DOI : 10.17577/IJERTV4IS070897
- Open Access
- Total Downloads : 356
- Authors : Bhakare Vishal Shamrao, Dharane . S. S
- Paper ID : IJERTV4IS070897
- Volume & Issue : Volume 04, Issue 07 (July 2015)
- DOI : http://dx.doi.org/10.17577/IJERTV4IS070897
- Published (First Online): 30-07-2015
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Experimental Performance, Mathematical Modeling and Development of Stress Block Parameters of R.C.C Beam with Rectangular Trough Shaped Reinforcement
.Bhakare Vishal S.
PG Student ,ME Civil (Structures), SVERIs College of Engineering Pandharpur,
Solapur University, India
Dharane S. S.
Assistant Professor,
SVERIs College of Engineering Pandharpur, Solapur University, India
Abstract In civil engineering design and construction the main steel in beams and slabs are used to take the tension and to increase the ductile behavior. In slabs the main and distribution steel is provided in horizontal form. Also the main steel in singly reinforced and doubly reinforced beams is in horizontal form only. But if we replace this horizontal form of the reinforcement in beams and slabs by Rectangular Trough shaped, reinforcement there will be increase in the moment carrying, shear carrying capacity with reduced deflection by enhancing its ductile behavior because of additional folded plate action and it takes in to account the reversal of loading thereby the sections becomes more strong in fatigue action.
Keywords Beams, Slabs, Main Reinforcement, Rectangular Trough shaped reinforcement and ductility.
-
INTRODUCTION
In civil engineering design and construction steel reinforcements are used to take mainly tension and increase ductility of the members such as beams, slabs, footing, etc. In case of doubly reinforced beams and columns the steel is also used to take compressive force along with increase in its ductility. In present theory the main steel used in the beam at top & bottom is in horizontal form. So instead of providing separately steel at top and bottom side of the beam if the steel is provided in the form of alternate Rectangular Trough shaped main and distribution steel the beam will also act as a folded plate up certain limit. Because of the action of folded plate in addition to bending and shear the load carrying capacity of the beam will also increase with reduced deflection.
Sidramappa Dharane & Archita Malge [1]: The paper of effect of shape of main reinforcement in slabs. The aim of the work is to study the effect of rectangular trough shaped reinforcement in beams, in this case, instead of horizontal bar at top & bottom in doubly beam, if we replace this horizontal form of the reinforcement in beams and slabs by rectangular trough shaped reinforcement, there will be increase in the moment carrying, shear carrying capacity with reduced deflection because of folded plate action and thereby the sections becomes more economical.
Sidramappa Dharane & Archita Malge [2]: The paper of civil engineering analysis and design software -limitations of end- users and feedback almost all structural engineers are assigning the live loads on all the floors /slabs/beams and analyzing the framed structures. But assigning the live loads on all the floors/beams is not being the critical case. Critical analysis is required for safe design. And to get the critical section it is necessary to improve the software which takes into account the number of trials i.e. various positions of live loads. The software should be improved in such a way that it should take automatically all positions of live load on floors/beams for critical analysis.
Mahmood et. al. [3]: Studied the effect of different numbers of wire mesh layers on the flexural strength of folded and flat ferrocement panels and to compare the effect of varying the number of wire mesh layers on the ductility and the ultimate strength of these types of ferrocement structure. Seven ferrocement elements were constructed and tested each having (600x380mm) horizontal projection and 20mm thick, consisting of four flat panels and three folded panels. The experimental and numerical results show the superiority of the folded to the flat panel in terms of ultimate strength and initiation of cracking.
.
Rao et. al. [4]: Studied the shear strength of 6 ferrocement panels having size 600x150mm with 25mm thick. Each panel was tested under various spans to depth ratios from 1 to 6. Based on the experimental result, a formula for predicting the shear strength of the ferrocement panel was proposed. The aim of the work was to study the effect of varying the number of steel wire layers on the flexural behavior of folded ferrocement panels and to compare cracking, ultimate flexural strength and load deflection behavior with that of the flat panels.
