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Lecture 2 - Reinforced Concrete(I) Topic: Introduction to reinforced concrete; structural behaviors of reinforced concrete; Calculation of shearing force and bending moment 1.0 Introduction to reinforced concrete Reinforced Concrete Concrete is a manmade material created by the proper mixing of coarse aggregate, such as fine aggregate, such as sand, and cement, with adequate and controlled amounts of water. The cement forms a slurry with the water, which under chemical action starts to set. The fine aggregate fills the interstices of the coarse aggregate and the slurry fills the interstices of the fine aggregates, coating all the particles and binding the whole together into a monolithic mass when set. This is concrete. Concrete is a very strong in compression, and ultimate strength is about 40 to 60N/mm2 can be obtained. However, concrete is very weak in tension, and so has to be adequately reinforced with steel rebars. The concrete then takes the compression, and the steel takes all the tension. Both materials work together as a composite material. This material is then called reinforced concrete. Failure of the beam due to bending, shear and deflection The beam will have to be strong enough to withstand the loads, stable enough not to fall over, and be stiff enough to meet the function requirement of serviceability. The stability of the beam can be achieved by giving it sufficient lateral bracing, which will hold it firmly in position, and the serviceability or function can be achieved by limiting the amount of the deflection. However, the strength will depend upon two forces, the force which will cause the beam to bend and the force which will cause the fibres to shear past each other. The first force could produce failure due to bending and the second, failure due to shear. Assuming then, there is sufficient stability, it is worth looking at a simple beam, and seeing how it may fail under bending, shear and deflection. Failure due to bending Consider the beam in Figure 1, which is supported at point A and point B. If a number of loads are placed on the beam, the beam will bend and deflect as shown in Figure 2. The fibres in the top of the beam will be in compression while the bottom fibres will be in tension. If the tension and compression exceed the natural strength of the material from which

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the beam has been made, then the beam will fail due to excessive bending stress.

Figure 1:-Simply supported beam

Figure 2

Figure 3:- Bending failure

Let us take the example of concrete beam (Figure 3.) the load on the beam will cause tension cracks in the bottom fibres. If the load is increased, the cracks will become larger and the beam will become fail. This is because concrete, although very strong in compression, has very little strength in tension, and cannot on its own withstand tension stresses. Where tension occurs, the concrete has to be reinforced with steel reinforcing bars. If the steel bars were placed at the bottom of the beam, then the steel will resist the tension stress and the beam will carry a much greater load before it fails under bending. For failure to occur, either the concrete in the top of the beam has to crush, or the steel in the bottom has to be pulled apart. Which will fail first will depend upon the amount of steel, then the steel will fail first, but if there is a large amount, then the concrete in the top will crush long before the steel will begin to yield.

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Failure due to shear

Figure 4:- Deep Beam

Figure 5:-Shear Failure Consider a deep beam spanning between two supports A and B (Figure 4). If a very heavy load is placed on the beam, it will punch a section out of the beam (Figure 5). This kind of failure is called shear failure, because the molecules on one sides of the shear line have sheared past the molecules on the other side of the line. The angle of the shear line is usually at about 45 degrees as shown in Figure 5. Take an example of a reinforced concrete beam, it is considered how this beam could be made strong enough to resist the shear forces. Remember that concrete has little strength in tension, so to resist the shear, steel reinforcement has to be placed across the 45 degree ‘shear crack’. This can be done in two ways. The normal way is to use stirrups or links fixed vertically along the beam. These links are vertical reinforcement bars which wrap around and hold the main reinforcement bars in position. Because these links are vertical, they cross over the shear line and prevent the concrete from cracking.

