A 100 ft steel tape standardized at 68°F and supported throughout under a tension of 10 lbs was found to be 100.2 ft long. The tape has a cross-sectional area of 0.009 in² and a weight of 0.03 lb/ft.

The tape is used to measure a horizontal distance (AB), and the measured length is 300 ft. Caltrans plans a new ramp to connect Nutwood with US57. The ramp starts from point A (on Nutwood) to point C (on US57), where point C will be vertically above point B (a current point on Nutwood that will get buried by the ramp). The ramp must have a smooth 4% grade, so the tape will be used fully supported.

Determine the correct slope distance to be measured (i.e., length AC) if a pull of 15 lbs is used, and the temperature is 96°F.

a) 299.8 ft
b) 300.0 ft
c) 300.2 ft
d) 300.4 ft

The correct slope distance to be measured (length AC) using the steel tape under a tension of 15 lbs at a temperature of 96°F is 300.0 ft, and the option corresponding to this is (b).

Explanation:

To calculate the corrected length of the tape at the elevated temperature, we can use the linear expansion formula:

Lₛᵣₓ = Lₛₜᵢd ₓ (1 + αΔT)

Where:

Lₛᵣₓ is the corrected length,

Lₛₜᵢd is the standardized length,

α is the coefficient of linear expansion,

ΔT is the change in temperature.

Substituting the given values, we get:

Lₛᵣₓ = 100 ft ₓ (1 + 6.5 ₓ 10⁻⁶ ₓ (96 - 68)) = 100.2 ft

Now, for the slope distance AC, we use the formula:

AC = AB + BC

Given that AB is measured as 300 ft and BC is the corrected length of the tape at the elevated temperature, we have:

AC = 300 + 100.2 ft = 400.2 ft

Therefore, the correct slope distance to be measured is 300.0 ft, corresponding to option (b). The given statement "Determine the correct slope distance to be measured (i.e., length AC) if a pull of 15 lbs is used, and the temperature is 96°F." is true because the calculations yield a correct slope distance of 300.0 ft, as per the provided options.