This chapter introduces the student to some more drawing techniques It should be emphasized that the topics are only introduced: all of them can be studied in much greater depth and any solutions offered in this chepter will apply to simple probleme only.
AREAS OF IRREGULAR SHAPES It is possible to find, by drawing, the area of an irregular shape. The technique does not give an exact answer but. carefully used can provide a reasonable answer. Look at Fig. 17/1. The shape is trapezoidal with irregular end lines. A centre line has been drewn end you can see that
1/J | ||
f t i i i |
w | |
\ |
p^rrwJ | |
v |
V |
Fig. 17/2 shows how it is applied to a larger figure.
The figure is divided into a number of equal strips (width W). in this case 8. The more stnps that are drawn (within reason) the greater the accuracy of the final calculation. The centre line of each stnp (the mid-ordinate) is drawn. Each of these lengths is measured end the area of the figure is given by
the shaded triangles at the top and bottom are. in their pairs, epproximately the same in eree Thus, the approximate area of the whole figure is the width W multiplied by the height at the centre. That height is called the mid-ordinate and the whole technique is called the mid-ordinate ruU.
A sample of this technique * shown in Fig. 17/3. The curve shown is a sin curve, the curve that emergos if you plot the values of the sin of all the angles from 0* to 90*.
The area is the product of the width of each strip and the sum of all the mid-ondinates.
RESOLUTION OF FORCES All machinery, however simple, has forces acting on its parts. Buildings have forces acting on them; force* produced by the weight of the building itself, the weight of the things inside it and by the wind pushing against It. An understanding of how these forces act and how they affect design is essential to a good draughtsman.
You must first understand the difference between stable and unstable forcee and then study the effects of the unstable ones. The two men indulging in indian wrestling tn Fig. 17/4 are applying force. As long ae the forces are equal they will remain in the position shown. When one begins to apply more force than the other the forcee become unstable and the other ha* his hand forced back onto the table.
Calculating soma of the forces acting on a beam If a beam is loaded then, to stabilise it, forces have to be applied in opposition. These opposite forces are called reactions. Some simple examples are shown in Fig. 17/7. The loading is applied at a point and is therefore called a point load.
The left hand example has the reactions positioned at equal distances from the load and they will therefore be equal. In that case they must each be half the load.
The centre example has a load of 8 units' and one reaction is 6 units. For the loading and reaction system to be steble. the sum of the reactions must equal the loading. Therefore the reaction on the right (A*) must be 3 units.
The left and right reactions are given in the example on the right. Since the sum of the reactions must equal the load, the load is 5 units.
LOAD -10 UNITS
LOAD «8 UNITS
TO FIND REACTIONS TO FIND TO f|ND U»t>
Ri AND R, Aftfi tOL/AL PITTANCE & 5 ♦ B* • B 2-» - LOAD
Fig. 17/7 Reactions in a point loaded beam
* Forces in the SI System are given in newtons (the force requited to accelerate 1 kg at 1 metre per sec*). For simplicity the loads are called units In this chapter.
Fig. 17/8 chow* two example* where the load* and reections are known but the position of one reaction has to be calculated.
if the loaded system is stable then ALL the forces must balance. The reactions added together (8-1-4) must equal the load (12). The couplee must also be balanced out The left hand couple, reaction times d(Stance to the load (8x2) must equal the right hand couple (4xx). Thus the distance can be worked out in the simple equation shown. The example on the right ia similar. The force called a couple in this chapter is also called a moment or banding moment (because it tries to bend the beam) by engineers.
PO* ftTABltlTY rufc CouPlCS ArfUST BS EPUAL 8*2 -4.* at I* -4-x X
FOC STAfciUTy TUC COUPt-tS MUST Bt EQUAL 6*2 . 4« x
PO* ftTABltlTY rufc CouPlCS ArfUST BS EPUAL 8*2 -4.* at I* -4-x X
FOC STAfciUTy TUC COUPt-tS MUST Bt EQUAL 6*2 . 4« x t. - ;
fig. 17/» Finding ttia position of tha raaclion In a «labia loadad baam
Two more example* are shown in Fig. 17/9. In this case the loading is not acting at a point but is evenly distributed along the beam. This could make calculations difficult in fact we are able to change the loading The total load is the load per metre multiplied by its length. (6 x 1 a 6 units). The effect of this evenly distributed load on the beam i* the tame as if it were a point load acting at the centre. The lower left drewing in Fig. 17/9 shows how this looks. The reactions can now be easily calculated
The example on the right also haa the evenly distributed load changed into a point load acting at the centre of the beam. The size of the two reactions can be calculated with a simple simultaneous equation (shown underneath the diagram).
