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Chapter 17 Flexible Mechanical Elements Shigley’s MED Solution Manual, Lecture notes of Mechanical Systems Design

Chapter 17 Exercise problems solved for Shigley's Mechanical Engineering Design 10th Edition, by Michael A. Latcha, PhD. ME 486 Oakland University (OU)

Typology: Lecture notes

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bg1
Shigley’s MED, 10
th
edition Chapter 17 Solutions, Page 1/39
Chapter 17
17-1 Begin with Eq. (17-10),
1
2exp( )
exp( ) 1
c i
f
F F F f
θ
θ
= +
Introduce Eq. (17-9):
1
1
exp( ) 1 2exp( ) 2 exp( )
exp( ) 1 exp( ) 1 exp( ) 1
exp( )
exp( ) 1
c c
c
f f T f
F F d F
f f d f
f
F F F f
θ θ θ
θ θ θ
θ
θ
+
= + = +
+
= +
Now add and subtract
exp( ) 1
c
f
Ff
θ
θ
1
exp( ) exp( ) exp( )
exp( ) 1 exp( ) 1 exp( ) 1
exp( ) exp( )
( ) exp( ) 1 exp( ) 1
exp( )
( ) exp( ) 1 exp( ) 1
( )exp( )
ex
c c c
c c c
c
c
c c
f f f
F F F F F
f f f
f f
F F F F
f f
f F
F F f f
F F f F
θ θ θ
θ θ θ
θ θ
θ θ
θ
θ θ
θ
= + +
= + +
= +
+
= . . .
p( ) 1 Q E D
f
θ
From Ex. 17-2:
θ
d
= 3.037 rad,
F = 664 lbf, exp( f
θ
) = exp[0.80(3.037)] = 11.35, and
F
c
= 73.4 lbf.
1
2 1
1
2
(73.4 664)11.35 73.4 802 lbf
(11.35 1)
802 664 138 lbf
802 138 73.4 396.6 lbf
2
1 1 802 73.4
ln ln 0.80 .
3.037 138 73.4
i
c
d c
F
F F F
F
F F
f Ans
F F
θ
+
= =
= = =
+
= =
= = =
______________________________________________________________________________
17-2 Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H
nom
= 2 hp, C = 9(12) =
108 in, velocity ratio = 0.5, K
s
= 1.25, n
d
= 1
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a
pf1b
pf1c
pf1d
pf1e
pf1f
pf20
pf21
pf22
pf23
pf24
pf25
pf26
pf27

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th

Chapter 17

17-1 Begin with Eq. (17-10),

1

2 exp( )

exp( ) 1

c i

f F F F f

θ

θ

Introduce Eq. (17-9):

1

1

exp( ) 1 2 exp( ) 2 exp( )

exp( ) 1 exp( ) 1 exp( ) 1

exp( )

exp( ) 1

c c

c

f f T f F F d F f f d f

f F F F f

θ θ θ

θ θ θ

θ

θ

 −^   +^   − 

Now add and subtract

exp( )

exp( ) 1

c

f F f

θ

θ

1

exp( ) exp( ) exp( )

exp( ) 1 exp( ) 1 exp( ) 1

exp( ) exp( ) ( ) exp( ) 1 exp( ) 1

exp( ) ( ) exp( ) 1 exp( ) 1

( ) exp( )

ex

c c c

c c c

c c

c c

f f f F F F F F f f f

f f F F F F f f

f F F F f f

F F f F

θ θ θ

θ θ θ

θ θ

θ θ

θ

θ θ

θ

 −^   −^   − 

 −^   − 

 −^  −

p( ) 1

Q E D

f θ −

From Ex. 17-2: θ d = 3_._ 037 rad, ∆ F = 664 lbf, exp( f θ ) = exp[0_._ 80(3_._ 037)] = 11_._ 35, and

Fc = 73_._ 4 lbf.

