Theory and design of Structures

If method 2 is used , integration is involved and it is a bit impractical for everyday use. In order to simplify this, stress block parameters αc and βc are derived. Both parameters depend on the amount of strain that the concrete is experiencing. Figure 'stress block parameters' shows the equations that can be used to determine αc and βc. @theoryanddesignofstructures @Cengtalk_bot

The force that the uncracked concrete can develop, symbolysed C, can be determined from 1. Equivalent Rectangular stress block 2. Parabolic rectangular stress block 3. Bi- linear stress block @theoryanddesignofstructures @Cengtalk_bot

In order to determine the moment any given section can Carry, we must first understand that the couple created by concrete and steel is responsible to create this resistance moment.

Analysis of sections is based on two equations: 1. Stress and strain compatibility 2. Equilibrium equations The first part is can be described as " stress at any point in a member must correspond to the strain at that point " And the second part as " the internal force must balance the external load " @theoryanddesignofstructures @Cengtalk_bot

Two things are common in RCC sections: 1. Design of sections 2. Analysis of sections When design of RCC sections is encountered, only geometry and load are known and all the other parameters, such as depth, width , area of reinforcement and cover , are determined. If analysis of sections is encountered, all the parameters are known, but the amount of moment, shear or any other internal action, the section can carry is not known and it is expected to be established through the analysis. Design of singly reinforced beam is discussed in previous sessions.and the output of the design process is shown in ' details of beam '. Here after some examples on section analysis would be done. @theoryanddesignofstructures @Cengtalk_bot

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Design chart based on ES EN 1992-1-1:2015

Design Table continued

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👉 beam section detail ✔️All dimensions are in centimetres ✔️The two bars at the top are nominal reinforcement to hang the stirrups( the blue lines).

Design step 8:- check for minimum and maximum longitudinal reinforcement Minimum reinforcement ✅As,min=0.26{fctm/fyk}bt*d. ( ES EN 1992-1-1:2015 equation 9.1 N) ✅From table 3.1 of ES EN 1992-1-1:2015 the value of fctm for C30/37 concrete is 2.9 Mpa. 👉As,min=0.26*{2.6/500}*250*452=152.776 mm² 👉As> As,min....provide As Maximum reinforcement ES EN 1992-1-1:2015 Article 9.2.1.1[3] " the cross sectional area of tension or compression reinforcement should not exceed 0.04Ac" and the national annex recommends the same value. As,max=0.04*250*500=5000 mm² As< As,max.....ok! Design step 9:- calculate the number of bars 👉# of bars=As/as Where 👉As calculated area of reinforcement 👉as is the area of one bar 👉as=(πφ²/4)=(π*20²/4)≈314 mm² # of bar=(739.65/314)=2.355≈3 ❤️provide 3φ20 bars .......All Good...... @theoryanddesignofstructures @Cengtalk_bot

Design step 5:-design strength of material Concrete(fcd) 👉fcd=αcc*fck/γc 👉for C30/37 concrete fck is 30 Mpa. 👉αcc=0.85 👉γc=1.5 👉fcd=(0.85*30/1.5)=17 Mpa Steel (fyd) 👉fyd=fyk/γs 👉γs=1.15 👉fyd=500/1.15=434.78 Mpa Design step 6:- design moment calculation 👉 for simply supported beam the maximum moment occurs at mid span and it's vale is ➡️Mmax=Msd=Pd*L²/8 ➡️Mmax=Msd=(29.72)*(6²)/8=133.73 KN-m Design step 6:- check depth for flexure 👉effective depth of beam in order to be singly reinforced is calculated from ✓d=sqrt(Msd/(μsd*fcd*b)) ✓Sqrt...[square root of ] For zero percent redistribution of moment the limiting μsd is 0.295. d=sqrt({133.73*10⁶}/{0.295*17*250})=326.59 mm And the provided effective depth: d=500-30-8-20/2=452 mm Since the provided depth is greater than calculated from flexure, depth is ok! Design step 7:- Reinforcement calculation Using design chart method ✓μsd={Msd*10⁶}/{fcd*b*d²) ✓μsd={133.73*10⁶}/{17*250*452²}=0.154 <0.295 ok! from design chart the value of kz is ✓kz=0.92 and z( moment arm , the distance between Cc and Ts) ✓Z=kz*d=0.92*452=415.84 mm Now the amount of reinforcement ✓As={Msd}/{Z*fyd} ✓As={133.73*10⁶}/{415.84*434.77}=739.65 mm² @theoryanddesignofstructures @Cengtalk_bot