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Figure 6:- Shear Reinforcement links

If the shear loads are very large, it may be necessary to increase the shear strength by bending the main reinforcing bars up at 45 degrees so that they cross the ‘shear cracks’ at right angle. (Figure 7)

Figure 7:- Shear reinforcement bent-up bars

Failure due to deflection Consider a shallow beam spanning between two supports A and B (Figure 8). A small load on this beam will cause it to deflect. If the load is increased then the deflection may be excessive, causing the beam to look unsafe although structurally it is not. In addition, it may be causing damage to the finishes such as plaster around the beam, and causing the floors to tilt. Therefore, from a serviceability or function point of view, the beam has failure due to excessive deflection. (Figure 9)

Figure 8:- Simply supported beam

Figure 9:- Deflection failure

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Bending moment and Shear Force If loads are applied to a beam, then the internal fibres in the beam have to resist these applied external loads by developing bending moments and shear forces within the beam. These moments and forces are a measure of what is happening inside the beam. Bending moment

Figure 10:- Moments in a beam

When a beam bends under the load (Figure 10), the horizontal fibres will change in length. The top fibres will become shorter and the bottom fibres will become larger. That is, the very top fibre will become the shortest and be under maximum compression, while the very bottom fibre will be the longest and be under maximum tension. from the top fibre to the central fibre, the compression gradually decrease until it is zero at the centre, which is called the neutral axis (N.A.). From the neutral axis to the bottom fibre, the fibres are in tension gradually increase from zero to a maximum at the bottom fibre.

Figure 11

The load on the beam is resisted by the compression force in the top and the tension force in the bottom of the beam. These forces form a ‘couple’, and this couple produces a moment to

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counterbalance the external bending moment, which in turn is generated by the applied loads. Now remember that a moment is a force times a distance, and a couple is two equal but opposed forces parallel to one another. A couple cannot resolved into one force, but it does produce a constant moment about any axis, perpendicular to the plane containing the forces. Thus, in Figure 11, the tension force and the compression force act as a couple, so that:- Force C = Force T Moment about A = Force C x Z Moment about B = Force T x Z

Figure 12 Thus the internal bending moment within a beam (Figure 12) is equal to the distance between the compression and tension forces times either the tension or compression force T or C. The forces are the summation of all the small forces in each fibre varying from zero at the centre to a maximum value at the top and bottom. These small forces can be diagrammatically represented by a triangular as shown in Figure 12 and the total forces will act at the centroid of a triangle is one-third the height above the base, and it can been seen from Figure 13, that if the depth of the beam is d, then the distance between the forces equals two-third d and the internal bending moment equal M:

zTzCM ×=×=

where .32 dz =

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Figure 13.

Shear force

Figure 14:-Vertical Shear Force

A small section of a beam is shown in Figure 14. on the left there is a vertical force pushing upwards, and on the right there is a vertical force pushing downwards. The fibres between the two forces will try to slide or shear over one another. These forces are described as the vertical shear forces and are generated by the loads applied on the beam and the corresponding support reactions, the reactions pushing upwards and the loads pushing downwards. The value of the shear force and bending moment will vary along the beam. As the beam has to be made large enough and strong enough to resist the maximum shear force and bending moment, it is useful to plot their values, so that the positions of the maximum values can be located. The diagram produced by plottings are called shear-force and bending-moment diagrams, and they are a direct reflection of ideal visual proportions for the beam. This is because the dept of the beam at any section is a function of the bending moment and the shear

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force. If the shear force is relatively small, then the depth should follow the pattern of the bending moment. This is very evident in the design of bridges and long span roofs (Figure 15), but it is not often very practical to vary the depth of a beam within a building.