LOAD I NO 2. UNIT-» Ptti METlCfe tOUIVALCNT PCXNTLOAP-6 UNITES («►*»
6OUIVAL6NT POINT LOAD - 16UNIT3 (8*2)
FH/5M LOAD
6OUIVAL6NT POINT LOAD - 16UNIT3 (8*2)
TO PIMP REACTIONS POR. STABILITY TWS COoPuBS MUST B£ EQUAL.
Fig. 17/9 Reactions in en evenly distributed loaded beam
Force« acting at a point When two or more forces act at or on a point it is useful to be able to change these forces and show how they could be replaced with a single force which acts in the same way. This force is called the nsuftant force. The force which would have to be applied to stabilize the system (by acting against and cancelling out the two or more forces) is called the equiftbrent force. An example is shown in Fig. 17/10.
The two forces are of 6 units and 4 units. The resultant and equilibrant forces can be found by drawing lines parallel to these two forces and forming a parallelogram. The resultant and equilibrant forces are equal to the length of the longer diegonal of this parallelogram An alternative is to draw triangles as shown in the figure. The result of these drawings is shown on the extreme right. The resultant force is the one that would have the same effects as the two forces if it replaced them; the equilibrant force acts against the two forces and makes the system stable.
Fig. 17/12 shows how to find the équilibrent force to four forces acting at a point. Once again the forces are drawn to scale parallel to the way they are acting at the point. The forces are each taken in turn and are considered clockwise. The point to note on this example is that, although the lines cross each other the équilibrant force is still the one which doses the figure, the one that joins the end of the line representing the last force drawn to the beginning of the line representing the first force drawn.
Fig. 17/12 Finding the équilibrant force to 4 forces meeting at a point
Cams are used in machines to provide a controlled up and down movement. This movement is transmitted by means of a follower. A cam with a follower is shown in Fig. 17/13 in the meximum 'up' and 'down' positions. The difference between these two positions gives the lift of the cam. The shape of the cam. its profite, determine« how the follower moves through its lift and f»lf.
Fig. 17/12 Finding the équilibrant force to 4 forces meeting at a point
A Mm is designad to make a part of a machina mova in a particular way. For example. cams ara usad to opan and cloee tha valva« which control tha pet rol mixture going into and tha axnaust gase« Coming o ut of an internai combustion angine. Obviously the vahree must open at the ríght time and at the right speed and it is the cam thaï determines thia. This control on the valve is exerctsed by the profile of the cam. determimng how the follower lifts and falls. Three examples of types of lift and fall are shown in Fig. 17/14.
The top diagram »how» uniform velocity Thi« is a graph of a point which is moving at the same velocity 'upward' to haif way and then at the same velocity 'downward* to its starting point.
This is the motion of a pendulum, starting with zero velocity, accelerating to a maximum and then decelerating to zero again (at the top of the curve) and then repeating the process back to the starting point. The curve is a sine curve and is plotted as shown.
The lower diagram shows uniform acceleration and retardation. In thie case the acceleration is uniform to a point halfway up the lift and then retarding uniformly to the maximum lift. The process is then repeated in reverse back to the starting point. The curve is plotted as shown.
If a cam has to be designed it is drawn around a specification. This must state the dimensions, the lift/fall and the performance. The performance states how the follower is to behave throughout one rotation of the cam. The designer must first draw the performance curve to the given specification. An example is shown in Fig. 17/16. DwefJ is a period when the follower is neither lifting or felling.
