WHEEL & DIFFERENTIAL AXLEgecdahod.tk/notices/manual_Mechanics_of_solids2017-… · · 2017-12-22Plot the graphs of load on X axis v/s (i) actual effort (ii) ideal effort ... is

Manual for Engineering Mechanics Laboratory

1

1. WHEEL AND DIFFERENTIAL AXLE OBJECT: To determine the mechanical advantage, velocity ratio, ideal effort, frictional effort, ideal load, frictional load and efficiency of a given wheel and differential axle. THEORY:

The wheel and differential axle acts on the differential principle. It is a very simple device to achieve very high velocity ratio by keeping very small difference in diameters of two axles. The axle is made up of two cylinders B and C of different diameters. The wheel A and the cylinders B and C turn about a common axis. One string is wound round the wheel A and to the one end of which the effort P is applied. The second string is wound round the two cylinders B and C. This string goes around a pulley to which the load W is attached. This string is wound on the two cylinders in such a way that as the cylinders turn, it unwinds on the smaller cylinder and winds at the same time on the larger cylinder, lifting the load W attached to the pulley. For one revolution of the wheel and axle, the displacement of the effort P is equal to the length of the string that unwinds from the wheel A = лD.

Effort Pulley

Weight

Support axle

axle

Support

wheel

D

d1 d2

Bigger

Smaller

Effort

x

Y

WHEEL & DIFFERENTIAL AXLE


2

The length of the string that unwinds from the cylinder C = лd2 The length of the string that winds on the cylinder B = лd1 The load string shortens by = лd1 - лd2, therefore, the displacement of the load W = (лd1 - лd2)/2 лD 2D VR = = (лd1 - лd2)/2 d1 - d2 For a larger velocity ratio, d1 and d2 are made nearly equal. APPARATUS: Wheel and differential axle, weights, calipers, meter rules etc. PROCEDURE: (Procedure to be written by students in passive voice) 1. Note the total load applied and corresponding effort required to raise the load with a slow

uniform motion. 2. Take six readings with uniform increase in load and corresponding effort required. 3. Plot the graphs of load on X axis v/s (i) actual effort (ii) ideal effort (iii) frictional effort and

(iv)efficiency on Y axis. Use same scale for all effort curves and different scale for efficiency. It is desirable to draw all graphs on same graph paper. Work out the law of machine P = mW + C where, P is actual effort required to raise the load

W, m and C are constants of a given machine where m is the slope of the graph and C is the initial effort required to start the machine. OBSERVATION: 1. Diameter of wheel (D) = 35.00 cm. 2. Diameter of bigger axle (d1) = 9.54 cm. 3. Diameter of smaller axle (d2) = 4.13 cm. 4. Velocity ratio VR = 2D / (d1-d2) 5. Weight of load pulley = ………. gms. OBSERVATION TABLE:

Sr. No.

Total Load

W

Total Effort

P

Mechanical Advantage

W/P

Ideal Effort W/VR

Frictional Effort

P-(W/VR)

Ideal Load P*VR

Frictional Load

P*VR – W

Efficiency [W/(P*VR)]*100

1 2 3 4 5

CALCULATION:

1. Give sample calculation for all quantities for any one reading of observation table. 2. For law of machine, P = mW + C from the graph of actual effort, determine the value of m. 3. Maximum M.A = 1/m = ____ 4. Maximum efficiency = 1/(m x v) = ______ CONCLUSION: * Nature of graphs * Which graph passes from origin and why? * Reversibility of this machine * Law of machine P = mW + c (write value of m & c)


3

2. SINGLE PURCHASE CRAB

OBJECT: To determine the mechanical advantage, velocity ratio, ideal effort, frictional effort, ideal load, frictional load and efficiency of a given single purchase crab. THEORY: The single purchase crab is a mechanical device to achieve velocity ratio for lifting machine using gearing. (Draw figure of this machine and write the working phenomenon)

APPARATUS: Single purchase crab, weights, callipers, meter rules etc. PROCEDURE: (Procedure to be written by student in passive voice) 1. Note the total load applied and corresponding effort required to raise the load with a slow

uniform motion. 2. Take six readings with uniform increase in load and corresponding effort required. 3. Plot the graphs with (i)actual effort (ii)ideal effort (iii)frictional effort and (iv)efficiency on

Y axis and load on X axis. Use same scale for all effort curves. Work out the law of machine P= mW + C where, P is actual effort required to raise the load W; "m" and "C" are constants of a given machine.

