# Exercises

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Modelling slabs

## Exercise 5.4.7-1

Consider the typical floor of a multistorey building, with the centre of mass located in position X=9.0, Y=9.0 and the seismic force applied at the ground floor is equal to 2000 kN. Calculate the displacements and stress resultants of columns C1, C3, C4, C6, C11, C13, C16 under seismic action in direction X. The ground floor height is 3.0 m. Concrete class C30/37.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 3,67 0,97 47,5 12,6 71,3 -71,3 18,8 -18,8 C3 3,67 -0,32 116,0 -10,2 174,0 -174,0 -15,3 15,3 C4 3,67 -0,97 47,5 -12,6 71,3 -71,3 -18,8 18,8 C6 4,31 0,32 136,4 10,2 204,6 -204,6 15,3 -15,3 C11 4,96 -0,32 156,9 -10,2 235,3 -235,3 -15,3 15,3 C13 5,60 0,97 72,6 12,6 108,9 -108,9 18,8 -18,8 C16 5,60 -0,97 72,6 -12,6 108,9 -108,9 -18,8 18,8
The sum of shear forces in each direction shall be equal to the corresponding seismic force, regardless of the stiffness distribution or the deviation of the CM from the CT, i.e. in this particular case Σ(Vx)=2000 kN and Σ(Vy)=0.

## Exercise 5.4.7-2

In the structure of the previous exercise, assuming that the number of storeys is 10 and the interstorey height is 3.0 m, calculate the displacement the centre of stiffness CT of each floor and the maximum displacement of the higher floor of the building. Consider two cases: (a) orthogonal and (b) triangular distribution of seismic forces.

## (a) Orthogonal distribution of seismic forces

Base seismic shear = 2000 kN, orthogonal distribution of seismic forces

Conclusions:

• ·    The maximum absolute displacement δ10=55a occurs at the 10th level, and the minimum δ1=10a at the 1st level. On the contrary, the maximum relative displacement δz1=10a occurs at the 1st level, and the minimum δz10=a at the 10th level.
• ·    The maximum stress is developed at the 1st level due to the maximum relative displacement, while the minimum at the 10th level.
• ·    Due to non-coincident center of mass CMand center of stiffness CT, the diaphragm moments are proportional to the shear forces and the corresponding displacements are proportional to the Cdisplacements.  Therefore the maximum displacement occurs at the 10thlevel in column c13 and is 55/10=5.5 times greater than the one of the ground floor, i.e. δxx,10,13=5.5·5.60=30.8 mm and δxy,10,13=5.5·0.97=5.3 mm.

## (b) Triangular distribution of seismic forces

Base seismic shear = 2000 kN, triangular distribution of seismic forces

Conclusions:

• ·     At the 1st level, all types of seismic force distribution, including the triangular and the orthogonal, give the same displacement and stress.
• ·     Given the same seismic shear, the displacement in the last floor δ10=70a according to the triangular distribution is greater than the respective value δ10=55a of the orthogonal one.
• ·     The area of the shear forces diagram and heights represents the total moment of the floors and is larger in the triangular distribution than in the rectangular one.
• ·     The maximum displacement, developed at the 10th level in column c13, is 70/10=7.0 times greater than the one of the ground floor, i.e. δxx,10,13=7.0×5.60=39.2 mm and δxy,10,13=7.0×0.97=6.8 mm

## The behaviour of the actual structure under orthogonal seismic force distribution

The actual structure of the building with the wireframe model

## The behaviour of the actual structure under orthogonal seismic force distribution

Seismic action with base shear 2000 kN and orthogonal distribution of seismic actions

## Exercise 5.4.7-3

In the floor of exercise 5.4.7-1 above, calculate the corresponding results if the columns C3, C5, C12, C14 are replaced by walls with cross-section 2000/300 parallel to each perimeter side.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5 for columns and k=1.5 for walls)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 1,76 0,26 22,4 3,3 33,5 -33,5 4,9 -4,9 C3 1,76 -0,09 564,5 -2,3 846,8 -846,8 -3,5 3,5 C4 1,76 -0,26 22,4 -3,3 33,5 -33,5 -4,9 4,9 C6 1,93 0,09 59,3 2,6 88,9 -88,9 3,9 -3,9 C11 2,10 -0,09 64,5 -2,6 96,8 -96,8 -3,9 3,9 C13 2,27 0,26 28,9 3,3 43,3 -43,3 4,9 -4,9 C16 2,27 -0,26 28,9 -3,3 43,3 -43,3 -4,9 4,9
In the presence of walls, the shear effect should be taken into account, because they affect significantly both the displacements magnitude and the stress resultants distribution.

