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

 Page 1


 
 
 
LIMIT STATE OF TORSION 
 
1. Equivalent Shear 
The equivalent shear is calculated by the following formula: 
?? ?? =?? ?? +1.6
?? ?? ?? 
 
Where, 
Ve = Equivalent shear force 
Vu = Shear force  
Tu = Torsional moment 
B = Width of the section 
2. Longitudinal reinforcement 
The longitudinal tension reinforcement should be designed to carry equivalent bending moment of 
?? ?? 1
=?? ?? +?? ?? 
Where, Mu = Flexural moment 
Mt = ?? ?? (
1+
?? ?? 1.7
) 
Tu = Torsional moment 
D = Overall depth of the section 
3. Transverse Reinforcement 
As per Is 456, transverse reinforcement is provided in the form of two legged closed hoops. The area 
of transverse reinforcement is obtained by the following formula: 
?? ????
=
?? ?? ?? ?? ?? 1
?? 1
(0.87?? ?? )
+
?? ?? ?? ?? 2.5?? 1
(0.87?? ?? )
 
Subjected to a maximum value of  
(?? ????
-?? ?? )?? ?? ?? 0.87?? ?? . 
 
Where, 
Tu = Torsional moment 
Vu = Shear force 
sv = Spacing of shear reinforcement 
b1 = centre to centre distance between corner bar in the direction of width 
d1 = centre to centre distance between corner bar in the direction of depth 
b = width of the member 
fy = Characteristics strength of stirrup reinforcement 
tve = equivalent nominal shear stress 
tc = shear strength of concrete  
Page 2


 
 
 
LIMIT STATE OF TORSION 
 
1. Equivalent Shear 
The equivalent shear is calculated by the following formula: 
?? ?? =?? ?? +1.6
?? ?? ?? 
 
Where, 
Ve = Equivalent shear force 
Vu = Shear force  
Tu = Torsional moment 
B = Width of the section 
2. Longitudinal reinforcement 
The longitudinal tension reinforcement should be designed to carry equivalent bending moment of 
?? ?? 1
=?? ?? +?? ?? 
Where, Mu = Flexural moment 
Mt = ?? ?? (
1+
?? ?? 1.7
) 
Tu = Torsional moment 
D = Overall depth of the section 
3. Transverse Reinforcement 
As per Is 456, transverse reinforcement is provided in the form of two legged closed hoops. The area 
of transverse reinforcement is obtained by the following formula: 
?? ????
=
?? ?? ?? ?? ?? 1
?? 1
(0.87?? ?? )
+
?? ?? ?? ?? 2.5?? 1
(0.87?? ?? )
 
Subjected to a maximum value of  
(?? ????
-?? ?? )?? ?? ?? 0.87?? ?? . 
 
Where, 
Tu = Torsional moment 
Vu = Shear force 
sv = Spacing of shear reinforcement 
b1 = centre to centre distance between corner bar in the direction of width 
d1 = centre to centre distance between corner bar in the direction of depth 
b = width of the member 
fy = Characteristics strength of stirrup reinforcement 
tve = equivalent nominal shear stress 
tc = shear strength of concrete  
 
 
Note: The distribution of transverse reinforcement should be such that the spacing should be a 
minimum value of x1, 
?? 1
+?? 1
4
 or 300 mm where x1 and y1 are short and long dimension of stirrup. 
 
x1 = b1 + Diameter of longitudinal bar + Diameter of stirrup 
y1 = d1 + Diameter of longitudinal bar + Diameter of stirrup