Eng. Hasan Shehada Eng. Ruba Awadsite.iugaza.edu.ps/.../10/Lecture-2-Design-1-revsision-1.pdf ·...

Eng. Hasan Shehada Eng. Ruba Awad


Steps For Designing Reinforced Concrete buildings:

The design of any concrete structure goes through several steps. These steps lead at the end to a

designed building that must fulfill the requirements (Safety, serviceability, economy). These

steps are:

1- The definition of material properties, soil and site properties, and the structural system

that will be used.

2- Structural system must be divided into its elements (Slab, columns, foundation). Each

element must be put in a suitable way so that we accomplish all the requirements.

3- Calculation of loads.

4- Design of elements according the load calculated in the previous step.

5- Drawing all designed members.

These steps will be discussed in this lecture in details.

Definition of Structural System (Skeleton):

It means that the type of structural system that will be used must be chosen in the beginning.

There are many types of structures, but the most common type is the traditional frame system

(whether it is a moment resisting frame or not). In this course, we are focusing on this type of

structures.

This type mainly contains three elements which are slabs, columns, and footing. The transfer of

loads through this system is as follows:

1- Slabs get external loads (from users, environment, furniture,…). Slabs can be either solid

or ribbed slabs, whichever is suitable and more economic.

2- Slabs transfer loads to secondary beams.

3- Secondary beams transfer loads to main beams.

4- Main beams transfer loads to columns.

5- Columns transfer loads to footings.

6- Footings transfer loads to soil.

The figure below shows these steps.


Loads Calculations:

The building must be designed to resist the loads applied on it, multiplied by factors given by the

ACI code. But first we must calculate those loads. Load are classified into:

1- Dead load:

Dead loads in buildings include own weight, covering materials, and equivalent partition

load. The calculation of these loads will be discussed in details in the example below.

2- Live Load:

To find the live load applied on a building, other codes like UBC and IBC which are the most

commonly used codes. The value of live load varies according to the usage of the building.

For example, for a regular residential building live load is 150 kg/m2 while for schools it is

250 kg/m2. If the same building is used for multiple purposes, the maximum value must be

used.

3- Other loads:

This include seismic load, wind load, and other dynamic loads. Our concern in this course is

to find live and dead loads only.


Example:

The drawing above shows a plan view of a residential building. The columns are distributed as

shown. The following are the properties of the building:

F’c = 240 kg/m2 , Fy = 4200 kg/cm

2 , Height of floor = 3 m

Length of 10 cm thick walls = 40.1 m

For plaster works, thickness of plaster 1.5 cm, ( )

For tiles works:

Thickness of sand = 7 cm, ( ) .

Thickness of mortar = 2.5 cm, ( ) .

Thickness of tiles = 2.5 cm, ( ) .


To design this slab, these steps must be followed:

1- Define the system of building. We are going to use tradition non-moment resisting frame

for this building.

2- Define the distribution of ribs in the building. The drawing below shows the distribution

of ribs, secondary beams, and main beams on the slab.

(Notice that the ribs are directed in the short direction)

3- After defining all elements, we can now calculate the loads applied on the building. The

main goal is to convert all weights and loads to a united force (weight) per square unit of

area (in this example t/m2. To do so, the loads are categorized as follows:

Own Weight of ribs:

The drawing below shows a typical rib.


As we can we can notice, the thickness of slab is 28 cm. This value is calculated later in

this lecture in ribs design section. The block used is 40x25x20 cm depth so the topping

slab will be 8 cm thick. The width of rib is assumed 12 cm.

To calculate the own weight of slab, take a representative part of the slab that contains

one block and one rib (half rib from each side) as shown in the drawings above.

Vtotal = 0.52×0.25×0.30 = 0.039 m3

Vblock = 0.4×0.25×0.22 = 0.022 m3

Vconcrete= 0.039-0.022 = 0.017 m3

Wconcrete= 0.017×2.5 = 0.0425 ton

Wblock= 22/1000 = 0.022 ton

Total weight = 0.022+0.0425 = 0.0645 ton

Weight per unit area = 0.0645/(0.52*0.25) = 0.496 = 0.5 t/m2

Dead Load Due To Own Weight = 0.5 t/m2

Covering materials:

Covering material includes tiles work and plaster weights. The figure below shows a

profile for the materials used for both works and their thicknesses.


Plaster coat density = 2.1 t/m3, thickness = 0.015 m

Weight of plaster =0.015 * 2.1 = 0.0315 ton/m2

Sand density = 1.8 t/m3, thickness = 0.07 m

Weight of sand = 0.07* 1.8 = 0.126 ton/m2

Mortar density = 2.1 t/m3, thickness = 0.025 m

Weight of mortar = 0.025 * 2.1 = 0.0525 ton/m2

Tiles density = 2.1 t/m3, thickness = 0.025 m

Weight of tiles = 0.025 * 2.1 = 0.0525 ton/m2

Total weight = 0.2625 ton/m2

So weight of covering materials = 0.2625 ton/m2

Equivalent Partition load:

Equivalent partition load is load of internal walls distributed on the area of slab without

including the area of staircase.

