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THESIS

DESIGN OF A 10 TON OVERHEAD
PU Nam CMO NA

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G. W. QUICK. L.A. PRATT

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Vee ace olf Dt ee mE

DESIGN OF A 10 TON OVERHEAD TRAVELING CRANE.

A Thesis Submitted to
The Faculty of the

MICHIGAN AGRICULTURAL COLLEGE
By

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rf i; fc a ie be

y yy

GW Quick LeA-Pratt
>,

Candidates for the Degree of

Bachelor of Science.

June, I9I7 «
THESIS
SPECIFICATIONS.

Blectric overhead traveling crane;
Cage operated;

Capacity--I0 tons;

Span-<22!';

Maximum Lift--12',

Mish-belly box girder g;

Speed of noiste-25'=-75'per min»
Speed of trolley--100'-150' per min-,;
Speed of bridge--250-300' per mine;
Source of power--500 volts AeC.

MOTOR SIZES «
HOIST MOTORS
Efficiency = gear eff. x sheave effe- x motor effe x drunteff
= 096 x 096 x96 x 096 x 085 x 090 3% 065 or 65%.~
Speed of hoist = 25'-75' per mine
HeP»>“ 10 x 2,000 x 25/33,000 x 65 5 22i.Pe
BRIDGE MOTOR.
Assume diae of wheels = 18"
Assume wt- of trolley = 6,000#
Total load = 2,000 plus 6,000# = 26,000
Speed of travel = 250'~-300' per mine

HePeo 2 105(42,000)( 002 plus 08 x 15)300/33,000 x 9 & 9
Use IOH -P.

TROLLEY MOTOR «
Assume diae of wheels & I2*

HePe 2 1025(26,000)(-002 plus 08 x I)150/6 x 33,000 2 2.02
Use 4 HePeo

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GEARS «

HOIST GEARS«

“Motor ased--20 Heb «+550 ReP Me

4 rope hoiste

Tnerefore, 20,000/4 = 5,000# on each ropes

Mean speed of lift = 50" per mine

Ends of rope winding on drwn must ‘travel 2 x as fast as the
pulley block or 2 x 37 = 75° per mine

Best rope to use = 9/16" diae--6 strands --37 wires euch.
(Kents hend book

Dise of drum to be used = (24-30)x diae of rope =
28 x 9/16 2 1568 Use 16" drum.
- 301416 x 16/12 2 402' cire of drum.
100/422 2 24 RePolle that the drum must have to give a lift
of I100' per mine
RePeMe of motor = 550

  

A B Cc R
Lock all x x x xX
Lock R XDB/CA -XD/C -xD/C @X 0
Totals X plusXBD/CA
Therefore X plus XBD/CA = 559 (Eqe I)

Assume diae A = 4"

Assume dia B = 6"
ome

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Assume dia- C a 4"

Assume dia- D= 14"

Subst.» in Ege I X(Iplue 6 x 1444 x 4) = 550
Xs 104 RP de

Assume dia- F = 18"

(104-24)/24 = 18/x

XS 505" diae of Re

Diae B 2(18-5.5)/2 2 64"

Therefore,

Gear A = 4° in diae

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Gear Ae
Assume D.Pe = 6 PeDes 4" Noo teeths 6x48 24.
Speed of pitch cire & 301416 x 4/12 x 550 = 580' per mine
Torque = 33,000 x H«Pe x I12/(2 x 301416 x RP Me
833,000 x 22 x 12/(2 x 31416 x 550) 2 2,520%,:
Let Ws wte on tooth a 2,520/2 # 1260#
Lewis formula,
v= W/fpy
We 12604
f @ 0955
y = 1300 (for steel)
PS 301416/6 & 2525
 

 

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b2 1260/0955 x 1300 x -525 = I-96"
Use 2" e3 width cf facee

Gear Be
PD. = 6"
Width of face = 2"
DePe = 6
Noe of teeth = 6 x 6 = 36
Gear C
Assume DeP. = 6 PeDe 3 4"
Ho» of teeth 2 6 x 4 2 24
ReP Me of Bw (X=XD)/C & (75=75 x 14)/4 © 339
Speed of pitch cire B 361416 x 263012 = 355' per min.
y = 0955
II ,500 (for steel)
» 3 301416/6 m 2525
ba W/tpy = 2390AX1,500 x 00955 x 0525) w@ 401" Use 4"

#

44

Gear D

PeDe wm I4"

DePe ss 6

Width of face a 4"

Noe of teeth 3 6 x 14 & 84
Gear R

PeDe 5 59"

DePe S 4

Noe of teeth 8 4 x 5¢& 22

Wa 6100

Speed of pitch cire = 323-3416 x $4 x 100/12 = 202" per mine

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= 15,000 (for cast iron)
3eI1416.4 = ~78
610CA-093 x 078 x 15,000 = 5-6 Use 53"

Gear E

PD. & 63"

DePe S 4

Noe of teeth s 4 x 6¢ = 25
Gear F

PeD. = 18"

DePe = 4

Noeof teeth = 4x 18 = 72

Width of face = 5$"

TROLLEY GEARS.