Effort wheel

Spur Wheel

Load drum

Pinion Wheel

Y X

SINGLE PURCHASE CRAB

D

d


4

OBSERVATIONS: 1. Diameter of wheel (D) = 24.20 cm. 2. Diameter of axle (d) = 6.84 cm. 3. No. of teeth on spur wheel (TS) = 100 4. No. of teeth on pinion wheel (TP) = 20 5. Velocity ratio VR = (D*TS)/(d*TP) OBSERVATION TABLE:

Sr. No.

Total Load

W

Total Effort

P

Mechanical Advantage

W/P

Ideal Effort W/VR

Frictional Effort

P-(W/VR)

Ideal Load P*VR

Frictional Load

P*VR – W

Efficiency [W/(P*VR)]*100

1 2 3 4 5

CALCULATION: 1. Give sample calculation for all quantities for any one reading of observation table. 2. For law of machine, P = mW + C select any two points on graph of actual effort and determine

the value of m. 3. Maximum mechanical advantage = 1 / m = ______________ 4. Maximum efficiency = 1 / (m x VR) = _______________

CONCLUSION: * Nature of graphs * Which graph passes from origin and why? * Reversibility of this m/c * Law of machine P = mW + c (write value of m & c)

.


5

3. EQUILIBRIUM OF COPLANAR, CONCURRENT FORCES OBJECT: To verify the principle of the polygon law of forces for a system in equilibrium, both analytically and graphically. THEORY: Concurrent forces in equilibrium can be represented by a closed polygon. The respective sides are being parallel and proportional to their respective forces. The closing side of polygon shows resultant of system of forces. APPARATUS: A force table with five attached pulleys. PROCEDURE: Five strings are radiating from a single point. These pass over smooth pulleys freely slidable along the circumference of a horizontal force table with graduations for measuring angles. Attach five known weights to five strings arranged at particular angles. Note down the readings when the central ring is at centre of force table. As this system is in equilibrium the resultant to be found out should be zero, but due to errors, the resultant force is obtained analytically as well as graphically which should be negligible in magnitude. Draw space diagram by drawing the angles as measured on force table and show the respective forces. Using Bow's notation draw force (vector) diagram, which will be the polygon. For the system is the equilibrium the sides of polygon should be five but due to error the closing side is reverse order will represent the resultant.

Board

E

P4

String

Weight P3

E

P 2

P 1


6

OBSERVATION TABLE: Sr. No.

P1 P2 P3 P4 1 2 3 4

1 2 3 4 5

(For one set of readings)

Force Value of P Components

Fx=Pcos Fy=Psin P1 P2 P3 P4

Fx= Fy= R = (Fx2+Fy2)= error CALCULATIONS: Resultant force is obtained analytically as well as graphically. CONCLUSIONS: State the law and comment on the verification of law.


7

4. EQUILIBRIUM OF COPLANAR, NON CONCURRENT, NON PARALLEL FORCES OBJECT: To determine the weight of a plate by equilibrium of coplanar, non-concurrent, non parallel forces. (Graphically and analytically)

THEORY: If for forces P1, P2, P3 & P4 are applied they have a resultant R, it must be equal, opposite and collinear with weight of the plate(W). As system is in equilibrium, W is equilibriant of applied system of forces. (State Polygon law of forces). APPARATUS: Board, sheet of paper, pulleys, strings, mild steel plate of unknown weight, known weights. PROCEDURE: (Write in passive form) 1. Attach the sheet of paper on a vertical board. 2. Adjust position of pulleys so that the strings passed through them and attached to four corners A,

B, C, & D of the plate remains in same vertical plane. 3. Apply different weights at the other ends of the strings to keep the plate in equilibrium at middle

portion of the attached sheet. 4. Note down the magnitude of weights P1, P2, P3 & P4 attached to string passing through

corners A,B,C,D and the centre of the plate.

Paper

Board

P4

Pulley

String

Weight P3

P 2

P 1

COPLANAR-NON CONCURRENT FORCE SYSTEM


8

5. Mark the projections of strings and plate perpendicular to the paper. 6. Remove the sheet, mark angle O1, O2, O3, O4 respectively with horizontal. 7. Find out the value of W analytically as well as graphically. 8. Change the position of lower two pulleys and repeat procedure with different weights and

changed position of plate. OBSERVATION TABLE:

Sr. No.

