Chapter 9, Problem 9.1.1P

To determine

(a)

The moment of inertia of the transformed section.

Answer to Problem 9.1.1P

$1761.00{\left(inches\right)}^4$

Explanation of Solution

Given:

A W 18 X 40 floor beam, the 28-day compressive strength of the concrete is $f_C^{\text{'}}$

$4ksi$, 4-inch-thick reinforced concrete slab with an effective width b of 81 inches.

Calculation:

We have the modulus of elasticity of concrete as follows:

$E_C={\left(w_C\right)}^{\frac{3}{2}}\sqrt{f_C^{\text{'}}}ksi$

Where, the modulus of elasticity of concrete is $E_C$,

unit weight of concrete is $w_C$ and

the 28-day compressive strength of concrete is $f_C^{\text{'}}$.

Substitute $145\frac{lb}{f{t}^3}$ for $w_C$ and $4ksi$ for $f_C^{\text{'}}$.

$\begin{array}{l}{E}_C={\left(145\frac{lb}{f{t}^3}\right)}^{\frac{3}{2}}\sqrt{4}ksi\\ E_C=1746.03\frac{lb}{f{t}^3}\times 2ksi\\ E_C=3492.06ksi.\end{array}$

Modular ratio by using the following formula:

$n=\frac{E_S}{E_C}$

Where, the modulus of elasticity of concrete is $E_C$,

the modulus of elasticity of steel is $E_S$,

and n is the modular ratio.

$n=\frac{E_S}{E_C}$

Substitute $3492.06ksi$ for $E_C$ and $29000ksi$ for $E_S$.

$\begin{array}{l}n=\frac{29000ksi}{3492.06ksi}\\ n=8.304\approx 8\end{array}$

Now the section will be transformed.

Since the modulus of elasticity of concrete can only be approximated, the usual practice of rounding n to the nearest whole number is sufficiently accurate. Thus,

The transformed width of the section is as following:

$b_t=\frac{b}{n}$

Where, $b_t$ is the transformed width,

$b$ is the width of the section and

$n$ is the modular ratio.

Substitute $81$ for $b$ and $8$ for $n$.

$\begin{array}{l}{b}_t=\frac{81}{8}\\ b_t=10.125in\end{array}$

Following is the section for the given set of conditions:

Data from the steel code:

DesignationImperial (in x lb/ft)	Depthh (in)	Widthw (in)	Web Thicknesstw (in)	Flange Thicknesstf (in)	Sectional Area (in2)	Weight (lbf/ft)	Static Parameters
							Moment of Inertia		Elastic Section Modulus
							Ix (in4)	Iy (in4)	Sx (in3)	Sy (in3)
W 18 x 40	17.9	6.02	0.315	0.525	11.8	40	612	19.1	68.4	6.4

The transformed section is shown in the above figure. The neutral axis is not known yet whether it lies in the steel or the concrete.

The location of the neutral axis can be found by applying the principle of moments with the axis of moments at the top of the slab. The computations are summarized in Table, and the distance from the top of the slab to the centroid is:

Component	$A{\left(inches\right)}^2$	$Y\left(inches\right)$	$AY{\left(inches\right)}^3$
Concrete	$40.5$	$2.0$	$81.0$
W 18 X 40	$11.8$	$12.95$	$152.81$
Sum	$52.3$		$233.81$

The distance of centroid of the section is as follows:

$\overline{Y}=\frac{{\displaystyle \sum A\times y}}{{\displaystyle \sum A}}$

Substitute $233.81{\left(inches\right)}^3$ for ${\displaystyle \sum A\times y}$ and $52.3{\left(inches\right)}^2$ for ${\displaystyle \sum A}$

$\begin{array}{l}\overline{Y}=\frac{233.81{\left(inches\right)}^3}{52.3{\left(inches\right)}^2}.\\ \overline{Y}=4.4705inches.\end{array}$

Since this is greater than 4 inches (the thickness of the slab) the neutral axis lies below the slab in the web. Applying the parallel axis theorem and tabulating the computations in table, we obtain the moment of inertia of the transformed section as:

Component	$A{\left(inches\right)}^2$	$Y\left(inches\right)$	$\overline{I}$${\left(inches\right)}^4$	$d=\overline{Y}-Y$$\left(inches\right)$	$I_{transformed}=\overline{I}+A{d}^2$${\left(inches\right)}^4$
Concrete	$40.5$	$2.0$	$54$	$2.4705$	$301$
W 18 X 40	$11.8$	$12.95$	$612$	$8.4795$	$1460$
Sum	$52.3$				$1761$

Conclusion:

Therefore, total moment of inertia of the transformed section is $1761.00$

${\left(inches\right)}^4$.

To determine

(b)

The stress at the top of the steel (indicate whether tension or compression), the stress at the bottom of the steel, and the stress at the top of the concrete.

Answer to Problem 9.1.1P

$\underset{\_}{f_s=0.927ksi\left(compressive\right)},\underset{\_}{f_s=34.44ksi\left(Tensile\right)}and\underset{\_}{f_s=1.104ksi\left(compressive\right)}.$

Explanation of Solution

Given:

Positive service load moment of 290 ft-kips.

Calculation:

For the computation of the stress at top of the steel is as following:

$f_s=\frac{My}{I_{transformed}}$

Where, the stress at the top of the steel is $f_S$, the moment acting on the composite section is M, the distance from neutral axis to the top of steel is Y and the moment of inertia of the transformed section is I.

The distance from neutral axis to top of steel section is as follows:

$\begin{array}{l}y=\overline{y}-t\\ y=4.4705-4\\ y=0.4705\end{array}$

Here the thickness of the concrete slab is t.

Compute the stress at top of the steel as:

$\begin{array}{l}{f}_s=\frac{M\left(\overline{y}-t\right)}{I_{transformed}}\\ f_s=\frac{\left(290\times 12\right)in.kips\left(0.4705\right)in}{1761.00i{n}^4}\\ f_s=0.927ksi.\end{array}$

As, the centroid lies below the top of the steel, the stress is compressive.

Now, the stress at the bottom of the steel is as follows:

$f_s=\frac{My}{I_{transformed}}$

The distance from the neutral axis to bottom of steel section.

$\begin{array}{l}y=d_w+t-\overline{y}\\ y=17.9+4-4.4705\\ y=\mathrm{17.43.}\end{array}$

Where, $d_w$ is the depth of the web.

$\begin{array}{l}{f}_s=\frac{M(d_w+t-\overline{y})}{I_{transformed}}\\ f_s=\frac{\left(290\times 12\right)in.kips\left(17.43\right)in}{1761.00i{n}^4}\\ f_s=34.44ksi.\end{array}$

As, the centroid lies above the top of steel, the stress is tensile.

Stress at the top of the slab is as follows:

$f_s=\frac{My}{n{I}_{transformed}}$

$y=\overline{y}$

Where, n is the modular ratio is n.

$\begin{array}{l}{f}_s=\frac{\left(290\times 12\right)in.kips\left(4.4705\right)in}{8\times 1761.00i{n}^4}\\ f_s=\frac{\left(15557.34\right)i{n}^2.kips}{14088.00i{n}^4}\\ f_s=1.104ksi.\end{array}$

As the concrete slab is above the neutral axis, hence the stress is compressive.

Conclusion:

Therefore, the stress at top of steel section is $f_s=0.927ksi$ (compressive), the stress at bottom of steel section is $f_s=34.44ksi$ (Tensile) and the stress at the top of the slab is $f_s=1.104ksi$ (compressive).