hw1 (solution) - fundamental concepts, flow lines, and shear and pressure forces

HW #1: Fundamental Concepts, Velocity Field, Flow Lines, Shear and Pressure Forces Due: Fri, Feb 10, 2012, at the ISE office.

1

HW #1: Fundamental Concepts, Velocity Field, Flow Lines, and Newton Viscosity Laws

Problem 1. The Principle of Dimensional Homogeneity

A valid physical relation for the time rate of energy transfer as heat Q [Energy/Time] due to convection over a flat

plate can be expressed as

ThAQ ,

where h is the heat transfer coefficient, A is the area of the plate [Area], and T (= fs TT ) is the temperature

difference between the plate surface and the flowing fluid above it [Temperature]. 1.1. What are the dimensions of the heat transfer coefficient h in terms of the primary dimensions

EAtT (E = Energy, A = Area, t = time, T = Temperature), and MLtT (M = Mass, L = Length, t = time, T = Temperature)?

1.2. What is the physical interpretation of the heat transfer coefficient h ?

Solution 1.1. From the valid physical relation

ThAQ ,

we have

TA

tE

eTemperaturArea

TimeEnergy

TA

Qh

//

]][[

][][

ANS

Tt

M

TL

tML

TL

ttLM

eTemperaturArea

TimeEnergy

TA

Qh

32

32

2

2 //)/(/

]][[

][][

ANS

1.2. The heat transfer coefficient h can be interpreted as the time rate of (thermal) energy transfer per unit area of the plate

and per unit temperature difference between the plate surface and the flowing fluid above it. ANS

For the same temperature difference T and the same area A if one plate has lower h (say, a smooth plate) and another plate has higher h (say, a rough plate), the time rate of (thermal) energy transfer of the one with higher h (rough plate) is more.

Q

sT

Flow, fT


2

Problem 2. Velocity Field and Flow Lines [Adapted from Fox et al., 2010, Problem 2.17, p. 45.] Consider the flow described by the velocity field

jCyiAtBxV ˆˆ)1(

,

where A , B , and C are constants and all numerical values in the above expression are dimensionless.

2.1. Is the velocity field steady? 2.2. What are the dimensions of A , B , and C in the system of primary dimensions MLtT? 2.3. Pathline:

2.3.1. Find the pathline of the particle that is located at the point ),( oo yx at time ot .

2.3.2. Find the velocity of the above particle at any time t . 2.3.3. Find the acceleration of the above particle at any time t .

2.4. Streamline: 2.4.1. Find the streamline that passes through the point ),( oo yx at time ot .

2.5. Use Excel to plot the pathline in 2.3 and the streamline in 2.4 for )1,1(),( oo yx and ot = 0 and 1 s for the case

where A = 0.5 s-1, and B = C = 1 s-1. Solution

2.1. From ),,(ˆˆ)1( tyxVjCyiAtBxV

, we see that the velocity field depends on time, 0/

tV . Hence, the

velocity field is not steady. ANS 2.2. According to the principle of dimensional homogeneity, for a valid physical relation all additive terms in the relation

must have the same dimensions. Thus, we have

jCyiBxAtiBxjCyiAtBxV ˆ)(ˆ)(ˆ)(ˆˆ)1(

1][][][ tCBA . ANS

2.3. 2.3.1. The pathline can be found as follows.

)(2

)(

22

2

)(

22

)(

)(2

)(

2

)(ln

)1(

)1(

oo

o

oo

ttA

ttB

o

oo

t

to

t

t

tx

x

extx

ttA

ttB

tA

tBx

tx

dtAtBx

dx

dtAtBx

udtdx

)(

)(

)(

)(

)(ln

v

o

o

oo

ttCo

o

t

to

t

t

ty

y

eyty

ttC

tCy

ty

Cdty

dy

Cydt

dtdy

Thus, the path line is given by

LDimensionjeyiexjtyitxtr ooo ttC

o

ttA

ttB

o

,ˆˆˆ)(ˆ)()( )()(

2)( 22

. ANS

2.3.2. The velocity of the particle in 2.3 can be found as follows.

Approach 1: Directly time differentiate the position vector to the pathline.


