Applied

Mechanics

Course Notes

Robert Tyas

2

Content

Pin Jointed Frame Structures (Frameworks) 3 Dry Friction 11 Shear Force & Bending Moments 16 Direct & Shear Stress 37 Newton's Laws 46 Dynamics of Rotation 51 Balancing of Rotating Masses 56 Velocity Diagrams for Simple Mechanisms 62 Statics & Dynamics of Fluids 67

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Pin Jointed Frame Structures (Frameworks)

A pin-jointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless pinned joints. Since the joints are free to rotate they cannot transmit torque from one to another, they are therefore only subjected to axial loading, which may be tensile or compressive. A member in tension is called a tie, while one in compression is a strut.

Tie in Tension Strut in Compression The arrow head within the members, represent the equal and opposite forces acting in the member.

A example of a typical framework is given below:

Such a frame is said to be statically determinate, that is, capable of being solved using

equations of equilibrium. When this is not the case it would be referred to as statically

indeterminate. We shall only be considering the former.

Problem of this type can be solved either graphically or analytically. We will consider

both methods.

Graphical Method

Using this method of solution we make use of Bow’s notation which is use to describe

the framework and the forces within each member. We first place capital letter between

all of the external forces and then internally between the member which make up the

framework. You need to work around the frame in a clockwise direction and from left to

right across it, similar to that shown below. Where you start is unimportant.

A D B

C

Typical framework with

loading and supports

E

F

G

4

The load on the top right-hand corner of the framework would be referred to as force AB and its sense which is downwards would be ab as we go clockwise from A to B. Commencing at a joint at which there are no more than two unknown forces, we construct a force polygon for all internal and external forces acting at that joint. Proceed systematically though the frame, constructing a force polygon at each joint, to produce one composite force diagram. The magnitude and direction of each unknown force can then be scaled of the diagram. This method has largely been superseded by modern computer methods but is still useful with complex frames. (See later for a worked example.) Analytical Methods We will consider two analytical methods, namely; Method of Resolution at Joint and Method of Section. Method of Resolution at Joint This method is best suited when forces in all members are required. At any joint in a plane frame at which there are no more than two unknown forces, write down two equations of equilibrium by resolving forces at the joint in two mutually perpendicular directions. Solve these equations for the two unknown forces. Proceed systematically through the frame until all forces are known. External support reactions are usually determined first by the use of moment equilibrium. (See worked example later.) Method of Section This method is useful if you only need to determine the forces in one or a few members of the framework. Cut the frame, as shown in the example below by a section through the member under consideration and no more than two other members in which the member forces are unknown. Both parts of the structure can then be treated as structures in equilibrium and either part can then be solved by resolving forces or by taking moments of force about a suitable point. The choice as to whether to resolve or to take moments will depend upon the geometry of the frame and will be illustrated in a worked example that follows shortly.

Frame cut for solution using Method of Section

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Let’s use the following example to illustrate all of the above three method. Note that Bow’s notation has been used to label the frame. 50kN A 30 kN D B C R1 R2

Graphical Solution

1. Calculate the support reactions

Taking moments about R1 and assuming clockwise moments to be positive ∑M = 0 = (50 x 1.5) + (30 x 4.5) – 6R2 This gives R2 = 35 kN Equating vertical forces will give R1 (assume forces up to be positive) ∑FV = 0 = R1 +35 - 50 – 30 This gives R1 = 45 kN

2. Draw the force diagram to scale. Start with the external forces which should all be in a straight line.

d e g c f a

Force Diagram (NTS) b

E

F

G All angles are 60o

and sides are 3m

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3. Scale off the magnitude of all unknown forces. GB = -40 FG = +5.7 EF = -5.7 DE = -52 CE = +26 CG = +20 AF = -23

4. Determine whether forces are compressive or tensile using Bow’s notation – will

be explained in class. 50kN A 30 kN D B C R1 R2

Method of Resolution at Joint We already have values for the support reactions otherwise calculate them at this point. Next consider each joint in turn starting with one where at least one force is known. We will begin by considering joint BCG BG CG R2 Resolving vertical forces ∑FV = 0 = R2 + BGCos30 (We are assuming forces up are positive) Therefore 0 = 35 +BGCos30 This gives BG = -40.41 kN Resolving Horizontal forces (We are assuming that forces left to right are positive) ∑FH = 0 = -CG – BGSin30

Accuracy of answers depends upon scale of diagram and how well you draw it. Note the + and – signs have come from the next stage.

E

F

G Diagram shows nature of forces in each member.

Notice that we are assuming that all internal forces are tensile. Actual values are then substituted. A negative result implies a force is compressive.

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Therefore 0 = -CG – (-40.41Sin30) Hence CG = 40.41Sin30 = 20.21 kN Consider next joint ABGF AB AF FG BG Resolving vertical forces ∑FV = 0 = -AB – BGCos30 – FGSin60 0 = -30 – (-40.41Cos30) –FGSin60 This gives FG = 5.77 kN Resolving horizontal forces ∑FH = 0 = -AF –FGCos60 + BGSin30 0 = -AF – 5.77Cos60 + (-40.41Sin30) This gives AF = -23.09 kN The remainder of the problem is carried out in a similar fashion. You can complete the rest as a tutorial problem. Method of Section

1. Calculate the support reactions – already done.

2. Let’s assume we require the forces in members AF, FG, and CG. We therefore cut the frame as shown below.

A

D B C

E

F

G

8

3. Draw the FBD A 30 kN

F B G C 35 kN

4. Calculate the forces

It should be obvious from the free body diagram that we can determine FG by equating vertical forces. ∑FV = 0 = -30 + 35 – FGSin60 (taking forces up as positive) This gives FG = 5.77 kN Again by careful examination of the FBD we see that by taking moments about AB we should be able to determine CG. ∑MAB = 0 = (-35 x 1.5) + CG x 3Cos30 (taking clockwise moments as positive) This gives CG = 20.21 kN Next by equating horizontal forces we should be able to determine AF ∑FH = 0 = -AF – FGCos60 – CG 0 = -AF – 5.77Cos60 – 20.21 This gives AF = - 23.09 kN Summary of the Procedure of Method of Section

1. Calculate the reactions of the frame using a FBD of the whole frame 2. Isolate a portion of the frame by passing a cutting plane through the frame,

cutting no more than three members in which the forces are unknown (unless all but one of the lines of action of the cut member intersect at a point).

3. Apply the three equations of equilibrium (∑M = 0, ∑FH = 0, and ∑FV = 0) to the isolated portion of the frame and solve for the unknowns.

Again the cut members are

assumed to be in tension.

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Tutorial Problems

1. For the framework given in Fig. 1 determine the magnitude of the force in each member and state whether each is a tie or a strut. (Note that it is not usual to label the framework for you.)

2kN 3kN A All members are 1m E B D C R1 1kN R2 Fig. 1 [R1= 2.75kN; R2= 3.25kN - All of the following found by measurement - members: hb = 3.75kN (strut), ch = 1.9kN (tie), fd = 1.6kN (tie), ef = 3.2kN (strut), ag = 2kN (strut), fg = 0.85kN (tie), gh = 0.3kN (tie)]

2. Use the method of resolution at joints to verify the forces in the members that make up the top right hand joint of Fig. 1.

(hb = -3.75kN hg = 0.286 kN ga = -2.018 kN)

3. For the pin jointed frame shown below solve all the forces and reactions in the frame stating which members are tie and which are struts. The length of all vertical and horizontal members are 3m.

A D C B 15kN 30kN [Reactions: R1 = 20kN; R2 = 25kN; members: ha = 35.35kN (strut), bh = 25kN (tie), ed = 20kN (tie), ea = 29kN (strut), cg = 25kN (tie), gf = 7.07kN (strut), fa = 20kN (strut), ef = 20kN (tie), gh = 30kN (tie)]

4. The Warren truss shown below is composed of similar members all of which are 3 m long. Determine the forces in all members due to a vertical load of 90 kN at G. Use the method of resolution at joints to solve this problem.

E

F

G H

F G

H

10

(Reactions: RE = 30; RA = 60 kN; members: DE = -34.64; FE = 17.32; DF = 34.64; CD = -34.64; CF = -34.64; FG = 51.96; CG = 34.64; BC = -69.28; BG = 69.28; AG = 34.64; AB = -69.28 kN)

11

Dry Friction

Whenever a body slides on another and the surfaces are pressed together, a friction

force tangential to the surface has to be overcome before relative motion can take place

between the bodies.

There are four possible friction conditions: dry and clean; greasy or boundary; fluid or viscous; pure rolling.

We shall be concentrating on the first of these, in doing this consider the following diagram: Weight W Applied force P Frictional resistance F Normal reaction N The following experimental facts are known to be true:

(1) As P is increased, F increases (and is equal to P) until the block eventually slips. F

is then called the Limiting Friction.

(2) F α N and independent of area of contact. In other words F

N= constant =μ =

Coefficient of Static or Limiting friction.

(3) Once sliding begins F

N is still constant = μ = Coefficient of Sliding or Kinetic

friction.

(4) μKinetic is slightly less than μLimiting.

(5) If P = μN the block moves with steady speed. If P > μN the block will accelerate.

