Short Notes: Limit State of Shear | Short Notes for Civil Engineering - Civil Engineering (CE) PDF Download

 Page 1


 
 
  
 
 
     
 
 
      
 
LIMIT STATE OF SHEAR 
 
1. NOMINAL SHEAR STRESS 
 
? The average shear stress can be calculated using the following formula: 
?? ?? =
?? ?? ????
 
Where, 
Vu = ultimate shear stress at the section 
b = width of the section 
d = effective depth of the section 
? For beams with varying depth 
( )
2u
v
V M d tan
bd
?? ??
?=??
??
??
 
Where, 
ß = inclination of flexural tensile force to the horizontal. 
Mu = factored bending moment at the section.  
  
Page 2


 
 
  
 
 
     
 
 
      
 
LIMIT STATE OF SHEAR 
 
1. NOMINAL SHEAR STRESS 
 
? The average shear stress can be calculated using the following formula: 
?? ?? =
?? ?? ????
 
Where, 
Vu = ultimate shear stress at the section 
b = width of the section 
d = effective depth of the section 
? For beams with varying depth 
( )
2u
v
V M d tan
bd
?? ??
?=??
??
??
 
Where, 
ß = inclination of flexural tensile force to the horizontal. 
Mu = factored bending moment at the section.  
  
 
 
2. DESIGN SHEAR STRENGTH 
The design shear strength of concrete depends upon two factors: 
(i) Grade of concrete 
(ii) Percentage tensile reinforcement 
The value of tc is given in table 19 of IS: 456. 
s
100 A
bd
 Concrete grade 
(1) 
M20 
(2) 
M25 
(3) 
M30 
(4) 
= 0.15 0.28 0.29 0.29 
0.25 0.36 0.36 0.37 
0.50 0.48 0.49 0.50 
0.75 0.56 0.57 0.59 
1.00 0.62 0.64 0.66 
1.25 0.67 0.70 0.71 
1.50 0.72 0.74 0.76 
 
The nominal shear stress should not exceed the maximum shear strength of concrete as given 
in table 20 of IS: 456 to avoid compression failure. Maximum shear strength is given in the 
following table: 
Concrete Grade M-15 M-20 M-25 M-30 M-35 M-40 and above 
tc max (N/mm
2
) 2.5 2.8 3.1 3.5 3.7 4.0 
 
3. MINIMUM SHEAR REINFORCEMENT 
Minimum amount of shear reinforcement should always be provided in RCC section to 
? Avoid sudden shear failure.  
? To hold the reinforcing bars together 
? To prevent cracks in concrete due to shrinkage, thermal stresses etc. 
? IS 456 specifies the following formula for calculation of minimum shear reinforcement. 
sv
vy
A 0.4
bS 0.87 f
??
??
? ? ?
??
??
??
??
??
 
Where Asv = total cross-sectional area of stirrup legs effective in shear. 
Sv = spacing of stirrups. 
b = breadth of the beam or breadth of web of flanged beams. 
fy = characteristic strength of stirrup reinforcement in N/mm
2
 which shall not be taken greater 
than 415 N/mm
2
. 
 
4. DESIGN OF SHEAR REINFORCEMENT 
Page 3


 
 
  
 
 
     
 
 
      
 
LIMIT STATE OF SHEAR 
 
1. NOMINAL SHEAR STRESS 
 
? The average shear stress can be calculated using the following formula: 
?? ?? =
?? ?? ????
 
Where, 
Vu = ultimate shear stress at the section 
b = width of the section 
d = effective depth of the section 
? For beams with varying depth 
( )
2u
v
V M d tan
bd
?? ??
?=??
??
??
 
Where, 
ß = inclination of flexural tensile force to the horizontal. 
Mu = factored bending moment at the section.  
  
 
 
2. DESIGN SHEAR STRENGTH 
The design shear strength of concrete depends upon two factors: 
(i) Grade of concrete 
(ii) Percentage tensile reinforcement 
The value of tc is given in table 19 of IS: 456. 
s
100 A
bd
 Concrete grade 
(1) 
M20 
(2) 
M25 
(3) 
M30 
(4) 
= 0.15 0.28 0.29 0.29 
0.25 0.36 0.36 0.37 
0.50 0.48 0.49 0.50 
0.75 0.56 0.57 0.59 
1.00 0.62 0.64 0.66 
1.25 0.67 0.70 0.71 
1.50 0.72 0.74 0.76 
 
The nominal shear stress should not exceed the maximum shear strength of concrete as given 
in table 20 of IS: 456 to avoid compression failure. Maximum shear strength is given in the 
following table: 
Concrete Grade M-15 M-20 M-25 M-30 M-35 M-40 and above 
tc max (N/mm
2
) 2.5 2.8 3.1 3.5 3.7 4.0 
 
3. MINIMUM SHEAR REINFORCEMENT 
Minimum amount of shear reinforcement should always be provided in RCC section to 
? Avoid sudden shear failure.  
? To hold the reinforcing bars together 
? To prevent cracks in concrete due to shrinkage, thermal stresses etc. 
? IS 456 specifies the following formula for calculation of minimum shear reinforcement. 
sv
vy
A 0.4
bS 0.87 f
??
??
? ? ?
??
??
??
??
??
 