P1 P2 P3 P4 1 2 3 4

1 2 3 4 5

(For one set of readings)

Force Value of P Components

Fx=Pcos Fy=Psin P1 P2 P3 P4

Fx= Fy= M@ A = CONCLUSION : * Thus the weight of the plate as determined by this practical is _____gms. * Which law is verified? State. * Comment about graphical value v/s analytical value (calculated).


9

5. SIMPLY SUPPORTED BEAM

OBJECT: To determine the reactions at supports and verify the condition of equilibrium for beam simply supported at ends. THEORY :

Beam is a structural element carrying transverse loads. Analytical condition of equilibrium (i) Algebric sum of horizontal forces must be equal to zero. i.e. H=0 (ii) Algebric sum of vertical forces must be equal to zero. i.e. V=0 (iii) Algebric sum of moments of forces about any point is zero. i.e M=0 Graphical conditions of equilibrium (i) Force polygon should close which indicates that there is no motion of translation. (ii) Funicular polygon should close which indicates that there is no motion of rotation APPARATUS : Simply supported beam , weights , meter rule . PROCEDURE: (to be written in passive voice) 1. Note down the self weight of the beam, weight of hooks and weight of hangers. 2. Put the known loads (w1, w2, w3) at various points along the length of the beam. 3. Measure the horizontal distances of these loads from one of the support. (x1, x2, and x3) 4. Observe the reactions in the compression spring provided at each support. OBSERVATIONS: 1) Self weight of the beam (WS) = 3030 gms 2) Weight of hook = 94/ / gms 3) Length of the beam = 110 cms

Beam

W1

L1

Spring balance L2

L3

W2 W3

SMPLY SUPPORTED BEAM


10

OBSERVATION TABLE ________________________________________________________________________ Sr. APPLIED LOADS OBSERVED REACTIONS DISTANCE OF LOAD No. W1 W2 W3 R1 R2 X1 X2 X3 1 2 3 4

CALCULATION TABLE FOR FORCES AND MOMENTS: ______________________________________________________________________ Sr No. R1+R2 W1+W2+W3+WS % Error M @ Moment due % Error In force Sup. 1 Reaction 1 in moment

1 2 3 4 CONCLUSION : Discuss - verification of equilibrium conditions in determining reactions.


11

7. CO-EFFICIENT OF STATIC FRICTION

OBJECT : To determine the co-efficient of static friction between Wood to Wood , Wood to Aluminium and Wood to Brass surfaces. APPARATUS : Inclined plane, Wood, Aluminum & Brass surfaces and Weights. THEORY :

When a body is lined over another body and is gently pushed, it doesn't move. The frictional force prevents the motion. This shows that the force of friction has increased itself, so as to become equal and opposite of the applied force. Thus the force of friction has a remarkable property of adjusting its magnitude so as to become exactly equal to applied force which tends to produce motion. There is however a limit beyond which the force of friction can not increase. If the applied force exceeds this limit, the force of friction cannot balance it and block begins to move, in the direction of the applied force. This maximum value of frictional force which comes on to play when the body just begins to slide over the surface of the other body, is known as limiting friction and the ratio of the limiting friction to the normal reaction between the two bodies is known as Co-efficient of Static Friction and is denoted by . F = tan = R Where, F = Frictional force acting upward along the plane R = Normal reaction acting at right angle to the plane = Angle of inclined plane PROCEDURE: (1) Fix the angle of inclined plane and clean the surface. (2) Put on the clean surface a block, whose co-efficient of static friction is required to be found out, at the lower end of inclined plane. Connect the box with a string which will pass over the frictionless pulley fixed at higher end. Connect pan with it. (3) Put a limiting value of weight which cause a uniform upward sliding of the surface box. (4) Find out the co-efficient of friction by the formula.

Weight box

Plane

String

Pan

Nob

Pulley

Scale

Support


12

OBSERVATION TABLE: ________________________________________________________________ Sr. Angle Wood to Wood Wood to Brass Wood to Aluminium No. M W M W M W 1. 3.