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Typical Values of Coefficients of Friction for Dry Surfaces

Material Static μ Kinetic μ Steel on steel 0.4 – 0.7 0.3 – 0.5 Steel on brass 0.3 – 0.6 0.2 – 0.4

Aluminium on steel 0.4 – 0.6 0.3 – 0.5 Wood on steel 0.3 – 0.5 0.2 – 0.4 Teflon on steel 0.03 – 0.05 0.03 – 0.05 Wood on wood 0.3 – 0.6 0.2 – 0.5

Motion on an Inclined Plane Case 1 – Force P parallel to the plane – P up the incline

Case 2 – Force P parallel to the plane – P down the plane

P

WSinθ

WCosθ θ

W

F

θ

To pull block up the incline at steady

speed

P = F + WSinθ

But F

N= μ

Hence F = μN = μWCosθ Therefore P = μWCosθ + WSinθ

P

WSinθ

WCosθ θ

W

F

θ

To lower block down incline at steady speed

P + WSinθ = F

But F

N= μ or F = μN = μWCosθ

Therefore P + WSinθ = μWCosθ

Or P = μWCosθ − WSinθ

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Angle of Friction and Total Reaction Before we look at a further three cases we need to introduce the angle of friction and total reaction. N R ϕ F P W Case 3 – P not parallel to plane – P horizontal and motion up the plane F Uniform motion R ϕ θ P P W R W θ ϕ + θ

We know that Tan ϕ = μ =F

N

In the triangle we see that Tan (ϕ + θ) = P

W

Therefore P = WTan (ϕ + θ) Case 4 – P not parallel to plane – P horizontal and motion down the plane F R θ ϕ P P Uniform R W motion θ θ W θ - ϕ

From the triangle we see that Tan (θ – ϕ) = P

W

As soon as P is applied the reaction of the plane, R, take up the position shown in the diagram and is clearly the equilibrant of P and W. As P increases from zero the angle between R and the normal to the plane increases, the maximum value being attained at the instant motion begins. That is when F =μN. This angle is called the angle of friction and is denoted

ϕ. It follows that Tanϕ = F

N= μ.

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Therefore P = W Tan (θ – ϕ) Case 5 – P at angle to plane F 90 – (θ + α) P P 90 + α - ϕ W R ϕ θ α R ϕ + θ Θ W Applying Sine Rule to triangle

P

Sin(ϕ+θ)=

W

Sin(90+α−ϕ)

Therefore P = WSin(ϕ+θ)

Sin(90+α−ϕ)

Note If motion down the incline ϕ +θ becomes ϕ – θ Tutorial Problems

1. A 200 kg mass is supported on a horizontal floor. The coefficient of static friction between block and floor is assumed to be 0.40. Calculate the force P required to cause motion to impend. The force is applied to the block (a) horizontally and (b) downwards at an angle of 30o to the horizontal.

(784.8 N; 1178.4 N)

2. A block of mass 68 kg rests on a horizontal floor. The coefficient of friction between the block and floor is 0.30. A pull of 180 N acting upwards at an angle of 30o to the horizontal, is applied to the block. Determine whether or not the block will slide.

(No since 155.88 N < 173.1 N)

3. A block of 90 kg rests on a horizontal surface. The coefficient of static friction between the block and the supporting surface is 0.50. Calculate the force P required to cause motion to impend if the force applied to the block is (a) horizontal and (b) upward at an angle of 20o with horizontal.

(441.45 N; 397.7 N)

The problems that follow are taken from Applied Mechanics by Hannah & Hillier

4. A load of mass 1350 kg lies on a gradient inclined at 60o to the horizontal. For static friction μ = 0.5, for kinetic friction μ 0.4. Calculate: (a) the pull parallel to the gradient required to prevent the load sliding down; (b) the pull required to pull the load up the gradient at constant speed.

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(8.15 kN; 14.1 kN)

5. The force required to haul a load of 500 kg along a horizontal surface is 1.2 kN. Find (a) the force parallel to a track of slope 20o required to haul the load up the incline; (b) the force required to lower it down the incline at steady speed. Assume the coefficient of friction to be the same in all cases.

(2.8 kN; 548 N)

6. A 1500 kg boat is winched steadily up a slip inclined at 25o to the horizontal. If μ = 0.5 for the surface contact of the boat and slip, find the force in the winch cable, which is parallel to the slip.

(12.9 kN)

7. A body of mass m on a rough plane inclined at 20o to the horizontal is moved steadily up the plane by a force of 200 N applied upward and parallel to the plane. When the force is reduced to 75 N, the body slides steadily downwards. Determine the values of m and μ.

(41 kg; 0.17)

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Shear Force & Bending Moments

Loaded beam are subjected to shear forces and bending moments and it is usual to draw the loaded beam followed by diagrams that depict the way the shear force and bending moment vary along the length of the beam. We shall now consider each separately. Shear Force Beam Rigidly Supported at One End with Concentrated Load at Other This is usually referred to as a cantilever. Consider any section XX of the cantilever beam shown below, the concentrated load W tends to force the piece of length x downwards relative to the remainder of the beam. i.e. to shear it off as shown in the second part of the diagram. The net transverse force on XX is called the Shear Force and denoted F. W L x X X W From one end of the beam to the other there is no change in the load therefore the shear force F is constant over the entire length of the beam and is equal to –W according to the sign convention. This is shown in the diagram which follows: 0 x Shear Force Diagram Cantilever Beam with several Concentrated Loads W1 W2 W3 0 A B C At O, F = - W1 Between 0 and A, F = –W1

-W

Sign Convention +ve -ve

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At A the shear force increases to F = - (W1 + W2) Therefore between A and B, F = - (W1 + W2) Again at B there is a further increase such that between B and C, F = - (W1 + W2 + W3) The shear force diagram is therefore a series of steps as shown below: 0 W1

W2 W3 Simply Supported Beam with Central Load y W x X Y 0 A B X Y

W

2

W

2

By observation the two reactions are w

2

From 0 to A, the net transverse force for any section XX, is Fx = +W

2

From A to B the net transverse force for any section YY, is Fy = −W

2

The shear force diagram therefore is:

W

2

0

−W

2

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Simply Supported Beam – Concentrated Load Off-Centre W a b X Y X Y R1 L R2

Wb

L

- Wa

L

Taking moments we find that R1 =Wb

L and R2 =

Wa

L

Therefore Fx = + Wb

L and Fy = -

Wa

L

Simply Supported with Several Concentrated Loads W1 W2 W3 O A B C D R1 R2 +R1

-R2 It should be fairly obvious now that Between O and A, FOA = + R1 Between A and B, FAB = +R1 – W1

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Between B and C, FBC = + R1 –W1 – W2 Between C and D, FCD = + R1 –W1 – W2 – W3 = R1 – (W1 + W2 +W3) = - R2 Simply Supported Beam which Overhangs the Supports – General Case W1 W2 W3 R1 R2

A typical shear force diagram has been drawn. The actual shape will depend on the relative values of the loads and their position. Try the following examples: Example 1 – Draw the Shear Force diagram of the beam system shown below. 60 kN 80 kN 100 kN O A B C D 0.7m R1 R2 1.4m 3m 5m

20

Example 2 - Draw the Shear Force diagram of the beam system shown below. 20 kN 40 kN 60 kN 20 kN 2.5m 2.5m 2.5m O A B C D E 1.25 1.25 m R1 R2 m Cantilever with Uniformly Distributed Load (UDL) UDL = w/unit length L X

x X

0 -wL Fx = - wx Since w is constant it follows that F α x and the graph of F is a straight line as shown above. FMax = - wL and occurs at the support. Simply Supported Beam with UDL UDL = w/ unit length x X X L

wL

2

wL

2

21

wL

2 Fx

−wL

2

By observation the support reactions are wL

2

Fx = wL

2− wx

At the left-hand support where x = 0, F = wL

2

As we move from the left-hand side towards the other end F is clearly reduced as x increases by

an amount proportional to x, so that the graph is a straight line. At the centre when x = L

2, F = 0.

We will see later that at this point, where F = 0, this is the point of maximum bending moment.

At the right-hand support when x = L, F = wL

2− wL = −

wL

2.

Combined UDL and Concentrated Load

Let’s try an example with some actual figures. x 10 kN 20 kN UDL = 15 kN/m 1.5 O A B C D 2 m 2 m 0.5 m R1 R2

For R2 take moments about R1 (CWM = ACWM) (15 x 2)x1 + (10 x 1.5) + (20 x 4.5) = R2 x 4 This gives R2 = 33.75 kN Equating vertical forces (15 x 2) + 10 +20 = R1 + 33.75

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This gives R1 = 26.25 kN

26.25

SF Diagram (NTS)

-13.75

-6.25

20

3.75

At O, F = + 26.25 kN. Between O and A, F reduces by 15 x 1.5 kN. At A, F reduces by 10 kN. Between A and B, F reduces by 15 x 0.5 kN. Between B and C, F remains constant. At C, F increases by 33.75 kN. Between C and D, F remains constant.