-
EXPERIMENTAL PROGRAMS
-
.Methodology-
The experimental programme consists of casting, testing of reinforced concrete beams, cubes with different shapes of reinforcement provided in the beams at various percentage of steel at top & bottom. The reinforcement provided in the beam is alternatively rectangular trough shaped. Total 38 reinforced high strength concrete beams were cast.
B Details of beam specimen
The specimens used were cubes, beam specimens
.Dimensions of each test specimen are as under: Cube: 150 mmx150 mm x150 mm,
Cube: 75 mmx75 mm x75 mm, Beam: 150 mm x 150 mm x 700 mm
Beam specimens were used to determine .
,flexural of frc beams .
Cubes of 150 mm x 150 mm size Cubes of 75 mm x 75 mm
were used to find the compressive strength.
Compressive strength of cubes are determined at 28days using compression testing machine (CTM) of capacity 2000 KN. FTM testing machine of capacity 250 KN was used to determine the flexural strength of beams.
C. Test Materials-
-
Cement Ordinary Portland cement whose 28 day compressive strength was 53Mpa was used.
-
Fine Aggregate – Natural River sand confirming with specific gravity is 2.65 and fineness modulus 2.33 was used.
-
Coarse Aggregate – Crushed Coarse aggregate of 20mm and 10 mm procured from local crusher grading with specific gravity is 2.63 was used.
-
Water- Portable water free from any harmful amounts of oils, alkalis, sugars, salts and organic materials was used for proportioning and curing of concrete.
-
Steel Diameter of steel is used 6mm.
-
Proportion- cement: sand: aggregate: water (1:1.5:3:0.38)
-
Ferrocement-
-
Welded Square Mesh-
-
Wire diameter 1 millimetres
-
Size of mesh opening 12.5 X 12.5 mm
-
Maximum use of 12 layers of mesh per 6 mm of thickness.
-
Maximum 10 square in transverse directions.
-
Maximum 55 square in longitudinal directions.
-
Poisson ratio is 0.3.
REINFORCEMENT PROFILE:-
1. RECTANGULAR TROUGH SHAPED ONE WAY DIRECTION REINFORCEMENT:-
In reinforced cement concrete beam the main reinforcement is provided along the longitudinal direction is straight bar and in transverse direction trough shaped reinforcement is provided at 60 mm spacing each interval. Because of trough shaped arrangement the load carrying capacity, moment resisting capacity, shear resisting capacity & ductility will increase with reduce deflection and self weight. Also it takes into account the reversal of loading. Due to inter locking of trough shaped reinforcement the bond between concrete and steel is very strong.
The fig shows the detailed Rectangular trough shaped arrangement.
Rectangular Trough Shaped Reinforcement Profile
2.RECTANGULAR TROUGH SHAPED TWO WAY DIRECTIONAL REINFORCEMENT:-
In reinforced cement concrete beam the main reinforcement is provided along the longitudinal direction is trough shaped and in transverse direction trough saped reinforcement is provided at 60 mm spacing each interval. Because of trough shaped arrangement the load carrying capacity, moment resisting capacity, shear resisting capacity & ductility will increase with reduce deflection and self weight. Also it takes into account the reversal of loading. Due to inter locking of trough shaped reinforcement the bond between concrete and steel is very strong. Due to folded plate action the beam takes reversal loading which is beneficial earthquake.
