**Key Topics**

Civil Engineering assignment based on strengthening concrete design.

### Introduction

Strengthening concrete design is the concept used to strengthen the structure/dream of an architect or client. The Euro code standards come up with general structural designs for every structure stands for traditional or advanced in nature. This standard is generally used in the Europe and being considered in several countries. Designing of concrete structures is based on environmental conditions, nature of the materials used, and natural disasters are clearly and evidently discussed in one of their kinds known “Euro code 2: Design of Concrete Structures”. EC 2 is explained with the modal created in SAP 2000 and analysed the same modal in Abacus software. This report emphasis the column & beam design of a curve – type office building (African Union Headquarters) using Euro code 2 standards. The aim is to understand the depth of EC 2 and the usage of SAP 2000 and Abacus software.

EC2 is a concept of structural design of concrete structures designates the procedures for the concrete design, strengthening and pre-stressed concrete with a help of limit state design concept. Compression and tensile loads are effectively acting over the structure that in turn failure occurs due to improper design solution. Concrete is the material that withstand compression load and steel is the material that withstand both compression and tensile loads. These materials can be combined and does technical magic in the structure strength. The procedure for the combination of concrete and steel are carried out using EC 2 standard. It is related to the requirements of serviceability, durability, and resistance of concrete structures. Thermal and sound insulation effects are not concerned under this standard.

EC2 is a concept of structural design of concrete structures designates the procedures for the concrete design, strengthening and pre-stressed concrete with a help of limit state design concept. Compression and tensile loads are effectively acting over the structure that in turn failure occurs due to improper design solution. Concrete is the material that withstand compression load and steel is the material that withstand both compression and tensile loads. These materials can be combined and does technical magic in the structure strength. The procedure for the combination of concrete and steel are carried out using EC 2 standard. It is related to the requirements of serviceability, durability, and resistance of concrete structures. Thermal and sound insulation effects are not concerned under this standard.

The model is created in SAP 2000, is an engineering tool used widely for the designing or analysis of different kind of structural systems. The fundamental and most-recent design, conversion of complex model from 2D geometry, can be designed, checked, and optimised with the help of user-friendly space that makes the engineering process to be efficient. It contains finite element analysis technique with analysis engine integration and advanced analysis feature are available with non-linear and dynamic considerations. A broad range of modelling options for the structures like buildings, bridges, landscape infrastructure, or special structures like dams, stadiums, prisons, and offshore systems are implemented. Under loading conditions, wind, snow, seismic, vehicle, wave and thermal forces will be generated automatically and assigned accordingly.

The analysis of structures ranging from buildings to roads to bridges and other infrastructures have to withstand loads directed by the occupants, equipments, vehicles, wind, artificial and natural disasters can be accurately carried out using Abacus software, to ensure safe and economical design. The optimisation is accurate and keen for the various type of analysis used globally to incorporate less failure to the structures or products.

### Literature review

**Geometric data of the structure:**

The structure consists of twenty-one levels (above ground level) with one underground level which lies under Terrain Category III (for wind actions).

The total plot area is 1236.2 m^2 and the height is 86.4 m and working life(100 years)

**Assumptions on materials:**

I – Concrete:

The characteristic compressive cylinder strength is

f_ck = 40MPa

The partial factor for fatigue load is

Sub-clause: 2.4.2.3(1), γ_(F,fat) = 1.0

For serviceability limit state, the partial factor is

sub-clause: 2.4.2.4(2), γ_c = 1.0

For ultimate limit state, the partial factor is

sub-clause: 2.4.2.4(1)

**Design Criteria**γ_c for concrete

Persistent & transient 1.5

Accidental 1.2

The coefficient strength classes (Class N)

s = 0.25

The ratio α = 2⁄3 for t ≥ 28 days.

