SSC JE Civil Engineering 30-Oct-2020 Shift 1 Paper

Question : 122 of 200

Marks: +1, -0

If design bond stress = 1.5 N/mm

^2

is assumed, then the development length of an Fe 500 HYSD bar of nominal diameter 12 mm - which is fully stressed in tension - will be:

Solution:

Given,

\ϕ=12 {\mm}

\τ_{{\bd}}=1.5 {\N} \/ {\mm}^{2}

So,

\τ_{\bd}=1.5 {\N} \/ {\mm}^{2}

Development length,

\L_{\d}={\ϕ×0.87 \f_{\y}}/{4 \τ{\bd}}

{\L}_{{\d}}={12 ×0.87 ×500}/{4 ×1.6 × 1.5}

= {5 4 3.75} {\mm} ≈{5 4 4} {\mm}

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