Sr.n o
Mix designation
Specification
28 th days Collapse Load
(KN)
Collapse load KN
Average Collapse
load KN
1
BC -1
RCC NO OF BAR AT TOP
&BOTTOM -2
84
83.33
2
BC -2
RCC NO OF BAR AT TOP &
BOTTOM 2
85
3
BC -3
RCC NO OF BAR AT TOP &
BOTTOM 2
72
4
BC-4
RCC NO OF BAR
AT TOP & BOTTOM 3
86
94
5
BC-5
RCC NO OF BAR
AT TOP & BOTTOM 3
98
6
BC-6
RCC NO OF BAR AT TOP &
BOTTOM 3
98
7
BC -7
RCC NO OF BAR
AT TOP & BOTTOM 4
104
114.66
8
BC- 8
RCC NO OF BAR AT TOP &
BOTTOM 4
110
9
BC -9
RCC NO OF BAR AT TOP &
BOTTOM 4
130
10
VSB- 1
ONE DRECTIONAL RECTANGULAR TROUGH SHAPED
REINFORCEMENT
97
96
11
VSB- 2
ONE DRECTIONAL RECTANGULAR TROUGH
SHAPED REINFORCEMENT
95
12
VSB- 3
TWO DRECTIONAL RECTANGULAR TROUGH
SHAPED REINFORCEMENT
98
96.5
13
VSB- 4
TWO DRECTIONAL RECTANGULAR TROUGH
SHAPED REINFORCEMENT
95
14
BC- 10
FERROCEMENT
NO.OF MESH 3
20
20.5
15
BC -11
FERROCEMENT
NO.OF MESH 3
21
The fig shows the detailed Rectangular trough shaped arrangement.
Rectangular Trough Shaped Reinforcement Profile
-
-
Proportion-cement:sand:water (1:2:0.38
-
-
III RESULT AND DISCUSSSION
A. .Figures and tables-
Results after casting, testing of concrete beams are shown in table 1.
Result of average collapse load for Rcc,Ferrocement & Rectangular Trough shaped reinforcement beam specimens shown in table
Table 1
LSM CALCULATINON-:
LSM CALCULATION-1. RCC NO OF BAR AT TOP & BOTTOM-2
By using LSM we can calculate the collapse load (W) of section
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm
,Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 2 X /4 X d 2= 2 X /4 X 6 2= 56.54 mm 2 In compression zone)
Assume Xu =0.4 d = 0.4 X 135 = 54 mm
-
Tensile force in concrete (Tu) = 0.87fy. Ast
=12.29×103 N
-
Compressive force in concrete (Cu) = 0.36fck X Xub + fsc X Asc – 0.446 fck X Asc =76.017 X10 3 N
-
Moment of Resistance (MR) = Ccu (d 0.416 Xu) + Csu (d-d) = 8.038 x106 KNM.
MR=BM
8.038 x106 = WL/6
… W = 80.38 KN.
Check for stress-
(6max)T = M/I x Y (6max)T = 16.67 N/mm2
16.67 N/mm2=17.833N/mm2 – Hence ok.
In this case the therotical (6max)T is matches with the experimental & ansys results, it means that the valu of 6max is equal to the actual value of stress in experimental & ansys results(6max)A
LSM CALCULATION-2. RCC NO OF BAR AT TOP & BOTTOM-3
By using LSM we can calculate the collapse load (W) of section
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm , Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20
-
Compressive force in concrete (Cu) = 0.36fck X Xub
+ fsc X Asc – 0.446 fck X Asc =76.011 X10 3 N
-
Moment of Resistance (MR) = Ccu (d 0.416 Xu) + Csu (d-d) =8.77 x106KNM.
MR=BM 8.77 x106= WL/6
… W = 87.76 KN
Check for stress-
(6max)T = M/I x Y
(6max)T = 18.21 N/mm2
18.21 N/mm2=19.995N/mm2 – Hence ok.
In this case the therotical (6max)T is matches with the experimental & ansys results, it means that the valu of 6max is equal to the actual value of stress in experimental & ansys results(6max)A
LSM CALCULATION-3. ONE DIRECTIONAL RECTANGULAR TROUGH SHAPED STRAIGTH BAR AT TOP-3 &BOTTOM -3
By using LSM we can calculate the collapse load (W) of section
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm , Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 3 X /4 X d 2= 2 X /4 X 6 2= 84.82 mm In compression zone
Assume Xu =0.4 d = 0.4 X 135 = 54 mm
-
Tensile force in concrete (Tu) = 0.87fy. Ast =18.44×103 N
-
Compressive force in concrete (Cu) = 0.36fck X Xub
+ fsc X Asc – 0.446 fck X Asc =76.011 X10 3 N
-
Moment of Resistance (MR) = Ccu (d 0.416 Xu) + Csu (d-d) =8.77 x106KNM.
