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Bar Bending Schedule, commonly referred to as “BBS” is a comprehensive list that describes the location, mark, type, size, length and number, and bending details of each bar or fabric in a Reinforcement Drawing of a Structure. ... In context of Reinforcement bars, it is called bar scheduling.

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Different mixes (M) come in various mix proportions of the various ingredients of cement, sand and coarse aggregates. For instance, M20 comes in the respective ratio of 1:1:5:3. You can see other examples below in the table.

 

Concrete Grade	Mix Ratio (cement : sand : aggregates)	Compressive Strength
		MPa (N/mm2)	psi
Grades of Concrete
M5	1 : 5 : 10	5 MPa	725 psi
M7.5	1 : 4 : 8	7.5 MPa	1087 psi
M10	1 : 3 : 6	10 MPa	1450 psi
M15	1 : 2 : 4	15 MPa	2175 psi
M20	1 : 1.5 : 3	20 MPa	2900 psi

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Difference between Short- and Long-term Deflections:

Short-term deflection means the immediate deflection after casting and application of partial or full service loads, while the long-term deflection occurs over a long period of time largely due to shrinkage and creep of the materials.

The following factors influence the short-term deflection of structures:

(a) Magnitude and distribution of live loads,

(b) Span and type of end supports,

(c) Cross-sectional area of the members,

(d) Amount of steel reinforcement and the stress developed in the reinforcement,

(e) Characteristic strengths of concrete and steel, and

(f) Amount and extent of cracking.

The long-term deflection is almost two to three times of the short-term deflection.

The following are the major factors influencing the long-term deflection of the structures:

(a) Humidity and temperature ranges during curing,

(b) Age of concrete at the time of loading, and

(c) Type and size of aggregates, water-cement ratio, amount of compression reinforcement, size of members etc., which influence the creep and shrinkage of concrete.

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Cement Concrete

Cement	Predetermined proportions.
Fine Aggregate
Coarse Aggregate
Water

 

When water is added, chemical reaction will occur, which is very much slow and requires time (setting time) and favourable temperature for its completion,

This setting time can be divided into 3 distinct phases known as:  

Initial set	30-60 minutes
Final set	5-6 hours)
Progressive hardening (increase in strength)

 

Concrete has enough strength in compression but has little strength in tension, so obviously concrete is weak in Bending, Shear and Torsion.

So, to use Cement Concrete in structures which will be having tensile loads (Beams, Slabs etc.) steel bars may be placed at tensile zones of the structure.

The steel bars (steel reinforcement) takes the tensile stresses & Concrete so obtained is known as reinforced cement concrete (R.C.C)

Grades of concrete:

IS 456:2000 specifies 15 grades of concrete.

Group	Grade Designation	Specified characteristic compressive strength of 150 mm cube at 28 days (N/mm2)
Ordinary concrete	M10	10
	M15	15
	M20	20
Standard Concrete	M25	25
	M30	30
	M35	35
	M40	40
	M45	45
	M50	50
	M55	55
High Strength Concrete	M60	60
	M65	65
	M70	70
	M75	75
	M80	80

 

Letter M refers as the Mix and the number to the Specified Characteristic Strength ( ) of 150 mm cube at 28 days (in N/mm2).

Characteristic Strength:

i. It is the strength of the material below which not more than 5 % of the test results are expected to fall.

ii. As per IS 456:2000 minimum grade of concrete shall not be less than M 20 in RCC.

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If you are preparing for the competitive exam like GATE, ESE etc then I would recommend Theory Of Structures (Vol. 1) by G. Pandit and S. Gupta. For Matrix analysis and some other topics which come under Analysis 2, you can buy Structural Analysis - A Matrix Approach by the same authors G. Pandit and S. Gupta. In case you need a book just for academic purposes then you can go for Theory Of Structures by S. Ramamrutham, it has so many questions for every topic. If your basics are weak then you can start with MECHANICS OF STRUCTURES VOL. I by H.J.Shah S.B.Junnarkar. and then for part II, it's second volume Mechanics of Structures-Vol-II ( H.J. Shah and S-B-Junnarkar). read less

Let us assume a member with uniform cross section A and of lenth l, be subjected to external axial load W as shown in the figure.

As the load is gradually applied the load is incresed from 0 to W, due to which the member is gradually extended by Δ.

Work done due to load is given by the product of avarage load and the displacement Δ.

   work done = 1/2 * W * Δ

Let the tenstion developed in the member be T, @ Equilibrium W = T

   Tensile stress (ƒ) = T/A

   Tensile strain(e) = ƒ/E = T/AE

   Δ = el = S/AE *l]

Strain Energy Stored = Work Done = 1/2*W*Δ = 1/2*T*Tl/AE = T²l/2AE

 

Strain Energy Stored per unit volume = (T²l/2AE)/V =T²/2A2E=ƒ2/2E)

                                        

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