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Structural Design of Thin Masonry Walls Based on the Assumptions
Augmenting the Present Requirements of DIN 1053
Eichstaedt, H. J.
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PREFACE
In this country and abroad there is a movement towards establishing the design of unit masonry on an analytical basis as an alternative to the older traditional methods governing heights and thicknesses. Leadership in this field has come from Europe and this translation is one of a number that have been made for the infor-mation of Canadian committees on masonry and the building industry in general.
The Division of Building Research extends its thanks to Mr. D.A. Sinclair of the Translations Section of the National Research Council who translated this paper and to Dr. P.T. Mikluchin, of Toronto, a member of the Revision Committee on Masonry
for the National Building Code 1965, who checked the translation.
ottawa June, 1967
R.F. Legget Director
NATIONAL RESEARCH COUNCIL OF CANADA
Technical Translation 1289
Title:
The structural design of thin masonry walls based on the
assumptions augmenting the present requirements of DIN 1053
iDie statische Berechnung dunner gemauerter Wande bei
Uberschreitung der Grenzen der DIN 1053)
Author:
H.J. Eichstaedt
Reference:
Bautechnik,
7:
244-249, 1964
THE
STRUCTURAL DESIGNOF
THIN セaセョnryWALLS
AASED ON THE ASSUMPTIONS AUGMENTING THE PRESENTセセMMMMBM
REQUIREMENTS OF DIN 1053 1. Int roduct ion
In the northwest German coastal region cavity walls consisting of two masonry shells, each one half brick thick, are used as exterior walls on account of the climate and especially on account of the frequent driving rains. Half-brick walls are used as interior walls in order to gain space. The design of walls with continuous air space is based on DIN 1053, number 5.1, as well as Tables 2 and
5,
and that of load bearing interior walls with thicknesses of less than 24 cm, on DIN 1053 number 2.1 and Tables I,2 and
6.
In at least half the residential and commercial buildings in the coastal region these regulations are not followed - with the tacit consent of the building inspection authorities. Construction with thin walls has "proved itself" in the coastal region for decades and longer for reasons of economy, and cannot be prevented except where structural considerations are paramount. On the other hand, the occasional liberality of the DIN requirements is unsatisfactory especially where the absenceof restrictions with respect to the safety of mortar joints against cracking cannot be justified.
The following suggestions show how the critical areas of walls may be quickly and simply designed, maintaining the required safety factor of
5.
2. Infractions in Practice Against DIN 1053
The DIN 1053 reference numbers are given in brackets. 2.1. 11.5 and 17.5 cm interior walls.
2.1.1. Openings in 11.5 cm walls for sliding doors are always> 1.25 m. Generally, 11.5 cm bricks + 7 cm air space + 5.5 em edge bricks are used
(cf. DIN 1053, Table I, line 5).
2.1.2. Doors in stiffening transverse walls are situated less than 50 cm from the "stiffened" 11.5 or 17.5 cm thick main wall, for example, in the vestibules of public apartment houses (cf. number 2.21).
2.1.3. 11.5 cm walls are used as end supports for slabs, e.g. around the stairwells of single-family houses (cf. Table I, line
5).
2.1.4. 11.5 cm and 17.5 em walls support portions of slabs with live loads in excess of 275 kp/m2
, e.g. staircase landings, where a live load
2.1.5. The centre-to-centre distance of the 11.5 cm wall exceeds the
4.5
m limit, because4.5
m wall-to-wall distance is taken. Spaces with 4.5 clear spans have become, so to speak, the rule. For 17.5 cm walls, correspondingly, the 6.0 m specifications has come to be regarded as clear span. (Table 2, line 1). Exceeding this limit by 50 cm is con-sidered as "inadmissible".2.1.6. For years 11.5 cm walls have been used to support one-way simply supported prefabricated floor units without anyone objecting. The same applies to wooden beam floors (Table 1, line 5). Occasionally a 17.5 cm wall is provided for this, but never a 24 cm wall. If this were required -due to lack of continuity - the prefabricated floors would hardly be competi-tive in the coastal region (cf. also explanation to 2.12, number 1).
