For a bar diameter 20 mm, design stress 150 N/mm² and bond stress 1.2 N/mm², development length Ld=φσ/(4τ) is

Why is this the answer?

Development length Ld = (phi * sigma_s) / (4 * tau_bd), where phi = bar diameter (mm), sigma_s = design tensile stress in steel (N/mm^2), tau_bd = design bond stress per IS 456 (depends on fck and bar type; tau_bd for M20 deformed bars = 1.6 N/mm^2; plain bars = 1.2 N/mm^2). Ensures bar force is fully transferred by bond before the critical section. Ld=φσ/(4τ)=20×150/(4×1.2)=625 mm.

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