design of structure-concrete part...

Design of Structure-Concrete Part /Lecture

Loads

Loads that act on structures can be divided into three broad categories; dead loads, live

loads and environmental loads.

Dead Loads

Dead loads are those that are constant in magnitude and fixed in location throughout the

lifetime of the structure. Usually the major part of the dead load is the weight of the

structure itself. This can be calculated with good accuracy from the design configuration,

dimensions of the structure, and density of the material. For buildings, floor fill, finish floors,

and plastered ceilings are usually included as dead loads, and an allowance is made for

suspended loads such as piping and lighting fixtures. For bridges, dead loads may include

wearing surfaces, sidewalks, and curbs, and an allowance is made for piping and other

suspended loads.

Live Loads

Live loads consist chiefly of occupancy loads in buildings and traffic loads in bridges. They

may be either fully or partially in place or not present at all, and may also change in location.

Their magnitude and distribution at any given time are uncertain, and even their maximum

intensities throughout the lifetime of the structure are not known with precision. The

minimum live loads for which the floors and roof of a building should be designed are

usually specified in the building code that governs at the site of construction.

Representative values of minimum live loads to be used in a wide variety of buildings are

found in Minimum Design Loads for Buildings and Other Structures.

Live loads for highway bridges are specified by the American Association of State Highway

and Transportation Officials (AASHTO) in its LRFD bridge Design Specifications. For railway

bridges, the American Railway Engineering and Maintenance-of-way Association (AREMA)

has published the Manual of Railway engineering, which specifies traffic loads.

Environmental Loads

Environmental loads consist mainly of snow loads, wind pressure and suction, earthquake

loads (i.e. inertia forces caused by earthquake motions),soil pressures on subsurface

portions of structures, loads from possible ponding of rainwater on flat surfaces, and forces

caused by temperature differentials. Like live loads, environmental loads at any given time

are uncertain in both magnitude and distribution.

Design Codes and Specifications

The design of concrete structures is generally done within the framework of codes giving

specific requirements for materials, structural analysis, member proportioning, etc. The


International Building Code is an example of consensus code governing structural design

and is often adopted by local municipalities. The responsibility of preparing material-specific

portions of the codes rests with various professional groups, trade associations, and

technical institutes.

The American Concrete Institute (ACI) has long been a leader in such efforts. As one part of

its activity, the American Concrete Institute has published the widely recognized Building

Code Requirements for Structural Concrete and Commentary, which serves as a guide in the

design and construction of reinforced concrete buildings. The ACI Code has no official status

in itself. However, it is generally regarded as an authoritative statement of current good

practice in the field of reinforced concrete. As a result, it has been incorporated into the

International Building Code and similar Codes, which in turn are adopted by law into

municipal and regional building codes that do have legal status. Its provisions thereby attain,

in effect, legal standing. Most reinforced concrete buildings and related construction in the

United States are designed in accordance with the current ACI Code. It has also served as a

model document for many other countries. The American Concrete Institute also publishes

important journals and standards, as well as recommendations for the analysis and design

of special types of concrete structures such as tanks.

Most highway bridges in the United States are designed according to the requirements of

the AASHTO bridge specifications which not only contain the provisions relating to the loads

and load distributions, but also include detailed provisions for the design and construction

of concrete bridges. Many of the provisions follow ACI Code provisions closely, although a

number of significant differences will be found.

The design of railway bridges is done according to the specifications of the AREMA Manual

of Railway Engineering. It, too, is patterned after the ACI Code in most respects, but it

contains much additional material pertaining to railway structures of all types.

No code or design specification can be construed (interpret in a particular way) as a

substitute for sound engineering judgement in the design of concrete structures. In

structural practice, special circumstances are frequently encountered where code provisions

can serve only as a guide, and the engineer must rely upon a firm understanding of the basic

principles of structural mechanics applied to reinforced or prestressed concrete, and an

intimate knowledge of the nature of materials.

Loads Supported by Beams

In a structural system, slabs are to be supported by beams at a regular spacing provided in

one or two mutually perpendicular directions. Each beam has to support a particular area

and width of the slab that determines the total load acting on that beam. If the load acting

on the beam is not UDL, the calculations become very lengthy especially in case of

continuous beams. In such cases, usually an equivalent UDL is calculated which gives the


same maximum bending moment as in case of the actual load for a simply supported beam.

For triangular loads with maximum ordinate at the midspan and trapezoidal loads, the

actual load on the beam is to be increased by some factor and is then distributed over the

length of the beam. This increased load may not give the true shear force for the beam.

