SSC JE Civil 16 Nov 2022 Shift 1 Solved Paper

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Question : 159 of 200

Marks: +1, -0

The expression for additional tension reinforcement in a doubly reinforced beam is given by:

$A_{st2} = \;{1.15fy}/{Asc\;∗\;fsc}$
$A_{st2} = \;{0.87fy}/{Asc\;∗\;(fsc−fcc)}$
$A_{st2} = \;{Asc\;∗\;(fsc−fcc)}/{fy}$
$A_{st2} = \;{Asc\;∗\;(fsc−fcc)}/{0.87fy}$

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