3

jeCyiexAtB

jCeyiAtBexjdt

tdyi

dt

tdx

dt

trdtV

ooo

ooo

ttCo

ttA

ttB

o

ttCo

ttA

ttB

o

ˆˆ1

ˆˆ1ˆ)(ˆ)()()(

)()(

2)(

)()(

2)(

22

22

Approach 2: Recognize that 1. the velocity field ),( txV

gives the velocity at any point in space at any time t , and

2. the point/location of the particle at any time t is given by the pathline.

Hence, the velocity of the particle )(tV

is the velocity at the point at which the particle currently

occupies, i.e., )(trx

. Thus,

jeCyiAteBx

jtyitxtrxjCyiAtBx

txVtV

ooo ttC

o

ttA

ttB

o

trx

trx

ˆˆ)1(

pathlineˆ)(ˆ)()(,ˆˆ)1(

),()(

)()(

2)(

)(

)(

22

Thus, VelocityDimensionjeCyiexAtBtV ooo ttC

o

ttA

ttB

o

,ˆˆ)1()( )()(

2)( 22

ANS

2.3.3. The acceleration of the particle in 2.3 can be found as follows.

onAcceleratiDimensionjeyCiexAtBAB

jeyCieAtBAtAeBx

dt

tVdta

ooo

ooooo

ttCo

ttA

ttB

o

ttCo

ttA

ttBttA

ttB

o

,ˆˆ)1(

ˆˆ1)1()(

)()(

)(2)(

2)(

2

)(2)(

2)()(

2)(

22

2222

ANS 2.4. . 2.4.1. The streamline can be found as follows.

essDimensionlx

x

y

y

y

y

Cx

x

AtB

tCy

dy

AtBx

dx

dy

u

dx

AtB

C

oo

oo

y

y

x

x oo

)1(

ln1

ln)1(

1

timefix,)1(

v

ANS


4

2.4.2.

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10

x (m)

y (m

)

Series1

Series2

Series3

Pathline

Streamline at t = 0 s


Pathline




5

Problem 3. Boundary Condition for Velocity Field, Velocity Profile, Shear Deformation, and Shear Stress [Adapted from 2141-365 – 2009, HW 1]

Qualitatively sketch the velocity profile along the transverse AB. Be mindful about physical constraints at the solid boundary. Assume for simplicity first that there is only one dominant velocity component.

For the marked surface (marked by solid dot) of the fluid element C (and D): a. state whether the shear deformation (or velocity gradient) dydu / (or drdu / ) at the point on that

surface is positive or negative, b. state whether yx (or r ) at the point on that surface is positive or negative, and

c. sketch the correct direction of the shear stress yx (or r ) on that surface at that point.

d. Also, state whether the fluid element C is being dragged forward (in x direction) or backward (in x direction) by the fluid element adjacent to it at that surface.

3.1. Flat plate in a uniform freestream (boundary layer)

3.2. Flow between two stationary, parallel plates

y

u , x Flow

A

B

Flow in a channel (Both plates stationary)

C

D

Marked surface on fluid element C

dydu / :

yx :

Forward/Backward C

Marked surface on fluid element D

dydu / :

yx :

Forward/Backward D


dydu / :

yx :

Forward/Backward

(Being dragged forward or backward by the fluid element adjacent to it at that

surface?)

y

V y

u , xA

B

C

C


6

3.3. Moving plate in atmosphere

For 3.3, if I am to set up the y coordinate such that it is positive pointing down,

will the sign of the numerical value of dydu / and yx change?

will the physical direction of the stress yx on the fluid element change?

3.4. Annular flow in concentric cylinders with very small gap relative to the radius of the inner cylinder (rotating inner

cylinder, stationary outer cylinder)

A B A C

re

e

C


drdu / :

r :

CW/CCW

(Being dragged clockwise (CW) or counter-clockwise (CCW) by the fluid

element adjacent to it at that surface?)

