Resultant force at elbow from water flow

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foo9008
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Homework Statement


water flows in an elbow was aligned horizontally , of angle 135 degree . If the water volume is section 1 and 2 is 0.2m^3 , the elbow is 12kg , flow rate is 0.4(m^3)/s , find the resultant force

Homework Equations




The Attempt at a Solution


i found x and y component of the force
for x-component , i have P1A1 - P2A2costheta -Fx = ρQ(v1-v2costheta)
(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-45) -Fx = (1000)(0.4)( 3.18- 12.7cos-45)
Fx= 19170N to the left
for y component ,
P2A2sintheta +Fy -12(9.81) = ρQ(v2 sin theta)
(90x10^3)(pi)[(200x10^-3 / 2 )^2 ](sin-45) +Fy -12(9.81) = (1000)(0.4)(12.7sin45)
Fy = 1711 N upwards
FR = 19246N
 

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Answers and Replies

  • #2
BvU
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Are there relevant equations ?

Is there a question ?

Strange you use ## -F_x## but ##+F_y##. I think like that you let the resultant force point the wrong way

Advice: with a bit of ##\TeX## this would look a lot more agreeable to the eye ... o0)
 
  • #3
foo9008
678
4
Are there relevant equations ?

Is there a question ?

Strange you use ## -F_x## but ##+F_y##. I think like that you let the resultant force point the wrong way

Advice: with a bit of ##\TeX## this would look a lot more agreeable to the eye ... o0)
it's okay to use use different sign for Fx and Fy , right ? when the ans that we get has different sign with the initial Fx and Fy , it means it's opposite direction to the initial direction of Fx and Fy that we assume , right ?
 
  • #4
foo9008
678
4
Are there relevant equations ?

Is there a question ?

Strange you use ## -F_x## but ##+F_y##. I think like that you let the resultant force point the wrong way

Advice: with a bit of ##\TeX## this would look a lot more agreeable to the eye ... o0)
other than the sign of Fx and Fy , is my equation okay ?
 
  • #5
foo9008
678
4
Are there relevant equations ?

Is there a question ?

Strange you use ## -F_x## but ##+F_y##. I think like that you let the resultant force point the wrong way

Advice: with a bit of ##\TeX## this would look a lot more agreeable to the eye ... o0)
it's okay to use use different sign for Fx and Fy , right ?
 
  • #6
BvU
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Do you get the correct direction, then ?
 
  • #7
foo9008
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Do you get the correct direction, then ?
Is my idea correct?? I don't have the ans, I hope experts like you can check for me...
 
  • #8
BvU
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Thank you for your confidence, but I was just trying to read what you wrote and I noticed a strange asymmetry...that has disappeared since I ##TeX##ed your working -- it appears you simply took negatives on both sides for the y equation ?
(do you see how much easier the typeset equations can be read ?)

water flows in an elbow was aligned horizontally , of angle 135 degree . If the water volume is section 1 and 2 is 0.2m^3 , the elbow is 12kg , flow rate is 0.4(m^3)/s , find the resultant force
In combination with the picture I translate/guess that they want us to calculate the force this elbow exerts on the pipe segment that comes in from the left ?

1. Homework Statement
see above; establish a coordinate system, e.g. x is to the right, z is upwards (as in your drawing).
##P_1, P_2##
##\vec A_1, \vec A_2##
##\vec v_1, \vec v_2##
##\theta = -135^\circ ## (why -45##^\circ##?)
##F_x##
etc

2. Homework Equations
I still miss relationships here: continuity, weight = mg, momentum balance ?

3. The Attempt at a Solution

for x-component , $$P_1A_1 - P_2A_2\cos\theta - F_x = ρQ(v_1-v_2\cos\theta)$$
(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-45) -Fx = (1000)(0.4)( 3.18- 12.7cos-45)
Fx= 19170N to the left
for y component ,
$$-P_2A_2\sin\theta -F_y +12(9.81) = ρQ(-v_2 \sin \theta)$$
(90x10^3)(pi)[(200x10^-3 / 2 )^2 ](sin-45) +Fy -12(9.81) = (1000)(0.4)(12.7sin45)
Fy = 1711 N upwards
FR = 19246N

I notice you do charge for the elbow weight, but not for the water in the elbow (0.2 m3 respectively -- meaning 0.2 and 0.2 I suppose, so some 400 kg !).

