I corrected some mistakes.

1. CHST Prep
The MATH AREAs
John Newquist
johnanewquist@gmail.com

2. Strategy
• Do math first. Start at #200 and work back.
• Math is demoralizing. Try for 70%.
• Only 10 questions usually.
• This will leave you 190 multiple choice questions.

3. 3OSHA Form 300

4. Calculation of the DART
• DART = Days Away, Restricted or
Transferred
• DART= # lost time cases x 200,000/ #hours
4

5. TCIR
• 3.0 million work-related injuries
and illnesses
• The national TCIR rate for the
private sector in 2013 was 3.3
per 100 workers (down from 3.7)
• Total Incident Case Rate = TCIR
• TCIR = (# lost time + medical treatment) cases x 200,000/ #hours

7. OSHA 300
• Summary
• Review OSHA 300
• Dart Exercise on Sample Prefilled
Form
• What is the DART of the
company listed on the handout?
• What is the DART of the
company listed on the handout?
• Hours = 100,000
• 4x200000/100000 = 8.0
7

8. DART and TCIR harder
• 14 lost time cases
• 16 medical treatment cases
• 18 first aid cases
• 20 property damage cases
• 22 auto claims
• 450 employees working and
average of 2050 hours a year
• What is the DART?
• 14x200000/(450x2050)
• What is the TCIR?
• (14+16)x200000/(450x2050)

9. Recordable Accident Rate and DART
• Year 2014 2015 2016
• First Aid Cases 5 3 2
• Medical Treatment 5 5 3
• Lost day cases 3 2 1
• Restricted day Cases 6 5 2
• Vehicle Accidents 4 7 9
• Hours worked 784,222 824,221 935,109
• DART – 2014 (3+6)x200000/784222
• Recordable Rate-2014 (3+5+6)x200000/784222

10. Losses
• A construction company
averages 2 ergonomic back
injuries yearly.
• The average cost of each injury
$50,000.00 per injury.
• If things go unchanged what will
the cost be after 12 years?
• ( Simple math )
• 2x50,000x12

11. Losses
• A construction company
averages 9 injuries yearly.
• The average cost of each injury
$19,000.00 per injury.
• If things go unchanged what will
the cost be after 5 years?
• ( Simple math )
• 9x19000x5

12. Excavation
Excavation
d = 8 feet
W = 12
Soil Type B
If sloping is used to comply, What will the top
width be?
8+12+8

13. Excavation
Excavation
d = 13 feet
W = 5
Soil Type A
If sloping is used to comply, What will the top
width be?
13x0.75 + 5 + 13x0.75

14. Excavation
Excavation
d = 10 feet
W = 10
Soil Type C
If sloping is used to comply, What will the top
width be?
10x1.5 + 10 + 10x1.5

15. Rigging
• A load of 500 pounds is
supported by a two legged sling.
If the slings make an angle of 60
degrees with the load, what
force is exerted by each leg of
the sling?
• 60 degrees is 14% more tension
• 1-0.14 = 0.86
• 500/0.86

16. Rigging
• The metal ring weighs 3000 lbs.
• The sling angle is at 45 degrees
from horizontal.
• What is the tension in each sling
leg? _______________ pounds.
3 legs so 3000/3 is 1000 pound.
45 degrees is 30% more tension
1-0.3 = 0.70
1000/0.7

17. Rigging
• The concrete bucket below
weighs 1000 lbs.
• The sling angle is 30 degrees
from the horizontal.
• What is the tension in each sling
leg?
• Two legs = 500 pounds each leg
• Loss at 30 degree 50%
• 500/0.50

18. Rigging
• The concrete vault weighs 2200
lbs.
• The sling angle is 60 degrees
from the horizontal.
• What is the tension in each sling
leg?
• Equal loading in each sling.
• 2200/4 = 550 lb
• 550lb / loss (1-.14)

19. Rigging
• The concrete vault weighs 2200
lbs.
• 2200/4 = 550 lbs each leg
• 60 degree is 14% loss or 0.86
• 550/0.86 = 639 lbs.
• Or
• 550 lb/sin 60 = 639 lbs.

