# Column Shear Design Force Example

## Column shear safety check of the 3rd floor S07 column:

Major and minor bending moments and beam bearing strength moments can be seen at the upper and lower joints in the column reinforced concrete dialogue.

Major 3 direction: Mra + Mrü = 37.874 tfm> 1.2 (Mri + Mrj) = 6.889 tfm is provided.

In the direction of minor 2: Mra + Mrü = 17.608 tfm> 1.2 (Mri + Mrj) = 13.624 tfm is provided.

### Major 3 axis:

M = 8.002 * 2.519 / (2.519 + 2.563) = 3.97 tfm

Ma = 8.002 * 0.677 / (0.677 + 0.956) = 3.32 tfm

Ln = floor height- beam height = 2.80- 0.50 = 2.30 m

And = (3.97 + 3.32) / 2.30

And = 3.17 tf

### Minor 2 axis:

M = 15.894 * 2.15 / (2.15 + 2.26) = 7.75 tfm

Ma = 15.894 * 0.451 / (0.451 + 0.524) = 7.35 tfm

Ln = floor height- beam height = 2.80- 0.50 = 2.30 m

And = (7.75 + 7.35) / 2.30

And = 6.56 tf

The sum of the shear force calculated from the earthquake increased with D with the vertical loads, Eq. If calculated by (7.5) is less than Ve, this shear force will be used instead of Ve. The upper bound value for ve is given in column V in the report.

Ve = 3.17 tf <V = 5.84 (used Ve = 3.17 tf)

Ve = 6.56 tf> V = 5.79 (used Ve = 5.97 tf)

And <0.85 root (fck) * Aw fcd

and

And <Vr

should be.

### Major Vr account

Vc = 0 (TBDY Article 7.3.7.6 is not provided)

Vw = (As * Coefficient / range) * fywd * b

Vw = (0.5cm * 3 / 8 cm) * 1/100 * 37241.81tf/m2 * 0.575m=40.15 tf

Vr = Vc + Vw = 0 + 40.15

Vr = 40.15 tf

And = 3.17 tf <Vr = 40.15 provided

And = 3.17 tf <0.85 Ack root (fck) = 0.85 * 300 * 600 * root (30) = 838015.513 KN = 85.45 tf provided

### Minor Vr account

Vc = 0 (TBDY Article 7.3.7.6 is not provided)

Vw = (As * Coefficient / range) * fywd * b

Vw = (0.5cm * 5 / 8 cm) * 1/100 * 37241.81tf/m2 * 0.275 m=32 tf

Vr = Vc + Vw

Vr = 0 + 32

Vr=32 tf

And = 5.97 tf <Vr = 40.15 provided

And = 5.97 tf <0.85 A root (fck) = 0.85 * 300 * 600 * root (30) = 838015.513 KN = 85.45 tf provided

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