The outcome still surprises me -- I expect the water to push to the right and upwards, not to the left and upwards, but maybe the mass term dominates and it's to the right and downwards...

My aim is not to do the exercise for you, but to ask and challenge (and tease) until you have put together an answer you yourself are convinced must be correct :smile:
 
  • #9
foo9008
678
4
Thank you for your confidence, but I was just trying to read what you wrote and I noticed a strange asymmetry...that has disappeared since I ##TeX##ed your working -- it appears you simply took negatives on both sides for the y equation ?
(do you see how much easier the typeset equations can be read ?)

In combination with the picture I translate/guess that they want us to calculate the force this elbow exerts on the pipe segment that comes in from the left ?

1. Homework Statement
see above; establish a coordinate system, e.g. x is to the right, z is upwards (as in your drawing).
##P_1, P_2##
##\vec A_1, \vec A_2##
##\vec v_1, \vec v_2##
##\theta = -135^\circ ## (why -45##^\circ##?)
##F_x##
etc

2. Homework Equations
I still miss relationships here: continuity, weight = mg, momentum balance ?

3. The Attempt at a Solution

for x-component , $$P_1A_1 - P_2A_2\cos\theta - F_x = ρQ(v_1-v_2\cos\theta)$$
(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-45) -Fx = (1000)(0.4)( 3.18- 12.7cos-45)
Fx= 19170N to the left
for y component ,
$$-P_2A_2\sin\theta -F_y +12(9.81) = ρQ(-v_2 \sin \theta)$$
(90x10^3)(pi)[(200x10^-3 / 2 )^2 ](sin-45) +Fy -12(9.81) = (1000)(0.4)(12.7sin45)
Fy = 1711 N upwards
FR = 19246N

I notice you do charge for the elbow weight, but not for the water in the elbow (0.2 m3 respectively -- meaning 0.2 and 0.2 I suppose, so some 400 kg !).

The outcome still surprises me -- I expect the water to push to the right and upwards, not to the left and upwards, but maybe the mass term dominates and it's to the right and downwards...

My aim is not to do the exercise for you, but to ask and challenge (and tease) until you have put together an answer you yourself are convinced must be correct :smile:
i mean measure from another direction , so i get 45 degree , is it wrong ?
 

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  • #10
BvU
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What would you fill in for a 45 degree elbow ? Doesn't the direction of the water flow change from 0 to -135 degrees ?

And: do you see how simple (simple) ##TeX## is ?
 
  • #11
foo9008
678
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What would you fill in for a 45 degree elbow ? Doesn't the direction of the water flow change from 0 to -135 degrees ?

And: do you see how simple (simple) ##TeX## is ?
why not the angle = positive 135 degree ? why gt negative sign ? and do you mean i also missed out the water weight in my calculation for Fy , right ? i have put the -12(9.81) means the elbow weight
 
  • #12
foo9008
678
4
What would you fill in for a 45 degree elbow ? Doesn't the direction of the water flow change from 0 to -135 degrees ?

And: do you see how simple (simple) ##TeX## is ?
sorry , i noticed that cos (-135) = cos 135
 
  • #13
foo9008
678
4
What would you fill in for a 45 degree elbow ? Doesn't the direction of the water flow change from 0 to -135 degrees ?

And: do you see how simple (simple) ##TeX## is ?

after making correction , my ans is
my (150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 3.18- 12.7cos-135)
so , 20848N - Fx = 4864N
so, Fx = 15984N act to right , is it correct ?
for vertical forces , i have
-12(9.81) +(90x10^3)(pi)( [(200x10^-3)/ 2 ]^2 )sin45 -Fy -1000(9.81)(0.2) = 1000(0.4)(12.7sin45)
-81 - Fy = 3592N
Fy = -3673N , so Fy = 3673N upwards
anything wrong with this ?
 
Last edited:
  • #14
foo9008
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is there anything wrong with the working above ?
 
  • #15
haruspex
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is there anything wrong with the working above ?
(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 3.18- 12.7cos-135)
The left hand side is the net force on the water in the elbow, with positive being to the right. With that sign convention, what is the expression for the rate of increase in momentum?

