4.11 Multistory shear wall

Back

The multistory shear wall in Fig. 4.39 can serve as an illustration for how the

results of a larger FE analysis can be checked with simple models [32].

The thickness of the wall is 25 cm. The modulus of elasticity is 30 000

MN/m2, and Poisson’s ratio is ν = 0.0. The load consists of gravity loads,

floor loads, and wind loads; see Fig. 4.39. The FE analysis is based on Wilson’s

element (Q4 + 2).

As a first check, the support reactions were calculated for an equivalent

two-span beam with equivalent stiffness

EI = E · h3 · b

12

=

30 000MN/m2 · (13.06 m)3 · 0.25m

12

= 1 392 225MNm2 (4.76)

placed on elastic supports (see Fig. 4.40)

cw = EA

h

=

30 000MN/m2 · 0.4m · 0.25m

4.74m

= 633MN/m (4.77)

and carrying a corresponding equivalent load. The results agree quite well

with the FE/BE results; see Table 4.7.

The internal actions were checked in two sections (see Fig. 4.41), x = 3 m

and x = 10 m; see Table 4.8 and Fig. 4.42. The normal force N = Nx and the

shear force V = Nxy were calculated by integrating the stresses σxx and σxy,

366 4 Plane problems

Fig. 4.37. Analysis of a shear wall with bilinear elements (Q4): a) the original load

case p, b) the FE load case ph consists of line loads and volume forces, c) principal

stresses

A B C

4.11 Multistory shear wall 367

-200 0 200 400 600 800

Spannungen xx [KN/m2]

0

1

2

3

4

Hцhe [m]

FEM(Bilineares Element)

FEM(Wilson-Element)

REM

-160 -120 -80 -40 0 40

Spannungen xy [KN/m2]

0

1

2

3

4

Hцhe [m]

FEM(Bilineares Element)

FEM(Wilson-Element)

REM

A

A

Fig. 4.38. Bending stresses σxx and shear stresses σxy in a vertical section; study

of three different results

Table 4.7. Support reactions (kN) of the shear wall in Fig. 4.39

support FE beam deviations % BE

AH 41 41 0.0 41

AV 1493 1536 2.90 1498

BV 1701 1589 6.60 1686

CV 1503 1570 4.50 1511

Table 4.8. Comparison of the resultant stresses in the shear wall in Fig. 4.39

x = 3.0 m FE beam deviations %

N (kN) 0 0 0

V (kN) 330 425 29

M (kN m) 4506 4898 8.7

x = 10.0 m FE beam deviations %

N (kN) 0 0 0

V (kN) −458 −590 28.8

M (kN m) 2461 2787 13.2

respectively. In the section x = 10.0, the compressive force C and the tensile

force T were ±584.6 kN. The distance between the two forces C and T was

4.21 m, so the moment is

M = 584.6kN· 4.21m = 2461.1kNm, (4.78)

which agrees quite well with the beam moment of 2787 kN m.

A check of the lintels above the doors is not simple, because it is not

clear what portion of the total shear force in the cross section is carried by

368 4 Plane problems

gravity load 6.25 kN/m2

50.80 kN/m

50.60 kN/m

50.60 kN/m

48.60 kN/m

39.50 kN/m

80.60 kN

144.00 kN

220.00 kN

410.00 kN

195.00 kN

313.00 kN 96.00 kN

12.90

17.80

Fig. 4.39. Multistory shear wall

an individual lintel. With regard to the lintel above the ground floor, it was

assumed that the concrete is cracked and that it therefore does not carry any

shear force. If it is assumed that all stories carry the same load, the shear

force grows linearly in the vertical direction. Hence the uppermost lintel on

the fourth floor carries a shear force V4, the next lintel a force 2 · V4, etc., so

that the total shear force across the height of the building is distributed as

follows:

V = V4 + V3 + V2 + V1

= V4 + 2 · V4 + 3 · V4 + 4 · V4 = 10 · V4 . (4.79)

Similar considerations can be applied to check the compressive and tensile

4.11 Multistory shear wall 369

V

M

1341 kN

- 1114 kN

163 kN

- 1474 kN

2426 kNm

5173 kNm

4189 kNm

center of opening

3278 kNm

3318 kNm

7.5 m 5 m

327 kN/m

2426 kNm

313 kN

195 kN 96 kN

Fig. 4.40. Equivalent

system

forces in the lintels. It is simply assumed that in the upper lintels—down to

the second floor—only compressive forces act, which are neglected because

they would reduce the reinforcement, while it is assumed that in the two

lower lintels tensile forces are acting. The lever arm z is equal to the distance

between the uppermost and lowermost lintel:

z = 13.06m − 1

2

0.7m − 1

2

0.26m = 12.58m. (4.80)

The tensile force T = M/z is split evenly into two parts, which are carried by

the two lintels

T1 = Tfloor =

1

2

M

z

. (4.81)

According to beam theory the moment M and the shear force V in the cross

section are (see Fig. 4.40)

M = 4189.2kNm V = −802.9kN, (4.82)

which yield the following forces:

V4 =

1

10

(−802.9 kN) = −80.3 kN (4.83)

V3 = −160.6kN, V2 = −240.9kN, V1 = −321.2 kN (4.84)

Z1st floor = Zground floor =

1

2

4189.2kNm

12.58m

= 166.5kN. (4.85)

370 4 Plane problems

Fig. 4.41. Principal stresses

Навигация