Week 6---Frames+Friction+Fluid

FRAMES/FRAMES//INTERNAL HINGESINTERNAL HINGES

FRAMESFRAMESFRAMES

Frames & MachinesFrames & MachinesFrames & MachinesFrames & MachinesMany structures such as the frame of a car &Many structures, such as the frame of a car &

the human structure of bones, tendons & muscles are not composed entirely of 2-force members & thus cannot be modeled as trusses

Such structures are called:Frames if they areFrames if they are

designed to remain stationary & support loadsloads

Machines if they are designed to move &designed to move & apply loads

FramesFramesFramesFrames

FramesFramesFramesFramesWh t l d b ttiWhen trusses are analyzed by cutting

members to obtain free-body diagrams of joints or sections, the internal forces acting at the “cuts” are simple axialacting at the cuts are simple axial forces

This is generally not true for frames & a g ydifferent method of analysis is necessary

EXAMPLEEXAMPLEEXAMPLEEXAMPLE100 kN100 kN

B D

15 kN/m

2.5 m 2.5 mC

15 kN/m5 m

A

Draw BMD, SFD, AFD.

BEAMSBEAMS withwith

INTERNAL HINGESINTERNAL HINGES

Class ProblemClass ProblemClass ProblemClass ProblemD BMD SFD AFDDraw BMD, SFD, AFD

8kN 10kN 5kNhinge

30o

2m 3m 3m 2m

FRAMESFRAMES withwith

INTERNAL HINGESINTERNAL HINGES

FramesFramesFramesFramesT b i l i fTo begin analysing a frame:

1 d a a f ee bod diag am of the enti e1. draw a free-body diagram of the entire structure (i.e. treat the structure as a single object) & determine the reactions at itsobject) & determine the reactions at its supports

2. then draw the free-body diagrams of individual members or selected combinations of members & apply the equilibrium equations tomembers & apply the equilibrium equations to determine the forces & couples acting on them

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a Frame

Th f i bj t d t 200 N l D t iThe frame is subjected to a 200 N-m couple. Determine the forces & couples on its members

(i e AFD SFD BMD)(i.e. AFD, SFD, BMD).

Fig. 6.36


StrategyStrategy1. draw a free-body diagram of the entire frame,

treating it as a single object & attempt to determine the reactions at the supportsreactions at the supports.

2 draw the free body diagrams of the individual2. draw the free-body diagrams of the individual members & use the equilibrium equations to determine the forces & couples acting on them.the forces & couples acting on them.

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a FrameSolutionSolutionSolutionSolutionDetermine the Reactions at the Supports:Draw the free body diagram of the entire frame:Draw the free-body diagram of the entire frame:

hinge

The term MA is the couple exerted by the fixed supportsupport.

Note: Hinge = Smooth Pin, M=0

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a FrameSolutionSolutionSolutionSolutionFrom the equilibrium equations:

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a FrameSolutionSolutionSolutionSolutionAnalyse the Members:“Disassemble” the frame to obtain the free-bodyDisassemble the frame to obtain the free body diagrams of the members:

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a FrameSolutionSolutionSolutionSolutionThe equilibrium equations for member BC are:

ExampleExample -- Analysing a FrameAnalysing a FrameExample Example -- Analysing a FrameAnalysing a FrameSolutionSolutionSolutionSolutionThe equilibrium equations for member AB are:


SolutionSolutionThis completes the solution:This completes the solution:

Your task:

Draw BMD, SFD and AFD.,


Critical ThinkingCritical ThinkingWe were able to solve the equilibrium equations for member BCq q

without having to consider the free-body diagram of member AB:We were then able to solve the equilibrium equations for member

ABAB

By choosing the members with the fewest unknowns to analyze 1st, y g y ,you will often be able to solve the sequentially:But in some cases you will have to solve the equilibrium equations

for the members simultaneouslyfor the members simultaneously


Critical ThinkingCritical ThinkingE th h bl t d t i th 4 ti &Even though we were unable to determine the 4 reactions Ax, Ay, MA &

C with the 3 equilibrium equations obtained from the free-body diagram of the entire frame, we were able to determine them from the free-body diagrams of the individual members:

By drawing the free body diagrams of the members we gained 3By drawing the free-body diagrams of the members, we gained 3 equations because we obtained 3 equilibrium equations from each member but only 2 new unknowns Bx & By

FRICTIONFRICTIONFRICTIONFRICTION

Tutorial ProblemsTutorial ProblemsTutorial ProblemsTutorial Problems

Chapter 9Chapter 9pp

FRICTION

Objective:

a) Understand thea) Understand the characteristics of dry frictionfriction.

b) Draw a FBD including f i ifriction.

c) Solve problems involving ) p gfriction.