C

y

x , u A

B

U

Plate is moving at the speed U .

Still air

C

D


dydu / :

yx :

Forward/Backward C


dydu / :

yx :

Forward/Backward D


7

Solution 3.1. Flat plate in a uniform freestream (boundary layer)

3.2. Flow between two stationary, parallel plates

y

u , x Flow

A

B

Flow in a channel (Both plates stationary)

C

D


dydu / : positive

yx : positive

Forward/Backward forward

C


dydu / : positive

yx : positive

Forward/Backward backward

D


dydu / : positive

yx : positive


(Being dragged forward or backward by the fluid element

adjacent to it at that surface?)

Cy

V y

u , xA

B

C

V


8

3.3. Moving plate in atmosphere

For 3.3, if I am to set up the y coordinate such that it is positive pointing down,

will the sign of the numerical value of dydu / and yx change? yes

will the physical direction of the stress yx on the fluid element change? no

3.4. Annular flow in concentric cylinders with very small gap relative to the radius of the inner cylinder (rotating inner

cylinder, stationary outer cylinder)

NOTE Here, we assume that the gap a is much smaller than the radius of the inner cylinder R , Ra [the figure may not accurately depict this] such that

the velocity profile is approximately linear, and

dr

dur

.

Otherwise, the full expression for r is

rr

u

rr

u

rr

1.

For 0ru , if Ra , we have

a

R

a

R

r

u

R

R

r

u

r

u

r

u

r

u

r

u

rr AAAAAA

r

0.

A B A C

re

e

C

AAR Marked surface on fluid element C

drdu / : negative

r : negative

CW/CCW CCW

(Being dragged clockwise (CW) or counter-clockwise (CCW) by the fluid

element adjacent to it at that surface?)

C R

a

y

x , u

A

B

U

Plate is moving at the speed U .

Still air

C

D

U


dydu / : negative

yx : negative


C


dydu / : negative

yx : negative

Forward/Backward forward

D


9

Problem 4. Newton’s Viscosity Law [Slightly adapted from Fox et al., 2010, Problem 2.64, p. 50.] A concentric-cylinder viscometer is shown. Viscous torque is produced by the annular gap around the inner cylinder.

Additional viscous torque is produced by the flat bottom of the inner cylinder as it rotates above the flat bottom of the stationary outer cylinder.

4.1. Obtain an algebraic expression for the viscous torque due to flow in the annular gap of width a . 4.2. Obtain an algebraic expression for the viscous torque due to flow in the bottom clearance gap of height b . 4.3. Prepare a plot showing the ratio ab / , required to hold the bottom torque to 1 percent or less of the annulus torque,

versus the other dimensionless geometric parameter. 4.4. From the result in 4.3, if we want to reduce the ratio of the bottom torque to the annulus torque for a given HR / ,

what should we do? Also explain how this is consistent with your ‘physical intuition.’ Solution Assumptions 1. Newtonian fluid. 2. Annular gap: erurVtxV ˆ)()(),(

[steady velocity field with only u component, and no ,z dependent].

Bottom gap: ezruzrVtxV ˆ),(),(),(

[steady velocity field with only u component, and no dependent, see

Note at the end]. 3. Gaps are small such that the velocity distribution can be approximated linear across the gap. 4. = constant.

4.1.

Infinitesimal area rrr edzRdedAAd ˆ)(ˆ

(1)

Infinitesimal force edFFd ˆ

: rRrr dAdF

re

e

redzRdAd ˆ)(

Tope view

d

RRru )(

R 0)( aRru

re

e

ze

rredAAd ˆ

z

dF

z = 0


10

: a

R

RaR

R

dr

du

RrRrr

)(

0

:

dzd

a

RdA

a

RdF r

2

edzd

a

RedA

a

RFd r ˆˆ

2

(2)

Infinitesimal torque zrz eRdFedFeRFdredTTd ˆ)(ˆˆˆ

dzda

RRdFdT

3

(3)

Total torque a

HRH

a

Rdzd

a

RdTT

H 33

0

2

0

3 22

The above algebraic expression for torque can be re-arranged more physically as

TorqueDimensionRRHa

R

a

HRT

sA

:

3

)2(2

. (4) ANS

4.2.

