Chapter 6, Problem 6.3.16P

A reinforced concrete slab (see figure) is reinforced with 13-mm bars spaced 160 mm apart at d = 105 mm from the top of the slab. The modulus of elasticity for the concrete is E_c= 25 GPa, while that of the steel is £_s = 200 G Pa. Assume that allowable stresses for concrete and steel arecr_ac = 9.2 MPa and u _s = 135 MPa.

l()5 mm

1. Find the maximum permissible positive bending moment for a l-m wide strip of the slab.

What is the required area of steel reinforcement, A^ if a balanced condition must be achieved? What is the allowable positive bending moment? (Recall that in a balanced design, both steel and concrete reach allowable stress values simultaneously under the design moment.)

  

To determine

The maximum permissible positive bending and balanced condition for both steel and concrete.

Answer to Problem 6.3.16P

The maximum permissible positive pending moment is$10.193KN.m.$

Explanation of Solution

Given: .

  $E_s$=200GP

  $E_c$ =25GPa

D =0.13m

Calculation: .

The modular ratio of the momentum is,

  $n=\frac{E_s}{E_c}$.

  $\begin{array}{l}n=\frac{200}{25}\\ n=8\end{array}$.

The number of reinforcement bar provided is,

  $N=\frac{b}{s}$.

Substitute b=100mm, s=160mm

  $\begin{array}{l}N=\frac{1000}{160}\\ N=6\end{array}$.

The molecular ratio of the moment is$n=8.$.

The area of reinforcement bar is stated as,

  $\begin{array}{l}{A}_s=N\times \frac{\pi \times D^2}{4}\\ A_s=6\times \frac{\pi \times {0.013}^2}{4}\\ A_s=7.964\times {10}^{-4}{m}^2\end{array}$.

Moment of area of concrete section = Moment of area of steel about neutral Axis..

  $\begin{array}{l}by\times \frac{y}{2}=n{A}_s\left(d-y\right)\\ \frac{1}{2}.b{y}^2=n{A}_s\left(d-y\right)\\ \frac{b{y}^2}{2}+n{A}_{s}y-nAd=0\end{array}$ ..

Using Quadratic formula,

  $x=\frac{-q\pm \sqrt{{\left(q\right)}^2}-\left(4pr\right)}{2p}$.

Substitute p=$\frac{b}{2}$, q=$n{A}_s$r=$n{A}_{s}d$.

  $y=\frac{-n{A}_s\pm \sqrt{{\left(n{A}_s\right)}^2}+\left(4\times \frac{b}{2}\times n{A}_{s}d\right)}{2\times \frac{b}{2}}$.

  $\begin{array}{l}y=\frac{\left(8\times 7.964\times {10}^{-4}\right)\pm \sqrt{\left(8\times 7.964\times {10}^{-4}\right)}+\left(4\times \frac{1}{2}\times 8\times 7.964\times {10}^{-4}\times 0.105\right)}{2\times \frac{1}{2}}\\ y=\frac{\left(-6.3712\times {10}^{-3}\right)+0.0371}{1}\\ y=0.0308m\end{array}$.

The location of neutral axis for the top layer, y =0.0308m.

Moment of inertia of the transformed section is,

  $I_T=\frac{1}{3}b{y}^3+n{A}_S{(d-y)}^2$.

  $\begin{array}{l}{I}_T=\frac{1}{3}\times 1\times {\left(0.0308\right)}^3+8\times 7.964\times {10}^{-4}{\left(0.105-0.0308\right)}^2\\ I_T=4.4817\times {10}^{-5}{m}^4\end{array}$.

The bending moment at the top layer of concrete..

  $M_c=\frac{{\sigma }_{ac}}{y}{I}_T$.

Substitute y=0.0308m,$I_T$ =$4.487\times {10}^{-5}$.

${\sigma }_{ac}$ ,=$\left(9.2\times {10}^6\right)$.

  $M_c=\frac{\left(9.2\times {10}^6\right)}{0.0308}\left(4.487\times {10}^{-5}\right)$.

  $M_c=13.41KN.m$.

The bending moment at the top layer of concrete..

  $M_s=\frac{{\sigma }_{as}}{n\left(d-y\right)}{I}_T$.

Substitute the value

  $M_s=\frac{\left(135\times {10}^6\right)}{8\times \left(0.105-0.0308\right)}\left(4.487\times {10}^{-5}\right)$.

  $M_s=10.193KN.m$.

Conclusion: .

The maximum permissible positive pending moment is$10.193KN.m.$

To determine

To find: The allowable value of moment in steel in positive pending.

Answer to Problem 6.3.16P

The allowable value of moment in steel in positive pending is$15.77KN.m.$

Explanation of Solution

Calculation: .

