1

Definition of Discontinuity Regions

 Structural members may be divided into portions
called B-regions (B=Beams, or BERNOULLI), in
which beam theory applies and other portions
called D-regions (D = Discontinuity), where beam
theory does not apply and whose load flow
should be examined differently.
 D-regions can be geometric discontinuities,
adjacent to holes, abrupt changes in cross
section, etc. or statical discontinuities, which
are regions near concentrated loads and
reactions

2
 Corbels, dapped ends, and joints (beam-column)
are affected by both statical and geometric
discontinuities.
 For many years, D-region design has been by
“good practice,” by rule of thumb, and
empirical.
 Three landmark papers by Prof. Schlaich (1982,
1987, 1991) changed this (according to
MacGregor)
 Will present rules and guidance for the design of
D-regions based largely on these and other
recent papers.

3

Saint
Venant’s Principle and Extent of D-
regions
 Saint Venant’s Principle suggests that the
localized effect of a disturbance dies out about
one member-depth from the point of the
disturbance. It may be the case, however, that
the whole structure represents a D-region such as
in deep beams or short walls.
 Figures (next slide) show D-regions in a number
of structures

4
 D-regions
B-region

B-region
D-regions

B-region

Fig. B-regions and

D-regions
D-regions B-region

5
 tension
compression

Fig. Principal stress-trajectories of a beam under concentrated load

6

The load path in the simple beam loaded
with a concentrated load can be illustrate by
means of the stress-trajectories (see fig in
previous slide)

The stress-trajectories at D-regions show a
“disturbed” flow with sharp curvatures

Stress-trajectories in corbels behave in a
similar fashion. So also at the point of
application of load such as through
prestressing at the end of a prestressed
beam (next slide)

7
 cracks

Stress free region

Fig. Principal stress-trajectories of a corbel with dapped beam

8

Strut-and-Tie Models
 Prior to any cracking, an elastic stress field
exists, which can be quantified with an elastic
analysis, such as a finite-element analysis.
 Cracking disrupts this stress field, causing a
major reorientation of the internal forces.
 After cracking, the internal forces can be
modeled via a strut-and-tie model consisting of
concrete compression struts, steel tension ties,
and joints referred to as nodal zones.
 If the compression struts are narrower at their
ends than they are at midsection, the struts may,
in turn, crack longitudinally.→ may lead to
failure if unreinforced
9
 On the other hand, struts with transverse
reinforcement to restrain the cracking can carry
further load and will fail by crushing (see figure
next slide)
 Failure may also occur by yielding of the tension
ties, failure of the bar anchorage, or failure of
the nodal zones. As always, failure initiated by
yield of the steel tension ties tends to be more
ductile and is desirable

10
 Compression
struts failing by
Compression fan crushing
in D-region

11

A strut-and-tie model for a deep beam is shown
in figure (next slide). It consists of two concrete
compressive struts, longitudinal reinforcement
serving as a tension tie, and joints referred to as
nodes.

The concrete around a node is called a nodal
zone.

The nodal zones transfer the forces from the
inclined struts to other struts, to ties and to the
reactions.

ACI Section 11.1.1 allows D-regions to be
designed using strut-and-tie model according to
the requirements in ACI Appendix A (2002).

12
13

A strut-and-tie model is a model of a portion of
the structure that satisfies the following:
 (a) it embodies a system of forces that is in
equilibrium with a given set of loads, and
 (b) the factored-member forces at every section in
the struts, ties, and nodal zones do not exceed the
corresponding factored-member strengths for the
same sections

The lower-bound theorem of plasticity states
that the capacity of a system of members,
supports, and applied forces that satisfies both
(a) and (b) is a lower bound on the strength of
the structure. (see alternative next slide)
14

Iffor a given external loading we can find an
internal force distribution (in the struts, ties,
and nodal zones) that satisfies the condition
of equilibrium and boundary condition at all
points, and the factored-member forces at
every section in the struts, ties, and nodal
zones do not exceed the corresponding
factored-member strengths for the same
sections, then the applied load represents a
lower bound. (recall parallel for internal
force distribution was moment field in slabs)

15

Forthe lower-bound theorem to apply, the
structure must have:
 (c) sufficient ductility to make the transition
from elastic behavior to enough plastic behavior
to redistribute the factored internal forces into a
set of forces that satisfy items (a) and (b)
 The combination of factored loads acting on the
structure and the distribution of factored
internal forces is a lower bound on the strength
of the structure, provided that no element is
loaded beyond its capacity.

