Construction Safety Math: Noise and Excavation Slopes questions for the CHST
CHST5 Domain 1: Hazard and Risk Identification and Control (K1.3 Excavations; K1.18 Noise exposure). 6 free questions below. Every answer quotes and links the formula source it rests on.
Question 1Calculation
A maintenance mechanic's 8-hour shift includes 3 hours at 97 dBA, 1 hour at 90 dBA, 4 hours at 102 dBA. Using OSHA's noise dose method (Table G-16a reference durations), what is the mechanic's daily noise dose?
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Answer: A. 379.2%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (3/3 + 1/8 + 4/1.5) = 379.2%. A dose of 50% or more is at or above the action level, so a hearing conservation program is required under 1910.95(c)(1).
Answer A rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“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(C1 / T1 + C2 / T2 + Cn / Tn), where Cn indicates the total time of exposure at a specific noise level, and Tn indicates the reference duration for that level as given by Table G-16a.”
Question 2Calculation
A maintenance mechanic's 8-hour shift includes 2 hours at 85 dBA, 6 hours at 88 dBA. Using OSHA's noise dose method (Table G-16a reference durations), what is the mechanic's daily noise dose?
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Answer: D. 69.1%
D = 100 x (C1/T1 + C2/T2 + ...), with T from Table G-16a: D = 100 x (2/16 + 6/10.6) = 69.1%. A dose of 50% or more is at or above the action level, so a hearing conservation program is required under 1910.95(c)(1).
Answer D rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“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(C1 / T1 + C2 / T2 + Cn / Tn), where Cn indicates the total time of exposure at a specific noise level, and Tn indicates the reference duration for that level as given by Table G-16a.”
Question 3Calculation
A noise dosimeter worn for a full 8-hour shift reads a dose of 400%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: A. 100.0 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(400/100) + 90 = 100.0 dBA. A 100% dose corresponds to a TWA of 90 dBA.
Answer A rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“The eight-hour time-weighted average sound level (TWA), in decibels, may be computed from the dose, in percent, by means of the formula: TWA = 16.61 log10 (D/100) + 90.”
Question 4Calculation
A noise dosimeter worn for a full 8-hour shift reads a dose of 132%. What is the employee's 8-hour time-weighted average (TWA) sound level?
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Answer: D. 92.0 dBA
TWA = 16.61 log10(D/100) + 90 = 16.61 x log10(132/100) + 90 = 92.0 dBA. A 100% dose corresponds to a TWA of 90 dBA.
Answer D rests on
29 CFR 1910.95, Appendix A(opens ecfr.gov in a new tab)“The eight-hour time-weighted average sound level (TWA), in decibels, may be computed from the dose, in percent, by means of the formula: TWA = 16.61 log10 (D/100) + 90.”
Question 5Calculation
A crew will dig a 14-foot-deep trench in Type B soil with a 6-foot-wide bottom, using a simple slope on both sides at the maximum allowable slope (not short-term). What is the minimum width at the top of the excavation?
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Answer: A. 34 ft
Type B soil allows a maximum simple slope of 1:1 (horizontal:vertical). Each side extends 14 x 1 = 14 ft horizontally, so top width = 6 + 2 x 14 ft = 34 ft.
Answer A rests on
29 CFR 1926 Subpart P, Appendix B (Figure B-1)(opens ecfr.gov in a new tab)“All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1:1.”
Question 6Calculation
A crew will dig a 17-foot-deep trench in Type C soil with a 8-foot-wide bottom, using a simple slope on both sides at the maximum allowable slope (not short-term). What is the minimum width at the top of the excavation?
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Answer: C. 59 ft
Type C soil allows a maximum simple slope of 1 1/2:1 (horizontal:vertical). Each side extends 17 x 1.5 = 25.5 ft horizontally, so top width = 8 + 2 x 25.5 ft = 59 ft.
Answer C rests on
29 CFR 1926 Subpart P, Appendix B (Figure B-1)(opens ecfr.gov in a new tab)“All simple slope excavations 20 feet or less in depth shall have a maximum allowable slope of 1 1/2 :1.”
26 more construction safety math: noise and excavation slopes questions in the full CHST bank
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