limit state of collapse flexure.pptx

1. Limit State of Collapse (Flexure)

2. Types of load

3. Stress Block Parameter

4. Actual Depth Of Neutral Axis Xu

5. Ultimate Moment Of Resistance

6. Modes of
Failure
Maximum Strain criteria has been accepted as
failure criteria
Strain in concrete in extreme fibre reach its
ultimate value
Actual Strain in steel at failure shall not be less
than 0.002 +0.87fy/Es
Failure modes depend upon relative position
of steel & concrete & % of steel in section

7. Types Of Failure
Balance Section
Under Reinforced section
Over Reinforced section

8. Balance section Failure
• Xumax=Xu
• Strain in steel & concrete reach their
maximum value simultaneously
• Percentage of steel in this section is
called as critical steel

9. Design Parameters For Balanced section
CONCRETE
GRADE
M15 M20
STEEL
GRADE
Fe250 Fe415 Fe500 Fe250 Fe415 Fe500
Kumax 0.53 0.48 0.46 0.53 0.48 0.46
Rumax 0.149fck 0.138fck 0.133fck 0.149fck 0.138fck 0.133fck
Pt%max 1.32 0.72 0.57 1.76 0.96 0.76

10. Under Reinforced
Failure
• Xumax > Xu
• Amount of steel is lesser than critical
percentage
• Strain in steel reach their maximum
value but collapse of beam occurs due
to crushing of concrete
• Failure occurs due to deflection &
extensive cracking of concrete giving
ample warning of impending failure
• From Economy point of view under
reinforced section are designed

11. Over Reinforced
Failure
• Xumax < Xu
• Amount of steel is lesser than critical
percentage
• Strain in concrete reach its ultimate
value first sudden failure of beam
occurs by crushing of concrete without
giving any warning
• Percentage of steel in this section is
more than critical steel
• IS code does not allow Over reinforced
section

12. Design of Singly Reinforced Rectangular
Section
ANALYSIS PROBLEM DESIGN PROBLEM

13. Analysis
Problem
• Given fck, Fy, b, d, Ast
• Calculate Xu & Xumax
• Compare Xu with Xumax
• Calculate Mur

14. Design of
Singly
Reinforced
Rectangular
Section
Analysis
Problem
• Calculate Xu
• Cu = Tu
• 0.36fckbXu = 0.87fyAst
• Xu = 0.87fyAst
0.36fckb

15. Design of
Singly
Reinforced
Rectangular
Section
Analysis
Problem
• Calculate Xumax
• Design Parameter for Balanced
Section
• Kumax = Xumax/ d
• Check for Under reinforced
section
• Xu < Xumax

16. Design of
Singly
Reinforced
Rectangular
Section
Analysis
Problem
• Calculate Mur
• Mur = 0.87 fy Ast (d-0.42 Xu )
• d = D – effective cover
• effective cover = nominal cover +
½ bar diameter

17. Details of
Beam for
Flexure
• Nominal Cover
• Minimum Steel
• Maximum Steel
• Diameter of Bar
• Spacing of Reinforcement

18. Beam
Reinforcement

19. Analysis Problem
Given b=230mm, d=450mm, Ast=3no-20mm dia,
Mur=?, M20 Grade concrete & Fe415 steel
Solution.
Ast=
Xu = 0.87fyAst/0.36fck b
=
Xumax= 0.48d
=
Compare Xu & Xumax
Mur=0.87FyAst(d-0.42Xu)
=

20. Analysis Problem
Given b=230mm, D=400mm, Ast=4no-12mm dia,
effective cover d’=31mm Mur=?, M20 Grade concrete & Fe415 steel
Solution. d =D-d”
d = 400-31
=
Xu = 0.87fyAst/0.36fck b
=
Xumax= 0.48d
=
Compare Xu & Xumax
Mur=0.87FyAst(d-0.42Xu)
=

21. Analysis Problem
L=3.5m
Mu=WuL2 /8
Mu=Mur

22. Analysis Problem
Given b=250mm, D=500mm, Ast=4no-16mm dia, effective cover=50mm, effective span =6m
Concentrated load P =? In addition to its self weight, M20 Grade concrete & Fe415 steel
Solution.
Ast=
Xu = 0.87fyAst/0.36fck b
=
Xumax= 0.48d
=
Compare Xu & Xumax
Mur=0.87FyAst(d-0.42Xu)
=