16

In most applications of strut-and-tie models
the internal forces, Fu, due to the factored
loads, and the struts, ties, and nodal zones
are proportioned using
 φFn ≥ Fu (Design value of the nominal strength of
the struts, ties, and nodal zones are greater than
the design value of the internal action effects)

17

In a strut-and-tie model, the struts represent
concrete compressive stress fields with the
compression stresses acting parallel to the strut.
Although they are frequently idealized as
prismatic or uniformly tapering members (see fig
next slide), struts generally vary in cross section
along their length (see figure).

The spreading of the compression forces gives
rise to transverse tensions in the strut which may
cause it to crack longitudinally. A strut w/o
transverse reinforcement may fail after this
cracking occurs. If adequate transverse
reinforcement is provided, the strength of the
strut will be governed by crushing
18
 (b)Bottle shaped strut

(a) Idealized prismatic strut

(c) Strut-and-tie model of a bottle-shaped

strut
19

1.3.1 Strut Failure by Longitudinal Cracking
 Figure (next slide) shows one end of a bottle-
shaped strut. The width of the bearing area is a,
and the thickness of the strut (into the page) is t.
 At midlength the strut has an effective width bef.
According to Prof. Schlaich, bef=l/3 but not less
than a, and l is the length of the strut from face
to face of the nodes.
 For short struts, the limit that bef not less than a
often governs.

20
 C/2 C/2
bef bef/4
a
C/2 C/2
a/4

1.5bef bef/2

(a) Bottle shaped (b) Strut-and-tie

region model

(c)Transverse tension and compressions

21
  According to the ACI, → bef = a + l/6 but not
more than the available width

Figs 17-4c and 17-5b show strut-and-tie
models for the bottle-shaped region. It is
based on the assumption that the
longitudinal projection of the inclined struts
is equal to bef/2.

The transverse tension force T at one end of
the strut is: [from tan θ = V/H = T/(C/2) =
((bef/4) – (a/4))/(bef/2)→ see FBD of the
upper node]

22
 C/2
X

θ T

θ bef/4
a/4
bef/2

T = (C/2)((bef/4)-(a/4))/(bef/2) or simplified

T = (C/4)(1-(a/bef)) (Eq. 17.3)

The force T causes transverse stresses in the
concrete, which may cause cracking.

The transverse tensile stresses are distributed as
shown by the curved line in Fig 17.5c.
23

Analysesby Adebar and Zhou suggest that
the tensile stress distributions at the two
ends of a strut are completely separated
when l/a exceeds about 3.5 and overlap
completely when l/a is b/n 1.5 and 2.

The minimum load, Cn, that cause cracking in
a bottle shaped strut with a/bef is derived as:
Cn = 0.55atfc’ , where a×t is the loaded area
at the end of the strut (see derivation in
Macgregor) (fc’= fck)

24

The maximum load on an unreinforced strut
in a wall-like member such as the deep beam
in Fig 17-3, if governed by cracking of the
concrete in the strut, is given by Eq. 17-3.