I'm not sure how to interpret the information regarding the volume of the elbow. I suspect it has gained in translation, and should say that the volume between cross sections 1 and 2 is 0.2m3. As it stands, it sounds like there are two sections (i.e. lengths of pipe) each containing 0.2m3.
 
  • #16
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Don't have my Bird Stewart & Lightfoot at hand here, so I'm still wondering what relationship is being worked out :smile: . Well, re-read post #8.
I can decipher $$
P_1A_1 - P_2A_2 \cos(-{3\over 4}\pi) -F_x = \rho Q(v_1-v_2\cos(-{3\over 4}\pi))
$$with
##P_1 = 1.5 \times 10^5## N/m2, ## \quad P_2## = 1.5 \times 10^5## N/m2,
##\vec A_{1, x} = 0.04\pi ## m2, ##\quad \vec A_{2,x} = 0.01\pi\cos(-{3\over 4}\pi)##
##\rho = 1000 ## kg/m3,
## Q = 0,4 ## m3/s,
##\vec v_{1, x}, \vec v_{2,x}## ? Do you want me to check those numbers ?
##- F_x## meaning: if ## F_x## if the left pipe has to pull on the elbow, ## F_x## comes out positive ?
etc

For the z direction I would indeed stick to the 0.2 m3 between A1 and A2 .

(It is seldom a good idea to rename given coordinate systems...)
 
  • #17
foo9008
678
4
Don't have my Bird Stewart & Lightfoot at hand here, so I'm still wondering what relationship is being worked out :smile: . Well, re-read post #8.
I can decipher $$
P_1A_1 - P_2A_2 \cos(-{3\over 4}\pi) -F_x = \rho Q(v_1-v_2\cos(-{3\over 4}\pi))
$$with
##P_1 = 1.5 \times 10^5## N/m2, ## \quad P_2## = 1.5 \times 10^5## N/m2,
##\vec A_{1, x} = 0.04\pi ## m2, ##\quad \vec A_{2,x} = 0.01\pi\cos(-{3\over 4}\pi)##
##\rho = 1000 ## kg/m3,
## Q = 0,4 ## m3/s,
##\vec v_{1, x}, \vec v_{2,x}## ? Do you want me to check those numbers ?
##- F_x## meaning: if ## F_x## if the left pipe has to pull on the elbow, ## F_x## comes out positive ?
etc

For the z direction I would indeed stick to the 0.2 m3 between A1 and A2 .

(It is seldom a good idea to rename given coordinate systems...)
since flow rate = 0.4(m^3) /s , Q= A1V1 0.4= 0.04\pi ## m2 (V1) , V1 = 3.18m/s
Q= A2V2 0.4= 0.01(pi) V2 , V2 = 12.7m/s ,
 
  • #18
foo9008
678
4
Don't have my Bird Stewart & Lightfoot at hand here, so I'm still wondering what relationship is being worked out :smile: . Well, re-read post #8.
I can decipher $$
P_1A_1 - P_2A_2 \cos(-{3\over 4}\pi) -F_x = \rho Q(v_1-v_2\cos(-{3\over 4}\pi))
$$with
##P_1 = 1.5 \times 10^5## N/m2, ## \quad P_2## = 1.5 \times 10^5## N/m2,
##\vec A_{1, x} = 0.04\pi ## m2, ##\quad \vec A_{2,x} = 0.01\pi\cos(-{3\over 4}\pi)##
##\rho = 1000 ## kg/m3,
## Q = 0,4 ## m3/s,
##\vec v_{1, x}, \vec v_{2,x}## ? Do you want me to check those numbers ?
##- F_x## meaning: if ## F_x## if the left pipe has to pull on the elbow, ## F_x## comes out positive ?
etc

For the z direction I would indeed stick to the 0.2 m3 between A1 and A2 .

(It is seldom a good idea to rename given coordinate systems...)
in my previous calculation, my Fx = 16182N to the left , anything wrong with it ?
 
  • #19
BvU
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Depends what you mean with Fx
 
  • #20
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It looks like there have been significant difficulties with getting the trigonometry correct in this problem. Whenever that happens to me, I opt for expressing everything in terms of unit vectors in the x and y directions. This simplifies things considerably. foo9008, what is the unit vector perpendicular to the pipe at cross section 2 in terms of the unit vectors in the x and z directions? The force balance can then be expressed vectorially in terms of the unit vectors.

