machine design manual final

CHHOTUBHAI GOPALBHAI PATEL INSTITUTE OF TECHNOLOGY

Mechanical Engineering Department (Diploma) Semester - V

Design of Machine Elements (020020502)

Experiment: 1 Aim: To study general consideration and factors influencing the design of machine

elements.

1. Define machine design. Explain factors affecting the design of machine element. 2. Define factor of safety and explain factors for F.O.S. 3. Classify stresses and explain each with example. 4. Find six standard speeds between 250 to 1400 rpm. 5. A punch can withstand a safe compressive stress of 300 N/mm2, it is used for punching

hole of 15mm diameter in plate material which has ultimate shear stress of 120 N/mm2. Calculate maximum thickness of plate material through which a hole can be punched.

6. Find minimum size of square hole which can be drilled in 10 mm thick steel plate. Ultimate shear stress (for steel plate material) =300 N/mm2 Permissible crushing stress

(for punch) = 150 N/mm2

7. An angle section having equal sides with 6 mm thickness is subjected to axial tensile

load of 30 kN. Determine the dimension of each side of the angle for the allowable

stress of 90 N/mm2.




Experiment: 2 Aim: To study design of different kinds of joints.

1. A cotter Joint is to resist an axial load of 40KN. The allowable stresses are

σT=70N/mm2 t =55N/mm2 and σcr =140N/ mm2

Find i) Diameter of Rod ii) Diameter of spigot rod and iii) width and thickness of

cotter. 2. A cotter joint is to be designed to connect two rods of equal diameter. The axial

load to be resisted is 60 KN. Assuming τ = 0.8σt σc = 2σt. Design the joint. 3. A Knuckle joint is to carry a tensile load of 20KN, permissible stresses in Tension,

shearing and bearing as 90 N/mm2 and 75 N/mm2, Design a Knuckle joint. 4. A Knuckle Joint is to carry a load of 30 KN. If the allowable stresses are σ = 75 N/mm2

, τ = 60 N/mm2 and bearing pressure for pin is limited to 25 N/mm2. Find (i) Diameter of the rod (ii) Diameter of the pin, take l/dp = 1.25. (iii) Check the pin

in shear failure.

5. A Double riveted lap joint is to be used to connect two M. S. plates 10 mm thick. The diameter and pitch of the rivets are 20 mm and 60 mm respectively. The

allowable stresses for the plates and riveters are σ =60 N / mm2, τ =50 N/ mm2

and σcr = 80 N /mm2. Find i) Strength of joint ii) Efficiency of joint

6. Design a double riveted chain type equal cover butt joint to connect two plates of

16mm thickness, The allowable stresses are s = 100 MPa, t =80 MPa, σcr = 150

MPa. 7. What may be the maximum efficiency of Screw Jack, and why? Obtain the equation

for the Efficiency of screw jack. 8. The details relates to a screw jack: Maximum load 30KN,Safe Compressive stress in

screw 85 N/mm2 , Pitch of single start square threads 5 mm, Safe bearing pressure

17.5 N/mm2 . Calculate the screw size, height of nut and number of threads in nut.

9. The compressive load on a screw jack is 50 KN. Determine the diameter of the screw

and the height of nut if the allowable compressive stress for screw material is75

N/mm2 and allowable bearing pressure for the bronze nut for low velocity is 16N/mm2.

Assume single start screw threads having 2 threads per cm.




Experiment: 3

Aim: To study design of various types of levers. 1. The bell crank lever of Hartnell governor has vertical arm of 120 mm length and

horizontal arm 100 mm. The maximum centrifugal force acting on the vertical arm is

1500 N. Design the lever completely with neat sketch. Take σt = 70 N/mm2 , σbr = 30 N/mm2.

2. A bell crank lever has horizontal short arm of 125 mm, and lift the load of 2250 N. A

vertical effort arm is 250 mm. Design fulcrum pin, if, τ =70 N/mm2 and Pb =10

N/mm2.