CALCULATION: M - W sin = -------------- W cos Where, = Inclination of plane W = Weight of sliding surface M = Weight in pan + Weight of pan CONCLUSION: Values of


13

8. BRINELL HARDNESS TEST

OBJECT: To perform hardness test on metal specimens and study their relative hardness. APPARATUS: Brinell hardness testing machine and microscope. THEORY :

Hardness is an important property of metal. It is generally defined as resistance offered to scratching, abrasion

or indentation. It is measured in an indirect form. This experiment pertains to indentation hardness. Indentation hardness test which are widely used and accepted, usually consist of pressing an indenter of standard shape and size into the material to be tested, under a given load to determine hardness number. The hardness number is based on depth of indentation produced by a given load. Hardness testing machines based on indentation principle, consist mainly of the following ports.

1. An indentor of definite shape and size. 2. A means of applying the known load. 3. A means of measuring the dimension of the impression made by indentor.

Widely used in indertation tests are Brinell, Vicker's and Rockwell Hardness test. LIMITATIONS OF BHN : 1. This test is generally not used for testing extremely hard materials (above 450 BHN)because the ball itself get deformed. 2. This test is also not satisfactory for testing thin pieces such as razor blades because the usual indentation may be greater than the thickness of the piece. 3. The center of impression cannot be closer to any edge or to the previous indentation, than a distance equal of 3d or else the results will be erroneous.


14

TEST PROCEDURE: 1. All the I. S. specifications regarding specimen should be satisfied. 2. Select appropriate load and ball diameter. 3. Put the specimen to be tested on anvil and bring the ball penetrator in contact with the surface of the specimen. 4. Apply the load gradually until the test load is attained. 5. Unload the specimen and measure the diameter of indentation with the help of microscope. OBSERVATION TABLE:

SR No.

METAL BALL

DIAMETER D [mm]

LOAD P [Kg]

DIA OF INDENTATION

d [mm]

BHN NUMBER

1 2 3 4 5

Mild steel Cast iron

Brass Copper

Aluminum

10 10 5 5 5

3000 3000 250 250 125

P

BHN =

D ----- [ D - ( D2 - d2 ) ] 2 CONCLUSION:


15

9. COMPRESSION TEST OBJECT: To determine the compressive strength of various materials such as M.S., C.I., Timber and study their behavior when subjected to compressive load up to fracture. MATERIALS: M.S., C.I., and Timber THEORY : Specimen for compression test usually fail by buckling if they are slender. So as to avoid buckling

failure, the length of the specimen should not be more than twice the minimum dimension in c/s. For ductile materials like M.S. or Copper the specimen bulges laterally under compressive load and takes a barrel shape. Finally failure occurs then cracks appear on the circumferential surface.

Brittle materials such as C.I. usually fail by shearing along plane inclined at 50* to 70* with longitudinal axis. But, sometimes a normal failure plane is not fully developed within the length of the specimen and in such case apparent strength is appreciably increased and crushing may occur. When there is a combination of high compression strength and unrestrained lateral expansion at the ends, the specimen fails by separation in to columnar fragment known as columnar fracture. Wood is not an isotropic material, means it has different strength in different directions. It is composed of series of parallel tubes along the grain. Due to this type of structure, its strength along the grain is considerably higher than that across the grain. For load normal to the grain, the load that causes lateral collapse of tubes is the significant load.


16

LIMITATIONS: 1. It is difficult to apply truly axial load. 2. Friction between the heads of the testing machine or bearing plates and the surface of the specimen due to Lateral expansion of the specimen may alter the results considerably. 3. To obtain a proper degree of stability of the piece, the c/s area should be large compared to its length. It is possible in two ways. (a) Providing large area for the same length, but this may require larger amount of load and larger machine capacity. (b) Providing smaller specimen length for the same c/s area, but this poses difficulty in measuring strain accurately. APPARATUS: Universal testing machine with an arrangement to test specimen in tension or compression, Vernier callipers, scale. TEST PROCEDURE : 1. Put the specimen of mild steel in the machine and apply the load gradually on to it. Stop the loading when first crack is observed on the surface of the specimen or it breaks. 2. Repeat (1) for cast iron and for timber (parallel to grain & perpendicular to grain) 3. Find out stresses on each specimen in N/mm2 OBSERVATION TABLE :

SR No.