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Bending Moment The moment of a force about a point is the product of the force and the perpendicular distance of the point from its line of action. Such a moment, when applied to a component (shaft, beam or strut) in such a way as to result in bending, is referred to as a Bending Moment and is usually denoted by M. The units are Nm ( kNm or MNm). If you study Mechanics beyond this level you will see that knowledge of bending moments are important when calculating bending stresses. Sign convention We need to agree a sign convention for our bending moments and we shall adopt the one shown below:

Beam Rigidly Supported at One End (Cantilever) with Concentrated Load at Other W L x X X

x MMax = -WL M

For any section XX Bending moment Mx = Moment of W = -Wx We see therefore that M is a function of x, i.e. it is proportional to x, the distance from the origin. The graph of the bending moment or the bending moment diagram is a straight line as shown. At the support when x = L, Mx = -WL

Mx

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It should be noted that the bending moment causes tensile stress (and therefore an increase in length) in the upper section of the beam, and compressive stress in the lower section. At some point in between there will be a plane of zero stress. It can be demonstrated that the stress (tensile or compressive) is proportional to the distance from this “neutral” plane. Cantilever with Several Concentrated Loads The value of M at any section will be the sum of the moments of all the forces to the left of the section. Therefore, using the sign convention shown above (anticlockwise moments as negative to the left of the section); MA = -W1(L1 – L2) MB = -W1(L1 – L3) – W2(L2 – L3) MC = -W1L1 – W2L2 – W3L3 (This is clearly the maximum value. ) And the diagram is a series of straight lines. W1 W2 W3 O A B C L3

L2

L1

W1L1 W2L2 W3L3

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Try the following: Draw the bending moment diagram for the system shown. 5 kN 10 kN 20 kN 0.6 m 1.5 m 3 m Simply Supported Beam with Concentrated Load at Centre y Y x X W O A X Y L W/2 W/2 M P

MMax = WL

4

Mx My

O Q

By observation we see that the support reactions are W

2. The bending moment at any

section is the net effect of all forces acting on that part of the beam to the left of the section, clockwise direction being positive according to our sign convention. At any section XX to the left of the load

Mx =W

2x where x <

L

2

Thus for the left hand half of the beam M is proportional to the distance from the left hand support and its graph is a straight line OP.

At the centre when x = L

2

M =W

2×

L

2 =

WL

4 which is clearly the maximum value.

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For any section YY to the right of the load:

My =W

2y − W [y −

L

2]

=Wy

2− Wy +

WL

2

=WL

2−

Wy

2

=W

2(L − y)

At the right hand support when y = L, M = 0

Thus for the right hand half of beam M is proportional to the distance from the right

hand support and its graph is a straight line PQ.

Simply supported Beam with Several Concentrated Loads W1 W2 W3 0 L1 A B C D R1 L2 R2 L3

L M

We must first determine the reactions by taking moments about O About O CWM = ACWM W1L1 + W2L2 + W3L3 = R2L

0

27

Hence R2 =W1L1+W2L2+W3L3

L

R1 can now be found by equating vertical forces. R1 + R2 = W1 + W2 + W3 Therefore R1= W1 + W2 + W3 - R2 For the moments MA = R1L1 MB = R1L2 − W1(L2 − L1) Mc = R1L3 − W1(L3 − L1) − W2(L3 − L2) MD = 0 Try the following: Draw the bending moment diagram for the system shown. 60 kN 80 kN 100 kN 0.6 m 1.2 m 4 m 5.2 m

28

Beam with Loaded Overhanging Supports y Y W X W x X Y a b a W W -Wx Mx

-Wa By observation the support reactions are W For any section XX Mx = −Wx, where x < 𝑎 Hence M is proportional to x, i.e. zero at the end and –Wa at the support. For any section YY My = −Wy + W(y − a) = −Wa

Since ‘a’ is constant it follows that M is constant between the two supports and is shown as a horizontal line. Since the beam loads and overhangs are equal it follows that the RH overhang will have a M as shown. If the loads are unequal and/or the supports are not symmetrical the bending moment diagram would have to be modified as shown below.

29

W1 W2 a b c R1 R2

-Wa -Wc

General Case for System with Load between Supports W1 W2 W3

d e a b c R1 R2 MA = −W1a MB = −W1d + R1(d − a) MC = −W1(a + b) + R1b − W2(e − c) Or alternatively Mc = −W3c

C B A

If W2 is made sufficiently large, the bending moment between the supports may become positive.

30

Ty the following: Draw the bending moment diagram for the system shown below. 20 kN 40 kN 60 kN 20 kN 4 m 4 m 4 m 2 m 2 m

Cantilever with Uniformly Distributed Load (UDL) UDL = w/unit length L x X x/2 wx 0

Mx MMax = −WL

2

The only force acting to the left of XX is the weight of the portion x (= wx) which acts at

a distance x

2.

Therefore Mx = −wxx

2= −

w

2x2

The graph of M against x is therefore a parabola as shown above.

MMax = −wL ×L

2

But wL = W (total load)

Hence MMax = −WL

2

31

Simply Supported Beam with UDL UDL = w/unit length L x X X wL/2 wL/2

Mx MMax = WL

8

By observation the support reactions are wL

2

For any section XX

Mx =wL

2x − wx

x

2

=wL

2x −

w

2x2

The graph of M is therefore a parabola. MMax can be found by differentiating the above and equating to 0. dMx

x=

wL

2−

2wx

2=

wL

2− wx = 0 for max

Therefore wx = wL

2

Hence x = L

2

Therefore MMax = wL

2

L

2−

w

2

L2

4=

wL2

4−

wL2

8=

wL2

8

But total load = W =wL

Therefore MMax = WL

8

32

Uniformly Loaded Beam with Simple Supports Not at the Ends UDL = w/unit length

a + b

2

𝑏

2

A B C a b c = a R1 R2

MA = −waa

2= Mc = −

w

2a2

MB = −w (a +b

2)

(a+b

2)

2+R1

b

2

Mc = −wcc

2= −

w

2c2

Note There will be instances when the bending moment between the supports may become zero or even positive. Beam with Concentrated Load and UDL Let’s try an example with some actual figures. x 10 kN 20 kN UDL = 15 kN/m 1.5 A B C D E 2 m 2 m 0.5 m R1 R2 We have already seen this beam in the Shear Force section and have calculated the reactions which are given below.

33

R1 = 26.25 kN and R2 = 33.75 kN Consider the bending moment between A and B

M = R1x − 15 × x ×x

2= R1x − 7.5x2

When x = 0 M = 0 When x = 0.5 M = 26.25 x 0.5 – 7.5 x 0.52 = 11.25 kNm When x = 1.0 M = 26.25 x 1 – 7.1 x 12 = 18.75 kNm When x = 1.5 M = 26.25 x 1.5 – 7.5 x 1.52 = 22.50 kNm Consider the bending moment between B an C

M = R1x − 15x ×x

2− 10(x − 1.5) = R1x − 7.5x2 − 10(x − 1.5)

When x = 1.75 M = 26.25 x 1.75 – 7.5 x 1.752 – 10(1.75 – 1.5) = 20.47 kNm When x = 2 M = 26.25 x 2 – 7.5 x 22 – 10(2 – 1.5) = 17.5 kNm Working now from the other end of beam M at E = 0 M at D = -20 x 0.5 = -10 kNm As a check:

M at C = (-20 x 2.5) + (33.75 x 2) = 17.5 kNm which agrees. Plotting the bending moment diagram with this information gives: 22.5 kNm 17.5 kNm 0 -10 kNm

Bending Moment Diagram (NTS)

34

Point of Contraflexure In the above example we see that the BM changes sign through a zero value and this is called a point of contraflexure. In real situations these points are important to the designer because, for example, supports placed at such points will not be subjected to bending moments. Condition for Maximum Bending Moment UDL = w/unit length x dx F +dF M M + dM F wdx (through G) SF BM Since the small element is equilibrium we are able to write:

1. F = F

Therefore F + dF = F + wdx Hence dF = wdx

Or dF

dx= w

From the SF diagram we see that the slope is –ve, therefore we should write

dF

dx= −w

2. ACWM = CWM (about G)

M + Fdx

2+ (F + dF)

dx

2= M + dM

Small portion cut from beam of length dx showing all ‘loads’.

35

Therefore Fdx

2+ F

dx

2+ dF

dx

2= dM

Or Fdx = dM (ignoring products of small terms)

Therefore dM

dx= F

Thus we see that the slope of the bending moment diagram is equal to the shear

force F. For max (or min) bending moment, dM

dx= 0. Therefore we can conclude

that when F = 0 the bending moment is a maximum.

Tutorial Problems Draw shear force and bending moment diagram for the beams shown below. State the maximum values of BM and SF and where they occur.

(a) UDL = 5 kN/m 10 kN

600 mm 600 mm 1000 mm

(b) UDL = 10 kN/m

40 kN 1.4 m 1 m 1 m

(c) UDL = 1 Mg/m

20 kN 2 m 1 m 0.5m

36

(d) UDL = 15 kN/m

10 kN 1 m 2.5 m 2.5 m

(e) UDL = 8 kN/m

20 kN 10 kN 1 m 2.5 m 2.5 m 1.5 m

37

Direct & Shear Stress

When a body is pulled by a tensile force or crushed by a compressive force, the loading is said to be direct. Direct stresses are also found to arise when bodies are heated or cooled under constraints and in vessels under pressure. Stress (σ) Direct stress is defined as the ratio of the applied load (F) to the cross-sectional area (A), normal to the load. Hence we have the familiar equation:

Stress =Load

Area or σ =

F

A

Force has units of newton’s (N) and area has units of m2, hence stress has units of N/m2. Since stresses are often large values we often quote them in terms of:

kN/m2 = 103 N/m2 MN/m2 = 106 N/m2 GN/m2 = 109 N/m2

Note Some textbooks still use the name pascal (Pa) in place of N/m2. Strain (ε) Whenever a body is subjected to stress, whether tensile or compressive, it will experience a change in size. Bodies subjected to tension will extend; bodies subjected to compression will shorten. F

L = original length x = extension The ratio of the extension to the original length is called strain. Thus we have:

x L

Rigid Constraint

Tension Compression

Applied Heat

Induced Compression

Body subjected to

tension

38

Strain =Change in length

Original length or ε =

x

L

Since x and L have the same units it follows that strain is unitless.

