Chapter 4.7, Problem 1P

To determine

Find the preliminary concrete moment frame system for the reinforced concrete frame with concrete masonry units’ infill shear walls on the exterior of office building.

Find the amount of concrete and steel saved when using concrete masonry units and also find the additional structural dead load added by the concrete masonry units.

Answer to Problem 1P

The shape and the size of the column is $\underset{\_}{\text{square}-\left(14\text{in}\text{.}\times 14\text{in}\text{.}\right)}$.

The dimension of the beam is $\underset{\_}{14\text{in}\text{.}\times \text{28}\text{in}\text{.}}$.

The dimension of the joist is $\underset{\_}{12\text{in}\text{.}\times \text{18}\text{in}\text{.}}$.

The slab thickness using in the project is $\underset{\_}{8\text{in}\text{.}}$.

The percentage saved in the concrete quantity is $\underset{\_}{48\%}$.

The percentage saved in the steel quantity is $\underset{\_}{22\%}$.

The structural dead load added by the CMU unit is $\underset{\_}{6,080\text{kips}}$.

Explanation of Solution

Given information:

The floor plan is L in shape.

The number of stories in the building is 4 with a roof.

The floor-to-floor height between the first and second stories is 24 ft.

The floor-to-floor height for the upper floors is 15 ft.

The thickness of reinforced block used in this project is 8 in..

The percentage covered by the reinforced block in the building’s exterior walls between the beams and columns is 60 %.

Calculation:

The estimation of diaphragm of the office building has done by calculating the diaphragm of the two rectangles in the given floor plan. The dimension of the first rectangle is $\left(100\times 210\right)\text{ft}$ and other dimension of second rectangle is $\left(80\times 140\right)\text{ft}$.

Determine the length to width ratio ${\left(\frac{l}{b}\right)}_{R1}$ of the first rectangle.

Substitute 100 ft for l and 210 ft for b.

$\begin{array}{c}{\left(\frac{l}{b}\right)}_{R1}=\frac{100}{210}\\ =\text{1:2}\text{.1}\end{array}$

Determine the length to width ratio ${\left(\frac{l}{b}\right)}_{R2}$ of the first rectangle.

Substitute 80 ft for l and 140 ft for b.

$\begin{array}{c}{\left(\frac{l}{b}\right)}_{R2}=\frac{80}{140}\\ =\text{1:1}\text{.75}\end{array}$

The value of length to width ratio of two rectangles is less than the maximum ratio required for an optimal diaphragm (1:3). Therefore, the diaphragm should work with the given condition.

Column–beam layout:

The floor plan has the irregularities resulting the column–beam layout will not produce equal structural bays. The aim of the design is to produce a floor plan with the fewest number of columns is suitable given that the floor-to-floor heights are sufficient to handle deeper beams.

The dimension of $\left(100\times 130\right)\text{ft}$ segment can be divided into three bays of 33 ft 4in. by three bays of 43 ft 4 in..

The dimension of $\left(100\times 80\right)\text{ft}$ segment can be divided into three bays of 33 ft 4in. by two bays of 40 ft.

The dimension of $\left(40\times 80\right)\text{ft}$ segment can be divided into one bay of 40 ft by two bays of 40 ft. At last the three concrete joists per primary beam will be used to support the floor.

Column size:

Refer Table 4.3, “Determinates for column dimensions” in the textbook.

As per Table 4.3, the project has the four stories and column spacing of 24 ft for first and second floor and 15 ft for other floors defines that the project is moderate project.

The project is falls under the large project due to the column spacing 39 ft is greater than the 36 ft and 4 stories makes it a moderate project.

Refer Table 1.3, “Minimum floor live loads” in the textbook.

Take the uniformly distributed load as 50 psf for offices, 100 psf for lobbies and first-floor corridors, and 80 psf for corridors above first floor.

The offices have the floor live load of 50 psf, which would be a modest project. Lobbies and first floor corridors have the floor live load of 100 psf, and corridors above first floor have the 80 psf, which would be a moderate project.

For the moderate project, take the moderate column size of 14 in. with the square shape.

Therefore, the shape and the size of the column is $\underset{\_}{\text{square}-\left(14\text{in}\text{.}\times 14\text{in}\text{.}\right)}$.

Beam, joist, and floor slab dimensions:

The width of beam is typically same as the width of the column. The beam width is taken as 14 in.

The depth of the concrete beam is approximately equal to 2 times the beam width. Hence, the depth of beam is $2\times 14=28\text{in}\text{.}$. The dimension of the beam is $14\text{in}\text{.}\times \text{28}\text{in}\text{.}$.

Therefore, the dimension of the beam is $\underset{\_}{14\text{in}\text{.}\times \text{28}\text{in}\text{.}}$.

The joist widths are usually 2 in. less the width of the secondary beam. The joist width is taken as $14-2=12\text{in}\text{.}$. The joist depth is approximately equal to 1.5 times its width. Hence, the joist depth is $1.5\times 12=18\text{in}\text{.}$.

Therefore, the dimension of the joist is $\underset{\_}{12\text{in}\text{.}\times \text{18}\text{in}\text{.}}$.

The three concrete joists per primary beam will be supporting the floor.

Take the slab thickness of 8 in. due to the usage of beam and joist.