3. Design fulcrum pin & lever cross section of Bell crank lever from following data:

(1) Vertical load acting at the end of long arm = 6 KN. (2) Arm length of lever = 450 mm & 150 mm

(3) τ (for pin) =70 N/mm2 & Pb (for pin) = 10N/mm2

(4) [σb] (for lever) = 60N/mm2

(5) height / breadth = h/b = 3 (for lever cross-section )

(6) L/dp = 1.25 (for pin) 4. Fulcrum pin of a bell crank lever is to be designed where 7000N. Load acting on longer

arm is to be lifted. The length of arms is 400 mm & 100 mm. τ = 50 N/mm2, σt = 70 N/mm2 & Pb = 15 N/mm2. Find diameter and length of fulcrum pin .do not

consider bending.




Experiment: 4 Aim: To study design of different types of leaf spring.

1. A semi – elliptical laminated spring 900 mm long and 55 mm wide is held together by

a central band 50 mm long. If the thickness of each leaf it 5 mm, Find the number of

leaves required to carry a load of 4500 N. Assume maximum working stress of 490

MPa. If the two leaves extend the full length of spring. Find the deflection of spring. Take E = 210 KN/mm2

2. A semi elliptical leaf spring has 12 leaves; in which first two are full lengths leaves and

the remaining are graduated leaves. The span of the spring is 1200 mm and leaves are

held together by a central band clip 80 mm wide. Maximum load on the spring is 8 KN

and the permissible bending stress is 300 MPa. If the ratio of the total thickness of the

leaves and its width is 1.5,

Find: (1) Thickness and width of a leaf.

(2) Maximum deflection of the spring. Take, E = 2 × 105 MPa. 3. Determine Bending Stress induced in a Semi-elliptical leaf spring from the following data.

(i) Central load = 10 KN (ii) Effective span = 1000 mm (iii) Width of leaves = 60 mm. (iv) Thickness of Leaves = 6 mm (v) Total number of leaves = 10, including 2 extra full

length leaves. Also find deflection of spring if E = 2 x 105 MPa. 4. A semi-elliptical leaf spring has a span of 1 meter. There are two extra full length

leaves and eight graduated leaves including a master leaf. The spring material is 55

Si2Mn90 steel having yield point stress of 1500 Mpa. If E = 2.07 × 105 Mpa and

central load is 30 KN. Find : (1) Width and thickness of leaves if b/h = 10 (2) Deflection of spring. Take factor of safety = 2.

5. Find out No. of leaves, width , thickness & curvature for semi-elliptical leaf spring

from following data : (1) Length of spring = 1000 mm

(2) Maximum load = 3600 N

(3) Maximum deflection = 75 mm

(4) [σb] = 360 N/mm

(5) width /thickness = b/t = 12

(6) E = 2× 105 N/mm2




Experiment: 5 Aim: To study design of shafts.

1. State the difference between shaft spindle and axle giving example. 2. Write the equation by which angle of twist for a shaft is calculated. 3. Write design procedure of the shaft.

4. Write design procedure of shaft for torsional rigidity. 5. Write design procedure of shaft for lateral rigidity. 6. A line shaft rotating at 200 r.p.m. is to transmit 20 kW. The shaft may be assumed to

be made of mild steel with an allowable shear stress of 42 MPa. Determine the diameter

of the shaft, neglecting the bending moment on the shaft. 7. A solid shaft is transmitting 1 MW at 240 r.p.m. Determine the diameter of the shaft

if the maximum torque transmitted exceeds the mean torque by 20%. Take the

maximum allowable shear stress as 60 MPa. 8. Find the diameter of a solid steel shaft to transmit 20 kW at 200 r.p.m. The ultimate

shear stress for the steel may be taken as 360 MPa and a factor of safety as 8. If a

hollow shaft is to be used in place of the solid shaft, find the inside and outside diameter

when the ratio of inside to outside diameters is 0.5. 9. A solid circular shaft is subjected to a bending moment of 3000 N-m and a torque of

10000 N-m. The shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa

and a ultimate shear stress of 500 MPa. Assuming a factor of safety as 6, determine

the diameter of the shaft. 10. A solid shaft is subjected to bending moment of 3.46 kN.m and torque of 11.5 kNm.