Where Asv = total cross-sectional area of stirrup legs effective in shear. 
Sv = spacing of stirrups. 
b = breadth of the beam or breadth of web of flanged beams. 
fy = characteristic strength of stirrup reinforcement in N/mm
2
 which shall not be taken greater 
than 415 N/mm
2
. 
 
4. DESIGN OF SHEAR REINFORCEMENT 
 
 
? When the nominal shear stress exceeds the design shear strength, extra shear reinforcement 
is provided in the form of 
• Vertical/Inclined stirrup 
 
• Bent up bars 
 
 
 
? The design shear stress is given by the following formula: 
Vus = (Vu – Vc) = (tv – tc) bd 
Where, 
Vu = total Shear Force. 
Vc = shear resisted by concrete 
Vus = shear resisted by reinforcements (Links or bent up bars) 
tv = nominal shear stress. 
tc = design shear stress of concrete  
5. Vertical/Inclined Stirrup 
? The spacing of vertical stirrup can be calculated by using the following formula: 
y sv
v
us
0.87 f .A .d
S
V
??
=
??
??
??
 
Where, 
Asv = total area of the legs of shear reinforcement. 
Sv = spacing of the links. 
d = effective depth of section. 
Page 4


 
 
  
 
 
     
 
 
      
 
LIMIT STATE OF SHEAR 
 
1. NOMINAL SHEAR STRESS 
 
? The average shear stress can be calculated using the following formula: 
?? ?? =
?? ?? ????
 
Where, 
Vu = ultimate shear stress at the section 
b = width of the section 
d = effective depth of the section 
? For beams with varying depth 
( )
2u
v
V M d tan
bd
?? ??
?=??
??
??
 
Where, 
ß = inclination of flexural tensile force to the horizontal. 
Mu = factored bending moment at the section.  
  
 
 
2. DESIGN SHEAR STRENGTH 
The design shear strength of concrete depends upon two factors: 
(i) Grade of concrete 
(ii) Percentage tensile reinforcement 
The value of tc is given in table 19 of IS: 456. 
s
100 A
bd
 Concrete grade 
(1) 
M20 
(2) 
M25 
(3) 
M30 
(4) 
= 0.15 0.28 0.29 0.29 
0.25 0.36 0.36 0.37 
0.50 0.48 0.49 0.50 
0.75 0.56 0.57 0.59 
1.00 0.62 0.64 0.66 
1.25 0.67 0.70 0.71 
1.50 0.72 0.74 0.76 
 
The nominal shear stress should not exceed the maximum shear strength of concrete as given 
in table 20 of IS: 456 to avoid compression failure. Maximum shear strength is given in the 
following table: 
Concrete Grade M-15 M-20 M-25 M-30 M-35 M-40 and above 
tc max (N/mm
2
) 2.5 2.8 3.1 3.5 3.7 4.0 
 
3. MINIMUM SHEAR REINFORCEMENT 
Minimum amount of shear reinforcement should always be provided in RCC section to 
? Avoid sudden shear failure.  
? To hold the reinforcing bars together 
? To prevent cracks in concrete due to shrinkage, thermal stresses etc. 
? IS 456 specifies the following formula for calculation of minimum shear reinforcement. 
sv
vy
A 0.4
bS 0.87 f
??
??
? ? ?
??
??
??
??
??
 
Where Asv = total cross-sectional area of stirrup legs effective in shear. 
Sv = spacing of stirrups. 
b = breadth of the beam or breadth of web of flanged beams. 
fy = characteristic strength of stirrup reinforcement in N/mm
2
 which shall not be taken greater 
than 415 N/mm
2
. 
 
4. DESIGN OF SHEAR REINFORCEMENT 
 
 
? When the nominal shear stress exceeds the design shear strength, extra shear reinforcement 
is provided in the form of 
• Vertical/Inclined stirrup 
 
• Bent up bars 
 
 
 
? The design shear stress is given by the following formula: 
Vus = (Vu – Vc) = (tv – tc) bd 
Where, 
Vu = total Shear Force. 
Vc = shear resisted by concrete 
Vus = shear resisted by reinforcements (Links or bent up bars) 
tv = nominal shear stress. 
tc = design shear stress of concrete  
5. Vertical/Inclined Stirrup 
? The spacing of vertical stirrup can be calculated by using the following formula: 
y sv
v
us
0.87 f .A .d
S
V
??
=
??
??
??
 
Where, 
Asv = total area of the legs of shear reinforcement. 
Sv = spacing of the links. 
d = effective depth of section. 
 
 
7. For inclined stirrup: 
( )
y st
us
v
0.87 f A d
V sin cos
S
= ? + ? 
Where, 
a = Angle of inclination of stirrup 
8. The spacing between two stirrup shall be minimum of following values: 
(?? ) ?? ??  >
?  (?? ?? )
??????  ?? h??????  ?????? (???? ) ?? ??  >
?  0.75??  (??????  ?? -???????????????? )
              >
?  ??  (??????  ????????????????  ???????????????? ) 
(?????? ) ?? ??  >
?  300 ????
}
 
 
 
 
 ??????  ?????????????? ???? ?? h?????? 
9. Bent Up Bars 
Vus = 0.87 fy Asb sin a  
Where, 
a = Angle of inclination of bar with horizontal 
Asb = Area of bent up bar