QUIZQU

1. A friction force always acts to the contact surface.1. A friction force always acts _____ to the contact surface.

A) normal B) at 45°

C) parallel D) at the angle of static friction

2 If bl k i t ti th th f i ti f ti it2. If a block is stationary, then the friction force acting on it is ________ .

A) ≤ μs N B) = μsN

C) ≥ μ N D) = μ NC) ≥ μsN D) = μk N

APPLICATIONSAPPLICATIONS

In designing a brake system forIn designing a brake system for a bicycle, car, or any other vehicle, it is important tovehicle, it is important to understand the frictional forces involved.involved.

For an applied force on the brake pads, how can we determine the magnitude and direction of the resulting friction force?

APPLICATIONS

Consider pushing a box as shown here. How can you determine if it will slide, tilt, or stay in static equilibrium?

Wh t h i l f t ff tWhat physical factors affect the answer to this question?

CHARACTERISTICS OF DRY FRICTION

Friction is defined as a force of resistance acting on a gbody which prevents or retards slipping of the body

relative to a second body.y

CHARACTERISTICS OF DRY FRICTION

Experiments show that frictional forces act tangent (parallel) to theforces act tangent (parallel) to the contacting surface in a direction opposing the relative motion oropposing the relative motion or tendency for motion.

F th b d h i th fi tFor the body shown in the figure to be in equilibrium, the following must be true:must be true:

F = P, N = W, and Wx = Ph.

CHARACTERISTICS OF FRICTION

To study the characteristics of the friction force F, let us assume that tipping does not occur (i e “h” is small or “a” is large)tipping does not occur (i.e., h is small or a is large).

Then we gradually increase the magnitude of the force P. Typically, experiments show that the friction force F varies with P as shown inexperiments show that the friction force F varies with P, as shown in the left figure above.

FRICTION CHARACERISTICS

The maximum friction force is attained just before the block begins to move

(a situation that is called “impending motion”). The value of the force is foundusing Fs = μs N, where μs is called the coefficient of static friction. The value of μs

depends on the materials in contact.depends on the materials in contact.

Once the block begins to move, the frictional force typically drops and is given byFk = μk N. The value of μk (coefficient of kinetic friction) is less than μs .

DETERMING μs EXPERIMENTALLYμs

A block with weight w is placed on an inclined plane The plane is slowly tilted until the blockplane. The plane is slowly tilted until the block just begins to slip.

Th i li i θ i d A l i f h bl kThe inclination, θs, is noted. Analysis of the block just before it begins to move gives (using Fs = μs N):+ ∑ Fy = N – W cos θs = 0

+ ∑ FX = μS N – W sin θs = 0

Using these two equations, we get

μs = (Wsinθs)/(W cos θs )= tan θs

This simple experiment allows us to find the μSbetween two materials in contact.

PROCEDURE FOR ANALYSISSteps for solving equilibrium problems involving dry friction:

1. Draw the necessary free body diagrams. Make sure that you show the friction force in the correct direction (it alwaysshow the friction force in the correct direction (it always opposes the motion or impending motion).

2. Determine the number of unknowns. Do not assume F = μS N nless the impending motion condition is gi enunless the impending motion condition is given.

3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns.q

IMPENDING TIPPING versus SLIPPING

For a givenW and h how canFor a given W and h, how can we determine if the block will slide first or tip first?slide first or tip first?

In this case, we have four unknowns (F, N, x, and P) and only three EofE.


Hence, we have to make an assumption to give us another equation.

Then we can solve for the unknowns using the three EofE.

Finally, we need to check if our assumption was correct.