Substitute y equation above

  $\frac{b\times \frac{-n{A}_s\pm \sqrt{{\left(n{A}_s\right)}^2}+\left(4\times \frac{b}{2}\times n{A}_{s}d\right)}{2\times \frac{b}{2}}\times {\sigma }_{ac}}{2}=A_s\times {\sigma }_{as}$.

  $\frac{b\times \frac{-8A_s\pm \sqrt{{\left(8A_s\right)}^2}+\left(4\times \frac{1}{2}\times 8A_{s}d\right)}{b}\times {\sigma }_{ac}}{2}=A_s\times {\sigma }_{as}$.

Substitute the value n=8,${\sigma }_{ac}$ =$9.2\times {10}^6$ , b=1m,d=0.105m,${\sigma }_{as}$ =$135\times {10}^6.$.

  $\frac{\frac{-8A_s\pm \sqrt{{\left(8A_s\right)}^2}+\left(4\times \frac{1}{2}\times 8A_s\times 0.105\right)}{}\times 9.2\times {10}^6}{2}=A_s\times 135\times 10$.

  $\frac{\frac{-8A_s\pm \sqrt{{\left(8A_s\right)}^2}+\sqrt{\left(1.68A_s\right)}}{}\times 9.2\times {10}^6}{2}=A_s\times 135\times {10}^6$.

  $\left[\pm \sqrt{{\left(8A_s\right)}^2}+\sqrt{\left(1.68A_s\right)}\right]=37.3478A_s$.

Take squaring on both sides,

  ${\left(8A_s\right)}^2+\left(1.68A_s\right)={37.3478}^2{A}_s{}^2$.

  $\begin{array}{l}{\left(1,330.8582A_s\right)}^2-\left(1.68A_s\right)=0\\ A_s=1.2623\times {10}^{-3}{m}^2\end{array}$.

Required area of steel reinforcement$A_s=1.2623\times {10}^{-3}{m}^2.$.

Moment of area of concrete section = Moment of area of steel about neutral Axis..

  $\begin{array}{l}by\times \frac{y}{2}=n{A}_s\left(d-y\right)\\ \frac{1}{2}.b{y}^2=n{A}_s\left(d-y\right)\\ \frac{b{y}^2}{2}+n{A}_{s}y-nAd=0\end{array}$.

Use quadratic equation..

  $x=\frac{-q\pm \sqrt{{\left(q\right)}^2}-\left(4pr\right)}{2p}$.

Substitute p=$\frac{b}{2}$, q=$n{A}_s$r=$n{A}_{s}d.$.

  $y=\frac{-n{A}_s\pm \sqrt{{\left(n{A}_s\right)}^2}+\left(4\times \frac{b}{2}\times n{A}_{s}d\right)}{2\times \frac{b}{2}}$.

  $y=$.

  $\frac{\left(-8\times 1.2623\times {10}^{-3}\right)\pm \sqrt{\left(8\times 1.2623\times {10}^{-3}\right)}+\left(4\times \frac{1}{2}\times 8\times 1.2623\times {10}^{-3}\times 0.105\right)}{2\times \frac{1}{2}}$.

  $y=\frac{\left(-0.01\right)+0.0471}{1}$.

  $y=0.0371m$.

The allowable positive pending moment in concrete is$15.77KN.m.$.

  $\begin{array}{l}{I}_T=\frac{1}{3}\times 1\times {\left(0.0371\right)}^3+8\times 1.2623\times {10}^{-3}{\left(0.105-0.0371\right)}^2\\ I_T=6.3579\times {10}^{-5}{m}^4\end{array}$.

The allowable positive pending moment in concrete is,

  $M_c=\frac{{\sigma }_{ac}}{y}{I}_T$.

Substitute the value y=0.0371m,${\sigma }_{ac}$ =$9.2\times {10}^6$ ,$I_T$ =$\left(6.3579\times {10}^{-5}\right)$.

  $\begin{array}{l}{M}_c=\frac{\left(9.2\times {10}^6\right)}{0.0371}\left(6.3579\times {10}^{-5}\right)\\ M_c=15.77KN.m\end{array}$.

The allowable positive pending moment in steel is,

  $M_c=\frac{{\sigma }_{as}}{ny}{I}_T$.

  $\begin{array}{l}{M}_c=\frac{\left(135\times {10}^6\right)}{8\times 0.0371}\left(6.3579\times {10}^{-5}\right)\\ M_c=28.92KN.m\end{array}$.

Conclusion: .

The lesser value of moment in concrete and steel as allowable value of moment..

Allowable positive bending moment in steel is$15.77KN.m.$