1

2 1

1

2

802 lbf (11.35 1)

802 664 138 lbf

73.4 396.6 lbf 2

ln ln 0.. 3.037 138 73.

i

c

d c

F

F F F

F

F F

f Ans θ F F

 −^  ^ − 

______________________________________________________________________________

17-2 Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, H nom = 2 hp, C = 9(12) =

108 in, velocity ratio = 0.5, Ks = 1.25, nd = 1

th

V = π d n / 12 = π (2)(1750) / 12 = 916.3 ft/min

D = d / vel ratio = 2 / 0.5 = 4 in

Eq. (17-1):

2sin 2sin 3.123 rad 2 2(108)

d

D d

C

θ π π

− −^ − ^ − 

Table 17-2: t = 0.05 in, d min = 1.0 in, Fa = 35 lbf/in, γ = 0.035 lbf/in

3 , f = 0.

w = 12 γ bt = 12(0.035)6(0.05) = 0.126 lbf/ft

(a) Eq. ( e ), p. 877:

2 2 0.126 916. 0.913 lbf. 60 32.17 60

c

V

F Ans g

w

90.0 lbf · in 1750

2 2(90.0) 90.0 lbf 2

nom s d

a

H K n T n

T F F F d

Table 17-4: Cp = 0.

Eq. (17-12): ( F 1 ) a = bFaCpCv = 6(35)(0_._ 70)(1) = 147 lbf Ans.

F 2 = ( F 1 ) a − [( F 1 ) aF 2 ] = 147 − 90 = 57 lbf Ans.

Do not use Eq. (17-9) because we do not yet know f

Eq. ( i ), p. 878:

0.913 101.1 lbf. 2 2

a i c

F F

F F Ans

Using Eq. (17-7) solved for f ′ (see step 8, p.888),

1

2

ln ln 0. 3.123 57 0.

a c

d c

F F

f θ F F

 −^  ^ − 

The friction is thus underdeveloped.

(b) The transmitted horsepower is, with ∆ F = ( F 1 ) aF 2 = 90 lbf,

Eq. (j), p. 879:

2.5 hp. 33 000 33 000

F V

H Ans

nom

f s s

H

n H K

Eq. (17-1):

2sin 2sin 3.160 rad 2 2(108)

D

D d

C

θ π π

− −^ − ^ − 

Eq. (17-2): L = [4 C

2 − ( Dd )

2 ]

1/

  • ( D θ D + d θ d )/

th

2 3 108 / 12 0. dip 0.222 in 2(68.9)

So, reducing Fi from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50,

with a corresponding dip of 0.222 in. Having reduced F 1 and F 2 , the endurance of the belt

is improved. Power, service factor and design factor have remained intact.


17-3 Double the dimensions of Prob. 17-2.

In Prob. 17-2, F-1 Polyamide was used with a thickness of 0.05 in. With what is available

in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let

b = 12 in, d = 4 in with n = 1750 rev/min, H nom = 2 hp, C = 18(12) = 216 in, velocity

ratio = 0.5, Ks = 1.25, nd = 1.

V = π d n / 12 = π (4)(1750) / 12 = 1833 ft/min

D = d / vel ratio = 4 / 0.5 = 8 in

Eq. (17-1):

2sin 2sin 3.123 rad 2 2(216)

d

D d

C

θ π π

− −^ − ^ − 

Table 17-2: t = 0.11 in, d min = 2.4 in, Fa = 60 lbf/in, γ = 0.037 lbf/in

3 , f = 0.

w = 12 γ bt = 12(0.037)12(0.11) = 0.586 lbf/ft

(a) Eq. ( e ), p. 877:

2 2 0.586 1833 17.0 lbf. 60 32.17 60

c

V

F Ans g

w

90.0 lbf · in 1750

2 2(90.0) 45.0 lbf 4

nom s d

a

H K n T n

T F F F d

Table 17-4: Cp = 0.

Eq. (17-12): ( F 1 ) a = bFaCpCv = 12(60)(0_._ 73)(1) = 525.6 lbf Ans.

F 2 = ( F 1 ) a − [( F 1 ) aF 2 ] = 525.6 − 45 = 480.6 lbf Ans.

Eq. ( i ), p. 878:

17.0 486.1 lbf. 2 2

a i c

F F

F F Ans

Eq. (17-9):

1

2

ln ln 0. 3.123 480.6 17.

a c

d c

F F

f θ F F

 −^  ^ − 

Shigley’s MED, 10

th edition

The friction is thus un

(b) The transmitted horsepower is,

f s

H Ans

n H K

Eq. (17-1):

Eq. (17-2): L =

= [4(

(c) Eq. (17-13):

______________________________________________________________________________

As a design task, the decision set on p. 8

A priori decisions:

  • Function: H nom = 60 hp,
  • Design factor: nd = 1
  • Initial tension: Catenary
  • Belt material. Table 17
  • Drive geometry: d =
  • Belt thickness: t = 0

Design variable: Belt width.