6ut =690 Mpa. Ultimate t = 516 Mpa. Find shaft diameter by equivalent torque if

factor of safety is 6. 11. A hollow shaft 200 mm inside diameter and 300 mm outside diameter transmits 2500

KW at 200 RPM. Calculate the stress induced in shaft. 12. A hollow motor car shaft is required to transmit 20KW at 300 rpm. If the inner diameter

is 0.6 times the outer diameter and the allowable shear stress for the shaft material is

55 MPa. Determine the thickness of the shaft. 13. A 25 KW motor transmits power at 1000 rpm. The motor has M.S. shaft & key. If shear and

crushing stress for M.S. shaft & key are 40 N/mm2 110 N/mm2 respectively. Find the width

& thickness of key for a length 60 mm.




Experiment: 6 Aim: To study design of various kinds of keys.

1. What do you understand by key? With the help of neat sketches, explain various types

of keys and state the application of each one. 2. Design the rectangular key for a shaft of 50 mm diameter. The shearing and crushing

stresses for the key material are 42 MPa and 70 MPa. 3. A pulley is fixed on a 100 mm diameter shaft with the help of 100 mm long key. The

shaft transmit 6000 Nm torque with this key. Find dimension of key if σuc = 500

N/mm2 and Ultimate τ = 250 N/mm2. Factor of safety is 5.

4. A 45 mm diameter shaft is made of steel with a yield strength of 400 MPa. A parallel

key of size 14 mm wide and 9 mm thick made of steel with a yield strength of 340 MPa

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 a factor

of safety of 2. 5. Design a rectangular key for transmitting 1400 Nm torque to the pulley mounted on it.

For shaft material, and key material, =60 N/mm2 and σc= 100 N/mm2. Consider

weakness factor for key way as 0.8 and over loading torque of 25%. Take width of

Key = 0.3 * Shaft diameter, and Thickness of key = 0.2 Shaft diameter.




Experiment: 7 Aim: To study design of different types of coupling.

1. A flange coupling is required to transmit 60KW at 250 rpm Find

i) Shaft diameter if τ = 60MPa

ii) Number and size of bolts if, τ bolt = 25 MPa

iii) Thickness and diameter of flange.

2. List various parts subjected to direct twisting movement. A flanged coupling is required

to transmit 60 KW at 250 rpm. τshaft = 60 MPa, τ bolt = 25 MPa. Find shaft diameter, bolt diameter and no. of bolts, flange thickness and flange diameter.

3. A flange coupling transmits 10KW power at 500 RPM. [τ] (for shaft) =60 N/mm2 , [τ] (for bolt) = 40 N/mm2

Find following considering 25% overload condition: (i) Shaft diameter (ii) No. of bolt (iii) Bolt diameter.

4. A muff coupling is to be designed to connect motor shaft and pump shaft of 35 mm. diameter, the motor is to transmit 8 KW at 500 RPM. Determine (i) size of muff (ii) shear stress induced in the shaft and muff.

5. A muff coupling is to be designed to connect two shafts 50mm diameter. The permissible

shear stresses for shaft and muff are 55 MPa and 10 Mpa respectively and given keyway

factor is 0.8.

Find: (i) Size of muff. (ii) Power transmitted by the coupling at 200 rpm. Considering

weakening effect of key way.




Experiment: 8 Aim: To study design of different types of types of helical spring.

1. Calculate the Spring wire diameter and active number of turns for a closely coiled

helical spring from the following data: Range of service load = 2 KN to 4 KN

Spring deflection = 7 m

Spring Index = 5

Spring shear stress = 400 N/ mm2

Modulus of rigidity = 8.3 x 104 N/mm2

2. A closed coil helical spring is to be designed for loads ranging from 4 KN to 4.5 KN. The axial compression of the spring is 8 mm and the spring index is 6. If the allowable

shear stress for spring material is 320 N/mm2 and modulus of rigidity is 80 KN/mm2, find: (1) Spring wire diameter (2) Number of active coils (3) Stiffness of the spring.