This assumes that the compression force
spreads in only one direction. If the bearing
area does not extend over the full thickness
of the member, there will also be transverse
tensile stresses through the thickness of the
strut that will require reinforcement through
the thickness as shown in figure (next slide)

25
26

1.3.2 compression failure of Struts

The crushing strength of the concrete in a
strut is referred to as the effective strength,
fcu = νfc’, where ν = the efficiency factor b/n
0 and 1. ACI replaces fcu with fce = the
effective compressive strength. → fce = νfc’
The major factors affecting the effective
compressive are:
 (1) The concrete strength→ concrete becomes
more brittle and ν tends to decrease as the
concrete strength increases

27
 (2) Load duration effects → The strength of
concrete beams and columns tends to be less
than the cylinder strength, fc’. Various reasons
are given for this including the observed
reduction in compressive strength under
sustained load. For struts, load duration effects
are accounted for by rewriting Eqn above as: fce
= 0.85βsfc’. For nodal zones βn is used
 (3) Tensile strains transverse to the strut
resulting from tensile forces in the reinforcement
crossing the cracks

28
 (4) cracked struts: Struts crossed by cracks
inclined to the axis of the strut are weakened by
the cracks. ACI presents the nominal compressive
strength of a strut as: Fns = fceAc (n=nominal,
s=strut, Ac=cross sectional area at the end of the
strut, and fce= 0.85βsfc’). For nodal zones:→ Fnn =
fceAn , where fce= 0.85βnfc’). Values of βs and βn
are given in Table 17-1 (next slide)

29
30

Explanation of Types of Struts Described in
Table 17-1
 Case A.3.2.1 applies to a strut equivalent to a
rectangular stress block of depth, a, and
thickness, b, as occurs in the compression zones
of beams or eccentrically loaded columns. In this
case βs = 1.0. The strut is assumed to have a
depth of a and the resultant compressive force in
the rectangular stress block, C=fceab, acts at a/2
as shown in figure (next slide)

31
fce=0.85βsfc’ = 0.85fc’

32
 Case A.3.2.2(a) applies to bottle-shaped struts
similar to those in Fig. 17-4b which contain
reinforcement crossing the potential splitting
cracks. Although such struts tend to split
longitudinally, the opening of a splitting crack is
restrained by the reinforcement allowing the
strut to carry additional load after the splitting
cracks develop. For this case βs = 0.75. If there is
no reinforcement to restrain the opening of the
crack, the strut is assumed to fail upon cracking,
or shortly after, and a lower value of βs is used.

33
 Case A.3.2.2(a) cont’d →
 The yield strength of the reinforcement required
to restrain the crack is taken equal to the
tension force that is lost when the concrete
cracks. This is computed using a localized strut-
and-tie model of the cracking in the strut as
shown in Fig. 17-4c. The slope of the load-
spreading struts is taken as slightly less than 2 to
1 (parallel to axis of strut, to ⊥ to axis): Tn =
(Cn/4){1-(a/bef)}.
 Setting Tn equal to Asfy gives the transverse
tension force Tn at the ends of the bottle-shaped
strut at cracking as: Asfy ≥ Σ (Cn/4){1-(a/bef)}.(*)
34
 Case A.3.2.2(a) cont’d →
 Where Cn is the nominal compressive force in the
strut and a is the width of the bearing area at
the end of the strut as shown in Fig 17-5a. The
summation Σ implies the sum of the values at the
two ends of the strut. If the reinforcement is at
an angle θ to the axis of the strut, As should be
multiplied by sin θ. This reinforcement will be
referred to as crack control reinforcement.
 In lieu of using a strut-and-tie model to compute
the necessary amount of crack control
reinforcement, ACI allows As determined from: Σ
(Asi/bsi ) sin γi ≥ 0.003 (**)
35
 Case A.3.2.2(a) cont’d →
 Where Asi refers to the crack control reinforcement
adjacent to the two faces of the member at an angle
γi to the crack, as shown in Figure (next slide)
 The ACI Eq for the crack control reinforcement is
written in terms of a reinforcement ratio rather than
the force to simplify the presentation. This is
acceptable for concrete strengths not exceeding 40
MPa. For higher concrete strengths the committee
felt the load-spreading should be computed. NB: A
tensile strain in bar 1, εs1, in Figure results in a
tensile strain of εs1 sinγ1 ⊥ to the axis of the strut.
Similarly for bar 2 strain ⊥ to the axis of the strut is
εs2 sinγ2, where γ1 + γ2 = 90°