Beams Supporting One-Way Slabs

If a slab panel has its length over width ratio greater than two, it is called a one-way slab. In

case of one-way slabs supported on beams; the width of slab supported by each interior

beam is equal to the spacing of the beams as shown in the figure. Similarly, the width

supported by an exterior beam in such cases will be equal to half of the spacing between

the beams.

𝑊𝑖𝑑𝑡ℎ 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑏𝑦 𝑎𝑛 𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑏𝑒𝑎𝑚 = 𝑙𝑥

Width supported by an exterior beam = 𝑙𝑥

2+ 𝑐𝑎𝑛𝑡𝑖𝑙𝑒𝑣𝑒𝑟 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 , 𝑖𝑓 𝑎𝑛𝑦

𝑤ℎ𝑒𝑟𝑒 𝑙𝑥 = 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚𝑠 𝑎𝑛𝑑

𝑤ℎ𝑒𝑟𝑒 𝑙𝑦 = 𝑠𝑝𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚𝑠

Fig.3.12. Width of One-Way Slabs Supported by Beams

Beams Supporting Two-Way Slabs

If a slab panel has its length over width ratio lesser than or equal to two, it is called a two

way slab. The beams are usually provided in the two mutually perpendicular directions for

such slab systems. The area contributing load to each beam may be assumed to have 45

degree angles from all the beams, as shown in the figure. This will cause some difference

from the actual situation in case one end is simply supported.


Fig.3.13.Width of Two-Way slabs Supported by Beams

Slab Width supported by interior small beams (B1)

The area, contributing to one interior small beam is as below.

Actual area Supported=𝑙𝑥

2

2

This area exerts a triangular load on the beam. To get an equivalent UDL on the beam that

gives the same maximum bending moment in simply supported conditions, it is multiplied

by the factor 4/3.

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 =

43

×𝑙𝑥

2

2𝑙𝑥

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 =2

3𝑙𝑥

For first interior beams, the above width is to be increased by 10 %, if exterior beams do not

have a significant cantilever slab portion.

Slab width Supported by exterior smaller beams (B2)

The width supported by the exterior beams will simply be half as that of the interior beams.

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 =1

3𝑙𝑥

Slab Width supported by interior longer beams (B3)

The area, contributing load to one interior longer beam, is shown in Fig. 3.13.

Actual area Supported=𝑙𝑥 × 𝑙𝑦 −𝑙𝑥

2

2


This area exerts a trapezoidal load on the beam and to approximately convert it into an

equivalent UDL, the load is to be increased by a factor given below.

𝐹 =1 −

𝑅2

3

1 −𝑅2

Where 𝑅 =𝑙𝑥

𝑙𝑦

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝐹 ×

(𝑙𝑥 × 𝑙𝑦 −𝑙𝑥

2

2 )

𝑙𝑦

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝐹 × (1 −𝑙𝑥

2𝑙𝑦) 𝑙𝑥

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝐹 × (1 −𝑅

2) 𝑙𝑥

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 =1 −

𝑅2

3

1 −𝑅2

× (1 −𝑅

2) 𝑙𝑥

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (1 −𝑅2

3) × 𝑙𝑥

Slab Width supported by exterior longer beams (B4)

The width supported by the exterior beams will simply be half as that of the interior beams.

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑠𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑓𝑜𝑟 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑏𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡

=1 (1 −

𝑅2

3 ) × 𝑙𝑥

2+ 𝑐𝑎𝑛𝑡𝑖𝑙𝑒𝑣𝑒𝑟 𝑤𝑖𝑑𝑡ℎ, 𝑖𝑓 𝑎𝑛𝑦

The effective slab width supported by a beam multiplied with the load per unit area of the

slab gives the load on the beam per unit length.

Walls Supported By Beams

The load of wall on the beams may easily be calculated by multiplying the unit weight of

brickwork with the volume of brick wall per unit length as follows.

𝑡𝑤 = 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙 𝑖𝑛 𝑚𝑚

𝐻 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑙𝑙 𝑖𝑛 𝑚

𝑈𝐷𝐿 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = 𝑢𝑛𝑖𝑡 𝑤𝑡. 𝑜𝑓 𝑤𝑎𝑙𝑙 × 𝑡𝑤 × 𝐻 × 𝐿


𝑈𝐷𝐿 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = (1930 ×9.81

1000 ) ×

𝑡𝑤

1000× 𝐻 × 𝐿

𝑈𝐷𝐿 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = (1930 ×9.81

1000 ) ×

𝑡𝑤

1000× 𝐻 × 𝐿

𝐴𝑠 𝑤𝑒 𝑎𝑟𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡ℎ 𝑤𝑒𝑖𝑔ℎ𝑡 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐿 = 1 𝑚

𝑈𝐷𝐿 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = 0.019 𝑡𝑤𝐻 𝑘𝑁𝑚⁄ (SI)

𝑈𝐷𝐿 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 = 0.12 𝑡𝑤𝐻 𝑘𝑓𝑡⁄ (FPS)

Calculation of Beam Loads

Slab Loads

The slab load is always calculated as the load acting on a unit area (1 m2 ) of the slab.