The co-efficient of compressive strength considering long-term effects and un-favourable effects, sub-clause: 3.1.6(1)P,α_cc = 1.0

The co-efficient of tensile strength considering long-term effects and un-favourable effects, sub-clause: 3.1.6(2)P, α_ct = 1.0

**II – Reinforcing steel:**

The yield strength of the strengthening is given in

sub-clause: 3.2.2(3)P, f_yk = 550MPa

For serviceability limit state, the partial factor is given in

sub-clause: 2.4.2.4(2), γ_s = 1.0

For ultimate limit state, the partial factor is given in

Sub-clause: 2.4.2.4(1)

**Design Criteria**

γ_s for reinforcing steel

**Persistent & transient**

1.15

Accidental 1.0

The assumed mean density value is 7850kg⁄m^3

The Young’s Modulus assumed for the strengthening bars is 200 GPa.

Material properties:

I – Concrete:

The compressive strength at time (t) is shown in the equation

f_ck(t) = f_ck = 40 MPa, for “t greater or equal to 28 days”

The compressive strength at various ages is shown by the equation

f_cm(t) = β_cc(t).f_cm = 48MPa

Where

β_cc(t) = co-efficient that depends on age t = e^([?s(1-(28⁄t)?^(1/2)])= 1.0

f_cm = mean compressive strength = f_ck + 8 MPa = 48 MPa

The development of tensile strength at time ‘t’ is shown in the equation

f_ctm(t) = ?[β_cc (t)]?^α.f_ctm = 3.51 MPa

f_ctm = mean tensile strength = 0.3.?f_ck?^(2⁄3) = 3.51 MPa

The modulus of elasticity is given by the equation

E_cm = 22.?{f_cm⁄10}?^0.3 = 35.2 GPa where f_cm is in MPa

Since, the limestone aggregate is used, E_cm is reduced by 10% , therefore, E_cm = 31.68 GPa

The modulus of elasticity is considered to be varying with time ‘t’ as given in the equation

E_vt(t) =?((f_cm (t))/f_cm )?^0.3.E_cm = 31.68 GPa

The tangency modulus is shown in the equation

E_c = 1.05E_cm = 1.05 x 31.68 = 33.26 GPa

Provided that the compressive stress should be less than 0.45f_ck(t_0)

The creep deformation of concrete at time (t) = ∞ is shown in the equation

ε_cc(∞,t_0) = φ (∞,t_0).[σ_c⁄E_c ] = 0.36

Where

φ (∞,t_0) = final creep co-efficient = 0.5

σ_c = 24 N/m^2 should not exceed 0.6f_ck

The equation shows the total shrinkage strain which is the summation of two types of strain

ε_cs = ε_cd + ε_ca

Where

The drying shrinkage strain is given in the equation, ε_(cd,∞)= K_h.ε_(cd,0) = 0.75 x 0.42 = 0.32

K_h = the co-efficient that depends onh_0 = 0.75

The assumed mean density value is 7850kg⁄m^3

The Young’s Modulus assumed for the strengthening bars is 200 GPa.

Material properties:

I – Concrete:

The compressive strength at time (t) is shown in the equation

f_ck(t) = f_ck = 40 MPa, for “t greater or equal to 28 days”

The compressive strength at various ages is shown by the equation

f_cm(t) = β_cc(t).f_cm = 48MPa

Where

β_cc(t) = co-efficient that depends on age t = e^([?s(1-(28⁄t)?^(1/2)])= 1.0

f_cm = mean compressive strength = f_ck + 8 MPa = 48 MPa

The development of tensile strength at time ‘t’ is shown in the equation

f_ctm(t) = ?[β_cc (t)]?^α.f_ctm = 3.51 MPa

f_ctm = mean tensile strength = 0.3.?f_ck?^(2⁄3) = 3.51 MPa

The modulus of elasticity is given by the equation

E_cm = 22.?{f_cm⁄10}?^0.3 = 35.2 GPa where f_cm is in MPa

Since, the limestone aggregate is used, E_cm is reduced by 10% , therefore, E_cm = 31.68 GPa

