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hotjohn
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Homework Statement
if we divide the force balance equation by 2pi r dx , we would get
(P_(x+ Δx) ) - (P_x) + ( τ_(x + Δx) ) - τ_(x) = 0 , am i right ? why the notes give different eqaution ?
Can you explain how do you get the same formula derivation with the book? I can't understand it...Chestermiller said:The analysis in the book looks correct to me. Why do you have the τ's evaluated at x and x + Δx? They are shear stresses on the shell element being analyzed at r and r + Δr.
There are four horizontal forces acting on the element: two opposing but slightly different pressures acting on the sides, and two opposing but slightly different shear forces acting above and below.hotjohn said:Can you explain how do you get the same formula derivation with the book? I can't understand it...
can you expalin why the area of τ is 2 pi r dx ? why not 2 pi rdr?haruspex said:There are four horizontal forces acting on the element: two opposing but slightly different pressures acting on the sides, and two opposing but slightly different shear forces acting above and below.
The shear force varies with radius, not as a function of x.
The element is a ring, radius r, rectangular cross section dxdr. The shear forces act on the surfaces parallel to the pipe. These are bands of width dx. The radius of one is r, the other r+dr. So the surface areas are 2 pi r dx and 2 pi (r+dr)dx.hotjohn said:can you expalin why the area of τ is 2 pi r dx ? why not 2 pi rdr?
The surfaces that the ##\tau##'s are acting upon are cylindrical. In terms of R and L, what is the curved surface area of a cylinder?hotjohn said:can you expalin why the area of τ is 2 pi r dx ? why not 2 pi rdr?
ok , I understand that the shear stress act on the area = 2 pi r dx , can you explain why the pressure act on the area = 2 pi r dr ?haruspex said:The element is a ring, radius r, rectangular cross section dxdr. The shear forces act on the surfaces parallel to the pipe. These are bands of width dx. The radius of one is r, the other r+dr. So the surface areas are 2 pi r dx and 2 pi (r+dr)dx.
It acts on the end of the shell, with cross sectional area ##\Delta A = \pi (r+\Delta r)^2-\pi r^2=\pi[(r+\Delta r)^2-r^2)=\pi \Delta r(2r+\Delta r)##. In the limit of small ##\Delta r## compared to r, this becomes ##dA=2\pi r dr##.hotjohn said:ok , I understand that the shear stress act on the area = 2 pi r dx , can you explain why the pressure act on the area = 2 pi r dr ?
in the pipe , why not the r = constant, which is r ? why there is delta r ?Chestermiller said:It acts on the end of the shell, with cross sectional area ##\Delta A = \pi (r+\Delta r)^2-\pi r^2=\pi[(r+\Delta r)^2-r^2)=\pi \Delta r(2r+\Delta r)##. In the limit of small ##\Delta r## compared to r, this becomes ##dA=2\pi r dr##.
why the shear stress not act on a round cross sectional area ? if it act on the round surface area , the area would be 2 pi r dr , am I right ? the dx is for the rectangular cross sectional area , right ?Chestermiller said:The surfaces that the ##\tau##'s are acting upon are cylindrical. In terms of R and L, what is the curved surface area of a cylinder?
r is not the pipe radius. r is the radial coordinate measured from the centerline.hotjohn said:in the pipe , why not the r = constant, which is r ? why there is delta r ?
The shear stress is acting on surfaces of constant r and is oriented in the x direction. At radial coordinate r, the area of the free body surface that the shear stress acts upon is ##2\pi r dx##. There is no shear stress acting on the end surfaces of the free body. (Only the pressure acts on the end surfaces).hotjohn said:why the shear stress not act on a round cross sectional area ? if it act on the round surface area , the area would be 2 pi r dr , am I right ? the dx is for the rectangular cross sectional area , right ?
do u mean the shear stress act on the cylindrical volume , but not on the end surface of free body ? since 2pi r dx = cylindrical volume ? 2 pi dx is the circumference , times dx , we would get the volumeChestermiller said:The shear stress is acting on surfaces of constant r and is oriented in the x direction. At radial coordinate r, the area of the free body surface that the shear stress acts upon is ##2\pi r dx##. There is no shear stress acting on the end surfaces of the free body. (Only the pressure acts on the end surfaces).
Chet
Cylindrical surface, not volume. And, yes, it doesn't act horizontally on the end surface of the free body.hotjohn said:do u mean the shear stress act on the cylindrical volume , but not on the end surface of free body ?
This is surface area, not volume.since 2pi r dx = cylindrical volume ?
2 pi r is the circumference, times dx, we get the surface area.2 pi dx is the circumference , times dx , we would get the volume
Shear force is like friction.hotjohn said:do u mean the shear stress act on the cylindrical volume , but not on the end surface of free body ? since 2pi r dx = cylindrical volume ? 2 pi dx is the circumference , times dx , we would get the volume
why 2 pi r dx is sufrace area ?Chestermiller said:Cylindrical surface, not volume. And, yes, it doesn't act horizontally on the end surface of the free body.
This is surface area, not volume.
