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For I-section, it is assumed that the bending moment is taken by flanges only. With this assumption,
the section modulus is given by
Z = Flange area × depth of section
The section of the arm is usually tapered from the fulcrum to the ends. The dimensions of the
arm at the ends depends upon the manner in which the load is applied. If the load at the end is applied
by forked connections, then the dimensions of the lever at the end can be proportioned as a knuckle
joint.
4. The dimensions of the fulcrum pin are obtained from bearing considerations and then checked
for shear. The allowable bearing pressure depends upon the amount of relative motion between the
pin and the lever. The length of pin is usually taken from 1 to 1.25 times the diameter of pin. If the
forces on the lever do not differ much, the diameter of the pins at load and effort point shall be taken
equal to the diameter of the fulcrum pin so that the spares are reduced. Instead of choosing a thick
lever, the pins are provided with a boss in order to provide sufficient bearing length.
5. The diameter of the boss is taken twice the diameter of pin and length of the boss equal to the
length of pin. The boss is usually provided with a 3 mm thick phosphor bronze bush with a dust proof
lubricating arrangement in order to reduce wear and to increase the life of lever.
Example 15.1. A handle for turning the spindle of a large valve is shown in Fig. 15.5. The
length of the handle from the centre of the spindle is 450 mm. The handle is attached to the spindle by
means of a round tapered pin.
Fig. 15.5
If an effort of 400 N is applied at the end of the handle, find: 1. mean diameter of the tapered
pin, and 2. diameter of the handle.
The allowable stresses for the handle and pin are 100 MPa in tension and 55 MPa in shear.
Solution. Given : L = 450 mm ; P = 400 N ; ∋
t
= 100 MPa = 100 N/mm
2
; ( = 55 MPa= 55 N/mm
2
1. Mean diameter of the tapered pin
Let d
1
= Mean diameter of the tapered pin, and
d = Diameter of the spindle = 50 mm (Given)
We know that the torque acting on the spindle,
T = P × 2L = 400 × 2 × 450 = 360 × 10
3
N-mm
(i)
Since the pin is in double shear and resists the same torque as that on the spindle, therefore
resisting torque,
T =
22
11
50
2() 2()55 N-mm
424 2
d
dd
&&
%(%!% %
= 2160 (d
1
)
2
N-mm
(ii)
From equations (i) and (ii), we get
(d
1
)
2
= 360 × 10
3
/ 2160 = 166.7 or d
1
= 12.9 say 13 mm
Ans.
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2. Diameter of the handle
Let D = Diameter of the handle.
Since the handle is subjected to both bending moment and twisting moment, therefore the design
will be based on either equivalent twisting moment or equivalent bending moment. We know that
bending moment,
M = P × L = 400 × 450 = 180 × 10
3
N-mm
The twisting moment depends upon the point of application of the effort. Assuming that the
effort acts at a distance 100 mm from the end of the handle, we have twisting moment,
T = 400 × 100 = 40 × 10
3
N-mm
We know that equivalent twisting moment,
T
e
=
2 2 32 32
(180 10 ) (40 10 )
MT
#! % #%
= 184.4 × 10
3
N-mm
We also know that equivalent twisting moment (T
e
),
184.4 × 10
3
=
33
55
16 16
DD
&&
%(% ! % %
= 10.8 D
3
) D
3
= 184.4 × 10
3
/ 10.8 = 17.1 × 10
3
or D = 25.7 mm
Again we know that equivalent bending moment,
M
e
=
22
11
()
22
e
MMT MT
∗+
##!#
,−
=
33
1
(180 10 184.4 10 )
2
%# %
= 182.2 × 10
3
N-mm
We also know that equivalent bending moment (M
e
),
182.2 × 10
3
=
33
100
32 32
b
DD
&&
%∋ % ! % %
= 9.82 D
3
(∵ ∋
b
= ∋
t
)
) D
3
= 182.2 × 10
3
/ 9.82 = 18.6 × 10
3
or D = 26.5 mm
Taking larger of the two values, we have
D = 26.5 mm
Ans.
