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ESE Civil 2014 Paper 1: Official Paper

Option 3 : 1.75%

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

\({\rm{Loss\;of\;stress\;}} = {\rm{Strain\;due\;to\;anchorage}} \times {\rm{Modulus\;of\;Elasticity\;}}\)

\({\rm{Strain\;due\;to\;anchorage}} = {\rm{\;}}\frac{{{\rm{Anchorage\;slip}}}}{{{\rm{length\;of\;beam}}}}\)

\({\rm{Percentage\;loss\;in\;pre}} {\rm{stress}} = \frac{{{\rm{loss\;in\;pre}} {\rm{stress}}}}{{{\rm{initial\;pre}}{\rm{stress}}}} \times 100\)

__Calculation:__

Given,

Anchorage slip = 3 mm; Beam length = 30 m; Modulus of Elasticity of steel = 2.1 × 10^{5} N/mm^{2},

initial pre-stress = 1200 N/mm^{2}

\({\rm{Strain\;due\;to\;anchorage}} = {\rm{\;}}\frac{3}{{30 \times 1000}} = {\rm{\;}}{10^{ - 4}}\)

Loss of stress = 10^{-4} × 2.1 × 10^{5} = 21 MPa

\({\rm{Percentage\;loss\;in\;pre}} - {\rm{stress}} = {\rm{\;}}\frac{{21}}{{1200}} \times 100\)