MR=BM 8.77 x106= WL/6
… W = 87.76 KN
Check for stress-
N/mm,2 Grade of steel (fy)= 250 N/ mm2,
(6max)T
= M/I x Y
Asc= 3 X /4 X d 2= 2 X /4 X 6 2= 84.82 mm In compression zone
Assume Xu =0.4 d = 0.4 X 135 = 54 mm
1.Tensile force in concrete (Tu) = 0.87fy. Ast =18.44×103 N
(6max)T = 18.21 N/mm2
18.21 N/mm2=20.535N/mm2 – Hence ok.
In this case the therotical (6max)T is matches with the experimental & ansys results, it means that the valu of 6max is equal to the actual value of stress in experimental & ansys results(6max)A
LSM CALCULATION-4. TWO DIRECTIONAL RECTANGULAR TROUGH SHAPED REINFORCEMENT.
By using LSM we can calculate the collapse load (W) of section
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm , Diameter of Bar (d)= 6mm ,
Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 4 X /4 X d 2= 2 X /4 X 6 2= 113.097 mm 2
In compression zone
Assume Xu =0.4 d = 0.4 X 135 = 54 mm
-
Tensile force in concrete (Tu) = 0.87fy. Ast =24.59×103 N
-
Compressive force in concrete (Cu) = 0.36fck X Xub
+ fsc X Asc – 0.446 fck X Asc =81.90X10 3 N
-
Moment of Resistance (MR) = Ccu (d 0.416 Xu) + Csu (d-d) =9.51 x106KNM.
-
Depth of N.A ( n) :-
Using equation to find the value of n=?
b x n 2 /2 + ( mc -1 ) Asc ( n d ) = m x Ast ( d n )-
————————— [1]
But, mc =1.5 m= 1.5 x 13.33 =19.995
Putting above values in equation (1) we get,
150x n 2 /2 + ( 19.995 1 ) x 56.54( n -15 ) =
13.33 x56.54 (135 n )
75 n2 +1.827 x10 3 n -117.84 x 10 3 = 0
Solving we get,
n= 29.28 mm.
-
Moment of Resistance (MR) = b x n x c/2 (d n/3 ) + ( mc
-1 ) Asc x C ( d d )———–[2]
But, C = C ( b d/ n ) = 7 ( 150 -15 /29.28 ) =32.27
Putting above values in equation (2) we get,
(MR) = 150 x 29.28x (7/2 ) (135- 29.28/3 ) + (19.995 1) x 56.54 x 32.27 x (135 15 )
=6.084 x 10 6 KNM.
… Moment of Resistance (MR) =6.084 x106 KNM.
Equating MR to BM MR=BM
BM = 6.084 x 10 6
But, BM formula for two point loading is = WL/6
… WL/6 = 6.084 x 10 6
WL = 6.084 x 10 6 x 6
MR=BM
9.51 x106= WL/6
… W = 95.14KN
Check for stress-
WL =36.504 x 10 6/600
… W =60.84 KN.
Check for stress-
(6max)T = M/I x Y
By using flexural strength formula we can calculate the maximum stress (6max ), which is matches the results of Ansys stress & experimental stress.