The modulus of elasticity is considered to be varying with time ‘t’ as given in the equation

E_vt(t) =?((f_cm (t))/f_cm )?^0.3.E_cm = 31.68 GPa

The tangency modulus is shown in the equation

E_c = 1.05E_cm = 1.05 x 31.68 = 33.26 GPa

Provided that the compressive stress should be less than 0.45f_ck(t_0)

The creep deformation of concrete at time (t) = ∞ is shown in the equation

ε_cc(∞,t_0) = φ (∞,t_0).[σ_c⁄E_c ] = 0.36

Where

φ (∞,t_0) = final creep co-efficient = 0.5

σ_c = 24 N/m^2 should not exceed 0.6f_ck

The equation shows the total shrinkage strain which is the summation of two types of strain

ε_cs = ε_cd + ε_ca

Where

The drying shrinkage strain is given in the equation, ε_(cd,∞)= K_h.ε_(cd,0) = 0.75 x 0.42 = 0.32

K_h = the co-efficient that depends onh_0 = 0.75

The value of ε_(cd,0)for RH 50% and C40/50

The value of K_h corresponding to h_0

h_0 = 2A_c/u = [(2){(3.14/4)?0.9?^2}] / [2(3.14)(0.45) = 450 mm

Drying shrinkage strain which is developed varies with time ‘t’ as shown in the equation

ε_cd (t) = β_ds (t,t_s).K_h.ε_(cd,0)= 0.16 x 0.75 x 0.42 = 0.05

β_ds = ((t-t_(s)))/((t-t_s )+0.04√(?h_0?^3 )) = 0.16

Where

t = 100 days = age at moment

t_s = 28 days = at the end of curing

The autogenously shrinkage is considered according to the equation

ε_ca (t) = β_as (t) ε_ca (∞) = ?6.45x10?^(-5)

β_as (t) = 2.5(f_ck-10)?10?^(-6) = ?7.5x10?^(-5)

ε_ca (∞) = 1 – e^( (?-0.2t?^0.5) )= 0.86

Therefore, ε_cs = 0.05 + (?6.45x10?^(-5)) = 0.05

The stress/strain relationship for the analysis of non-linear structures can be defined by following equation:

σ_c/f_cm = (Kn-n^2)/(1+.(K-2)n)

In which:

n = ε_c⁄ε_c1 = 1.46 / 2.3 = 0.63

ε_c1 – The strain at peak stress = 2.3

K = 1.05E_cm {|ε_c1| /f_cm} = (1.05 x 31.68 x [2.3 / 48]) = 1.6

Therefore, ε_c = 1.46 which lies 0 < |ε_c| < |ε_cu1| where ε_cu1 = 3.5

The design compressive strength is defined by the equation

f_cd = (α_cc f_ck)⁄γ_c = ((1 x 40))⁄1.2 = 33.33 MPa

The design tensile strength is defined by the equation

f_ctd = (α_ct f_(ctk,0.05))⁄γ_c = ((1 x 3.51))⁄1.2 = 2.93 MPa

The cross-section design from stress/strain relationship is given by the equation

σ_c = f_cd [1 – {1 - ?ε_c/ε_c2 }?^η]for 0 ≤ε_c≤ ε_c2

= 33.33 [1 – {1 – ?1.46/2}?^0.63]

= 18.72 KN / m^2

The effective height of the zone compressed is defined by the factor

λ = 0.8 for f_cklesser or equal to 50 MPa

The effective strength is defined by the factor

η = 1.0 for f_ck ≤ 50 MPa

II – Reinforcement:

The ductility of the strengthening depends up on the tensile strength, the maximum force during elongation and the yield stress. It is given by the equation

K = f_tk⁄f_yk and ε_uk

According to Class C (Seismic condition), the value of k should lies between 1.15 ≤k <1.35 for the satisfaction of the design.

