machine design & industrial drafting

1. Guided By :-
 Rudra sir

2.  Lever :- Lever is a rigid rod or bar pivoted at a
point called “fulcrum” and used either to overcome
a load by the application of small effort.
 Arms of lever :- In lever a perpendicular
distance of the effort and load from the fulcrum are
known as effort arm and load arm respectively.

3. Leverage:- The ratio of the effort arm to the load
arm is called leverage.
leverage =
𝒍 𝒑
𝒍 𝒘
Why levers are tapered ?
• In lever, the bending moment is maximum at the
fulcrum and minimum at the ends. Therefore, the
levers are usually tapered from the fulcrum to the
ends. This would help in saving the material at the
ends.

4. Mechanical Advantage:-
Let,
w = load to be overcome
p = effort applied
lw = load arm
lp = effort arm
θ = angle between lines of action of load and effort.
• Taking moment at falcrum
p * lp = w * lw
𝑤
𝑝
=
𝒍 𝒑
𝒍 𝒘
= Mechanical Advantage

5. • Types of lever :-
1. Lever with fulcrum between load and effort.
• In such lever M.A. is equal to one.
• Example:- Rocker arm in I.C. engine, handle of hand
pump.

6. 2. Lever with load between fulcrum and effort
• In this case the effort arm is longer than load arm so
M.A. is grater then one.
• Example :- Lever of boiler safety valve.

7. 3. Lever with effort between fulcrum and load.
• Since effort arm is smaller then load arm so M.A. is
less then one.
• Such type of lever are not used in engineering
application.
• Example:- forceps

8. 4. Angular lever
• A lever in which the load arm and effort arm are
inclined at an angle other than 1800 which is called
angular lever.
• Generally the angle between the two arm is 1350
• Example :- Rocker arm’s in I.C. engine , Break lever

9. 5. Bell Crank lever
• When the angle between the load arm and effort
arm is 900 then the lever is called as bell crank
lever.
• Example :- Lever in railway signal

10. 6. One Arm Lever
• It has only one arm and that is effort arm
• Example :- Foot lever , hand break

11. ROCKER ARM OR ANGULAR
LEVER

12. ROCKER ARM.

13. USE AND WORKING ROCKER ARM.
A rocker arm (in the context of an internal combustion
engine of automotive, marine, motorcycle and reciprocating
aviation types) is an oscillating lever that conveys radial
movement from the cam lobe into linear movement at the
poppet valve to open it.
One end is raised and lowered by a rotating lobe of the
camshaft (either directly or via lifter and pushrod) while the
other end acts on the valve stem.

14. USE AND WORKING ROCKER ARM.
When the camshaft lobe raises the outside of
the arm, the inside presses down on the valve
stem, opening the valve.
When the outside of the arm is permitted to
return due to the camshafts rotation, the inside
rises, allowing the valve spring to close the valve.

15. USE AND WORKING ROCKER ARM.

16. ROCKER ARM.
For Ic engines the rocker arms are
generally steel stampings, providing a reasonable
balance of strength, weight and economical cost.
Truck engines (mostly diesel) use stronger and
stiffer rocker arms made of cast iron (usually
ductile), or forged carbon steel.

17. ROCKER ARM.
Because the rocker arms are, in part, reciprocating
weight, excessive mass especially at the lever ends
limits the engine's ability to reach high operating
speeds.

18. ROCKER ARM.
Many lightweight and high strength alloys,
and bearing configurations for the fulcrum, have
been used in an effort to increase the RPM
limits for high performance applications,
eventually lending the benefits of these race
bred technologies to more high-end production
vehicles.

19. ASSEMBLY OF ROCKER ARM

20. Design of angular lever

21. 1. Determination of effort
• The bell crank lever is shown in Fig. First of
• all, let us find the centrifugal force (or the effort P)
required
• at the ball end to resist the load at A.
• We know that the maximum load on the roller arm
at A,
• Taking moment at fulcrum
. . p
p w
w
l
l p l w M A
l
 

22. 2. Determination of fulcrum reaction
• Fulcrum reaction is found by parallelogram.
• We know that reaction at the fulcrum F,
2 2
2 cosfR w p wp   

23. 3. Dimensions of fulcrum pin
• The fulcrum pin supports the lever and allows it to
oscillate hence there is a relative motion between
the pin and the lever.
• i) bearing pressure
d = Diameter of the fulcrum pin,
and
l = Length of the fulcrum pin = 1.25 d
f
b
R
P
A

f
b
b
R
P
l d


24. • ii) direct shear stress: the fulcrum pin is subjected
to double shear.
2
2
4
f
f
R
A
R
d