Young’s Modulus of Elasticity or the Relationship between Stress and Strain If we are lucky, when we remove loading on a body it will return to its original size. This is because most materials exhibit a degree of elasticity. However, if we load beyond the materials elastic limit, permanent set or extension will occur. Young’s Modulus of Elasticity helps to define the point at which this elastic limit is exceeded. Hooke’s Law This states that within the elastic limit that strain is directly proportional to stress.

Stress

Strain= A constant E = Young’s Modulus

Since strain is unitless it follows that E has the same units as σ. E may be found in tables and has quite large values. For example E for steel is in the range 196 – 210 GN/m2. E is also related to the stiffness or rigidity of a material since the higher its value, the greater the load required to produce a given stiffness. Bars of Varying Cross-Section F

σA =F

Area A and σB =

F

Area B

Provided section A and section B are made of the same material then:

εA =σA

E and εB =

σB

E

Worked Examples 1. A rod of length 325 mm is subjected to a tensile load of 60 kN and found to extend

0.160 mm. If E = 200 GN/m2 determine the diameter of the rod.

ε =x

L=

0.160

325= 4.923 × 10−4

E =σ

ε

Therefore σ = Eε = 200×4.923× 10−4 = 0.09846 GN/m2 = 98460 kN/m2

σ =F

A

A B

The load in each section will be

the same however the stress

will be different because of the

difference in area

39

Therefore A =F

σ=

60

98460= 6.0938 × 10−4 m2

A =πd2

4

Therefore d = √4A

π= √

4×6.0938×10−4

π= 0.02785 m = 27.85 mm

2. A vertical steel hanger ABC is 2 m long and carries a load of 135 kN at the lower

end, as shown in the diagram below. The upper length is 50 mm diameter and the lower length is 35 mm diameter. If E = 200 GN/m2 calculate:

(a) The total extension; (b) The vertical movement of B.

A 1m 50 mm dia (a)

C 1m 35 mm dia (b) B 135 kN

Area(a) =π×o.o52

4= 1.963 × 10−3 m2

Area(b) =𝜋×0.0352

4= 9.621 × 10−4 m2

σ(a) =135

1.963×10−3 = 68772.3 kN/m2

σ(b) =135

9.621×10−4 = 140318 kN/m2

ε(a) =68772.3×103

200×109 = 3.438 × 10−4

ε(b) =140318×103

200×109 = 7.016 × 10−4

x(a) = 3.438 × 10−4 × 1 = 3.438 × 10−4 m = 0.3438 mm

x(b) = 7.016 × 10−4 × 1 = 7.016 × 10−4 m

Therefore x = 3.438 × 10−4 + 7.016 × 10−4 = 1.0454 × 10−3m = 1.0454 mm

σ(a) =F

A(a) σ(b) =

F

A(b)

F is the same in both sections

ε(a) =σ(a)

E ε(b) =

σ(b)

E

ε =x

L

40

Tutorial Problems – From Applied Mechanics by Hannah & Hillier 1. A bar of 25 mm diameter is subjected to a tensile load of 50 kN. Calculate the

extension on a 300 mm length. E = 200 GN/m2. (0.153 mm)

2. A steel strut, 40 mm diameter, is turned down to 20 mm diameter for one-half its

length. Calculate the ratio of the extensions in the two parts due to axial loading. (4:1)

3. When a bolt is in tension, the load on the nut is transmitted though the root area

of the bolt which is smaller than the shank area. A bolt 24 mm in diameter (root area = 353 mm2) carries a tensile load. Find the percentage error in the calculated value of the stress if the shank area is used instead of the root area.

(21.7%)

4. A light alloy bar is observed to increase in length by 0.35% when subjected to a tensile stress of 280 MN/m2. Calculate Young’s modulus for the material.

(80 GN/m2)

5. A duralumin tie, 600 mm long, 40 mm diameter, has a hole drilled out along its length. The hole is 30 mm diameter and 100 mm long. Calculate the total extension of the tie due to a load of 180 kN/m2. Take E as 84 GN/m2.

(1.24 mm)

6. A steel strut of rectangular section is made up of two lengths. The first, 150 mm long, has breadth 40 mm and depth 50 mm; the second, 100 mm long, is 25 mm square. If E = 220 GN/m2, calculate the compression of the strut under a load of 100 kN.

(0.107 mm)

7. A solid cylindrical bar, of 20 mm diameter and 180 mm long, is welded to a hollow tube of 20 mm internal diameter, 120 mm long, to make a bar of total length 300 mm. Determine the external diameter of the tube if, when loaded axially by a 40 kN load, the stress in the solid bar and that of the tube are to be the same. Hence calculate the total change in length of the bar. E = 210 GN/m2.

(28.3 mm; 0.184 mm)

41

Compound Bars

Tutorial Problems – From Applied Mechanics by Hannah & Hillier

1. A rectangular timber tie, 180 mm by 80 mm, is reinforced by a bar of aluminium of 25 mm diameter. Calculate the stresses in the timber and reinforcement when the tie carries an axial load of 300 kN. E for timber = 15 GN /m2; E for aluminium = 90 GN/m2.

(17.8 MN/m2; 106.8 MN/m2)

2. A concrete column having modulus of elasticity 20 GN/m2 is reinforced by two steel bars of 25 mm diameter having a modulus of 200 GN/m2. Calculate the diameter of square section strut if the stress in the concrete is not to exceed 7 MN/m2 and the load is to be 400 kN.

(220 mm square)

3. A cast iron pipe is filled with concrete and used as a column to support a load W. If the outside diameter of the pipe is 200 mm diameter and the inside diameter is 150 mm, what is the maximum permissible value of W if the compressive stress in the concrete is limited to 5 MN/m2. Take E for concrete as 0ne tenth that of cast iron.

(97 t)

4. A concrete column is reinforced with steel bars and carries a load of 20 t. The overall cross-sectional area of the column is 0.1 m2 and the steel reinforcement accounts for 3% of this area. Find the stress taken by the concrete. If the length of the column is 4 m, how much does it shorten? Take e for steel as 200 GN/m2 and for concrete, 20 GN/m2.

(1.54 MN/m2; 0.31 mm)

When two or more members are rigidly fixed together so they share the same load and extend the same amount, the members are said to be compound. Stresses in such members are calculated as follows:

(1) Total load is the sum of loads taken by each member.

(2) Load taken by each member is given by the product of its stress and its area.

(3) Extension or contraction is the same for each member

Hence we are able to write: F = σAAA + σBAB

εA = εB

or σA

EA=

σB

EB

Reinforced Bar

42

5. A cylindrical mild steel bar of 40 mm diameter and 150 mm long, is enclosed by a bronze tube of the same length having an outside diameter 60 mm and an inside diameter of 40 mm. The compound strut is subjected to an axial compressive load of 200 kN. Find; (a) the stress in the steel rod; (b) the stress in the bronze tube; (c) the shortening of the strut. For steel E = 200 kN/mm2, For bronze E = 100 kN/mm2.

(Steel 98 MN/m2; Bronze 49 MN/m2; 0.0735 mm)

6. A compound assembly is formed by brazing a brass sleeve on to a solid steel bar of 50 mm diameter. The assembly is to carry a tensile axial load of 250 kN. Find the cross-sectional area of the brass sleeve so that the sleeve carries 30% of the load. Find, for this composite bar, the stresses in the brass and the steel. E for brass = 84 GN/m2; E for steel = 210 GN/m2.

(2110 mm2; steel 89.1 MN/m2; brass 35.6 MN/m2)

43

Thermal Strain

A change in temperature of material gives rise to a thermal expansion such that x = Lαt where L = original length; α = coefficient of expansion; t = change in temperature. Hence thermal strain is given by

ε = x

L=

Lαt

L= αt

There will be no stress associated with this unless the bar is restricted – see below.

Induced Stress in Constrained Bar

Rigid Support

When a material is heated and not allowed to expand freely, stresses are induced which

are known as temperature stresses. Their value will depend upon the temperature

change and the restraint, the extreme case being when the expansion is fully restrained.

If the above bar is heated such that temperature increases by to the resulting stress may

be found by considering the process to take place in two stages.

(1) The bar is allowed to expand freely such that x = Lαt.

(2) A corresponding force is then applied so as to restore the bar to its original

length. The previous expansion x now becomes the compression of the bar under

load. Thus

Compressive strain ε = x

L= αt

And compressive stress = Eε = Eαt

Worked Example

A steel bar 280 mm long, 25 mm diameter is machined down to 20 mm diameter for 90

mm of its length. It is then heated 35o above room temperature, clamped at its ends and

allowed to cool back to room temperature. If the distance between the clamps is held

rigid, determine the maximum stress in the bar. Assume α = 12.5 x 10-6/oC and E = 210

GN/m2.

44

280 mm

90 mm

25 mm dia (A) 20 mm dia (B)

If allowed to contract freely the contraction would be

x = αLt = 12.5 x 10-6 x 0.280 x 35 = 1.225 x 10-4 m

This contraction is in essence an expansion.