And the characteristic strain at maximum force should be ε_uk ≥7.5 and assume it to be 8

The fatigue stress is defined by the class C as 150 MPa

Durability & cover to reinforcement:

Relating to environmental conditions, the exposure class is assumed as

XC4

The summation of minimum cover and an allowance for deviation is called nominal cover and given in the relation

C_nom= C_min+ ??C?_dev

The following equation shows the minimum concrete cover

C_min = max o {C_(min,b) ; C_(min,dur)+ ??C?_(dur,γ)- ??C?_(dur,st.)- ??C?_(dur,add) ;.10mm}

C_(min,b) = It is the required cover gap in support of the need of bond

C_(min,dur) = the required cover gap in support of the need of environment = 40 mm

??C?_(dur,γ) = additionof safety element = 0 mm

??C?_(dur,st)= the use of stainless steel for the reduction = 0 mm

??C?_(dur,add) = the use of additional protection for the reduction = 0 mm

??C?_dev = 10 mm

Therefore,

C_nom= 40 + 10 = 50 mm

Structural analysis:

Imperfection on geometry is considered only in ULS

Imperfections are represented by an inclination, θ_1= θ_0.α_h.α_m

Where

θ_0 =1⁄200

α_h = the height or length reduction factor = 2I√I for 2⁄3 ≤ α_h ≤1

α_m = the number of section reduction factor = √(0.5.(1/m+1)) = 1

The value of K_h corresponding to h_0

h_0 = 2A_c/u = [(2){(3.14/4)?0.9?^2}] / [2(3.14)(0.45) = 450 mm

Drying shrinkage strain which is developed varies with time ‘t’ as shown in the equation

ε_cd (t) = β_ds (t,t_s).K_h.ε_(cd,0)= 0.16 x 0.75 x 0.42 = 0.05

β_ds = ((t-t_(s)))/((t-t_s )+0.04√(?h_0?^3 )) = 0.16

Where

t = 100 days = age at moment

t_s = 28 days = at the end of curing

The autogenously shrinkage is considered according to the equation

ε_ca (t) = β_as (t) ε_ca (∞) = ?6.45x10?^(-5)

β_as (t) = 2.5(f_ck-10)?10?^(-6) = ?7.5x10?^(-5)

ε_ca (∞) = 1 – e^( (?-0.2t?^0.5) )= 0.86

Therefore, ε_cs = 0.05 + (?6.45x10?^(-5)) = 0.05

The stress/strain relationship for the analysis of non-linear structures can be defined by following equation:

σ_c/f_cm = (Kn-n^2)/(1+.(K-2)n)

In which:

n = ε_c⁄ε_c1 = 1.46 / 2.3 = 0.63

ε_c1 – The strain at peak stress = 2.3

K = 1.05E_cm {|ε_c1| /f_cm} = (1.05 x 31.68 x [2.3 / 48]) = 1.6

Therefore, ε_c = 1.46 which lies 0 < |ε_c| < |ε_cu1| where ε_cu1 = 3.5

The design compressive strength is defined by the equation

f_cd = (α_cc f_ck)⁄γ_c = ((1 x 40))⁄1.2 = 33.33 MPa

The design tensile strength is defined by the equation

f_ctd = (α_ct f_(ctk,0.05))⁄γ_c = ((1 x 3.51))⁄1.2 = 2.93 MPa

The cross-section design from stress/strain relationship is given by the equation

σ_c = f_cd [1 – {1 - ?ε_c/ε_c2 }?^η]for 0 ≤ε_c≤ ε_c2

= 33.33 [1 – {1 – ?1.46/2}?^0.63]

= 18.72 KN / m^2

The effective height of the zone compressed is defined by the factor

λ = 0.8 for f_cklesser or equal to 50 MPa

The effective strength is defined by the factor

η = 1.0 for f_ck ≤ 50 MPa

II – Reinforcement:

The ductility of the strengthening depends up on the tensile strength, the maximum force during elongation and the yield stress. It is given by the equation

K = f_tk⁄f_yk and ε_uk

According to Class C (Seismic condition), the value of k should lies between 1.15 ≤k <1.35 for the satisfaction of the design.