This step must be done for easier calculation of loads applied on slab.


For 10 cm thick blocks:

No. of blocks per m2 = 12.5 blocks

Weight of blocks per area = 12.5 *10= 125 kg/m2 = 0.125 t/m²

Weight of plaster per area = 2*0.02*2.1 = 0.084 t/m²

Total weight = 0.125 + 0.084 = 0.209 t/m²

Total weight (t/mꞌ) = weight/m2*height of floor = 0.209*3 = 0.627 t/mꞌ

Total length = 40.1 m

Total weight (t) = 0.627 *40.1 = 25.2 t

Total area of building = 27.2*17.4 = 473.28 m2

Weight per unit area = 12.6027/473.28 = 0.0532 t/m2

Live Load:

Live Load can be brought from UBC for residential buildings = 150 kg/m2 = 0.15 t/m

2

Loads combination:

After finding the total dead load and live load, ACI uses load combination to enlarge the load

before the design.

Total Dead Load = 0.5 + 0.2625 + 0.0532 = 0.8157 t/m2

Total Factored Load = 1.2*DL + 1.6LL

Total Factored Load = 1.2*0.8157 + 1.6*0.15 = 1.2188 t/m2

This load is the applied load that the design will be based on.


Ribs Design:

After finding the value of total load, the elements are designed. Based on the mechanism of load

transfer, the ribs are the first elements to take the load applied. The design of it is based on three

requirements that must be fulfilled (deflection, shear, and flexure).

To fulfill deflection requirement, the following table that shows minimum thicknesses for ribs

and beams is used. When choosing this thickness, then deflection requirement is accomplished.

Element Simple supported O.E.C B.E.C Cantilever

Min. thickness L/16 L/18.5 L/21 L/8

O.E.C (one end continuous) B.E.C (both end continuous)

Where L is the length from center to center of rib or beam.

In the example:

Ribs at ends = 4.5/18.5 = 0.2432 = 24.32 cm

Ribs in the middle = 4.5/21 = 0.2143 = 21.43 cm

The largest is 24.32 cm

We can use a thickness of 25 cm, but we used 28 cm.

For shear design of ribs, we first find the load per meter length of it and then use one of the

structural analysis software to find the shear force and bending moment on rib.

To find the load per meter length, take a splice a shown in the figure above.

Load per length = 0.52*load = 0.52*1.2188 = 0.6338 t/m


Using SAP2000 analysis, we find the maximum shear and bending moments as follows:

Vmax = 1.653 ton

Mmax = 1.12 t.m

Now we find the value of shear that can be resisted by concrete (ΦVc):

√

D = 28 – 2 – 0.7 = 25.3 cm

√

ΦVc > Vu ok

NOTE:

If the maximum shear is greater than ΦVc, we have two choices:

1- Increase the depth of the slab, which is uneconomic.

2- Change the direction of blocks at maximum shear area so that the width of rib is

increased. The figure below describes this solution.


For Flexural Design:

Minimum ratio of reinforcement is given by:

√

√

So minimum reinforcement ratio = 0.0033

Maximum reinforcement ratio is given by:

( )

Now using the maximum moment, the ration of reinforcement is:

[ √

]

[ √

]

( ) ( )

Use 2Φ12 mm for each rib.

Check For spacing is ok.


Beams Design:

Beam design is the same as ribs design except that if concrete is not adequate for resisting shear

force, stirrups are used.

The following example shows a complete beam design (it is not related to the previous example).

The beam is 70 x 30 cm cross section. Design the beam for shear and moment.

- Effective depth d = 30 – 2.5 – 0.8 – 0.8 = 25.9 cm

Design for flexure:

layer Bottomin 149 Use

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Design for shear:

Checking beam strength for beam shear:-

Assuming Φ14 mm reinforcement bars and Φ 8 mm stirrups.

Vu max= 20.41 ton, but Vu critical= 17.68 ton.

Φ Vc = dbfc w '53.0

. 16.11

1000

9.257024053.075.0tons

Φ Vc < Vu…So is it not ok.

Ok isit ..So 591000

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i.e. Cross section dimension are adequate in terms of preventing brittle mode of failure.


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Though shear reinforcement is required, use 2 Ø8 mm per 15cm.

Eng. Hasan Shehada Eng. Ruba Awadsite.iugaza.edu.ps/.../10/Lecture-2-Design-1-revsision-1.pdf ·...

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