25. 4. Diameter of the boss of the lever
 3 3
0
12
b p w
b
i
M
M l p l w
Z
I
Z
Y
l
I d d
   

 
0 . .
.i
d out dia of theboss
d inner dia of theboss



26.  
0
3 3
0
0
0
3 3
0
2
6
6
b i
p
b
b i
d
Y
l d dI
Z
Y d
l p dM
Z l d d


 
   
 

 


27. • As there is a relative motion between the fulcrum
pin and boss some time a brass bush of about 3mm
thickness is pressed into the boss of the lever.
6id d 

28. 5. Dimension of lever c/s
• The lever arm is subjected to the bending moment.
0
2
0
2
2
6
2
6
b
p
p
b
M
Z
d
M p l
I bd
Z
Y
d
p l
M
bdZ



 
  
 
 
 
 
  

29. 0
2
6
2
p
d
p l
bd
 
 
 σb

30. EX: The rocker arm is used for operating the exhaust valve of
an I.C. Engine. The maximum force required for opening the
valve is 2000N. The rocker arm oscillates about a pin whose
axis is 200 mm away from the valve axis. The two arms of the
rocker are equal and make an included angle of 160º. The
permissible stresses for lever and pin materials in shear and
tension are 40 N/mm^2 and 70 N/mm^2 respectively. The
bearing pressure on the pin is to be limited to 13 N/mm^2.
Assuming the lever cross section as b*3b and fulcrum pin
length as 1.25 times diameter, design the rocker arm.
W = 2000N lp = lw =200mm
h =3b

31. lb= 1.5d
•Determination of fulcrum reaction (Rf):

32. •Dimension of fulcrum pin: Let, d= diameter of the fulcrum pin, mm Lb= length of the fulcrum pin,
mm = 1.25d
•Bearing Pressure:
Considering the bearing pressure at fulcrum pin,
•Direct shear stress:
so design is safe.

33. •Diameter of boss of the lever (db):
Let db = Outside diameter of the boss, mm
di = inside diameter of the boss, mm
•Consider a brass bush of 3mm thickness is pressed into the boss.
•The bending moment acting on the boss cross section is,

34. •The maximum bending stress included in the boss is,
•Solving above equation by trial and error,
•The minimum outside diameter of boss is,
•The outside diameter of boss is taken as the larger of two values given by Equation 1 And
2
•Dimensions of lever cross section:
•The maximum bending moment on the lever cross section is,

35. •The maximum bending moment stress induced in a lever cross section is,

36. DESGN OF BELL CRANK LEVER

37. 1. Determination of effort
• The bell crank lever is shown in Fig. First of
• all, let us find the centrifugal force (or the effort P)
required
• at the ball end to resist the load at A.
• We know that the maximum load on the roller arm
• at A,
• Taking moment at fulcrum
. . p
p w
w
l
l p l w M A
l
 

38. 2. Determination of fulcrum reaction
• Fulcrum reaction is found by parallelogram.
• We know that reaction at the fulcrum F,
2 2 0
90fR w p   

39. 3. Dimensions of fulcrum pin
• The fulcrum pin supports the lever and allows it to
oscillate hence there is a relative motion between
the pin and the lever.
• i) bearing pressure
d = Diameter of the fulcrum pin,
and
l = Length of the fulcrum pin = 1.25 d
f
b
R
P
A

f
b
b
R
P
l d


40. • ii) direct shear stress: the fulcrum pin is subjected
to double shear.
2
2
4
f
f
R
A
R
d






41. 4. Diameter of the boss of the lever
 3 3
0
12
b p w
b
i
M
M l p l w
Z
I
Z
Y
l
I d d
   

 
0 . .
.i
d out dia of theboss
d inner dia of theboss



42.  
0
3 3
0
0
0
3 3
0
2
6
6
b i
p
b
b i
d
Y
l d dI
Z
Y d
l p dM
Z l d d


 
   
 

 


43. • As there is a relative motion between the fulcrum
pin and boss some time a brass bush of about 3mm
thickness is pressed into the boss of the lever.
6id d 

44. 5. Dimension of lever c/s
• The lever arm is subjected to the bending moment.
0
2
0
2
2
6
2
6
b
p
p
b
M
Z
d
M p l
I bd
Z
Y
d
p l
M
bdZ