METAL SIZE mm

AREA mm²

MAXIMUM LOAD

kN

MAXIMUM STRESS N/mm²

1 2 3 4

Cast iron Mildsteel

Timber [Parallel to grain] Timber [Perpendicular to

grain]

CONCLUSION :


17

10. IMPACT TEST OBJECT : To obtain the information about the behavior of materials under impact. APPARATUS : Izod impact testing machine. MATERIALS : M.S., Copper, Aluminum, Brass. THEORY :

Impact loading is suddenly applied force which must be absorbed by material. This test is particularly applicable for the ductile materials, as they can easily absorb the impact energy in a better way and in more amounts, Whereas brittle materials are susceptible to break under impacts. Hence the test is not suitable for such materials. At ordinary temperature, a plain bar of ductile metal will not fracture under impact load in flexure. So a notched specimen is used. The use of notch causes high-localized stress concentration, which artificially reduces ductility and makes all specimens fail from the same portion. Impact blow may be delivered through the use of (i) Dropping weight (ii) Swinging pendulum (iii) Rotating fly wheel. TEST PROCEDURE : 1. Check the specimen for I.S. requirements. 2. Fix up the specimen tightly in the anvil such that the longitudinal axis of the piece lies in the plane of swinging of hammer. 3. The notch will be positioned such that the plane of symmetry of notch coincides with the top surface of the grips. 4. Set the pointer to read the energy of the blow of the pendulum. 5. Release the pendulum and allow it to strike the test piece. Read indicator and from this reading, deduct the energy lost and thus find out the impact value of specimen.


18

6. Note the temperature (atmospheric) 7. Study the fracture. LIMITATION : 1. To determine the impact resistance, temperature should be between 32c to 38c, because temperature has a very marked effect on impact resistance of notched bar. 2. Here energy lost due to friction between hammer and air, and bearing friction in indicator and pendulum is neglected. REFERENCE : IS:1985-1960 Method for izod impact test on steel OBSERVATION TABLE :

SR No.

METAL IMPACT VALU

MODE OF FAILURE

1 2 3 4

Mild steel Copper Brass

Aluminum

CONCLUSION :


19

11. TENSION TEST

OBJECT: To study stress strain curve and its characteristics with the help of tensile test on ductile and brittle materials. MATERIALS: Ductile material-Mild steel, Brittle material-Cast iron APPARTUS: Universal testing machine with an arrangement to test specimen in tension, vernier calipers & extension meter. THEORY:

In Tension test, a standard specimen is subjected to gradually increasing static (i.e. not changing with time), uniaxial load until failure occurs. Deformation is measured over the gauge length. Gauge length is the prescribed part of the cylindrical portion of the test piece on which elongation is measured at any moment during the test. SCOPE: Tension tests are generally conducted on rolled or forged ferrous and non-ferrous metals and alloys

which are required to resist tensile forces. The static tension test is the simplest and most common among all the mechanical tests. This test helps us in evaluation of fundamental properties for use in design like

1. Elastic or Plastic behavior. 2. Limit of proportionality and elasticity. 3. Yield stress and proof stress. 4. Modulus of elasticity. 5. Ultimate strength & breaking strength. 6. Toughness. 7. Percentage elongation. 8. Percentage reduction in area. 9. Ductile or brittle characteristics based on type of fracture surface.

Ultimate stress

Breaking rupture stresss

Fracture

Stress –strain Linear relation

Nome ductile Or Brittle

Stress

Yield stress Fracture

Necking

B

Strain hardening

Yielding

Strain

C

F

E

D

A

B

A-Limit of Propotionality

B-Elastic Limit

C-Upper Yield Point

D- Lower Yield Point

E- Ultimate stress

F-Fracture


20

LIMITATIONS: This test is not sufficient to provide information about the performance of material under all loading condition like impact load, fatigue load. TEST PROCEDURE : 1. Check whether the specimen fulfills I.S. requirements. 2. See that the specimen is symmetrical with respect to its longitudinal axis throughout the length. 3. Measure the diameter and its length before applying the load. 4. Mark the gauge length at 5d where d = diameter of bar. 5. Fit extensometer on the specimen and fix the specimen in universal testing machine. 6. Apply tensile load at regular intervals. 7. Measure elongation with the help of extensometer up to yield point. Note down the ultimate load & breaking load. 8. Measure the reduced diameter & increased length and note down the type of fracture. 9. See whether the fracture lies within gauge length or out side. If it is out side, comment on validity of test. 10. Plot a graph of load v/s. deformation and stress v/s strain. Determine the stress-strain characteristics and from fracture comment on the type of material. NOTE : After yield load ,extensometer should be removed to avoid damage to the same. OBESRVATION: 1. Original diameter of cross section :__________ 2. Original gauge length lo :____________ 3. Final gauge length lu :_____________ 4. Minimum c/s dimension after fracture du. :__________ 5. Load at yield point :_____________ 6. Ultimate load :_____________ 7. Ultimate or breaking load :_____________ 8. Type of fracture :_____________

SR No.