(6max)T = 19.73 N/mm2
19.73 N/mm2=20.842N/mm2 – Hence ok.
In this case the therotical (6max)T is matches with the experimental & ansys results, it means that the valu of 6max
M/I = 6/Y
Where,
M = Bending moment I = Moment of inertia
is equal to the actual value of stress in experimental & ansys results(6max)A
WSM CALCULATIONS 1- RCC NO OF BAR AT TOP -2 & BOTTOM-2:
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm
,Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 2 X /4 X d 2= 2 X /4 X 6 2= 56.54 mm 2 (In compression zone)
6cbc = c = 7 , and m =13.33——————- for M20 grade concrete ,Table no 2.1, Page No-42
Y= Depth of N.A to top of beam, ( i.e d/2)
-
Bending moment (M ) = WL/6
= 60.84 x10 3x 700/6
= 7.098 x10 6 knm.
-
Moment of inertia ( I ) = bd3/12
= 150 x 150 3/12
=42.18 x 106 mm4
-
Depth of N.A to top of beam (Y)= d/2
= 150/2
= 75 mm
-
Maximum stress ( 6max ) = M/I x Y
-
= 7.098 x10 6/42.18 x 106 x75
(6max ) = 12.62 N/mm2
In this case the therotical 6max is not approximately matches with the experimental & ansys results, it means that the value of 6max is less than actual value of stress in experimental & ansys results.So that the variation of stress block is not goes parabolic.
(6max )T< (6max)A—-Hence not ok.
12.62 N/mm2 < 17.833N/mm2—Hence not ok. Hence we can not use this stress varitation diagram for new stress block diagram approch.
WSM CALCULATIONS 2- RCC NO OF BAR AT TOP -3 & BOTTOM-3:
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm
,Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 3 X /4 X d 2= 3 X /4 X 6 2= 84.82 mm 2 (In compression zone)
6cbc = c = 7 , and m =13.33——————- for M20 grade concrete ,Table no 2.1, Page No-42
-
Depth of N.A ( n) :-
Using equation to find the value of n=?
b x n 2 /2 + ( mc -1 ) Asc ( n d ) = m x Ast ( d n )-
—————————- [1]
But, mc =1.5 m= 1.5 x 13.33 =19.995
Putting above values in equation (1) we get,
150x n 2 /2 + ( 19.995 1 ) x 84.82 ( n -15 ) =
13.33 x84.82 (135 n )
75 n2 +2.741 x10 3 n -176.79 x 10 3 = 0
Solving we get,
n= 33.60 mm.
-
Moment of Resistance (MR) = b x n x c/2 (d n/3 ) + ( mc
-1 ) Asc x C ( d d )———–[2]
But, C = C ( b d/ n ) = 7 ( 150 -15 /33.60 ) =28.12
Putting above values in equation (2) we get,
(MR) = 150 x 33.60x (7/2 ) (135- 33.60/3 ) + (19.995 1) x 84.82 x 28.12 x (135 15 )
=7.62 x 10 6 KNM.
… Moment of Resistance (MR) =7.62 x106 KNM.
Equating MR to BM MR=BM
BM = 7.62 x 10 6
But, BM formula for two point loading is = WL/6
… WL/6 = 7.62 x 10 6
WL = 7.62 x 10 6 x 6
WL =45.72 x 10 6/600
… W =76.20 KN.
Check for stress-
By using flexural strength formula we can calculate the maximum stress (6max ), which is matches the results of Ansys stress & experimental stress.
M/I = 6/Y
Where,
M = Bending moment I = Moment of inertia
Y= Depth of N.A to top of beam, ( i.e d/2)
-
Bending momen (M ) = WL/6
= 76.20 x10 3x 700/6
= 8.89 x10 6 knm.
-
Moment of inertia ( I ) = bd3/12
= 150 x 150 3/12
=42.18 x 106 mm4
-
Depth of N.A to top of beam (Y)= d/2
= 150/2
= 75 mm
-
Maximum stress ( 6max ) = M/I x Y
-
= 8.89 x10 6/42.18 x 106 x75
(6max ) = 15.80 N/mm2
In this case the therotical 6max is not approximately matches with the experimental & ansys results, it means that the value of 6max is less than actual value of stress in experimental & ansys results.So that the variation of stress block is not goes parabolic.
