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1. DESIGN OF STEEL AND
RCC STRUCTURES
CHAPTER 4: DESIGN OF SHEAR REINFORCEMENT AND BOND

2. Shear in a loaded beam is caused by the variation of bending moment
along the span.
The variation in bending stress at two cross sections in the same fibre
gives rise to shear stress.A combination of shear stress and bending stress
produces principal tension at an angle of 45 degrees and is called diagonal
tension.
As concrete is weak in tension , large diagonal tensile stresses can
produce cracking and even failure in concrete members.
Hence if shear stresses are large, steel in the form of vertical stirrups
or bent up bars should be provided to take up shear stresses.

4. Nominal shear stress ( ‫ح‬v): The nominal shear stress ‫ح‬v in beams of uniform depth can be
obtained by the equation ‫ح‬v =
𝑉𝑢
𝑏𝑑
Where Vu is the shear force due to design loads. b =breadth of beam
d = effective depth

5. Design shear strength of concrete ( ‫ح‬c) :
A)Without shear reinforcement: The shear strength of concrete in beams without shear
reinforcementis given in the table.
The value of ( ‫ح‬c) for intermediate percentage of steel can be obtained by interpolation
‫ح‬c = ‫ح‬c +
‫ح‬c 2
−
‫ح‬c1
𝑝𝑡2
−𝑝𝑡1
(pt –pt1)
B)For solid slabs: The design shear strength for concrete shall be ‫ح‬c x k where k has the
following values:
C)Maximum shear stress in concrete with shear reinforcement ( ‫ح‬cmax):
The nominal shear stress even with shear reinforcement cannot exceed ‫ح‬cmax
Overall depth of
slab in mm
300 or
more
275 250 225 200 175 150 or
less
k 1.0 1.05 1.1 1.15 1.20 1.25 1.30
Concrete grade
in N/mm2
M15 M20 M25 M30 M35 M40
‫ح‬cmax in N/mm2 2.5 2.8 3.1 3.5 3.7 4.0

6. D) For solid slabs: The nominal shear stress shall not exceed half the approximate
values given in the table
ie ‫ح‬v shall not be greater then ½ ‫ح‬cmax
Maximum spacing of shear reinforcement:
The maximum spacing of shear reinforcement shall not exceed :
0.75d for vertical stirrups and
d for inclined stirrups at 45 degrees
Or 300mm
where d is the effective depth of the section.

7. Minimum shear reinforcement:
Minimum shear reinforcement in the form of stirrups shall be provided such that:

8. When ‫ح‬v exceeds ‫ح‬ c , shear reinforcement shall be provided in the foll forms:
a) Vertical stirrups
b) Bent up bars along with stirrups
c) Inclined stirrups
Contribution of bentup bars should not be more then half of the total shear
reinforcement.
The strength of shear reinforcement can be calculated as below:
a) For vertical stirrups :
Vus =
0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑
𝑆𝑣
a) For inclined stirrups or a series of bars bentup at different cross sections:
Vus =
0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑 (𝑆𝑖𝑛α+𝐶𝑜𝑠α)
𝑆𝑣

9. c) For single bars or single group of parallel bars (All parallel bent up bars are at
the same cross section}:
Vusb = 0.87 fy Asb Sinα
Where :
Asv is the total cross sectional area of stirrups.
Asb is the total cross sectional area of bentup bars
Sv is the spacing of stirrups or bentup bars
‫ح‬v =Nominal shear stress
‫ح‬c = Design shear strength of concrete
b = breadth of the member (For flanged beams, b= bw)
fy = characteristic strength of stirrups or bent up reinforcements which shall not
be greater then 415 N/mm2
α = angle between the inclined stirrups or bent up bars and the axis of the
members , which shall not be less then 45 degrees.
d = effective depth
If a combination of Fe415 stirrups and Fe500 stirrups are used , then fy shall be
taken as 415 N/mm2

10. Uses of bentup bars:
1)It enhances economy since unwanted tension bars are bentup and used as shear
Reinforcement.
2) In most cases, half of the shear reinforcement is contributed by bentup bars.
3) Cracks due to diagonal tensile stresses can be resisted by bent up bars.