2.1.7. Continuous slabs frequently have a bay with span greater than 4.5 m - nevertheless, the forbidden averaging (explanation to 2.12) is being practised.
2.1.8. Concentrated loads from the attic storey are supported on solid slabs or prefabricated units partly in the bay, partly in the vicinity of walls or directly over walls, and thus violate the explanation 2.12, numbers
7
and8.
Distributing beams are omitted.2.1.9. Slots in 11.5 and 17.5 cm thick walls are made 3 cm deep (the DIN regulations are so interpreted). Bearing walls for kitchen and bathroom installations are sometimes entirely chiselled out afterwards by the build-ing contractor, partly for space-savbuild-ing reasons and partly for appearance's sake (cf. number 2.5).
2.1.10. The building inspection authorities tacitly permit various exceptions from Table 1, line 1, 2 and 5 (number 2.122, paragraph 2). There are many different reasons for this 'liberality:
1. On the basis of experience these infractions 。ァ。ゥョウセ various require-ments of DIN 1053 do not incur any risk of collapse and hence do not endanger the public safety. In such a situation the authorities are not legally
responsible (cf. "Responsibility of the Building Licensing Authorities", Federal Building Gazette 1963, Issue 9, pages 431 to 432).
2. The designs and calculations for many of the simpler types of
building come from designers who are not familiar with the technical problems, but may be good practitioners.
3.
Architects, housing construction companies and customers all have an interest in keeping wall thicknesses to a minimum. Since financing with public funds is based on the number of cubic metres reconstructed space, they wish to produce as much floor area as possible. At the same time, of course, they wish to keep cost down.MUセ
4.
Many infractions cannot be recognized from the 1 : 100 bUilding plans (for example slots or concentrated loads). They cannot be seen until the construction is complete.5.
Due to the lack of simple, but reliable design rules for thin walls similar to thew
method for concrete, wood and steel (unfortunately slender-ness Table6
is not adequate), people think they must rely on "experience".2.2. 28 cm walls
For 28 cm walls (inside shell backing brick 11.5 cm, outside shell 10.5 cm, air space
6
cm) conditions are indeed given in Number 5.1 of DIN 1053 for storey height and stiffening, but not for openings (windows). Thus, questions like the following arise:2.2.1. Can a window be made any size?
2.2.2. Is a heating element recess under a window of no importance? Concerning 2.2.1. and 2.2.2.
As a result of the window opening especially where there is a heating element recess, wall sections are created on either side of the opening which are now supported only along three sides. The net distance up to the lintel (which doesn't even extend from one transverse wall to the other) is perhaps 0.8 + 1.5
=
2.3 m. If we merely apply the criterion "stiffening every 4.5 m" to such a wall, as was formerly done, we barely reach a limit that is definitely too favourable. If we took the degree of slenderness of the wall sections next to the opening as a basis (A=
ゥゥセU
=
20), such a wall with window and recess for heating element would just be unacceptable in 28 cm thickness and would have to be increased to 35 cm thickness (inside shell 17.5 cm). However, this would apply to most of the residentialconstruction in the coastal region. The one is thus too favourable, and the other too unfavourable.
2.2.3. What is the situation with window posts between two windows and heating element recesses beneath the windows?
2.2.4. What is the maximum floor span at which the floor may still be supported on the 11.5 cm inside shell (i.e. on one side)?
2.2.5. Can stiffening uprights of steel or concrete be used in place of transverse walls?
2.2.6. What regulation prevents the frequently observed separation of the gable triangle?
2.2.7. What is the maximum distance that wall sections can extend beyond the masonry without the use of end columns (e.g. for balconies)?
-6-2.2.8. Can the outside shell be supported with the reinforcing steel in joints only, without a lintel, even when the window extends to the transverse wall?
Concerning 2.2.3.