(6max )T< (6max)A ——Hence not ok.
15.80 N/mm2 < 19.995N/mm2—Hence not ok.
Hence we can not use this stress varitation diagram for new stress block diagram approch.
WSM CALCULATIONS 3- ONE DIRECTIONAL RECTANGULAR TROUGH SHAPED AND OTHER DIRECTIONAL STRAIGTH BAR AT TOP &BOTTOM -3
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm
,Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 3 X /4 X d 2= 3 X /4 X 6 2= 84.82 mm 2 (In compression zone)
6cbc = c = 7 , and m =13.33——————- for M20 grade concrete ,Table no 2.1, Page No-42
-
Depth of N.A ( n) :-
Using equation to find the value of n=?
b x n 2 /2 + ( mc -1 ) Asc ( n d ) = m x Ast ( d n )-
—————————- [1]
But, mc =1.5 m= 1.5 x 13.33 =19.995
Putting above values in equation (1) we get,
150x n 2 /2 + ( 19.995 1 ) x 84.82 ( n -15 ) =
13.33 x84.82 (135 n )
75 n2 +2.741 x10 3 n -176.79 x 10 3 = 0
Solving we get,
n= 33.60 mm.
-
Moment of Resistance (MR) = b x n x c/2 (d n/3 ) + ( mc
-1 ) Asc x C ( d d )———–[2]
But, C = C ( b d/ n ) = 7 ( 150 -15 /33.60 ) =28.12
Putting above values in equation (2) we get,
(MR) = 150 x 33.60x (7/2 ) (135- 33.60/3 ) + (19.995 1) x 84.82 x 28.12 x (135 15 )
=7.62 x 10 6 KNM.
… Moment of Resistance (MR) =7.62 x106 KNM.
Equating MR to BM MR=BM
BM = 7.62 x 10 6
But, BM formula for two point loading is = WL/6
… WL/6 = 7.62 x 10 6
WL = 7.62 x 10 6 x 6
WL =45.72 x 10 6/600
… W =76.20 KN.
6cbc = c = 7 , and m =13.33——————- for M20 grade concrete ,Table no 2.1, Page No-42
-
Depth of N.A ( n) :-
Using equation to find the value of n=?
b x n 2 /2 + ( mc -1 ) Asc ( n d ) = m x Ast ( d n )-
—————————- [1]
Specimens
Averag e Collaps e load by(EX P) (KN)
Colla pse load by (AN SYS) (KN)
Colla pse load by LSM
(KN)
Collaps e load by WSM
(KN)
Stress Results
Expe rime ntal resul t
Mpa
Ansy s resul t Mpa
New approc h LSM
Results Mpa
New approc h WSM
Results Mpa
RCC NO OF BAR AT TOP -2& BOTTOM-2
83.66
82.8
80.38
60.84
17.35
/td>
17.83
16.67
12.62
RCC NO OF BAR AT
TOP -3& BOTTOM-3
94
93.4
87.76
76.20
19.49
19.99
18.21
15.80
RCC NO OF BAR AT TOP -4& BOTTOM-4
114.66
112.2
95.14
89.93
23.64
23.23
19.73
18.65
ONE DIRCTIONA L RECTANGU LAR TROUGH
SHAPED
96
95.1
87.76
76.20
19.91
20.53
18.20
15.98
TWO DIRCTIONA L RECTANGU LAR TROUGH
SHAPED
96.5
95.4
95.14
89.99
20.04
20.84
19.73
18.94
Check for stress-
By using flexural strength formula we can calculate the maximum stress (6max ), which is matches the results of Ansys stress & experimental stress.
M/I = 6/Y
Where,
M = Bending moment I = Moment of inertia
Y= Depth of N.A to top of beam, ( i.e d/2)
-
Bending moment (M ) = WL/6
= 76.20 x10 3x 700/6
= 8.89 x10 6 KNm.
