Maximum moment that can be resisted by a given section is termed as Moment of resistance.
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-67-1024x439.png)
a) Balanced Section
MOR = C x LA or, T x LA = MoR,lim
MoR,lim = (0.36 fck Xu,lim ) b (d-0.42 Xu,lim)
= 0.36 fck Xu,lim (d-0.42 Xu,lim) x b x d2/d2
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-68.png)
It can also written as,
MoR, lim = Q bd2
Or,
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-69.png)
Note : For Balanced Section,
Ast = Ast,lim
C= T
(0.36 fck Xu,lim ) b = 0.87 fy Ast,lim
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-70.png)
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-71.png)
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-72-1024x297.png)
b) Under-Reinforced Section
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-73-1024x463.png)
c) Over-Reinforced Section
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-74-1024x446.png)
Note : Here issue is actual depth of Neutral axis is computed by equating compressive force with tensile force.
C= T
(0.36 fck Xu,lim ) b = 0.87 fy Ast,lim
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-65.png)
Here, there is two unknowns for over reinforced section. instead of 0.87fy, fst is for for this section. Hence to obtain depth of Neutral axis for over reinforced section following procedure is adopted (Hit and Trial Method)
Step I) Assume some value of Xu (Xu>Xu,lim) and calculate strain in steel.
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-75-1024x634.png)
Step II) Calculate the stress in steel corresponding the above strain from stress-strain diagram .
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-76-1024x666.png)
Step III) Calculate of New depth of Neutral axis
![](https://ervivekshah.com.np/wp-content/uploads/2024/04/image-77.png)
if new Xu is equal to Xu, assumed (Xu,a) then OKAY, Repeat the process till Xu =Xu,a otherwise.
- Section is always fails due to crushing.
- Steel never fails, it always yield.
- Balanced failure is that where steel & concrete yields simultaneously.
- Tension failure / Ductile failure / Secondary compression failure is the failure initiated by yielding of steel but takes place due to crushing of concrete (Always preferred due to sufficient warning).
- Brittle failure /primary compression failure is that which takes place due to crushing of concrete (Not preferred)