36
Crack Axis of strut

γ1
Strut

As1(bar 1)

s2

γ2

As2(bar 2)
s1

37
 Case A.3.2.2(b) In mass concrete members such
as pile caps for more than two piles it may be
difficult to place the crack control
reinforcement. ACI Section A.3.2.2(b) specifies a
lower value of fcu in such cases. Because the
struts are assumed to fail shortly after
longitudinal cracking occurs, βs is multiplied by
the correction factors, λ, for light weight
concrete when such concrete is used. Values of λ
are given in ACI Section 11.7.4.3. It is 1.0 for
normal- weight concrete.

38
 Case A.3.2.3 The value of βs in Section A.3.2(3)
is used in proportioning struts in strut-and-tie
models used to design the reinforcement for the
tension flanges of beams, box girders and the
like. It accounts for the fact that such cracks will
tend to be wider than the cracks in beam
webs(nicht so eindeutig!).
 Case A.3.2.4 The value of βs in Section A.3.2(4)
applies to all other types of struts not covered
above. This includes struts in the web of a beam
where more or less parallel cracks divide the web
into parallel struts. It also includes struts likely
to be crossed by cracks at an angle to the struts
(recent publication in ACI)

39

1.3.3 Design of struts

Once the strut-and-tie model has been laid out,
the strength of a strut is computed as follows:
 (1) If there is no transverse reinforcement in the
strut, the strength should be taken as the
compression causing cracking, computed as: Ccr =
0.85(0.60λ)f’cat (see ACI Section A.3.2.2(b)), where a
and t, respectively are the width and thickness of the
nodal zone.
 (2) If the strut is crossed by reinforcement satisfying
Eq (*) and (**), the nominal strength of the strut is
determined based on the smallest cross-sectional
area of the strut and on the applicable concrete
strength from Table 18-1

40

The 2nd major component of a strut-and-tie
model is the tie. A tie represents one or
several layers of reinforcement in the same
direction.

Design is based on: φFnt ≥ Fut; where t refers
to “tie,” and Fnt is the nominal (strength)
resistance of the tie, taken as: Fnt = Astfy +
Aps(fse + ∆fp). The 2nd term on the RHS is for
prestressed ties.

ACI Section A.4.2 requires that the axis of
the reinforcement in a tie coincide with the
axis of the tie

41

In the layout of the strut-and-tie model, ties
consist of the reinforcement plus a prism of
concrete concentric with the longitudinal
reinforcement making up the tie.

The width of the concrete prism surrounding
the tie is referred to as the effective width
of the tie, wt.

ACI sections A.4.2 and R.A.4.2 give limits on
wt. The lower limit is a width equal to twice
the distance from the surface of the
concrete to the centroid of the tie reinforcement

42

In a hydrostatic C-C-T nodal zone, the
stresses on all faces of the nodal zone should
be equal. As a result the upper limit on the
width of a tie is taken equal to wt,max =
Fnt/(fcub)

The concrete is included in the tie to
establish the widths of the faces of the nodal
zones acted on by ties. The concrete in a tie
does not resist any load. It aids in the
transfer of loads from struts to ties or to
bearing areas through bond with reinforcement

43

The concrete surrounding the tie steel
increases the axial stiffness of the tie by
tension stiffening. Tension stiffening may be
used in modeling the axial stiffens of the ties
in a serviceability analysis.

Ties may fail due to lack of end anchorage.
The anchorage of the ties in the nodal zones
is a major part of the design of a D-region
using the strut-and-tie model. Ties are shown
as solid lines in strut-and tie models.

44

The points at which the forces in the struts-and-
ties meet in a strut-and-tie model are referred
to as nodes. Conceptually, they are idealized as
pinned joints.

The concrete in and surrounding a node is
referred to as a nodal zone.