And the characteristic strain at maximum force should be ε_uk ≥7.5 and assume it to be 8

The fatigue stress is defined by the class C as 150 MPa

Durability & cover to reinforcement:

Relating to environmental conditions, the exposure class is assumed as

XC4

The summation of minimum cover and an allowance for deviation is called nominal cover and given in the relation

C_nom= C_min+ ??C?_dev

The following equation shows the minimum concrete cover

C_min = max o {C_(min,b) ; C_(min,dur)+ ??C?_(dur,γ)- ??C?_(dur,st.)- ??C?_(dur,add) ;.10mm}

C_(min,b) = It is the required cover gap in support of the need of bond

C_(min,dur) = the required cover gap in support of the need of environment = 40 mm

??C?_(dur,γ) = additionof safety element = 0 mm

??C?_(dur,st)= the use of stainless steel for the reduction = 0 mm

??C?_(dur,add) = the use of additional protection for the reduction = 0 mm

??C?_dev = 10 mm

Therefore,

C_nom= 40 + 10 = 50 mm

Structural analysis:

Imperfection on geometry is considered only in ULS

Imperfections are represented by an inclination, θ_1= θ_0.α_h.α_m

Where

θ_0 =1⁄200

α_h = the height or length reduction factor = 2I√I for 2⁄3 ≤ α_h ≤1

α_m = the number of section reduction factor = √(0.5.(1/m+1)) = 1

l= it is the factor of depth by stretch.

“m” = number of vertical sections

isolated member(Column)

l – the actual length = 3.96 m and m = 1

“m” = number of vertical sections

isolated member(Column)

l – the actual length = 3.96 m and m = 1

the isolated member (Beams)

l = 8.76 and m = 1

l = 7.07 and m = 1

l = 7.5 and m = 1

l = 10.36 and m. = 1

l = 7.87 and m = 1

l = 11.96 and m. = 1

l = 8.7 and m = 1

l = 13.84 and m. = 1

l = 9.67 and m = 1

l = 9.15 and m = 1

l = 9.33 and m = 1

l = 11.21 and m = 1

l = 8.76 and m = 1

l = 7.07 and m = 1

l = 7.5 and m = 1

l = 10.36 and m. = 1

l = 7.87 and m = 1

l = 11.96 and m. = 1

l = 8.7 and m = 1

l = 13.84 and m. = 1

l = 9.67 and m = 1

l = 9.15 and m = 1

l = 9.33 and m = 1

l = 11.21 and m = 1

For imperfections, the eccentricity effect will be e_i= (θ_i l_0)⁄2 where l_0= effective length.

The transverse force shows the position at which maximum moment occurs.

For braced system, H_i=2θ_i N where N is the axial load

The effect of inclination by transverse forces, H_i = θ_i (N_b- N_a) where N_b and N_a are the longitudinal forces.

For columns which are isolated in braced system, e_i= l_0⁄400

Idealisation of structure:

Beams: the span > Three times the section depth.

Columns: the section depth must be less than 4 times its width and the height should be near to 3 times the section depth.

The effective span of beams can be calculated using given equation,

l_eff= a_1+ a_2+ l_nwherel_nis the clear distance

a_i = max {1/2h; 1/2t} where t is the width.

To determine the action effects, the linear analysis is carried out by the assumption of

Un-cracked cross-section

Stress/strain relationship

Mean to modulus of elasticity

Continuous beams that are subjected to flexure should have the adjacent span length ratio in the range 0.5 to 2

The redistributed bending moment to the elastic flexural moment ratio is given by the equation, δ≥ k_1+ k_2 X_u/d for f_ck lesser than or equal to 50 MPa

Where k_1 = 0.44 andk_2=1.25(0.6+0.0014/ε_cu2) and d - the effective depth

To design the columns, with no such redistribution, the elastic moment from frame action can and may be used.

