DARSHAN INSTITUTE OF ENGG. & TECH. · PDF fileWhat is stress concentration? ... Determine the minimum size of a circular hole that can be ... Take the following permissible stress

DARSHAN INSTITUTE OF ENGG. & TECH.

Department of Mechanical Engineering

B.E. Semester – IV

Machine Design & Industrial Drafting (2141907)

Batch: ____________ Roll No.: ___________

List of Assignments

Sr.

No. Title Start Date End Date Sign Remark

1. Design Against Static Load

2. Design of Cotter and Knuckle Joints

3. Design and Analysis of Levers

4. Design of Beams

5. Design of Columns

6. Design of Shaft, Keys and Couplings

7. Design of Threaded Joints

8. Design of Welded Joints

9. Design of Riveted Joints

10. Introduction of Limits, Fits and

Tolerances

Machine Design and Industrial Drafting (2141907)

B.E. Semester IV Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1

ASSIGNMENT – 1 DESIGN AGAINST STATIC LOAD

Theory

1. What is stress concentration? Explain methods to relieve stress concentration? Explain

any two stresses with simple sketches

2. Define factor of safety and state the important factors affecting the factor of safety.

3. Define following:(1) Proof Resilience (2)Preferred number (3)Principle stress

4. Explain the following principle theories of elastic failure:

[1] Max. Principle stress theory (Rankine theory)

[2] Max. shear stress theory (Coulomb theory or Tresca and Guest theory)

[3] Distortion energy theory (Von Mises and Hencky theory)

[4] Selection and use of failure theories

5. Distinguish clearly between bending and bearing stress.

6. Explain the following terms with neat sketches:(1)Tensile stress(2) Compressive

stress(3)Principle Stress (4) Bearing pressure

7. Classify the different types of load & explain each In brief.

8. Differentiate between (with neat sketch): (1) crushing and compressive stresses (2)

torsional and transverse shear stress

Examples

1. Determine the thickness of a 120 mm wide uniform plate for safe continuous operation

of the plate is to be subjected to tensile load that has maximum value of 250 KN and

minimum value of 100 KN. The properties of the plate material are as follows:

Endurance limit=225 N/mm2,

Yield point stress=300 N/mm2, Factor of safety=1.5

2. Determine the minimum size of a circular hole that can be punched in a M.S. plate, 5 mm

thick and having ultimate shear strength of 300 MPa. Take compressive strength of

punch as 360 MPa.

3. A mild steel link is as shown in Fig. 1, which is subjected to a tensile load of 80 kN. Find

the b. The permissible tensile stress is 70 MPa.



Fig. 1

4. An offset link, Fig. 2 has a clear swing of 10 mm. the radius of curvature of curved

part is more compare to swing of link. The cross section is shown, where the

thickness (t) of link is 26 mm. Find the load corresponding to a maximum fiber stress

of 100 N/mm2.

Fig. 2

Fig. 3



5. A wall bracket with rectangular cross section is shown in following Fig. 3. The depth

of the cross-section is twice of the width. The force P acting on the bracket at 300 to

the horizontal is 5 KN. The bracket is made gray cast iron FG 200 (Sut = 200 N/mm2)

and factor of safety = 3.5 Determine the dimensions of the cross section of the

bracket.



ASSIGNMENT – 2 DESIGN OF COTTER AND KNUCKLE JOINT

Theory

1. Explain the design process for spigot and socket cotter joint with neat sketch.

2. What are the uses of cotter joint? Why is taper provided on the cotter? What is the

purpose of clearance in Cotter Joints?

3. Write the advantages of cotter joint. State the different applications of the cotter joint.

4. Explain the design process for knuckle joint with neat sketch.

5. Write the advantages of knuckle joint. State the different applications of the knuckle

joint.

Examples

1. Design a socket and spigot joint to resist a tensile load of 28 KN. All the parts of the joint

are made from same material with following allowable stresses: σt=50 N/mm²,

σc=60N/mm², τ=35 N/mm², σb =50 N/mm².

2. It is required to design a cotter joint to connect two steel rods of equal diameter. The

permissible stresses for the rods, spigot end and socket end are σt=96 N/mm²,

σc=134N/mm², τ=45 N/mm².For cotter, σt=80 N/mm², τ=40 N/mm². Each rod is

subjected to an axial tensile force of 80 KN.

Calculate the following dimensions:

(1.) Diameter of spigot

(2.) Width & thickness of cotter

(3.) Thickness of socket collar

3. It is required to design a cotter joint to connect two steel rods of equal diameter. Each

rod is subjected to an axial tensile load of 50 kN. The permissible stresses are 67 MPa in

tension, 34 MPa in shear and 134 MPa in crushing for all the parts.

4. Two rods of 50mm diameter are to be joined by a cotter joint, with thickness of cotter

as 12.5mm. If the joint is withstand an axial pull of 6000KN find the various dimensions

required. The permissible stresses are 300N/mm2 in tension, 200N/mm2in shear

and450N/mm2 crushing.