In a planar structure, 3 or more forces must
meet at a node for the node to be in
equilibrium. This requires that: ΣFx = 0; ΣFy = 0;
and ΣM = 0. The condition ΣM = 0 implies that the
lines of actions of the forces must pass through a
common point, or must be resolved into forces that
act thru a common pt (SNS)

45
 C C C

Forces acting T
on nodes
C C
C

46

Nodal zones are classified as C-C-C if 3
compressive forces meet (Fig b), and as C-C-T if
one of the forces is tensile (Fig c)

C-T-T joints may also occur.

1.5.1 Hydrostatic Nodal Zones

Two common ways of laying out nodal zones are
shown in Figure 17-10 and 17-11 (SNS). The
prismatic compression struts (SPS) are assumed
to be stressed in uniaxial compression. A section
⊥ to the axis of a strut is acted on only by
compression stresses, while sections at any other
angle have combined comp and shear stresses.
47
Fig 17-10 Hydrostatic nodal
zones in planar structures

48
Fig 17-10 Hydrostatic nodal
zones in planar structures
49
Fig 17-10 Hydrostatic nodal
zones in planar structures

50
51
Fig 17-11 Extended
nodal zones

52

One way of laying out nodal zones is to
orient the sides of the nodes at right angles
to the axes of the struts or ties meeting at
that node (see Fig. 17-10) and to have the
same bearing pressure on each side of the
node.

When this is done for a C-C-C node, the ratio
of the lengths of the sides of the node,
w1:w2:w3, is the same as the ratio of the
forces in the 3 members meeting at the
node, C1:C2:C3 as shown in Fig 17-10(a).

53

Nodal zones laid out in this fashion are
sometimes referred to as hydrostatic nodal
zones since the in-plane stresses in the node
are the same in all directions. In such a case,
the Mohr’s circle for the in-plane stresses
reduces to a point.

If one of the forces is tensile, the width of
that side of the zone is calculated from a
hypothetical bearing plate on the end of the
tie, which is assumed to exert a bearing
pressure on the node equal to the
compressive stress at that node as shown in
Fig 10-7(b).
54

Alternatively, the reinforcement may extend
through the nodal node to be anchored by bond,
hooks, or mechanical anchorage before the
reinforcement reaches point A on the RHS of the
extended nodal zone as shown in Fig 17-10(c).

Such a nodal zone approaches being a
hydrostatic C-C-C nodal zone. However, the
strain incompatibility resulting from the tensile
steel strain adjacent to the compressive
concrete strain reduces the strength of the nodal
zone. Thus, this type of joint should be designed
as a C-C-T joint with βn = 0.80

55

1.5.2 Geometry of Hydrostatic Nodal Zones

Because the stresses are equal or close to
equal on all faces of a hydrostatic nodal zone
that are ⊥ to the plane of the structure,
equations can be derived relating the lengths
of the sides of the nodal zone to the forces
in each side of the nodal zone.

Figure 17-10a shows a hydrostatic C-C-C
node. For a nodal zone with a 90° corner, as
shown, the horizontal width of the bearing
area is w3 = lb.

56

The height of the vertical side is w1 = wt. The
angle b/n the axis of the inclined strut and the
horizontal is θ. The width of the 3rd side, the
strut, w2 = ws, can be computed as:

ws = wt cos θ + lb sin θ.

This equation can also be applied to a C-C-T
node, as shown in Figure 17-10b.

If the width of the strut, ws, computed from the
strut force is larger than the width calculated
using the above eqn, it is necessary to increase
either wt or lb or both, until the width equals or
exceeds the width calculated from the strut
forces.

57

1.5.3 Extended Nodal Zones

The use of hydrostatic nodes can be tedious
in design, except possibly for C-C-C nodes.
More recently, the design of nodal nodes has
been simplified by considering the nodal
zone to comprise that concrete lying within
extensions of the members meeting at the
joint as shown in Fig 17-11(SNS) (J.
Schlaich/FIP).

This allows different stresses to be assumed
in the struts and over bearing plates, for eg.