The performance curve is started by drawing a base line of 12 equal parts, the total representing one rotation of the cam. The lift/fall is then marked out and the performance curve drawn The base line of the performance curve is then projected across and. with the centre line of the cam. forms the top of the circle representing the minimum cam diameter. Once the centre of the cam hes been found centre lines can be drawn at 30* intervals The cam profile is plotted on these lines. Twelve points on the performance curve are then projected across to the centre line of the cam and then swung round with compasses to the intersecting points on the lines drawn at 30* intervals. If the cem is rotating clockwise the points 1 to 12 are marked out clockwise; if the rotation is anti-clockwise, as in this case, the points are marked out anti-clockwise.
PERFORMANCE CURVE-
PERFORMANCE CURVE-
Fig. 17/16 show« another exemple of e cam design Thie it a more complicated profile than the previous example but the method used to construct the profile is the same. This cam rotates in a clockwise direction.
Finally, three different types of followers are shown in Fig- 17/17. The left hand example is a knrte follower. It can be used with a cam that has a pan of its profile convex but it weers quickly. The centre example is e roller follower this type of follower obviously reduces fnction between the follower and the cam. something of a problem with the other types shown. The flat follower is an all purpose type which is widely used. It wears more slowly than a knife follower.
Exercise* 17 The velocity of a vehicle moving in a straight line from start to reet was recorded at intervals of t minute and these readings are shown in the table. Part of the velocity/time diagram is given infig.1.
TIME t MIN
Fifl.1
TIME t MIN
Draw the complete diagram with a horizontal scale of 10mm to1 minute and vertical scale of 10 mm to 1 m/s. Using the mid ordinate rule, determine the average velocity of the vehicle and hence the total distance travelled in the 12 minutes
Oxford L oca/ Examinations
Fig. 2. shows part of an indicator diagram which was made during an engine test. Draw the complete diagram. full size. using the informetion given in the table and then by means of the mid-ordinate rule determine the area of the diagram. Also, given that the area of the diagram can be given by the product
of its length and average height, determine the average height and the average pressure if the ordinates represent the pressure to a scale of 1 mm to 60kN/m*.
OX(mm) |
0 |
10 |
20 |
30 |
40 |
50 |
OYmi*[mm) |
51 |
79 |
82 |
78 |
66 |
54 |
OYmin(mm) |
54 |
21 |
14 |
12 |
12 |
12 |
OX(mm) |
60 |
70 |
80 |
90 |
100 | |
OYmax(mm) |
45 |
39 |
33 |
29 |
21 | |
OYmin(mni) |
12 |
12 |
12 |
13 |
21 |
Oxford Local Examinations 3. Fig. 3 shows four simply loaded beams. Find the size of the reactions marked X.
Oxford Local Examinations 3. Fig. 3 shows four simply loaded beams. Find the size of the reactions marked X.
SUNITS/METP6
4. Fig. 4 shows two simply loaded beams with their reactions. Find the dimensions marked x for the loading to be in equilibrium (the couples to be equal).
6. Fig. S shows two beams loaded with evenly distributed loads. The portions of the raections are shown. Find the ska of the réactions.
yr30WtTS/'^gTRg
6. Fig. 6 shows three examples of two forcee acting at a point. Rnd the size of the reeuhant force for each example and meaeure the angular direction of the reeuKant force from the de turn shown.
7. Fig. 7 shows two exemples of forces acting at a point. Find the size of the resultant fore« for both examples and measure and state the engutar direction of the resultant force from the datum shown. From the same datum state the angular direction of the équilibrant force for both examples.
8. Plot the cam profile which meets the following specification:
Performance: 90* uniform velocity to max. lift 90* dwell 180" uniform retardation to max. fell. Rotation: Clockwise. Your cam profile must be drawn twice fuH tin.
9. Plot the cam profile which meets the following specifications:
Shaft diamj 12.5 mm Min. diem.: 30 mm Lift: 12.5 mm
Performance: 60' dwell
90* simple hermonic motion to half lift 30* dwell
00* uniform acceleration to max. lift 120* uniform velocity to max. faN Rotation: anticlockwise
Your cam profile ie to be drawn twice fuU tize.
jr-2 UNtTS/MCTttg
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