3. A valve spring having inner diameter of a coil 40 mm, deflects for 40 mm at the

maximum axial load of 500 N. Find wire diameter and number of turns for the spring. [τ] = 300 N/mm2, spring index=6 and G = 0.82 x 105 N/mm2

4. For Closed coil helical spring, find spring wire diameter & no. of active turns of coil

from following data: (i) Load Range = 2.25 KN to 2.75 KN (ii) Spring Deflection for

above load range = 6 mm (iii) Spring Index = 5 (iv) Shear Stress for spring = 420

N/mm2 (v) Modulus of Rigidity, G = 84 KN/mm2

5. Design a compression helical spring to take load of 700 N. The maximum compression

on the spring is 25 mm. The spring index is 8. take , [τ] = 325 MPa and G = 84000

MPa.

6. Following data refers to a valve of 4-stroke diesel engine. - Length of spring when the valve is closed = 60 mm

- Length of spring when the valve is open = 50 mm. - Load on the spring when the valve is closed = 225 N. - Load on the spring when the valve is open = 365 N.




- Spring Index = 8

- Allowable shear stress for spring = 325 N/mm2

- Modulus of rigidity = 83 x 103 N/mm2.

Find: i. Spring wire diameter. ii. Nos. of coils if ends are squared and

ground. iii. Stiffness of spring.

7. Find out spring wire diameter & number of total coils for Helical spring from following

data : (1) Inner diameter of Helical spring = 50 mm (2) Maximum load = 500 N (3) Spring deflection = 40 mm (4) Spring Index = 6 (5) Modulus of Rigidity = 0.82 ×

105 N/mm2 (6) τ (for spring) = 300 N/mm2




Experiment: 9 Aim: To study different machine elements subjected to eccentric loading.

1. A rectangular strut is 150 mm wide and 120 mm thick. It

carries a load of 180 kN at an eccentricity of 10 mm in a

plane bisecting the thickness as shown in Fig.1. Find the maximum and minimum intensities of stress in the

section.

2. Define eccentric loading. The frame of a “C” clamp has

a regular section of 90mm x 45mm. A maximum

clamping load of 30KN is acting at a distance of 155mm

from the inner edge of the frame. Find the maximum and

minimum stresses induced in the frame section.

3. A “C” clamp carries a vertical load of 25 KN. With

rectangle cross section. Eccenticity is 150mm. Take safe

tensile stress of 100 n/mm^2. Find section of clamp body if h= 2b

4. A maximum clamping load of 12 KN is acting at the end of a “C-Clamp” having

rectangular cross-section. The distance between the neutral axis of the cross-section

and the load axis is 120 mm. maximum safe stress for the clamp is 100 M Pa. Find

out the dimensions of rectangular cross-section of the clamp. Take the height of the

cross-section twice its width. 5. A frame of C-clamp has rectangular cross-section of 60mm×20mm. A maximum

clamping load of 30KN is acting at a distance of 60mm from the centre of axis of the

frame. Determine the maximum and minimum stresses in the frame section. 6. A “C” clamp carries a vertical load of 20 KN. The body of the “C” clamp is of

rectangular cross- section. The distance between the load line and neutral axis of

section is 150 mm. Take safe tensile stress of 100 N/mm2 Find the section of the clamp

body. Assume height two times breadth.




Experiment: 10 Aim: To study various types of load acting on bolts.

1. The cover of hydraulic press cylinder is secured by means of three steel bolts. The inner

diameter of cylinder is 80mm and the maximum water pressure is 12MPa. If the

allowable stress for the bolt material is 45Mpa, find size of the bolts. 2. A Pillar crane is fastened to the foundation by 4 bolts (two on – XX- axis and two on

YY-axis). Equally spaced on a circle of 2000 mm diameter. The diameter of flange is

2500 mm. Determine the diameter of bolts to lift a load of 50 KN at a radius of 7500

mm.Take σt = 100 MPa for bolt. 3. The cover of a hydraulic press cylinder is secured by means of 6 chromium bolts. The

inner diameter of the cylinder is 100 mm and the maximum fluid pressure is 15 M Pa. The initial tightening load on each bolt is 10 KN. If the allowable stress for the bolt

material is 45 M Pa, find the size of bolts. Take, overall stiffness co-efficient = 0.6. 4. A cover of a steam engine cylinder is secured by means of 10 bolts. The inner diameter

of cylinder is 260 mm and maximum steam pressure is 1.20 MPa. If the allowable

tensile stress for the bolt material is 60 MPa, find the size of bolt. 5. Find size of bolt for cover of steam engine cylinder using following data :

