Download Wall Calculation Example EC2 PDF

TitleWall Calculation Example EC2
File Size245.6 KB
Total Pages15
Document Text Contents
Page 1

Worked Examples for Eurocode 2


Draft Version















All advice or information from The Concrete Centre is intended for those who will evaluate
the significance and limitations of its contents and take responsibility for its use and
application.

No liability (including that for negligence) for any loss resulting from such advice or
information is accepted by the Concrete Centre or their subcontractors, suppliers or
advisors.

Readers should note that this is a draft version of a document and will be subject to
revision from time to time and should therefore ensure that they are in possession of the
latest version.

Page 2

WE 6 Walls v7a chg 17 Sep 07.doc 17 Sep. 07 Page 2 of 15

6 Walls

6.1 General
Walls are defined as being vertical elements whose lengths are four times greater than
their thicknesses. Their design does not differ significantly from the design of columns in
that axial loads and moments about each axis are assessed and designed for.

Generally, the method of designing walls is as follows:

1. Determine design life. <BS EN 1990 & NA Table NA 2.1>
2. Assess actions on the column. <BS EN 1991 (10 parts) & UK

NAs>
3. Determine which combinations of actions

apply.
<BS EN 1990 & NA Tables NA
A1.1 & NA A1.2(B)>

4. Assess durability requirements and determine
concrete strength.

<BS 8500–1>

5. Check cover requirements for appropriate fire
resistance period.

<Approved Document B
BS 1992–1–2>

6. Determine cover for fir e, durability and bond. <BS EN 1992–1–1 Cl. 4.4.1>
7. Analyse structure for critical combination

moments and axial forces.
<BS EN 1992–1–1 Section 5>

8. Check slenderness and determine design
moments.

<BS EN 1992–1–1 Section 5.8>

9. Determine area of reinforcement required. <BS EN 1992–1–1 Section 6.1>
10. Check spacing of bars <BS EN 1992–1–1 Sections 8 & 9>

Example 6.2 shows the design of a simple linear shear wall as typically used in medium rise
buildings. Similar principals may be applied to walls that are shaped as C, L, T, Z and rectangles
in-plan but issues of limiting flange dimensions and shear at corners need be addressed. The
example shows only ULS design as, apart from minimum areas of steel to control cracking, SLS
issues are generally non-critical in medium-rise structures. For shear walls in high-rise
structures, reference should be made to specialist literature (ref to CIRIA R102 Design of shear
wall buildings).



6.2 Shear wall (Wall A)
Wall ‘A’ is 200 mm thick and in addition to providing ve rtical support to 200 mm flat slabs at
roof level and floors 1 to 3, it helps to provide lateral stability to the four storey office block.
Assuming the stair itself provides no lateral stability, the wall is to be designed for the critical
section at ground and fi rst floor level using BS EN 1990 Exp. (6.10). The concrete is C30 / 37.
The wall is supported on pad foundations an d the ground floor is ground bearing.



Figure 6.1
Typical floor plan

Page 7

WE 6 Walls v7a chg 17 Sep 07.doc 17 Sep. 07 Page 7 of 15

Wall A takes 51% of wind load
so characteristic wind load on wall A,
Wk = 51% wk Lx = 51% 1.1 30.7 = 17.2 kN / m


at just above ground floor, characteristic in-
plane moment in wall A, Mk

= 17.2 1412 / 2 = 1709.8 kNm

Resolving into couple using 1 m either end of wall‡,
characteristic wind load in each end, Wk

= 1709.8 / 3.4 = 502.9 kN











Figure 6.6 Wall A – wind loads N–S





6.2.5 Effects of global imperfections in plane of wall A



Figure 6.7 Global imperfections


‡ For medium rise shear walls there are a number of methods of design. Cl. 9.6.1 suggests strut-and-
tie (see Section xx). Another method [ref to Concrete Buildings Design manual] is to determine elastic
tensile and compression stresses from NEd/bL +/– 6MEd/bL

2
and determine reinforcement

requirements based on those maxima. The method used here assumes a couple, consisting of 1.0 m of
wall either end of the wall. The reinforcement in tension is assumed to act at the centre of one end and
the concrete in compression (with a rectangular stress distribution) acts at the centre of the other
end. The forces generated by the couple add or subtract from the axial load in the 1 m ends of the
walls. The method is useful for typical straight shear walls of say 2.5 to 5.0 m in length.