5. Design a knuckle joint to connect two rods subjected to tensile force of 50 KN. The rods

and pin are made of plain carbon steel 30C8. The permissible stresses are σt = σc = 80

MPa and τ = 40 MPa.



6. It is required to design knuckle joint to connect two mild steel roads of equal diameter.

Each rod is subjected to an axil tensile load 50 KN. The permissible stresses are 80 MPa

in tension and crushing and 40 MPa in shear for all parts.

7. Design a knuckle joint to connect two mild steel bars under a tensile load of 25 kN. The

allowable stresses are 65 MPa in tension, 50 MPa in shear and 83 MPa in crushing.

Standard diameter of solid bars are 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 mm. Check

failure of knuckle pin in shear, failure of rod end & forked end in tension, shearing and

crushing.

8. Design a knuckle joint for a tie rod of a circular section to sustain a max. pull of 70kN.

The ultimate strength of the material of the rod against tearing is 420N/mm2. The

ultimate tensile and shearing strength of the pin material are 510N/mm2 and

396N/mm2 respectively. Determine the tie rod section and pin section. Take F.S. = 6.



ASSIGNMENT – 3 DESIGN OF LEVER

Theory

1. Explain the basic types of levers with the help of neat sketches & examples.

2. Define following : (1)Arm of lever,(2) Leverage, (3)Displacement ratio (4) mechanical

advantage

3. Differentiate between simple and compound lever. Why a boss is generally needed at

the fulcrum of the levers?

4. State the application of hand and foot levers.

5. State and explain the different functions of levers. Why the levers are generally made

tapers?

6. What is lever? Explain the principle on which it works.

7. Briefly explain general procedure for lever design.

Examples

1. Design a bell crank lever to apply a load of 5 kN (vertical) at the end A of an horizontal

arm of length 400 mm. The end of the vertical arm C and the fulcrum B are to be fixed

with the help of pins inside forked shaped supports. The end A is itself forked.

Determine the cross-section of the arms and the dimensions of the pins. The lever is to

have mechanical advantage of 4 with a shorter vertical arm BC. The ultimate stresses in

shear and tension for the lever and pins are 400 MPa and 500 MPa respectively. The

allowable bearing pressure for the pins is 12 N/mm2. Assume a factor of safety as 4 and

the cross-section of the lever as rectangular with depth (b) as three times the thickness

(t).

2. A bell crank lever is to be designed to raise a load of 5 KN at the short arm end. The arm lengths are 150 mm and 500 mm. The permissible stresses for lever and pin materials in shear and tension are 60 MPa and 90 MPa respectively. The bearing pressure on the pin is to be limited to 12 MPa. Assume the lever cross section as t x 4t and fulcrum pin length as 1.25 times pin diameter.

3. Design a right angled bell crank lever having one arm 500 mm & the other 150 mm long. The load of 5 kN is to be raised acting on a pin at the end of 500 mm arm & the effort is applied at the end of 150 mm arm. The lever consists of steel forgings, turning on a point at the fulcrum. The permissible stresses for the pin & lever are 84 MPa in tension & compression & 70 MPa in shear, the bearing pressure on the pin is not to exceed 10 N/mm2.



4. A right angle bell crank lever is shown in Fig-1. The load W = 4.5KN. The lever consists of forged steel material and a pin at the fulcrum. Take the following permissible stress for the pin and lever material. Safe stress in tension = 75MPa, Safe stress in shear = 60MPa, Safe bearing pressure on pin = 10N/mm2. The length of fulcrum pin is 1.25 times the diameter of fulcrum pin. Calculate the following: (1.) Reaction at fulcrum pin (2). Fulcrum pin dimensions (3). Lever dimensions

Fig. 1

5. Design a lever of a lever loaded safety valve based on following data: Steam pressure acting on the valve = 1.2 MPa Valve diameter = 60 mm. Width to thickness ratio for lever = 3:1 Length to diameter ratio for pins = 1.25:1 The material used is forged steel with t s =80MPa, t =50 Mpa, s =100 Mpa, Pb = 20 Mpa The lever has a rectangular cross section. The distance between the fulcrums and the dead weights on lever is 800 mm and distance between the fulcrum and the pin connecting the spindle of the valve to the lever is 100 mm. Calculate: (i) the length and the diameter of the pin connecting the valve spindle to the lever (ii) the lever cross sectional dimension

6. The lever of a lever loaded safety valve shown in Fig. 2. The valve is 80 mm and valve

has to blow off at a pressure of 1.25 MPa. The permissible stress in tension, shear and crushing are 70 MPa, 20 MPa and 50 MPa respectively. The permissible bearing pressure for the pin may be taken as 20 MPa. Design the pins and the lever; assume rectangular cross section of the lever with height equal to three times the thickness.