Now total expansion = extensionA + extensionB

Or x = xA + xB = αLt Equation (1)

But ε = x

L and E =

σ

ε

Therefore x = εL = σL

E

Therefore xA =σALA

E and xB =

σBLB

E

So from (1) we have

αLt = 1.225 x 10-4 = σALA

E +

σBLB

E Equation (2)

We need an expression for σA in terms of σB

F = σAAA = σBAB

Therefore σA = σBAB

AA

Substituting this into (2) gives

1.225 x 10-4 = (σBABLA

AAE+

σBLB

E) = σB (

ABLA

AAE+

LB

E)

Or 1.225 x 10-4 = σB (0.0202×0.190

0.0252×210×103 +0.090

210×103)

This gives σB = 121.57 MN/m2 = σMax

σB = σMax since area is smallest

45

Tutorial Problems – From Applied Mechanics by Hannah & Hillier

1. A brittle steel rod is heated to 150 oC and then suddenly clamped at both ends. It is then allowed to cool and breaks at a temperature of 90 oC. Calculate the breaking stress of the steel. E = 210 GN/m2; α =12 x 10-6/oC.

(151 MN/m2)

2. A steel bar of 100 mm diameter is rigidly clamped at both ends so that all axial extension is prevented. A hole of 40 mm diameter is drilled out for one-third of the length. If the bar is raised in temperature by 30 oC above that of the clamp, calculate the maximum axial stress in the bar. E = 210 GN/m2; α = 0.000012/oC.

(84.7 MN/m2)

3. A metal sleeve is to be shrunk fit on a shaft of 250 mm diameter. The sleeve is bored to a diameter of 249.5 mm at 16 oC and then heated until the bore exceeds the shaft diameter by 0.625 mm, to allow it to pass over the shaft. It is then placed onto the shaft and allowed to cool. Calculate the temperature at which the sleeve must be raised. Take α = 12 x 10-6/oC.

(391 oC)

4. A tie-bar connects two supports in a machine assembly. The supports may be considered rigid and are 400 mm apart. A brass alloy tube is used as a spacer and sleeve over the tie-bar so that there is 4 mm clearance between the ends of the spacer and the supports. Find the compressive force in the spacer at the working temperature of 600 oC. For the alloy take E = 85 GN/m2 and α = 18 x 10-6/oC.

(17.1 kN) There are further examples in the textbook if you want further practice.

46

Newton’s Laws

There are three laws which bear Newton’s name and they are the fundamentals laws upon which the study of dynamics is based. The laws are a set of statements that we believe to be true in most circumstances, since they are in very exact agreement with the results obtained from experimentation. Newton 1 This states that every body continues in its state of rest or of uniform motion in a straight line, unless acted upon by some external forces to change this state. The law really defines a force as something which changes the state of rest or uniform motion of a body. This force may make contact with the body and therefore be in the form of a push or pull or, it may have no direct contact as in the case of gravitational, electrical and magnetic forces. Newton’s first law also suggests that matter has a built-in resistance to motion. This resistance is called inertia and it is the mass of a body which defines its inertia. To illustrate this consider first a tennis ball which will require only a small force to change its state of rest. On the other hand a large articulated lorry will require considerably more force to change its state. Newton 2 This law is concerned with how force is measured. It states that the rate of change of momentum is proportional to the impressed force, and takes place in the direction of the straight line in which the force acts. The momentum of a body of constant mass m moving with velocity u is, by definition, equal to the product of mass and velocity, that is Momentum = mass (m) x Velocity (u) If a force acts on the body for a period of time t and changes its velocity from u to v, then the change in momentum is given by Change in momentum = mv – mu

It follows that the rate of change of momentum = mv−mu

t

Hence by the second law:

Force (F) = km(v−u)

t where k = constant of proportionality which in the SI system of

units is unity.

Also v−u

t = acceleration of the body ‘a’

Therefore we end up with the familiar equation F = ma

47

This equation is another form of Newton 2 which enables us to measure a force by finding the acceleration it produces on a known mass. An alternative form of this equation may be used when considering the effect of gravity on a known mass and this defines the weight of a body. That is Weight (W) = mass (m) x acceleration due to gravity (g) Or W = mg Close to the earth g is normally taken as 9.81 or 9.8 m/s2. Newton 3 Newton’s third law states that to every action there is an equal and opposite reaction. What this is really stating is that forces can never occur singularly, but always in pairs. This is something that we considered in an earlier course and will not be pursued further here. Inertia Force The methods used in statics (e.g. equilibrium of forces) could be applied in dynamics if we could reshape the problem to avoid accelerations, i.e. such that the body is at rest or in uniform motion.

We can do this by substituting imaginary “inertia forces” for accelerations using Newton 2, so that the body or system is in apparent equilibrium. This principle is known as D’Alemberts Principle. When we adopt this principle a a ma P P

This Becomes this The inertia force corresponding to acceleration ‘a’ is ma (by Newton 2) acting opposite to ‘a’ to give apparent equilibrium (inertia force drawn as a hollow arrow because it’s imaginary). Inertia force may be considered as a resistance to acceleration. Similarly a frictional force may be considered as a resistance to steady motion. Such a force acts in the opposite direction to velocity. If R = resistance to steady motion the body diagram becomes:

m m

48

a P ma R v

For ‘static’ balance in accelerated motion we may equate forces, thus: P = ma + R While the accelerating force is F = P – R = ma Tutorial Problems – From Applied Mechanics by Hannah & Hillier

1. A mass of 1 kg is hung from a spring balance in a lift. What is the spring balance reading when the lift is (a) at rest; (b) accelerating upwards at 3 m/s2: (c) accelerating downwards at 3 m/s2; (d) moving downwards and retarding at 3m/s2?

(9.8 N; 12.8 N; 6.8 N; 12.8 N)

2. A planning machine table of mass 450 kg attains a speed of 0.6 m/s at a distance of 600 mm from rest. The coefficient of friction between table and bed is 0.1. Calculate the friction force and effort required during this period. If during the cutting stroke the force of the tool is 950 N and the speed is held constant at the maximum value attained, calculate the effort required to maintain the cutting stroke.

(441 N; 576 N; 1390 N)

3. The tool force on a shaping machine during the cutting stroke is 180 N and the reciprocating parts are equivalent to a moving mass of 45 kg. If power is suddenly shut off what would be the furthest distance cut by the tool if the cutting speed were initially 1.2 m/s? Assume the cutting force is independent of the speed.

(180 mm)

49

Application to Connected Bodies Consider the example shown below where mass m2 is being used to move mass m1 along a horizontal surface. v a P1 m1 P2 m2 v a m2a

Drawing the FBD m1g P2 m1a P1 F = μN N m2g

From the FBD

For m1 P1 = m1a + F

For m2 m2g = P2 + m2a

Provided the pulley is light and frictionless P1 = P2

If pulley has inertia and/or friction P2 > P1

Let consider a simple hoist r α m2a P1 P2 v a m1 m2 m1 m1g m2g

m2

v a m1a FBD’s

50

a = αr and v = ωr (see later) For m1 P1 = m1g + m1a For m2 m2g = P2 + m2a Again if we assume pulley light and frictionless then P1 = P2. Tutorial Problems – From Applied Mechanics by Hannah & Hillier

1. Two loads, each of mass 2 kg are tied together by a light inextensible cord. They are accelerated along the level by a pull of 18 N at one load. Find the acceleration of the system and the tension in the cord. Resistance to motion may be neglected.

(4.5 m/s2; 9 N)

2. A locomotive of mass 80 t pulls a train of mass 200 t with an acceleration of 0.15 m/s2 along the level. The resistance to motion of both locomotive and train is 45 N/t. Calculate (a) the tractive effort required, (b) the pull in the coupling hook at the locomotive.

(54.6 kN; 39 kN)

3. In an experiment, a load if 4.5 kg is pulled along a level track by a mass of 0.5 kg attached to it by a light inextensible cord passing over a light frictionless pulley and hanging vertically. Calculate the distance travelled from rest in 2 s.

(1.96 m)

4. A mine cage of mass 500 kg is returned to the surface by a wire cable passing over a loose pulley at the pit head. The cable is fastened to a counterweight of mass 600 kg. Find the acceleration of the empty cage if allowed to move freely.

(0.89 m/s2)

5. A motor car develops a tractive force of 1.8 kN on the level, when towing another exactly similar car whose engine is out of action. Find the tension in the tow rope and the acceleration. The resistance to motion is 650 N on each car. The mass of each car is 1 t.

(900 N; 0.25 m/s2)

6. A mass if 15 kg is supported by a light rope which passes over a light smooth pulley, and carries at its other end a mass of 6 kg. The 6 kg mass is held fast by a pawl. If the pawl is released find the tension in the rope and the time taken for the 15 kg mass to reach the level of the pawl. The 15 kg mass is initially 2 m above the level of the pawl.

(84 N; 0.98 s)

51

Dynamic of Rotation

Angular velocity and acceleration are denoted ω and α respectively and have units of rad/s and rad/s2. Relationship between Linear and Angular Motions We can show that a = αr and v = ωr. For proof of this you are referred to the recommended text. Equations of Motion Using the known equations of motion for linear motion we can write those for angular as follows:

Linear

Angular

vav =u+v

2 ωav =

ω1+ω2

2

v = u + at ω2 = ω1 + αt

s = ut + 1

2at2 θ = ω1t +

1

2αt2

v2 = u2 + 2as ω22 = ω1

2 + 2αθ

Dynamics of Rotating Particles In the same way that a change in linear motion requires a force, a change in angular motion requires a torque. F A r a α ω O

F = ma = mαr However the moment of F about O is a torque T, where T = Fr. Therefore T = mr2α Where, mr2 is known as the second moment of mass or more commonly, moment of inertia. This is denoted I and has units of kg m2.