In window posts in thin cavity walls cracks are sometimes observed which, in my opinion, are caused by bending and compression in which case the
tensile strength of the masonry is exceeded so that the joints open right to the centre. Apart from the fact that cracks in the interior are ugly, cracks in the outside masonry are harmful. During frost, water gets in, freezes and then in the course of time destroys the post. Such cracks arise (a) from rotation due to deflection of the floor slabs at supports;
(b) from wind suction. (The reduced brick strength due to the degree of
slenderness is not involved, otherwise a collapse would already have occurred.) Concerning 2.2.4.
The magnitude of the angle of rotation of the floor slab and hence of the wall, depends on the floor slab span. For example, with a
4.5
m floor span and 1 cm deflection due to elastic and plastic deformation, the angle of rotation is 1/90. This imparts a moment to the wall which is not taken into account in the calculation. DIN 1053 takes this unfavourable circum-stance into account for inside walls through the requirements of Table 1, line5,
"Only admissible as intermediate support of continuous slabs less than or equal to4.50
m".In the case of exterior walls, however, no limitation is prescribed by DIN, even though in this case besides one-sided slab support there are additional unfavourable circumstances, namely unlimited window opening and loading by wind suction. The liberality of DIN 1053 with respect to the cavity wall is justified in the explanation to 5.11, namely, that the provi-sion of an outside shell connected with wire ties in front of an inside bearing shell definitely contributes to the improvement of the bearing capacity of the 'latter. The improvement of the resistance to buckling (how great?) is unquestionable, but whether this justifies neglect of the "end supports of ceilings", "size of windows" and "wind" is uncertain.
Concerning 2.2.5.
Stiffening columns in outside walls are readily accepted by architects in 28 cm walls with wall length greater than
4.5
m (up to 9.0 m), because the mentioned savings of space and cost are desirable. However: are these columns イ・。セャケ equivalent to cross walls? First the execution: about 3-7-encased in cement mortar (protection against rust). At the building sites, certainly, the variation of mortar qualities is rarely considered. Then the slab deflection: whereas in the region of a true cross wall the slab deflection, and hence deflection of the loaded wall, is 0, the stiffening column, e.g. 17/24 in reinforced concrete (flush with the wall), will scarcely prevent the deflection on account of its low section modulus. The entire length of the wall, with all its windows, thus bends under the slab supported on one side. Finally, the wall is divided into two parts and no longer acts as a continuous wall. In my opinion such a stiffening column is not equivalent to a toothed in stiffening wall.
Where there are no calculation guides in addition to the arrangement prescriptions of DIN 1053, taking into account the bearing behaviour in an approximate way, the designer and building inspectors have no criteria by which to assess and prevent violations of the spirit of DIN 1053, and specifically the observance of the uniform safety factor in all bearing masonry parts.
3. Suggestions for Supplementing DIN 1053
For cases where the regulations of DIN 1053 cannot be observed, or if violations of DIN 1053 must be prevented, an exact structural design has to be carried out. Number 2.22 provides the method (in number 2.122 the second sentence should be deleted where observance of the conditions of Table 1, numbers 1, 2 and 5 is basically required). The following guides for the calculation of 11.5 cm and 17.5 cm thick interior walls and cavity walls in the coastal region are presented for this purpose. Their application should be permitted by a supplement to DIN 1053 or by official promulgation. A condition for the application of these guides is the undisplaceability of floor slabs at every point.
Guides for the dimensioning of thin walls
Here b = width of masonry section to be calculated, h floor to floor distance,
d
=
wall thickness.1. Under fixed ends of slabs for allowable pressure a according
1
to DIN 1053, Table 5 is to be employed.
2. Wall sections supported on two sides are to be designed for A = hid, even when they are joined by wall sections supported on three sides, i.e. starting from a ratio blh = 1.6 (admissible pressure
a).
-8-and a , taking into account the
ec-3
They can be obtained from
3.
Walls supported on three sides, between wall stiffening and an opening (width B) should be calculated with a slenderness ratioA
= 0.-hid. Support on three sides is present up to a ratio b/h = 0.8 (admissible pressure a ).3
4. The allowable stresses a
2 centric floor load and a = a
(40 -
A)/35.