Figure 15

2.0 Shearing Force (SF) and Bending Moment (BM) Introduction and Classification of BEAMS Whenever a HORIZONTAL BEAM is loaded with VERTICAL LOADS, sometimes it bends due to the action of the loads. The amount with which a beam bends, depends upon the amount and the type of the loads, length of the beam, elasticity and types of the beam. Classification of Beams (a) Cantilever Beam

Figure 16 A beam FIXED at one end and free at the other hand. (Figure 16)

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(b) Simply Supported Beam (Figure 17)

Figure 17

Support A is a HINGED or PINNED (FIXED) Support B is on a Roller (c) Overhanging Beam

Figure 18 A beam having its end portion extended in the form of cantilever beyond the support (Figure 18)

Figure 19

An overhanging beam may be overhanging on one side or on both sides. (Figure 19) (d) Rigidly Fixed Beams or Built-in Beams(Figure 20)

Figure 20

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A beam whose both ends are RIGIDLY FIXED or BUILT-in WALLS (e) Continuous beam (Figure 21)

Figure 21

A beam supported on more than TWO supports. (Note: A continuous beam may or may not be an OVERHANGING beam) Types of loads A beam may be subjected to the following TYPES OF LOADS: (a) Concentrated or Point Load (Figure 22) A load acting at a point on a beam is known as CONCENTRATED or POINT LOAD

Figure 22

(b) Uniformly Distributed Load (Figure 23)

Figure 23

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A load which is spread over a beam in such a manner that each unit length is loaded to the same extent is known as a UNFORMLY DISTRIBUTED LOAD (U.D.L) For calculations the total UDL may be assumed to act the centre of gravity of the load (c) Uniformly Varying Load

Figure 24

A load which is spread over a beam with uniform variation is known as UNIFORMLY VARYING LOAD. Shearing Force * Sign Convention

Figure 25

External force UPWARDS to LEFT of a section or DOWNWARDS to the RIGHT of a section are defined as positive. (Figure 25)

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Figure 26

External force DOWNWRDS to LEFT of a section or UPWARDS to the RIGHT of a section are defined as positive. (Figure 26)

A Shearing Force Diagram is one which shows the variation of S.F. along the length of a beam.

Figure 27

A beam carried loads 321 ,, WWW and is simply supported at two points (C and B) – Figure 27 R1 and R2 are the reactions at support C and B. The shearing force at any section (X-X) of a beam represent the tendency for the portion of the beam to one side of the section to shear laterally relative to the other portion. The resultant of the loads and reactions to left of the section X-X is ‘F’ vertically upwards. At the point X-X,

211 WWRFSF −−== Since the whole beam is in equilibrium, the resultant of the forces to the right of X-X must also be F, acting downwards,

23 RWFSF −==

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Bending Moment (B.M.) Sign convention

Figure 28- Positive (+) B.M. [Sagging Moment]

Clockwise BM to the LEFT of a section X-X and Counterclockwise (Anti-clockwise) B.M. to the RIGHT of a section are POSITIVE. (Figure 28 and Figure 29)

Figure 29:-Negative(-) BM [Hogging Moment]

In addition to shear, every section of the beam will be subjected to bending. For example, to a resultant BM, which is net effect of the moment of each of the individual loads. (Figure 30)

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w Figure 30

Again, for equilibrium, the values on either side of the section X-X must have equal values. “M” is called the BENDING MOMENT (BM) at section X-X. BM is defined as the algebraic sum of the moment about the section X-X of all the forces acting on either sides of the section The Bending Moment (B.M.) of the Loads and reaction to left of the section X-X is

221111 awbRawM −+−= Since the whole beam is equilibrium, the BM of the force to the right of X-X must also be

3322 awbRM −= Diagrams which show their variation in the BM and SF along the length of a beam or structure for any fixed loading condition are termed BM and SF diagrams. They are therefore graphs of BM and SF value drawn on the beam as a base and they clearly illustrate in the early design stages the positions on the beam which are subjected to the greater SHEAR or BENDING STRESSES and hence they may require further consideration or strengthening. Example 1: A cantilever of length ‘L’ carried a uniformly distributed load of intensity ‘w kN/m’. Draw the SF and BM diagram.