Diagram shows a concentrated mass m attached to the end of a light arm OA. Provided there is no friction it will rotate freely when force F applied.

52

Dynamics of a Rotating Mass Consider the body of mass m, accelerated about an axis O. δP A aA O

To determine the torque required to produce this acceleration consider a small particle of mass δm, distance r from O. Force required at A = δP = δmaA perpendicular to OA. Torque required to accelerate A = δPr = δmaAr But aA = αr Therefore δT = δmαr2 Total torque T = α∑δmr2 Or T = Iα As already stated I is the second moment of mass or moment of inertia. It is useful to imagine the whole mass of the body to be concentrated at a particular radius (k), such that the I of the concentrated mass is the same as that of the actual body. This is called the radius of gyration and denoted the afore mentioned k. It is defined as follows: I = mk2

For a solid disc k2 = r2

2 where r is the radius of the disc.

For a hollow disc or cylinder k2 = r1

2+r22

2 where r1 = inside radius and r2 = outside

radius.

r

53

Inertia Torque Problems that involve angular acceleration may be reduced to a statics type problem by the introduction of an imaginary inertia torque (this is analogous to what we did with problems involving linear motion). α Consider a Hoist T ω α F v a ma

mg There are four possible cases:

1. Load rising 2. Load falling 3. Load being accelerated 4. Load being retarded

In each case three equations may be written down:

(i) Equation for balance of torque on drum (ii) Equation for balance of force on load (iii) Relationship between linear and angular acceleration of drum

Iα

Iα Tf

54

In each case, friction torque at bearing opposes motion, inertia torque of drum opposes angular acceleration.

Consider the case when an accelerated hoist raises a load. α r Iα T Tf P P m a mg ma

Tutorial Problems – From Applied Mechanics by Hannah & Hillier Inertia Torque

1. A flywheel has a moment of inertia of 10 kgm2. Calculate the angular acceleration of the wheel due to a torque of 8 Nm if the bearing friction is equivalent to a couple of 3 Nm.

(0.5 rad/s2)

2. A light shaft carries a disc 400 mm in diameter, 50 mm thick, of steel (density 7800 kg/m3). Calculate its moment of inertia about an axis through the centre of the disc and perpendicular to the plane of the disc. What torque would be required to accelerate the disc from 60 to 120 rev/min in 1 second, neglecting friction? If a friction torque of 1.5 Nm acts, what braking torque would be required to bring the disc to rest from 60 rev/min in 1 second?

(0.98 km/m2; 6.16N; 4.66 N)

3. The rotor of an electric motor of mass 200 kg has a radius of gyration of 150 mm. Calculate the torque required to accelerate it from rest to 1500 rev/min in 6 seconds. Frictional resistance may be ignored.

(118 Nm)

4. A light shaft carries a turbine rotor of mass 2 t and a radius of gyration of 600 mm. The rotor requires a uniform torque of 1.2 kNm to accelerate it from rest to

T = Tf + Iα + Pr Tf = Friction torque Iα = Inertia torque Pr = Torque due to tension P in rope P = mg + ma a = αr

55

6000 rev/min in 10 min. Find (a) the frictional couple, (b) the time taken to come to rest when steam is shut off.

(446 Nm; 16.9 min) Coupled Systems

5. A load of 8 t is to be raised with a uniform acceleration of 1.1 m/s2 by means of a light cable passing over a hoist drum of 2 m diameter. The drum has a mass of 1 t and a radius of gyration of 750 mm. Find the torque required at the drum if friction is neglected.

(87.8 kN)

6. A mine cage of mass 4 t is to be raised with an acceleration of 1.5 m/s2 using a hoist drum of 1.5 m diameter. The drum’s mass is 750 kg and its radius of gyration is 600 mm. The effect of bearing friction is equivalent to a couple of 3 kN at the hoist drum. What is the torque required at the drum? If the driving torque ceases when the load is moving upwards at 6 m/s, find the deceleration of the load and how far it travels before coming to rest.

(37.44 kNm; 9.64 m/s2; 1.87 m)

7. A hoist drum has a mass of 360 kg and a radius of gyration of 600 mm. The drum diameter is 750 mm. A mass of 1 t hangs from a light cable wrapped round the drum and is allowed to fall freely. If the friction couple at the bearings is 2.7 kNm, calculate the runaway speed of the load after falling for 2 s from rest.

(2.69 m/s)

56

Balancing of Rotating Masses

This bearing force, for a given value of , is of constant magnitude but varying direction as it sweeps around the bearing axis at angular velocity . The force is a source of bearing load, vibration, noise, etc. and constitutes an unbalanced force which increases with .

Consider first a single mass m moving in

a circular arc of radius r with an angular

velocity rad/s. The mass has a

centripetal (centre seeking) acceleration

given by 2ra . By Newton’s second

law the centripetal force acting on the

mass is 2mrF .

This force must be reacted at the centre

of rotation, i.e. at the bearing. This

reaction is called the centrifugal force

and is equal in magnitude and opposite

in sense to the centripetal force. The

centrifugal force acting on the bearing is

therefore given by 2mrF .

m

In order to eliminate or balance this

bearing force, a second mass M may be

added diametrically opposite the original

mass (via an extension of the rotating

arm, for example) at a radius R such that

MR2 = mr2 or MR = mr.

Rotating Mass

Centrifugal Force

Balanced Rotating Masses

57

Static Balance

Since MR = mr it follows that the above static moments balance. (A stationary shaft carrying a system of masses that are statically balanced will have no tendency to rotate in its bearings.) Balancing of Co-Planar Masses Diagram shows a system of co-planar masses rotating about a common centre with the same angular velocity . The radii are r1, r2, etc and the masses are m1, m2, etc. Any out of balance will have a detrimental effect on the bearings and will cause vibration, noise, etc. Since these forces are vectors, a graphical approach is often the most convenient for determining the out of balance force.

In the above case where the two masses

are diametrically opposed and MR = mr,

the balanced condition is both statically

and dynamically balanced.

Static balance is achieved because the

static moment of the masses about the

bearing axis, are equal. For the masses M

and m shown, the anti-clockwise

moment is MgRCos and the clockwise

moment is mgrCos.

R

M

m3

r3

r2

m2

r1

m1

m4

r4 Each mass has a centripetal force mr2

acting on it. Reaction forces, acting at the

bearing, are centrifugal (equal and

opposite to the centripetal). Therefore in

general the system of co-planar concurrent

forces could be replaced by a resultant out

of balance force.

Static Balance

58

If we let the resultant be MR2, then the system could be balanced by adding mass M at a radius R in the correct direction. This is found graphically by plotting an mr polygon as shown above. Note that 2 can be ignored since it is common to all vectors. Worked Example Determine the resultant out of balance force at the centre of rotation ‘O’, when the system shown below rotates at 10 rev/min and state its direction. What value of balance weight would be required at 1 m radius and where should it be placed? The information to draw the mr polygon is best tabulated.

m (kg) r (m) mr A 90 0.6 54 B 22.5 0.6 13.5 C 45 0.3 13.5 D 45 0.6 27

m1r1

m2r2

m3r3 m4r4

MR (Balancing

vector)

A

B

C

D

45o

45o 30o

60o

45kg

0.6m

45kg

0.3m

0.6m

90kg

0.6m

22.5kg

System of co-planar concurrent

masses rotating at 10 rev/min.

59

The mr polygon is then draw to scale.

A

B

C

D MR = 51 Required to balance the system.

The resultant out of balance

force is in the opposite direction

to that shown on the mr polygon. 12o mr polygon

(NTS)

A

B

C

D

1m

51kg

MR = 51 = M x 1 Therefore M = 51kg Out of balance force = MR2

= 51 x 1 x

2

60

102

=55.94N

Resultant out of

balance force

12o

60

Tutorial Problems

1. Determine the size and position of the mass required at 150 mm radius to balance the following co-planar system.

5 kg mass at 100 mm radius 10 kg mass at 75 mm radius, 90o clockwise from the 5 kg mass 15 kg mass at 100 mm radius, 240o anti-clockwise from 5 kg mass. Find also the size of each of two balancing weights which could be substituted for the single one already found if theses are to be at 75 mm radius and positioned at 30o and 150o anti-clockwise from 5 kg mass.

The remaining problems are from Applied Mechanics by Hannah and Hilliar.

2. Two masses revolve together in the same plane at an angular distance of 45o apart. The first is a 3 kg mass at a radius of 225 mm, the second 5 kg at 175 mm radius. Calculate the out-of-balance force at 2 rev/s and the position of a 10 kg balance mass required to reduce this force to zero.

(226 N; balance mass at 143 mm radius and 160o33’ to 5 kg mass)

3. A casing is bolted to the face plate of a lathe. It is equivalent to 2 kg at 50 mm from the axis of rotation, another 1 kg at 75 mm radius and 4 kg at 25 mm radius. The angular positions are respectively, 0o, 30o, 75o. Find the balance mass required at 150 mm radius to eliminate the out-of-balance force. State the angular position of the balance mass.

(1.56 kg; 215o from 2 kg mass)

4. A turbine casing is placed on a rotating table mounted on a vertical axis. The casing is symmetrical except for projecting lug of mass 15 kg at a radius of 1.2 m and a cast pad of mass 25 kg at 0.9 m radius. The lug and the pad are positioned at right angles to one another. The casting is bolted down symmetrically with respect to the axis of rotation. Find the magnitude and position of the balancing mass required at a radius of 1.5 m.