I also noticed that the Bernoulli equation has not been applied to this problem yet.
 
  • #21
foo9008
678
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Depends what you mean with Fx
force of pipe acting on water , is it correct ?
 
  • #22
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It looks like there have been significant difficulties with getting the trigonometry correct in this problem. Whenever that happens to me, I opt for expressing everything in terms of unit vectors in the x and y directions. This simplifies things considerably. foo9008, what is the unit vector perpendicular to the pipe at cross section 2 in terms of the unit vectors in the x and z directions? The force balance can then be expressed vectorially in terms of the unit vectors.

I also noticed that the Bernoulli equation has not been applied to this problem yet.
Hi Chet. Is there room for Bernoulli ? ( since P1 and P2 are given already )
 
  • #23
BvU
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force of pipe acting on water , is it correct ?
I think that is correct, yes. I was confused why you wrote ##-F_x## in your first post, but with ##{\Delta p\over \Delta t} - F = 0## it's sensible.
 
  • #24
foo9008
678
4
I think that is correct, yes. I was confused why you wrote ##-F_x## in your first post, but with ##{\Delta p\over \Delta t} - F = 0## it's sensible.
How about the whole working for fx? Anything wrong with it?
 
  • #25
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Hi Chet. Is there room for Bernoulli ? ( since P1 and P2 are given already )
Oh. Sorry. I didn't notice that the pressures were given. I wonder how closely the Bernoulli equation is satisfied for the case given. Maybe there is also supposed to be frictional pressure loss. I'm too lazy to check.

Chet
 
  • #26
foo9008
678
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I think that is correct, yes. I was confused why you wrote ##-F_x## in your first post, but with ##{\Delta p\over \Delta t} - F = 0## it's sensible.
my (150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 12.7cos(-125) -3.18)
so , 20848N - Fx = -4864N
so, Fx = 25348N act to left , is it correct ?
 
Last edited:
  • #27
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Here's how I would set up this problem using unit vectors:

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 1 is ##\vec{i}_x##

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 2 is ##\frac{\vec{i}_x}{\sqrt{2}}+\frac{\vec{i}_y}{\sqrt{2}}=\frac{(\vec{i}_x+\vec{i}_y)}{\sqrt{2}}=0.707(\vec{i}_x+\vec{i}_y)##

The pressure force exerted on the fluid in the control volume by the fluid behind it (at cross section 1) is ##P_1A_1\vec{i}_x##

The pressure force exerted on the fluid in the control volume by the fluid ahead of it (at cross section 2) is ##P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)##

The force that the pipe bend exerts on the fluid in the control volume is ##(F_x\vec{i}_x+F_y\vec{i}_y)##, where ##F_x## is the component of the force in the +x direction, and ##F_y## is the component of the force in the +y direction.

The rate of momentum entering the control volume is ##\rho Qv_1\vec{i}_x##

The rate of momentum exiting the control volume is ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)##

The rate of change of momentum of the fluid in the control volume = ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x##

So, the momentum balance on the fluid in the control volume is:
$$P_1A_1\vec{i}_x+P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)+(F_x\vec{i}_x+F_y\vec{i}_y)=-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x$$
So the component of the momentum balance in the x direction is:
$$P_1A_1+(0.707)P_2A_2+F_x=-(0.707)\rho Qv_2-\rho Qv_1$$
And the component of the momentum balance in the y direction is:
$$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$

Chet
 
  • #28
foo9008
678
4
Here's how I would set up this problem using unit vectors:

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 1 is ##\vec{i}_x##

The inwardly directed unit normal vector (i.e., directed into the control volume) at cross section 2 is ##\frac{\vec{i}_x}{\sqrt{2}}+\frac{\vec{i}_y}{\sqrt{2}}=\frac{(\vec{i}_x+\vec{i}_y)}{\sqrt{2}}=0.707(\vec{i}_x+\vec{i}_y)##

The pressure force exerted on the fluid in the control volume by the fluid behind it (at cross section 1) is ##P_1A_1\vec{i}_x##

The pressure force exerted on the fluid in the control volume by the fluid ahead of it (at cross section 2) is ##P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)##

The force that the pipe bend exerts on the fluid in the control volume is ##(F_x\vec{i}_x+F_y\vec{i}_y)##, where ##F_x## is the component of the force in the +x direction, and ##F_y## is the component of the force in the +y direction.