Beam model with design load

Design Step 1: depth from deflection( deformation) control 👉Since C30/37 concrete is used the limiting span to depth ratio from table 7.4 N Is 14 ( the beam is assumed to be highly stressed with ρ=1.5 %). 👉The limiting depth to prevent excessive deformation is: 👉d=le/14 ....where le is effective length of beam. 👉d=6000 mm/14=428.57 mm Design step 2:- cover to reinforcement 👉Exposure class is assumed to be XC 1 and the value of Cmin,dur becomes 15 mm. 👉 for separate bar, the value of Cmin,b is equals to the diameter of bar( φ). Assuming φ=20 mm, and Cmin,b=20 mm. ➡️Cₘᵢₙ= max {Cₘᵢₙ,b ; Cₘᵢₙ,dᵤᵣ; 10 mm } ➡️Cₘᵢₙ=max{20 mm,15 mm, 10 mm} ➡️Cₘᵢₙ=20 mm And the nominal cover becomes ➡️Cnom=Cmin+10=20+10=30 mm cover. Design step 3:- overall depth For stirrup assume 8 mm bars. 👉D=d+cover +φₛ+φₗ/2 👉D=428.57mm+30mm+8+20/2=476.57 mm For ease of construction take D=500 mm ( nearest 50 ). Design step 4:- Load calculation Dead load 👉 self weight of beam Unit weight of concrete(γ)= 25 KN/m³ Weight per meter:- ➡️γ*b*D ➡️25*0.25 *0.5=3.125 KN/m 👉 total dead load Superimposed+self weight=10KN/m+3.125 KN/m=13.125 KN/M. Design Load Pd=1.35Gk+1.5Qk Pd=1.35*13.125+1.5*8=29.72 KN/m. @theoryanddesignofstructures @Cengtalk_bot

Beam section Dimensions in millimeters

✳️ Beam design( flexure) example according to ES EN 1992-1-1: 2015 ➡️ Material Data Concrete grade:- C30/37(fck=30 Mpa). Steel grade:- S500 ( fyk=500 Mpa). ➡️ Beam geometry Type:- Simply supported beam Effective length of beam:- 6 m. Beam width(b):- 250 mm. ➡️ Load Dead load(Gk):- self weight of Beam plus 10 KN/m super imposed load. Live Load(Qk):- 8 KN/m Depth is not limited by any requirement( singly reinforced beam is possible). @theoryanddesignofstructures @Cengtalk_bot

Figure 8.3

Table 8.2

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✔️ design anchorage length lbd =α₁α₂α₃α₄α₅lb,req>= lb,min (eqn. 8.4) α₁,α₂,α₃,α₄ and α₅ are coefficients given in table 8.2. ( see picture with caption ' table 8.2 '). Where :- α₁ is the effect of the form of the bars assuming adequate cover. α₂ is the effect of concrete minimum cover. α₃ is the effect of confinement by transverse reinforcement α₄ is for the influence of one or more welded transverse bars along the design anchorage length. α₅ is for the effect of the pressure transverse to the plane of splitting along design anchorage length. 🔑The product α₂α₃α₅>= 0.7 🔑 the minimum anchorage length if no other limitation is applied: 👉 for anchorage in tension lᵇᵐⁱⁿ>={0.3l b,req; 10φ ;100 mm} 👉for anchorage in compression lᵇᵐⁱⁿ>={0.6l b,req; 10φ ;100 mm} @theoryanddesignofstructures @Cengtalk_bot