2 , 3 1
Table I or Figure
3.
The allowable load should be compared to the maximum edge pressure P/F+
M/W (i.e. no. w method).5.
For floor load on one side, a triangular stress distribution is assumed over the entire supporting width (Figure 1), taking into account the elastic and plastic behaviour of the floor. It may be assumed that the floor load on one side already acts uniformly on the floor beneath it.The simplified calculation method is applicable only to masonry. For lintels, as always, centric loading is assumed and where necessary two-way load distribution is considered (trapezoidal and triangular loads with or without effect of continuity).
6.
In boundary sections which extend less than50
cm in front of the transverse wall, the unreduced DIN stresses can be accepted (no danger of buckling here).7.
Single loads beside openings. The support pressure underneath a lintel, given great rigidity in bending, can be regarded as uniformly distributed. In the sections of wall underneath it, depending on the de-formation of the masonry and in agreement with DIN1053
number 2.6, the assumption of a 60° load distribution is admissible - but one-sidedly, not uniformly distributed (Figure 2). Stresses over the corner do not need to be taken into account. The most unfavourable edge pressure must always be used.8.
Rules 1 to7
are applicable equally to inside walls and inside shells of cavity walls.4.
Admissible stresses a in kp/cm2 for11.5
and17.5
cm thick walls instress analysis, taking into account eccentric floor loading
The stress a -
40-A .
a is derived from the admissible stress a of- 35
1 1DIN
1053,
Table5
for slenderness ratio5.
From this point on the values drop linerarly to zero for a slenderness ratio of40.
This curve is derived from "Reports of the Advisory Council on Building Research in the Federal Ministry for Residential Construction, Number8":
Uber die Tragfahigkeit von Mauerwerk, insbesondere von stockwerkshohen Waden", by Professor Otto Graf,1952,
Franckh'sche Verlagshandlung Stuttgart-O, and from the publication of the Section Chief of the Eidg. Materialprufungs - und Versuchsanstalt-9-Zurich, Dipl. - Ing. P. Haller, "DeI' Ziegelbau in der Schweiz", Die Ziegelindustrie, August 1953, Figure 15 (here Figure 3).
It is worth noting that in Switzerland a 」ッョウゥ、・セ。「ャケ greater number of factors is or can be taken into consideration. For example, they con-sider not only brick strength and mortar quality, but also the masonry quality (kind of bond, absorption, position of point of attack, length of indentation). A safety factor of
4
is required. When only brick and mortar quality are taken into account, the safety factor is increased to5.
Walls and ceilings are even calculated as a framework system, e.g. according to Cross, and the total end moments at the supports are distributed according to rigidities of floors and walls. With such careful design approach it is understandable that much larger fields of application are opened up to walls of < 24 cm. There are tall buildings in Basle, for example, with 15 cmthick walls designed and approved as bearing walls 10 storeys high. For German conditions such detailed methods are ruled out for reasons of time
(shortage of engineers, overtaxing of offices, etc.) - without discrediting the fundamental German experiments, on which Dr., Brocker, Chairman of the FN Construction Committee on "Masonry", reported.
The following is a comparison of wall strength with "admissible" stress in Mz 289/111 according to Haller, Number 8, cf. Figure 29 of the above-cited work.
From Figure
4
it is evident that:1. The wall strength in the lower part of compression is about 1/3 the brick strength.
2. The allowable stresses for the masonry is about 1/5 the wall strength (safety factor n =
5).
3.
The suggested linear curve for a in the region ofA
= 5 toA
= 102 , 3
gives less favourable values than DIN 1053, Table
6,
and morefavourable values in the region above 10; in no case, however, does it fall short of the safety factor.
4. For eccentric loading at the kern boundary the wall strength should have decreased to half the strength for centric loading. In prac-tice, however, it is at least 20% higher. Therefore, if walls are calculated for eccentric loading and based on the suggested stress curve as admissible, the safety factor still lies above
5.
5.
Examples of ApplicationWe shall illustrate the application of the calculation suggestions outlined above with a few examples. A floor to ceiling distance of 3.0 m
-10-and masonry quality Mz ャセoOii is assumed. Examples (1) to
(5)
are for 11.5 cm walls, examples(6)
to (10) for 17.5 cm walls. The numbers in the sketches are intended to indicate whether slabs are supnorted on one, two or three sides. Accordingly, stresses a should be calculated "without risk1
of buckling" and stresses a and a "with risk of buckling". (In brackets
2 3
the numbers of DIN 1053 for comparison.) All walls in Mz 150/11.
1.
2.
3.
11.5 cm outside wall without heating element recess
11.5 cm interior wall with door width less than 1.25 m
11.5 cm outside wall with heating element recess beneath large window A 3 a 3 DIN 1053 Table 5 Number 8.114 ... 12 kp/cm2 *) Window edge
A
=
125/11.5=
11 a = 10 kp/cm2 0.8 '300 _ 11.5 - 20.8 ... 6.6 kp/cm2DIN 1053 Table 6, Line 2 ... 8 kpz cm" *) Door edge
A
200/11.5 = 17.5 a=
3.5 kp/cm2 a as(1) ...
6.6
kp/cm2 3 DIN 1053, Table 6 as "wall" (?) 12 kp/cm2*)
Edge as post A = 240/11.5 Application unclear > 20 inadmissible a as (l) 36.6
kp/cm 2 A 2 a 2 a 3 4.11.5 cm outside wall with heating element recess for two windows
DIN 1053 as at (3) Application unclear 300 - 6 11.5 - 2 .1. ..
=
4.8 kp/cm2 as (1) 6.6 kp/cm25.
6.
7.
8.9.
10.11.5 cm inside wall greater than 4.5 m with no opening
TWijJ
i
1U
t
セZHQセ
17.5 cm outside wall < 6.0 m without heating element recess
17.5 cm outside wall < 6.0 m with heating element recess
セQQキセQ
1--16.0---I17.5 cm outside wall> 6.0 m without heating element recess
Mz 150/11
17.5 cm outside wall> 6.0 m with corner window
jセ
.L1:::::l...LLI
I-->6.0---j
17.5 cm inside wall> 6.0 m with vertical slots
-11-DIN 1053 as wall inadmissible as "pillar" likewise
0..
> 20) 0 as(1)
6.6 kp/cm2 3 0 as ( 4 ) 4.8 kp/cm2 2 DIN 1053 as "wall" 12 kp/cm2 Edge as postA
= 125/17.5 < 10 0.8'300 _A
3 = 17.5 - 13·7 ... o=
9.0 kp/cm2 3 DIN 1053 as "wall" 12 kp/cm2 As postA
= 240/17.5 = 13.7 o=
6.3 kp/cm2 o as(6)
9.0 kp/cm2 3 DIN 1053 A = 17.5 -300 _ 17.1. .. o = 3.5 kp/cm2 A = 17.1. .. 2 0 = 7.8 kp/cm2 2 0 as (6 ) 9.0 kp/cm2 3 DIN 1053 as ( 8) 3.5 kp/cm2 0 as (8) 7.8 kp/cm2 2 0 as (6) 9.0 kp/cm2 3 DIN 1053 as (8) 3.5 kp/cm2 o=
12 kp/cm2 1 (Suggestion 6) o as (8) 7.8 kp/cm2 2
-12-6.
Explanation gf the Examples6.1. Examples (1) and (2):
The walls correspond to DIN 1053 Table 1 with respect to length, heights of door or window: they can be subjected to stresses of 8 and 11 kp/cm2
,
respectively, without reference to the slenderness ratio. When using the
40 - A 6 6 2
3tress
°
= 35 . 01 a lower value is obtained, namely . kp/cm. Thewall can be executed with sufficient safety (n >
5).
If Table6
were used,300
at a slenderness ratio of 11.5 = 26.1, no value could be obtained from the table and the wall could therefore not be executed.
If this wall were loaded by only one floor span, this would contravene DIN 1053, Table 1. According to rules (1) to (5), however, it could be calculated - even though the floor span exceeded the 4.5 m limit and the floor live load exceeded the 275 kp/m2 limit.
For example, let the load from the ceiling be
(0.4 + 0.35) . 5
25
=
2.1 Mp/m from wall load 1.2 Mp/mtaking into account a window 1.25 m wide, then in example (2) we get
( 1.2 2'2.1) 4.5 63
M /
2 66°3
=
0.115 + 0.115 . 3.25=
P
m < •In such outside walls loaded on three sides with a width ratio of less than 0.8 wind load need not be taken into account.
6.2. Example (3):
The application of DIN 1053 for this case of an outside wall is unclear, because the heating element recess occupying the full width of the window cannot be taken into account. If in order to get around this the outside wall is treated as "inside wall", then according to DIN 1053 the opening greater than 1.25 m is critical. The wall would have to be thickened to
24 cm. According to the above rules, however,
°
=6.6
kp/cm2 is admissible. 3Where the top storey is involved, for example, the following loads apply:
Roof load 0.2 . 5.0
=
1.0 Mp/rnWeight of wall and lintel
=
0.3 Mp/rn1.3 Mp/m
Floor load 0.75 . 2.7 2.1 Mp/m
Thus, for a window width of approximately 2.0 m
_ (1.3 2'2.1) 4.5 - 86
-
-13-Thickening is therefore necessary, since Mz 250/111 is unusual and expensive, and therefore not considered. The inside shell is therefore thickened to 17.5 cm, giving o 3 6.3. Example (4): = 86· 11.5 = 57 Mp/m2<90 (cf. example
(7»
17.5This "wall" can also be executed in all probability according to the wording of DIN 1053 as a 2 x 11.5 cm cavity wall, although there are two openings instead of one. Calculation of the 50 cm wide posts with the load-ing of example
(3)
first gives 11.5 cm thickness for the inside shell, if the windows are each 1.4 m wide:- (1.3
+
2'2.1) 1.9 - 181 M / 2 48O 2 - 0.115 0.115' 0.5 - p m > .
First try strengthening to 17.5 cm thickness and mortar group III: o = 181 117 Mp/m2>105 (Table 1).
If thickening of the post to 24 cm is not possible, e.g. for architectural reasons, then a reinforced concrete column would have to be employed.
With walls supported on two sides and for posts, the wind effect must also be taken into account. Once again, example (4) with 11.5 cm thick posts for the purpose of estimating the order of magnitude. Simplified assumption: posts fixed at the bottom, hinged at the top; dynamic pressure up to 8 m heyght, 50 kp/m2
M (50.1.9)'
ゥTセ[
= 60 kpm W= 50'11.52 = 1100 cm3b
from wind pressure
o
=
±
セセセセ
=
±
5.6 kp/cm2 from wind suctiono =
+
5.6'20/50 =+
2.2 kp/cm2 •Perhaps the moment of resistance of the second shell can be activated and twice the moment of resistance applied if the outside shell is forced to take on the same deflection as the inside shell using wire ties (5 ties per m2) . Strictly speaking, in this case the smaller thickness of the
-14-outside shell
(10.5
cm hard-baked or clinker format) and the greater or lesser width depending on inside or outside estimates should be taken into account.Superimposition of the pressures from eccentric floor load a occurs
2
only for the case of wind suction. The order of magnitude of
22
or even only11
Mp/m2 will certainly become noticeable in dimensioning for pressure. The tensile stress on the masonry outside of the inside shell from wind is usually overshadowed by a sUfficiently large central vertical load.6.4.
Example5:
This does not fundamentally make it possible to exceed the
4.5
m lilnit in11.5
cm thick masonry. In residential construction, for example,generally greater loads occur than with the adopted a . 2
Reference to a standard example:
Loading by floor
(0.75
Mp/m2) at4.8
m (living room) and2.3
m(vestl-bule) span and by wall load above it of
2.1
Mp/ma 2
2.1
0.75'4.8 - 49 6 M /
248
0.115
+0.115
-
.
pm> .(The load on the vestibule acts in accordance with Figure 1).
Garages, on the other hand, even at depths of
5 - 6
m are now possible in11.5
cm thickness (according to practice), provided the height remains small, so that wind pressure can be resisted.Example:
Wall height
2.25
m, roof span3.0
m, roof load0.5
Mp/m2,from eccentricity from wind pressure
a =
f
=0.5'1.5
o
F
0.115
6.5
Mp/m2 ±6.5
Mp/m2from wind suction
hence inside: outside:
0.05'2.25
2 / 1 4 . 2 =8.1
Mp/m20.115
2/68.1'20/50
=3.3
Mp/m26.5
+6.5 + 3.3
16.3
Mp/m2 (compression)6.5
6.5
3.3
=3.3
Mp/m2 (tension).On account of the small horizontal joint tension
(0.33
kp/cm2) cement
-15-The assumption that the ceiling must be stabilized and secured against displacement or rotation must be born in mind here.
In steel construction it is a rule that bracing with half-brick walls may not be greater than 16 m2 without analysis (Stahl in Hochbau, 12th edition, page 582), e.g. 2.7 x 6.0 m2 • To be sure these are walls which
only have to carry their own weight and are prevented from horizontal displacement in all directions by the steel of frame (columns and beams). When applying the relative dimensions dealt with hitherto there are no objections to the retention of this rule or its inclusion in DIN 1053.
In the construction of barns and stables, 11.5 cm cavity walls without stiffening every 4.5 m are also being built and thus violate the letter of DIN 1053. With a floor to ceiling height of about 2.5 m, a roof slab is thus supported on the wall in the vicinity of ,the cattle stall. The ceiling span would be 5 m.
250
Applying
A
= 11.5 = 21.8, the stability of the structure is assured according to what has been said above:a
=
ᄚoセセゥUP
=
35 Mp/m2±
wind<62 (Table 1).Again, bear in mind the stabilization of the roof slab. The stability of the structure in such cases is achieved by the transverse or gable wall on which the roof, in spite of one way load distribution, is supported. For construction reasons, of course, independently of this calculation,
intermediate stiffenings will be provided, if not at intervals of 4.5 m, then e.g. every 5.0 to 6.0 m. However, one cannot expect too much from these intermediate stiffeners, unless reinforced concrete columns are provided which are fixed in the foundation.
7.0 Mp/m 2.0 Mp/m2 63.0 Mp/m2 6.0 Mp/m2 69.0 Mp/m2<78
=
0.05.3.02 / 1 4 . 2=
± 0.1762/6 a 2from wind pressure a
6.5. Concerning Examples (8) to (10):
The admissible stress of 3.5 kp/cm2 for Mz 150/11 for 17.5 cm walls according to DIN 1053 does not do justice to the considerably higher resistance compared with the 11.5 cm wall; higher admissible stresses
would appear in order, even when the stiffening limit of 6.0 is not observed. Example for Region 2:
2 floors
+
Rセ walls centrically 1 floor eccentrically= 7.0 2·2.0 0.175
+
0.175-16-Example for Region 1:
b
=
50 cmstandard a
=
120 Mp/m2 (instead of 8 kp/cm2 according to DIN 1053). 1In this way, therefore, the inadmissible vertical slots according to DIN 1053 number 2.5 are also covered. Finally, door and window openings closer than 50 cm from the transverse wall can also be covered and repre-sented.
7. Lintels in Cavity Walls
The lintels in clinker are produced in the coastal region by incor-poration in the lowermost 1, 2 or 3 courses in the cement mortar of 2
セ
61 1 1 in each. No calculations of this are prescribed or demanded, even when the opening is several metres wide or when there is only 1 m of masonry above the lintel, or the window is close to the end of the building. To be sure, there is no ceiling loading (the lintels over the inside shell are obviously always produced in concrete or steel) and certainly number 7.11 of DIN 1053, according to which the isosceles triangle is to be placed over the beam, often applies. However, how much safety is there in such a "beam" when the assumptions for the applicability of Section 7.11 no longer apply, or only partly apply, that is to say, when there are interfering openings above the beam and the load bearing surfaces which interfere with the in-tended arching effect? In my opinion tests should be devised to provide the local building authorities with guides whereby any possible danger may be assessed.8. Stability
Departures from the requirements concerning stiffening of DIN 1053 (11.5 cm walls to be stiffened every 4.5 m, 17.5 cm walls every 6.0 m, and 24 cm walls every 8.0 m) mean that greater attention has to be paid to the stability of a structure as a whole. The provision of frames or fixed supports is not always helpful, for example, when the resulting deformations are incompatible with the deformations of the rest of the system (hori-zontal displacement and rotation cf the ceiling slabs). Garages, for example, where the entire front facade is open, conform even without front
,
corner columns to the known principle: resistance to wind force transverse to the wide side by the rear wall panel, resistance to the rotations from eccentricity of the wind attack centre as far as the rear wall by the moment of the longitudinal wall (for more accurate determination of the centre of inertia in other cross-sectional forms see, for example, Hahn,-17-of course, for all brick residential and commercial buildings. The build-ing inspectors of various cities, e.g. Hamburg, in general have insisted that every ッー・ョMキ。ャセ end (except for doors and windows) must be stiffened by a special supporting member. The above considerations show, however, that stability can be obtained in other ways.
On the other hand, the principles described cannot be applied in-descriminently. This is not solely a question of deformation and safety; this can be clarified by tests as far as necessary. Rather, the possibility must be kept in mind that ultimately someone unaware of the static require-ments will remove the stabilizing transverse walls and back walls and thus endanger the building. In every building, however, there are enough
transverse walls which may safely be expected to. remain, e.g. stairwell walls. In such cases the stability evidence should 「セ accepted, and be admissible even without concrete or steel columns at the opened end.
9.
SummaryWith rules (1) to
(8)
in conjunction with the explanations to the calculated examples, it becomes possible to calculate all wall parts very easily. In this way the statically unharmful infractions against DIN 1053, which is felt in this region to be inadequate, can be avoided and at the same time the statically harmful infractions (for example, of window posts) can and shall be effectively treated. In the residential building heat and sound insulation must be considered along with the structural problems. In agricultural buildings and garages, on the other hand, where sound and heat insulation is less important, savings can almost always be made by carrying out the stress analysis.It is again emphasized that the stress determined according to the cited dimensioning rules does not represent the actual stress distribution in the walls, but that these do give sufficient assurance against wrong dimensioning on the one hand and unnecessary strict requirements on the other. For better establishment of the various still unproven calculation assumptions, tests would be desirable with completely fixed or elastically fixed wall sections, employing the formats and brick qualities Mz 150 and Mz 250 used in the northwest German coastal region alone. Wherever possible, the more recent and latest experimental results of the Swiss Material
Inspection and Testing Institute, Zurich, with respect to stiffening of walls and bearing capacity of walls should be made use of.
.,
M7. 50/11 100111 Iso/II 25011I 3:.0/11 Ii [Ill [3]J iiセLiii セ セ 10 6,0 7,7 10,0 1.1,7 18,9 12 5.r. 7,2 9,6 12,8 17,6 14 5,2 6,7 8,9 11,9 16,3 16 4,8 (,,2 B,2 11,0 15,11 18 4,4 G,6 7,5 10,0 !J,B 20 4,0 5,1 6,9 9,1 12,6 22 3,6 4,6 6.1 8,2 11,3 24 3,2 4,1 5,6 7,3 10,0 26 2,8 3,6 4,R (.,4 B,R 30 2,0 2,2 3,4 4,(, 6,3 ----...O,/b---0, I,---,-,-,--+---
It---1
Fig. 1 Fig. 2-19-Fig. 3
Fig.