Example 15.2.
A vertical lever PQR, 15 mm thick
is attached by a fulcrum pin at R and to a horizontal
rod at Q, as shown in Fig. 15.6.
An operating force of 900 N is applied horizontally
at P. Find :
1. Reactions at Q and R,
2. Tensile stress in 12 mm diameter tie rod at Q
3. Shear stress in 12 mm diameter pins at P, Q
and R, and
4. Bearing stress on the lever at Q.
Solution. Given : t = 15 mm ; F
P
= 900 N
1. Reactions at Q and R
Let R
Q
= Reaction at Q, and
R
R
= Reaction at R,
Taking moments about R, we have
R
Q
× 150 = 900 × 950 = 855 000
) R
Q
= 855 000 / 150 = 5700 N
Ans.
Fig. 15.6
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A Textbook of Machine Design
Fig. 15.7
These levers are used to change railway tracks.
Since the forces at P and Q are parallel and opposite as shown in Fig. 15.7, therefore reaction at R,
R
R
= R
Q
– 900 = 5700 – 900 = 4800 N
Ans.
2. Tensile stress in the tie rod at Q
Let d
t
= Diameter of tie rod = 12 mm (Given)
) Area, A
t
=
2
(12)
4
&
= 113 mm
2
We know that tensile stress in the tie rod,
∋
t
=
Q
Force at ( )
5700
Cross - sectional area ( ) 113
t
QR
A
!
= 50.4 N/mm
2
= 50.4 MPa
Ans.
3. Shear stress in pins at P, Q and R
Given : Diameter of pins at P, Q and R,
d
P
= d
Q
= d
R
= 12 mm
) Cross-sectional area of pins at P, Q and R,
A
P
= A
Q
= A
R
=
4
&
(12)
2
= 113 mm
2
Since the pin at P is in single shear and pins at Q and R are in
double shear, therefore shear stress in pin at P,
(
P
=
P
P
900
113
F
A
!
= 7.96 N/mm
2
= 7.96 MPa
Ans.
Shear stress in pin at Q,
(
Q
=
Q
Q
5700
2 2 113
R
A
!
%
= 25.2 N/mm
2
= 25.2 MPa Ans.
and shear stress in pin at R,
(
R
=
R
R
4800
2 2 113
R
A
!
%
= 21.2 N/mm
2
= 21.2 MPa
Ans.
4. Bearing stress on the lever at Q
Bearing area of the lever at the pin Q,
A
b
= Thickness of lever × Diameter of pin = 15 × 12 = 180 mm
2
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565
) Bearing stress on the lever at Q,
∋
b
=
Q
5700
180
b
R
A
!
= 31.7 N/mm
2
= 31.7 MPa
Ans.
15.4 Hand Levers
A hand lever with suitable dimensions and proportions is shown in Fig. 15.8.
Let P = Force applied at the handle,
L = Effective length of the lever,
∋
t
= Permissible tensile stress, and
( = Permissible shear stress.
For wrought iron, ∋
t
may be taken as 70 MPa and ( as 60 MPa.
In designing hand levers, the following procedure may be followed :
1. The diameter of the shaft ( d ) is obtained by considering the shaft under pure torsion. We
know that twisting moment on the shaft,
T = P × L
and resisting torque, T =
3
16
d
&
%(%
From this relation, the diameter of the shaft ( d ) may be obtained.
Fig. 15.8. Hand lever.
2. The diameter of the boss (d
2
) is taken as 1.6 d and thickness of the boss (t
2
) as 0.3 d.
3. The length of the boss (l
2
) may be taken from d to 1.25 d. It may be checked for a trial
thickness t
2
by taking moments about the axis. Equating the twisting moment (P × L) to the moment
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A Textbook of Machine Design
of resistance to tearing parallel to the axis, we get
P × L =
2
22
2
t
dt
lt
#
./
∋
01
23
or l
2
=
22
2
()
t
PL
tdt
%
∋#
4. The diameter of the shaft at the centre of the bearing (d
1
) is obtained by considering the shaft
in combined bending and twisting.
