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Section of curved beam

Location of neutral axis and cenrroidul axis

NO

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r,=

[& &I1 +

-

4 d

R= I

I

!

'3.+2

i

rt8= -

(bL+ h,, )7 J

9,b = R", =

C

(G +TI )c2 C

m e resultant reactions at bearings 2 and 3 are

R, =

,I~~~[RZZ

R, =

JkJTiQ

(a) Design of crank pin: Ii

FG=- Dip 4

Gas force,

Bending moment at the center of the crank pin, @ =

Ry,x B2

ad= -M C-& n 3 -xd, 32

We know that,

Check the crank pin for bearing pressure

where,

I, - Length of the crank pin d, - Diameter of the crank pin

6)~

& n ofcrank web:

The dimensions of the crank web can be found empirically as follows: Width,

b=(1.1 to 1.2)d,

Thickness,

1 = (0.6 to 0.75)dc

The crank web at TDC is subjected to bending moment and direct Maximum bending momrnf M= -

Bending stress, a,=

M 6M

-z7

Z

FG Direct dress, o = br

bt

compressive stress

Resultant a e s s = oa + o < o . # d i e

he resultant mess should be less than the permissible s k s in bending. 2.3.6.2. Crank at an angle of maximum torque

The centre crank at maximum torque has been shown in Figure 2.41

Centre crank at an angle o f m i m u m torque

Figure 2.41

The connecting md t h m t (F,), tangential and radial force can be found in the similar

I

way as that ofoverhung crank. (a) Design ojcrankpin:

Bending moment,

M, = ~ X B ,

Twisting moment,

4 T =-

Equ~valenttorque,

T,=

We know that,

T.= -d:r

where,

I

2

r

2

.,

...... (i) 77

16

......(ii)

r = 3 1 to 40~/mrn'

By equating the (i) and (ii), the diameter of crank pin can be determined. Compare the value of d, at the two crank positions, and take the maximum value of d,.

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(5) Derign ofshap at the junclion of right-hmd crank web:

Bendingmoment, where,

Twisting moment,

M=R,

c

j

-+'+-

: I:

-F

-+-

RI =Resultant reaction of bearing 1

T = F, x r

.,,.,. (iii)

Equivalent torque,

=

We know that,

T. = -d: 16

77

...... (iv)

r

By equating (iii) and (iv), the diameter (dj) of shaft at the junction of crank web can be determined.

: I

I 1

(c) Derign of crank web:

Bending moment due to radial force

Bending skess due to F,

Mar - Mbr - 6 M r

u*,

=-T(b,2,

Bending moment due to tangential force

F, Stress due to direct campression, o = 2bt

I

Maximum compressive stress,

o, = ob, + o* + o

br2

Twbting moment on the arm, T =

Shear stress,

'1

'('

t-

T

_

2 2 2

I=--

4.5T

br2

Z,

Maximum normal stress,

l&cG,.

--l+-

2

2

The value ofam, should be less than the allowable value. (@Design o/sha/r under thefly whee[:

The shaft under the flywheel is subjected IO nibnirnurn bending moment at the flbwheel location. Let d,= diameter of the shaft under the flywhecl Horimntal bending moment at the flywheel location due to belt tension

MH=RH,x CI =RH2x C 2 = ( 4 +T>)C, C We know that vertical bending moment due to the f l y h e e l weight,

M v = R ' v , X C , = R V , ~ C I =wc,c, C

Therrlore rrsultant bending momcnc

M= , J=

...... (i)

MR= ;(dry%

...... (ii)

We also know that

From equations (i) and (ii), we can calculate the diameter of the shaFt under the flywheel.

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