Chapter 13 (cont..10) - Shear resistance offered by Inclined Stirrups

In the previous section we completed the discussion on 'the design of vertical stirrups'. Now we will see inclined stirrups.

Design shear resistance offered by inclined stirrups

The fig.13.49 below shows the inclined stirrups in a beam. The stirrups are inclined at an angle α to the horizontal.

Fig.13.49

Inclined stirrups

The force in each of the stirrups at the ultimate state will be same as in Eq.13.35 which is equal to 0.87f_y A_sv. But here, this force is inclined. we can find the vertical component of the force as shown in fig.13.50 below:

Fig.13.50

Vertical component of the force

In triangle oqr, sinα = qr/oq . So we get:

Eq.13.40:

qr = op = the vertical component =oq sinα. Thus:

Eq.13.41:

Vertical component =0.87f_y A_sv sinα

This is the vertical force in one inclined stirrup. This force multiplied by the total number of stirrups n intercepting the crack will give the total vertical force offered by the inclined stirrups. To find n, we can use the following fig.:

Fig.13.51

No. of stirrups intercepting the crack

In the fig.,

• the crack is inclined at an angle θ with the horizontal.

• the spacing S_v of the inclined stirrups is measured parallel to the horizontal

We will add some more details to the above fig., to determine the horizontal projection of the crack. The modified fig.13.52 is shown below:

Fig.13.52

Calculation of horizontal projection

In the fig., bc is drawn vertically. cd is drawn parallel to the inclined stirrups. Thus two triangles abc and bcd are formed. Note that the horizontal projection ab is not sufficient for our calculations because, the no. of stirrups which intercept the crack fill the space from a to d.

Considering Δabc, tanθ =bc/ab. So

Eq.13.42: ab =bc / tanθ =bc cotθ =d cotθ

Considering Δbcd, tanα =bc/bd. So

Eq.13.43: bd =bc/tanα =bc cotα =d cotα

Adding 13.42 and 13.43 we get

Eq.13.44: ab + bd = ad = d(cotθ + cotα)

So we get the required horizontal projection. Dividing this by the spacing S_v, we get n, the number of stirrups intercepting the crack. Thus:

Eq.13.45:

Now we can multiply the force in a single inclined stirrup (given by eq.13.41) by the total number of stirrups intercepting the crack, to get the total resistance force. But this total resistance from the stirrups is denoted as Vus. So we get

Eq.13.46:

If for convenience, we assume that the crack is inclined at θ =45^o to the horizontal, the above Eq.13.46 will be simplified to the form:
Eq.13.47:

This is the same equation given in cl.40.4.c.b of the code.

The derivation of Eq.13.47 from 13.46 is shown below:

This completes the discussion on the resistance offered by 'vertical stirrups' and 'inclined stirrups'. In the next section we will discuss about the resistance offered by 'bent-up bars'.

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