ReP Me of motor = 940

RP M of wheel shaft = ?

Assume dia of wheel <= I2"*

Cire of wheel = 3-I4I16 x 12/12 = 3.1416!
Mean travel speed = 125' per mine

125/3I416 & 40 ReP eM. of wheel shaft.

3/14 2 X0940
X S 201 RePeMe of B and Ce

I4/y = 201/40
ys 22" Use 2666"

 
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Gear A
T = 33,000 x 12 x 4/940 x 2 x 301416 = 270%
Material used ----cast iron.
270/15 = r80# or W
D.P. 3 §
P.D. = 3"
S peed of pitch cire w 3eI14I6 x 3 x 940/12 = 740" per mine
y = 083
f 2 3500 (for cast iron)
P= 3-1416/6 = 525
b = T80/Q500 x -083 x -525) = 1-19 Use It" as width of A.
Hear B
PeDe @ 14*
Width of face s& I2"
Torque = 180 x 7 = 1260%#
DePo = 6
Gear C
PeDe 5 20666"
We 1260/1333 = 915¢
RPeMe Ft about 200
Speed of pitch cire S 301416 x 20666 x 200/12 = 140' per min
f 27,000 (for cast iron)
Noe of teeth & 20666 x 6 & 16

y = 0077

Pes £525

bd @ 915/(7,000 x 0077 x 0525) © 362 Use 3¢"
Gear D

Width of face = 3}"
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PD. -14"
BRIDGE GEARS |
Diae of bridge wheels = i"
R.P.Me of wheel = 275 x 12/3.J4b6 x 18 = 5865 Use 56 R-PoMe

A Assume:
D = 16", C = 6", and A = 3"

 

Let X = RePeM. of mein shaft.
58/X = 6/16
x = IDR -P Me

AYB = 1°5/600
4/8 - 155/800
B = 20" diae of gear Be
on
Gear A-
T = 33,000 x 17 ~ “/2x 361416 x 800 = 793"F
793/2 = 396 or We
Speed of pitch circle = 301416 x 4 x 800/12 = 840" per min-
F - 3000(for cast iron)
y = 096
Ps 9525
Aseume D.Pe- = 6
PeD. = 4
Noe of teeth = 24
b = 396/096 x 0525 x 3000 = 206 Use 2a /2"
Gearbe
Width of face - 2=1/2"
DePe ow 6
P.D. = 20°
Torque = 39665 x 10 = 3965"#
Gear Ce
DP« =6
PD. = 6
Noeof teeth=- 36
Speed pitch circles 155 x 301416 x 6 = 244'per min.
f =» 5300 (for cast iron)
= 106
3965/3 = 1322f

= 9825

y
W
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| 1322/0106 x 0528 x 5300 = 4045 Use 4-1/2"
Gear JD.

Width of “-cr = 4-1/2
 

 

 

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Nop of teeth - 96

SHAFTSand PINS.
HOIST GEAR PINS.

= Pin for Gear PB and C. or Ae

  

Check for torsione

Te: Px 301416 x A¥t6

3780 x 16/ 15,000 x 361416 = d3= 103
d= I.I Use 1-1/4" pin.

Checked safely for shear and bending.
Pin Be

Torsion on drum G = 6100 x 9 = 54,900%¥

54,900/5 875 = 9350#

 
10

Mos PL/4 = 9350 x 7/4 = 16350"#
M = PI/e = p-098D3

p3. 16350/-093 x 8000 = 2-08

D 210255 Use 1=3/8"

HOIST MOTOR SHAFT.

 

Pencire

w= PL/4 = 10,000 x 23/4 = 57,500%#
p> = Me/pl = 57,500/13000 x +098 = 45
D=- 3-2 Use 3°

Checked for torsion and sheare

Small and of hoist motor shafte
Torque = 1260 x 2 = 2520"#

T = pex 34426 x d?/16

d = 1-157 lyse I=I/2"

TROLLEY WHEEL SHAFTS.

