Question

A post-tensioned beam of span 25 m is prestressed with 20 numbers of 40 mm diameter cables, each stressed to 1500 MPa, with eccentricity e = 0 at supports and e = 500 mm at midspan, varying parabolically. If the shear force at the support section due to externally applied load is 4500 kN, what is the nearest magnitude of the shear force resisted by the stirrups?

• A. 3060 kN
• B. 4540 kN
• C. 250 kN
• D. 1480 kN

Answer 1

The correct option is D 1480 kN


$\begin{array}{cc}Since,\frac{w{l}^2}{8}& =Pe\\ \Rightarrow w& =\frac{8Pe}{l^2}=\frac{8\times 1500\times 20\times \frac{\pi }{4}\times (40{)}^2\times 500}{(25000{)}^2}\\ \Rightarrow w& =241.152N/mm=241.152kN/m\\ \therefore & \text{upward load due to prestressing force}\\ & =241.152\times 25=6028.8kN\\ \text{Reaction due ot this at support}& =3014.4kN.\text{Net shear force created at support}\\ & =4500-3014.4=1485.6kN\\ & \text{This is the shear force that has to be}\\ & \text{approximately resisted by the stirrups.}\end{array}$