Fig. 2

7. A lever loaded safety valve is 70mm in diameter and is to be designed for a boiler to blow off at pressure of 1N/mm2 gauge. Design a suitable mild steel lever of rectangular cross section. For mild steel: Permissible tensile stress =70MPa, Shear stress = 50MPa, Bearing pressure intensity = 25N/mm2. The pin is also made of mild steel. The distance from the fulcrum to the weight of the lever is 880mm and the distance between the fulcrum and pin connecting the valve spindle links to the lever is 80mm.

8. Design a rocker arm lever having equal arms of 160 mm length inclined at 135 for an exhaust valve of a gas engine subjected to a maximum force 2500 N at roller end. Consider – I cross section 6t x 2.5t x t size (where t = thickness of web and flange) for lever. The permissible stresses for the lever material are 80MPa in tension and design bearing pressure is pin 6 MPa for pin.


B. E. Semester – IV Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1

ASSIGNMENT – 4 DESIGN OF BEAMS

Theory

1. Distinguish between beams, columns and strut giving suitable examples.

2. Explain types of beams with neat sketch.

3. Explain types of supports (or end conditions) of beam with neat sketch.

4. Explain types of loads on beam with neat sketch.

5. Define following with reference to beam with neat sketch: (i) Deflection, (ii) Slope

and (iii) Flexural rigidity.

6. List the equations for slope and deflection for following types of beams with

different loading conditions:

(i) Cantilever beam with point load at free end

(ii) Cantilever beam with UDL on the entire span

(iii) Cantilever beam with point load at free end and UDL on the entire span

(iv) Simply supported beam with central load

(v) Simply supported beam with UDL on the entire span

(vi) Simply supported beam with central load and UDL on the entire span

Examples

1. A 2 metres long cantilever beam is having 100 mm width and 200 mm depth,

carrying point load at free end. If deflection at free end is 6 mm, calculate point load

at free end. Take E = 2 x 105 N/mm2.

2. A cantilever beam of span 1.5 metres carries a point load of 20 kN at its free end.

Find maximum slope of beam. Flexural rigidity (EI) = 2 x 104 kN.m2.

3. A hollow rectangular section 200 mm x 450 mm external and 15 mm thickness is

used for 2.7 metres cantilever beam, subjected to UDL of 64 kN/m and point load of

60 kN at free end, both downward. Find maximum slope and deflection. Take E =

200 GPa.

4. A simply supported beam 3 metres in span is subjected to UDL of 10 kN/m over

entire span with central point load 5 kN. The cross section of beam is 150 mm wide x

300 mm depth. Calculate the maximum slope & deflection for the beam.

5. A cantilever beam 120 mm x 200 mm is 2.5 metres long. What UDL should the beam

carry to produce a deflection of 5 mm at free end? Take E = 2 x 105 N/mm2.



6. Cross section of wooden beam is 100 mm x 240 mm. It is simply supported with 4

metres span. Find out UDL that can be placed on its full span so that deflection at

centre is 6 mm. Take E = 0.11 x 105 N/mm2.

7. A steel tube of external diameter 60 mm and 8 mm thickness is used as simply

supported beam of span 4 metres. If it deflects 10 mm due to a central load, find

magnitude of the point load. Take E = 2 x 105 N/mm2.


B. E. Semester – IV

Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1

ASSIGNMENT – 5 DESIGN OF COLUMNS

Theory

1. What is Slenderness ratio? State the assumptions used in Euler’s Column theory.

2. Describe the use of Johnson’s formula & Euler’s formula by graph.

3. Explain & derive the equation for Rankine formula for column.

4. Explain theory about design of connecting rod. Also derive that Ixx

Iyy= 3.2 for I-section

of connecting rod.

Examples 1. Calculate the diameter of a piston rod for a cylinder of 1.5 m diameter in which the

greatest difference of steam pressure on the two sides of the piston may be assumed

to be 0.2 N/mm2. The rod is made of mild steel and is secured to the piston by a

tapered rod and nut and to the crosshead by a cotter. Assume modulus of elasticity

as 200 kN/mm2 and factor of safety as 8. The length of rod may be assumed as 3

metres. (RSK)

2. A connecting rod of length l may be considered as a strut with the ends free to turn

on the crank pin and the gudgeon pin. In the directions of the axes of these pins,

however, it may be considered as having fixed ends. Assuming that Euler’s formula

is applicable, determine the ratio of the sides of the rectangular cross – section so

that the connecting rod is equally strong in both planes of buckling. (RSK)

3. Determine the dimensions of an I – section connecting rod for a petrol engine from

the following data:

Diameter of the piston = 110 mm

Mass of the reciprocating parts = 2 kg

Length of the connecting rod from centre to centre = 325 mm

Stroke length = 150 mm

R. P. M. = 1500 with possible over speed of 2500

Compression ratio = 4: 1

Maximum explosion pressure = 2.5 N/mm2 (RSK)

4. A 25 x 50 mm bar of rectangular cross – section is made of plain carbon steel 40C8

(Syt = 380 N/mm2 and E = 207 000 N/mm2). The length of the bar is 500 mm. The

two ends of the bar are hinged and the factor of safety is 2.5. The bar is subjected to

axial compressive force.