Use a method of trials. Initially

2 2

nom

12 12(0.042)(6)(0.13) 0.393 lbf/ft

c

dn V

bt

V

F

g

H K n T n

π π

γ

w

w

Chapter 17

The friction is thus underdeveloped.

The transmitted horsepower is, with ∆ F = ( F 1 ) aF 2 = 45 lbf,

nom

2.5 hp. 33 000 33 000

1

s 2(1.25)

F V

H Ans

H

H K

2sin 2sin 3.160 rad 2 2(216)

D

D d

C

θ π π

− −^ − ^ − 

= [4 C

2 − ( Dd )

2 ]

1/

  • ( D θ D + d θ d )/

= [4(216)

2 − (8 − 4)

2 ]

1/

  • [8(3.160) + 4(3.123)]/2 =

2 2 3 3(216 / 12) (0.586) dip 0.586 in. (^2) i 2(486.1)

C

F

w

______________________________________________________________________________

the decision set on p. 885 is useful.

= 60 hp, n = 380 rev/min, C = 192 in, Ks = 1_._ 1

= 1

Catenary

. Table 17-2: Polyamide A-3, Fa = 100 lbf/in, γ = 0.042 lbf/in

= D = 48 in

= 0_._ 13 in

ign variable: Belt width.

Use a method of trials. Initially, choose b = 6 in

2 2

nom

4775 ft/min 12 12

12 12(0.042)(6)(0.13) 0.393 lbf/ft

77.4 lbf

10 946 lbf · in 380

H K n s d

n

π π = = =

Chapter 17 Solutions, Page 5/

2sin 2sin 3.160 rad 2 2(216)

− ^ − 

(3.123)]/2 = 450_._ 9 in Ans.

dip 0.586 in Ans.

______________________________________________________________________________

= 0.042 lbf/in

3 , f = 0.

10 946 lbf · in

th

17-5 From the last equation given in the problem statement,

0 2

0 2

0 2

0 2

exp 1 2 / [ ( ) ]

1 exp 1 ( )

exp exp 1 ( )

1 2 exp

exp 1

f T d a a b

T

f d a a b

T

f f d a a b

T f b a a d f

φ

φ

φ φ

φ

φ

 −^  =

  =^ −

 ^ 

But 2 T/d = 33 000 Hd/V. Thus,

0 2 (^ )

1 33 000 exp

... exp 1

H d f b Q E D a a V f

φ

φ

 ^ ^ 

______________________________________________________________________________

17-6 Refer to Ex. 17-1 on p. 882 for the values used below.

(a) The maximum torque prior to slip is,

63 025 (^) nom 63 025(15)(1.25)(1.1) 742.8 lbf · in. 1750

H K n s d T Ans n

The corresponding initial tension, from Eq. (17-9), is,

exp( ) 1 742.8 11.17 1 148.1 lbf. exp( ) 1 6 11.17 1

i

T f F Ans d f

θ

θ

 −^  ^ − 

(b) See Prob. 17-4 statement. The final relation can be written

min (^2)

2

1 33 000^ exp

(12 / 32.174)( / 60) [exp 1]

100(0.7)(1) [12(0.042)(0.13)] / 32.174 (2749 / 60) 2749(11.17 1)

4.13 in.

a

a p

H f b F C C t V V f

Ans

θ

γ θ

v

This is the minimum belt width since the belt is at the point of slip. The design must

th

round up to an available width.

Eq. (17-1):

1 1

1 1

2sin 2sin 2 2(96)

3.016 511 rad

2sin 2sin 2 2(96)

3.266 674 rad

d

D

D d

C

D d

C

θ π π

θ π π

− −

− −

 −^ ^ ^ − 

 −^ ^ ^ − 

Eq. (17-2):

[4(96) (18 6) ] [18(3.266 674) 6(3.016 511)]

230.074 in.

L

Ans

(c)

247.6 lbf 6

T

F

d

1 1

2 1

2 2

1 2

( ) 4.13(100)(0.70)(1) 289.1 lbf

289.1 247.6 41.5 lbf

12 12(0.042)4.13(0.130) 0.271 lbf/ft

17.7 lbf 60 32.17 60

17.7 147.6 lb 2 2

a a p

c

i c

F bF C C F

F F F

bt

V

F

g

F F

F F

γ

v

w

w

f

Transmitted belt power H

nom

20.6 hp 33 000 33 000

15(1.25)

fs s

F V

H

H

n H K

Dip:

2 2 3 3(96 / 12)^ 0. 0.176 in (^2) i 2(147.6)

C

dip F

w

( d ) If you only change the belt width, the parameters in the following table change as

shown.

