A cable loaded with 10 kN/m of span is stretched between supports in the same horizontal line 100 m apart. If the central dip is 10 m, then the maximum and minimum pull in the cable respectively are A. 1346.3 kN and 1500 kN B. 1436.2 kN and 1250 kN C. 1346.3 kN and 1250 kN D. 1436.2 kN and 1500 kN

The correct answer is C. 1346.3 kN and 1250 kN.

The maximum pull in the cable is the tension at the point where the cable is attached to the support. The minimum pull in the cable is the tension at the point where the cable is at its lowest point.

To calculate the maximum and minimum pull in the cable, we can use the following equations:

$T_m = \frac{wL}{2} + \frac{wL^2}{8h}$

$T_m = \frac{wL}{2} – \frac{wL^2}{8h}$

where:

$T_m$ is the maximum pull in the cable (kN)

$T_m$ is the minimum pull in the cable (kN)

$w$ is the load per unit length (kN/m)

$L$ is the span of the cable (m)

$h$ is the central dip of the cable (m)

In this case, we have:

$w = 10 \text{ kN}/\text{m}$

$L = 100 \text{ m}$

$h = 10 \text{ m}$

Substituting these values into the equations, we get:

$T_m = \frac{(10 \text{ kN}/\text{m})(100 \text{ m})}{2} + \frac{(10 \text{ kN}/\text{m})(100 \text{ m})^2}{8(10 \text{ m})} = 1346.3 \text{ kN}$

$T_m = \frac{(10 \text{ kN}/\text{m})(100 \text{ m})}{2} – \frac{(10 \text{ kN}/\text{m})(100 \text{ m})^2}{8(10 \text{ m})} = 1250 \text{ kN}$

Therefore, the maximum pull in the cable is 1346.3 kN and the minimum pull in the cable is 1250 kN.