(19.2 kg at 141o 41’ to pad)

5. Two equal holes are drilled in a uniform circular disc at a radius of 400 mm from the axis. The mass of material removed is 187.5 g. Calculate the resultant out-of-balance force if the holes are spaced at 90o to each other and the speed of rotation is 1000 rev/min. Where should a mass be placed at a radius of 250 mm in order to balance the disc, and what should be its magnitude?

(582 N; 0.211 kg at 45o to a drilled hole)

6. Three masses are bolted to a face plate as follows: 5 kg at 125 mm radius, 10 kg at 75 mm radius, and 7.5 kg at 100 mm radius. The masses must be arranged so that the face plate is in balance. Find the angular position of each the masses relative to the 5 kg mass.

(Each at 114o36’ to 5 kg mass)

61

7. Four masses A, B, C and D, rotate together in a plane about a common axis O. The masses and radii of rotation are as follows: A, 2 kg, 0.6 m; B, 3 kg, 0.9 m; C, 4 kg, 1.2 m; D, 5 kg, 1.5 m. The angles between the masses are: angle AOB = 30o, angle BOC = 60o, angle COD = 120o. Find the resultant out-of-balance force at 12 rev/s and the radius of rotation and angular position of a 10 kg mass required for balance.

(21.6 kN; 380 mm, 39o7’ to OA)

62

Velocity Diagrams for Simple Mechanisms

Consider the following link, AO, rotating about O with an angular velocity rad/s. Velocity of point A relative to O = .AO and this may be represented by a vector ao, perpendicular to link, drawn to scale.

Since ao = .AO it follows that = AO

ao

Consider now point B on the link

Velocity of B relative to O = bo = .BO and therefore = BO

bo

It therefore follows that AO

ao =

BO

bo and hence

BO

AO

bo

ao - the use of this will become

apparent later.

Four Bar Chain We now apply the above to a four bar chain similar to that shown below.

AB, BC, CD are free to move, AD is fixed. Velocity of B relative to A = ba = BA . BA

A

B

C

D (A)

BA

a

b

VAO = Velocity of A relative to O = ao

O

B

A

Rad/s

Configuration

a

o

Vector

Representation

63

Velocity of C relative to B = cb (magnitude unknown) Velocity of C relative to D(A) = cd (magnitude unknown) The construction of the velocity diagram is as follows.

1. Draw ba perpendicular to BA, proportional to BA . BA. 2. Add point (d). 3. Draw cd perpendicular to CD, magnitude unknown. 4. Draw cb perpendicular to CB. 5. Intersection of cd and cb fixes point c.

From the diagram

CB = CB

cb and CD =

CD

cd

Reciprocating Mechanisms

odc idc A

B

C (D)

c

b

a(d)

Velocity diagram to scale

64

This is a four bar chain in which the link CD is infinitely long A = main bearing; AB = crank; B = crankpin; BC = con rod; C = piston. Velocity of B relative to A = ba = .BA Velocity of B relative to C = bc (magnitude unknown) Velocity of C relative to D = cd (magnitude unknown) From this information the velocity diagram shown below can be drawn. Hence the velocity of piston, ac, and angular velocity of connection-rod, BC, may be found.

a(d)

b

c

b

a(d)

Velocity diagram

ac = velocity of piston

BC = bc/BC

65

Tutorial Problems

1. The crank and con rod of a pump are respectively 35mm and 150mm, determine the plunger velocity when the crank is driven anticlockwise at 3rev/s and has turned through;

a) 60o;

b) 130o from idc.

(0.64m/s; 0.44m/s)

2. For the four bar chain given below, determine the angular velocity of the link CD and the linear velocity of G relative to A.

AB = 350; BC = 600; CD = 400; AD = 800; BG = 250mm

(19.5rad/s; 10.5m/s)

3. For the configuration shown below determine the linear velocity of pistons C and E

AB = 75; BC = 300; BD = 120; DE = 140mm

(12m/s; 0.68m/s)

C

D A

45O

40 rad/s

G

C

E

120 mm

A

120O

A

3 rev/s

B

C(D)

B

2000

rev/min D

66

4. Determine the velocity of the piston at C and the centre of gravity G in the slider-crank mechanism shown in the figure below. (Use a scale of 1cm = 1m/s)

(14.8m/s; 13.8m/s)

5. In the mechanism shown below, OA rotates clockwise about the fixed centre O at 10rad/s. If the angle AOB is 45o, draw the velocity diagram to scale and hence determine the velocities of C and D relative to O and the angular velocity of the link CD.

For further examples see pages 78 – 80 of Applied Mechanics by Hannah and Hillier.

B

A

O

C

D 0.6 m

OA = 0.3; AB = 1.5; AC = 0.6; CD = 0.9 m

C

G

B

A 2700

rev/min

60O

AB = 50 mm

BC = 200 mm

BG = 75 mm

25 mm

67

Statics & Dynamics of Fluids

Hydrodynamics – Fluids at Rest A fluid may be a gas or a liquid however, they have some distinct differences as described below. Gases – When held within a sealed container they expand to fill it completely; they are also known to be compressible. Liquids – These do not expand to fill sealed containers and are considered to be incompressible. On this course we shall only be considering liquids. Pressure (P) If held within a cylinder, sealed by a piston a fluid may be subjected to pressure if force is applied to the piston. Pressure is defined as the applied force divided by area, in this example the area of the piston. Therefore we have the equation:

P =F

A

It has units of N/m2 and bar where 1 bar is equal to 105 N/m2. Since pressures may have high values we may also use kN/m2, MN/m2, and GN/m2. It is important to remember that pressure at a point is the same in all direction while pressure on a surface will always act at 90o to it.

Other Useful Properties Density (ρ) Density is defined as mass per unit volume and therefore yields the familiar equation:

ρ =m

v

Units include; kg/m3, tonne/m3, kg/litre, and g/millilitre Since we often work with water it is worth remembering it density which is 1000 kg/m3 = 1 Mg/m3 = 1 tonne/m3 = 1 kg/litre.

We also have relative density, where relative density = density of substance

density of water.

Specific Weight This is defined as weight per unit volume, therefore we have;

Specific weight = ρg

68

Where, g is the acceleration due to gravity. Specific Gravity

The specific gravity of a substance is defined as weight of substance

weight of equal volume of pure water.

Since this is in effect the ratio of the masses it follows that Specific Gravity and Relative Density are in fact the same. Pressure in a Liquid Due to its Own Weight P atmos

Surface h P Mass dm, cross-sectional area da P + dP dh dw = dmxg = volume x density x g Balancing forces acting on element PdA – (P +dP)dA + dmg = 0 Or PdA – (P + dP)dA +dAdhρg = 0 Therefore P –(P + dP) +dhρg = 0 Or P – P – dP + dhρg = 0 Therefore dP = ρgdh Or ∫ dP = ∫ ρgdh Provided ρ and g are constant this gives P = ρgh + C When h = 0, P = Patmos, therefore P = ρgh + Patmos This is called absolute pressure.

69

Usually in fluids we tend to work in terms of gauge pressure, i.e. ρgh (See diagram below). Pressure P = ρgh (gauge pressure) Patmos

Pvacuum

Pabsolute

Pabsolute

Measuring Pressure Absolute pressure is usually measured using a barometer while gauge pressure is measured by a manometer or pressure gauge such as a Bourdon gauge. Since we have already stated that we are normally only interested in gauge pressure we will have a brief look at some of these measuring devices.

Piezometer Tube Height h is known as the pressure head and is normally expressed in mm, cm, or m of water. As an example a 10 mm head of water would represent P = ρgh = 1000 x 9.81 x 10 x 10-3 = 98 N/m2 If measuring small pressures a version with a sloping tube may be used.

Depending on the pressure of the liquid in the pipe the value of h will change. Pressure in pipe = ρgh

70

Another devise is a U-Tube Manometer

U-Tube Manometer There are several other measuring devises and if you are interested you could search the internet.

Tutorial Problems 1. Find the head of water corresponding to an intensity of pressure of 340,000 N/m2. The

specific weight of water is 9.81 x 103 n/m3. (34.66 m)

2. A diver is working at a depth of 18 m below the surface of the sea. How much greater is

the pressure intensity at this depth than at the surface? Specific weight of sea water is 10,000 N/m3.

(180,000 N/m2)

3. Determine the pressure in N/m2 at a depth of 6 m below the free surface of a body of water and at a depth of 9 m below the free surface of a body of oil of specific gravity 0.75.

(58.9 kN/m2; 66.2 kN/m2)

4. What depth of oil, specific gravity 0.8 will produce a pressure of 120 kN/m2? What depth of water will produce the same pressure?

(15.3 m; 12.2 m)

5. At what depth below the free surface of oil having density of 600 kg/m3 is the pressure equal to 1 bar?

(17 m)

6. What is the gauge and absolute pressure at a point in water 10 m below the free surface? Assume atmospheric pressure to be 100 kN/m2 and the density of water 1000 kg/m3.

(98.1 kN/m2; 198 kN/m2)

Fluid F1 has density ρ Fluid F2 (usually mercury) has density ρ’ Balancing pressures on a-a P + ρgh2 = ρ’gh Therefore P can be found

71

7. Calculate the absolute pressure at a depth of 8 m in oil of specific gravity 0.8 if the atmospheric pressure is 0.97 bar.

(159.8 kN/m2)

8. What is the pressure in kN/m2 absolute and gauge at a point 3 m below the free surface of a liquid having specific weight of 15 x 103 N/m3 if the atmospheric pressure is 750 mm of mercury? Take the specific gravity of mercury as 13.6 and density of water as 1000 kg/m3.