Shigley’s MED, 10

th edition

Find the resultant of F

1

1 2 1 2

1 2 1 2

sin 2

sin 2

cos 1 2 2

cos cos ( ) 1.

sin sin ( ).

x

y

D d

D d

C

R F F F F Ans

R F F F F Ans

α

α

α

α α

α α

From Ex. 17-2, d = 16 in,

1 2

sin 2.

(940 276) 1 1214.4 lbf

(940 276) 34.6 lbf

( ) (940 276) 5312 lbf · in

x

y

R

R

T F F

α

− = =

______________________________________________________________________________

17-9 This is a good class project. Form four groups, each with a belt to design. Once each

group agrees internally, all four should report their designs including the forces and

torques on the line shaft. If you give them the pulley locations, they could

shaft.


Chapter 17

F 1 and F 2 :

2

2

1 2 1 2

1 2 1 2

cos cos ( ) 1. 2 2

sin sin ( ). 2

D d

C

D d

C

D d

C

D d R F F F F Ans C

D d R F F F F Ans C

α α

α α

 ^  

= 16 in, D = 36 in, C = 16(12) = 192 in, F 1 = 940 lbf,

1 o

2

1 2

sin 2. 2(192)

(940 276) 1 1214.4 lbf 2 2(192)

(940 276) 34.6 lbf 2(192)

( ) (940 276) 5312 lbf · in 2 2

d T F F

− ^ − 

______________________________________________________________________________

This is a good class project. Form four groups, each with a belt to design. Once each

group agrees internally, all four should report their designs including the forces and

torques on the line shaft. If you give them the pulley locations, they could

______________________________________________________________________________

Chapter 17 Solutions, Page 10/

R F cos F cos ( F F ) 1 Ans.

= 940 lbf, F 2 = 276 lbf

( ) (940 276) 5312 lbf · in

______________________________________________________________________________

This is a good class project. Form four groups, each with a belt to design. Once each

group agrees internally, all four should report their designs including the forces and

torques on the line shaft. If you give them the pulley locations, they could design the line

______________________________________________________________________________

th

17-10 If you have the students implement a computer program, the design problem selections

may differ, and the students will be able to explore them. For Ks = 1_._ 25, nd = 1_._ 1, d = 14

in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do ( b min = 6_._ 58 in)


17-11 An efficiency of less than unity lowers the output for a given input. Since the object of

the drive is the output, the efficiency must be incorporated such that the belt’s capacity is

increased. The design power would thus be expressed as

nom . eff

s d d

H K n H = Ans

______________________________________________________________________________

17-12 Some perspective on the size of Fc can be obtained from

2 2 12

c

V bt V F g g

  γ   = (^)   =      

w

An approximate comparison of non-metal and metal belts is presented in the table below.

Non-metal Metal

γ, lbf/in

3 0.04 0.

b , in 5.00 1.

t , in 0.20 0.

The ratio w / wm is

m^ 12(0.28)(1)(0.005)

w

w

The second contribution to Fc is the belt peripheral velocity which tends to be low in

metal belts used in instrument, printer, plotter and similar drives. The velocity ratio

squared influences any Fc / ( Fc ) m ratio.

It is common for engineers to treat Fc as negligible compared to other tensions in the

belting problem. However, when developing a computer code, one should include Fc.


17-13 Eq. (17-8):

1 2 1 1

exp( ) 1 exp( ) 1 ( ) exp( ) exp( )

c

f f F F F F F F f f

θ θ

θ θ

Assuming negligible centrifugal force and setting F 1 = ab from step 3, p. 889,

min

exp( ) (1) exp( ) 1

F f b a f

θ

θ

th

exp( ) exp[0.35(3.010)] 2.