-
Moment of inertia ( I ) = bd3/12
= 150 x 150 3/12
=42.18 x 106 mm4
-
Depth of N.A to top of beam (Y)= d/2
= 150/2
= 75 mm
-
Maximum stress ( 6max ) = M/I x Y
= 8.89 x10 6/42.18 x 106
x75
(6max ) = 15.98 N/mm2
In this case the therotical 6max is not approximately matches with the experimental & ansys results, it means that the value of 6max is less than actual value of stress in experimental & ansys results.So that the variation of stress block is not goes parabolic.
(6max )T< (6max)A –Hence not ok.
15.98 N/mm2 < 20.535N/mm2–Hence not ok.
Hence we can not use this stress varitation diagram for new stress block diagram approch.
WSM CALCULATIONS 4- TWO DIRECTIONAL RECTANGULAR TROUGHS SHAPED REINFORCEMENT
Given data:
Width of section (b)=150 mm, Depth of section (D)= 150 mm
,effective cover (d)=15 mm, effective depth (d)=135 mm
,Diameter of Bar (d)= 6mm ,Grade of concrete (fck)=20 N/mm,2 Grade of steel (fy)= 250 N/ mm2,
Asc= 4 X /4 X d 2= 3 X /4 X 6 2= 113.097 mm 2 (In
compression zone)
But, mc =1.5 m= 1.5 x 13.33 =19.995
Putting above values in equation (1) we get,
150x n 2 /2 + ( 19.995 1 ) x 113.097 ( n -15 ) =
13.33 x113.097 (135 n )
75 n2 +3.655 x10 3 n -235.73 x 10 3 = 0
Solving we get,
n= 36.76 mm.
-
-
Moment of Resistance (MR) = b x n x c/2 (d n/3 ) + ( mc
-1 ) Asc x C ( d d )———–[2]
But, C = C ( b d/ n ) = 7 ( 150 -15 /36.76 ) =25.70
Putting above values in equation (2) we get,
(MR) = 150 x 36.76x (7/2 ) (135- 36.76/3 ) + (19.995 1) x 113.097 x 25.70 x (135 15 )
=8.99 x 10 6 Knm.
… Moment of Resistance (MR) = 8.99 x106 Knm.
Equating MR to BM MR=BM
BM = 8.99 x 10 6
But, BM formula for two point loading is = WL/6
… WL/6 = 8.99 x 10 6
WL = 8.99 x 10 6 x 6
WL =53.96 x 10 6/600
… W =89.99 KN.
-
Check for stress-
Sr no
Mix proport ion
Specificat ion
Stress Results
Remark
Experi mental result Mpa
Ansys result Mpa
New approa ch LSM
Results Mpa
New approc h WSM
Results Mpa
1
M20
RCC NO OF BARS
2
17.351
17.833
16.67
12.62
LSM IS APPLIC ABLE
2
M20
RCC NO OF BARS
3
19.496
19.995
18.21
15.80
3
M20
RCC NO OF BARS
4
23.644
23.237
19.73
18.65
4
M20
ONE DIRECTI ONAL RECTA NGULAR TROUGH SHAPED REINFO RCEMEN T
19.991
20.535
18.20
15.98
5
M20
TWO DIRECTI ONAL RECTA NGULAR TROUGH SHAPED REINFO RCEMEN T
20.04
20.842
19.73
18.98
By using flexural strength formula we can calculate the maximum stress (6max ), which is matches the results of Ansys stress & experimental stress.
Where,
M = Bending moment I = Moment of inertia
M/I = 6/Y
Y= Depth of N.A to top of beam, ( i.e d/2)
-
Bending momen (M ) = WL/6
= 89.99×10 3x 700/6
= 10.59 x10 6 knm.