Assume: Slipping occurspp g

Known: F = μs N

Solve: x, P, and N

Check: 0 ≤ x ≤ b/2


OrOr

Assume: Tipping occurs

Known: x = b/2

Solve: P, N, and FSolve: P, N, and F

Check: F ≤ μs N

EXAMPLEGiven: A uniform ladder weighs 100 N. The vertical wall is smooth (no friction). The floor is ( )rough and μs = 0.8.

Find: The minimum force P needed to move ( tip ( por slide) the ladder.

Plan:

a) Draw a FBD.a) Draw a FBD.

b) Determine the unknowns.

) M k f i ti tic) Make any necessary friction assumptions.

d) Apply EofE (and friction equations, if appropriate ) to solve for the kunknowns.

e) Check assumptions, if required.

NB A FBD of the ladderEXAMPLE

2 m

B

2 mP

100 N

2 mN

NFA

1.5 m 1.5 mNA

EXAMPLENB A FBD of the ladder

P100100 N

2 m

F

1.5 m 1.5 mNA

FA

QUIZ1. A 100 N box with wide base is pulled by a

force P and μs = 0.4. Which force μsorientation requires the least force to begin sliding?

A) A

P(A)P(B)P(C)

100 NA) A

B) B

P(C)

C) C

D) Can not be determined

2. A ladder is positioned as shown. Please indicate the direction of the friction force on h l dd B

B

the ladder at B.

A) ↑ B) ↓

A

C) D)

GROUP PROBLEM SOLVINGGiven: Drum weight = 500 N, μs = 0.5, a = 0 75 m and b = 1ma = 0.75 m and b = 1m.

Find: The smallest magnitude of P that will cause impending motion p g(tipping or slipping) of the drum.

Plan:

a) Draw a FBD of the drum.

b) Determine the unknowns.

c) Make friction assumptions, as necessary.

d) Apply EofE (and friction eqn. as appropriate) to solve for the unknowns.

e) Check assumptions, as required.

GROUP PROBLEM SOLVING

35

0.375 0.375 P

4 m m500 N 1500 N 1 m

A FBD of the drum:

X

0F

X N

GROUP PROBLEM SOLVING5P

34

50.375 m

0.375 m

P

A FBD of the drum:

m m500 N 1 m

0

X

0F

N

QUIZ

1. A 10 N block is in equilibrium. What is

Q

0 3the magnitude of the friction force between this block and the surface?

μ S = 0.32 N

A) 0 N B) 1 N

C) 2 N D) 3 NC) 2 N D) 3 N

2 The ladder AB is postioned as shown What is the2. The ladder AB is postioned as shown. What is the direction of the friction force on the ladder at B.

A) B)

BA) B)

C) ← D) ↑ AC) ← D) ↑ A

HYDROSTATICHYDROSTATICHYDROSTATIC HYDROSTATIC FORCESFORCESFORCESFORCES

fluid staticsfluid staticsfluid staticsfluid statics

T t i lT t i lTutorial Tutorial ProblemsProblemsProblemsProblems

10 7310 73 –– 10 9210 9210.73 10.73 –– 10.9210.92

FLUIDSFLUIDSFLUIDSFLUIDSFluid Statics and Fluid Dynamics form the two Fluid Statics and Fluid Dynamics form the two

constituents of Fluid Mechanics. constituents of Fluid Mechanics.

Fluid Statics deals with fluids at rest while Fluid Fluid Statics deals with fluids at rest while Fluid Dynamics studies fluids in motion. Dynamics studies fluids in motion.

A fluid at rest has no shear stress. A fluid at rest has no shear stress. Consequently, any force developed is only due to Consequently, any force developed is only due to

l il i e ee enormal stresses i.e, normal stresses i.e, pressurepressure. .

Such a condition is termed theSuch a condition is termed theSuch a condition is termed the Such a condition is termed the hydrostatic conditionhydrostatic condition. .

In fact, the analysis of hydrostatic systems In fact, the analysis of hydrostatic systems , y y y, y y yis greatly simplified when compared to that is greatly simplified when compared to that

for fluids in motionfor fluids in motionfor fluids in motion.for fluids in motion.

Though fluid in motion gives rise to manyThough fluid in motion gives rise to manyThough fluid in motion gives rise to many Though fluid in motion gives rise to many interesting phenomena, fluid at rest is by no interesting phenomena, fluid at rest is by no

means less important. means less important.