Therefore, the slab thickness using in the project is $\underset{\_}{8\text{in}\text{.}}$.

The first and second floor has the column height of 24 ft and upper floor has the column height of 15 ft.

Find the area of the column using the relation.

$A_c=b_c\times d_c$

Here, $b_c$ is the width of the column and $d_c$ is the depth of the column.

Substitute 14 in. for $b_c$ and 14 in. for $d_c$.

$\begin{array}{c}{A}_c=14\times 14\\ =196\text{in}{\text{.}}^{\text{2}}\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =1.36{\text{ft}}^{\text{2}}\end{array}$

Find the area of the beam using the relation.

$A_b=b_b\times d_b$

Here, $b_b$ is the width of the beam and $d_b$ is the depth of the beam.

Substitute 14 in. for $b_b$ and 28 in. for $d_b$.

$\begin{array}{c}{A}_b=14\times 28\\ =392\text{in}{\text{.}}^{\text{2}}\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =2.72{\text{ft}}^{\text{2}}\end{array}$

Find the area of the joist using the relation.

$A_j=b_j\times d_j$

Here, $b_j$ is the width of the joist and $d_j$ is the depth of the joist.

Substitute 12 in. for $b_j$ and 18 in. for $d_j$.

$\begin{array}{c}{A}_j=12\times 18\\ =216\text{in}{\text{.}}^{\text{2}}\times {\left(\frac{1\text{ft}}{12\text{in}\text{.}}\right)}^2\\ =1.5{\text{ft}}^{\text{2}}\end{array}$

Consider the height of the column in the ground floor and the above the roof.

Determine the total length of the column using the relation.

$\begin{array}{c}\text{Total}\text{length}=\left(24\times 2+15\times 4\right)\\ =108\text{ft}\end{array}$

The number of primary beams is 5.

Determine the total length of the beam using the relation.

$\begin{array}{c}\text{Total}\text{length}=4\times 210+1\times 80\\ =920\text{ft}\end{array}$

The number of joist is 10.

Determine the total length of the joist using the relation.

$\begin{array}{c}\text{Total}\text{length}=10\times 100+4\times 40\\ =1,160\text{ft}\end{array}$

Quantity of concrete and steel:

Show the estimation of concrete and steel quantities as in Table (1).

Element	Quantity of Concrete $\left({\text{ft}}^{\text{3}}\right)$	Average % of steel	Quantity of steel $\left({\text{ft}}^{\text{3}}\right)$
Columns	$\begin{array}{l}\left\{\begin{array}{l}1.36{\text{ft}}^{\text{2}}\times 108\text{ft}\times \\ 27\text{columns}\end{array}\right\}\\ =3,966\end{array}$	2 %	79 (17 tons)
Beams	$\begin{array}{l}\left\{\begin{array}{l}2.72{\text{ft}}^{\text{2}}\times 920\text{ft}\times \\ 5\text{floors}\end{array}\right\}\\ =12,512\end{array}$	4 %	501(112 tons)
Joists	$\begin{array}{l}\left\{\begin{array}{l}1.5{\text{ft}}^{\text{2}}\times 1,160\text{ft}\times \\ 5\text{floors}\end{array}\right\}\\ =8,700\end{array}$	4 %	348 (78 tons)
Slabs	$\begin{array}{l}\left\{\begin{array}{l}0.67\text{ft}\times 24,200{\text{ft}}^2\times \\ 5\text{floors}\end{array}\right\}\\ =81,070\end{array}$	0.25 %	202.7 (45 tons)
	$106,248{\text{ft}}^{\text{3}}$ (7,240 tons)		1,130 (252 tons)
CMU Reinforced blocks	$\begin{array}{l}\left\{\begin{array}{l}0.67\text{ft}\times \frac{60}{100}\times 24,200{\text{ft}}^2\times \\ 5\text{floors}\end{array}\right\}\\ =48,642\end{array}$	0.25 %	122
Total quantity	$\begin{array}{l}106,248{\text{ft}}^{\text{3}}-48,642{\text{ft}}^{\text{3}}\\ =57,606{\text{ft}}^{\text{3}}\left(3,926\text{tons}\right)\end{array}$		$\begin{array}{l}1,130-122\\ =1,008\left(225\text{tons}\right)\end{array}$

Refer Table 4.6, “Estimates of concrete and steel quantities for Example 3” in the textbook.

Refer Table 1.

Determine the percentage saved in the concrete quantity using the relation.

$\begin{array}{c}{P}_c=\frac{111,420-57,606}{111,420}\times 100\\ =48\%\end{array}$

Therefore, the percentage saved in the concrete quantity is $\underset{\_}{48\%}$.

Determine the percentage saved in the steel quantity using the relation.

$\begin{array}{c}{P}_s=\frac{1,285-1,008}{1,285}\times 100\\ =22\%\end{array}$

Therefore, the percentage saved in the steel quantity is $\underset{\_}{22\%}$.

Determine the structural dead load added by the CMU unit using the relation.

$\begin{array}{c}DL=48,642\times 125{\text{lb/ft}}^{\text{3}}\\ =6,080,250\text{lb}\times \frac{1\text{kip}}{1,000\text{lb}}\\ =6,080\text{kips}\end{array}$

Therefore, the structural dead load added by the CMU unit is $\underset{\_}{6,080\text{kips}}$.