# Example 2.

Preliminary Information:

C35/B420C fck= 3059.149 tf/m2 fcd=2039.432 tf/m2

bw = 30 cm, h =50 cm

ln = 350 cm

fywd=37241.81 tf/m2

stirrups available = fi 8/20/9 (diameter / middle / wrap)

Left support upper reinforcement pieces of the beam 3fi16 + 3fi14

Left support lower reinforcement pieces of the beam 2fi14 + 3fi14

Right support top reinforcement units of the beam 3fi14 + 4fi16

Right support lower reinforcement pieces of the beam 2fi114 + 3fi16

### Determination of beam bearing moment moments:

The beam bearing strength moments are the value corresponding to Mu from the moment curvature diagram according to the DGT material model.

Mri(-)=12.252 tfm

Mri(+) =16.53 tfm

Mrj(-)=14.42 tfm

Mrj(+) = 19.32 tfm

### Finding Mp values from Mr values

Mpi (-) = 12.252 * 1.4 = 17.15 tfm

Mpi (+) = 16.53 * 1.4 = 23.14 tfm

Mpj(-)=14.42*1.4=20.06 tfm

Mpj(+)=19.32*1.4=27.04 tfm

### Calculation of the shear force Ve to be taken as basis in the transverse reinforcement calculation:

**According to Article 7.4.5.1 of TBDY** , the shear force V _{e to} be taken as basis in calculating the transverse reinforcement in beams should be found by **Equation 7.9 in** a way to give unfavorable results separately for the directions of earthquake .

Vdy value for left support Vdi = 3.18 tf

Vdy value for right support Vdi = 4.91 tf

### Finding the largest Ve value

Left support 1st value → Ve = 3.18 + (17.15 + 27.04) /3.50 = **15.81 tf**

Left support 2nd value → Ve = 3.18 - (17.15 + 27.04) /3.50 = **-9.44 tf**

Left support 3rd value → Ve = 3.18 + (20.06 + 23.15) /3.50 = 15.52 tf

Left support 4th value → Ve = 3.18 - (20.06 + 23.15) /3.50 = -9.16 tf

Right support 1st value → Ve = 4.91 + (27.04 + 17.15) /3.50 = **17.54 tf**

Left support 2nd value → Ve = 4.91 - (27.04 + 17.15) /3.50 = **-7.72tf**

Right bracket 3rd value → Ve = 4.91 + (20.06 + 23.15) /3.50 = 17.25 tf

Right bracket 4th value → Ve = 4.91 - (20.06 + 23.15) /3.50 = -7.44 tf

**The largest Ve = 17.54 tf is chosen.**

**The value of ve = 17.84**

The shear force calculated from the earthquake effects increased with the Resistance Excess Coefficient D together with the vertical loads, 27.90 tf for the left end and 29.33 tf for the right end. These values are not less than **Ve = 17.54 tf** . Therefore, the shear force to be used in reinforcement design is V _{e} = **17.54 tf** .

**7.4.5.2 TBDY substance** according to her, **7.4.5.1** 'calculated according to the shear force, V _{e} , **Eq. (7.10)** shall satisfy the conditions given.

However, in **TBDY Article 7.4.5.3** , the contribution of concrete to shear strength will be taken as zero (V _{c} = 0).

**Vr = Asw/s * fywd * d = 2* 0.50 cm/ 9cm * 3.724 tf/cm2 * 455cm =18.26 tf**

**Since Vr = 18.26 tf> Ve = 17.54tf** , cutting security is provided.

**0.85 * 300 * 450 * root (30) = 635495.096 N = 64.80 tf> Ve = 17.54 tf** provided

### Control of TS 500 Condition

For loading combinations that do not contain earthquake loads, the upper limit of shear force control given in **TS500 Equation 8.7** and calculated to prevent crushing of the body concrete is done with the following steps.

Vd<=0.22 fcd bw d

Shear forces generated in the beam due to the 1.4G + 1.6Q loading combination, V _{d} ,

V _{d} = 4.55 tf for left (i) end

**V _{d} = 7.06 tf** for right (j) end

calculated as. In this case,

0.22 * 2039.432 tf / m2 * 0.3m * 0.455m = **62.24 tf> V = 7.06 tf** provided.

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