Bolts are tightened with automatic wrenches on a production line. Each bolt has a mean twisting-off strength of 25 N.m with a standard deviation of 1.5 N.m. If the automatic wrenches have a standard deviation of 2 N.m, calculate the mean value of the wrench torque setting that results in an estimated 2 bolts in 500 twisting off during assembly.

To achieve a reliability of 99.8 percent, what would be the new value of the standard deviation of twisting-off strength with no other change?

The mean value of wrench torque setting that results in an estimated 2 bolts in 500 twisting off during assembly is 26.5 N-m. To achieve a reliability of 99.8 percent, the new value of the standard deviation of twisting-off strength with no other change is 0.7 N-m.

The probability that a bolt will twist off during assembly is given by the following equation:

P(twist) = (mean of wrench torque setting - mean of twisting-off strength)/(standard deviation of wrench torque setting + standard deviation of twisting-off strength)

We are told that the probability of a bolt twisting off is 2 in 500, or 0.04. We are also given the mean and standard deviation of the twisting-off strength of the bolts, and the standard deviation of the wrench torque setting. We can use these values to solve for the mean value of the wrench torque setting.

Code snippet

0.04 = (mean of wrench torque setting - 25)/(2 + 1.5)

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Code snippet

mean of wrench torque setting = 26.5 N-m

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To achieve a reliability of 99.8 percent, the probability of a bolt twisting off must be 0.002. We can solve for the new standard deviation of the twisting-off strength using the following equation:

Code snippet

0.002 = (26.5 - 25)/(2 + new standard deviation of twisting-off strength)

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Code snippet

new standard deviation of twisting-off strength = 0.7 N-m

Use code with caution.

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