A simply supported rolled-steel joist under a uniformly distributed load shows a maximum midspan deflection of 10 mm and an end slope of 0.002 rad; determine the span length of the joist.

Introduction / Context:
For a simply supported beam carrying UDL, classic formulae relate maximum deflection and end slope to load, span, and flexural rigidity. With both deflection and slope known, their ratio eliminates E and I, allowing span determination directly.

Given Data / Assumptions:

• Beam: simply supported, UDL over entire span.
• Maximum deflection: delta_max = 10 mm = 0.010 m.
• End slope: theta_end = 0.002 rad.
• Linear elastic behavior; small deflection theory.

Concept / Approach:
For UDL w on span L: delta_max = (5*w*L^4)/(384*E*I) and theta_end = (w*L^3)/(24*E*I). Taking the ratio cancels w, E, and I, yielding a simple proportionality between delta_max, theta_end, and L.

Step-by-Step Solution:
1) Write ratio: (delta_max)/(theta_end) = [(5*w*L^4)/(384*E*I)] / [(w*L^3)/(24*E*I)] = (5*L/16).2) Solve for L: L = (16/5) * (delta_max / theta_end).3) Substitute values: delta_max = 0.010 m, theta_end = 0.002 → delta_max/theta_end = 5.4) L = (16/5) * 5 = 16 m.

Verification / Alternative check:
Dimensional consistency: delta/theta has units of length, so L emerges in meters as expected. Cross-check with typical steel spans—16 m is a reasonable magnitude for the given small slope and 10 mm deflection under UDL.

Why Other Options Are Wrong:

• 10, 12, 14, 18 m: Do not satisfy the ratio (5*L/16) = 0.010/0.002 = 5.

Common Pitfalls:

• Using the point-load formula instead of UDL relations.
• Forgetting to convert 10 mm to meters.

Final Answer:
16 m.