 
II

Bending

Ns 6500 x 12-1/2 = 31, 250%#

Ms p x 098D°

TP~ 81,250/8000 x 0098 = 1004

Deo 22% Use 2=1/4" shaft.
TROLLEY GEAR: BINS.

 

Pin for gear B and Ce

Torsione

T - p x 301416 x D/I6

Di = 126A50x 301416) = 08

D= 093 Use I® pine

Checks safely for shear and bending.
Pin for gear De

Tortione

T= 915 x7 = 6405

6405 = 8000 x 361416 x D/16

DB - 4.08

D=Ie6 Use I-I/2" shaft.

MAIN SHAFT ALONG BRIDGER.
Noraque = 396-5 x 10 = 3965°#
T = p x 301416 xD/16

Dp = 4000# vbecause of the shaft being so long we use an extra
large factor of safety.

 
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p36 3965 x 16/( 4000 x 31416) = 5
B= about 2 Use 2-1/2" shaft
BRIDGE WHEFL PINS.

 

Bendinge
Mie PL/4 = 16,000 x II-1/2/4 = 46,000
D3- 46,000/8000 x 098 = 58

D2 308 Use 3-3/4" pine
WHEELS «
BRIDGE WHYELS
Diae of wheel assume ss 18"
Let bd © breadth of wheel-e
» = W/500D = 16375/500 x 18 = 1685 Use 2"
Use a 40# rail.
TROLLEY WHEELS «
Assume diae = I2"
b = 6500/5900 « 12 2£ 13/12 211" use 1-3/4"
Use a 25# raile
DRUM

 

 
“ t
or P<.

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13

R - radius of rope + 1/32 = 9/Iéplus 1/32 = 19/32"
a = 3/16"

P - QR plus a = 2 x 19/32 plus 3/16 = 1-3/8"
Distance between centers of pulleys in sheat-% 5"
t $15/I16 plus 3/8 = 14/16"

1 $16 x i 3/8 plus 3 plus 2 - 27

MOTOR FRAME.

   

Section in square hollow cross sectione

Ms PL/4 = 13000 x 37/4 = 120,000%F
Ife = 1/6n(Btor \/

120,000 = 4000(Btb 9/68

B- 8"

b = 7"

BRIDGE DETAILS.
ERIDGE GIRDER «
Take deflection e025" per ft+ span.

6750# weight of girder.
I4

6000# wte of trolley.
20, 000fwt~+ of live poad

3375i%-- wte of 1/2 girderD.Le

3000#-- wte of I/2 trolley.

10000#-- wt- of 1/2 live load.
16,375#-- wheel load on bridge wheelse

3000#--wt. I/2 troley-
10000#--wte 1/2 live loads
13000#--wte on bridgeedue to TeLe and Lele

13000/2 s 6500#--trolley wheel load.

 

Fer shear on ende

6500 plus (6500 X 1892/22) = 12b00#

22-3208 = 18.92

Calculated from section wte of one girder about equal to dead
loed = 13500/2 = 6750#

6750 plus I2I100 - 18,850#

Wed area = 18,850/10,500 = 168 sqe ine

| Flange areae

A - M/sa

A = area of one flange which is in tension or compression.

M - bending moment in ine lds.

S = allowable stress in lbse per sqe ine This must be

allowable average stress at extreme fibre and equals I2,500#

 
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d = effective depth = 305-4 = 2665"

Therefore the working stress = 12,500 x 26 05/3065 = 10,8507
per 3qe ine

 

Assume section and check for resisting moment of girdere
Me FI/C Tx
I, for 14" plas =.2( +02 plus I 53375 x 35)= 1,640

4m Ig plus uF

I, for 30" plse = 2(562-5) 2 16225
I, for Angles = 4(1-5 plus 14-14 plus 1.44) -1,157 om
Total I - 3,922
Wp = 10,850 x 3,922A30075 x 12) = II5,500'#
allow 10% for rivets = I1,550'#
| My 2 103,950'#

 
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16

 

Rh.
Ry = 13,000 x 102/22 = 6,020#
therefore M © 6,030 x 1002 = 61,600'#

M(D.Le) at center = 1/8 x 13,500 x 22 x 10-2/II »
34,500'#

612600 plus 34,500 = 96,300'# or 1,158,000" total Me

This moment checks very closely to the resisting moment of

A = 1,358,000/(10,850 « 2665) = 4203 sqe ine
Allow I/8 of web as required area.