58
59
Fig 17-11 Extended
nodal zones

60

Two examples are given in Fig 17-11. Fig 17-
11a shows a C-C-T node. The bars must be
anchored within the nodal zone or to the left
of point A, which ACI Section A.4.3.2
describes as “the point where the centroid of
the reinforcement in the tie leaves the
extended nodal zone.”

The length, ld, in which the bars of the tie
must be developed is shown.

The vertical face of the node is acted on by
stress equal to the tie force T divided by the
area of the vertical face.

61

Thestresses on the three faces of the node
can all be different, provided that
 1. the resultants of the 3 forces coincide
 2. the stresses are within the limits given in Table
17-1
 3. the stress is constant on any one face

Eq17-15 is used to compute the widths ⊥ to
the axis of the struts of inclined struts in
extended nodal zones, as shown in Fig (SNS),
even though these equations are derived for
hydrostatic nodal zones.

62
63

An extended nodal zone consists of the node
itself, plus the concrete in extensions of the
struts, bearing areas, and ties that meet at a
joint. Thus in Fig 17-11a the darker shaded
region indicates the nodal zone extends into
the area occupied by the struts and ties at
this node.

This layout of a nodal zone contains much of
the concrete stressed in compression over
the reaction.

64

The advantage of the nodal zones in Fig 17-
11a and b comes from the fact that ACI
Section A.4.3.2 allows the length available
for bar development to anchor the tie bars to
be taken out to point A rather than point B at
the edge of the bearing plate.

This extended anchorage length recognizes
the beneficial effect of the compression from
the reaction and the struts improving bond
b/n the concrete and the tie reinforcement

65

1.5.4.Strength of Nodal Zones

Nodal zones are assumed to fail by crushing.
Anchorage of the tension ties is matter of design
consideration.

A tension tie is anchored in a nodal zone tends to
weaken the nodal zone (strain incompatibility).

ACI Section A.5.1 limits the effective concrete
strengths, fce, for nodal zones as: Fnn = fceAn;
where An is the area of the face of the node that
the strut or tie acts on, taken ⊥ to the axis of
the strut or tie, or the area of a section through
the nodal zone and fce = 0.85βnfc’
66

ACISection A.5.1 gives the following 3 values
for nodal zones (See also Table 17-1)
 1. βn = 1.0 in C-C-C nodal zones bounded by
compressive struts and bearing areas
 2. βn = 0.8 in C-C-T nodal zones anchoring a
tension tie in only one direction
 3. βn = 0.60 in C-T-T nodal zones anchoring
tension ties in more than one direction

67

Tests of C-C-T and C-T-T nodes reported in lit
developed βn = 0.95 in properly detailed
nodal zones

1.5.5 Subdivision of Nodal Zones

Frequently it is easier to lay out the size and
location of nodal zones if they are subdivided
into several parts, each of which is assumed
to transfer a particular component of the
load through the nodal zone. (SNS-Fig 17-
11b)

68
69

The reaction R has been divided into 2
components R1, which equilibrates the
vertical component of C1, and R2, which
equilibrates C2.

Generally, subdivision simplifies the layout of
the struts and nodes. Subdivision is useful in
dealing with the dead load of a beam which
can be assumed to be applied as a series of
equivalent concentrated loads, each of which
is transferred to a reaction by an individual
strut as shown in Fig (SNS)

70
71

In the figure the DL is transferred to the
support by 4 inclined struts, which are
supported by the portion of the support
nodal zone labeled as Vs.

The horizontal width of the part of the nodal
zone labeled Vs is the sum of the horizontal
widths of the 4 struts that support the DLs on
this half of the beam.

72

The vertical truss members represent the
subdivided DLs and stirrup forces, i.e. the DL
to the right of the main strut is hang up to
the compression face by the stirrups and
delivered to the 4 struts that transfer this
and the remaining DL to the left of the main
strut to the support.

73

1.5.6 Resolution of Forces Acting on a
nodal Zone

If more than 3 forces act on a nodal zone in a
planar strut-and-tie model, it is
advantageous to subdivide the nodal zone, so
that only 3 forces remain on any part of the
node.