Hence, the dead load may be calculated by multiplying the thickness with the unit weight of

various materials exerting load on the slab. The following two examples explain the

calculation of the roof loads:

Top Roof: For 125 mm R.C. slab, 100 mm earth filling and brick tiles (38 mm thick), the slab

load may be calculated as under:

𝑅. 𝐶. 𝑆𝑙𝑎𝑏: 125

1000× 2400 = 300

𝑘𝑔

𝑚2

𝐸𝑎𝑟𝑡ℎ 𝑓𝑖𝑙𝑙𝑖𝑛𝑔: 100

1000× 1800 = 180

𝑘𝑔

𝑚2

𝐵𝑟𝑖𝑐𝑘 𝑇𝑖𝑙𝑒𝑠: 38

1000× 1930 = 74

𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑: = 554 𝑘𝑔

𝑚2

𝐺𝑖𝑣𝑒𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑: = 200 𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1.2 𝐷𝐿 + 1.6 𝐿𝐿

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1.2 (554) + 1.6(200)

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 984.8 𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 984.8 × 9.81/1000𝑘𝑁

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 9.66𝑘𝑁

𝑚2


Intermediate Floor: For 150 mm R.C.slab, 75 mm brick ballast,40 mm thick P.C.C.,20 mm

thick terrazzo finish, and the occupancy type of an office, the slab load may be calculated as

under:

𝑅. 𝐶. 𝑆𝑙𝑎𝑏: 150

1000× 2400 = 360

𝑘𝑔

𝑚2

𝐵𝑟𝑖𝑐𝑘 𝑏𝑎𝑙𝑙𝑎𝑠𝑡: 75

1000× 1800 = 135

𝑘𝑔

𝑚2

𝑃. 𝐶. 𝐶 + 𝑡𝑒𝑟𝑟𝑎𝑧𝑧𝑜: 60

1000× 2300 = 138

𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑: = 633 𝑘𝑔

𝑚2

𝐺𝑖𝑣𝑒𝑛 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑: = 250 𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1.2 𝐷𝐿 + 1.6 𝐿𝐿

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1.2 (633) + 1.6(250)

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1159.6 𝑘𝑔

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 1159.6 × 9.81/1000 𝑘𝑁

𝑚2

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 11.38 𝑘𝑁

𝑚2

Calculation of Self Weight of Beams

After selection of trial dimensions, self weight per unit length may be approximated as

follows:

𝑆𝑒𝑙𝑓 𝐿𝑜𝑎𝑑 = 2400 ×𝑏𝑤

1000×

ℎ

1000×

9.81

1000 𝑘𝑁

𝑚⁄

𝑀𝑜𝑠𝑡 𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 𝑜𝑓 𝑢𝑛𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 = 2400 ×𝑙

18×

𝑙

12×

9.81

1000 𝑘𝑁

𝑚⁄

Bar Bending Schedule

After calculation of steel requirement for bending moment, shear force and torque at

important sections of a beam, detailing for reinforcement is carried out. Detailing means

selection of the bar sizes, number of bars, bar spacing, bar positions and bar cut offs and

showing the results on neat sketch. A bar bending schedule is then prepared which helps

the cutting, bending and placing of bars. The number, length, diameter, steel grade and


shape of each bar are separately written. Each type of bar is given a separate designation on

the drawings and the corresponding values are entered in the bar-bending schedule. This

table gives the total amount of steel required for a particular project. The columns of a

typical bar bending schedule are given in the Table 3.3

The details of bent-ups and standard hooks are given below:

𝐸𝑥𝑡𝑟𝑎 𝑙𝑒𝑛𝑔𝑡ℎ 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑎𝑛 𝑖𝑛𝑐𝑙𝑖𝑛𝑒𝑑 𝑝𝑎𝑟𝑡 = √2 ℎ − ℎ = 0.414 ℎ

𝐸𝑥𝑡𝑟𝑎 𝑙𝑒𝑛𝑔𝑡ℎ 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑎 90°ℎ𝑜𝑜𝑘 ≅ 18 𝑑𝑏

𝐸𝑥𝑡𝑟𝑎 𝑙𝑒𝑛𝑔𝑡ℎ 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝑎 180°ℎ𝑜𝑜𝑘 ≅ 5 𝑑𝑏 + 15 𝑑𝑏 = 20 𝑑𝑏

