Cable with uniform horizontal load: A cable carries w = 0.5 tonne per horizontal metre over a 400 m span between level supports with central dip (sag) of 20 m. What is the minimum tension in the cable and where does it occur?

Introduction / Context:

Cables under uniformly distributed load per horizontal span (e.g., suspension bridges) take a parabolic shape. The key design quantity is the horizontal component of tension, which attains its minimum value at the lowest point of the cable and increases toward the supports due to vertical reactions.

Given Data / Assumptions:

• Uniform load per horizontal length w = 0.5 tonne/m.
• Span L = 400 m between level supports.
• Midspan sag (dip) y = 20 m.
• Self-weight of cable is included in w; effects like temperature and secondary bending neglected.

Concept / Approach:

For a parabolic cable under uniform horizontal load, the horizontal tension H is given by H = w * L^2 / (8 * y). The minimum cable tension occurs where slope is zero (at the lowest point), and equals T_min = H. Support tensions exceed H due to the vertical component from reactions.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional sanity: tonnes per metre * metre^2 / metre yields tonnes; trends make sense—greater sag lowers H, smaller sag raises H.

Why Other Options Are Wrong:

• 200-tonne options misapply the formula or confuse units.
• “500 tonnes at each support” is incorrect; support tensions are larger than H due to vertical components and are not equal to the midspan minimum.

Common Pitfalls:

• Using w as load per cable length (catenary case) instead of per horizontal length; here the given w is per horizontal metre.

Final Answer:

500 tonnes at the centre