(i) Determine the slenderness ratio;

(ii) Which of the two equations – Euler’s or Johnson’s will you apply to the bar?

(iii) What is the safe compressive force for the bar? (VBB)

5. A trunnion mounted hydraulic cylinder is shown in figure below. The internal

diameter of the cylinder is 75 mm and the maximum operating pressure in the


B. E. Semester – IV

Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 2

cylinder 25 N/mm2. The piston rod is made of steel 40Cr1 (Syt = 530 N/mm2 and E =

207000 N/mm2). For buckling considerations, the effective length of the piston rod

is considered as the distance between the trunnion and the clevis – mount, when the

piston rod is extended to its full working stroke and this distance is 1000 mm.

Determine the diameter of the piston rod, if the factor of safety is 2.5. (VBB)

Figure: Hydraulic cylinder

6. A column of hollow rectangular cross – section and made of steel plates is shown in

the figure below. The thickness of the section is 2.5 mm throughout. The plate

material is steel 30C8 (Syt = 400 N/mm2 and E = 207000 N/mm2). The end fixity

coefficients can be taken as 1.5 and 1 for bending about long and short axes

respectively. The effective length of the column is 1 m. Determine the load capacity

of the column from buckling consideration. (VBB)

7. It is required to design the screw of a screw jack by buckling consideration. One end

of the screw is fixed in the nut and the other end supports a load of 20 kN. The

length of the screw between the fixed and free ends is 500 mm, when the load is

completely raised. The screw is made of steel 40C8 (Syt = 380 N/mm2 and E =

207000 N/mm2). Assuming a factor of safety 2.5, determine the core diameter of the

screw. (VBB)

8. A piston rod of rectangular cross – section, with both ends hinged, is shown in the

figure below. It is made of steel 40C8 (Syt = 380 N/mm2 and E = 207000 N/mm2) and

subjected to an axial compressive force of 15 kN. Determine the ratio of (b/d) for

equal buckling strength in either plane. Also determine the dimensions of cross –

section, if the factor of safety is 4. (VBB)

Figure: Piston rod



ASSIGNMENT – 6 DESIGN OF SHAFTS, KEYS AND COUPLINGS

Theory

SHAFTS

1. Explain functions and classification of shaft.

2. Define – Shaft, Axle and Spindle. Also state the difference between shaft, axle and

spindle.

3. Explain the design of shaft based on strength basis.

4. Explain torsional rigidity and lateral rigidity in detail.

5. Explain the ASME code for shaft design.

6. Explain critical speed of shaft in details.

KEYS

1. What are the basic functions of the key? Explain different types of keys with its

applications.

2. What are the different types of Sunk key? Explain each with application.

3. Derive strength equations of sunk key based on shear and crushing (or

compression) failures. Show that square key is equally strong in shearing and

crushing compare to rectangular key.

4. What is splined shaft? State the applications of splined shaft. Explain the design of

splined shaft.

COUPLINGS

1. Explain the purpose, requirements and applications of shaft coupling.

2. How does the working of a clamp coupling differ from that of a muff coupling?

3. Differentiate between flexible coupling and rigid coupling.

4. Draw a neat sketch of a protected type flanged coupling and write the design

procedure with the design equations for different failure criteria.

Examples SHAFTS

1. Compare the weight, strength and rigidity of a hollow shaft of same external

diameter as that of solid shafts, both the shafts are made of same material. Assume

that diameter ratio for the hollow shaft is di/do = 0.6. (Refer R. S. Khurmi – Ex. 14.22)



2. A steel spindle transmits 4 KW at 800 r.p.m. The angular deflection should not

exceed 0.25⁰ per metre of the spindle. If the modulus of rigidity for the material of

the spindle is 84 x 103 N/mm2, find the diameter of the spindle and the shear stress

induced in the spindle. (R. S. Khurmi – Ex. 14.21)

3. Determine the diameter below which the angle of twist of a shaft is the controlling

factor in design of solid shaft in torsion. The allowable shear stress is 56 MPa and

the maximum allowable twist is ¼ degree per meter. Take G = 84 GPa.

4. Find the diameter of a solid shaft to transmit 30 kW at 230 rpm. The shear stress is

50 MPa. If a hollow shaft is to be used in place of solid shaft, find the inside and

outside diameters when the ratio of inside to outside diameter is 6:8.

(Refer R. S. Khurmi – Ex. 14.3)

5. A line shaft is driven by means of a motor placed vertically below it. The pulley on

the line shaft is 1.5 meter in diameter and has belt tensions 5.4 kN and 1.8 kN on the

tight side and slack side of the belt respectively. Both these tensions may be

assumed to be vertical. If the pulley be overhang from the shaft, the distance of the

centre line of the pulley from the centre line of the bearing being 400 mm, find the

diameter of the shaft. Assume maximum allowable shear stress of 42 MPa.