Table 3.3 Bar Bending Schedule. Steel Grade_________________

S.NO. Bar Designation

No.of Bar Length of Bar Dia of Bar

Weight of Bars Shape of Bar

No.10 No.15 No.20 No.25

∑ =

Fig.3.18. Details of Bent-Ups and Standard Hooks

Approximate amount of steel in beams = 0.0002 b h kgs/meter (±20%)

Example 3.3: Prepare bar bending schedule for the reinforcement of the beam shown in

the fig.3.19.Clear cover is 40 mm and the bar sizes are in SI units.


a) Mid-Span Cross Section

b) Longitudinal Section

Solution:

C/C Bar length=Beam clear Span+2 × Wall width-2 × clear cover- Bar Dia

Bar No.1 = 4000+2 x 228 - 2 x 40 – 20 = 4356

Bar No.2 = 4000+2 x 228 - 2 x 40 – 15 = 4361

Bar No.3 = 4000+2 x 228 -2 x 40 -10 = 4366

Bar No.4 = (Beam Width- 2 x clear cover –Bar Dia) x(Beam Depth – 2 x clear Cover –Bar Dia)

Bar No.4 = ( 228 – 2 x 40 -10) x( 375 -2 x 40 -10) = 138 x 285

Total Length of Bar Calculation

Bar No.1= C/C Bar Length – Bar Dia + 2 x 18 db due to 90 deg hooks

Bar No.1= 4356 – 20 + 2 x 18 x 20 = 5056 mm

Bar No.2= C/C Bar Length – Bar Dia + 2 x 18 db due to 90 deg hooks+2 x 0.414 bar bent up

length c/c height(375-40 x 2- 2 x Ring Dia-Bar Dia= 375-80-2x10-15=)

Bar No.2= 4361– 15 + 2 x 18 x 15+2 x 0.414 x 260= 5101 mm

Bar No.3= C/C Bar Length – Bar Dia + 2 x 18 db due to 90 deg hooks

Bar No.3= 4366 – 10 + 2 x 18 x 10 = 4716 mm


Bar No.4 = 2x (C/C Bar Length along breadth – Bar Dia + C/C Bar Length along Depth +Bar

Dia) + 2 x 18db

Bar No.4 = 2x (138 – 10 + 285 +10) + 2 x 18 x 10 = 1206 mm

Formula for weight calculation in kg per unit length = 3.14

4𝐷2 × 7850

Sr.No. Dia.mm Wt=kgs./m

1 20 2.465

2 15 1.387

3 10 0.616

Table 3.4. Bar Bending Schedule For Example 3.3. Steel Grade: 420

S.NO. Bar Designation

No.of Bar

Length of Bar(m)

Dia of Bar(mm)

Weight of Bars(kgs.) Shape of Bar

No.10 No.15 No.20 No.25

1 M-1 2 5.056 20 24.92

2 M-2 1 5.101 15 7.08

3 M-3 2 4.716 10 5.81

4 S-1 25 1.206 10 18.57

Total Weight 24.38 7.08 24.92

5 % Wastage

∑ =

1.219 0.354 1.246

Total Steel Required=59.2 kg

4356

4366

4361

138 x 285


Continuous Beams

Continuous beams are those beams that have two or more spans built monolithically. These

beams are indeterminate and require detailed procedures for plotting shear force and

bending moment diagrams. The beams in frames are to be analyzed for various load

combinations of different loads such as dead, live, wind and earthquake loads. For each

combination of these loads involving live load, pattern loading is to be considered to get

maximum force effect at any point. In pattern loading, live load may be applied in adjacent

panels or alternate panels in order to produce critical magnitude of a particular moment or

shear. This means that the analysis is to be performed many times for each combination

involving different pattern loads. Maximum values at various points are then picked out of

these results.

ACI Moment Coefficients For Continuous Beams

ACI Code 8.3 includes expressions that may be used for the approximate calculation of

maximum moments and shears in continuous beams and one way slabs. The expressions for

moment take the form of a coefficient multiplied by 𝑤𝑢𝑙𝑛2 where 𝑤𝑢 is the total factored

load per unit length on the span and 𝑙𝑛 is the clear span from face to face of supports for

positive moment, or average of two adjacent clear spans for negative moment. Shear is

taken equal to coefficient multiplied by 𝑤𝑢𝑙𝑛

2.The coefficients, found in the ACI Code 8.3.3,

are given in the table 12.1.

The ACI moment coefficients were derived by elastic analysis, considering alternative

placement of live load to yield maximum negative or positive moments at the critical

sections. They are applicable within the following limitations.

1) There are two or more spans

2) Spans are approximately equal, with the longer of two adjacent spans not greater

than the shorter by more than 20 percent.

3) Loads are uniformly distributed.

4) The unfactored live load does not exceed 3 times the unfactored dead load.

5) Members are prismatic.