23. Analysis Problem
Mu=WuL2 /8 + PuL/4
Mu=Mur

24. Analysis Problem
Given b=200mm, d=400mm, Ast=4no-16mm dia,
Mur=?, M20 Grade concrete & Fe415 steel
Solution.
Ast=
Xu=0.87FyAst/0.36Fckb
Xu=
Xumax=0.48d
Xumax=

25. Analysis Problem
Mur = 0.87FyAst(d-0.42Xu)
Mur=
Murmax=Rumax b d2
Murmax=
Xumax=
Ptmax=
Astmax=

26. Analysis Problem
Ptmax%=0.36fck/0.87fy x Kumax

27. Find Xumax Mumax & Ptmax for M20 Fe415

30. Design
Problem
• Given fck, Fy, b, b/d & Mu
• Calculate D, Ast

31. Design of
Singly
Reinforced
Rectangular
Section
Design
Problem
• Calculate Xumax
• Calculate Murmax=Rumaxbd2
• Parameter for Balanced Section
• Check for Under reinforced section
• Mu < Murmax
• Calculate Ast

32. Design Problem
Given b=200mm, D=400mm, Mu=60KNm,
Ast=?, M20 Grade concrete & Fe500 steel & effective cover = 40mm
Solution.
d=
Xumax=0.48d
Xumax=
Murmax=
Check for under Reinforced section
Calculate Ast

33. Design Problem
Astmin=
Reinforcement Detailing

34. Design Problem
Given L=4.5m, b=230mm, D=?mm, w=30KN/m
Ast=?, M20 Grade concrete & Fe250 steel & Load Factor = 1.5
Design smallest concrete section
Solution.
Wu=
Mu=WuL2/8
Murmax=Mu
Xumax= d
Xumax=
Calculate Ast

35. Design Problem
Astmax=0.36FckbXumax/0.87Fy
b=
D=
Reinforcement Detailing

36. Design Problem
Given b=230mm, d=450mm, M=160KNm,
Ast=?, M20 Grade concrete & Fe415
Solution.
Mu=
Xumax=0.48d
Xumax=
Murmax=
Check for under Reinforced section
Calculate Ast

37. Design Problem
Redesign section
Mu=Murmax
d=
Astmax=
Reinforcement Detailing

40. Doubly Reinforced Section
• Section of a beam is restricted due to head room space appearance.
Applied moment exceeds moment resisting capacity of a singly reinforced
beam.
• If high B.M exists over a relatively short length of beam only than cost of
additional steel is small as compared to cost of singly reinforced beam of
constant cross section
• Compression steel is sometimes provided to reduce deflection or to
increase the stiffness of section

41. Doubly Reinforced Section

42. Behavior of Doubly Reinforced Section

43. Properties of Doubly Reinforced Rectangular
Section
Section1-A singly reinforced section in which compression in concrete
Cu1 is balanced by tension steel Ast1, this section resists part moment
Mu1 out of total moment Mu.
Section2- A section with compression steel Asc & additional tension
steel Ast2 resisting balanced moment Mu2=Mu-Mu1

44. Analysis Problem
Depth of Neutral Axis
Total Compression=Total Tension
C1+C2=T1+T2

45. Analysis Problem
Ultimate Moment of Resistance
Mur=Mu1+Mu2

46. σsc stress in
steel in
compression
Fe250

47. In case of Fe250 stress-
strain relationship is
linear & hence relation
between Xu & σsc can
be obtained
stress=strain x modulus
of elasticity

48. Idealised
stress strain
curve of
Fe250

49. σsc Fe250
 σsc=0.87fy Substitute in Xu Equ & calculate Xu
 Calculate σsc =700(1-dc/Xu) using calculated Xu
 If σsc calculated > 0.87fy than σsc=0.87fy
assumption is correct
 Use Xu & σsc =0.87fy to calculate Mur
 If σsc calculated < 0.87fy
 Than recalculate Xu using σsc =700(1-dc/Xu) in
Xu equ

51. Doubly Reinforced Analysis Problem
b=250mm, d=450mm, Ast=4-25dia, Asc=2-16dia, dc=50mm
Mur=? Use M20 & Fe250
Solution
Calculate Xu
Compare with Xumax