(R. S. Khurmi – Ex. 14.8)

6. A 600 mm diameter pulley transmits 16 kW power at a speed of 400 rpm. Pulley is

cantilever at a distance of 200 mm from the nearest bearing. The weight of the

pulley is 1500 N. It is driven by a horizontal belt drive. The co-efficient of friction

between belt and pulley is 0.3 and the angle of lap 180⁰. Take the fatigue and shock

factors as Kb = 2.0 and Ks = 1.5. Determine the shaft diameter. The allowable shear

stress in the shaft may be taken as 50 MPa.

7. Design a shaft to transmit power from an electric motor to a lathe head stock

through a pulley by means of a belt drive. The pulley weighs 200 N and is located at

300 mm from the centre of the bearing. The diameter of the pulley is 200 mm and

the maximum power transmitted is 1 KW at 120 RPM. The angle of lap of the belt is

180 degree and coefficient of friction between the belt and the pulley is 0.3. The

shock and fatigue factors for bending and twisting are 1.5 and 2.0 respectively. The

allowable shear stress in the shaft may be taken as 35 MPa. (R. S. Khurmi –Ex. 14.13)

8. A belt driven C. I. pulley of 0.9 m diameter overhangs the bearing by 0.2 m as shown

in below figure. The pulley is driven from the bottom by a belt. The angles of lap and

tension on tight side are 180° and 2600 N respectively. The weight of pulley is 600

N. Assume co-efficient of friction between pulley and belt is 0.25. Shaft is made up of

30C8. σyt = 400 N/mm2, σut = 500 N/mm2. Determine the shaft diameter according to

ASME code. Take Ks=1.0, Kb=1.5. (Refer R. S. Khurmi – Ex. 14.13)



9. A shaft transmits 75 kW power at 300 rpm and load is gradually applied. It is also

subjected to B .M. of 500 Nm; shear stress in shaft material should not exceed 40

N/mm2. Shaft must not twist more than 2⁰ per meter length. Modulus of rigidity of

shaft material is 0.8 × 105 N/mm2. Find the diameter of solid shaft. If the shaft is

chosen is hollow with inside to outside diameter ratio = 0.5, find the size of the

hollow shaft. What is the percentage saving in material by using hollow shaft instead

of solid shaft?

10. The armature shaft of a 40 kW, 720 r.p.m. electric motor, mounted on two bearings

is as shown in below figure. The total magnetic pull on the armature is 7 kN and it is

assumed to be uniformly distributed over a length of 700 mm midway between the

bearings. The shaft is made-up of steel with an ultimate tensile strength of 770 MPa

and yield strength of 580 MPa. Determine the shaft diameter using ASME code if, Kb

= 1.5 and Kt = 1.0. Assume that the pulley is keyed to the shaft.

(V. B. Bhandari – Ex. 9.7)

11. A horizontal shaft AD supported in bearings at A and B and carrying pulleys at C and

D is to transmit 75 kW at 500 r.p.m. from drive pulley D to off-take pulley C, as

shown in below figure. Calculate the diameter of shaft. The data given is: P1 = 2 P2

(both horizontal), Q1 = 2 Q2 (both vertical), radius of pulley C = 220 mm, radius of

pulley D = 160 mm, allowable shear stress = 45 MPa.



12. The lay shaft driven by pulley ‘D’ from an electric motor is shown in below figure.

Another belt drive from pulley ‘C’ is running a compressor. The permissible shear

stress of the shaft material is 85.5 N/mm2. The belt tensions for pulley C are T3 =

1500 N and T4 = 600 N. The shock and fatigue factors are: Kb = 1.75 and Kt = 1.25.

The ratio of belt tension for pulley ‘D’ = 3.5, Diameter of pulley C = 150 mm,

Diameter of pulley D = 480 mm. Determine the diameter of the shaft.

KEYS & COUPLINGS

1. A 45 mm diameter shaft is made of steel with yield strength of 400 N/mm2. A

parallel key of size 14 mm wide and 9 mm thick made of steel with yield strength of

340 N/mm2 is to be used. Find the required length of key, if the shaft is loaded to

transmit the maximum permissible torque. Use maximum shear stress theory and

assume factor of safety of 2. (R. S. Khurmi –Ex. 13.2)

2. Design a muff coupling which is used to connect two steel shafts transmitting 40 kW

at 350 r.p.m. The material for the shafts and key is plain carbon steel for which

allowable shear and crushing stresses may be taken as 40 MPa and 80 MPa

respectively. The material for the muff is cast iron for which the allowable shear

stress may be assumed as 15 MPa. (R. S. Khurmi –Ex. 13.4)



