CASE STUDY – Combined Loading
• Drill bit isolators for underground mining – all
metal components and elastomer must be
safe!
• Design requirement: 8,000 lb thrust load and
300 lb-ft (3,600 lb-in) torque – these are
basically the cutting forces.
• Detailed FE analysis but ALWAYS verify with
hand calculations!!
Bit
Isolator
Drill Depth
up to 48”
Chuck
Isolator
2010 SME Annual Meeting & Exhibit
Phoenix, Arizona
35 mm (1 3/8 in) Isolator Load Requirements:
Drill Bit
Drill Bit Isolator (
Thrust Load = 8,000 to
10,000 lb
Bending Load= 125 lbs)=
K*d
Drill Rod
Torsion Load= 300 lb-ft
Chuck
Isolator
HOT SPOT!
2010 SME Annual Meeting & Exhibit
Phoenix, Arizona
Results showed that in combination, the bit isolator
and the drill chuck isolator provided a 7 dB(A)
reduction in sound pressure level at the operator
position.
2010 SME Annual Meeting & Exhibit
Phoenix, Arizona
A-weighted Sound Level Spectrum
110
108.9 dB(A)
Baseline:
T2 - 45 NR BISO, 45 NR CISO: 104.4 dB(A)
105
A-weighted Sound Level (dB)
T5 - 55 NR BISO, 55 NR CISO: 102.7 dB(A)
100
T7 - 70 NR BISO, 70 NR CISO: 104.4 dB(A)
95
90
85
80
75
70
65
Overall
Phoenix, Arizona
10000
2010 SME Annual Meeting & Exhibit
8000
1/3-Octave-Band Center Frequency (Hz)
6300
5000
4000
3150
2500
2000
1600
1250
1000
800
630
500
400
315
250
60
1 3/8” Drill Bit Isolator – Design Review:
Drill Bit isolator:
Chuck isolator:
2010 SME Annual Meeting & Exhibit
Phoenix, Arizona
1 3/8" DBI, R. Michael, PE
6
Critical Area!!
Look at metal components: Where would you expect
max stress to occur?
4/13/2015
1 3/8" DBI, R. Michael, PE
7
Critical component – inner member, thin wall, stress concentration!
Results: FE Analysis 1 3/8” DBI
All cases, thurst load =
8,000 lb, torque = 300 lb-ft
(3,600 lb-in)
HOT SPOT!
4/13/2015
1 3/8" DBI, R. Michael, PE
9
Max Principal, s1
Min Principal, s1
Max shear stress, tmax
VERIFY FEA RESULTS WITH HAND CALCS!!!!
Verify Principal Stresses with
Hand Calculations or Mohr’s Circle!!
Thrust Load = 8,000
to 10,000 lb
Torsion Load= 300 lb-ft
HOT SPOT!
y
x
HAND CALCS:
sy 
t xy 
P
 8,000lbs

 23,895 psi
A  (0.88in) 2  (0.59in) 2
4

From Axial Load
(Compression)

Tc
3,600lbin(0.44in)

 33,720 psi
J  [(0.44in) 4  (0.295in) 4 ]
2
From Torque
s y  23,895 psi
s x  0 psi
t xy  33,720 psi
HAND CALCS Cont’d:
Principal Stresses:
s 1, 2 
s x s y
2
 s x s y 
  t xy2 
 /  
 2 
2
=-11.95 +/- 35.8 ksi
0  23.895
 0  23.895
2
 / 
  33.720
2
2


2
s1  23.85ksi
 s x s y 
  t xy2  35.8 ksi
t max  
 2 
2
s 2  47.75ksi
How do these
values compare
with FEA
results?
Next: use Mohr’s circle to resolve