Infinitesimal area zzz edrrdedAAd ˆ)(ˆ

(negative surface area) (5)

Infinitesimal force edFFd ˆ

: zzz dAdF0

: b

r

b

r

dz

du

zzz

)(0

0

00

:

drd

b

rdA

b

rdF z

2

edrd

b

redA

b

rFd z ˆˆ

2

(6)

Infinitesimal torque zrz erdFedFerFdredTTd ˆ)(ˆˆˆ

drdb

rrdFdT

3

(7)

Total torque 24

2 44

0

2

0

3 R

b

R

bdrd

b

rdTT

R

The above algebraic expression for torque can be re-arranged more physically as

zzedAAd ˆ

Z = 0

Z = -b

re

e ze

dFd r


11

TorqueDimensionRRb

RR

bT

bA

:

24

)(22

. (8) ANS

4.3 b

a

H

R

aHR

bR

T

T

side

bottom

4

1

/)2(

)2/()(3

4

For side

bottom

T

T, we must have

H

R

a

b

b

a

H

R

T

T

side

bottom

4

1

4

1 .

For 01.0 , we must have H

R

a

b25 .

That is, in order to hold the bottom torque bottomT to 1 percent or less of the annulus torque sideT , we must have

the ratio of the bottom gap to the annulus gap ( ab / ) being equal to or greater than twenty five times the ratio of the radius to the height ( HR / ) of the cylinder. The plot is shown below. ANS

0

10

20

30

40

50

0 0.5 1 1.5 2

R/H

b/a b/a

Tbottom < (0.01) Tside

Tbottom > (0.01) Tside

b/a = 25 (R/H)Tbottom = 0.01 Tside

4.4. For a given HR / , we can reduce the ratio of the bottom torque to the annulus torque by reducing the ratio ba / . This can be achieved by either reducing a or increasing b , or both.

All other parameters being equal, widening b decreases the shear at the bottom, hence reduces the bottom torque, and reducing a increases the shear on the side, hence increases the annulus side torque. ANS

Note

Velocity field in the bottom gap Since the velocity u at the bottom of the cylinder is linear with respect to the radial coordinate r ,

rzru )0,( [no-slip condition]

and we also assume that ay any r the velocity u is linear across the gap, i.e., linear wrt the axial coordinate

z ,


12

rb

bzzru

b

r

bz

zru

b

r

bz

zru

b

bruru

bz

bruzru

),(

),(

)(0

0

)(

0),(

)(0

),()0,(

)(

),(),(

Thus, in effect, we assume that the velocity field u is given by rb

bzzru

),( ,

and the velocity field in the bottom gap by erb

bzezruzrV ˆˆ),(),(

.

rru )0,( RRu )0,(

bz

0z

r

),( zru

0),( bru0),( bRu


13

Problem 5. Static Fluid [Fox et al., 2010, Problem 3.18, p. 82.] A partitioned tank as shown contains water and mercury. 5.1. What is the gage pressure in the air trapped in the left chamber? 5.2. What pressure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces

to the same level?

Solution Assumptions

1. Static fluid 2. Fluids each has constant specific weight . w = 9,810 N/m3 , M = 132,925 N/m3.

Basic Equation: )( 1212 zzppdz

dp

Pamm

Nm

m

N

zzzzpp

zzzzpp

zzpp

zzpp

bcMabwca

bcMabwac

bcMbc

abwab

483,3)1.0(925,132)1(9810:

)3()()(:

)()()2()1(

)2()(

)1()(

33

The gage pressure in the air trapped in the left chamber is 3.48 kPa. ANS (1.1)

If the free surfaces are at the same level, we have ca zz .

Since the volume of water is conserved, we still have ba zz = 1 m.