Page 8

WE 6 Walls v7a chg 17 Sep 07.doc 17 Sep. 07 Page 8 of 15

Global imperfections can be represented by forces Hi at floor level where
Hi = i(Nb – Na) <Exp. (5.4)>
where
i = (1 / 200) h m

where

<5.2(1),
5.2(5), 5.2(8)
& NA>

h = 0.67 2 / l
0.5 1.0

= 0.67 2 / 14.70.5 1.0
= 0.67 0.52 1.0
= 0.67
m = [0.5(1 + 1 / m)]

0.5
where
m = no. of members contributing to the total effect
= 25 vertical elements on 4 floors
= 100
m = 0.71
i = 0.67 0.71 / 200
= 0.0024
Nb, Na = axial forces in members below and above
(Nb – Na) = axial load from each level

At roof level
Area = 30.4 14.5 – 1.3 2.5 – 3.6 4.8 = 420.3 m3
Perimeter = 2 (30.4 + 14.5) = 89.8 m
(Na – Nb) = axial load from roof level
= 420.3 (7.05 + 0.6) + 89.8 0.9 4.0 = 3286.4 + 252.2 kN

At 3rd floor
(Na – Nb) = 420.3 (5.78 + 2.5) + 89.8 3.3 4.0 = 3615.7 + 1050.8 kN

At 2nd floor
(Na – Nb) = 3615.7 + 1050.8 kN

At 1st floor
(Na – Nb) = 3615.7 + 1050.8 kN

HiR = 0.0024 (3286.4 + 252.2) = 7.9 + 0.6 = 8.5 kN
Hi3 = Hi2 = Hi1 = 0.0024 (3615.7 + 1050.8) = 8.7 + 2.5 = 11.2 kN
Characteristic design moment at ground floor
Mk = 8.5 13.2 + 11.2 (9.90 + 6.60 + 3.30)
= 112.2 + 221.8 = 334.0 kNm
As before, wall A resists 51% of this moment. Resolving into couple using 1 m either end of wall,

GkH
§ = 0.51 334.0 / 3.4 = 50.1 kN

i.e. GkH = 50.1 kN / m



§ As Hi derives mainly from permanent actions its resulting effects are considered as being a
permanent action too.

Page 14

WE 6 Walls v7a chg 17 Sep 07.doc 17 Sep. 07 Page 14 of 15

6.2.14 Check for tension at top of foundation
Permanent and variable (see Section6.2.2) <Section 6.2.2>
Gk = 1021.0 / 4.4 = 232.0 kN / m
Qk = 225.1 / 4.4 = 51.2 kN / m
Wind <Section 6.2.4>
Mk = 17.2 14.1 [14.1 / 2 + 0.6] = 1855.3 kN / m
Resolved into couple 1 m either end of wall
Wkw = 1855.3 / 3.4 = +/- 545.7 kN / m
Global imperfections(see Section 6.2.5) <Section 6.2.5>
Mk = 8.5 13.8 + 11.2 (10.5 + 7.2 + 3.9 + 0.6)
= 365.9 kNm
GkH = 365.9 0.51 / 3.4 = 54.9 kN / m
At ULS for maximum axial tension Wk is lead variable action:
NEd = 1.0 232.0 – 1.35 54.9 – 1.5 545.7 + 0 51.2
= –660.7 kN / m
MEd = nominal = e2NEd = 0.02 660.7 <6.1.4>
= 13.2 kNm / m
As before
NEd MEd


As = fyk / m
+

(d – d2)fyk / m
= 660.7 103 / 434.8 + 13.2 106 / [(157 – 43) 434.8]
= 1520 + 266
= 1786 mm2 i.e. not critical
Use 6 no. H16 @ 200 cc b.s. for at least 1 m either end of wall.

6.2.15 Check stability
Assume base extends 0.3 m beyond either end of wall A, i.e. is 5.0 m long
and is 1.2 m wide by 0.9 m deep



Overturning moments <BS EN 1990 Table
A1.2(A) & NA>

Wind (see Figure 6.6)
Mk = 0.51 x 17.2 14.1 [14.1 / 2 + 1.5] = 1057.5
Global imperfections (see Figure 6.7)
Mk = 0.51 x [8.5 14.7 + 11.2 (11.4 + 8.1 + 4.8 + 1.5)]
= 0.51 x [125.0 + 11.2 25.8]
= 0.51 x 414.0
= 211 kNm




Restoring moment
Mk = (1021.0 + 5.0 x 1.2 x 0.9 x 25 + 0 x 225.1 ) x (0.3 + 2.2)

= 2890 kNm

At ULS of EQU,
Overturning moment
= fn( Q,1Qk1 + G,supGk)
= 1.5 x 1057.5 + 1.1 x 211.0 = 1818.4 kNm

<BS EN 1990 Table
A1.2(A) & NA>

Restoring moment
= fn( G,infGk)
= 0.9 x 2890 = 2601 kNm i.e. > 1818.4 kNm

<BS EN 1990 Table
A1.2(A) & NA>

OK

Page 15

WE 6 Walls v7a chg 17 Sep 07.doc 17 Sep. 07 Page 15 of 15


6.2.16 Design summary





Figure 6.13 Wall design summary

Similer Documents