DEFLECTIONS AN EXAMPLE CALCULATION

DEFLECTIONS: AN EXAMPLE CALCULATION

The Instruction for the design and installation of structural concrete beam and pot floors made with prefabricated components (EFHE) establishes that:

► Total deflection at infinite time shall not exceed the smaller of the values

L/250 and L/500 + 10 mm, where L is span length in millimetres [1.6 times the overhang, in the case of overhangs]

► The specific active deflection shall not exceed the smaller of the values L/500

and L/1000 + 5 mm, in slabs that support walling or partition walls and envelopes

EFHE also specifies that it is not compulsory to verify whether the deflections

meet the foregoing conditions in: ► Joist slabs with spans smaller than 7 m ► Decks of prestressed hollow core slabs with spans smaller than 12 m

provided that the overloads are no larger than 4 kN/m2 and slab edge h is thicker than hmin, calculated from the expression:

L/C h 21min ⋅⋅= δδ where:

δ1: is a factor that depends on the total load q (kN/m2) and is equal to 7q

δ2: is a factor that has the value 4

1

6L⎟⎠⎞

⎜⎝⎛

L: is the slab calculation span in m C: is a constant according to the type of slab, load, and stretch as set out in the following table,

which is usually expressed as h ≥ 4.141

Lq

CL 4⋅⋅

C COEFFICIENTS

Type of section Types of slabs Types of loads Insulated End Internal

With partitions or walls 17 21 24 Reinforced joists Roofs 20 24 27 With partitions or walls 19 23 26 Prestressed joists Roofs 22 26 29 With partitions or walls 36 - - Prestressed hollow

core slabs[1] Roofs 45 - - [1] Prestressed pieces designed in such a way that, for an infrequent combination, they do not go beyond the cracking moment In the case of overhangs, C = 6 when the slab supports the load of partitions or walls; in other cases, C = 9.

Concrete beam and pot floors 1 Deflections: an example calculation

For example, for an isolated stretch of prestressed joist slab that supports partitions, with a total load of q = 6.5 kN/m2 and span of L = 5 m,

h ≥ m 0.24 4.141

56.5 195 4

=⋅

⋅

The following diagram

expresses the span/edge ratio in the consideration of hmin for the three types of slabs envisaged in the C coefficient table: (1) reinforced joists, (2) prestressed joists, and (3) decks of hollow core slabs, with the constraints in the span curves at 7 m for joist slabs and at 12 m for decks of hollow core slabs.

Value of hmin as a function of the span and type of slab For the calculation of deflection by the general method, Instruction EFHE refers

to sub-section 50.2.1 of the Instruction on Structural Concrete EHE; however, it provides a simplified method based on the calculation of instant deflection, considering for each slab span a single inertia value equivalent to Ie, which takes into account the cracking effect.

A case study follows from the book Forjados (Slabs) by Luis Felipe Rodríguez:

Reinforced joist slab with a span of 350 cm and 26-cm edge, with type IV-07 joists, with 2Ø6 + 1Ø12 below and 1Ø12 above, and uniformly distributed linear service load of 7.0 kN/m.

The data on the prefabricated joists, cracking moments, total stiffness, and

cracked stiffness, in both positive and negative bending, for 1-m bands, have been drawn from the document ‘Authorisation for Use’.


Bending Cracking moment

[kN·m] Total stiffness

[kN/m] Cracked stiffness [kN/m]

Positive Negative

6.82 14.83

13,690 12,240

2,600 1,590

First, it is necessary to verify that the equivalent stiffness (product of the

modulus of longitudinal deformation E and the moment of equivalent inertia Ie is smaller than or equal to the total stiffness (product of E and the gross moment of inertia Ib), according to the expression:

bf

3

0a

0fb

3

0a

0fe I E I E

M - MM - M

- 1 I E M - MM - M

I E ⋅≤⋅⋅⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⋅⋅⎥

⎦

⎤⎢⎣

⎡=⋅

where: Mf: is the cracking moment Mo: is the flector moment associated with the situation of zero curvature of the

section, with a value equal to P·e·β-Mv·(β-1), where P is the absolute value of the prestressed force, if it exists, which may be taken as 90% of the initial prestressed force; e is the eccentricity of the prestressed equivalent tendon, in the section being studied, in absolute value, with respect to the centre of gravity of the joist; and β is the relation between the gross inertia of the slab section, in the construction phase in which one calculates the deflection and gross inertia of the section of the prefabricated piece, larger than or equal to one. In non-shored construction, when deflection is calculated under the own weight of the prefabricated piece or of the poured concrete in the building work, β=1