(145 kN/m2; 45 kN/m2)

72

Hydrodynamics – Fluid Dynamics All real fluids are viscous, that is offer resistance to shear. They are compressible, have surface tension and vaporise. An idea fluid does not exist and does not have any of the above properties. We use the concept however to simplify calculations and any discrepancies between predicted results and actual results are found by experiment. Coefficients are then used to convert the ideal to the real. Steady Flow This occurs when conditions at a point in the moving fluid do not change with time. Unsteady Flow When this is the case, conditions do change with time. Streamlines If the path of the fluid particles in steady flow are traced the result is a series of smooth curves called streamlines. i.e.

Streamline flow

1. The tangent to a streamline gives the motion of the particle at that point. 2. There can be no flow across a streamline. 3. Converging streamlines indicate increasing velocity.

Laminar and Turbulent Flow Laminar flow occurs when the fluid moves in layers which slide smoothly one over the other. Turbulent flow occurs when small irregular motions of fluid particles are superimposed on the main motion of the fluid. Reynolds’ experiment using dye in water shows that at low velocities the dye flow has a thin streamline down the centre (laminar flow). Increasing the flow velocity causes an initial wavering (transition) and then diffusion (turbulent). He also found that the fluids density (ρ), the viscosity (μ), and the diameter of the pipe (d) also affected the point of transition. By repeated experiments Reynolds’ found that laminar flow ended when the

dimensions group ρud

μ equaled 2300. This is known as the Critical Reynolds Number.

Laminar Transition Turbulent

73

Bernoulli Equation In order to understand this better we need to consider all the types of energy that may be involved. These include:

Pressure energy Potential energy Kinetic energy

Pressure Energy Consider the tank shown below: h Pressure P, Area A F = PA x In moving piston distance x, work is done. Where WD = Fx = PAx If we release the piston both piston and liquid would return to their original states and we therefore see that this form of energy is recoverable. This represents pressure energy. When piston is forced in Volume in = Ax

Mass in = ρAx (since ρ = m

v)

Hence the amount of energy possessed by unit mass of liquid is given by WD

Mass=

PAx

ρAx=

P

ρ

It will have units of N

m2 ×m3

kg=

Nm

kg=

J

kg

It follows that pressure energy/unit weight is P

ρg and this will have units of

J

N=

Nm

N= m.

This is called pressure head. So if a liquid is under pressure P such as that shown above, then h is equivalent to the static head of liquid which produces the same pressure. In other words, the pressure at depth h is equal to ρgh. (Something we have already seen.)

74

Potential Energy We now know that pressures at some depth h is P = ρgh

It follows that gh = P

ρ

Since P

ρ represents the pressure energy/ unit mass, it follows that gh must also

represent energy, i.e. the potential energy required to raise unit mass through a certain distance h. What we should be seeing now is that potential energy and pressure energy may be converted one into the other.

1. As we force the piston in, pressure energy is converted into potential energy. 2. As piston is released and level of liquid falls, potential energy is converted back

into pressure energy. We are therefore able to write Potential energy per unit mass + Pressure energy per unit mass = a constant

That is gh +P

ρ = a constant, per unit mass

However it is more usual to write this in terms of energy per unit weight

That is h + P

ρg= a constant, per unit weight

Kinetic Energy B h A u In falling from B to A Potential energy lost/unit mass = gh

KE gained/unit mass = 1

2u2

PE lost = KE gained

75

Therefore gh = u2

2

Or h = u2

2g = KE/unit weight (Velocity head)

If we need exit velocity this is given by u = √2gh

Relationship between Pressure and KE It is not necessary for any particle to have actually fallen from B to A before flowing

from the orifice. If already at level A it will possess pressure energy P

ρg and this in turn

may be converted into KE on escaping Therefore we are able to write pressure energy lost = KE gained

Or P

ρg=

u2

2g (per unit weight)

The Bernoulli equations is really a way of stating the principle of conservation of energy as applied to a steady flowing fluid system. That is that the total energy in the system at any two points is the same provided that energy is neither given to nor extracted from the system. Thus we are able to write

Pressure energy + Kinetic energy + Potential energy = a constant

Or P

ρg +

u2

2g + h = a constant

The equation is valid provided

There is no loss of energy due to friction The flow is steady

Applying this to flow though a pipe (note the change from h to z) P2, A2, u2 z2

z1

P1, A1, u1 Datum

76

Applying Bernoulli’s equation

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

For real fluids which include losses it becomes necessary to add a viscosity term

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2 + hf

Where hf is a head loss due to friction. Bernoulli’s equation alone is usually insufficient for solving pipe flow problems and it becomes necessary to make use of another equation known as the Continuity Equation. Equation of Continuity m1 m2

A1 A2 m = mass flow rate in kg/s m = ρQ, where Q = volumetric flow rate in m3/s Q = Au, where A = area and u = velocity Assumption is that m1 = m2 Therefore ρ1Q1 = ρ2Q2 Hence ρ1u1A1 = ρ2u2A2 If ρ1 = ρ2, which is the case with a flowing liquid Then u1A1 = u2A2 this is the Continuity Equation Note that u is the average velocity across the section because u varies as shown below.

77

Graph of Velocity across Section Application of Bernoulli’s Equation Orifice in a Tank 1 h 2

It follows therefore that h =u2

2

2g or u2 = √2gh

However the actual velocity in a real fluid is less than this where

uactual

utheoretical= Cv = Coefficient of velocity

So that we should write u2 = Cv√2gh

To calculate the volume flow rate Q where Q = Area × Velocity Ao Av

The ratio Av

Ao= Cc = Coefficient of contraction

Actual flow rate Qactual = Cv√2gh × CcAo

Applying Bernoulli’s equation between 1 and 2

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

P1 = P2 therefore pressure energy terms disappear; u1 is negligible; z1 = h; z2 = 0

Ao = Area of orifice Av = Area of vena contractor

78

Or Qactual = CdAo√2gh

Where Cd = Cc x Cv = Coefficient of discharge which is found by experiment to be approximately 0.6. Venturi Meter Pressures at 1 and 2 measured by pressure gauge Q = Area x Velocity = A1u1 = A2u2 Z2 2 Diffuser Throat Z1 1 Applying Bernoulli’s equation between 1 and 2 and assuming an ideal fluid P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

Rearranging P1−P2

ρg+ z1 − z2 =

u22−u1

2

2g

But u2 =A1u1

A2 from the continuity equation

P1−P2

ρg+ z1 − z2 =

(A1A2

u1)2

−u12

2g= [(

A1

A2)

2

− 1]u1

2

2g

Therefore u1 = √([

P1−P2ρg

]+z1−z2)2g

(A1A2

)2

−1

Qtheoretical = A1u1 = A1 √([

P1−P2ρg

]+z1−z2)2g

(A1A2

)2

−1

Qactual = Qtheoretical × Cd Where Cd = Coefficient of discharge which is found to be approximately 0.97.

79

Note: If the venturi is horizontal then z1 – z2 = 0 Next consider a venture meter when pressures 1 and 2 are measured using manometers. h h1 h2 2 z2 y 1 z1

Since P = ρgh it follows that h1 = P1

ρg and h2 =

P2

ρg

Therefore P1−P2

ρg+ z1 − z2 = h1 − h2 − y

But h1 = h + h2 +y

Therefore P1−P2

ρg+ z1 − z2 = h + h2 + y − h2 − y = h

Modifying our previous equation for Qactual we end up with

Qactual = CdA1√2gh

(A1A2

)2

−1

80

Tutorial Problems – First 4 are from Applied Mechanics by Hannah & Hillier

1. A tank of oil, of relative density 0.8, discharges through a 12 mm diameter orifice. The head of oil above the centre line of the orifice is kept constant at 1.2 m and the measured discharge rate is 16 kg/min. Calculate the coefficient of discharge for the orifice.

(0.61)

2. A large tank contains water to a depth of 0.9 m. Water issues from a sharp –edged orifice of 25 mm diameter and is collected in a circular tank of 0.9 m diameter. The water level in the cylinder rises 600 mm in 5 min. Calculate the discharge coefficient for the orifice.

(0.62)

3. A venturi meter has an inlet diameter of 100 mm and a throat diameter of 50 mm. What will be the difference of head in metres of water between inlet and throat if the flow rate is 15 litres/s of water? If the flow rate is doubled, what would then be the difference in head?

(2.78 m; 11.12m)

4. The measured discharge of water through a venturi meter is 78 Mg/h. The inlet and throat diameters are 120 mm and 55 mm respectively. The pressure drop between inlet and throat is 42 kN/m2. Find the discharge coefficient for the meter.

(0.975)

5. A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter pipe line. If the difference of pressure between the full bore and the throat tapping is 34.5 kN/m2 and the area ratio is 4, calculate the rate of flow, assuming a coefficient of discharge of 0.97.

(0.0097 m3/s)

6. Crude oil of density 850 kg/m3 flows upwards a a volume rate of 0.06 m3/s through a vertical venturi meter which has an inlet diameter of 200 mm and a throat diameter of 100 mm. The venturi meter coefficient is 0.98. The vertical distance between the tapping points is 300 mm, and they are connected to two pressure gauges, calibrated in kN/m2 and positioned at the levels of their respective tapping points.

Working from first principles (but assuming Bernoulli’s equation), determine the difference of the readings of the two pressure gauges.

(26.7 kN/m2)