20. Rigging
• A two piece block and tackle
with 5 part of rope to lift a
weight of 3,000 pounds requires
how much force for lift (no
friction)?
___________________ lbs.?
• 3000/5

21. Rigging
• A two piece block and tackle
with 4 part of rope to lift a
weight of 1,000 pounds requires
how much force for lift (no
friction)?
___________________ lbs.?
• 1000/4

22. Fleet Safety
• 4 million miles driven,
• 12 accidents.
• What is the accident rate per million miles driven
• 12/4

23. Fleet Safety
• 2 million miles driven,
• 17 accidents.
• What is the accident rate per million miles driven
• 17/2

24. LEL
• LEL on gas meter to PEL
• The gas detector is detecting
• 0.5% of Xylene in air
• Xylene PEL is 100 ppm
• Xylene LEL: . . 0.9-1.1% (10% LEL, 900-1,100 ppm)
• Can a fire start?
• Is the worker over exposed if working in the room all day?
• 0.5% = 5000 ppm so yes

25. LEL
• LEL on gas meter to PEL
• The gas detector is reading 8% of the LEL for methyl ethyl ketone
(MEK)
• MEK is 200 ppm
• MEK LEL: . . 1.4% (10% LEL, 1,400 ppm)
• Can a fire start?
• Is the worker over exposed if working in the room all day?
• 1.4% is 14000 ppm x 0.08 = 1120 ppm so no. under 1400 ppm

26. Heat
Stress

27. Heat Stress
• You take the temperatures inside
• Wet bulb temperature - 66F
• Globe temperature – 110F
• Dry bulb – 92F
• What is the current wet bulb globe
temperature?
• Equation for this is WBGT = 0.7 TWB +0.3 GT
• Wbgt = 0.7x66 + 0.3x110
TWB is the wet bulb temperature, which indicates humidity
GT is the globe temperature, which indicates radiant heat
TD is the ambient air (dry) temperature (solar radiation)

28. Heat Stress
• You take the temperature
outside
• Wet bulb temperature - 70F
• Dry bulb – 91F
• Globe temperature – 100F
• What is the current wet
bulb globe temperature?
• Equation is WBGT =
• 0.7 TWB +0.2 GT + 0.1TD
• 70x0.7+ 0.2x100+0.1x91

29. Heat Stress
• You take the temperatures inside
• Wet bulb temperature - 92F
• Globe temperature – 100F
• Dry bulb – 96F
• What is the current wet bulb globe
temperature?
• Equation for this is WBGT = 0.7 TWB +0.3 GT
• Wbgt = 0.7x92+0.3x100
TWB is the wet bulb temperature, which indicates humidity
GT is the globe temperature, which indicates radiant heat
TD is the ambient air (dry) temperature (solar radiation)

30. Noise
• TABLE D-2 - PERMISSIBLE NOISE EXPOSURES
• __________________________________________________
• | Sound level | Duration per day, hours | dBA slow
• | response
• ___________________________________|______________
• 8..................................| 90 or 100% dose
• 6..................................| 92
• 4..................................| 95 or 200%
• 3..................................| 97
• 2..................................| 100 or 400%
• 1 1/2............................| 102
• 1..................................| 105
• 1/2..............................| 110
• 1/4 or less...................| 115
• Dose DBs
• 50 ………………………….. 85.0
• 75 ……………………………. 87.9
• 100 ............................ 90.0
• 110 ............................ 90.7
• 115 ............................ 91.1
• 120 ............................ 91.3
32 hours – 80 dba – 25% dose
16 hours – 85 dba – 50% dose

31. Noise
• When the workshift noise
exposure is composed of two or
more periods of noise at
different levels, the total noise
dose over the work day is given
by:
• D = 100 (C(1)/T(1) + C(2)/T(2) +
... + C(n)/T(n))
• 85 dbs – 16 hours
• 80 dbs - 32 hours
Suppose that a worker has been assigned to operate
3 machines in an 8-hour day. The machine noise
levels are 85, 95, and 80 dBA, respectively. Also, the
work durations at each machine are 2, 2, and 4
hours, respectively. We can estimate this workers 8
hour TWA as follows.