2 pi r is the circumference, times dx, we get the surface area.
when we take thickness multiply by the 2 pi r , we eou;ld get vplume , right ? can you explain further ? or can you attach a diagram ?haruspex said:Shear force is like friction.
Imagine pulling on a long tight sock. The radius of your leg is r, the thickness of the sock dr, its length x. The normal pressure is P.
The surface area of the sock contacting your leg is ##2\pi r x##, so the frictional force is ##2\pi r x P \mu##. No dr in there anywhere.
That's only multiplying two distances, so it can give area but not a volume.hotjohn said:when we take thickness multiply by the 2 pi r ,
do u mean the shear stress act on the pipe inward and outward of tha paper? while the pressure act on the surface area of pipe from the right adn the left ?haruspex said:That's only multiplying two distances, so it can give area but not a volume.
Look at http://keisan.casio.com/exec/system/1340330749.
For the shear force, we are interested in the cylindrical surfaces. In the image, these have areas ##2\pi r_1 h## and ##2\pi r_2 h##.
No, all four forces act horizontally right and left.hotjohn said:do u mean the shear stress act on the pipe inward and outward of tha paper? while the pressure act on the surface area of pipe from the right adn the left ?
##2\pi r_1 h## is the lateral area right ? shear force must act perpendicular to the lateral area , right ? so it the shear force should be inward and outward of the book , right ?haruspex said:That's only multiplying two distances, so it can give area but not a volume.
Look at http://keisan.casio.com/exec/system/1340330749.
For the shear force, we are interested in the cylindrical surfaces. In the image, these have areas ##2\pi r_1 h## and ##2\pi r_2 h##.
No, shear force acts along the surface. That's why it is called shear force, not normal force.hotjohn said:is the lateral area right ? shear force must act perpendicular to the lateral area , right ? so it the shear force should be inward and outward of the book , right ?
can someone try to explain why the shear stress force at down and above in diagram 8-11 is opposite in direction ? why in diagram 8-12 , they are in the same direction ?Chestermiller said:
8-11 considers two radii r and r+Δr on the same side of the central axis. Because of the velocity (fastest in the cental axis) the relative velocities create opposite drags either side of the elemenrthotjohn said:can someone try to explain why the shear stress force at down and above in diagram 8-11 is opposite in direction ? why in diagram 8-12 , they are in the same direction ?
for 8-11, since the velocity profile is drawn from left to right ( same direction) for both r and r+Δr ,m why it will create opposite drags?haruspex said:8-11 considers two radii r and r+Δr on the same side of the central axis. Because of the velocity (fastest in the cental axis) the relative velocities create opposite drags either side of the elemenrt
8-12 considers an entire disc of radius r, so the opposite sidea of it are on opposite sides of the central axis. The same velocity profile leads to drag in the same direction on opposite sides.
See Chet's diagram above. The two τ(r+Δr) are in the same direction, but in opposite direction to τ(r).
The velocity is greatest at the central axis. The annulus just inside radius r is moving faster than that between r and r+Δr, so drags it in the forward direction. The annulus just beyond r+Δr Is moving more slowly than that between r and r+Δr, so drags it in the backward direction.hotjohn said:for 8-11, since the velocity profile is drawn from left to right ( same direction) for both r and r+Δr ,m why it will create opposite drags?
i still don't understand why the resultant drag is in opposite direction , could you explain further?haruspex said:Δ
The velocity is greatest at the central axis. The annulus just inside radius r is moving faster than that between r and r+Δr, so drags it in the forward direction. The annulus just beyond r+Δr Is moving more slowly than that between r and r+Δr, so drags it in the backward direction.
If τ is the shear force per unit area of interface and μ is the viscosity then that equation comes straight from the definition of viscosity.hotjohn said:one more question , why the τ is given by - μ / ( du /dr ) ? why y = R -r ? where is r measured from ? where is y measured from ? it's not shown in the diagram
An object rests on a block on a carpet. You yank the carpet. The box slides on the carpet. The carpet's drag on the box is towards you. The object slides on the block. The box's drag on the object is towards you, so the object's drag on the box is away from you. The two drags on the box are in opposite directions.hotjohn said:i still don't understand why the resultant drag is in opposite direction , could you explain further?
why there is i assume you mean an object rest on a box on a carpet...haruspex said:An object rests on a block on a carpet. You yank the carpet. The box slides on the carpet. The carpet's drag on the box is towards you. The object slides on the block. The box's drag on the object is towards you, so the object's drag on the box is away from you. The two drags on the box are in opposite directions.
It might move with the box, or it might slip. Either way, the box drags the object towards you, so by action and reaction the drag on the box from the object must be away from you.hotjohn said:why there is i assume you mean an object rest on a box on a carpet...
but , i can't visualize the situation although i have tried out myself . Based on my observation , the object will moves with the box ?
ok , back to the topic . by taking the example aforementioned as anology , which is box ? which is object ? which is carpet? r ? r+ del (r) ?haruspex said:It might move with the box, or it might slip. Either way, the box drags the object towards you, so by action and reaction the drag on the box from the object must be away from you.