A uniform prismatic pile (length L) is to be lifted at two points so that the maximum hogging moment at supports equals the maximum sagging moment at midspan. The optimal distance of each suspension point from either end is closest to:

Introduction / Context:
Long members such as piles or pipelines are often lifted or transported using two-point suspension. To minimize maximum bending stress, the suspension points are positioned so that hogging (over supports) and sagging (at midspan) moments are equal in magnitude, giving the most economical arrangement.

Given Data / Assumptions:

• Uniform self-weight w per unit length.
• Two identical suspension points located symmetrically at distance x from each end.
• Static, simply supported behavior between the lifting points.

Concept / Approach:
For a uniformly loaded beam with overhangs of length x and central span of L − 2x, bending moments include hogging at supports (negative) and sagging at the center span (positive). Setting these maxima equal yields the “economic” lifting distance.

Step-by-Step Solution:
1) Let reactions be equal due to symmetry: R = wL/2.2) Max sagging in central span (simply supported) occurs at midspan: M_sag = R*(L/2 − x) − w*(L − 2x)^2/8.3) Max hogging at each support (due to overhang) occurs just inside the support: M_hog = Rx − wx^2/2.4) Set |M_hog| = |M_sag| and solve for x/L, which yields approximately 0.207.

Verification / Alternative check:
Published lifting tables for uniform beams and piles list the two-point suspension distance as about 0.207 L from each end for equal extreme moments, confirming the calculation.

Why Other Options Are Wrong:

• 0.107 L / 0.307 L / 0.407 L / 0.500 L: These do not produce equality of peak hogging and sagging; they either over-stress midspan or the supports.

Common Pitfalls:
Ignoring overhang effects or assuming the best position is at quarter points (0.25 L), which is not optimal for equalizing extremes with uniform load and two-point lift.

Final Answer:
0.207 L