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MPSC AE CE Mains 2019 Official (Paper 1)

- \(\frac{{{S_v}}}{{bA_{s_v}}} = \frac{{0.87\;{f_y}}}{{0.4}}\)
- \(\frac{{{A_{{s_v}}}}}{{b{S_v}}} = \frac{{0.4}}{{0.87\;{f_y}}}\)
- \(\frac{{{A_{{s_v}}}}}{{0.87\;{f_y}}} = \frac{{0.4}}{{b{S_v}}}\)
- \(\frac{{b{S_v}}}{{0.4\;}} = \frac{{{A_{{s_v}}}}}{{0.87{f_y}}}\)

Option 2 : \(\frac{{{A_{{s_v}}}}}{{b{S_v}}} = \frac{{0.4}}{{0.87\;{f_y}}}\)

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CT 1: Current Affairs (Government Policies and Schemes)

54434

10 Questions
10 Marks
10 Mins

__Concept:__

Minimum transverse reinforcement is provided in the form of stirrups

According to clause 22.4.3.1 of IS 1343 - 1980

\(\frac{{{{\rm{A}}_{{\rm{sv}}}}}}{{{\rm{b}} \;\times\; {{\rm{s}}_{\rm{v}}}}} = \frac{{0.4}}{{0.87\; \times\; {{\rm{f}}_{\rm{y}}}}}\)

A_{sv}** – **total cross-sectional area of stirrup legs effective in shear

s_{v} – stirrup spacing along the Length of the member.

b – breadth of the member which for T, I and L beams should be taken as the breadth of ribs b_{w}

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