The magnitude of the maximum displacement is equal to 2.27 mm compared to 5.60 mm of the frame type structure of exercise 5.4.7-1 thus requiring smaller seismic joints, when necessary. The resulting wall displacements are smaller leading to less important or nil damages in the event of an earthquake. Moreover smaller stresses are developed in the columns and less reinforcement is required.

## Exercise 5.4.7-4 (a)

In the storey of exercise 5.4.7-1 calculate the results for the following cases:

(a) replacement of columns C5, C12, C14 with walls of cross-section 2000/300 parallel to the respective sides of the perimeter.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5 for columns and k=1.5 for walls)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 3,365 -0,370 42,8 -4,7 64,3 -64,3 -7,1 7,1 C3 3,365 0,123 103,6 3,8 155,4 -155,4 5,7 -5,7 C4 3,365 0,369 42,8 4,7 64,3 -64,3 7,1 -7,1 C6 3,118 -0,124 96,0 -3,8 144,0 -144,0 -5,7 5,7 C11 2,872 0,123 88,4 3,8 132,6 -132,6 5,7 -5,7 C13 2,625 -0,370 33,4 -4,7 50,1 -50,1 -7,1 7,1 C16 2,625 0,369 33,4 4,7 50,1 -50,1 7,1 -7,1

## Exercise 5.4.7-4 (b)

(b) replacement of the columns C12, C14,  with walls of cross-section 2000/300 parallel to the respective sides of the perimeter.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5 for columns and k=1.5 for walls)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 3,620 -0,795 46,1 -10,1 69,1 -69,1 -15,2 15,2 C3 3,620 -0,045 111,4 -1,4 167,1 -167,1 -2,1 2,1 C4 3,620 0,330 46,1 4,2 69,1 -69,1 6,3 -6,3 C6 3,245 -0,420 99,9 -12,9 149,8 -149,8 -19,4 19,4 C11 2,870 -0,045 88,3 -1,4 132,5 -132,5 -2,1 2,1 C13 2,495 -0,795 31,8 -10,1 47,7 -47,7 -15,2 15,2 C16 2,495 0,330 31,8 4,2 47,7 -47,7 6,3 -6,3

## Exercise 5.4.7-4 (c)

(c) replacement of the columns C5, C12, with walls of cross-section 2000/300 parallel to the corresponding sides of the perimeter.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5 for columns and k=1.5 for walls)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 4,438 0,407 56,5 5,2 84,7 -84,7 7,8 -7,8 C3 4,438 -0,136 136,6 -4,2 204,9 -204,9 -6,3 6,3 C4 4,438 -0,407 56,5 -5,2 84,7 -84,7 -7,8 7,8 C6 4,709 0,136 144,9 4,2 217,4 -217,4 6,3 -6,3 C11 4,980 -0,136 153,3 -4,2 229,9 -229,9 -6,3 6,3 C13 5,251 0,407 66,9 5,2 100,3 -100,3 7,8 -7,8 C16 5,251 -0,407 66,9 -5,2 100,3 -100,3 -7,8 7,8

## Exercise 5.4.7-4 (d)

(d) replacement of the column C5 with wall of cross-section 2000/300 in parallel to the respec-tive side of the perimeter.

## Calculation using file “diaphragm_ortho.xls” displacements and stress resultants (it is assumed k=5 for columns and k=1.5 for walls)

 δx δy Vx Vy Mx,1 Mx,2 My,1 My,2 (mm) (mm) (kN) (kN) (kNm) (kNm) (kNm) (kNm) C1 4,113 0,406 52,4 5,2 78,5 -78,5 7,7 -7,7 C3 4,113 -0,509 126,6 -15,7 189,9 -189,9 -23,5 23,5 C4 4,113 -0,967 52,4 -12,3 78,5 -78,5 -18,5 18,5 C6 4,570 -0,052 140,7 -1,6 211,0 -211,0 -2,4 2,4 C11 5,028 -0,509 154,8 -15,7 232,1 -232,1 -23,5 23,5 C13 5,485 0,406 69,8 5,2 104,8 -104,8 7,7 -7,7 C16 5,485 -0,967 69,8 -12,3 104,8 -104,8 -18,5 18,5