Fig (SNS) shows a hydrostatic nodal zone that
is in equilibrium with 4 strut forces meeting
at point D.

The nodal zone for pt D can be subdivided
(See fig b)

74
Fig 17-14: Resolution
of forces acting on a
nodal zone

75

For subnode EFG, the two forces acting on
faces EF and EG can be resolved into a single
inclined force (50.6 kN) acting b/n the 2
subnodes.

That inter-nodal force must also be in
equilibrium with the forces acting on faces
AB and BC of subnode ABC

The overall force equilibrium for node D is
demonstrated in Fig c.

76

1.5.6 Anchorage of Ties in a Nodal Zone

A challenge in design using strut and tie
models is the anchorage of the tie forces in
the nodal zones at the edges or ends of a
strut-and-tie model. This pm is independent
of the type of analysis used in design. It
occurs equally in structures designed by
elastic analysis or strut-and-tie models. In
fact, one of the advantages of strut-and-tie
models comes from the attention that the
strut-and-tie model places on the anchorage
of ties as described in ACI, Section A.4.3.

77

1.5.7 Nodal zones Anchored by a Bent Bar

Refer

1.5.8 Strut Anchored by Reinforcement

Sometimes, diagonal struts in the web of a
truss model of a flexural member are
anchored by longitudinal reinforcement that,
in turn, is supported by a stirrup (SNS)

It is recommended that the length of
longitudinal bar able to support the strut be
limited to 6 bar dia each way from the
center of the strut
78
Struts anchored by stirrups and longitudinal bars

79

Design of a wall loaded and supported by
columns

The 350 mm thick wall shown in Fig (SNS)
supports a 350 mm by 550 mm column
carrying unfactored loads of 440 kN DL 730 kN
LL, plus the wt of the wall. The wall supports
this column and is supported on 2 other
columns which are 350 mm by 350 mm. The
floor slabs (not shown) provide stiffness
against out-of-plane buckling. Design the
reinforcement. Use fc’=20 MPa and fy= 420MPa

80
 1696 kN 1696 kN
D1

948 kN 948 kN

2500 mm D2
848 kN 848 kN

848 kN 848 kN
2500 mm
189kN 943kN 943kN

D3 943kN 943kN
2500 mm
1054 kN
1054 kN

D4 D5

943kN 943kN

81

Solution
 1. Isolate the D-Regions. The loading
discontinuities (wall has 2 statical discontinuities
at the top and bottom) dissipate in a distance
approximately one member dimension from the
location of the discontinuity. → wall is divided
into 2 D-regions separated by a B-region. There
are 3 more D-regions at the ends of the columns
which have little effect on the wall → not
considered.
 2. Compute factored loads. U=1.2×440 + 1.6×730
= 1696 kN

82

Subdivide
the boundaries of the D-regions
and compute the force resultants on the
boundaries of the D-region
 For D2, we can represent the load on the top
boundary by a single force of 1696 kN at the
center of the column, or as two forces of 848 kN
acting at the quarter pts of the width of the
column at the interface with the wall. Strut-and-
tie is drawn using one force. The bottom
boundary of Region D2 is divided into 2 segments
of equal lengths, b/2 each with its resultant
force of 848 kN acting along the struts at the
quarter pts. This gives uniform stress on the
bottom of D2.
83

Lay out the strut-and-tie models
 Two strut-and-tie models are needed, one in
each of D2 and D3. The function of the upper
strut-and-tie model of D2 is to transfer the
column load from the center of the top of D2 to
the bottom of D2. The compression stresses fan
out from the column (2:1), approaching a uds at
the ht where the struts pass through the quarter
pts of the section. In D2 this occurs at level B-L.
Below this level struts B-C and L-K are vertical
and pass through the quarter pts of the width of
the section. This gives uniform compression
stresses over the width
84
  For D3, similarly, the load on top of D-region D3
will be represented by struts at the quarter pts
of the top of the D-region. The strut-and-tie
model in D3 transfers the uniformly distributed
loads, including the dead load of the wall, from
the top of D3 down to the 2 concentrated loads
where the wall is supported by the columns.