Equation (3) can be evaluated to give

Pamm

Nm

m

N

zzzz

zzzzpp

baMabw

bcMabwca

115,123)1(925,132)1(9810

)()(

)()(

33

In order to bring the water and mercury free surfaces to the same level, the air on the left need to be pumped to 123 kPa gage. ANS (1.2)

a

b c z

1 m a

b

c

(1.1) (1.2)


14

Problem 6. Resultant Pressure Force on A Flat Surface [Fox et al., 2010, Problem 3.53, p. 85.] A plane gate of uniform thickness holds back a depth of water as shown. Find the minimum weight (of the gate)

needed to keep the gate closed.

Solution

Assumptions

1. Static fluid 2. Constant specific weight . w = 9,810 N/m3.

The condition of the minimum weight occurs when there is no reaction force N on the gate at the floor end.

Hence, no reaction force N on the FBD. Since the net pressure force (forces on both surfaces of the gate) on the FBD of the gate is required for the

equilibrium of the gate, we use the gage pressure distribution on the FB as demonstrated in the derivation above. Approach 1

Net pressure force on both surfaces of the gate:

Adgh

AdghpAdp

AdpAdpFd

oo

lluu

)(

)()(

)()(

Hence, result in the use of gage pressure distribution in the FBD for the equilibrium of the gate above.

x

y

W

xR

yR

F

o

l

y

W

xR

yR o

gage pressure distribution gh

uAd

AdAd l

L/2 h

y

AdghFd

)(


15

FBD and Equilibrium of the gate

0cos2

:0 ydFL

WM o

tan3

23

sin

cos

2

3

sin:

,sin,sin:

,:

,:

cos

2

2

3

3

0

2

gwL

gwL

LW

gwL

wdydAyhdyygw

ghpyhdAg

dApdFdAypydF

ydFL

W

L

g

gg

Thus,

N

mmm

N

gwLW

o

966,67

30tan32810,93

2

tan3

2

223

2

Thus, the minimum weight needed to keep the gate closed is 68.0 kN. ANS Approach 2 FBD and Equilibrium of the gate

0cos2

:0 FlL

WM o (1)

cos

2

L

FlW (2)

Net Pressure Force For a flat gate above, we have the net pressure force due to the pressures on both sides of the gate given by

)3(sin2

AL

g

AghApF cc

where cc ghp is the gage pressure at the centroid of the gate area, ch is the depth of the centroid from free

surface, and A is the total area of the gate.

[Note that the force kNmm

m

NA

LgF o 15.44630sin

2

3810,9sin

22

3 .]

The net force F acts through the centroid of the volume of pressure distribution. Thus, we have 3/2Ll . (4)

Equations (2)-(4) then give

N

mmm

N

gwLL

LLwgL

L

FlW

o

966,67

30tan32810,93

2

tan3

2

cos

1

3

2

2

)(sin2

cos

2

223

2

Thus, the minimum weight needed to keep the gate closed is 68.0 kN. ANS


16

NOTE: Hydrostatic Force on A Flat Surface: Volume of Pressure Distribution

AC = Centroild of the area

VC = Centroild of the volume of pressure distribution

pC = Center of pressure

Magnitude of the force F

ApdFd

)ˆ(ˆ kdApkdF dVpdAdF ,

where pdAdV : is the infinitesimal volume of pressure distribution at Ad

.

VdVpdAFVA

,

where pdAdVVV

: is the volume of pressure distribution.

Direction of the force F

Compressively normal to the surface.

Line of action of F

dVrV

r

kdVrVrdVrVr

kdVrVr

VdVpdAFkdVrkVr

dVpdAkdVrkpdArkFr

FdrFr

VA

1

ˆ toparallenot for,

,0ˆ

,ˆˆ

:,ˆ)(ˆ)(ˆ

x

o op

g

pC

ApdFd

p

pdAdV

A

pdAV

VC

The height of the pressure volume is equal to the pressure p.

pC

AC

AC

y

y

z

o

r

r


17

Thus, the line of action of F

passes through the centroid of the volume of pressure distribution VC .