We know that bending moment on the shaft,
M = P × l
and twisting moment, T = P × L
) Equivalent twisting moment,
T
e
=
22 2 2 22
()( )
MT Pl PL PlL
#! %#% ! #
We also know that equivalent twisting moment,
T
e
=
3
1
()
16
d
&
%(
or
22 3
1
()
16
Pl L d
&
#!%(
The length l may be taken as 2 l
2
.
From the above expression, the value of d
1
may be determined.
5. The key for the shaft is designed as usual for transmitting a torque of P × L.
6. The cross-section of the lever near the boss may be determined by considering the lever in
bending. It is assumed that the lever extends to the centre of the shaft which results in a stronger
section of the lever.
Let t = Thickness of lever near the boss, and
B = Width or height of lever near the boss.
We know that the bending moment on the lever,
M = P × L
Section modulus, Z =
2
1
6
tB
%%
We know that the bending stress,
∋
b
=
2
2
6
1
6
MPL PL
Z
tB
tB
%%
!!
%
%%
The width of the lever near the boss may be taken from 4 to 5 times the thickness of lever, i.e.
B = 4 t to 5 t. The width of the lever is tapered but the thickness (t) is kept constant. The width of the
lever near the handle is B/2.
Note: For hand levers, about 400 N is considered as full force which a man is capable of exerting. About 100 N
is the mean force which a man can exert on the working handle of a machine, off and on for a full working day.
15.5 Foot Lever
A foot lever, as shown in Fig. 15.9, is similar to hand lever but in this case a foot plate is
provided instead of handle. The foot lever may be designed in a similar way as discussed for hand
lever. For foot levers, about 800 N is considered as full force which a man can exert in pushing a foot
lever. The proportions of the foot plate are shown in Fig. 15.9.
Example 15.3. A foot lever is 1 m from the centre of shaft to the point of application of 800 N
load. Find :
1. Diameter of the shaft, 2. Dimensions of the key, and 3. Dimensions of rectangular arm of the
foot lever at 60 mm from the centre of shaft assuming width of the arm as 3 times thickness.
The allowable tensile stress may be taken as 73 MPa and allowable shear stress as 70 MPa.
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Solution. Given : L = 1 m = 1000 mm ; P = 800 N ; ∋
t
= 73 MPa = 73 N/mm
2
;
t = 70 MPa = 70 N/mm
2
1. Diameter of the shaft
Let d = Diameter of the shaft.
We know that the twisting moment on the shaft,
T = P × L = 800 × 1000 = 800 × 10
3
N-mm
We also know that the twisting moment on the shaft (T),
800 × 10
3
=
33
70
16 16
dd
&&
%(% ! % %
= 13.75 d
3
) d
3
= 800 × 10
3
/ 13.75 = 58.2 × 10
3
or d = 38.8 say 40 mm
Ans.
We know that diameter of the boss,
d
2
= 1.6 d = 1.6 × 40 = 64 mm
Thickness of the boss,
t
2
= 0.3 d = 0.3 × 40 = 12 mm
and length of the boss, l
2
= 1.25 d = 1.25 × 40 = 50 mm
Now considering the shaft under combined bending and twisting, the diameter of the shaft at the
centre of the bearing (d
1
) is given by the relation
3
1
()
16
d
&
%( =
22
Pl L
#
3
1
70 ( )
16
d
&
%%
=
22
800 (100) (1000)
#
(Taking l = 2 l
2
)
or 13.75 (d
1
)
3
= 804 × 10
3
) (d
1
)
3
= 804 × 10
3
/ 13.75 = 58.5 × 10
3
or d
1
= 38.8 say 40 mm
Ans.
Fig. 15.9. Foot lever.
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A Textbook of Machine Design
2. Dimensions of the key
The standard dimensions of the key for a 40 mm diameter
shaft are :
Width of key, w = 12 mm
Ans.
and thickness of key = 8 mm Ans.
