# 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. Next Topic JavaScript errors detected