916.3 ft/s 12 12

175 kpsi

d

y

f

dn V

S

θ

π π

Eq. (17-15):

6 0.407 6 6 0.407 3 S (^) y 14.17 10 N (^) p 14.17 10 10 51.212 10 psi

− − = = =

From selection step 3, p. 889,

6 3 2 2

1

16.50 lbf/in of belt width

f

a

Et a S t d

F ab b

ν

 −^   − 

For full friction development, from Prob. 17-13,

min

exp( )

exp( ) 1

45.38 lbf 2

d

d

F f b a f

T

F

d

θ

θ

So

min

4.23 in 16.50 2.868 1

b

Decision #1 : b = 4_._ 5 in

1 1

2 1

1 2

( ) 16.5(4.5) 74.25 lbf

74.25 45.38 28.87 lbf

51.56 lbf 2 2

a

i

F F ab

F F F

F F

F

Existing friction

1

2

nom

ln ln 0. 3.010 28.

1.26 hp 33 000 33 000

1(1.2)

d

t

t fs s

F

f F

F V

H

H

n H K

θ

′ =^ =^   =

This is a non-trivial point. The methodology preserved the factor of safety corresponding

to nd = 1_._ 1 even as we rounded b min up to b.

th

Decision #2 was taken care of with the adjustment of the center-to-center distance to

accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this.

Remember to subsequently recalculate θ d and θ D.


17-15 Decision set:

A priori decisions

  • Function: H nom = 5 hp, N = 1125 rev/min, VR = 3, C  20 in, Ks = 1_._ 25,

Np = 10

6 belt passes

  • Design factor: nd = 1_._ 1
  • Belt material: BeCu, Sy = 170 kpsi, E = 17 Mpsi, ν = 0_._ 220
  • Belt geometry: d = 3 in, D = 9 in
  • Belt thickness: t = 0_._ 003 in

Design decisions

  • Belt loop periphery
  • Belt width b

Preliminaries :

nom 5(1.25)(1.1)^ 6.875 hp

63 025(6.875) 385.2 lbf · in 1125

H (^) d H K ns d

T

Decision #1 : Choose a 60-in belt loop with a center-to-center distance of 20.3 in.

1

1

2sin 2.845 rad 2(20.3)

2sin 3.438 rad 2(20.3)

d

D

θ π

θ π

For full friction development:

exp( ) exp[0.32(2.845)] 2.

883.6 ft/min 12 12

56.67 kpsi

d

f

f

dn V

S

θ

π π

From selection step 3, p. 889,

th

Five groups of students could each be assigned a belt thickness. You can form a table

from their results or use the table given here.

t , in

0.002 0.003 0.005 0.008 0.

b 4.000 3.500 4.000 1.500 1.

CD 20.300 20.300 20.300 18.700 20.

a 109.700 131.900 110.900 194.900 221.

d 3.000 3.000 3.000 5.000 6.

D 9.000 9.000 9.000 15.000 18.

Fi 310.600 333.300 315.200 215.300 268.

F 1 439.000 461.700 443.600 292.300 332.

F 2 182.200 209.000 186.800 138.200 204.

nf s 1.100 1.100 1.100 1.100 1.

L 60.000 60.000 60.000 70.000 80.

f ′ 0.309 0.285 0.304 0.288 0.

Fi 301.200 301.200 301.200 195.700 166.

F 1 429.600 429.600 429.600 272.700 230.

F 2 172.800 172.800 172.800 118.700 102.

f 0.320 0.320 0.320 0.320 0.

The first three thicknesses result in the same adjusted Fi , F 1 and F 2 (why?). We have no

figure of merit, but the costs of the belt and pulleys are about the same for these three

thicknesses. Since the same power is transmitted and the belts are widening, belt forces

are lessening.


17-17 This is a design task. The decision variables would be belt length and belt section, which

could be combined into one, such as B90. The number of belts is not an issue.

We have no figure of merit, which is not practical in a text for this application. It is

suggested that you gather sheave dimensions and costs and V-belt costs from a principal

vendor and construct a figure of merit based on the costs. Here is one trial.

Preliminaries : For a single V-belt drive with H nom = 3 hp, n = 3100 rev/min, D = 12 in,

and d = 6_._ 2 in, choose a B90 belt, Ks = 1_._ 3 and nd = 1. From Table 17-10, select a

circumference of 90 in. From Table 17-11, add 1.8 in giving

Lp = 90 + 1_._ 8 = 91_._ 8 in

Eq. (17-16 b ):

2 2 0.25 91.8 (12 6.2) 91.8 (12 6.2) 2(12 6.2) 2 2

31.47 in

C

π π

 ^ ^  

th

2sin 2.9570 rad 2(31.47)

θ (^) d π

exp( ) exp[0.5123(2.9570)] 4.