The length of the key (l
1
) is obtained by considering the
shearing of the key.
We know that twisting moment (T),
800 × 10
3
=
1
2
%%(%
d
lw
1
40
12 70
2
!% % %
l
= 16 800 l
1
) l
1
= 800 × 10
3
/ 16 800 = 47.6 mm
It may be taken as equal to the length of boss (l
2
).
) l
1
= l
2
= 50 mm
Ans.
3. Dimensions of the rectangular arm at 60 mm from the
centre of shaft
Let t = Thickness of arm in mm, and
B = Width of arm in mm = 3 t (Given)
) Bending moment at 60 mm from the centre of shaft,
M = 800 (1000 – 60) = 752 × 10
3
N-mm
and section modulus, Z =
22
11
(3 )
66
tB tt
%% ! %
= 1.5 t
3
mm
3
We know that the tensile bending stress (∋
t
),
73=
33
33
752 10 501.3 10
1.5
M
Z
tt
%%
!!
) t
3
= 501.3 × 10
3
/73 = 6.87 × 10
3
or t = 19 say 20 mm
Ans.
and B = 3 t = 3 × 20 = 60 mm Ans.
The width of the arm is tapered while the thickness is kept constant throughout. The width of the
arm on the foot plate side,
B
1
= B / 2 = 30 mm
Ans.
15.6 Cranked Lever
A cranked lever, as shown in Fig. 15.10, is a hand lever commonly used for operating hoisting
winches.
The lever can be operated either by a single person or by two persons. The maximum force in
order to operate the lever may be taken as 400 N and the length of handle as 300 mm. In case the lever
is operated by two persons, the maximum force of operation will be doubled and length of handle
may be taken as 500 mm. The handle is covered in a pipe to prevent hand scoring. The end of the shaft
is usually squared so that the lever may be easily fixed and removed. The length (L) is usually from
400 to 450 mm and the height of the shaft centre line from the ground is usually one metre. In order
to design such levers, the following procedure may be adopted :
Accelerator and brake levers inside
an automobile.
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1. The diameter of the handle ( d ) is obtained from bending considerations. It is assumed that
the effort (P) applied on the handle acts at
2
3
rd of its length (l).
Fig. 15.10. Cranked lever.
) Maximum bending moment,
M =
22
33
l
PPl
% !%%
and section modulus, Z =
3
32
d
&
%
) Resisting moment = ∋
b
× Z =
3
32
b
d
&
∋% %
where ∋
b
= Permissible bending stress for the material of the handle.
Equating resisting moment to the maximum bending moment, we have
3
32
b
d
&
∋% %
=
2
3
Pl
%%
From this expression, the diameter of the handle ( d ) may be evaluated. The diameter of the
handle is usually proportioned as 25 mm for single person and 40 mm for two persons.
2. The cross-section of the lever arm is usually rectangular having uniform thickness through-
out. The width of the lever arm is tapered from the boss to the handle. The arm is subjected to
constant twisting moment, T =
2
3
Pl
%%
and a varying bending moment which is maximum near the
boss. It is assumed that the arm of the lever extends upto the centre of shaft, which results in a slightly
stronger lever.
) Maximum bending moment = P × L
Since, at present time, there is insufficient information on the subject of combined bending and
twisting of rectangular sections to enable us to find equivalent bending or twisting, with sufficient
accuracy, therefore the indirect procedure is adopted.
We shall design the lever arm for 25% more bending moment.
) Maximum bending moment
M = 1.25 P × L
Let t = Thickness of the lever arm, and
B = Width of the lever arm near the boss.
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) Section modulus for the lever arm,
Z =
2
1
6
tB
%%
Now by using the relation, ∋
b
= M / Z, we can find t and B. The width of the lever arm near the
boss is taken as twice the thickness i.e. B = 2 t.
After finding the value of t and B, the induced bending stress may be checked which should not
exceed the permissible value.