EXTENSION IN DIVISION

EXTENSION IN mm

DIVISION * 0.01 LOAD IN KG

CALCULATIONS (FROM GRAPH) : 1. Yield Stress = Yield load / Original area 2. Ultimate Stress = Ultimate load / Original area 3. Breaking Stress = Breaking load / Original area 4. Actual Breaking Stress = Breaking load / Area after fracture 5. Percentage elongation in gauge length = (lu - lo)/lo * 100 6. Percentage reduction in area = (Ao-Au)/Ao * 100 7. Stress = load / Original area N/mm2 8. Strain e = change in length/original length 9. Modulus of elasticity E = stress/strain N/mm2 RESULT TABLE:

SR STRESS STRAIN


21

No. N/mm²

QUESTIONS : (1) What is the importance of gauge length ? Discuss. (2) What is the maximum permissible percentage elongation & reduction for structural steels? (3) Explain the phenomenon of necking. CONCLUSION :

.


22

12.SHEAR TEST OBJECT : To determine the shear strength of mild steel bar in single shear, double shear and average shear stress up to distortion. MATERIALS: Mild steel round bar APPARATUS: Shear tool for steel, universal testing machine. THEORY:

A shearing stress is one that acts parallel to a plane as distinguished from tensile and compressive stress that acts

normal to a plane. Some examples of loading that produce shear condition of principle interest are as follows: 1. The resultant of parallel but opposed forces act through the centroid of section that are spaced "infinitesimal" distance apart. 2. If the above-applied forces are at finite distance apart, then in addition to shear there will be a

bending moment also. So the governing force which cause failure depending upon the relative magnitude of shear force and bending moment.

PURE SHEAR: Practically it is not possible to obtain a beam segment under pure shear without bending

stress. But to keep the bending stress to a minimum, most of the beam-span is kept supported and only a small central segment is kept unsupported. The test is useful to design riveted joints, crank pins, which are subjected to shearing stresses. Determination of shear strength is also useful to design the members which are subjected to axial, bending and shear stress for example Beam, machine shaft etc.

TEST PROCEDURE : 1. Measure the diameter. of M.S. bar that is to be tested.

SINGLE SHEAR DOUBLE SHEAR

P P/2

P/2 P

P

SHEAR BOX

M.S.BAR

Load

SHEAR BOX

M.S.BAR

Load

Before Failure Before Failure

After Failure After Failure


23

2. Fix the bar in the assembly for single shear. 3. Apply the load until bar will fail at the connected surface of the assembly. 4. Note down the load at this instant. 5. Calculate the shear stress from this value. 6. Fix the bar in the assembly for double shear. 7. Repeat step 3 to 5. 8. Evaluate average shear stress. LIMITATIONS : 1. In the direct (transverse) shear test as discussed in theory, it is usual procedure to clamp or support a prism of

material so that bending stress are minimized across a plane along which shearing load is applied. But due to bending or friction between parts of tool or both, it gives an approximation and not the exact value of shearing stress.

2. The transverse test is useless for determination of elasticity or modulus of rigidity owing to impossibility of measuring strains.

OBSERVATION TABLE :

SR No.

MATERIAL TYPE DIA mm

C/S AREA mm²

LOAD N

SHEAR STRESS N/mm²

M.S bar Single shear

M.S bar Double shear

CALCULATION : 1. Shear stress in single shear = C/S area / Load 2. Shear stress in double shear = 2 x C/S area / Load 3. Average shear stress = CONCLUSION :

Documents

WHEEL & DIFFERENTIAL AXLEgecdahod.tk/notices/manual_Mechanics_of_solids2017-… · · 2017-12-22Plot the graphs of load on X axis v/s (i) actual effort (ii) ideal effort ... is