4.13 Shear wall and horizontal load

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The shear wall in Fig. 4.47 must sustain four different loadings. First gravity

loads, then the live load of the floor plates. In the third load case (see

376 4 Plane problems

Fig. 4.47. Shear wall: a) displacements, b) load, c) principal stresses, and

d) vertical reinforcement (designed for a total of 4 load cases)

4.13 Shear wall and horizontal load 377

Fig. 4.48. Detail a) reinforcement

as-x in cm2/m

and b) reinforcement as-y

in cm2/m in the columns

=280

kNm on the upper edge, which is represented by an antisymmetric line load

p = ±95.24 kN/m; see Fig. 4.47 b. The fourth load case is the same load case,

except that the forces act in the opposite direction.

In the third load case the four columns must sustain a shear force V =

180 + 100 = 280 kN. If it is assumed that only the three short columns carry

the load then each column carries a shear force V of 93.3 kN. The height of

the columns is 1.5 m, so the bending moment becomes

M = 93.3kNm· 1.5m

2

= 70kNm. (4.96)

The moment resulting from the antisymmetric line load on the upper edge is

split into two vertical forces:

C = T = ±280 kNm

3.9m

= 72kN z = 3.9m. (4.97)

Hence the design moment for the column with the heaviest load is

Ms = 70kNm− 72 kN · 0.1m = 62.8kNm, (4.98)

Fig. 4.47 b), the wall is subject to horizontal forces and to a moment M

378 4 Plane problems

and with2

kh =

25

_

62.8/0.3

= 1.73 → ks = 4.5 (4.99)

we have

As = 4.5 · 62.8

25

+

72

28.6

= 13.8 cm2 . (4.100)

A check of Fig. 4.48 shows that the reinforcement

As =

1

2

· (102.7 cm2/m + 3.0cm2/m) · 0.3m = 15.8 cm2 (4.101)

is sufficient. The same holds for the shear reinforcement.

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