3. Design a clamp coupling to transmit 30 kW at 100 r.p.m. The allowable shear stress,

for the shaft and key, is 40 MPa and the number of bolts connecting the two halves

are six. The permissible tensile stress for the bolts is 70 MPa. The coefficient of

friction between the muff and the shaft surface may be taken as 0.3. Take width of

key = Shaft diameter/4 and thickness of key = Shaft diameter/6. Assume number of

bolts = 4. (R. S. Khurmi –Ex. 13.5)

4. Design a cast iron split muff coupling to transmit a power of 10 kW at 250 rpm.

Consider an overload of 25%. The allowable shear stress in the shaft and key is 36

MPa and for the muff 16 MPa. Take the co-efficient of friction 0.3 and the tensile

strength of the high tensile bolts 150 MPa. (Refer R. S. Khurmi –Ex. 13.5)

5. Design a cast iron protective type flange coupling to transmit 15 kW at 900 r.p.m.

from an electric motor to a compressor. The service factor may be assumed as 1.35.

The following permissible stresses may be used :

Shear stress for shaft, bolt and key material = 40 MPa

Crushing stress for bolt and key = 80 MPa

Shear stress for cast iron = 8 MPa

Standard shaft diameter: 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 mm.

Take number of bolts are 3.

(R. S. Khurmi –Ex. 13.6)

6. Design a protective type cast iron flange coupling for a shaft transmitting 15 KW

power at a speed of 720 rpm. The permissible stresses are:

For the shafts, keys, and bolts material,

Permissible tensile stress (σt) = 133.33 MPa

Permissible compressive stress (σc) = 200 MPa

Permissible shear stress (τ) = 66.67 MPa

For flanges material,

Permissible shear stress (τ) = 16.67 MPa

The keys have square cross-section.

(Refer R. S. Khurmi –Ex. 13.6)

7. Design a protective type of cast iron flange coupling for a steel shaft transmitting 15

kW at 200 r.p.m. and having an allowable shear stress of 40 MPa. The working stress

in the bolts should not exceed 30 MPa. Assume that the same material is used for

shaft and key and that the crushing stress is twice the value of its shear stress. The

maximum torque is 25% greater than the full load torque. The shear stress for cast

iron is 14 MPa. (R. S. Khurmi –Ex. 13.7)

8. Design and draw a cast iron flange coupling for a mild steel shaft transmitting 90 kW

at 250 r.p.m. The allowable shear stress in the shaft is 40 MPa and the angle of twist

is not to exceed 1° in a length of 20 diameters. The allowable shear stress in the



coupling bolts is 30 MPa. Take width of key = shaft diameter/4 and thickness of key

= shaft diameter/6. Assume no. of bolts = 4. (R. S. Khurmi –Ex. 13.8)

9. The shaft and the flange of a marine engine are to be designed for flange coupling in

which the flange is forged on the end of the shaft. The following particulars are to be

considered in the design: Power of the engine = 3 MW, Speed of the engine = 100

r.p.m., Permissible shear stress in bolts and shaft = 60 MPa, Number of bolts used = 8

and Pitch circle diameter of bolts = 1.6 × Diameter of shaft.

Find: 1. diameter of shaft; 2. diameter of bolts; 3. thickness of flange; and 4. diameter

of flange. (R. S. Khurmi –Ex. 13.11)

10. A flexible coupling as shown in below figure is used to transmit 15 kW power at 100

rpm. There are six pins and their pitch circle diameter is 200 mm. The effective

length of the bush, the gap between the two flanges and the length of pin in contact

with the right hand flange are 35, 5 and 23 mm respectively. The permissible shear

and bending stress in the pin are 35 N/mm2 and 152 N/mm2 respectively. Calculate

pin diameter by shear & bending considerations. (V. B. Bhandari Ex – 9.23)

11. Design a bush pin type protected flexible flange coupling to connect the output shaft

of an electric motor to the shaft of a centrifugal pump. The motor delivers 20 kW

power at 720 r.p.m. The starting torque of motor can be assumed to be 150 % of the

rated torque. The permissible stresses are as under :

95 MPa in shear for shaft,

100 MPa in shear and 300 MPa in crushing for key,

200 MPa in tension and 35 MPa in shear for pin

17 MPa in shear for flanges

Consider the No. of pins = 6.

(V. B. Bhandari Ex – 9.23)


B.E. Semester – IV Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1

ASSIGNMENT – 7 Design of Threaded Joints Theory

1. Explain the different types of screw threads used in power screw stating their

applications.

2. What do you understand by the single start and double start threads? Define

following terms: (a) Major diameter, (b) Minor diameter, (c) Pitch and (d) Lead.

3. Derive an equation for torque required to raise (lift) load by square threaded screw.

4. Derive an equation for torque required to lower load by square threaded screw.

5. Derive an equation for efficiency of square threaded screw and maximum efficiency

of a square threaded screw.

6. What is self-locking and over-hauling of power screw? What is significance of these

properties? Show that the efficiency of self-locking screws is less than 50%.

