Examples

Example 2.11.1 Continuous beam in a domestic structure

Determine the appropriate load combination for a continuous beam in a domestic structure supporting a 175 mm slab at 6 m centres. gk = 51 kN/m and gk = 9.0 kN/m.

Figure 2.8

Continuous beam in a domestic structure

Figure 2.8

Continuous beam in a domestic structure

Actions: kN/m

Permanent action, gk

Self weight, 175 mm thick slabs = 26.3

E/o self weight downstand 800 * 225 = 4.5

Finishes and services = 3.0 Dividing wall 2.40 * 4.42 (200 mm dense blockwork with plaster = 10.6

both sides) _ 510

Total

Variable action, qk

Imposed, dwelling @ 1.5 kN/m2 Ultimate load, n

Assuming use of Exp. 6.10, n = 1.35 * 51 + 1.5 * 9.0 = Assuming use of worst case of Exp. (6.10a) or Exp. (6.10b) Exp. (6.10a): n = 1.35 * 51 + 0.7 * 1.5 * 9.0 = Exp. (6.10b): n = 1.25 * 51 + 1.5 * 9.0 =

In this case Exp. (6.10a) would be critical1 and

Example 2.11.2 Continuous beam in mixed use structure

Determine the various arrangements of actions and magnitude of actions for ULS verification of a continuous beam supporting a 175 mm slab @ 6 m centres. Note that the variable actions are from two sources: Office use: 2.5 kN/m2, y0 = 0.7;shopping use: 4.0 kN/m2, y0 = 0.7

<BS EN 1991-1-1 6.3.1.1 & NA> < BS EN 1990 A.1.2.2. & NA>

1 This could also be determined from Figure 2.1 or by determining that gk > 4.5qk WE 2 Basis v7c 17 Sep 07.doc 17-Sep-07

Figure 2.9

Continuous beam in mixed use structure

Figure 2.9

Continuous beam in mixed use structure a) I oad combination

Load combination Exp. (6.10a) or Exp. (6.10b) will be used, as either will produce a smaller total load than Exp. (6.10). It is necessary to decide which expression governs.

Actions:

Permanent action

As before, Example 2.8.1 Variable action

Office @ 2.5 kN/m2 Shopping @ 4.0 kN/m2 Ultimate load, n For office use:

Exp. (6.10a): n = 1.35 * 51 + 0.7 * 1.5 * 15.0 Exp. (6.10b): n = 1.25 * 51 + 1.5 * 15.0 For shopping use:

Exp. (6.10a): n = 1.35 * 51 + 1.5 * 0.7 * 24.0 Exp. (6.10b): n = 1.25 * 51 + 1.5 * 24.0

kN/m

By inspection Exp. (6.10b) governs in both cases*

b) Arrangement of actions i) Actions

As the variable actions arise from different sources, one is a leading variable action and the other is an accompanying variable action. The unit loads to be used in the various arrangements are:

Actions: Permanent

1.25 * 51.0 Variable

Office use as leading action, yQQk = 1.5 * 15 = as accompanying action, ^0yQQk = 0.7 * 1.5 * 15 =

kN/m

1 This could also be determined from Figure 2.1 or by determining that gk < 4.5qk

Shopping use as leading action, yaQk = 1.5 x 24 = = 36.0

as accompanying action, ty0yaQk = 0.7 x 1.5 x 24 = = 25.2

Total variable actions =163.25

ii) For maximum bending moment in span AB

The arrangement and magnitude of actions of loads are shown in Figure 2.10. The variable load in span AB assumes the value as leading action and that in span CD takes the value as an accompanying action.

Figure 2.10

For maximum bending moment in span AB

Figure 2.10

For maximum bending moment in span AB

iii) For maximum bending moment in span CD

The load arrangement is similar to that in Figure 2.10, but now the variable load in span AB takes its value as an accompanying action (i.e. 15.75 kN/m) and that in span CD assumes the value as leading action (36 kN/m).

Figure 2.11

For maximum bending moment in span CD

Figure 2.11

For maximum bending moment in span CD

iv) For maximum bending moment at support B

The arrangement of loads is shown in Figure 2.12. As both spans AB and BC receive load from the same source, no reduction is possible (other than that for large area2).

Permanent action = 63-0 kN/m

A B

C

0

For maximum bending moment at support B

Variable actions may be subjected to reduction factors: aA, according to A area supported (m2), aA = 1.0 - A / 1000 > 0.75. < BS EN 1991-1-1 6.3.1.2 (10) & NA>

v) For maximum bending moment at support D

The relevant arrangement of loads is shown in Figure 2.13. Comments made in iv) also apply here.

The relevant arrangement of loads is shown in Figure 2.13. Comments made in iv) also apply here.

Figure 2.13

For maximum bending moment at support D

Figure 2.13

For maximum bending moment at support D

vi) For critical curtailment and hogging in span CD The relevant arrangement of loads is shown in Figure 2.14.

vi) For critical curtailment and hogging in span CD The relevant arrangement of loads is shown in Figure 2.14.

Figure 2.14

For curtailment and hogging in span CD

Figure 2.14

For curtailment and hogging in span CD

Eurocode 2 requires that all spans should be loaded with either yGsupp or yGinf (as per Table 2.6). As illustrated in Figure 2.14, using yGjnf, = 1.0 might be critical for curtailment and hogging in spans.

Example 2.11.3 Propped cantilever

Determine the Equilibrium, ULS and SLS (deformation) load combinations for the propped cantilever shown in Figure 2.15. The action P at the end of the cantilever arises from the permanent action of a wall.

Figure 2.15

Figure 2.15

For the purposes of this example, the permanent action P is considered to be from a separate source than the self-weight of the structure so both Yg,sup and <BS EN 1990 Table Yen need to be considered. 1.2(B), Note 3>

a) Equilibrium limit state (FQU) For maximum uplift at A_

a) Equilibrium limit state (FQU) For maximum uplift at A_

Figure 2.16

EQU: maximum uplift at A

Figure 2.16

EQU: maximum uplift at A

b) Ultimate limit state (ULS)

i) For maximum moment at B and anchorage of top reinforcement BA.

yGkjnf gk = 1-0 gk may be critical in terms of curtailment of top bars BA.

Figure 2.17

ULS: maximum moment at B

yGkjnf gk = 1-0 gk may be critical in terms of curtailment of top bars BA.

Figure 2.17

ULS: maximum moment at B

ii) For maximum sagging moment AB

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Responses

• Michael
Is 8 m cantilever possible in concrete building structures?
7 years ago