3. The induced shear stress in the section of the lever arm near the boss, caused by the twisting
moment, T =
2
3
Pl
%%
may be checked by using the following relations :
T =
2
2
9
Bt
%%%(
(For rectangular section)
=
3
2
9
t
%%(
(For square section of side t)
=
2
16
Bt
&
%%%(
(For elliptical section having major axis B
and minor axis t)
4. Knowing the values of ∋
b
and (, the maximum principal or shear stress induced may be
checked by using the following relations :
Maximum principal stress,
∋
b(max)
=
22
1
() 4
2
bb
∗+
∋# ∋ # (
,−
Maximum shear stress,
(
max
=
22
1
() 4
2
b
∋#(
5. Since the journal of the shaft is subjected to twisting moment and bending moment, there-
fore its diameter is obtained from equivalent twisting moment.
We know that twisting moment on the journal of the shaft,
T = P × L
and bending moment on the journal of the shaft,
M =
2
3
l
Px
./
#
01
23
where x = Distance from the end of boss to the centre of journal.
) Equivalent twisting moment,
T
e
=
2
22 2
2
3
l
MTP x L
./
#! # #
01
23
We know that equivalent twisting moment,
T
e
=
3
16
D
&
%(%
From this expression, we can find the diameter (D) of the journal.
The diameter of the journal is usually taken as
D = 30 to 40 mm, for single person
= 40 to 45 mm, for two persons.
Note: The above procedure may be used in the design of overhung cranks of engines.
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Example 15.4. A cranked lever, as shown in 15.10, has the following dimensions :
Length of the handle = 300 mm
Length of the lever arm = 400 mm
Overhang of the journal = 100 mm
If the lever is operated by a single person exerting a maximum force of 400 N at a distance of
1
3
rd length of the handle from its free end, find : 1. Diameter of the handle, 2. Cross-section of the
lever arm, and 3. Diameter of the journal.
The permissible bending stress for the lever material may be taken as 50 MPa and shear stress
for shaft material as 40 MPa.
Solution. Given : l = 300 mm ; L = 400 mm ; x = 100 mm ; P = 400 N ; ∋
b
= 50 MPa
= 50 N/mm
2
; ( = 40 MPa = 40 N/mm
2
1. Diameter of the handle
Let d = Diameter of the handle in mm.
Since the force applied acts at a distance of 1/3 rd length of the handle from its free end,therefore
maximum bending moment,
M =
12 2
1 400 300 N-mm
33 3
Pl Pl
./
∃ %!%%!% %
01
23
= 80 × 10
3
N-mm (i)
Section modulus, Z =
3
32
d
&
%
= 0.0982 d
3
) Resisting bending moment,
M = ∋
b
× Z = 50 × 0.0982 d
3
= 4.91 d
3
N-mm
(ii)
From equations (i) and (ii), we get
d
3
= 80 × 10
3
/ 4.91 = 16.3 × 10
3
or d = 25.4 mm
Ans.
2. Cross-section of the lever arm
Let t = Thickness of the lever arm in mm, and
B = Width of the lever arm near the boss, in mm.
Since the lever arm is designed for 25% more bending moment, therefore maximum bending
moment,
M = 1.25 P × L = 1.25 × 400 × 400 = 200 × 10
3
N-mm
Section modulus, Z =
22
11
(2 )
66
tB tt
%% ! %
= 0.667 t
3
(Assuming B = 2t)
We know that bending stress (∋
b
),
50 =
33
33
200 10 300 10
0.667
M
Z
tt
%%
!!
) t
3
= 300 × 10
3
/50 = 6 × 10
3
or t = 18.2 say 20 mm
Ans.
and B =2 t = 2 × 20 = 40 mm Ans.
Let us now check the lever arm for induced bending and shear stresses.
Bending moment on the lever arm near the boss (assuming that the length of the arm extends
upto the centre of shaft) is given by
M = P × L = 400 × 400 = 160 × 10
3
N-mm
and section modulus, Z =
22
11
20 (40)
66
tB
%% ! %
= 5333 mm
3
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