Mv: is the moment due to the loads that act on the prefabricated piece before working jointly with the concrete in situ. Its value is: • For non-shored construction, the moment due to the own weight of the

prefabricated piece and to the weight of the concrete poured in situ • For shored construction, zero if the piece is reinforced, and the moment

due to its own weight if it is prestressed • Zero in the end sections subject to negative moments

Ma: is the maximum flector moment that the section considered has, historically, been able to display, including the phase under study

In the present example: Mf = 6.82 kN·m E·Ib = 13,690 m2 kN/m E·If = 2,600 m2 kN/m

8L q M

2

a ⋅= , where q is the uniform unit load and L is the slab span in metres;

in the present example: q = 7 kN/m and L = 3.5 m, so that

mkN 10.72 8

3.5 7 M2

a ⋅=⋅=


M0 = 0, since P = 0 (reinforced concrete joists) and Mv = 0 since the structure lies on straining pieces

kN/m 3,6901 kN/m 5,456 600,2 10.726.82 - 1 3,6901

10.726.82 I E

33

e ≤=⋅⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+⋅⎥⎦

⎤⎢⎣⎡=⋅

As a result, the instant deflection (finst) can be calculated from the formula:

e

4

inst I ELq

3845 f

⋅⋅

⋅= , where:

E·Ie = 5,456 kN/m, the equivalent stiffness calculated in the previous step q = 7 kN/m, the linearly distributed load L = 3.5 m, the slab span

mm 2.51 m 102.51 456,53.57

3845 f 3-

4

inst =⋅=⋅

⋅=

Once the instant deflection for this reinforced concrete beam and pot floor of

half-joists, with a 3.5-mm span and 26-cm edge, and uniform single load of 7 kN/m has been calculated, the total deflection at infinite time can be calculated from the expression:

Total deflection (fTOTAL) at infinite time = Instant deflection + Delayed deflection The delayed deflection (fdel) produced by long-lasting loads may be estimated by

multiplying the instant deflection by a factor λ, with the following value (head 50.2.2.3 of Instruction EHE):

ρ50 1ξ λ

′⋅+= , where:

ρ': Geometric quantity of the compression reinforcement As' relative to the compression area of the useful section b0·d, in the reference section

dbA

ρo

s

⋅′

=′

ξ: Coefficient that depends on load duration and has the values indicated in the table

Load duration 2 weeks 1 month 3 months 6 months 1 year 5 years

or more

Value of ξ 0.5 0.7 1.0 1.2 1.4 2.0

In the present example: ξ = 2.0, 0.00105 22.412

0.28 db

A ρo

s =⋅

=⋅′

=′

1.9 0.00105 50 1

2.0 ρ50 1

ξ λ =⋅+

=′⋅+

=

fTOTAL = finst + fdel = finst + λ·finst = finst·(1+λ) = 2.51·(1+1.9) = 7.28 mm


which meets the requirements of Instruction EFHE, since the total deflection at infinite time should not exceed the smaller value of:

mm 14 250

3,500 250L

== (L is given in millimetres)

mm 17 10 500

3,500 mm 10 500L

=+=+

In this example, fTOTAL = 7.28 < 14 mm If partitions with a load of 4 kN/m2 are built on this slab 14 days after the slab

has been installed, the correction factor for total stiffness of the prefabricated joists may be obtained from the document ‘Authorisation for Use’, this being 0.89. Where:

• Total stiffness E·I = 13,690 kN/m

• Maximum moment mkN 6.13 8

3.54 8L q M

22

a ⋅=⋅=⋅=

• [E·I]c = 0.89·13,690 = 12,184.1 m2 kN/m Instant deflection will be:

