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Okmarks here on the bookmarks bar. Import bookmarks now Answer all questions Each question is worth...

Question:

okmarks here on the bookmarks bar. Import bookmarks now Answer all questions Each question is worth 15 marks (Total Marks 60) Q. 1 Design the transverse shear reinforcement in terms of N10 ties for the shear force at the support for the beam shown below. fe 32 MPa and steel yield strength- 500MPa. (15 Marks) q 750 kN 60° 2 N28 g- 20 kN/m 762 790 4 m 8 N28 4 m 80 52 300

okmarks here on the bookmarks bar. Import bookmarks now Answer all questions Each question is worth 15 marks (Total Marks 60) Q. 1 Design the transverse shear reinforcement in terms of N10 ties for the shear force at the support for the beam shown below. fe 32 MPa and steel yield strength- 500MPa. (15 Marks) q 750 kN 60° 2 N28 g- 20 kN/m 762 790 4 m 8 N28 4 m 80 52 300

Answers

loads for the shear force calculations -

live load = 20 KN/m

self weight of beam = 25*0.3*0.87 = 6.525 KN/m

total load = 6.525 + 20 = 26.525 KN/m

maximum shear force = 26.525 * 8 / 2 = 106.1 KN

nominal shear stress \tau _{v} = v / bd = 106.1 / 300*700 = 0.505 N / mm^{2} < \tau _{cmax} ( \tau _{cmax} = 0.62\sqrt{f _{ck}} = 0.62 \sqrt{ 32 } = 3.5072 N / mm^{2} )

A_{stmin} = 0.85 bd / f_{y} = 0.85*300*700 / 500 = 357 mm^{2}

using 12 mm \phi bars .

A_{st} = A_{stmin} = 357 mm^{2}

no of bars = 357 / \pi /4*(12^{2}) = 4 bars

hence provide 4 bars of 12 mm \phi

for transverse reinforcement - using 8 mm \phi 2 legged stirrups for transverse reinforcement -

area of cross section -

As lsu w = 2*\pi*8^{2} / 4 = 100.50 mm^{2}

A_{sv} = \tau _{u}s_{v}\div b_{1}d_{1}\left ( .87f_{y} \right ) + V_{u}S_{V} \div2.5d_{1}\left ( 0.87f_{y} \right )

.

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