52. Doubly Reinforced Analysis Problem
Check σsc =700(1-dc/Xu)> 0.87fy
Calculate Mur=

53. Doubly Reinforced Analysis Problem
b=200mm, d=350mm, Ast=1600mm2, Asc=1245mm2,dc=50mm
Mur=? Use M20 & Fe250
Solution
Calculate Xu
Compare with Xumax

54. Doubly Reinforced Analysis Problem
Check σsc =700(1-dc/Xu)> 0.87fy
Calculate Mur

57. σsc in Fe415
HYSD Bar
In case of Fe415 HYSD bar no direct relation between stress &
strain is not available. So the inter relation σsc & Xu cannot be
established
So trial & error procedure required to obtain the depth of
neutral axis
Stress strain relationship is assumed linear for stress less than
or equal 0.8fy, for stress greater than 0.8fy the curve is non
linear
For HYSD the stress of 0.87fy reaches only at a strain of 0.0038
for Fe415 & 0.00417 for Fe500
Stress in compression steel never reaches a value of 0.87fy
prior of crushing of concrete because concrete fails at a strain
of 0.0035

58. Idealised
stress strain
curve for
HYSD

59. Steps to find σsc
• Calculate dc/d
• From table for dc/d ratio & grade of steel we get σsc
• Calculate Xu by using σsc
• Calculate Strain in steel
• From stress strain curve of HYSD, using calculated strain in steel calculate σsc
• This σsc calculated & σsc from dc/d ratio must be same

60. Values of σsc in
N/mm2
dc/d 0.050 0.075 0.100 0.125 0.150 0.175 0.2
Fe415 355 354 353 347.5 342 335.2 329
Fe500 424 418 412 403.5 395 382.5 370

61. Doubly Reinforced Analysis Problem HYSD
b=300mm, D=450mm, Ast=6-20diamm, Asc=4-20diamm, dc=40mm,
Mur=?, M20, Fe415?
Solution
Ast=
Asc=
d=
dc/d=

62. Doubly Reinforced Analysis Problem HYSD
σsc =
Calculate Xu=
Calculate Strain in steel using Xu

63. Doubly Reinforced Analysis Problem HYSD
σsc calculate from stress strain curve=
Check
If reqd second trial
Check for under reinforced section

64. Doubly Reinforced Analysis Problem HYSD
Calculate Mur=

65. Doubly Reinforced Analysis Problem HYSD
b=300mm, d=600mm, dc=30mm, fck=30N/mm2, Fy=500N/mm2,
Asc=2235mm2, Ast=4025mm2, Mur=?

66. Doubly Reinforced Analysis Problem HYSD
σsc = Calculate Xu= Calculate Strain in steel
using Xu

67. Doubly Reinforced Analysis Problem HYSD
σsc calculate from stress strain curve=
Check
If reqd second trial
Check for under reinforced section

68. Doubly Reinforced Analysis Problem HYSD
Calculate Mur=

69. Doubly Reinforced Analysis Problem HYSD
b=300mm, D=500mm, dc=40mm, fck=20N/mm2, Fy=415N/mm2, Asc=2-
12mm dia, Ast=4-25mm dia, M20 Fe415 Mur=?

70. Doubly Reinforced Analysis Problem HYSD

71. Doubly Reinforced Analysis Problem HYSD
Xumax=220mm,
Calculate Mur=

72. Using Ku
b=300mm, D=450mm, Ast=6-20diamm, Asc=4-20diamm, dc=40mm,
Mur=?, M20, Fe415?
Solution
Ast=
Asc=
d=
assume
Ku=

73. σsc from
table
Strain in
steel=
Calculate
Ku=0.3x(Ast-
Asc)/Ast1
Ast1=
Calculate

74. Calculate Xu=
Strain in steel=
σsc in steel from table=

76. Stress in HYSD bar

78. Design of
Doubly
Reinforced
Section

79. Design of Doubly Reinforced
Section
In Design of Doubly Reinforced section is kept balanced to make full
utilization of resistance of concrete means Ast1 is worked for
balanced section
Mu=Mu1+Mu2
Ast=Ast1+Ast2
Calculate design constant Kumax, Xumax, & Rumax
Murmax=Rumax x b x d2
Ast1=
Mu2=Mu-Murmax

80. Design of Doubly Reinforced Section
Ast2=
Strain in steel Fe250/HYSD
Cu2=Tu2
Asc=

81. Design Problem DRS
b=250mm, D=600mm, Mu=310KNm, M20, Fe250, dc=40mm, d’=55mm
Calculate amount of steel required ?
Calculate Design constant
Kumax=
Xumax=
Rumax=
Murmax=