Wed area = 2 x 30 x 1/4 = 15 sqe ine

1/8 x 15 = 1287

2 angles = 365 x 295 x 025 = 2288
I web plate - 14 x 025 = 305

Total = 8425
Deduct for rivet holes 88

Remaining area = 7237 sqe ine

 
 

 

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8
Check for bending.
Mie W/8L(1-e/2) plus G1/8 plus wte of motor x 1/4
= 26,000/(8 x 264) x (264= 37/2) plus 18,000 x
264/(2 x 8) plus 800 x 264/4 = 90,100'#

RIVETS.

Rivets are in double shear; the allowable shearing fibre str-
ess is 9,000# per sqe ine

Shearing value - 1094 x 9 = 9,820#

Bearing value = 656 x 9 = 5,900#

Therefpre it is evident that the rivet will stand 5,900#
pressure, this being the lower of the two values.
BRID GE TRUCK GIRDER.

Bridge wheel base of 10' 6".
Maximim wheel load is 16,375#

 

Bending moment Mp s 16375 x 37 = 605,000"
Fibre stress - 10,690# per sq."

~

 

 
 

r*

vt 4

 
Assume section as shown in figure and check by finding
resisting moment, VW,

Two 10" channels at 20# -- I50" = 300

Moment of inertia for flats neglected.

Area flats x h’=| ( 2 x 10 x 025) x 5125 - I3Z 23

Total = 300 plus 13163 = 43143

Section modulus = Ife = 4313/5025 = 822

Mp 2 82.2 x 10,850 = 892,000"# which is enough over the
bending moment to allow for riveting and overload.

CRANE BLOCK.

SHEAVE .

d=idiae of rope = 9/16""

R = a/2 plus 1/32 = 9/32 plus 1/32 = 312"

ae 105 a= 2a

a= 1675 x 9/16 = -988 Use I"
b = d/3 plus 3/16 = 0374 Use 3/8"
c = d/6 plus 3/16 = »2199"

PULLEY BLOCK SIDES
Load on one side = 1/2 x 20,000 = 10,000#

 

P = 10,000#
Assume p - 8,000#
A = P/p = 10,000%/ 8,000 = 1225 sqeine Area in shear.
Assume thickness ( t ) = 3/8"
Therefore 3/8 x 2X = 1625
X = 4x 1625/3 - 1667 Use 1-3/4" pius

radius of hook pin as radius for metal below hook pin center.

 
‘3
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ane
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19

HOOK
Crane hook of soft steel forging

 

The following points to be observed;
Ie The point of the hook to be carried well up to prevent
the sling from slipping off the hook.

2+ For chain slings , a = d-- 15d

ad = chain dia.

a = radius of the inside circle of the hook.

Design the hook to carry 20,000#

Allow 8,000 fibre stress per sqe ine on the section at the
root of the screw threads and 12,000# per sq- ins on the
extreme fibres as a result of combined tension amd flexure.

P/p = 20,000/8,000 = 265 sqeine

Referring to table of standard screw threads. A bolt 2-1/4"
in diae will be required.

do = 1e25d is 1625 x 2625 = 2081 Use 29"

Now considering the section of maximum bending a - 205/

Taking h = 205 x a = 6425"

dz - 6625/1433 = 4%"

by/o2 -(n/a) plus I + 265 plus I = 365

C by tables for trapezoidal section ia 5 206

Therefore by = P/(p x a x C) = 20,000 x5 -06/(12000 x 265) ‘5.
3037"

 
@.

 
20

bo = 3037/3+5 = 963 Use 1"
SWIVELLING CRANE HOOK WITH TRUKNION SUSPRaSTON

Hook to carry load of 20,000#

Balls---B - Kd~ K = 4,200

P = 20,000/30 = 6666

a*. 666 06/4,200 d= 399 Use 7/16"

Bolt dias thru trunnion © 2025 x 1el25 © 2e5I Use 2-1/2"
TRUNNION

Assume the load and width ~ 8 - 2 = 6*

M« Pl/e

M =< 20,000 x 10/4 = 8,000 x vn*/6

since C/e = vn-/6 n@ 2 20,000 x 10 x 6/(4 x 8,000 x 4)

h = 3-04 Use 3"
TRUNNION BEARINGS -

On the assumption of I500# per sqe- in - the projected should
be I10,000/1500 = 666 sqe ine

If the journal is now assumed as having a width equal to its
diae the diae will be determined for streugth as a cantilever
as follows;

M« Pi/e

= 10,000 x ad /2 for a circle Ife « 3-1416a732 or

approximately a3/z0

Taking P © 8,000# per sq ine

Mee (30,000 x d)/2 =(8,000 xa>)/10

a? = 10,000 x 10/( 2 x 8,000) 22.5"

The area 2-5 x 265 = 625 aqe ine, checxs very closely to
6066 sae ine

 
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