[ ] mm 0.6 m 106.4 12,184.1

3.5 6.13485

IELM

485 f 4-

2

C

2a

inst =⋅=⋅

⋅=⋅⋅

⋅=

Delayed deflection after 14 days will be:

0.46 0.00105 50 1

0.5 ρ50 1

ξ λ ≅⋅+

=′⋅+

=

fdel = λ·finst = 0.46·0.6 = 0.3 mm

So that the total deflection after 14 days will be:

fTOTAL = finst + fdel = 0.6 + 0.3 = 0.9 mm

Active deflection (fact), i.e. the deflection produced after the construction of the

partitions, will be the total deflection at infinite time minus the deflection that already existed when the partitions were constructed.

fact = 14 – 0.9 = 13.1 mm

According to Instruction EFHE, this active deflection would not be

permissible because it is larger than mm 7 500

3,500 500L

== (also larger than

mm 8.5 5 3.5 mm 5 1000

L=+=+ ).


A further calculation example follows that is perhaps closer to current Spanish reality in apartment buildings:

Beam and pot floor with prestressed joists of reinforced concrete, in a section

resting on two supports with a 5-m span and 30-cm edge, which supports a characteristic total load (including its own weight) of 7.0 kN/m2. Once the cracking moment has been calculated [Mf = 37.988 kN·m], document ‘Authorisation for Use’ of the prestressed joist indicating that the stiffness of its homogenised section is E·I = 18,755 kN/m, it may be verified whether the deflection at infinite time meets the EFHE requirement.

The maximum applied flector moment Ma is calculated from the formula

8L q M

2

a ⋅= , where q is the uniform unit load and L is the span in metres, 7 kN/m and 5

m, respectively

mkN 21.875 85 7 M

2

a ⋅=⋅=

Since Ma is smaller than the cracking moment [Mf = 37.988 kN·m], the slab will

not be cracked under those conditions and the instant deflection (finst) can be calculated from the expression:

mm 3 m .0030 18,755

5 21.875485

IELM

485 f

22a

inst ==⋅

⋅=⋅

⋅=

If λ is calculated for 5 years or more, the delayed deflection (fdel) can be

calculated:

fdel = λ·finst = 2·3 = 6 mm [setting λ = 2]

The total deflection at infinite time fTOTAL = finst + fdel = 3 + 6 = 9 mm, which conforms to the requirements of Instruction EFHE, since it is smaller than

mm 20 250

5.000 250L

== and than mm 20 10 500

5.000 mm 10 500L

=+=+ .

Assuming this slab is loaded with partitions two weeks after slab installation,

with a load of 4 kN/m2, knowing that the correction factor for total stiffness after 14 days is 0.89 (from the document ‘Authorisation for Use’), this gives:

• Total stiffness after 14 days: 0.89·18,755 = 16,692 kN/m

• Maximum moment for that load: mkN 12.5 8254

8L q M

2

a ⋅=⋅=⋅=

mm 2 16,692

25 12.5485

IELM

485 f

2a

inst =⋅

⋅=⋅

⋅=

The delayed deflection after 14 days [with λ = 0.5] will be: fdel = 0.5·2 = 1 mm The total deflection after 14 days will be fTOTAL = finst + fdel = 2 + 1 = 3 mm


As a result, the active deflection (fact) after construction of the partition walls will be the difference between the total deflection at infinite time (ftotal·p.i) and the total deflection after 14 days: (ftotal 14 d)

fact = ftotal p.i - ftotal 14 d = 9 - 3 = 6 mm

which meets the requirements of the EFHE Instruction because it is smaller than

mm 10 500

5.000 500L

==

mm 10 mm 5 1000

L=+

The following reproduces the tables of specific weights, weights per unit surface

area, and weights of the extracted construction elements from Annex C of the Basic Document on Structural Safety – Actions in Building Construction [DB SE-AE], as well as the table of characteristic values of service overloads [Table 3.1, DB SE-AE, page 5].





Documents

DEFLECTIONS AN EXAMPLE CALCULATION