In a RC beam of rectangular cross section the percentage of tensi

| In a RC beam of rectangular cross section the percentage of tension steel reinforcement provided is 0.5 and effective depth is 500 mm. The quantity of steel reinforcement provided such that the section behaves as under reinforced section (Working stress method) and stress induced in the steel is 140 Mpa. The corresponding stress induced in compression M 20 concrete is:

A. 5 MPa

B. 4.8 MPa

C. 4.2 MPa

D. 4.6 MPa

Please scroll down to see the correct answer and solution guide.

Right Answer is: D

SOLUTION

Calculation:

Given,

p_t = 0.5

d = 500 mm, t = 140 mm, M20

For under reinforced section: Steel will yield.

∴ σ_st = 140 MPa and let the stress in concrete is c.

For M20, m = 13.33

Actual Neutral axis distance:-

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E30! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E31! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E31! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)\(\frac{{{\sigma _{st}}}}{m} = \frac{c}{k}\left( {1 - k} \right)\)

\(\begin{array}{l} \frac{{140}}{{13.33}} = \frac{c}{k}\left( {1 - k} \right)\\ \frac{c}{k}\left( {1 - k} \right) = 10.503 \end{array}\)

Now, using under-reinforced beam equation,

\(\begin{array}{l} c = \frac{{2{\sigma _{st}}}}{{kbd}}\left( {\frac{{{p_t}}}{{100}}bd} \right)\\ c = \frac{{2 \times 140}}{k}\left( {\frac{{0.5}}{{100}}} \right) = \frac{{1.4}}{k}\\ k = \frac{{1.4}}{c} \end{array}\)

Putting k value in Eqn. \(\frac{c}{k}\left( {1 - k} \right) = 10.503\), we get

\({c^2} - 1.4c - 14.704 = 0\)

Solving for c, we get c = 4.6 MPa