7. Discuss on bolts of uniform strength giving examples of practical applications of

such bolts.

8. Explain the purpose of a turn buckle (or Coupler) with neat sketch and describe its

design procedure.

Examples

1. The mean diameter of the square threaded screw having pitch of 10 mm is 50 mm. A load of 20 KN is lifted through a distance of 170 mm. The external and internal diameters of the bearing surface of the loose head are 60 mm and 10 mm respectively. The coefficient of friction for the screw and the bearing surface may be taken as 0.08. Find the work done in lifting the load and the efficiency of the screw, when

a) The load rotates with the screw, and b) The load rests on the loose head which does not rotate with the screw.

2. The lead screw of a lathe machine has single start trapezoidal threads of 52 mm nominal diameter and 8 mm pitch. The screw is required to exert an axial force of 2 kN in order to drive the tool carriage, during turning operation. The thrust is carried on a collar of 100 mm outer diameter and 60 mm inner diameter. The values of co-efficient of friction at the screw threads and the collar are 0.15 and 0.12 respectively. The lead screw rotates at 30 rpm. Calculate:

a) The power required to drive the lead screw, b) The efficiency of the screw. 3. A power screw having double start square threads of 25 mm nominal diameter and

5 mm pitch is acted upon by an axial load of 10 kN. The outer and inner diameters of screw collar are 50 mm and 20 mm respectively. The coefficient of thread friction and collar friction may be assumed as 0.2 and 0.15 respectively. The screw rotates at 12 r.p.m. Assuming uniform wear condition at the collar and allowable thread bearing pressure of 5.8 N/mm2, find: a) The torque required to rotate the screw; b) The stress in the screw; and c) The number of threads of nut in engagement with screw.



4. A machine vice as shown in Fig. has single start square threads with 22 mm nominal diameter and 5 mm pitch. The outer and inner diameters of the friction collar are 55 mm and 45 mm respectively. The values of coefficient of friction for threaded collar are 0.15 and 0.17 respectively. The machinist can comfortably exert a force of 125 N on the handle at a mean radius of 150 mm. Assuming uniform wear for the collar calculate:

a) Clamping force developed between the jaws b) The overall efficiency of the clamp

5. A triple threaded power screw, used in a screw jack, has a nominal diameter of 50 mm and a pitch of 8 mm. The threads are square and the length of nut is 48 mm. The screw jack is used to lift a load of 7.5 KN. The coefficient of friction at the threads is 0.12 and collar friction is negligible. Calculate: (i) the principal shear stress in the screw body, (ii) the transverse shear stresses in the screw and the nut, (iii) the unit bearing pressure. State whether the screw is self-locking or not.

6. The screw shown in fig is operated by a torsional moment applied at the lower end. The nut is loaded and prevented from turning by guides. The outside diameter of the screws is 50 mm, pitch of 8 mm and the thread is acme triple start. The coefficient of friction of the threads is 0.15. Assume the friction in ball bearing as negligible. If the torsional moment Mt is 45 Nm.

a) Determine the load which could be raised b) Would the screw be overhauling? c) Determine the average bearing pressure between the screw and nut thread

surfaces.



7. The pull in the tie rod of an iron roof truss is 50 kN. Design a suitable adjustable Screwed joint (turnbuckle). The permissible stresses are 75 MPa in tension, 37.5 MPa in shear and 90 MPa in crushing.

8. Design a turnbuckle for a capacity of 40 kN, which is used for adjusting tension in a v-belt drive of a machine tool. The permissible stresses for rods and nut are 80 MPa in tension, 50 MPa in shear and 80 MPa in crushing.



ASSIGNMENT – 8 Design of Welded Joints Theory

1. What do you understand by the term welded joint? Explain advantages and

disadvantages of welded joints over riveted joints.

2. Classify and explain the types of welded joints with neat sketches and weld symbols.

3. Discuss the standard location of elements of a welding symbol.

4. Derive equations of strength for transverse and parallel fillet welded joints with neat

sketches.

5. What do you mean by eccentric loaded welded joint? Write the detail design

procedure for designing such a joint.

Examples

1. A plate 60 mm wide and 80 mm thick. It is welded with another plate by means of single transverse and double parallel fillet welds. Find the length of each parallel fillet if allowable tensile and shear stresses in the weld material are 80 and 60 MPa respectively.

2. A circular shaft, 75 mm in diameter, is welded to the support by means of a circumferential fillet weld. It is subjected to a torsional moment of 3000 N-m. Determine the size of weld, if the maximum shear stress in the weld is not to exceed 70 N/mm2.

3. A welded joint as shown in figure, is subjected to an eccentric load of 2 KN. Find the size of weld, if the maximum shear stress in the weld is 25 N/mm².

4. A shaft of rectangular cross-section is welded to a support by means of fillet welds,

as shown in fig. Determine the size of the welds, if the permissible shear stress in the weld is limited to 75 N/mm2.



