# Corrosion effects on strength of structure with example

Reinforced concrete structures are the most popular structures in society also they have the more share on total infrastructures due to its easiness in availability of materials to cast in any required shape. Although this is the mix of many materials mainly reinforcing bars, cement, sand, aggregate, water and admixtures. We have to get attention on every step to get the desired safety and life of structure. There are so many agents that they excite for the corrosion of rebar that will reduce the capacity of the section drastically. Chloride ion is one main agent for deterioration of structure if it crosses the threshold value of it. So for the structures to be built on severe chloride ion environment we have to consider some technical matters as below.

Design stage: During the design of any RC structures in severe chloride ion environment we have to think that how it will attack to the rebar so we can design the structure against the action of chloride ions. We can use the non-corroding reinforcement bars during the design such as stainless steel or fiber reinforcing bars also for case of concrete we can specify the high performance concrete. We have to provide sufficient cover thickness during the design. We can also apply the cathodic protection design.

Construction stage: Not only during the design stage we have to give more attention at the time of construction. Construction is the implementation time of design so if we design well but not constructed as per design and specification it does not mean. During the construction mainly we have to consider followings for reducing the corrosion environment

- Sufficient cover thickness of concrete as per design.
- Selection of cement, sand, aggregate, water and admixtures as per specification.
- We can use the galvanized or epoxy coated reinforcement bars.
- Well compaction of concrete that avoid the presence of pores and can reduce the passage of the chloride ion from outside to the reinforcement. Also we can use self-compacting concrete.
- Curing must have to do in right way considering the temperature and cement, admixtures used for the strong and impermeable cover concrete.

Maintenance stage: maintenance is also the major stage for caring the structure. In severe chloride ion environment we have to apply the routine maintenance work including the inspection and then fast action of remedial measures to protect form the future more deterioration. We can apply any suitable methods of maintenance as like surface coating or cathodic protection or others.

##
Example calculation of corrosion effect on reinforcement.

*Given condition,*

*i)*

*For initial condition without corrosion*
Sound concrete strength (f’

_{c })_{ }=24 N/mm^{2 }
Sound yield strength of steel (f

_{sy })_{ }=300 N/mm^{2 }
Ultimate strain (Ɛ

_{sf})_{ }=0.2
Area of tension steel (Ast) =5-25φ = 2455.36 mm

^{2 }
Area of compression steel (Asc) =4-16φ = 804.57 mm

^{2}

*ii)*

*Corrosion of concrete by 50%*
Concrete strength (f’

_{c })_{ }=12 N/mm^{2 }
Yield strength of steel (f

_{sy })_{ }=300 N/mm^{2 }
Ultimate strain (Ɛ

_{sf})_{ }=0.2
Area of tension steel (Ast) =2455.36 mm

^{2 }
Area of compression steel (Asc) =804.57 mm

^{2}

*iii)*

*Corrosion of tension steel by 50%*
Concrete strength (f’

_{c })_{ }=24 N/mm^{2 }
Yield strength of steel (f

_{sy })_{ }=300 N/mm^{2 }
Ultimate strain (Ɛ

_{sf})_{ }=0.2
Area of tension steel (Ast) =1227.68 mm

^{2 }
Area of compression steel (Asc) =804.57 mm

^{2}

*iv)*

*Corrosion of compression steel by 50%*
Concrete strength (f’

_{c })_{ }=24 N/mm^{2 }
Yield strength of steel (f

_{sy })_{ }=300 N/mm^{2 }
Ultimate strain (Ɛ

_{sf})_{ }=0.2
Area of tension steel (Ast) =2455.36 mm

^{2 }
Area of compression steel (Asc) =402.29 mm

^{2}
Total depth (D) =650mm

Width of beam section (b) =400mm

Cover (d’) =50mm

Here this is the doubly reinforced section,

d’/d= 50/600 =.082 <0.1 so

Here

k

_{i}= 1.0-0.003f’_{c}
=0.964 >0.85 so take k

_{i}=0.85
Ɛ’

_{cu }= (155-fc)/30000
=0.00476>0.0035 so take Ɛ’

_{cu }=0.0035
β = 0.52+80* Ɛ’

_{cu}
= 0.8

a = 0.8*x

####
**Calculation for the position of neutral axis (NA)**

Find out a by equating the tension and compression forces

C

_{c}+C_{s}= T
k

_{i}*f’_{c}*a*b + Asc*f`_{sy}= Ast*f_{sy}
Considering compression steel not yields

