Question

In the PSC beam shown, $f_{ck}=45MPa$ and it supports a UDL of 15 kN/m including self weight. It is prestressed by a parabolic cable carrying an effective prestress of 200 kN. The shear resistance of uncracked section at the support will be

• A. 93.8 kN
• B. 94.5 kN
• C. 94.4 kN
• D. 95.4 kN

Answer 1

The correct option is B 94.5 kN

$\begin{array}{cc}& \text{Slope of the cable at supports,}\\ \theta & =\frac{4(e_0-e_s)}{L}\\ e_0& =100mm\\ e_s& =25mm\\ \therefore \theta & =\frac{4\times (100-25)}{8000}\\ & =0.0375radians\\ & \text{Vertical component of prestressing force at support}\\ & =200\times 0.0375=7.5kN\\ & \text{As per IS: 1343-1980 the ultimate resistance of}\\ & \text{a section uncracked in flexure is}\\ V_{co}& =0.67bD\sqrt{f_t^2+0.8f_{cp}{f}_t}\\ & \text{Maximum principal tensile stress,}\\ f_t& =0.24\sqrt{f_{ck}}=0.24\times \sqrt{45}\\ & =1.61N/m{m}^2\\ b& =150mm\\ D& =300mm\\ & \text{Compressive stress at centroidal axis,}\\ f_{cp}& =\frac{P}{bD}=\frac{200\times {10}^3}{150\times 300}=4.44N/m{m}^2\\ \therefore V_{co}& =0.67\times 150\times 300\times \sqrt{(1.61{)}^2+0.8+1.61\times 4.44}\\ & =86918N=86.9kN\\ \therefore & \text{Shear resistance}=86.9+7.5\\ & =94.4kN\end{array}$