32. Noise
• Another way
• 2 hours - 85 dbs,
= 2/16 hours = .125 dose
• 2 hours – 95 dbs,
= 2/4 hours = .5
• 4 hours - 80 dbs
4/32 = .125 dose
Total .125 + .5 + .125 = .75 dose = 75% or 87.5dbs (close)

33. Noise
• 4 hours at 95dbs,
• 3 hours at 92dbs,
• 1 at 80 what is dose?
• 80 – 32 hours – 1/32 +
• 92 – 6 hours – 3/6
• 95 – 4 hours – 4/4
• (1/32+3/6+4/4) x100

34. Noise
• 4 hours at 95dbs,
• 3 hours at 92dbs,
• 1 at 80 what is dose?
• 4 hours /4hours = 1
• 3 hours /6hours = 0.5
• 1 hour/32 hours = 0.031
= 1.53 of dose = ~93 dbs

35. Noise
• 1 hours at 100dbs,
• 3 hours at 92dbs,
• 4 hour quiet
• 100 – 1 hours – 1/2 = 50%
• 92 – 6 hours – 3/6 = 50%
• Or 50% + 50% = 100%

36. Noise
• 1 hours at 100dbs,
• 3 hours at 92dbs,
• 4 hour quiet
• 1 hours /1hours = 1.0
• 3 hours /6hours = 0.5
= 150% of dose

37. Electrical Basic
• 7.2 kV is _________ volts 7200
• 34.5 kV is _________ volts 34500
• 1A is ___________ mA 1000
• 200mA is ________ A 0.2
• 5K is ____________ Ohm - 5000

38. Ohm Law
• V= I x R
• V = Voltage
• I = Current in Amperes
• R = Resistance in Ohm
• 10=2x(R+2)
Solve for R

39. Ohm Law
V = 5000x0.01

40. PEL
• Solvent is 0.3% in room.
• PEL is 50 ppm.
• Over or under or safe?
• 0.3% - 3000 ppm unsafe

41. PEL – Very hard
• An ironworker was exposed to lead in three samples
• The PEL of Lead is 50 mcg/m3
• #1 – 43 mcg/m3 for 3 hour 57 minutes
• #2 – 29 mcg/m3 for 90 minutes
• #3 – 10 mcg/m3 for 10 minutes
• Rest of the times the exposure was zero.
• What was his TWA?

42. PEL
• First, we put every thing in minutes
• Second, we determine the TWA % of each sample
• Third, we compared to the PEL of 50 mcg/m3 for a full day or 480 minutes
• The PEL of Lead is 50 mcg/m31 f/cc
• Lead Action level is 30 mcg/m3
• #1 – 43 mcg/m3 for 3 hour 57 minutes
• TWA of #1 – 43 mcg/m3 x (237 minutes/480 minutes) = 21 mcg/m3
• #2 – 29 mcg/m3 for 90 minutes
• TWA of #2 – 29 mcg/m3 x (90 minutes/480 minutes) = 5.4 mcg/m3
• #3 – 10 mcg/m3 for 10 minutes
• TWA of #3 – 10 mcg/m3 x (10 minutes/480 minutes) = 0.2 mcg/m3
• TWA = 21 + 5.4 + 0.2 = 26.6 mcg/m3

43. PEL
• An asbestos worker was exposed to asbestos in two samples
• The PEL of Asbestos is 0.1 f/cc
• #1 – 0.043 f/cc for 3 hour 57 minutes
• #2 – 0.029 f/cc for 3 hours 10 minutes
• Rest of the times the exposure was zero.
• What was his TWA?

44. PEL
• First, we put every thing in minutes
• Second, we determine the TWA % of each sample
• Third, we compared to the PEL of 0.1 f/cc for a full day or 480 minutes
• The PEL of Asbestos is 0.1 f/cc
• #1 – 0.043 f/cc for 3 hour 57 minutes
• #2 – 0.029 f/cc for 3 hours 10 minutes
• #1 – 0.043 f/cc for 237 minutes
• TWA of #1 – 0.043 f/cc x (237 minutes/480 minutes) = 0.021 f/cc
• #2 – 0.029 f/cc for 190 minutes
• TWA of #2 – 0.029 f/cc x (190 minutes/480 minutes) = 0.011 f/cc
• Rest of the times the exposure was zero.
• What was his TWA? = (0.021 + 0.011) f/cc = 0.032 f/cc
• They were 0.032/0.100 = 32% of the PEL