Short Rectangular Column (30 cm × 20 cm): Two opposite eccentric loads along a principal section parallel to the longer side produce the same extreme stress—find P1 : P2 when e1 = 4 cm and e2 = 8 cm

Introduction:
This problem uses the superposition principle for combined axial load and bending in a short rectangular column. Two concentric-but-opposite eccentricities along the same principal axis create counteracting moments. The question asks for the ratio P1 : P2 such that the maximum extreme fiber stress on each side is the same in magnitude.

Given Data / Assumptions:

• Section: 30 cm × 20 cm (dimensions ensure “short column” assumptions).
• Principal section considered is parallel to the longer side (30 cm).
• Eccentricities: e1 = 4 cm on one side; e2 = 8 cm on the opposite side.
• Linear elastic behavior; superposition valid; loads are axial with small eccentricities.

Concept / Approach:
Resultant stress at an extreme fiber is: sigma_extreme = (P/A) ± (M/Z). Here, M = Σ(Pi * ei) with sign based on sides. If the “extreme intensity on either side is the same,” the magnitudes of stress at the two opposite extremes must be equal. This condition is met when the net bending moment is zero, i.e., the opposite moments cancel each other.

Step-by-Step Solution:
Step 1: Write the moment balance about the centroidal axis along the given principal direction.Step 2: For equal extreme stresses on opposite faces, require net bending moment M = P1e1 − P2e2 = 0 (opposite sides give opposite signs).Step 3: Solve for ratio: P1e1 = P2e2 ⇒ P1/P2 = e2/e1.Step 4: Substitute e2 = 8 cm and e1 = 4 cm ⇒ P1/P2 = 8/4 = 2.Step 5: Therefore, P1 : P2 = 2 : 1.

Verification / Alternative check:
If M = 0, the section carries uniform axial stress only, so both extremes have the same intensity, satisfying the condition without needing Z or the exact dimensions.

Why Other Options Are Wrong:
1 : 1 — Would only work if e1 = e2.1 : 2 — Reverses the needed proportion for cancellation.4 : 1 and 1 : 4 — Over/under-compensate the moment, leaving unequal extremes.

Common Pitfalls:
Trying to compute with section modulus; it cancels for the equality condition. Forgetting sign convention for opposite sides. Mixing up which load should be larger.

Final Answer:
2 : 1.