A steel reinforcing rod of diameter D is subjected to a tensile (bar) stress of 1400 kg/cm^2, and the permissible bond stress in concrete is 6 kg/cm^2. In reinforced concrete anchorage design, determine the required development (bond) length expressed as a multiple of the bar diameter D.

Introduction / Context:
Development (bond) length is the embedment needed so that bond stress between concrete and steel can safely transfer the bar's tensile force into the surrounding concrete. This question checks the classic anchorage derivation that equates the tensile force in the bar to the circumferential bond resistance along a straight length.

Given Data / Assumptions:

• Tensile (steel) stress, sigma_s = 1400 kg/cm^2.
• Permissible bond stress, tau_bd = 6 kg/cm^2.
• Bar diameter = D (cm).
• Straight bar, uniform bond, full development from zero to full stress.

Concept / Approach:
The tensile force in the bar must be equilibrated by bond around its perimeter over the development length L_d. For a round bar, bar force T = sigma_s * A_bar and bond capacity = tau_bd * perimeter * L_d. Set these equal and solve for L_d in terms of D.

Step-by-Step Solution:
1) Bar area: A_bar = (pi * D^2) / 4.2) Bar tension: T = sigma_s * A_bar = sigma_s * (pi * D^2 / 4).3) Bond resistance: R = tau_bd * (perimeter) * L_d = tau_bd * (pi * D) * L_d.4) Equate T = R ⇒ sigma_s * (pi * D^2 / 4) = tau_bd * (pi * D) * L_d.5) Cancel pi and one D: L_d = (sigma_s * D) / (4 * tau_bd) = 1400 D / (4 * 6) = 1400 D / 24 = 58.33 D ≈ 59 D.

Verification / Alternative check:
Dimensional consistency holds (length ∝ D). The numeric factor 58.33 matches code-based approximate tables for plain, straight bars under the stated stresses.

Why Other Options Are Wrong:

• 30 D, 40 D, 50 D, 53 D: All are less than the computed 58.33 D and would not fully develop the bar under the given stress.

Common Pitfalls:

• Forgetting the 1/4 factor from area = pi D^2/4.
• Mixing up bond stress units or using bar stress instead of allowable stress.

Final Answer:
59 D.