5. A bracket is fillet welded to a structure as shown in figure, which is subjected to a

load of 50 kN. Find the size of weld required if allowable shear stress is not to exceed

75 MPa. Take polar moment of inertia J =𝑡 (b+𝑙)3

6

6. A welded joint has to support a load of 80 kN. Suggest a suitable size of fillet weld if

the safe shear stress for the weld material is 80 MPa. Refer the given figure.

7. A bracket carrying a load of 15 KN is to be fillet welded as shown in Fig. Find the size of weld required if the allowable shear stress is not to exceed 80 MPa.

Take polar moment of inertia J = t 𝑙 (3b2+𝑙2)

6



8. Figure shows 12 mm thick plates loaded by the forces of 100 KN applied eccentrically. Determine the required lengths L1 & L2 of the fillet welds so that they will be equally stressed in shear. Take working stress in shear for side fillets to be equal to 80 N/mm2.

9. A bracket is welded to the side of a column and carries a vertical load P, as shown in

Fig. Evaluate P so that the maximum shear stress in the 10 mm fillet welds is 80 MPa.



ASSIGNMENT – 9 Design of Riveted Joints Theory

1. Define riveted joints. Classify and explain the different types of riveted joints with

neat sketches.

2. Explain the following terms related to riveted joints:

a) Pitch, b) Margin, c) Diagonal pitch and d) Transverse pitch.

3. Explain caulking & fullering in terms of riveted joint.

4. Discuss the different types of failures in riveted joint (or the various ways in which

a riveted joint may fail).

Examples

1. Find the efficiency of the double riveted lap joints with zig-zag riveting is to be designed for 13 mm thick plates. Assume 80 MPa, 60 MPa and 120 MPa in tension, Shear and crushing respectively. Also calculate pitch of rivets.

2. Design a double riveted, double strap, chain type butt joint for plates having 10 mm thickness. Also find efficiency of the joint. Take σt = 95 N/mm², σc =155 N/mm² and τ =80 N/mm².

3. A double riveted double cover butt joint in plates 20 mm thick is made with 25 mm diameter. Rivets at 100 mm pitch. The permissible stress are σt =120 N/mm2, Shear stress= 100 N/mm2, σc= 150 N/mm2. Find the Efficiency of joint, taking the strength of the rivets in double shear as twice than that of single shear.

4. Design a double riveted butt joint with two cover plates for the longitudinal seam of a boiler shell 1.5 m in diameter subjected to a steam pressure of 0.95 N/mm2. Assume joint efficiency as 75%, allowable tensile stress in the plate 90 MPa, Compressive stress 140 MPa & shear stress in the rivet 56 MPa.

5. A bracket is supported by means of four rivets of same size as shown in Fig. Determine the diameter of rivet if the maximum shear stress is 140 MPa.

6. A bracket is to be attached to a wall with the help of six rivets. The different arrangements in which the bracket can be attached to the wall with these rivets are shown in figure. The maximum allowable stress in shear is 60 N/mm2. Determine the way in which the rivets should be arranged so that the design is economical. The



bracket is required to support a load of 90 KN with an eccentricity of 200 mm. Also determine the diameter of rivet for the selected arrangement.

7. An eccentrically loaded lap riveted joint is to be designed for a steel bracket as

shown in figure. The bracket plate is 25 mm thick. All rivets are to be of same size, load on the bracket P = 50 KN, rivet spacing c =100 mm, load arm e = 400 mm, permissible shear stress is 64 MPa, and crushing stress is 120 MPa. Determine the size of the rivets to be used for the joint.

8. A bracket is subjected to a load of 32 KN which is joined to a structure by means of 8

numbers of rivets as shown in Fig. Find the size of the rivets if the permissible shear stress is 80 MPa.


B. E. Semester IV Department of Mechanical Engineering Darshan Institute of Engineering and Technology, Rajkot 1

ASSIGNMENT 10 – INTRODUCTION TO LIMITS FITS AND

TOLERANCES

Theory

1. Define the following:

(i) Limit

(ii) Basic size

(iii) Tolerance

(iv) Allowance

(v) Deviation

(vi) Clearance

(vii) Fit

2. Explain types of tolerances with neat sketch and explain applications for it.

3. Explain types of deviations with neat sketch and explain applications for it.

4. Explain types of clearances with neat sketch and explain applications for it.

5. Explain types of fits with neat sketch and explain applications for it.

6. What is surface roughness? How it is indicated on drawing with the help of various

symbols? Explain importance of it.

7. Explain the various machining symbols with all parameters.

8. Explain hole-based and shaft based limit system with neat sketch. Give appropriate

examples also.

9. Give symbols for following various geometrical tolerances and explain it:

Straightness, Flatness, Circularity, Parallelism, Perpendicularity, Cylindricity,

Symmetry, Angularity and Concentricity

10. Explain maximum metal condition (MMC) and least metal condition (LMC).

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