The rotational capacity, θ_s for continuous beams, the length should maintain atleast 1.2 times the depth of sections.

θ_(pl,d) - Allowable plastic rotation that depends on shear (λ. = 3.0). For multiple values of slenderness under shear, the plastic rotation is multiplied with a correction factor. i.e., θ_(pl,d) k_λ

k_λ= √(λ⁄3) ,

λ= M_Sd/(V_Sd d)

Slenderness criterion for isolated members at where effects of second order should be rejected if the λ which is should be under the range ofλ_lim

λ_lim= 20ABC⁄√n

Where:

A = 0.7 =1⁄((1+0.2φ_(ef )))

B = √(1+2ω) = 1.1 when ω is not given

C = 1.7-r_m = 0.7 when r_m is not given

φ⁄efis effective creep ratio

ω =(A_s f_yd)⁄((A_c f_cd))

The normal force, n = N_Ed⁄((A_c f_cd))

Moment ratio, r_m = M_01⁄M_02

|M_02 |≥|M_01 |

Slenderness as well as the effective length of isolated members where λ is

λ =l_(0.)⁄(.i)

I= radius of gyration

l_0= total length

For compression members, the slenderness ratio must be cross-checked by the effectual extent

l_0 = 0.5l√((k_1⁄(.0.45+k_1 )+1)(1+.k_2⁄(0.45+.k_2 )))

k_1,k_2Are the virtual elasticity of the rotating constraints at ends 1 as well as 2 correspondingly.

k = (El⁄l)(θ⁄M)

θ - Restraining members’ revolution in support of flexural Moment

EI - composition member’s bending moment stiffness

l - The clear height of compression member

In a node, the adjacent compression member is let to meet the rotation at buckling, then (EI⁄l)

Here, k is [?(EI⁄l)?_b+?(EI⁄l)?_a],

The effect of creep shall be considered only where φ_efin conjunction with the nominal load, this leads to creep deformation for the quasi-enduring weight

The transverse force shows the position at which maximum moment occurs.

For braced system, H_i=2θ_i N where N is the axial load

The effect of inclination by transverse forces, H_i = θ_i (N_b- N_a) where N_b and N_a are the longitudinal forces.

For columns which are isolated in braced system, e_i= l_0⁄400

Idealisation of structure:

Beams: the span > Three times the section depth.

Columns: the section depth must be less than 4 times its width and the height should be near to 3 times the section depth.

The effective span of beams can be calculated using given equation,

l_eff= a_1+ a_2+ l_nwherel_nis the clear distance

a_i = max {1/2h; 1/2t} where t is the width.

To determine the action effects, the linear analysis is carried out by the assumption of

Un-cracked cross-section

Stress/strain relationship

Mean to modulus of elasticity

Continuous beams that are subjected to flexure should have the adjacent span length ratio in the range 0.5 to 2

The redistributed bending moment to the elastic flexural moment ratio is given by the equation, δ≥ k_1+ k_2 X_u/d for f_ck lesser than or equal to 50 MPa

Where k_1 = 0.44 andk_2=1.25(0.6+0.0014/ε_cu2) and d - the effective depth

To design the columns, with no such redistribution, the elastic moment from frame action can and may be used.

The rotational capacity, θ_s for continuous beams, the length should maintain atleast 1.2 times the depth of sections.