Its importance becomes apparent when we Its importance becomes apparent when we note that the atmosphere around us can benote that the atmosphere around us can benote that the atmosphere around us can be note that the atmosphere around us can be

considered to be at rest and so are the considered to be at rest and so are the oceans lakesoceans lakesoceans, lakes. oceans, lakes.

The simple theory developed here finds its The simple theory developed here finds its application in determining pressures at application in determining pressures at

different levels of atmosphere and in many different levels of atmosphere and in many pressurepressure--measuring devices. measuring devices.

Further, the theory is employed to calculate Further, the theory is employed to calculate force on submerged objects such as ships, parts force on submerged objects such as ships, parts

of ships and submarines. of ships and submarines.

The other application of the theory is in the The other application of the theory is in the calculation of forces on dams and other calculation of forces on dams and other

hydraulic systemshydraulic systems..hydraulic systemshydraulic systems. .

A practical example of a distributed loadA practical example of a distributed loadA practical example of a distributed load A practical example of a distributed load is the force imparted by water on the is the force imparted by water on the

side of a pool tank or other container orside of a pool tank or other container orside of a pool, tank, or other container or side of a pool, tank, or other container or on any submerged object.on any submerged object.

A hydrostatic analysis reduces a A hydrostatic analysis reduces a h i l t t i l t l dh i l t t i l t l dphysical system to equivalent loads physical system to equivalent loads

or forces that can be handled using or forces that can be handled using ggfamiliar techniques.familiar techniques.

(although hydrostatic concepts can be (although hydrostatic concepts can be generalised to other liquids we limit ourgeneralised to other liquids we limit ourgeneralised to other liquids, we limit our generalised to other liquids, we limit our

discussion to water)discussion to water)

The surface of the water in contactThe surface of the water in contactThe surface of the water in contact The surface of the water in contact with the atmosphere is called the with the atmosphere is called the

FREE SURFACEFREE SURFACE

The pressure at the surface is that of the The pressure at the surface is that of the atmosphere and absolute pressure patmosphere and absolute pressure p is theis theatmosphere, and absolute pressure patmosphere, and absolute pressure pabsabs is the is the sum of the gage pressure psum of the gage pressure pgg and atmospheric and atmospheric

pressure ppressure ppressure ppressure patmatm..

ppabsabs = = ppgg + + ppatmatm

Pressure in a pool or tank (river, ocean, etc.) Pressure in a pool or tank (river, ocean, etc.) increases with dept and can be found using increases with dept and can be found using

the relationthe relation

pp == γγhhp p γγhhpp == gage pressuregage pressure

γ = ρ gp p = = gage pressuregage pressureγγ = = unit weight of waterunit weight of water (9.8 kN/m(9.8 kN/m33))γγ ggh h = = the depth below the free surfacethe depth below the free surface

The pressure, and therefore the hydrostatic The pressure, and therefore the hydrostatic force, will always act only perpendicularforce, will always act only perpendicularforce, will always act only perpendicular force, will always act only perpendicular

(or normal) to any submerged surface or object)(or normal) to any submerged surface or object)

EXAMPLE 1EXAMPLE 1EXAMPLE 1EXAMPLE 1A 4m x 4m square plate is held at the bottom of aA 4m x 4m square plate is held at the bottom of aA 4m x 4m square plate is held at the bottom of a A 4m x 4m square plate is held at the bottom of a

tank with 8m of water in it, as shown. tank with 8m of water in it, as shown. Find the equivalent hydrostatic force on the plate.Find the equivalent hydrostatic force on the plate.Find the equivalent hydrostatic force on the plate.Find the equivalent hydrostatic force on the plate.