Draw the strut-and-tie models
 Recall recommendation that load-spreading
struts at a (2 to 1) slope relative to the axis of
the applied load → θ = arctan (1/2) = 26.6°

85

Compute the forces and strut widths
 The calculations are given in tables below
Calculation of forces and strut widths in regions D2
D- Member V- H- Axial fce Width
of strut
Region compo compo force or nodal
zone
D2 Node A 1696 0 1696 12.8 504.8
A-B 848 424 948 12.8 282.1
B-C 848 0 848 13.6 237.5
A-L 848 424 948 12.8 282.1
L-K 848 0 848 13.6 237.5
B-L 848 424 424 13.6 118.8

86
Calculation of forces and strut widths in region D3

D- Member V- H- Axial fce Width of

strut or
Region compo compo force nodal
zone
D3 D-E 943 0 943 13.6 264.1
EPF 943 471 1054 13.6 295.2
F-G 0 471 471 13.6 131.9
G-H 943 471 1054 13.6 295.2
H-J 943 0 943 13.6 264.1
E-H 0 471 471 13.6 131.9

87

D-region D2
 (i) Node A and struts A-B and A-L: Treating Node
A as a hydrostatic node, either the node or one
of the struts A-B and A-L will control.

Node A: βn = 1.0 (node is compressed on all in-plane
faces)→ fce = 0.85×1.0×20 = 17 MPa

Struts A-B and A-L: Bottle shaped struts → βs = 0.75 →
fce = 0.85×0.75×20 = 12.8 MPa → governs → an area of
(1696×1000)/(0.75×12.8) = 176667 mm2 is required.
Column area = 350 ×550 = 192500 mm2 → ok (capacity
reduction factor φ = 0.75)
 (ii) Minimum dimensions for Nodes B and L:
These are C-C-T nodes → fce = 0.85×0.80×20 =
13.6 MPa

88
 These will control the base dimension of Nodes B
and L because struts B-C and L-K are prismatic
struts that can be designed by using βs = 1.0. →
The base dimension of node B and the width of
strut B-C is:

ws = 848000/(0.75×13.6×350) = 237.5 mm.
 This is much less than b/2 (1250 mm), so the node
easily fits within the dimensions of the wall
 The height of node B is of interest for tie B-L. So,

wt = 424000/(0.75×13.6×350) = 118.8 mm.
 This is a very small dimension and the reinforcement
for tie B-L will be spread over a larger distance and
The dimensions of nodes B and L will be much larger
than the minimum values calculated here

89
 (iii) Required area of reinforcement

Tie force Tu = 848/tanθ = 424 kN

Required As = Tu/φfy = 424000/(0.75×420) = 1346 mm2
 Essentially a band of transverse steel having this
area should be provided across the full width of the
wall extending about 25% of the width of the wall
above and below the position of tie B-L so that the
centroid of the areas of the bars is close to tie B-L.
See fig. Both ends of each bar should be hooked.
 Use 8 N0. 16 M bars (As = 1592 mm2) at a vertical
spacing of 300 mm, half in each face and hooked
at both ends (See figure). This spacing provides a
tie width of approx 1200 mm, as recommended
above.

90

D-region D3
 Nodes F and G are C-C-T nodes, similar to nodes B
and L. → Use fce = 13.6 MPa, to determine the min
dimensions for all struts, ties, and nodes in D-region
D3. → See result in column (7) of Table. All element
dimensions fit within the wall and supporting column
dimensions
 (i) Required area of reinforcement for tie F-G

Tie force Tu = 471 kN

Required As = 471000/(0.75×420) = 1495 mm2
 Thus, use 6 No. 19M bars, As = 1704 mm2, placed in
two layers of three bars per layer. This would put the
centroid of these bars approximately at mid-height of
tie F-G whose height (width) is given in table above. All
of these bars must be hooked at the edges of the wall
as shown in figure.
91