5031.8 ft/min 12 12

f d

dn V

θ

π π

Table 17-13:

Angle θ θ (^) d (2.957 rad) 169. π π

The footnote regression equation of Table 17-13 gives K 1 without interpolation:

K 1 = 0. 143 543 + 0. 007 468(169. 42°) − 0. 000 015 052(169. 42°)

2 = 0_._ 9767

The design power is

Hd = H nom Ksnd = 3(1_._ 3)(1) = 3_._ 9 hp

From Table 17-14 for B90, K 2 = 1. From Table 17-12 take a marginal entry of H tab = 4,

although extrapolation would give a slightly lower H tab.

Eq. (17-17): Ha = K 1 K 2 H tab = 0_._ 9767(1)(4) = 3_._ 91 hp

The allowable ∆ Fa is given by

25.6 lbf ( / 2) 3100(6.2 / 2)

a a

H

F

n d

The allowable torque Ta is

79.4 lbf · in 2 2

a a

F d T

From Table 17-16, Kc = 0_._ 965. Thus, Eq. (17-21) gives,

2 2

0.965 24.4 lbf 1000 1000

c c

V

F K

At incipient slip, Eq. (17-9) provides:

exp( ) 1 79.4 4.5489 1 20.0 lbf exp( ) 1 6.2 4.5489 1

i

T f F d f

θ

θ

  ^ +^    + 

   −^    − 

Eq. (17-10):

th

1 2

1 2 10

9.99 lbf/belt, 24.4 lbf/belt

( ) 92.9 lbf/belt, ( ) 48 lbf/belt

133.7 lbf/belt, 88.8 lbf/belt

2.39(10 ) passes, 605 600 h

i c

b b

P

F F

F F

T T

N t

Initial tension of the drive:

( Fi )drive = NbFi = 2(9_._ 99) = 20 lbf


17-18 Given: two B85 V-belts with d = 5_._ 4 in, D = 16 in, n = 1200 rev/min, and Ks = 1_._ 25

Table 17-11: Lp = 85 + 1_._ 8 = 86_._ 8 in

Eq. (17-17 b ):

2 2 0.25 86.8 (16 5.4) 86.8 (16 5.4) 2(16 5.4) 2 2

26.05 in.

C

Ans

π π

 ^ ^  

Eq. (17-1) in degrees:

180 2sin 156. 2(26.05)

θ (^) d

− ^ − 

From Table 17-13 footnote:

K 1 = 0. 143 543 + 0. 007 468(156. 5°) − 0. 000 015 052(156. 5°)

2 = 0_._ 944

Table 17-14: K 2 = 1

Belt speed:

1696 ft/min 12

V

π = =

Use Table 17-12 to interpolate for H tab_._

Eq. (17-17) for two belts: H (^) a = K K N H 1 2 b tab = 0.944(1)(2)(2.31) =4.36 hp

Assuming nd = 1,

Hd = KsH nom nd = 1_._ 25(1) H nom

For a factor of safety of one,

tab

1.59 (1696 1000) 2.31 hp/belt 2000 1000

H

th

nom

nom

3.49 hp.

H a Hd

H

H Ans

______________________________________________________________________________

17-19 Given: H nom = 60 hp, n = 400 rev/min, d = D = 26 in on 12 ft centers.

From Table 17-15, a conservative service factor of KS = 1_._ 4 is selected.

Design task: specify V-belt and number of strands (belts). Tentative decision : Use D

belts.

Table 17-11: Lp = 360 + 3_._ 3 = 363_._ 3 in

Eq. (17-16 b ):

2 2 0.25 363.3 (26 26) 363.3 (26 26) 2(26 26) 2 2

140.8 in (nearly 144 in)

C

π π

 ^ ^ ^  

, , exp[0.5123 ] 5.0,

2722.7 ft/min 12 12

d D

dn V

θ π θ π π

π π

Table 17-13: For θ = 180°, K 1 = 1

Table 17-14: For D360, K 2 = 1_._ 10

Table 17-12: H tab = 16_._ 94 hp by interpolation

Thus, Ha = K 1 K 2 H tab = 1(1_._ 1)(16_._ 94) = 18_._ 63 hp / belt

Eq. (17-19): Hd = H nom Ks nd = 60(1_._ 4)(1) = 84 hp

Number of belts, Nb

d b a

H

N

H

Round up to five belts. It is left to the reader to repeat the above for belts such as C

and E360.