incline is designed to have two passengers per step. Find number of persons moved per hour for the design if velocity is 50cm/mins and the step tread is 600mm.

Answers

Answer:

The escalator disposes 58 passengers each hour.

Explanation:

The velocity diagram of the escalator is shown in the attached figure

We can obtain the vertical distance that the escalator moves in 1 hour as

$D_(v)=vsin(\theta )* 1hr\n\nD_(v)=50cm/min* sin(35^(o))* 60mins\n\n\therefore D_(v)=1720.73cm$

Thus in 1 hour the last thread moves 1720.73 cm

Now it is given that 1 thread = 600 mm =60 cm

Thus the number of times the last thread moves equals

$n=(1720.73)/(60)=28.68times$

Since each time last thread moves it disposes 2 passengers thus the number of passengers disposed when the thread moves 28.68 times equals

$N=28.68* 2=57.36$

Thus the escalator disposes 58 passengers each hour.

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

Vertical slope, = 2
Horizontal slope = 3
Clockwise moment = 330 ft-lb
Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

the A coordinate = (0, 5)
the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

$= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)$

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2, 2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

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