82. Design Problem DRS
Ast1=
Mu2=
Ast2=
Ast=Ast1+Ast2

83. Design Problem DRS
σsc =
Asc=
Reinforcement Detailing

85. Design Problem DRS
b=250mm, d=500mm, M=160KNm, M20, Fe415, dc=50mm, d’=50mm
Calculate amount of steel required ?
Calculate Design constant
Kumax=
Xumax=
Rumax=
Murmax=

86. Design Problem DRS
Ast1=
Mu2=
Ast2=
Ast=Ast1+Ast2

87. Design Problem DRS
σsc =
Asc=
Reinforcement Detailing

89. Design Of Singly Reinforced T & L Beam

90. Effective width of Flange
• For T beam
• For L beam

91. Design Of Singly
Reinforced T & L Beam
• 23.1 T-Beams and L·Beams
• 23.1.1 General
A slab which is assumed to act as a compression
flange of a T-beam or L-beam shall satisfy the
following:
• a) The slab shall be cast integrally with the web,
or the web and the slab shall be effectively
bonded together in any other manner; and
• b) If the main reinforcement of the slab is parallel
to the beam, transverse reinforcement shall be
provided as in Fig. 3; such reinforcement shall
not be less than 60 percent of the main
Reinforcement at mid span of the slab.

92. Design Of Singly
Reinforced T & L Beam
• Case1 Xu < Df
• Case2 Xu > Df
• Case2a 3Xu/7 < Df
• Case2b 3Xu/7 > Df

93. Case1 Xu<Df

94. Case2 Xu>Df

95. Case2a 3Xu/7<Df

96. Case2b 3Xu/7 > Df

99. Properties of Balanced Section
• For Df/d < 0.2, than equation for case 2b shall be considered taking Xu =
Xumax

100. Properties of Balanced Section
• For Df/d > 0.2, than equation for case 2a shall be considered taking Xu =
Xumax

102. Analysis Problem
Steps
Assume Xu<Df if assumption is correct
calculate Mur
If assumption is wrong
Assume Xu>Df, 3Xu/7>Df if assumption is
correct calculate Mur
If assumption is wrong
Assume Xu>Df, 3Xu/7<Df if assumption is
correct calculate Mur

104. Analysis Problem T Beam
Mur=?, T, bw=240mm, bf=740mm, Df=80mm, d=400mm, Ast=5no-20mm
dia, M15, Fe415
Ast=
Assume Xu<Df
Calculate Xu=

107. Analysis Problem T Beam
Mur=?, T, bf=1200mm, d=600mm, bw=300mm, Df=110mm M20 Fe500
Ast=4-25mm dia

110. Analysis Problem T Beam
Mur=?, T, bf=1200mm, d=600mm, bw=300mm, Df=110mm M20 Fe500
Ast=7-25mm dia

112. Design Problem T &
L Beam
Steps
Calculate Mu1 for Xu = Df in Xu < Df
moment Equation
If Mu1 > Mu than Xu < Df ok
Calculate Ast
Check Xu < Xumax

113. Design Problem
If Mu1 < Mu than Xu > Df ok
Calculate Mu2 for Xu = 7Df/3 in moment Equation
If Mu2 > Mu than 3xu/7 < Df
Calculate Xu by using moment equation
Check for Xumax

114. Design Problem
Calculate Ast
If Mu2 < Mu than 3xu/7 > Df
Calculate Xu by using moment
equation
Check for Xumax
Calculate Ast

115. Design Problem
A T beam bf=1100mm, Df=120mm, d=600mm, bw=275mm, Ast=?, Mu=380KNm, M15 & Fe415
Assume Xu=Df
Calculate Mu1

117. Design Problem
A T beam bf=1100mm, Df=120mm, d=600mm, bw=275mm, Ast=?,
Mu=520KNm, M15 & Fe415
Assume Xu=Df
Calculate Mu1

120. Design Problem
A T beam bf=1100mm, Df=120mm, d=600mm, bw=275mm, Ast=?,
Mu=563KNm, M15 & Fe415

123. Design Problem
A T beam L=7m, c/c of beam=3m, Df=100mm, w=30KN/m, bw=230mm,
D=600mm, M20 Fe415 Ast=?