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 had 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 comes out to be 300 ft.

CALTRANS plans a new ramp to connect Nutwood with US57. The ramp starts from point A (on Nutwood) to 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, thus 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.

To determine the correct slope distance (length AC) under the given conditions, we can use the formula for the apparent length of a tape in tension due to a change in temperature and the applied force (tension):

Apparent Length (LA) = Standard Length (LS) + ΔL

Where:

- LS is the standard length of the tape.

- ΔL is the change in length due to tension and temperature.

ΔL can be calculated using the following formula:

ΔL = (T × ΔT × L0) / (A × W)

Where:

- T is the tension applied (15 Ibs).

- ΔT is the change in temperature in degrees Fahrenheit (ΔT = 96°F - 68°F = 28°F).

- L0 is the original length of the tape (100.2 ft).

- A is the cross-sectional area of the tape (0.009 in², which needs to be converted to square feet).

- W is the weight of the tape per unit length (0.03 Ib/ft).

First, convert the area to square feet:

A = 0.009 in² * (1 ft / 12 in)² = 0.0000625 ft²

Now, calculate ΔL:

ΔL = (15 Ibs × 28°F × 100.2 ft) / (0.0000625 ft² × 0.03 Ib/ft)

ΔL ≈ 15.12 ft

Now, calculate the apparent length (LA):

LA = LS + ΔL

LA = 100.2 ft + 15.12 ft

LA ≈ 115.32 ft

Now that we have the apparent length (LA) of the tape, we can use it to find the correct slope distance AC. Given that the measured length AB is 300 ft, we can use the proportion:

AB / LS = AC / LA

Solving for AC:

AC = (AB * LA) / LS

AC = (300 ft * 115.32 ft) / 100.2 ft

AC ≈ 345.66 ft

So, the correct slope distance AC to be measured is approximately 345.66 feet when a pull of 15 Ibs is used, and the temperature is 96°F.

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