-
Moment of inertia ( I ) = bd3/12
= 150 x 150 3/12
=42.18 x 106 mm4
-
Depth of N.A to top of beam (Y)= d/2
= 150/2
= 75 mm
-
Maximum stress ( 6max ) = M/I x Y
= 10.59 x10 6/42.18 x 106 x75
(6max ) = 18.94 N/mm2
In this case the therotical 6max is not approximately matches with the experimental & ansys results, it means that the value of 6max is less than actual value of stress in experimental & ansys results.So that the variation of stress block is not goes parabolic.
(6max )T< (6max)A —–Hence not ok.
18.94 N/mm2 < 20.842N/mm2–Hence not ok.
Hence we can not use this stress varitation diagram for new stress block diagram approch.
Collapse load results & Stress distribution Results:-
By calculating the above theoretical values of load & stress, the experimental collapse load values is nearly matches with the values of limit states of collapse load so we can draw the new stress block diagram for limit state of collapse, but the stress distribution by the ansys is matches with the limit state of stress diagram i.e. the variation of stress distribution using the ansys stress diagram is going straight at top fiber & then decreasing with parabolic to the neutral axis at zero point.
Stress distribution Results:-
The interpretation of results of all the stress block parameters is goes parabolic & decrease from top fibre compression zone to the neutral axis is at zero point, so that the limit state theory is applicable for such stress distribution of stress block parameters. The stress block distribution of all the beams is as shown in the Table no.6.3
New Stress Distribution Diagram : Rcc- no of bar At top-2 & bottom-2
The variation of stress distribution of rcc no of bar At top-2 & bottom-2 as shown in table and fig. below
Stress Results
Experimental
result Mpa
Ansys result Mpa
New approach LSM
Results Mpa
17.351
17.833
16.67
New Rcc Stress Block Diagram No of bar At top-2 & bottom-2
Deflection of beam having 2 bar at top and bottom
Von-Mises stress of beam having two bar at top and bottom.
New Stress Distribution Diagram : Rcc- no of bar At top-3 & bottom-3
The variation of stress distribution of rcc no of bar At top-3 & bottom-3 as shown in table and fig
Stress Results
Experimental result Mpa
Ansys result Mpa
New approach LSM Results Mpa
19.496
19.995
p>18.21 New Rcc Stress Block Diagram No of bar At top-2 & bottom-2
.Deflection of beam having three bar at top and bottom.
Von-Mises stress of beam having three bar at top and bottom
New Stress Distribution Diagram : rectangular trough shaped- no of bar At top-3 & bottom-3
Stress Results
Experimental
result Mpa
Ansys result Mpa
New approach LSM
Results Mpa
20.535
New rectangular trough shaped Stress Block Diagram
(One Way Directional)
Deflection of beam having one directional rectangular trough shaped reinforcement
Von-Misses stress of beam having one directional rectangular trough shaped reinforcement.
Deflection of beam having two directional rectangular trough shaped
Von-Mises stress of beam having two directional rectangular trough shaped reinforceme
Stress Results
Experimental
result Mpa
Ansys result Mpa
New approach
LSM Results Mpa
20.014
20.842
New Stress Distribution Diagram: rectangular trough shaped- no of bar in longitudinal direction -3
New rectangular trough shaped Stress Block Diagram (Two Way Directional)
PROJECT PHOTO:-
No of beams casting
Flexural Testing Machine
Beam Testing with dial gauge
IV. CONCLUSION
The following conclusions can be drawn from the outcome of this project study:
-
The variation of ANSYS stress results is approximately matches with the stress results of theoretical calculation by limit state method of stress, so that the stress variation of stress block diagram is goes parabolic to the neutral axis zero.
-
Increase in moment resisisting , shear force, ductile behaviour capacity of the beams.
-
Increase in stiffness of the beams and reduces the deflections.
-
Increase in durability of the beams.
-
-
ACKNOWLEDGMENT
I am ineffably indebted to Prof.S.S.Dharane, Dr.Prashant Pawar, for conscientious guidance and encouragement to accomplish this project.
REFERENCE
Initial cracking
Failure pattern of reinforced concrete beams
-
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