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Solution Shearing Force (SF) SF at X-X, wxF −= (linear variation) When 0=x , 0)0( =−= wF When Lx = , wLF −= Bending Moment (BM)

BM at X-X, ( ) 222wxxwxM =−= (Parabolic curve)

When 0=x , 0=M When Lx = , ( ) 2/2wLM −=

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Example 2: A cantilever beam of length ‘L’ carries a concentrated load ‘w’ at its free end. Drawing the shear force (SF) and Bending moment (BM) diagram.

Solution: Shearing Force (SF) SF at X-X, wF −= This is constant along the whole beam Bending Moment (BM) BM at X-X, wxM −= When 0=x , 0=M When Lx = , wLM −=

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Relationship between SHEARING FORCE and BENDING MOMENT

Figure 31

Figure 31 shows a short length xδ of a loaded beam at a distance ‘x’ from a fixed origin ‘O’. Let the shearing force at ‘x’ br F, And the shearing force at xx δ+ be FF δ+ Similarly let the Bending moment at ‘x’ be ‘M’ and the bending moment at xx δ+ be

MM δ+ . Let w = uniform distributed load on the length xδ The total load = xwδ =acting approximately (exactly, if uniformly distributed) through the centre ‘C’ The element must be equilibrium under the action of these force and moment. Take moment about C: (clockwise as positive)

( ) ( ) ( ) 02/2/ =+−⋅++⋅+ MMxFFxFM δδδδ Neglecting the product xF δδ ⋅

MxF δδδ =⋅

Taking the limits, then dx

dMF =

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Resolving vertically upwards ( ) 0=+−− FFxwF δδ

dxdFw −=

In other words, the maximum or minimum Bending Moment occurs when 0=dxdM but

dxdMF =

Thus When Shear force is being zero, Bending Moment is Maximum or Minimum.

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Appendix: Resultant of Distributed Load A resultant force of distributed load should satisfy the following criteria.

a. The magnitude of the resultant force equals the total force of the distributed load. b. Take moment about any point, the moment of the resultant equals to the moment of the distributed load.

1.0 Resultant of uniformly distributed load (UDL) Refer to Figure A1, the resultant of UDL equals load intensity, q, times the length of the UDL, i.e. R = q L. The resultant acts at the middle of the UDL, Figure A1(b) refer.

Figure A1:- Resultant of UDL

Example Determine the resultant force for the UDL in Figure A2

Figure A2

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Example Calculate the moment of force of the UDL in Figure A3 with respect to point A.

Figure A3 Solution Since the moment of force of distributed load equals to the moment of its resultant, determine the resultant first. The magnitude of the resultant force is 16 (3) = 48kN (download) The moment of the resultant force with respect to point A is 48 (2.5) = 120kNm (clockwise) Therefore, the moment of the UDL with respect to A is 120kNm (clockwise) 2.0 Resultant of triangle distributed load (TDL) Refer to Figure A4, the intensity of the triangle distributed load (TDL) varies from 0 to a

maximum value, 0q . The magnitude of resultant of TDL is 20 LqR = acting at position

shown in Figure A4(b).

Figure A4:- Resultant of triangular distributed load

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Example Calculate the moment of force of the TDL in Figure A5.

Figure A5 Solution: Magnitude of resultant force is (12 x 6 ) /2 = 36Kn (download) Resultant acts at 6m x 2/3 = 4m from the end of zero intensity Refer to Figure A5(b), moment of force of the TDL with respect to A is 36 x 7 = 252kNm (clockwise) 3.0 Resultant of general distributed load Example Calculate the total force and moment of force with respect to point A for the distributed load in Figure A6.

