Question

A prestressed concrete simply supported beam of span 6 m has a rectangular cross-section 300 x 600 mm and unit weight 24 kN/m${}^3$. The resultant prestressing tendon has zero eccentricity at ends and an eccentricity, e, at midspan, and linear profile. Under initial condition, the allowable stresses are 20 N/mm${}^2$ in compression and 2 N/mm${}^2$ in tension. Using this condition, determine the prestressing force, P, and the eccentricity, e, at midspan. The imposed loading is a single concentrated load, W, at midspan and the tendon undergoes loss in prestress of 25%. Determine W, if the final allowable stresses in compression and tension are not to exceed 15 N/mm${}^2$ and 1 N/mm${}^2$ respectively.

Answer 1

$Z=\frac{b{d}^2}{6}=\frac{300\times {600}^2}{6}=1.8\times {10}^{7}m{m}^3$

D.L $=A\times {\gamma }_c=24\times 0.3\times 0.6=4.32$ kNm

$M_d=\frac{w_d\times l^2}{8}=\frac{4.32\times 6^2}{8}=19.44$ kNm
In initial condition tension will be at top and compression will be at bottom
$\therefore f_{top}=\frac{P}{A}-\frac{Pe}{Z}+\frac{M_d}{Z}$
$-2=\frac{P}{1.8\times {10}^5}-\frac{Pe}{1.8\times {10}^7}+\frac{19.44\times {10}^6}{1.8\times {10}^7}$ ...(i)
$f_{bottom}=\frac{P}{A}+\frac{Pe}{Z}-\frac{M_d}{Z}$
$20=\frac{P}{1.8\times {10}^5}+\frac{Pe}{1.8\times {10}^7}-\frac{19.44\times {10}^6}{1.8\times {10}^7}$ ...(ii)
from (i) and (ii)
$P=1.62\times {10}^6$ N
Substituting in (ii)
9 + 0.09 e - 1.08 = 20
e = 134.22 mm
Loss in prestress = 25%
$\therefore$ $P=0.75\times 1.62\times {10}^6$
$=1215\times {10}^3$ N
In final condition compression will be at top and tension will be at bottom.
$f_{top}=\frac{P}{A}-\frac{Pe}{Z}+\frac{M_d}{Z}+\frac{M_L}{Z}$
$1.5=\frac{1215\times {10}^3}{1.8\times {10}^5}-\frac{1215\times {10}^3\times 134.22}{1.8\times {10}^7}+1.08$
$+\frac{w\times 6\times {10}^6}{4\times 1.8\times {10}^7}$
$\therefore w=194.75kN$
$-1=\frac{1215\times {10}^3}{1.8\times {10}^5}+\frac{1215\times {10}^3\times 134.22}{1.8\times {10}^7}-1.08-\frac{w\times 6\times {10}^6}{4\times 1.8\times {10}^7}$
$f_{bottom}=\frac{P}{A}+\frac{Pe}{Z}-\frac{M_d}{Z}-\frac{M_d}{Z}+\frac{M_L}{Z}$
W = 188.75 kN
$\therefore$ The concentrated load at mid span = 194.75 kN.