11. Procedure for shear design in beams:
1) Calculate shear force V =
𝑤𝐿
2
Calculate factored shear force Vu = V x 1.5
2) Calculate nominal shear stress ‫ح‬v=
𝑉𝑢
𝑏𝑑
It should be less then ‫ح‬cmax
3) Calculate pt and find the value of shear strength of concrete ‫ح‬c
4)If ‫ح‬v < ‫ح‬c Provide nominal shear reinforcement
If ‫ح‬v > ‫ح‬c shear reinforcement is required.
5)Calculate shear force for which shear reinforcement is required
Vus = Vu - ‫ح‬c bd
6) If bentup bars are used, shear is carried by bent up bars
Vusb = 0.87 fy Asb Sinα and it should be less then equal to 0.5 Vus
Asb is the total cross sectional area of bentup bars
α = angle between the bent up bars and the axis of the members , which shall not be
less then 45 degrees.
Sin α = 0.707 for α = 45 degrees
If bentup bars are not provided, then delete this step.

12. 7)Calculate shear force to be resisted by vertical stirrups
Vusv = Vus – Vusb ( It has to be greater or equal to Vus)
Or
Vusv = Vus when bent up bars are not provided
Calculate spacing Sv of stirrups of area Asv
Sv =
0.87 𝑓𝑦 𝐴𝑠𝑣 𝑑
𝑉𝑢𝑠𝑣
Spacing should be less then or equal to 0.75d 0r 300mm

13. Numericals:
1) A rectangular beam of effective size 300mm x 900 mm has a span of 6m . It is
loaded with 35 KN/m over the entire span which is inclusive of self weight.Tensile
steel consists of 5 bars of 25 mm diameter . Use M20 and Fe415. Design shear
reinforcement using limit state method . For M20 grade of concrete, ‫ح‬cmax = 2.8 Mpa
Given table :
2)Design the shear reinforcement for a reinforced concrete beam 300 mm wide and 450 mm
deep effective for the foll data
a)M20 grade concrete and Fe415 steel is used for main steel
b)Bent up bars are not provided
c) 5 bars of 20 mm are used as tension steel
d)Effective span of 6 m and udl of 60 KN/m
e)10mm bar is used for shear reinforcement
f) ‫ح‬cmax = 2.8 Mpa
pt ‫ح‬c
0.5 0.48
0.75 0.56
1.0 0.62
1.25 0.67

14. 3) An RC beam 300 mm wide and 600 mm deep (overall) is reinforced with four bars of
25 mm diameter . It carries a udl of 75 KN/m over a span of 6.3 m . Design the shear
reinforcement . Use M20 grade of concrete and Fe415 grade steel . Assume two bars are
bent up . Use stirrups of 8mm diameter . Take cover = 40 mm. Given ‫ح‬cmax = 2.8 Mpa
4)Design the shear reinforcement for a RC beam 230 mm wide and 450 mm deep effective
For the foll data:
a)M25 and Fe500 is used for main steel. b)5 bars of 20 mm are used as tension steel
c)Factored shear force = 210 KN d)Fe415 – 8 mm bars are used as shear reinf
e) ‫ح‬cmax = 3.1 Mpa. Assume 2 bars are bentup
Given :
pt ‫ح‬c
0.5 0.48
0.75 0.56
1.0 0.62
1.25 0.67
pt ‫ح‬c
0.75 0.57
1.0 0.64
1.25 0.70

15. 5) A 230 mm x 500 mm (effective) beam is subjected to a factored shear force of 200 KN.
Design 8 mm diameter two legged stirrups . The shear strength of concrete is 0.52 Mpa.
Use steel of grade Fe415.