The rate of momentum entering the control volume is ##\rho Qv_1\vec{i}_x##

The rate of momentum exiting the control volume is ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)##

The rate of change of momentum of the fluid in the control volume = ##-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x##

So, the momentum balance on the fluid in the control volume is:
$$P_1A_1\vec{i}_x+P_2A_2(0.707)(\vec{i}_x+\vec{i}_y)+(F_x\vec{i}_x+F_y\vec{i}_y)=-\rho Qv_2(0.707)(\vec{i}_x+\vec{i}_y)-\rho Qv_1\vec{i}_x$$
So the component of the momentum balance in the x direction is:
$$P_1A_1+(0.707)P_2A_2+F_x=-(0.707)\rho Qv_2-\rho Qv_1$$
And the component of the momentum balance in the y direction is:
$$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$

Chet
(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 12.7cos(-135) -3.18)
18849 +1999 -Fx = -4666N
Fx = 25514N to the right
is this correct ?
for the component of the momentum balance in the y direction , why didnt you include the elbow weight and water weight , which is
-elbow weight + $$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$ -ρgV
-12(9.81)
+(90x10^3)(pi)( [(200x10^-3)/ 2 ]^2 )sin45 -Fy -1000(9.81)(0.2) = 1000(0.4)(12.7sin45)
-81 - Fy = 3592N
Fy = -3673N , so Fy = 3673N upwards ?
 
  • #29
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(150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 12.7cos(-135) -3.18)
18849 +1999 -Fx = -4666N
Fx = 25514N to the right
is this correct ?
What do you calculate from the equation that I gave ? (You are aware that we used opposite signs for components of the force, correct?)

for the component of the momentum balance in the y direction , why didnt you include the elbow weight and water weight , which is
-elbow weight + $$(0.707)P_2A_2+F_y=-(0.707)\rho Qv_2$$ -ρgV
-12(9.81)
+(90x10^3)(pi)( [(200x10^-3)/ 2 ]^2 )sin45 -Fy -1000(9.81)(0.2) = 1000(0.4)(12.7sin45)
-81 - Fy = 3592N
Fy = -3673N , so Fy = 3673N upwards ?
The weight of the elbow should not be included in the force balance on the fluid.

All I'm saying is that it is easier to guarantee that you get all the signs right in a problem like this if you use unit vectors.
 
  • #30
foo9008
678
4
What do you calculate from the equation that I gave ? (You are aware that we used opposite signs for components of the force, correct?)


The weight of the elbow should not be included in the force balance on the fluid.

All I'm saying is that it is easier to guarantee that you get all the signs right in a problem like this if you use unit vectors.
yes , for (cos-135), i would get -0.707 , since there is also negative sign in front of 90x10^3 , -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) , i got positive 1999N

how about the weight of water , should it be included in the calculation ? -1000(9.81)(0.2) ?
 
  • #31
22,232
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yes , for (cos-135), i would get -0.707 , since there is also negative sign in front of 90x10^3 , -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) , i got positive 1999N

how about the weight of water , should it be included in the calculation ? -1000(9.81)(0.2) ?
I interpreted the problem statement as saying that the entire section of pipe was horizontal, so that the weight of the water would be in the vertical z direction.

How do you feel about the approach I used in terms of unit vectors? Do you feel that it would make things easier, or no particular advantage?
 
  • #32
foo9008
678
4
I interpreted the problem statement as saying that the entire section of pipe was horizontal, so that the weight of the water would be in the vertical z direction.

How do you feel about the approach I used in terms of unit vectors? Do you feel that it would make things easier, or no particular advantage?
you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?
 
  • #33
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you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?
I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.
 
  • #34
foo9008
678
4
you mean the pipe is placed in x( left to right ) and y direction(inside or outside the book) only ?
I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.

can you draw a 3d diagram so that i can understand better ? thanks in advance
 
  • #35
foo9008
678
4
I don't know what you mean. What I'm saying is that, from the problem statement, I thought the centerline of the entire bend is horizontal.
so , for the forces along the y -axis , it should be 0.707(90x10^3)(pi)[(200x10^-3 /2 )^2 ] + Fy = -0.707(1000)(0.4)(12.7) , Fy = 5591N pointing outwards of the book) ?
 

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