Introduction of Reinforced Concrte Design, Advantages and DIsadvantages of RC, & Sample Computation of Loads

1. REINFORCED CONCRETE DESIGN
Concrete:
Concrete is a stone like substance obtained by permitting a
carefully proportioned mixture of cement, sand and gravel or
other aggregate and water to harden in forms of the shape and
of dimensions of the desired structure.
Reinforced cement concrete:
Since concrete is a brittle material and is strong in compression.
It is weak in tension, so steel is used inside concrete for
strengthening and reinforcing the tensile strength of concrete.
When completely surrounded by the hardened concrete mass
it forms an integral part of the two materials, known as
"Reinforced Concrete".

2. Advantages and Disadvantages of Reinforced Concrete
Flexural Strength of Concrete
Advantages of reinforced concrete
• It has relatively high compressive strength
• It has better resistance to fire than steel
• It has long service life with low maintenance cost
• In some types of structures, such as dams, piers and footings, it is most economical
structural material
• It can be cast to take the shape required , making it widely used in pre-cast structural
components
• It yields rigid members with minimum apparent deflection
• Yield strength of steel is about 15 times the compressive strength of structural concrete
and well over 100 times its tensile strength
• By using steel, cross sectional dimensions of structural members can be reduced.
Disadvantages of reinforced concrete
• It needs mixing, casting and curing, all of which affect the final strength of concrete
• The cost of the forms used to cast concrete is relatively high
• It has low compressive strength as compared to steel (the ratio is about 1:10 depending on
material) which leads to large sections in columns/beams of multi-storey buildings Cracks
develop in concrete due to shrinkage and the application of live loads

3. Reinforced Cement Concrete Design Philosophy and Concepts
“The design of a structure may be regarded as the process of selecting
proper materials and proportioned elements of the structure, according to
the art, engineering science and technology. In order to fulfil its purpose,
the structure must meet its conditions of safety, serviceability, economy
and functionality.”

4. Design Methods
• Strength Design Method
• Working Stress Design
• Limit State Design

5. Loads
-Forces for which a structure should be proportioned. Loads that act on
structure can be divided into three categories.
Dead Loads:
• Dead loads are those that are constant in magnitude and fixed in
location throughout the lifetime of the structure. It includes the weight
of the structure and any permanent material placed on the structure,
such as roofing, tiles, walls etc. They can be determined with a high
degree of accuracy from the dimensions of the elements and the unit
weight of the material.
Live loads:
• Live loads are those that may vary in magnitude and may also change in
location. Live loads consists chiefly occupancy loads in buildings and
traffic loads in bridges. Live loads at any given time are uncertain, both
in magnitude and distribution.
Environmental loads:
• Consists mainly of snow loads, wind pressure and suction, earthquake
loads. Soil pressure on subsurface portion of structures, loads from
possible pounding of rainwater on flat surfaces and forces caused by
temperature differences. Like live loads, environmental loads at any
given time are uncertain both in magnitude and distribution.

6. 1.Required Strength (Factored Load) U
1.1 To resist dead load & live load:
U=1.4DL + 1.7LL
1.2 If resistance to structural effects of specific wind load
U= 0.75(1.4DL+1.7LL+1.7W)
U=0.9DL+1.3W not less than 1.4DL+1.7LL
1.3 If resistance to specified earthquake loads
U= 0.75(1.4DL+1.7LL+1.87E)
U=0.9DL+1.43E not less than 1.4DL+1.7LL
1.4 If resistance to specified earth pressure
U= 1.4DL+1.7LL+1.7H
U=0.9DL not less than 1.4DL+1.7LL
1.5 Where structural effects T of differential settlement,
creep, shrinkage or temperature change are significant.
U= 0.75(1.4DL+1.4T+1.7LL) not less than 1.4(DL+T)

7. Sample Problem
1. CE BOARD 2012
• A singly reinforced T-beam with a cross
sectional area shown in the figure, span of
5m is subjected to support dead load and
live load of 10KN/m and 28KN/m
respectively. Find the factored load
required in ff;
a. using DL and LL only.
b. With additional wind pressure,
w=5KN/m.
b.1. use liveload
b.2. No liveload
c. With additional Earthquake load,
E=40KN/m
c.1. use liveload
c.2. No liveload

8. Solution:
Area of the T-Beam = (0.3x0.12)+(.220x.150) = .0069
Deadload of T-Beam = 25KN/m3 x .0069 = 1.725 KN/m
DLtotal = DL + DLT-Beam = 10 + 1.725 = 11.725KN/m
- Combination of DL & LL
a.) U=1.4DL + 1.7LL = 1.4(11.725) + 1.7(28) = 21.175KN/m
- with Wind Load = 5KN/m,
b.1) U= 0.75(1.4DL+1.7LL+1.7W) = 0.75(1.4(11.725)+1.7(28)+1.7(5)) =
54.39KN/m
b.2) U=0.9DL+1.3W = 0.9(11.725)+1.3(5) = 17.05, Use: = 21.175KN/m
- with Earthquake Load = 40KN/m,
c.1) U= 0.75(1.4DL+1.7LL+1.87E) = 0.75(1.4(11.725)+1.7(28)+1.87(40)) =
104KN/m
c.2) U=0.9DL+1.43E = 0.9(11.725)+1.43(40) = 67.75 KN/m

9. 2. Determine the Axial stress
acting on a circular column
shown in the figure.
Solution:
U = 0.75(1.4DL+1.7LL+1.87E)
= 121.725N
Stress = Force/Area
= (121.725)/π(0.125)2
= 2,479.76N/m2
Or 2.48KPa
DL = 20N
LL = 35N
D = 250
E = 40N

10. Thank you!