shear stress(1).ocw

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Hydroelectric Energy

Hydroelectric Energy

According to the United Nations, two thirds of the world's hydroelectric potential is being used, especially in North America and Europe. China is The world's largest producer ofhydroelectricity (95,000 MWinstalled), followed by the UnitedStates, Canada, and Brazil

SO; WHAT IS THE HYDROELECTRIC ENERGY?
SIMPLY; THE KINETIC ENERGY OF THE RIVERS IS TRANSFORMED INTO MECHANICAL ENERGY BY TURBINES THEN TRANSFORM THE MECHANICAL ENERGY INTO ELECTRICAL ENERGY.
SO WE NEED :

1-Turbine Room: The place where the kinetic energy of the rivers is transformed into mechanical energy by turbines and later into electrical energy by generators.
2-Generator: It transforms the mechanical energy of the turbines into electrical energy.

In brief :
1-Water enters the powerhouse under pressure and is injected into the turbine.
2-The force of the water on its blades causes the turbine to turn.
3-The turbine makes the generator turn, thereby producing electric energy. The water is returned to the river.
 

Plants :

1-Bypass Plant: Does not have a reservoir. It simply takes advantage of the available flow of water and thus is at the mercy of seasonal variations in water flow. It also cannot take advantage of occasional surplus water.

click to enlarge.

2-Plants with Reservoirs: The presence of a reservoir, formed by a containment dam, guarantees a constant flow of water—and, therefore, of energy—independent of variations in water level.

click to enlarge .

The water enters the powerhouse and turns the turbines. The generators produce electricity

For Pumping Plant it has also two reservoirs that In off-peak hours, thewater is pumped tothe first reservoir to be reused.

"Now about 20 percent of the world's electricity is generated by the force of rivers through the use of hydroelectric power plants. This technology, used since the 19th century, employs a renewable, nonpolluting resource, although the technology's impact on the environment is high."

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Notes "15" - Shear Stress - Part 2

Notes "15" - Shear Stress - Part 2

Now after we knew the main style to calculate the shear stress at any layer of the cross section we need to determine the shear stress for more problems to with more complex sections of Structure elements. Now let’s have the first problem.

Problem (1)  :

For the following determine the shear force that applied on the nail assuming that we apply distributed shear force (w) that any cross have shear force element (s) that the cross section is symmetric round the shear force element (s) direction.

Solution :

  
At first we should be certain that the section is symmetric round the shear force the apply on the section , For our case the cross section is symmetric round (S) so we could use our previous style of solving this package problems.

The shear force that the nail will feel is due to the shear stress between the section layers at the mid layer of the section .
So we should determine the shear stress at the mid layer .

The shear force apply to the nail not only depend on the the shear stress of the mid layer but also the area that the nail carry its load this area =( the width of the layer *distance between two nails) .

by knowing the diameter of the nail we could also calculate the shear stress of the nail

Note : The nail is also symmetric round the cross section

Another problem :

Calculate the shear force that apply on any nail of the following assuming that the distance between those 4 nails and the following group of nails at that structural element is (d).S .

Solution :

we use the same relation ….the same way

Assuming that the diameter of the nails is very small .

 

(we learned before how to calculate the second moment of area)
now we should ask which length will represent b in shear stress relation .

(b) Here will be the h as the shear stress will be at the connecting length



 

It is the time for the latest problem for today with our nails.

press to enlarge .

we need to calculate the shear force on the nails’ sections.
It seem to be hard problem with the first look but be patient and feel with nails .
If we replace the nails with ourselves what should we feel? That is the physics of the problem ,

ok let’s solve it .
 

For the upper two nails it seem to be symmetric . yes we will use our great relation .

And will get τ as the latest problem .
Note : I for this cross section will be calculated as we learned in previous lectures how to calculate I for any cross section.
Ok I think it is easy to calculate the force of the upper nails
 

for the other nail
if we separated that part.

Q: will be of separated part related to the centroid of the whole cross section .
b: will be of the connecting length without the tip as the separated part is free from the tip; it only connecting with the left & right sides .

S&I  known
Simply use :

 

The shear force on the nail Fnail