16. Bond: It refers to the interaction between reinforcing steel bars and the surrounding
concrete due to combined effect of adhesion, friction and bearing.
The adhesion between concrete and steel is provided by concrete during setting.
Friction is provided by gripping of bars due to shrinkage of concrete.
The bearing strength of concrete is related to the compressive strength and the ratio of
total surface area to load bearing area.
Types of bond:
1) Anchorage bond: It is developed in order to transfer the force from the bar to the
surrounding concrete.
2) Flexural bond: It is a bond developed between steel and concrete to resist the
variation in bending moment
3) Development bond: It is developed to transfer the force from the surrounding
concrete to the reinforcing bar.

17. Bond Failure: It is likely to occur near the ends of beams.

18. Anchorage bond and development length:
Anchorage bond stress arises when a bar carrying a certain force has to be terminated.
Anchorage length is the length of the bar Ld required to transfer the force from the bar
to the surrounding concrete . It is also called Development length Ld.
Where Ld =
0.87𝑓𝑦ɸ
4 ‫ح‬bd
Where ɸ is the nominal diameter of bar
‫ح‬bd is design bond stress
For deformed bars, the value of ‫ح‬bd shall be increased by 60 %
For bars in compression, the value of ‫ح‬bd shall be increased by 25 %.
Grade of concrete M20 M25 M30 M35 M40 and
above
Design bond stress ‫ح‬bd
in N/mm2
1.2 1.4 1.5 1.7 1.9

19. Lap length should be :
1) In tension: Ld or 30 ɸ whichever is greater
2) In compression: Ld or 24 ɸ whichever is greater
Anchoring reinforcing bars:
i)Anchoring bars in tension:
1)Deformed bars may be used without end anchorages provided development length
required is satisfied. Hooks should normally be provided for plain bars in tension.
2)Bends and hooks: The bends and hooks should conform to IS 2502
a) Bends : 45 deg = 4ɸ subjected to a max of 16 times the diameter of the bar.
90 deg = 8ɸ
b)U type hook = 16 ɸ
ii)Anchoring bars in compression: The anchorage length of straight bars in compression
Should be taken as Ld.
Ld =
0.87 𝑓𝑦 ɸ
4 ‫ح‬𝑏𝑑
where ‫ح‬𝑏𝑑 is the bond stress and ɸ is the diameter of the bar.

20. Available Development length:
Ldavail =
1.3 𝑀1
𝑉
+Lo
Where M1 = Moment of resistance of the bars continued into the support.
V = Total shear force at the section due to design loads
Lo = Sum of anchorage beyond the centre of the support and
the equivalent anchorage value of any hook.

21. 1)Calculate the anchorage value of 45 degrees and 90 degrees bend for 25 mm
diameter bar.
2)Calculate development length required for a bar of 32 mm dia in tension and
Compression. Use M20 grade concrete and Fe500 steel.
Given ‫ح‬𝑏𝑑 = 1.2N/mm2 for M20 grade concrete.
3)Find the lap length required for a bar of 25 mm diameter in tension and compression
for M20 grade concrete with mild steel and Fe500 steel.
Given ‫ح‬𝑏𝑑 = 1.2N/mm2 for M20 grade concrete.
4)A RC beam 230 mm wide and 500 mm deep effective is reinforced with 5 bars of 16mm at
midspan and 4 of these bars are continued into the support. The beam is 4m in span and carries
A load of 50 KN/m. Calculate development length at the supports . Assume M15 grade of concrete
And Fe415 steel . Given ‫ح‬𝑏𝑑 = 1.0 N/mm2 .
5) A simply supported beam is 5m in span and carries a characteristic load of 65 KN/m . If 5 nos
of 25 mm diameter bars are provided at the centre of the span and 4 of these bars are continued
into the support. Check the development length at the supports assuming M20 grade of concrete
and Fe415 steel. Given ‫ح‬𝑏𝑑 = 1.2N/mm2 for M20 grade concrete.