6) The beams must be in a braced frame without significant moments due to lateral

loads ( not stated in the ACI Code, but necessary for the students to apply)


Table 12.1

Moment and Shear values using ACI coefficients ∗

Positive moment End Spans

If discontinuous end is unrestrained 1

11𝑤𝑢𝑙𝑛

2

If discontinuous end is integral with supports 1

14𝑤𝑢𝑙𝑛

2

Interior spans 1

16𝑤𝑢𝑙𝑛

2

Negative moment at exterior face of first interior support

Two spans 1

9𝑤𝑢𝑙𝑛

2

More than Two spans 1

10𝑤𝑢𝑙𝑛

2

Negative moment at other faces of interior supports 1

11𝑤𝑢𝑙𝑛

2

Negative moment at face of all supports for (1)slabs with spans not exceeding 3 m and (2) beams and girders where ratio of sum of column stiffness to beam stiffness exceeds 8 at each end of span

1

12𝑤𝑢𝑙𝑛

2

Negative moment at interior faces of exterior supports for members built integrally with their supports

Where the support is a spandrel beam or girder

1

24𝑤𝑢𝑙𝑛

2

Where the support is a column 1

16𝑤𝑢𝑙𝑛

2

Shear in end members at first interior support 1.15𝑤𝑢𝑙𝑛

2

Shear at all other supports 𝑤𝑢𝑙𝑛

2

∗ 𝑤𝑢 =Total factored load per unit length of beam or per unit area of slab.

𝑙𝑛 =Clear span for positive moment and shear and the average of the two adjacent spans

for negative moment

Approximate Curtailment of Bars

In place of locating the actual cut-off points of bars by using moment envelopes and

development length evaluation procedures, approximate but safe curtailment of bars may

be carried out by using thumb rules. There are two methods of reducing the steel where it is

not required. The first method is curtailment of bars in portions where these are not

required for strength or serviceability on the same face of the beam where these are

originally placed, Fig 4.19 (a).The second method is the use of bent-up bars, where the

positive bars are bent up near the supports to act as negative reinforcement, Fig

4.19(b).Further, the bent up bars can increase the shear capacity of a local area in which the

actual bent-up is present.

According to ACI 12.11.1, at least one-third of the positive reinforcement in simple members

and one-fourth the positive reinforcement in continuous members must extend along the

same face of the member in to the support. However, to use the approximate curtailment


methods, design engineers prefer to extend one-half or two-third of the positive steel into

the support. According to the code, such reinforcement carried into the supports must

extend at least 150 mm into the support. However, it is better to fully anchor this steel to

utilize it in case of hogging due to lateral loads or as compression reinforcement.

In case adjacent spans are different, the distance from which bent-up or curtailment starts is

considered according to that clear span from where the bars originate. However, the

extension of the bars on the top in both directions is provided according to the larger of the

adjacent clear spans.

Fig:ACI moment and Shear Coefficients Pictorial Represenation.


a) Curtailed Bars for Beams

b) Curtailed Bars for Slabs

Distances are same as for Beams

c) Bent-Up Bars

Fig.4.19. Approximate Curtailment of Steel in Beams and Slabs.


Example 4.5

A roof system consists of 228 × 228 mm columns at a spacing of 5 m and 3.5 m in two

mutually perpendicular directions, as shown in the figure 4.20.Beams run in both

directions over the columns having more than two spans in each direction. Design the

first interior long beam (B2) if the factored slab load is 12 𝑘𝑁

𝑚2 ,thickness of slab is 125

mm,E-20 concrete, Grade 420 steel, and≤ live load 3× dead load. Select US customary

bars.

Fig.4.20. Beam Layout

Solution:

𝑙𝑦 = 5𝑚

𝑙𝑥 = 3.5𝑚

𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑆𝑙𝑎𝑏 𝐿𝑜𝑎𝑑 = 12𝑘𝑁

𝑚2

𝑏𝑤 = 228 𝑚𝑚, 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑖𝑧𝑒 𝑜𝑓𝑡ℎ𝑒 𝑐𝑜𝑙𝑢𝑚𝑛

ℎ𝑓 = 125 𝑚𝑚

�́�𝑐 = 20 𝑀𝑃𝑎

𝑓𝑦 = 420 𝑀𝑃𝑎

Beam B2 is to be designed.

𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑆𝑙𝑎𝑏 𝐿𝑜𝑎𝑑 = 12.00𝑘𝑁

𝑚2

𝐵𝑒𝑎𝑚 𝑆𝑒𝑙𝑓 𝑊𝑒𝑖𝑔ℎ𝑡 = 1.2 𝐷𝐿


𝐵𝑒𝑎𝑚 𝑆𝑒𝑙𝑓 𝑊𝑒𝑖𝑔ℎ𝑡 = 1.2 × ( 2400 ×228

1000× 𝑑𝑤 ×

9.81

1000)

𝑑𝑤 =𝐵𝑒𝑎𝑚 𝐿𝑒𝑛𝑔𝑡ℎ

12− ℎ𝑓 =

5000

12− 125 = 291.66 𝑚𝑚 ≅ 0.29166 𝑚

𝐵𝑒𝑎𝑚 𝑆𝑒𝑙𝑓 𝑊𝑒𝑖𝑔ℎ𝑡 = 1.2 × ( 2400 ×228

1000× 0.29166 ×

9.81

1000)

𝐵𝑒𝑎𝑚 𝑆𝑒𝑙𝑓 𝑊𝑒𝑖𝑔ℎ𝑡 = 1.9 𝑘𝑁/𝑚

Equivalent Width of Slab Supported By Beam B2

𝑅 =𝑙𝑥

𝑙𝑦=

3.5

5= 0.7

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑆𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 = (1 −𝑅2

3) × 𝑙𝑥

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑆𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 = (1 −0.72

3) × 3.5

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑆𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 = 2.93 𝑚

For first interior beams, the above width is to be increased by 10 %, as exterior beams do

not have a significant cantilever slab portion.

𝐴𝑑𝑗𝑢𝑠𝑡𝑒𝑑 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑆𝑙𝑎𝑏 𝑤𝑖𝑑𝑡ℎ 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 = 1.1 × 2.93 = 3.223 𝑚

∴ 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑆𝑙𝑎𝑏 𝐿𝑜𝑎𝑑 𝑜𝑛 𝑏𝑒𝑎𝑚 𝐵2 = 3.223 × 12 = 38.6 𝑘𝑁/𝑚

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑 = 𝐵𝑒𝑎𝑚 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐷𝐿 + 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑

𝑇𝑜𝑡𝑎𝑙 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝐿𝑜𝑎𝑑𝑤𝑢 = 1.9 + 38.6 = 40.5 𝑘𝑁/𝑚

Total Factored Bending Moments

𝑙𝑛 = 𝐵𝑒𝑎𝑚 𝐶𝐶⁄ 𝑆𝑝𝑎𝑛 − 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 5000 −

228

2−

228

2= 4.772 𝑚

𝐸𝑛𝑑 𝑆𝑝𝑎𝑛 (𝐼𝑓 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑒𝑛𝑑 𝑖𝑠 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑤𝑖𝑡ℎ 𝑆𝑢𝑝𝑝𝑜𝑟𝑡𝑠)𝑀𝑢+ =

1

14𝑤𝑢𝑙𝑛

2 =1

14× 40.5 × 4.7722 = 65.9 𝑘𝑁. 𝑚

𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑆𝑝𝑎𝑛𝑠(𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑀𝑜𝑚𝑒𝑛𝑡)𝑀𝑢+ =

1

16𝑤𝑢𝑙𝑛

2 =1

16× 40.5 × 4.7722 = 57.7 𝑘𝑁. 𝑚

𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝐹𝑎𝑐𝑒 𝑜𝑓 𝐹𝑖𝑟𝑠𝑡 𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑆𝑢𝑝𝑝𝑜𝑟𝑡 𝑀𝑢− =

1

10𝑤𝑢𝑙𝑛

2 =1

10× 40.5 × 4.7722 = 92.23 𝑘𝑁. 𝑚

𝑂𝑡ℎ𝑒𝑟 𝐹𝑎𝑐𝑒𝑠 𝑜𝑓 𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑆𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑀𝑢− =

1

11𝑤𝑢𝑙𝑛

2 =1

11× 40.5 × 4.7722 = 83.9 𝑘𝑁. 𝑚

𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝐹𝑎𝑐𝑒 𝑜𝑓 𝐸𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑆𝑢𝑝𝑝𝑜𝑟𝑡 𝑀𝑢− =

1

16𝑤𝑢𝑙𝑛

2 =1

16× 40.5 × 4.7722 = 57.7 𝑘𝑁. 𝑚


Beam Depth Selection

Many different options are available for sizing the beam. One option may be to size the

beam as a singly reinforced section for exterior span positive moment and then design

doubly reinforced for negative moments. Second option may be to proportion the section

for maximum negative moment, which may significantly increase the beam depth. Third

option may be to select dimensions in-between the earlier stated two options.

