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

Answer 1

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 .

= 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 $mm^{2}$

$A_{sv} = \tau _{u}s_{v}\div b_{1}d_{1}\left ( .87f_{y} \right )$ + $V_{u}S_{V}$ $\div$$2.5d_{1}$$\left ( 0.87f_{y} \right )$