18

Problem 7. Resultant Pressure Force on A Curved Surface [Adapted from Fox et al., 2010, Problem 3.75, p. 87.] A dam is to be constructed across the Wabash River using the cross-section shown. Assume the dam width w = 50 m.

For water height H = 2.5 m, calculate the resultant force vector due to fluid pressure at the water-dam interface (marked as red line). Note: Here, the resultant force in question is referred to the physical force that actually acts on that surface (and that

surface only).

Solution

mx

mx

mmm

mA

y

Bx

2.2

2.2

76.04.05.2

9.0

3

2

2

11

my

my

my

0

5.0

5.2

3

2

1

Assumptions

1. Static fluid 2. Constant specific weight . w = 9,810 N/m3.

3. Atmospheric pressure op = 101.325 kPa.

x

y

F

3

2

1

jYiXX ˆˆ

jYiXX ˆˆ

is the position vector that points

along the line of action of the resultant force F

.

x

y

Ad

ApdFd

Ax

Bxy

)(

H = 2.5 m

A = 0.4 m B = 0.9 m2

h

xd

3

2

1 op

jyixx ˆˆ

jyixx ˆˆ is the position vector that points to

the infinitesimal force on the infinitesimal area

element Ad


19

Resultant Force Vector Integrate from 1 3

jdyidxxd ˆ)(ˆ)(

jwdxiwdyxwdkAd ˆ)(ˆ)(

jwdxiwdyghpAdghpApdFd ooˆ)(ˆ)()()(

plate]flat verticalfrom component - theon tocontributi no[,0:

ln)()(:

))((::

ln)()(:

)ln()(:

,)()(:

,)()(:

)(:

))((::

:

)(2

1)()(:

))((::

)(2

1)()(:

2/)(:

,)()(:

)(:

))((::

:

ˆ))(())((ˆ))(())((

32

2

32323

2

1

21212

12

12

12

12

1

21

22

232323

2

21

221212

212

12

12

1

21

23

23

12

1231

3

2

2

1

2

1

2

1

2

1

2

1

3

2

2

1

2

1

2

1

2

1

3

2

2

1

3

2

2

1

yxx

Ax

AxBxxHgwxxwp

wdxghpF

Ax

AxBxxHgwxxwp

AxBHxgwxxwp

Ax

Bydx

Ax

BHgwxxwp

yHhdxyHgwxxwp

hdxgwxxwp

wdxghpF

FFF

yyyyHgwyywp

wdyghpF

yyyyHgwyywp

yHygwyywp

yHhdyyHgwyywp

hdygwyywp

wdyghpF

FFF

jwdxghpwdxghpiwdyghpwdyghp

ApdApdApdFdF

o

x

x

oy

o

xxo

x

x

o

x

x

o

x

x

o

x

x

oy

yyy

o

y

y

ox

o

yyo

y

y

o

y

y

o

y

y

ox

xxx

componenty

x

x

o

x

x

o

componentx

y

y

o

y

y

o

surfaceflatthefromonContributisurfacecurvedthefromonContributi

Hence, the resultant force vector is given by


20

jAx

AxBxxHgwxxwpiyyyyHgwyywp

jAx

AxBxxHgwxxwp

iyyyyHgwyywpyyyyHgwyywp

jFiFF

oo

o

oo

yx

ˆln)()(ˆ)(2

1)()(

ˆln)()(

ˆ)(2

1)()()(

2

1)()(

ˆˆ

1

21212

21

231313

1

21212

22

232323

21

221212

w

hich is evaluated to be

MN

mmmmm

kNmmkPa

yyyyHgwyywpF ox

2.14

)5.20(2

1)5.2(5.25081.9)5.2(50325.101

)(2

1)()(

223

21

231313

MN

m

mmmmm

m

kNmmkPa

Ax

AxBxxHgwxxwpF oy

35.8

)4.076.0(

)4.02.2(ln9.0)76.02.2(5.25081.9)76.02.2(50325.101

ln)()(

23

1

21212

Thus, the resultant force vector is given by MNjiF ˆ35.8ˆ2.14

. ANS


21

Line of Action of F

: Resultant moment about the origin oM

)(3

1)(

2)(

2

1ln)()(

2)(

2

1

)(3

1)(

2)(

2

1)(

3

1)(

2)(

2

1

ln)()(2

)(2

1

)(3

1)(

2)(

2

1)(::