θ_(pl,d) - Allowable plastic rotation that depends on shear (λ. = 3.0). For multiple values of slenderness under shear, the plastic rotation is multiplied with a correction factor. i.e., θ_(pl,d) k_λ

k_λ= √(λ⁄3) ,

λ= M_Sd/(V_Sd d)

Slenderness criterion for isolated members at where effects of second order should be rejected if the λ which is should be under the range ofλ_lim

λ_lim= 20ABC⁄√n

Where:

A = 0.7 =1⁄((1+0.2φ_(ef )))

B = √(1+2ω) = 1.1 when ω is not given

C = 1.7-r_m = 0.7 when r_m is not given

φ⁄efis effective creep ratio

ω =(A_s f_yd)⁄((A_c f_cd))

The normal force, n = N_Ed⁄((A_c f_cd))

Moment ratio, r_m = M_01⁄M_02

|M_02 |≥|M_01 |

Slenderness as well as the effective length of isolated members where λ is

λ =l_(0.)⁄(.i)

I= radius of gyration

l_0= total length

For compression members, the slenderness ratio must be cross-checked by the effectual extent

l_0 = 0.5l√((k_1⁄(.0.45+k_1 )+1)(1+.k_2⁄(0.45+.k_2 )))

k_1,k_2Are the virtual elasticity of the rotating constraints at ends 1 as well as 2 correspondingly.

k = (El⁄l)(θ⁄M)

θ - Restraining members’ revolution in support of flexural Moment

EI - composition member’s bending moment stiffness

l - The clear height of compression member

In a node, the adjacent compression member is let to meet the rotation at buckling, then (EI⁄l)

Here, k is [?(EI⁄l)?_b+?(EI⁄l)?_a],

The effect of creep shall be considered only where φ_efin conjunction with the nominal load, this leads to creep deformation for the quasi-enduring weight

φ_((∞,t0))is the resulting co-proficient of creep.

M_0Eqpis the bending moment for first order in the quasi-permanent load combinations

M_0Ed=First order flexural moment

For a slender compression member, the nominal stiffness for any cross section

EI = ?K_s E_s I_s+K?_c E_cd I_c

Where

E_cd - modulus of elasticity (concrete), value used in design

I_c - moment of inertia subjected with cross section of the concrete

E_s - modulus of elasticity (reinforcement), value used in design

I_s - second moment of area of steel reinforcement

K_c = (k_1 k_2.)⁄((φ_ef.+1))

M_0Eqpis the bending moment for first order in the quasi-permanent load combinations

M_0Ed=First order flexural moment

For a slender compression member, the nominal stiffness for any cross section

EI = ?K_s E_s I_s+K?_c E_cd I_c

Where

E_cd - modulus of elasticity (concrete), value used in design

I_c - moment of inertia subjected with cross section of the concrete

E_s - modulus of elasticity (reinforcement), value used in design

I_s - second moment of area of steel reinforcement

K_c = (k_1 k_2.)⁄((φ_ef.+1))

K_c - factor for cracking effect, creep effect and more.

K_s–factor, provided ρ ≥ 0.002 = 1

Ρ =A_(S.)⁄A_C

k_1= factor for the concrete strength = √(f_(ck.)⁄20) (MPa)

k_2= factor for the axial force and slenderness = n λ/170 is less than or equal to 0.20

N =relative force in axial direction, N_Ed⁄((A_c f_cd))

If λ is unavailable, k_2 should be taken as

k_2 = n 0.30 is less than or equal to 0.20 , provided ρ ≥ 0.01

K_s = 0

K_c = 0.3⁄((0.5φ_ef+1))

K_s–factor, provided ρ ≥ 0.002 = 1

Ρ =A_(S.)⁄A_C

k_1= factor for the concrete strength = √(f_(ck.)⁄20) (MPa)

k_2= factor for the axial force and slenderness = n λ/170 is less than or equal to 0.20

N =relative force in axial direction, N_Ed⁄((A_c f_cd))

If λ is unavailable, k_2 should be taken as

k_2 = n 0.30 is less than or equal to 0.20 , provided ρ ≥ 0.01

K_s = 0

K_c = 0.3⁄((0.5φ_ef+1))

The stiffness needs to be based on a modulus of total concrete

E_(cd,eff) = E_(cd.)⁄((1+.φ_ef))