Figure A6

Solution: The load in (a) is divided into a UDL and a TDL as indicated in Figure A7. Resultant UDL = 3 x 6 = 18kN (downward) Resultant TDL = (5 x 6) / 2 = 15kN (downward)

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Total force = 18 + 15 = 33kN (downward)

Figure A7

Refer to Figure A7, moment of force of the distributed load is 15 x 4 + 18 x 3 = 114kNm (clockwise)

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Tutorial Question For the following beams: (1) Calculate the reactions (2) Calculate the shear force and bending moment for drawing the shear force and bending moment diagrams Question (a)

Question (b)

Question (c)

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Question (d)

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Solution to Tutorial Question Question (a) Take moment about point B,

kNRR

R

A

A

A

16808010

)10()8(10)4(20

=→+=

=+

kNRRR

B

BA

142010

=→+=+

Consider a section X-X at a distance ‘x’ from A and considering the shear force,

When ,0=x kNFS 16.. +=

When ,20 ≤< x kNFS 16.. +=

When ,62 ≤< x kNFS 61016.. +=−+=

When ,106 << x kNFS 14201016.. −=−−+=

When ,10=x kNFS 14.. −=

Consider a section X-X at a distance ‘x’ from A and considering the bending moment, When ,0=x kNmBM 0= When ,20 << x kNmxxRBM A )(16+=⋅= When ,2=x kNmBM 32)2(16 +== When ,62 << x )2(10 −−= xxRBM A When ,6=x kNmBM 56)4(10)6(16 +=−+= When ,106 << x )6(20)2(10 −−−−= xxxRBM A When ,10=x kNmBM 0=

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Question (b) Take moment about B,

( )kNR

R

A

A

802

1)8)(8(20)8(

=→

=

and )8(20=+ BA RR

kNRB 80=→ Consider a section X-X at a distance ‘x’ from A and considering the shearing force, When ,0=x kNFS 80.. += When ,80 << x xRFS A ⋅−= 20.. When ,8=x kNFS 80.. −= Consider a section X-X at a distance ‘x’ from A and considering the bending moment, When ,0=x kNmBM 0+= When ,80 << x

220 xxxRBM A ⋅⋅−=

When ,8=x kNmBM 0+=

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Question (c) Take moment about B,

( )

kNRR

R

A

A

A

40)23(80)10(

243)4(20)343(

=+=

+=++

kNRR

RR

B

B

BA

404080

)4(20

=→−==+

Consider a section X-X at a distance ‘x’ from A and considering the shearing force, When ,0=x kNRFS A 40.. +== When ,30 ≤< x kNRFS A 40.. +== When ,73 ≤< x xxxRFS A 20100)3(2040)3(20.. −=−−=−−= When ,107 << x kNRFS A 408040)4(20.. −=−=−= When ,10=x kNRFS B 40.. −== Consider a section X-X at a distance ‘x’ from A and considering the bending moment, When ,0=x kNmRBM A 0+== When ,30 ≤< x xxRBM A 40+=⋅= When ,73 ≤< x ( )2

3)3(20 −−−⋅= xxxRBM A

When ,107 << x xxRBM B 40+=⋅+= When ,10=x kNmRBM a 0+=+=

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Question (d) Take moment about B,

( ) ( )

NR

RR

NRA

R

B

BA

A

33332

)5(800)5(800

2667

031)5(20002

1)5(4000)5(

+=→

+=+

+=→

=−−

xyxy 1605800 =→=Q

Consider a section X-X at a distance ‘x’ from A and considering the shearing force, When ,0=x NRFS A 2667.. +== When ,50 << x 28080026672

1800.. xxxyxRFS A −−=−−=

When ,5=x NRFS B 3333.. −== Consider a section X-X at a distance ‘x’ from A and considering the bending moment, When ,0=x kNmRBM A 0+== When ,50 << x ( )( )( ) ( ) 32

6160

2800266731

21

2800 xxxxxyxxxRBM A −−=−⋅−⋅=

When ,5=x kNmRBM a 0+=+= Since the max bending moment is where the shear force is equal to zero, Let 0808002667 2 =−− xx Then )(6.1264.2 rejectedorx −= Maximum bending moment at x = 2.64 = 7040.88 – 2787.84 -490.66 = 3762Nm

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