A technician is servicing an auxiliary equipment mount secured with a 14\text{ mm} hex-hea… — Armed Services Vocational Aptitude Battery (ASVAB)

A technician is servicing an auxiliary equipment mount secured with a $14\text{ mm}$ hex-head bolt. If the technician attempts to substitute a standard fractional (SAE) wrench, which size will slip over the bolt head with excessive clearance and risk rounding off the fastener corners?

A
$1/2\text{ in}$
$9/16\text{ in}$ Answer
C
$5/8\text{ in}$
D
$7/16\text{ in}$

Answer

9/16\text{ in}

9/16\text{ in}

wrench measures

0.5625\text{ in}

, which equals approximately

14.29\text{ mm}

. Because this opening is only

0.29\text{ mm}

larger than a

14\text{ mm}

(

0.5512\text{ in}

) bolt head, the wrench will slide over the fastener but fit loosely. Under torque, this excess clearance causes the wrench jaws to ride over and round off the hex corners.

Step-by-Step Solution

Convert the metric bolt size to its exact inch decimal equivalent.

14\text{ mm} \div 25.4\text{ mm/in} \approx 0.5512\text{ in}

Converting metric dimensions to decimal inches allows direct comparison with fractional SAE sizes.

Calculate the decimal inch and millimeter equivalents of candidate fractional SAE wrenches.

7/16\text{ in} = 0.4375\text{ in} = 11.11\text{ mm}

;

1/2\text{ in} = 0.5000\text{ in} = 12.70\text{ mm}

;

9/16\text{ in} = 0.5625\text{ in} = 14.29\text{ mm}

;

5/8\text{ in} = 0.6250\text{ in} = 15.88\text{ mm}

Identifying the exact physical opening size of each SAE wrench determines whether it can fit over the bolt.

Evaluate which wrench size fits onto the head but creates dangerous slop.

9/16\text{ in}

0.29\text{ mm}

(

0.0113\text{ in}

) larger than

14\text{ mm}

, allowing it to fit over the hex head but with enough loose play to slip and round the corners.

Wrenches smaller than

14\text{ mm}

will not fit at all, while wrenches that are slightly oversized fit loosely and ruin the fastener under load.

Key Concept

Wrench Sizing and Metric-to-SAE Incompatibility

Alternative Method

Multiply the fractional inch by

25.4\text{ mm/in}

to determine millimeter width:

(9 \div 16) \times 25.4 = 14.2875\text{ mm}

, which reveals the slight

0.29\text{ mm}

oversize compared to

14\text{ mm}

Estimated Time:1m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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