Free surface symbol

water 8 msymbol

water4m x 4m plate

4 m

EXAMPLE 1EXAMPLE 1EXAMPLE 1EXAMPLE 1SOLUTION:SOLUTION:We first calculate the pressure on the plate. (note: We first calculate the pressure on the plate. (note: this will be a uniform pressure since the height of this will be a uniform pressure since the height of

ate is constantate is constantwater is constantwater is constant

pp == γ γ h = 9 8 (8) = 78 48 kN/mh = 9 8 (8) = 78 48 kN/m22p p γ γ h 9.8 (8) 78.48 kN/mh 9.8 (8) 78.48 kN/mPa = Pascal = N/m2

Pressure is defined as force per unit area (Pressure is defined as force per unit area (pp = F/A); = F/A); therefore, we can find force therefore, we can find force FF fromfrom

F = pAF = pA= (78 48) ( 4 x 4)= (78 48) ( 4 x 4) (78.48) ( 4 x 4) (78.48) ( 4 x 4)= 1255.7 kN = 1255.7 kN

EXAMPLE 1EXAMPLE 1EXAMPLE 1EXAMPLE 1

R = 1255.7 kN

4m

4m

4m

4m

Pressure on the Pressure on the Pressure on the Pressure on the submerged objectsubmerged objectwalls of a tankwalls of a tank

NONUNIFONONUNIFORM RM

PRESSUREPRESSUREPRESSURE PRESSURE DISTRIBUTIDISTRIBUTI

ONONONON

UNIFORMUNIFORMUNIFORM UNIFORM PRESSURE PRESSURE DISTRIBUTIDISTRIBUTI

UNIFORM UNIFORM PRESSURE PRESSURE

ONON DISTRIBUTIDISTRIBUTIONON

EXAMPLE 2EXAMPLE 2EXAMPLE 2EXAMPLE 2Water completely fills the tank shown below. Water completely fills the tank shown below.

Find the equivalent hydrostatic resultant force on aFind the equivalent hydrostatic resultant force on aFind the equivalent hydrostatic resultant force on a Find the equivalent hydrostatic resultant force on a typical 1m width of wall ABCD.typical 1m width of wall ABCD.

Free Free surfacesurface

4 m

Hydrostatic Hydrostatic pressurepressurepressurepressure

1 m

EXAMPLE 2EXAMPLE 2EXAMPLE 2EXAMPLE 2SOLUTION:SOLUTION:SOLUTION:SOLUTION:Since p = Since p = γγh, the pressure increases with depth, and h, the pressure increases with depth, and this creates a nonuniformly distributed load on thethis creates a nonuniformly distributed load on thethis creates a nonuniformly distributed load on the this creates a nonuniformly distributed load on the vertical wall of the tank. The maximum pressure is at vertical wall of the tank. The maximum pressure is at the bottom of the tank and has a magnitude of the bottom of the tank and has a magnitude of gg

p p = = γ γ h = 9.8 (4) = 39.2 kN/mh = 9.8 (4) = 39.2 kN/m22pp γγ ( )( )Recognising that we are analysing a 1m wide strip of Recognising that we are analysing a 1m wide strip of

ll id hi b l d i i fll id hi b l d i i fwall, we can consider this to be a load intensity of wall, we can consider this to be a load intensity of 39.2kN/m (per meter). The non39.2kN/m (per meter). The non--uniformly distributed uniformly distributed line then appears as a triangleline then appears as a triangleline then appears as a triangle.line then appears as a triangle.

EXAMPLE 2EXAMPLE 2EXAMPLE 2EXAMPLE 2SOLUTION:SOLUTION:SOLUTION:SOLUTION:We next find the magnitude of the resultant force by We next find the magnitude of the resultant force by calculating the area of the triangular load:calculating the area of the triangular load:calculating the area of the triangular load:calculating the area of the triangular load:

RR (39 2) (4)/ 2 78 4 kN ( t )(39 2) (4)/ 2 78 4 kN ( t )R R = (39.2) (4)/ 2 = 78.4 kN (per meter)= (39.2) (4)/ 2 = 78.4 kN (per meter)

The location of the resultant force is at the centroid of The location of the resultant force is at the centroid of the triangle, onethe triangle, one--third of the height above the base:third of the height above the base:

y y = (1/3) (4) = 1.33m= (1/3) (4) = 1.33m

EXAMPLE 2EXAMPLE 2EXAMPLE 2EXAMPLE 2

4 m R = 78.4 kN

1.33 39.2 kN/m2 39.2 kN/m

Nonuniformly Nonuniformly Nonuniformly Nonuniformly RESULTANTRESULTANTdistributed distributed

loadloaddistributed distributed

line loadline load

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Week 6---Frames+Friction+Fluid