Firstly find out the strain induced on compression steel when concrete have the ultimate strain

From Fig: A,

Ɛ’

_{s}= (x-50)*0.0035/x
Now

f`

_{sy}= Ɛ’_{s }*Es

*i)*

*For initial condition without corrosion*
0.85*24*0.8*x*400 + 804.57* (x-50)*0.0035/x*2*10

^{5 }= 2455.36*300
6528 x

^{2}-173407.1 x – 28160000 = 0
By solving the above equation

X= 80.29 mm

*ii)*

*Corrosion of concrete by 50%*
0.85*12*0.8*x*400 + 804.57*(x-50)*0.0035/x *2*10

^{5 }= 2455.36*300
3264 x

^{2}-173407.1 x – 28160000 = 0
By solving the above equation

X= 123.17 mm

*iii)*

*Corrosion of tension steel by 50%*
0.85*24*0.8*x*400 + 804.57*(x-50)*0.0035/x *2*10

^{5 }= 1227.68*300
6528 x

^{2}+194896.4 x – 28160000 = 0
By solving the above equation

X= 52.42 mm

*iv)*

*Corrosion of compression steel by 50%*
0.85*24*0.8*x*400 + 402.29* (x-50)*0.0035/x*2*10

^{5 }= 2455.36*300
6528 x

^{2}- 455007.1 x – 14080000 = 0
By solving the above equation

X= 92.91 mm

####
**Checking of compression steel stress (f’**_{sy}<300)

_{sy}<300)

f’

_{sy}= (x-50)*0.0035/x*2*10^{5}

*i)*

*For initial condition without corrosion*
f’

_{sy }=264.08 < 300 hence compression steel not yield

*ii)*

*Corrosion of concrete by 50%*
0.85*12*0.8*x*400 + 804.57* (x-50)*0.0035/x*2*10

^{5 }= 2455.36*300
f’

_{sy }=415.84 > 300 hence compression steel yield

*iii)*

*Corrosion of tension steel by 50%*
f’

_{sy }=32.39 < 300 hence compression steel not yield

*iv)*

*Corrosion of compression steel by 50%*
f’

_{sy }=323.30 > 300 hence compression steel yield####
**Calculation of Moment capacity Mu**

Mu = (Ast*f

_{sy}– Asc*f’_{sy})*(d-0.8*x/2) + Asc*f’_{sy}*(d – d’) for compression steel not yield condition
Mu = (0.85*f

_{ck}*b*0.8*x )*(d-0.8*x/2) + Asc*f_{sy}*(d – d’) for compression steel yield condition again we have to find out location of neutral axis by using following condition
k

_{i}*f’_{c}*0.8*x*b + Asc*f_{sy}= Ast*f_{sy}

*i)*

*For initial condition without corrosion*
Compression steel not yield

Mu = (2455.36*300 – 804.57*264.08)*(600-0.8*80.29/2) + 804.57*264.08*(600 – 50)

=

*414.51 KN-m*

*ii)*

*Corrosion of concrete by 50%*
Compression steel not yield, so

*X=(2455.36-804.57)*300/(0.85*12*0.8*400)*
x = 128.96 mm

Mu = (0.85*12*400*0.8*128.97)*(600-0.8*128.97/2) + 804.57*300*(600 – 50)

=

*363.60 KN-m*

*iii)*

*Corrosion of tension steel by 50%*
Compression steel not yield

Mu = (1227.68*300 – 804.57*32.40)*(600-0.8*52.43/2) + 804.57*52.42*(600 – 50)

=

*212.50 KN-m*

*iv)*

*Corrosion of compression steel by 50%*
Compression steel not yield, so

X=(2455.36-402.28)*300/(0.85*12*0.8*400)

x = 80.20 mm

Mu = (0.85*24*400*0.8*80.20)*(600-0.8*80.20/2) + 402.28*300*(600 – 50)

*= 363.70 KN-m*####
**Comparison of Moment capacity Mu**

We can calculate the ratio of the moment capacity of corroded section with original section.

For corrosion of concrete

r=363.70/414.51

=0.877

For corrosion of tensile steel

r=212.50/414.51

=0.513

For corrosion of compression steel

r=363.60/414.51

=0.877

Hence from the above ratios we can say that

**for the moment capacity reduction on degradation of materials. Compression steel and concrete have almost equal effect on the capacity reduction. So we have to protect the tensional rebar from the degradation agents.***tension steel is more sensitive*
## No comments: