Example 5: Beam with indirect support and loading 145

Example 5 · Beam with indirect support and

loading

Wiryanto Dewobroto

Karl-Heinz Reineck

Synopsis
Inadequate design of indirect supports resulted in a lot of structural damage and
near failure of structural concrete beams. Most codes, including ACI 318, do not
properly cover this case. However, strut-and-tie models almost automatically lead
to correctly reinforcing these critical discontinuity regions. This example com-
bines indirect supports as well as indirectly applied loads and demonstrates the
application of strut-and-tie models following Appendix A of ACI 318-2002.

Wiryanto Dewobroto is a lecturer in civil engineering at the University of Pelita

Harapan, Indonesia. He obtained BS in Civil Engineering from University of
Gadjahmada in 1989. After that he worked in consulting engineer firm and has
experience in design and supervision of many structures especially for high-rises,
industrial buildings and bridges. In 1998 he obtained his MS in Structural Engi-
neering from the University of Indonesia. From May to July 2002 he was a guest
researcher at the University of Stuttgart, Germany.

Karl-Heinz Reineck received his Dipl.-Ing. and Dr.-Ing. degrees from the Uni-
versity of Stuttgart. He is involved in both research and teaching at the Institute
for Lightweight Structures Conceptual and Structural Design, University of Stutt-
gart. His research covers design with strut-and-tie models, shear design and
detailing of structural concrete. He is chairman of ASCE-ACI Committee 445-1
and member of the .fib Task Group "Practical Design."
146 Example 5 : Beam with indirect support and loading

1 Geometry and loads

The T-beam shown in Fig. 5-1 is indirectly supported at support B by means of a

transfer beam shown in section II-II. Likewise the loads are not directly applied to
the web but are transferred by the beams shown in section 1-1. The loads are sym-
metrically applied so that no torsion is induced.

Elevation
300 2000 _ _lQ0"-0c----ci
(11.81 in) I (78.74 in) (78.74 in) c
,-I r-11
gl· eCi
...-
-----=rc:
or~ 81~
!:f,~ ~ coj ~
A -1 ____2QQ -1_1 11 , B-· -~-
2200 "-------'--=""--
(55.121n)
~_ __._.18=0=0
(70.87 in)
-t'-: 200
(86.61 in)

Section 1-1 Section 11-11

~!)50 ·n70055Q_t __...2QQ _ ....1.Q.Q._,
0 :? I I (7.87 1n) (27.56 in)
I
~~; 1 {u
I Fu 1 · _ I

~r; I
-
5 E
775 ' I 250
I gl i TFu
__,__j_.____F
.......,.._! ~'J
(30.51 in) (9.84 in) : Fu
200
~- 2200 200 -~1~8~0":'0-~----+-=~
(86.61 in) (7~87ini-· (70.87 in) 7.87 in)

Fig. 5-I: Member with indirect support and factored loads (I mm = 0.03937 in)

Design specification :
Factored load : F,= 160 kN (36 kips)
Concrete: f,= 31.6 MPa (4,580 psi)
Reinforcement : fy = 500 MPa (72,500 psi)
Example 5: Beam with indirect support and loading 147

2 Design procedure

The design is based on Strut-and-Tie-Model according to Appendix A of the ACI

318-02 and is carried out in the following steps:
Step 1: Analysis
Step 2: Flexural design of main beam and calculation of internal lever arm
Step 3: Stirrup design for main beam and calculation of strut angle in web
Step 4: Check of anchorage length at node A and B
Step 5: Design of beam transferring the loads to the main beam
Step 6: Design of beam supporting the main beam
Step 7: Arrangement of reinforcement
Editorial note: The calculations are carried out in SI units; primary results are
given in English units in brackets.

3 Design calculations

3.1 Step 1: Analysis

The shear force and moment diagram of the main beam are shown in Fig. 5-2.

2000 1600 2000

-c~-~~~---1
1' (78.74 in) T (63 in) (78.74 in)

320 B
A (72 kips) 320
(72 kips)
I Vu (kN)

~-:~~6-=-40'-.,---,...,.--.-·~·
(5669.3 kips-in)
8
Mu(kNm)

Fig. 5-2: Shear forces and flexural moment in the main beam
(I kN = 0.2248 kips; 1 kN-m =8.8496 kips-in)

Beams transferring the load to the main beam (Sec. 1-1 in Fig. 5-1):
Mu = 144 kN-m (1275 kips-in)
Yu = 160 kN (36 kips)

For the beam supporting the main beam at support B (Sec. 11-11 in Fig. 5-1) the
same values apply.
148 Example 5 : Beam with indirect support and loading

3.2 Step 2: Flexural design of main beam

Use equivalent rectangular concrete stress distribution based on Sec.10.2.7:
fc = 31.6 MPa 7 (3, = 0.85 - 0.005 (31.6-27 .6) I 6.89 = 0.82

= 0.75 = 0.638 r; (31 600 = 0.018

Pmax Pb fy 600+fy

M,. ;::>: ~" = ~~~ = 711.1 kN- m (6290 kips- in)

Make the assumptions that the compression zone is within the flange (Fig 5-3) and
steel yields.

-u··
.L
. -
700

i
d=525
-g·
~I
L..l__ _

Strain
- o.asrc

z
._..F_s_.__
Stress

Fig. 5-3: Distribution of strains and stress at midspan of main beam

(1 mm = 0.03937 in)

M 711.1 *10 6
-"-= 0.117 7 m=O.l26 < mmax=0.285
bd 2 f;
700 * 525 2 * 31.6

A =mr; bd=0.126* 3 1. 6 *700*525=2926mm 2 (4.54in 2)

' fy 500

use 6 0 25 A,= 2945 mm 2 (4.56 in 2 )

Check the assumption:

A, ·fy 2926* 500
a=-----'--- ------=79mm < hr =150mm OK
0.85. (. b 0.85 * 31.6 * 700

A, ·fy . .
a= =3.1lm < hr =5.91m
0.85·(. b

The arrangement of the main bars is shown in Fig. 5-4. (1 mm = 0.03937 in)
Example 5: Beam with indirect support and loading 149

--1Q_
~~
; g 6 025
(\jr
lO
It!e _- •
t;T I L::•::::::-~·:;:-~~·~·~
-o~ 250 .CJ
C");
Fig. 5-4 : Proposed arrangement of bars for main beam (1 mm =0.03937 in)

The updated values for the effective depth and the internal lever arm are as fol-
lows:
The updated lever arms are as follows:
d = h- d, =600-72.5 =527.5 mm (20.78 in)> 525 mm (20.67 in)
z =jd =d- 0.5a =527.5- (0.5 * 79) =488 mm (19.21 in)
j = z I d = 0.925

3.3 Step 4: Stirrup design for main beam

The dimensioning of the stirrups follows from chapter II of ACI 318. Subse-
quently the angle 8 is derived for the inclined struts in the web so that the strut-
and-tie model for the D-regions can be determined.
d.\! ;:>:
'I" n
v u

V <': V" = 320 = 427 kN (96 kips)

n r/J 0.75

V. = V, + V,
Eq. 11.3.1.1 for members subjected to shear and flexure only gives :

V, =0.17-.J( bw d

V, = 0.17*-J31:6 *250*527.5 = 126024 N = 126kN (28.3 kips)

Alternative Eq. 11.3.2.1 gives:

V, =( O.J6.j( +17 Pw ~ud}w d 0.29~(bwd

:<::;

where
V, d = 320 * 0.5275 = 0 _2637 :.::; I.O
M, 640
A, 2945
Pw = bw d = 250*527.5 0.0 223
so that
V, =(O.I6-J31:6+17*0.0233*0.2637p50*527.5 :<::; 0.29-J31:6*250*527.5
V, = 132 kN (29.67 kips) :<::; 215 kN (48.32 kips)
150 Example 5 : Beam with indirect support and loading

Use V, from the alternative equation for designing the stirrups:

V=V-V
' " ' (66.3 kips)
V, = 427 - 132 = 295 kN
< 0.68~(bwd = 504 kN (113.3 kips)

> 0.34-Jf:bwd = 252 kN (56.6 kips)~ s ~ d/4 =131 mm (5.16 in)

~=l= 295 •000 =1.12mm' =JJ20mm' (0.0441'"')

S fy·d 500*527.5 mm m m

use stirrups 0 10 at 125 mm spacing ~ (~) = 1250 m:;:' (0.0492 •::)

S prov

The angle 8 of the inclined struts in the web of the truss model can be derived now
because the amount of stirrups is known. The free-body diagram shown in Fig. 5-
5 shows that the shear force in the B-region has to be taken by the stirrup forces
over the length (z cot8):
V, = (Aj s) f, z cote
and from this the angle 8 can be calculated as follows:

cot8=~·~=-1-* 427•000 =1.5625 ~ e =32.6"

A. fyz 1.12 500*488

z =488mm

mrnnrrnrrq d,=...z2§
(A,Is)fy z cote

Fig. 5-5: Compression stress fields for the inclined struts (I mm =0.03937 in)
Example 5: Beam with indirect support and loading 151

3.4 Step 4: Check the anchorage length at nodes A and B

3.4.1 Check the development length of longitudinal rebars
According to Sec. 12.11 ACI 318-02 the development of rebars should satisfy the
following requirement (see also Fig. 5-6):
I. At least one-third the longitudinal reinforcement shall extend along the same
face of member into the support at least 150 mm.
2. At simple supports the diameter of the reinforcement should be small enough
so that computed development length 1. of the bar satisfies the following condi-
tion:

where
M. is nominal moment strength assuming all reinforcement at the section
(at support) to be stressed to the specified yield strength f,.
Therefore from Step. 2, it can be calculated as
M. =A,f,z
M. = 2945 * 500 * 488 =719.106 N-mm = 719 kN-m.
v. is factored shear force at the section (V. = 320 kN).
l, at the support shall be the embedment length beyond the center of the
support.

(a). Direct support (Node A) (b). Indirect support (Node B)

Fig. 5-6: Development length of the positive bar (1 mm = 0.03937 in)

Here for simplicity all bars are extended to the support.

The beam section at the support is similar to Fig. 5-4, the diameter of rebar is
25 mm, therefore according to Sec. 12.2.2 the development length can be calcu-
lated as follows:

ld = [ fya ~]db (in)

20vfc
152 Example 5 : Beam with indirect support and loading

This equation applies to normal weight concrete (A.= 1.0), uncoated reinforcement
((3=1.0) and bar diameters larger than bar No.7, or 22 mm (a= 1.0), for
( = 4,583 psi and f,= 72,500 psi:

(I) =( 725 00*I.O*I.O*I.O)d =54d =1350mm (53.15in)

d req 20.J 4583 b b

Check node A:
(I) =1.3*M./ +I =1.3*715,000/ +280=3185mm (98.98 in)
d pro' /Vu /320 '
0

(IJpro' = 3,185mm (98.98 in)>(ld),.q 7 OK

Check node B:

([) =M./ +I =715,000/ +95=2234mm (87.95in)>(ld)-q70K

d prm jV,, /320 0 ' ••

The requirements of Sec. 12.11 regarding the development of the longitudinal

reinforcement have been satisfied, however this does not represent a check of the
anchorage length directly at the support.

3.4.2 Check the anchorage length at direct support A

Use Standard hooks to terminate the rebars at the support. The check of the an-
chorage length at the direct support is carried out according to Sec. 12.5.2 for a
standard hook according to A.4.3.2. According to ACI A.4.3.2 the development
length starts at the point where the centroid of the reinforcement in the tie leaves
the extended nodal zone. For simplicity here the inner face of the support is taken,
and as shown in Fig. 5-7 the available development length (ldh) computed in this
way is 380 mm.

i .i :
f t ,.,z · E-::r"Y" fE•f T t .To

I I I
I
i I
'

~l
f I
II
'
I ''
12d.
II I
I
!
g. ,'
L ~~~ I I I ,
I: 3025
! :! 3025

280
ldh = 380
t 100

Fig. 5-7: Development length at direct support (Node A) (I mm = 0.03937 in)

Example 5: Beam with indirect support and loading 153

According to section 12.5.2 the development length is:

l.., = (0.0213 A f, I ..Jt) d,
where
13 and A are 1.0 for normal weight concrete and uncoated rebar
f/ = 4,580 psi
f, = 72,500 psi
therefore
l"" = (0.02 * 1.0 * 1.0* 72,500 I '1/4,580) d, = 21 d,
According to Sec.l2.5.3 where anchorage or development for f, is not specifically
required, reinforcement in excess of that required by analysis can be multiplied by
l.., = (A,.req I A •.,mJ 21 d,
In order to calculate A •req' the tension force (F,.) shown in Fig. 5-8 is required.
Based on the FIP Recommendations (1999), Sec.6.5.2.1, the angle e A for the
resultant of the fan-shaped compression field follows the geometry of the
fan (Fig. 5-8):
cote A = [0.5a, I z + (d, lz + 0.5) cote]
cote A = [ 125 I 488 + (72.5 I 488 + 0.5) 1.5625 ] = 1.2695 -7 e A= 38.2"

F,. = V. cote A= 427 * 1.2695 = 542 kN (122 kips)

Au"'= F,,./ f, = 542,000 I 500 = 1084 mm 2 (1.68 in 2 )
A,,,,.,= 6 0 25 (2945 mm 2) (4.56 in2 )

Therefore: /""""=(A,,._.., I A,.,,.J 21 db

1..,,.,. = ( 1084 12945) 21 d, = 7.73 d, = 193 mm < /d~pm• = 280 mm
1..,·"" = 7.60 in < l..,,,ro, = 11.02 in

The anchorage length at direct support (Node A) is adequate.

z =488mm
I
-j
~72~

0.5a,. =125 Vn
'
Fig. 5-8: Tension force (F"')to be anchored at the support (I mm = 0,03937 in)
154 Example 5 : Beam with indirect support and loading

3.4.3 Hanger reinforcement at indirect support

At an indirect support the strut-and-tie model in the web is the same as that for
direct supports [Reineck ( 1996)]. However, at node B the reaction of the main
beam must be transferred to the supporting beam by hanger reinforcement
(Fig. 5-9). The area of hanger reinforcement required can be calculated as follows:
A,= V,/ fy= 427,000 I 500 = 854 mm' (1.32 in')
Use 4 stirrups 0 12 A,= 905 mm'

3.4.4 Check the development length at indirect support B

In order to place the hanger reinforcement below the standard hooks at the indirect
support, the main beam should be extended I 00 mm beyond the edge of transfer
beam (see Fig. 5-9). Otherwise the stirrup anchorage of the main reinforcement is
not sufficient.
According to Sec. A.4.3.2 the development length can be calculated from the
point where the centroid of the reinforcement in a tie leaves the extended nodal
zone. As shown in Fig. 5-9 the development length (/,) computed in this way is
318 mm, which is larger than use /"' requirement from the previous calculation,
therefore the development length at the indirect support is adequate.
Note: It is evident that in ACI 318 there is no difference between the
requirements for the anchorage length for direct or indirect supports.
However, the stress condition at an indirect support is clearly more
unfavorable, because the transverse tensile stresses reduce the bond
strength and thus the required anchorage length should be larger than for
the case at a direct support.
Example 5: Beam with indirect support and loading !55

3.5 Step 5 : Beam transferring the load to the main beam

3.5.1 Strut-and-tie model
The entire deep beam is aD-region. To satisfy equilibrium at point C in Fig. 5-l 0
for the tie forces in the upward direction, hanger reinforcement is required. These
stirrups must be located within the intersection of the web of the main beam and
the beam transferring the load.
The strut-and-tie model may be assumed as shown in Fig. 5-10:
1800 200

A '• I 0 0 •' h
··· ..e =29.89" 1 1 ••• •••
···... ··~·ll ....··
• {.. Fu= 160 kN
z=485 h=600

'--------"-=--'-"---'-"-'-c.c.__-' ---·-~

Fig. 5-l 0: Strut-and-tie configuration of beam transferring the load

(I mm = 0.03937 in; kN = 0.2248 kips)

The angle of the strut is larger than 25°, therefore acceptable. However, a certain
minimum amount of transverse reinforcement is required according to Appendix
A.3.3 of ACI 318-2002 as calculated in the following. Because of that it may be
stated that this model as such is not fully transparent (see also Schlaich et al
(1987)), because it does not show the necessity to provide any transverse rein-
forcement. Therefore it may be considered to use the strut-and-tie model for point
loads near a support used in the FIP Recommendations (1999) "Practical design of
structural concrete", and this was also proposed by and MacGregor in part 2 of
this Special Publication.

3.5.2 Design hanger reinforcement

The up-ward forces in C-D must be resisted by hanger reinforcement as calculated
already in Section 3.4.3.

3.5.3 Design ties

Calculate the horizontal tie force (T.) at node A through node B as follows
T" = F" I tan8 = 160 I tan29.9° = 278.2 kN (62.5 kips)
T,;::.: T" I¢ = 278.2 I 0.75 = 371 kN (83.4 kips)
A,= TJ f, = 371,000 I 500 = 742 mm' (1.15 in 2)
Use 3 0 19, A,= 850 mm 2 (1.31 in2 )
156 Example 5 : Beam with indirect support and loading

3.5.4 Strength of nodal zone

The nominal compression strength of a nodal zone shall be
Fnn =fcu An
where
f'" = 0.85 13. (
13,= 0.8
f'" = 0.85 * 0.8 * 31.6 = 21.5 MPa (3.12 ksi)
A,= area of the nodal zone taken perpendicular to the resultant force, mm'

Fig. 5-11: Nodal zone at the load point (I mm = 0.03937 in)

Check nodal zone under the bearing force (F.) as shown in Fig 5-11:
A, = b t. = 200 * 200 = 40,000 mm 2 (62.0 in2 )
so that
F,, =f.,. A,= 21.5 * 40,000 = 860,000 N = 860 kN (193 kips)
$F.. = 0.75 * 860 kN = 645 kN (145 kips) > F.= 160 kN ~OK
Check nodal zone under the strut action (C. )
c.= F.f sine= 160 I sin29.9° = 321 kN (72.2 kips)
A, = b w, = b (w, cos8 + /b sin8)
= 200 (120 cos29.9° + 200 sin29.9°) = 40,745 mm 2 (63.15 in2 )
so that
F.. = f.,. A,= 21.5 * 40,745 = 876,000 N = 876 kN (197kips)

$F.. = 0.75 *876 = 657 kN (148 kips) > c.= 321 kN ~OK

For the anchorage of the reinforcement hooks are provided without further check.
Example 5: Beam with indirect support and loading 157

3.5.5 Strength of inclined struts

The nominal compressive strength of a strut shall be taken as

where
A, = cross-sectional area at one end of the strut
A,= b w, = b (w, cose + lb sinS)
f"' = 0.85 (3, (
(3,= 0.75 (bottle-shaped with reinforcing satisfying ACI 318-02 Sec. A.3.3)
therefore
(, = 0.85 * 0.75 * 31.6 = 20.1 MPa (2.92 ksi)
A,= 200 (120 cos29.9 + 200 sin29.9) = 40,700 mm' (63.09 in2)
F.,= 20.1 * 40,700 = 819,000 N = 819 kN (184 kips)
<j> F.,= 0.75 * 819 = 614 kN (138 kips) > C, ~OK

Design the rebar crossing the diagonal strut to satisfy Appendix A.3.3 (Fig. 5-12):

r-,

The reinforcement will be placed in one direction (vertical only) at an angle y to

the axis of diagonal strut so that y shall not be less than 40".
r = 90- e = 9o- 29.9 = 60.1"
158 Example 5 : Beam with indirect support and loading

For ( ~ 41.4 MPa (6000 psi) the amount of reinforcement is calculated as

follows:

( A,; J 0.003--..!?_
~ prov ~ siny
so that

(A,;) =0.003*~=0.692mm' =692mm' (0.0272;~)

S; m;n sin60.J mm m

Try stirrup 0 10 at 200 mm (8 in) spacing (A,;)si prov

= 785 m:,' (0.031 i~n')

3.6 Step 6: Beam supporting the main beam

The design is similar to that of the previous beam. Note that the model (Fig. 5-13)
is inverted relative to the beam transferring the load.

--

_,~--cj f" . . .
- t3 '::29 ago I I ......
""* \ D 0 ........

A B

·~ 1800
·~ 200
Fu Fu
Fig. 5-13: Strut-and-tie model for the beam supporting the main beam
(1 mm =0.03937 in; 1 kN = 0.2248 kips)

4 Reinforcement layout

The reinforcement layout is shown in Fig. 5-14 to Fig. 5-16

 'TI
<!Q' ~
Y' ~~ ~ t2gol 3
~~
~~ = ~
~
r:-:-:-
--
'"0
~
3 ;n -
~
;.....,;.......
-
Vl

3 ~
.'
! f I
1:0
II < - ,__j. __
·- (j)

0
.
I'>
c.
0
--
---- ...-+1
r--+-t
'. bottom rebar
----
--·
3
8:::. 1-- ,...,....
' ~ I. ~
\0~'>
...,:::. I stirrup 010-125 §:
-..jO. I -- :;·
lci<::]~
-· "'
::l (j)
_o I I : I 1 1 T-' I J T j ~ I I . -,j...:_.j ·-- ·-1 ·I I I~ l·l_l_ I i I ·1 I l l : !· nt 0.
'-"~ ~-
i5' ... ~
::l I ... ''.
rn
0
..... ' e rebar not shown '-1' "'
:g"'

~
·-~

0"
(j)
I'> : i i L stirrup 010-250
~j-
,..,... --
0
:4
3
~
.
•
l
!
(
!
.
it'
rn
'
----
-
I'>
::l

s: 700
stirrup 08-125 :-r-:
~ ..::....:::. 0
0.

:;·
0. ,_ r;~ ~/an View of Beam with Indirect Suooort 100]
I'>
9:
::;· r: I"'c ::l
(JQ
(j)

"'"'
'"0
8 1"~
1
'
stirrup 010-250/125 (in support zone)
'"0
0
:4
l
L25Qj
hanger reinforcement 40·
hanger reinforcement hanger reinforcement
1. stirrup 010-125 c
-- l
. fr-
_ 1 2022 24_"15;
-~
-··n - 1I r T
iftf1
1 .
I
~·HIt
r~- I·
~W+- r1'-
s~+ardhoc I i ;-
-·i-
_
. I I II' I
I s I I 1
J'
'"" 1QQ J
c Vl
\0
3025 _j

Elevation View of Beam with Indirect Support

160 Example 5 : Beam with indirect support and loading

hanger
reinforc~ment4012

Section B Section 0

Fig. 5-15: Elevation and section of beam transferring load (Node A)

(1 mm = 0.03937 in)

r::;::;:::=:::::::;::::;:::;:::;~::::::;::l;::::;:::::~ l
~ 0
'<!JQ_,
2016

stirrup 08-125

3019

Section E

Fig. 5-16: Elevation and section of beam supporting the main beam (Node B)
(1 mm = 0.03937 in)

5 Summary

Indirect supports are presently dealt with by rules for the shear design although
they represent a critical D-region. By using strut-and-tie models it becomes obvi-
ous that hanger stirrups have to be provided for the full support force. The shear
design of the adjacent web is the same like for a direct support.
A critical issue is also the anchorage of the longitudinal reinforcement at an indi-
rect support because at this TIC-node the transverse tensile stresses reduce the
bond capacity. Therefore, longer anchorage lengths are required than at a direct
support, a TCC-node, where transverse compression favorably influences the
bond stresses and thus the required anchorage length.
Example 5: Beam with indirect support and loading 161

Notation
(' specified compressive strength of concrete
f, specified yield strength of nonprestressed reinforcement
d distance from extreme compression fiber to centroid of tension
rebar (effective depth)
d. nominal diameter of bar
F. == factored force acting in a strut, tie, bearing area, or nodal zone in
a strut-and-tie model kN
c. factored compression strut in a strut-and-tie model kN
T. factored tension tie in a strut-and-tie model kN
A, area of the face of the nodal zone that F. acts on, taken perpen-
dicular to the line of action of F •• or the resultant force on the
section, mm'
A, effective cross-sectional area at one end of a strut in a strut-and-
tie model, taken perpendicular to the axis of the strut, mm'
s, == spacing of reinforcement in the i"' layer adjacent to the surface of
the member, mm
w, effective width of strut, mm
w, == effective width of tie, mm
[3, factor defined in Sec.! 0.2. 7.3 of ACI 318-02
13. factor to account for the effect of cracking and confinement rein-
forcement on the effective compressive strength of a nodal zone
y angle between the axis of a strut and the bars crossing that strut
8 angle between the axis of a strut or compressive field and the
tension chord of the member
$ strength reduction factor
l. width of bearing, mm
z inner lever arm

References
American Concrete Institute (2002) : Building Code Requirements for Structural
Concrete (ACI 318-02) and Commentary (ACI 318R-02), Appendix A.
FIP Recommendations (1999): Practical Design of Structural Concrete.
PIP-Commission 3 "Practical Design", Sept. 1996.
Pub!.: SETO, London, Sept. 1999. (Distributed by: fib, Lausanne)
Reineck, K.-H. (1996): Rational Models for Detailing and Design, p. 101-134,
in: Large Concrete Buildings: Rangan, B.V. and Warner, R. F. (Ed.).,
Longman Group Ltd., Burnt Mill, Harlow, England, 1996
Schlaich, J.; Schafer, K; Jennewein, M. (1987): Toward a Consistent Design for
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