 
  
 
 
 
 
  

45. 0
2
6
2
p
d
p l
bd
 
 
 σb

46. • Ex: design a bell crank lever having one arm 500
mm and other 150 mm long. The load of 5KN is
raised acting on a pin at the end of 500 mm arm
and the effort is applied at the end of 150 mm arm.
The lever consist of steel for turning on a point at a
fulcrum.
90
60
10
t
b
Mpa
Mpa
P Mpa






48. Given data:
W=5KN
90
60
10
t
b
Mpa
Mpa
P Mpa





500
150
w
p
l mm
l mm



49. 1. determination of effort
150 500 5000
16666.67
p wl p l w
p
p N

  

2. fulcrum reaction
   
2 2 0
2 2
90
16666.67 5000
17400.51
fR w p
N
  
 


50. 3. dimension of fulcrum pin:
2
17400.51
10
1.25
38
f
b
b
R
P
l d
d
d mm


  
2
2
2
4
17400.51
2 38
4
7.6752
.
f
f
R
A
R
d
Mpa
so designis safe







 


51. 4. diameter of the boss of the lever.
 
 
 
0
3 3
0
0
3
0
3
0 0
0
0
6
6
5000 500 6
90 44
1.25 38 85184
3508.77 85184 0
70
2
76
p
b i
b i
i
l p dM
d d
Z l d d
d
d mm
d
d d
d mm
d d
mm consider


   

  
 
  
  




52. 5. Dimension of c/s of lever
 
 
0
3
3
3
2
4
8
5000 500 38 3
90
8
22
4 88
p
b
d
p l
M
b b
Z b
b
b mm
h b mm

 
  
   



 

53. Design of Lever loaded safety valve

54.  Procedure
• Weight due to steam pressure
Ps =
𝑊
𝐴
• Determination of effort
P × Lp = W × Lw
• Determination of pin at valve spindle
Pb=
𝑊
𝑙 𝑏
∗𝑑
Shear stress =
𝑊
2𝐴

55. 4. Determination of boss of lever
Bending stress =
6∗𝑝∗(𝑙𝑝−𝑙𝑤−
𝑑𝑜
2
)
3∗𝑏3
• Inner diameter
di = do + 6
• Outer diameter
do = 2d
• Take larger diameter
5. Dimensions of lever arm
- Rf + W - P=0
Rf = W-p
Bending stress =
6∗𝑝∗(𝑙𝑝−𝑙𝑤−
𝑑𝑜
2
)
3∗𝑏3

56. EX :- A lever loaded safety valve of 70 mm diameter
is required to flow off at a pressure of 1mpa. Design a
lever of rectangular cross section (b × 3b) and pin at
valve spindle distance between the weight and
fulcrum is equal to 880 mm , distance between the
fulcrum and valve spindle is 80 mm. Tensile stress is
70 Mpa.

57. Given data
Lw = 80 Mpa
Ps = 1 Mpa
Lp = 880 Mpa
Pb = 25 Mpa
Dv= 70 mm
Tensile stress = 70 Mpa
Shear stress = 50 Mpa
1. Weight due to steam pressure
Ps =
𝑊
𝐴
W = Ps × A
W = 3846.5 N

58. 2. Determination of effort
Lw× W = Lp× P
P = 349.68 N
3. Determination of pin at valve spindle
Lb = 1.25 d = 1.25 × 12 = 15 mm
• Bering pressure
Pb=
𝑊
𝑙 𝑏
∗𝑑
d = 11.0945 mm ~ 12 mm
• Double shear stress
Shear stress =
𝑊
2𝐴
= 17.01 Mpa< 50 Mpa

59. 4. Dimensions of the boss
di = d + 6 = 12+ 6 = 18 mm
M = p ( Lp - Lw)
• Shear stress =
6∗𝑝∗(𝑙𝑝−𝑙𝑤−
𝑑𝑜
2
)
3∗𝑏3
do ^3 - 1598.54 do - 5832 = 0
do = 42 mm
do = 2× d = 2×12 = 24 mm
• Dimensions of the lever arm
- Rf + W - P = 0
Rf = 3496.9 N

60. • The value of Rf is nearest about W so
dimensions is same
• tensile stress =
𝑝∗ 𝑙 𝑝
−𝑙𝑤−
𝑑0
2
∗18
27∗𝑏∗𝑏∗𝑏
b = 13.72 mm
or b = 14 mm
h = 3×b = 42 mm