(1) No of bolts fitted = 12 (using soft copper gasket) (2) Initial Tightening load on each bolt = 10 KN

(3) Internal diameter of cylinder = 500 mm

(4) Maximum steam pressure = 2 N/mm2

(5) σt (for bolt) = 50 MPa

(6) Overall stiffness co-efficient K = 0.50.




Experiment: 11 Aim: To study design of pressure vessels.

1. A cylinder with 150 mm inside diameter and 15 mm plate thickness is subjected to

internal pressure of 5 N/mm2. find hoop stress, longitudinal stress and max. shear stress

in the cylinder.

2. A ram of a hydraulic cylinder having 200 mm internal diameter is subjected to oil

pressure of 10 MPa. If the permissible stress for the cylinder material is 28 MPa. Find

outside diameter of the cylinder.

3. 200 mm internal diameter and 3 mm thick cast iron pipe has 1.2 N/mm2 water pressure

inside pipe. Find hoop stress, longitudinal stress and max. Shear stress.

4. A thin cylindrical shell of 4500 litre capacity is to be designed. The internal

pressure is 1.6 MPa and allowable hoop stress is not to exceed 70 MPa for this

shell material. Assuming joint efficiency for the shell as 85%. Find the

thickness of the spherical shell.

5. Find hoop stress, longitudinal stress & maximum shear stress for a cylindrical vessel

from following data: (1) External diameter = 600 mm (2) Maximum internal pressure

= 2 N/mm2 (3) Thickness of vessel = 20 mm. And also state type of vessel as per D/t ratio.

6. The maximum internal pressure of 8 N/mm2 exerts in a cylindrical vessel having 180

mm external diameter and 15 mm plate thickness. Find Hop stress, Longitudinal stress

and Maximum Shear stress. Also, state the type of this cylinder as per D/t ratio. Take,

the Joint Efficiency = 80 %.

7. The air receiver tank consists of cylinder 1.2 m inside diameter which is closed by

hemispherical ends. The pressure of compressed air inside the cylinder is not to exceed

2.5 N/mm2. If the tank is of steel whose ultimate stress is 300 N/mm2, Calculate the

wall thickness of cylindrical and hemispherical portions. Take factor of safety as 3.




Experiment: 12 Aim: To study design procedure of different types of bearings.

1. A deep groove ball bearing having SKF 6309 number is rotating at 1500 rpm. the

bearing is subjected to radial load of 8500 N and thrust load of 5500 N. The inner race

of the bearing rotate with shaft. find the rating life and average life of the bearing. the

dynamic capacity of the bearing is 41500 N and the bearing is in continuous service. Take X= 0.56 & Y=1.3

2. A radial ball bearing has a basic dynamic load capacity of 50 KN. If the rating life of

the bearing is 6000 hrs, what equivalent load can the bearing carry at 500 RPM.

3. A ball bearing under 4 KN redial load and 5 KN thrust load with 2600 rpm working for

5 years, 300 days and 10 hours/day. Find basic dynamic load rating.

4. A ball bearing is required to operate for 6 years at 8 hours per day at 1000

rpm. If the equivalent load on the bearing is 5 KN. Find the basic dynamic

capacity of bearing. Take 300 working days in a year. Take bearing constant

k = 3 & rev. life L10 = 106.

5. The main bearing of a diesel engine is subjected to a load of 20 KN, at 700

rpm. The ratio d/c = 1000. If 0.25 KW power is lost in friction, determine the

viscosity of lubricating oil. Take journal diameter as 85 mm and bearing

length 105 mm.

machine design manual final

Documents

machine design

screw material is75

allowable stresses

efficiency of joint

knuckle joint

cotter joint

allowable compressive

compressive load