E_cd - Modulus of elasticity, the number used in the calculation of design

φ_ef - Effective creep ratio

The total design moment, including second order moment

M_Ed = M_0Ed [β/((N_B⁄N_Ed )-1)+1]

M_0Ed - first order moment

β - factor for the distribution of 1^st and2^nd order moments

β = π^2⁄c_0 for the axial load & cross-section (isolated members)

N_Ed - axial load, the value used in design

N_B - buckling load

c_0 - Coefficient for the first order flexural moment distribution

c_0 = 9.6 (parabolic)

c_0 = 8 (constant first order moment), and 12 (symmetric triangle)

The design bending moment is

M_Ed=M_2+M_0Ed

M_0Ed – Moment of first order

M_2 - Nominal moment for 2^nd order

The nominal second order moment M_2 is

M_2 = N_Ed e_2

e_2 - deflection = ??(l?_o?^2⁄c )((1)⁄r)

1⁄r – The length of the curvature

c - Factor for the curvature distribution, = 10

For members with uniform symmetrical cross sections, the given expression is:

1⁄r = K_r K_φ (1.)⁄r_0

Where

K_r = alteration aspect which depends up on axial weight

K_φ = aspect in support of consideration of creep

1⁄r_0 = E_(yd.)⁄((0.45d))

E_yd = f_(yd.)⁄E_S

d = effective depth

• If the entire steel strengthening is not limited to conflicting surfaces, however a certain percentage is similar to the flat surface of flexion, d is given as follows:

d = i_s+((h.)⁄2)

wherei_s is the radius of gyration of area of the steel reinforcement

K_r = ((n_u-n))⁄((n_u-n_bal)) ≤ one.

Where

n = N_(Ed.)⁄((A_c f_cd)) , which is the relative axial force

N_Edis used as the design value of force in axial direction

n_u = ω+1

n_balis the value of n at utmost moment resistance; 0.4 is generally used as it’s value

ω = (A_s f_yd)⁄((A_c f_cd))

A_sis the whole part of cross section for strengthening

A_cis the cross sectional area for the concrete

The influence of creep which needs to be considered through the given aspect

K_φ=.1+ β_(φ_ef ) ≥1

Where

φ_efis the ratio for effective creep.

β = f_ck⁄.200 - λ⁄150 + 0.35

λ= slenderness ratio

Ultimate limit states (ULS)

For members under compression loads the minimum eccentricity must be considered as, e_0=h⁄30as well as shouldn’t be less than 20mm, where h is considered as the depth of the member.

The subsequent characters are utilized in support of the shear force conflict computations.

V_(Rd,c)is the shear conflict in support of plan of the associate exclusive of shear strengthening

V_(Rd,s)is the value shear power in support of planning which the yielding shear strengthening can maintain.

V_(Rd,max) is the design cost of the utmost shear force that the element is able to withstand through squeezing the compression struts.

The shear resistance of an associate provided by the shear strengthening is equivalent to

V_Rd= ?V_td+V?_(Rd,s)+?.V?_ccd

In parts of the section whenV_Ed≤V_(Rd,c)we do not need to calculate shear reinforcement.V_Edis calculated only after considerations of pressurising.

to be continoued...

For members under compression loads the minimum eccentricity must be considered as, e_0=h⁄30as well as shouldn’t be less than 20mm, where h is considered as the depth of the member.

The subsequent characters are utilized in support of the shear force conflict computations.

V_(Rd,c)is the shear conflict in support of plan of the associate exclusive of shear strengthening

V_(Rd,s)is the value shear power in support of planning which the yielding shear strengthening can maintain.

V_(Rd,max) is the design cost of the utmost shear force that the element is able to withstand through squeezing the compression struts.

The shear resistance of an associate provided by the shear strengthening is equivalent to

V_Rd= ?V_td+V?_(Rd,s)+?.V?_ccd

In parts of the section whenV_Ed≤V_(Rd,c)we do not need to calculate shear reinforcement.V_Edis calculated only after considerations of pressurising.

to be continoued...