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑏𝑒𝑎𝑚 𝑓𝑜𝑟 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑓𝑜𝑟 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑝𝑎𝑛𝑒𝑙 =𝑙

18.5=

5000

18.5= 270 𝑚𝑚

𝐴𝑠𝑠𝑢𝑚𝑒𝑑 𝑑𝑒𝑝𝑡ℎ =5000

12= 417 𝑚𝑚

𝑑𝑚𝑖𝑛 𝑓𝑜𝑟 𝑀𝑢+𝑖𝑛 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑝𝑎𝑛𝑒𝑙 = √

𝑀𝑢

0.205 × �́�𝑐 × 𝑏

𝑑𝑚𝑖𝑛 𝑓𝑜𝑟 𝑀𝑢+𝑖𝑛 𝑡ℎ𝑒 𝑒𝑥𝑡𝑒𝑟𝑖𝑜𝑟 𝑝𝑎𝑛𝑒𝑙 = √

65.9 × 106

0.205 × 20 × 228= 266 𝑚𝑚

ℎ𝑚𝑖𝑛 = 𝑑𝑚𝑖𝑛 + 75 = 266 + 75 = 341 𝑚𝑚

𝑑𝑚𝑖𝑛 𝑓𝑜𝑟 𝑀𝑢−, max = √

𝑀𝑢

0.205 × �́�𝑐 × 𝑏

𝑑𝑚𝑖𝑛 𝑓𝑜𝑟 𝑀𝑢−, max = √

92.3 × 106

0.205 × 20 × 228= 315 𝑚𝑚

ℎ𝑚𝑖𝑛 = 𝑑𝑚𝑖𝑛 + 75 = 315 + 75 = 390 𝑚𝑚

Assuming a value slightly higher than 315 mm lets assume dmin= 350 mm

∴ ℎ = 350 + 75 = 425 𝑚𝑚

Maximum Capacity as Singly Reinforced Rectangular Section at Support

For negative moment at support,𝜌𝑚𝑎𝑥 corresponding to an extreme tensile strain of 0.0075

may be used.

𝜌𝑚𝑎𝑥 = 0.85𝛽1

2

7

�́�𝑐

𝑓𝑦

(𝐹𝑟𝑜𝑚 𝑝𝑎𝑔𝑒 81 𝑍𝐴) = 0.85 × 0.85 ×2

7×

20

420= 0.00983

𝐴𝑠1 = 𝜌𝑚𝑎𝑥 𝑏 𝑑 = 0.00983 × 228 × 350 = 785 𝑚𝑚2

𝑎 =𝐴𝑠1𝑓𝑦

0.85�́�𝑐𝑏=

785 × 420

0.85 × 20 × 228= 85 𝑚𝑚


𝑀1 = ∅𝑏𝑀𝑛 = ∅𝑏𝐴𝑠1𝑓𝑦 (𝑑 −𝑎

2) = 0.9 × 785 × 420 ×

(350 −852

)

106= 91.2𝑘𝑁. 𝑚

Effective Flange Width For T-Beam Behavior

The effective flange width,b,is the minimum of the following three dimensions

1)𝑙

4=

5000

4= 1250 𝑚𝑚

2)16 ℎ𝑓 + 𝑏𝑤 = 16 × 125 + 228 = 2228 𝑚𝑚

3) S= Center to Center spacing of the Beams=3500 mm

𝑏 = 1250 𝑚𝑚

𝜌𝑚𝑖𝑛 =1.4

𝑓𝑦=

1.4

420= 0.00333

𝐴𝑠,𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛𝑏𝑤𝑑 = 0.00333 × 228 × 350 = 266 𝑚𝑚2

Design For Positive Moment in Exterior Span

For positive moment, the flange of the T-Beam will be in compression. However, it is more

likely that for this smaller moment the N.A. will lie within the flange. The beam will act like a

rectangular section of dimensions 1250 x 425 mm.

Assume a=75 mm

𝐴𝑠 =𝑀𝑢

∅𝑏𝑓𝑦 (𝑑 −𝑎2)

=65.9 × 106

0.9 × 420 × (350 −752 )

= 558 𝑚𝑚2

𝑎 =𝐴𝑠𝑓𝑦

0.85�́�𝑐𝑏=

558 × 420

0.85 × 20 × 1250= 11 𝑚𝑚 ≤ 𝛽1ℎ𝑓 = 0.85 × 125 = 106 𝑚𝑚

∴ 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡.