)(3

1)(

2)(

2

1:

32)(

2

1:

)()(2

1:

)(2

1)(::

,0:

ln)()(2

)(2

1)(::

ln)()(2

)(2

1:

)ln(2

)(2

1:

)()(2

1:

,)()(2

1:

)(2

1)(::

)()()()(

)()(

ˆ

ˆˆˆˆ

,

31

33

21

23

21

23

1

212

21

22

21

22

32

33

22

23

22

23

31

32

21

22

21

22

1

212

21

22

21

22

32

33

22

23

22

234

31

32

21

22

21

22

3221

22

21

22

21

223

32

2

323

22

23

22

232

1

212

21

22

21

22

221

22

21

22

21

22

21

221

3

2

2

1

2

1

2

1

2

1

3

2

2

1

2

1

2

1

2

1

2

1

3

2

2

1

3

2

2

1

yyyyH

gwyywpAx

AxBAxxBxx

Hgwxxwp

yyyyH

gwyywpyyyyH

gwyywp

Ax

AxBAxxBxx

HgwxxwpM

yyyyH

gwyywpwydyghpM

yyyyH

gwyywp

yyHgwyywp

ydyyHgwyywp

hydygwyywpwydyghpM

xx

Ax

AxBAxxBxx

HgwxxwpwxdxghpM

Ax

AxBAxxBxx

Hgwxxwp

AxAxBxH

gwxxwp

xdxAx

BHgwxxwp

yHhxdxyHgwxxwp

hxdxgwxxwpwxdxghpM

wydyghpwydyghpwxdxghpwxdxghpM

wdyghpywdxghpx

ydFxdFdM

kydFxdF

jdFidFjyix

FdxMd

oo

oo

oo

o

y

y

oo

o

y

y

o

y

y

o

y

y

o

y

y

oo

o

x

x

oo

o

x

xo

x

x

o

x

x

o

x

x

o

x

x

oo

y

y

o

y

y

o

x

x

o

x

x

oo

oo

xyo

xy

yx

o

which is evaluated to be mMNM o 62.4 .

The line of action is then given by


22

mXXY

F

MX

F

FXY

MYFXF

kYFXFFXMFX

x

o

x

y

oxy

xyo

)33.059.0()(

)(

ˆ)(,

ANS


23

Excel Sheet for Problem 4

gamma (N/m^3) = rho g = 9810po (Pa) = 101325

w (m) = 50H (m) = 2.5

A (m) = 0.4B (m^2) = 0.9

x1 (m) = 0.76 y1 (m) = 2.5x2 (m) = 2.2 y2 (m) = 0.5x3 (m) = 2.2 y3 (m) = 0

x2 - x1 = 1.44 x22 - x1

2 = 4.2624 y2 - y1 = -2 y22 - y1

2 = -6x3 - x2 = 0 x3

2 - x22 = 0 y3 - y2 = -0.5 y3

2 - y22 = -0.25

x1 - A = 0.36 y23 - y1

3 = -15.5x2 - A = 1.8 y3

3 - y23 = -0.125

x3 - A = 1.8

Fx1 (MN) = -11.11Fx2 (MN) = -3.08Fx (MN) = -14.20

Fy1 (MN) = -8.35Fy2 (MN) = 0.00Fy (MN) = -8.35

Mo1 (MN-m) = -12.49Mo2 (MN-m) = 0.00Mo3 (MN-m) = 16.34Mo4 (MN-m) = 0.77Mo (MN-m) = 4.62

Fy/Fx = 0.59Mo/Fx = -0.33

Y(X) = 0.59 x + 0.33

x = 0 Y (0) = 0.33Y = 0 X = -0.55

hw1 (solution) - fundamental concepts, flow lines, and shear and pressure forces

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