𝐴𝑠 =𝑀𝑢


=65.9 × 106

0.9 × 420 × (350 −112 )

= 506 𝑚𝑚2


0.85�́�𝑐𝑏=

506 × 420

0.85 × 20 × 1250= 10 𝑚𝑚

𝐴𝑠 =𝑀𝑢


=65.9 × 106

0.9 × 420 × (350 −102 )

= 505 𝑚𝑚2 > 𝐴𝑠,𝑚𝑖𝑛

𝑈𝑠𝑒 2 ≠ 16 + 1 ≠ 13


Design For Positive Moment in Interior Span

𝑀𝑢+ = 57.7 𝑘𝑁. 𝑚

Assume a=10 mm

𝐴𝑠 =𝑀𝑢


=57.7 × 106

0.9 × 420 × (350 −102 )

= 442 𝑚𝑚2


0.85�́�𝑐𝑏=

442 × 420

0.85 × 20 × 1250= 8.7 𝑚𝑚

𝐴𝑠 =𝑀𝑢


=57.7 × 106

0.9 × 420 × (350 −8.72 )

= 442 𝑚𝑚2 > 𝐴𝑠,𝑚𝑖𝑛

𝑈𝑠𝑒 2 ≠ 16 + 1 ≠ 13 𝑓𝑜𝑟 𝑐𝑜𝑚𝑝𝑎𝑡𝑖𝑏𝑖𝑙𝑖𝑡𝑦.

Design for Negative Moment at Exterior Support

𝑀𝑢− = 57.7 𝑘𝑁. 𝑚

The T-Beam will act like a rectangular section of dimensions 228 x 425, because the flange

comes under tension.

We know that

𝑅 =𝑀𝑢

𝑏𝑑2 & 𝜔 = 0.85

𝑓𝑐̇

𝑓𝑦

𝜌 = 𝜔 [1 − √1 −2𝑅

0.765𝑓�́�]

𝑅 =57.7 × 106

228 × 3502= 2.066 𝑀 𝑃𝑎

𝜔 = 0.85 ×20

420= 0.0404

𝜌 = 0.0404 × [1 − √1 −2 × 2.066

0.765 × 20] = 0.0059 < 𝜌𝑚𝑎𝑥 = 0.00983 (𝑂𝐾)

𝐴𝑠 = 𝜌𝑏𝑑 = 0.0059 × 228 × 350 = 471 𝑚𝑚2 > 𝐴𝑠,𝑚𝑖𝑛

Design for Negative Moment at First Interior Support

𝑀𝑢− = 92.3 𝑘𝑁. 𝑚 > 𝜑𝑏𝑀𝑛 = 91.1 𝑘𝑁. 𝑚


Design as a doubly reinforced rectangular section of dimensions 228 x 425.

𝐿𝑒𝑡 𝑑′ = 60 𝑚𝑚

𝑀2 = 𝑀𝑢 − 𝑀1

𝑀2 = 92.3 − 91.1 = 1.2 𝑘𝑁. 𝑚

Assuming Compression steel to be yielding

𝐴𝑠′ =𝑀2

𝜑𝑏𝑓𝑦(𝑑 − 𝑑′)=

1.2 × 106

0.9 × 420(350 − 60)= 11 𝑚𝑚2

𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠′ = 784 + 11 = 795 𝑚𝑚2

A=85 mm ( as calculated before)

𝑓𝑠′ = 600𝑎 − 𝛽1𝑑′

𝑎= 600

85 − 0.85 × 60

85= 240 𝑀 𝑃𝑎

< 𝑓𝑦 , 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 𝑖𝑠 𝑛𝑜𝑡 𝑦𝑖𝑒𝑙𝑑𝑖𝑛𝑔

𝐴𝑠′, 𝑟𝑒𝑣𝑖𝑠𝑒𝑑 = 𝐴𝑠′, 𝑡𝑟𝑖𝑎𝑙 ×𝑓𝑦

𝑓𝑠′= 11 ×

420

240= 20 𝑚𝑚2

Design For Negative Moment At Interior Supports

𝑀𝑢− = 83.9 𝑘𝑁. 𝑚 < 𝜑𝑏𝑀𝑛 = 91.1 𝑘𝑁. 𝑚

Design as a Singly reinforced rectangular section of dimensions 228 x 425

We know that

𝑅 =𝑀𝑢

𝑏𝑑2 & 𝜔 = 0.85

𝑓𝑐̇

𝑓𝑦

𝜌 = 𝜔 [1 − √1 −2𝑅

0.765𝑓�́�]

𝑅 =83.9 × 106

228 × 3502= 3.004𝑀 𝑃𝑎

𝜔 = 0.85 ×20

420= 0.0404

𝜌 = 0.0404 × [1 − √1 −2 × 3.004

0.765 × 20] = 0.00891 < 𝜌𝑚𝑎𝑥 = 0.00983 (𝑂𝐾)

𝐴𝑠 = 𝜌𝑏𝑑 = 0.00891 × 228 × 350 = 712 𝑚𝑚2


Top steel available at exterior support = 3 ≠ 13 (387 𝑚𝑚2)

𝐸𝑥𝑡𝑟𝑎 𝑠𝑡𝑒𝑒𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 471 − 387 = 84 𝑚𝑚2(1 ≠ 13)

design of structure-concrete part...

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